Advanced Minesweeper Techniques with Examples

Advanced Minesweeper Solving Techniques

Scenario types, play principles, and practice puzzles with worked solutions

Introduction

I’ve written this guide because I found myself in need of one that dealt with advanced situations that are a bit rare in normal gameplay (most guides focus on nothing harder than Expert).  One which can cover the difference between “play the basic patterns with perfect logic”, and truly optimized play that can eke out an edge when those patterns have nothing left to give.  A set of techniques I could use in pursuit of high-difficulty wins, where I can’t just click “hint” 20 times a game, and where I needed some better strategies and fewer errors.  So I’ve developed a few heuristics and tactics, explained below.  But I’ve also compiled some puzzle “problem sets” for most of these scenarios, so you can train yourself to spot them and solve them, by challenging your understanding and then having full explanations available.  Some of these situations might be rarely encountered “in the wild”.  But if you’re going for (say) a 100k-difficulty win, or mastery, or high-level arenas, sooner or later you will need these in your toolbox.

This guide does not cover efficiency play, nor does it help much with speed.  But it will make you a better player who’s more likely to avoid mistakes, and likelier to win any given game.  And that alone can make you faster, and can train some of the brain muscles that efficiency needs, too.

This guide is thorough.  Giving advanced players new tactics and ways of thinking takes a lot of explaining, because they are starting from a strong base already.  Believe me when I say it’s faster to read than it was to write.  It looks much longer than it really is, too, because 2/3s is just the example puzzles and solutions, and much of the rest is just screenshots of game boards.

How to Use This Guide

Use the field manual beside the article (or above it on mobile) to jump directly to the topic or practice set you need. This is not a novel, so you do not have to read it straight through.

This guide offers 10 intermediate-to-advanced techniques, each with their own things to look for and thought processes to incorporate into your game.  It’s divided into 3 parts:

1. General Gameplay - 7 more-basic, universal topics on how to read a board or situation, diagnose what you’re seeing, strategic ideas, and other good sources of advice.
2. Rules for Guessing - 6 techniques you can try when you run out of logic moves to make, depending on what the board offers you and what phase of gameplay you’re in (early, mid or end-game).  The first one can help you avoid guessing entirely.
3. Deep Tactics - 4 ways you can sometimes find solutions to tough situations, and even if they fail, use the thinking that went into trying them to find optimal guesses.

Players who consider themselves fairly advanced already may prefer to skip Part 1, or just skim through it, and go directly to the more nuanced topics in Part 2 and Part 3.

Most chapters in Part 2 and Part 3 are accompanied by “problem sets” of puzzles, like a textbook, which give you opportunities to practice thinking the way the chapter asks you to, and learn to spot the situations where they can be used.  Then, in the corresponding Solutions section, you can find in-depth explanations and diagrams illustrating the right answer and how to arrive at it. The guide offers over 100 puzzles with solutions, of varying difficulty levels.  Use them to help yourself work on areas of your own game where you might want to improve, or need some thorough explanation to help something click into place for you.  Or because it’s fun!

Terminology

This guide was prepared for the Minesweeper Online community (formerly “World of Minesweeper”, aka “WoM”), so I’m using much of its lingo.  The tactics here, however, apply to Minesweeper gameplay anywhere.  I will also use a few terms of my own:

“Front” or “Frontline”: this refers to the area of the board where unknown cells border known cells, and thus are the portion that you are currently able to reason and make deductions on.  AKA the set of “exposed cells”.  The part of the board where we know more than nothing, but we do not yet know everything and thus have it solved and marked.  We’re using the term in its military sense, as distinguished from your forces’ rear; or if you prefer, in the weather sense, as you roll through a board like a stormfront.  Places we’ve cleared already are old news, while parts of the board that we’re nowhere near yet are just a pool of dangerous risk taking (aka “terra incognita” - unknown land - or for a clearer term, the “untouched” squares[1]).  We can’t start working there except by some very risky guessing.  So the operative part of the board, the parts that we have some information about but still have work left to do, is our “front”.

“Background Rate”: the percentage chance of any untouched square on the board (those about which we have no information) being a mine.  If you’re stuck and have no moves left to make from logic, and you just click somewhere random on the board (such as a remote corner), then your chance of blasting on that guess equals the background rate of the board.  If you use a hint on WoM that shows you the percentages on the board, it’s the number shown in gray text on some cells.  The term alludes to the universe’s background radiation rate.  On a totally new board, the background rate equals the board’s total mine density.  If you’ve played a bit and had an easier time than average, clearing (what happened to be) a lower-density part of the board, then the background rate goes up, because the density of the remainder must necessarily be higher than you started with.  Conversely, if you’ve cleared a bit and it was hard going, a higher-density area, then the background rate might now be lower than the board’s total density (which makes guesses into terra-incognita more favorable choices, sometimes even optimal).  The rate is also affected by the number of mines we know exist right along our frontline (but haven’t been marked yet), e.g. a mine on these 2 squares, 2 mines on these 3 squares, etc - and generally front areas will have higher mine probability than the background rate.  So the true background rate of a board takes into account the mines we know must be located somewhere on your front, and is only calculated off the number of remaining mines that must be among the cells that we know nothing about, and the quantity of those untouched cells.

“Progress”: An ability to continue playing a board using logic-based moves, rather than making further guesses that put your survival at risk and have a chance of blowing you up.  The top computer solver defines progress percentage as “the chance that a particular guess will lead to 100%-safe moves (from logic / patterns) being available next move”.  When you reach the end of what logic can do for you (other than for 50-50s) and you need to guess, certain guesses might yield a lot of new information, via moves that are now available because of the guess result and its following conclusions; it has resulted in you being able to continue making progress.  Whereas, some other guesses might be so limited in what they do for you that being correct lets you mark a single mine, and then that’s it, you can’t make further progress on the board, you’re back to making another guess.  When choosing guesses, a likelihood of further progress is an essential consideration, even more important than small differences in the safety percentages on a given guess square.  So the concept will frequently be relevant to analysis.

“Isolated”: A section of your front is said to be “isolated” if the placement of mines outside that section do not affect any scenarios or odds of placements within that section.  Usually, sections of the front will become isolated by having mines be interspersed between them, such that none of the possible configurations from one side of those mines on the front line touch the other side of it.  As a simple case, an opening puts distance between sections of a front, and they thereby become isolated from each other - and the same can happen with smaller gaps, too.

“Dependent”: Two (or more) cells on a frontline are said to be dependent if they determine each other’s status, i.e. if one being safe necessarily implies the other is safe, and conversely if one being a mine necessarily implies the other is a mine.  Long linear fronts will often have dependency chains, where every 3rd square has the same status as the one 3 to its left or right (or up/down).  Cells or groups of cells that collectively do not determine each other are thus said to be “independent”, particularly ones adjacent to each other, which may look like they might determine each other, but in fact do not.  Independence is sometimes a useful intermediate observation for revealing the applicability of a guessing technique.  Cells can also be “partly dependent”, in that a mine in one will determine status in the other, but the reverse is not true; or, something is determined only if one cell is a mine, but not if it’s a safe cell, and so on.

Lastly, as a matter of writing style, I will generally use number characters (1, 2, 3) when referring to the face value of number squares on a board; however I will usually type out the number words (one, two, three) when talking about a quantity of mines, or of scenarios or the like.  So if you see an actual number character, it’s referring to a square showing that number, at least for 1 through 5 or 6.  I’ll do this particularly when talking about specific game situations, such as in the puzzle solutions.  I hope this will cut down on confusion between the different contexts in which we discuss numbers in minesweeper.

General Gameplay

Before we get into advanced techniques, here are a few broad pieces of advice-in-general, guidelines on how to think about common game situations.  Some are reviews of or extensions to what we might consider intermediate-level skills and tactics, but they’re ones I want to emphasize for their importance, or for their relevance as building blocks to deeper ideas later.

“Effective” Number Squares

It’s not the number shown on the square that really matters, it’s how many mines are left still-unmarked around it.  What’s important is how many mines are remaining unknown.  That’s what determines which logic pattern applies in a given spot.  Xilef / DardHong terms this “reduction”, i.e. “reducing” the number of mines shown by the number of mines known.  The term “reduction” is also used in the official WoM patterns guide.  And this gets applied in practice based on one essential rule: Mine patterns behave in accordance with their “effective” numbers, not their initial or “face” numbers.

When looking at number squares in a game, I’ve trained myself to barely even see the number itself (how many mines are in all surrounding 8 squares), and instead ask “how many mines are remaining unmarked?”  So a square might show a 5, but if I’ve marked three of those already, it is an “effective 2”.  It is effectively the same as a 2, for purposes of logic.

Let’s illustrate with some examples, because we’ll use this terminology a lot, e.g. “that square was an effective-2”.  Here’s a basic 1-2-1 pattern, where the mines are under the 1s…

 →

Well, with other mines in the mix, the below may not at first glance appear equivalent… but it is!  Below, the right-most 2 is actually an “effective-1”, because one of its mines is already marked:

 →

So the 1-2-2 across that middle front is actually a “1-2-1”, solved the way we normally do with the first example.  This is seen constantly in games.  At right is another instance: the 2-3-2 front is effectively just 1-2-1, because each front square has one mine marked already.

Another example: Below is your basic 1-2-2-1 pattern, right?  The mines touch the 2s; the adjoining squares are safe.

                    …Well, so is this:

➤

       →    

It says “2-2-2-3” down along the left front there, but the top and bottom each touch other mines, so the front is effectively 1-2-2-1, and we mark it the same way as above.

Here’s a slightly fancier example:

(1)   (2)   (3)

Looking at the region in Step 1, we see the bottom front 3-4 “reduces” to a 1-1 pattern; the 4 says there’s one mine in the blue, the 3 says there’s one in yellow, therefore the lower-left square is safe.  On the right in Step 3, the “2-2-4” reduces to 1-1-2 and we have our 1-2 pattern: there’s a mine in the lower-right (teal), and then another safe square back above (purple).

You don’t even need a definite mine location, necessarily, to have your squares’ effective numbers reduced, and the appropriate pattern determined.  From here at 0:32:

 → →

The 3 at left implies there is a mine in the blue “group”, i.e. in one of the two squares in that blue box.  This means that the yellow-ringed 2 here is an effective-1.  Likewise, the mine to the right of the pink-ringed 2 also makes it an effective-1.  So the mines at the top here effectively make a 1-2-1 pattern, with known mines and safe squares.  Also, on the right, the lower-right 2 implies a mine in the teal group - we don’t yet know which square has the mine, but that’s enough to reduce the 2 above it (in purple) to being fully satisfied, and offer safe squares above-right.

The truth of this “effective numbers” rule is non-obvious.  It might not be apparent to a beginner.  However, it’s also not a particularly advanced technique, in fact it’s listed in the Patterns help page on WoM (as “Reduction”).  So why mention it, why does this matter?  For a few reasons:

1. It’s a lot faster to fit what you’re seeing into existing patterns, then to try to reason out the mine positions from first principles every time.  You’ll never be that fast if you aren’t thinking along these lines, marking things according to the pattern and moving on.
2. Failure to stay aware of what a front line’s effective numbers are can very easily result in you applying the wrong logic pattern, opening the wrong square, and blowing up.  This alone probably accounts for ~50% of my game losses that came from a logic error.[2]
3. As mines get denser and logic gets more complicated, it’s important to know that this equivalence is provable and trustworthy, rather than it just being likely, or true only under limited circumstances.  The fact that the patterns behave according to their effective numbers, rather than their on-square numbers, helps you play with confidence.
4. This way of thinking is required continuously on advanced boards.  One final example:

If you’re presented with this sort of situation, what you have to learn to notice is that the 2-3 pairs in the middle of the front are all effective-1s.  The 3s each have two mines already marked, and the 2s next to them each have one mine already marked.  And, importantly, the open squares for the 3s entirely overlap those for the 2s.  This means that the remaining mine for the 2s might not be able to be determined yet, but we DO know that the squares on the “back sides” of those 2s, must therefore be safe.

We can’t determine whether it’s the orange square, or the two yellow squares, that are mines.  Those are the two possibilities to satisfy the 3s.  However, because all four of the circled numbers are effective-1s, and because the remaining unknown squares for the 3s are entirely included within the remaining unknown squares for their neighbor 2s, we can infer, as a simple two-step reasoning, that the green squares - all 6 of them - are all safe, and can be opened.  The hope would then be that as a result, we gain enough information underneath those green squares to choose between the yellows or the orange, so we can complete the markings for those 2s and 3s.  But sometimes, we’ll have to come back around later, after getting through a bunch more moves and then finally being able to return with new information in that area.

Types of 50-50s

Because we will need to discuss them (and things that resemble them, and might-or-might-not be them) frequently, let’s first take a moment to lay out the primary types of 50-50s that we will encounter in the wild.  In general:

1. A 50-50 is a forced guess, where no logic or new information can discern between two mine layout possibilities, and so you have to accept a coin flip to lose the game and just guess it. It’s a common beginner mistake to assume a situation is a 50-50 when it’s not.
2. There are two broad categories of simple, common 50-50 situations, which I will call the “box-and-2” type, and the “line-and-1” type.
3. These two categories have variations that appear on the edges and at the corners of the board, where the wall takes the place of some of the otherwise-necessary mines (serves as the “lack of information” coming from that direction, really), so we will look at the “edge variants” and “corner variants” as well.
4. Because of #1, you should always guess 50-50s as soon as they are confirmed to be one.  There is no tactical value to waiting, just more potential time wasted.

“Box-and-2” 50-50s are where two mines are laid diagonally across a 2-by-2 box of four squares, which is framed by mines or walls on all four of its corners. Because we don’t have access to information from those corners, we can’t tell which diagonal the two inside mines are on.  We do have information from the sides in between those corners, but without a safe (number-square) corner, that’s not enough to solve it.

(in the following diagrams, the As and Bs, in color, are unknown squares.  The pairs of Cs and Ds, in black text, you should imagine are safe/known: at least one square of each pair is safe.)

This is the basic version of a box-and-2 50-50.  You have a 4-square box, in which the information coming from each side tells you that ONE of A or B is a mine - but you can’t tell whether it’s the two As or the two Bs that are mines.  And you can’t tell that because what would distinguish them - the 4 corners - are unavailable to provide that information.  In the middle of the board, that unavailability comes from the corners being mines.  For the sides, one (or both) of the two squares needs to be safe, or else it could be a different situation.

This is the “edge variant” of the box-and-2.  We still have the 4-square box there, and the information from the edges of that square (the Cs and Ds) tells us that either the As or the Bs are mines.  But because the two interior corners are mines, we can’t tell which it is.  And the wall prevents discernment from below.

Here is the “corner variant” of the box-and-2.  We know, from either direction (the pairs of squares labeled C), that one of A/B are a mine.  We don’t have the interior corner to tell us whether A is or isn’t one.  And importantly, to be sure this is a 50-50, we also have to know that there are exactly 2 mines here (from mine count), because one or three would be solvable (more on that later).  But if the count is two, then we don’t know if it’s the As or the Bs that are mines.

“Line-and-1” 50-50s are situations where there is one mine across two squares, and the edges of those squares (on the end of the “row” of 2 squares) are all either mines or edge-of-board wall.  So we can’t tell which of the two squares is the mine, because we lack information from any square that only sees one or the other.

Here is the basic version of the line-and-1 50-50.  We know, from the C squares, that one of A or B is a mine.  But the C squares can’t tell us which one - and because we have a row of mines above the A/B row, and also below the A/B row, we lack information from any square that only sees (touches) one or A or B, but not the other.

Here is the line-and-1 edge variant - vertical.  The basic layout above requires two framing rows of three mines, but if we move the formation to the wall, then that wall can replace one of the two rows.  So we need a wall of 3+ mines on the interior, three rows from the wall, and we need those two squares in the middle, A and B, to have one mine between them.  The Cs can tell us there’s one mine there, but can’t tell us which one it’s on.  (Rarely, the Cs might tell you that both A and B are mines, in which case there’s no guess, you just mark it and go on)

This, meanwhile, is the line-and-1 edge variant - horizontal, which I call the “half-full bucket” version of this 50-50.  So named because we have sides to the formation, two walls made of mines (but only 2 long, because we’re on the edge), and in the “bottom” of the bucket, the A/B squares, one mine in 2 squares.  As this requires more mines (5) than the vertical edge variant (4), it is consequently rarer.

Finally, this is the line-and-1 corner variant, which compresses the original formation in both the vertical and the horizontal directions.  Instead of one of the two rows of three mines, we have the wall at bottom; instead of needing a row of three on the other side, we have a row of two, an “overhang” if you like, and then one mine among A and B.  The Cs can’t tell us which it is.  This is the most common 50-50 type to see in games, as it only requires three mines.

The general theory of 50-50s is, they are a class of unsolvable situation in which there are two possible mine configurations, and because of mine placement or wall border, we have no information source that can distinguish which is true. At right is a less-common “corner L” type of 50-50: mines are either on A-C, or on B-D, but no square can help us to say which.  Minecount can tell us whether the two untouched squares are mines, but even if safe, those can’t rescue us from the dilemma either.  Just gotta guess.

There are more 50-50 mine layouts than could be catalogued here.  Some are discussed in Scar’s guide, and a few more are shown in the appendix below just to illustrate some of the principles of how 50-50s can get constructed, beyond the basic situations above.  But these two main categories, with their variations, make up the overwhelming majority of 50-50s that you’ll see in gameplay, probably 90%+.

It’s important to know what conditions are necessary and sufficient for a situation to be a true 50-50.  Sometimes situations will look a lot like a 50-50 guessing situation, they’ll be set up very similarly, but in fact do not meet all the criteria and do have a solution.  Some common bad assumptions like this include having a framed 4-square box that looks a lot like a box-and-2 50-50, but actually isn’t one (see Chapter 7, on the 4-square box), or a formation that looks like a line-and-1 50-50 but in fact has squares that can be cleared or marked as mines.  Unless you know these types, and the rules that determine whether they’re truly forced guesses or not, you’ll end up losing some games you could have won.

At right and below: Some members of the “Wait!  Don’t guess, think first!” hall of fame.

Play Evil NG A Lot

This is my single best piece of advice for intermediate-to-advanced players.  Playing Evil NG made me a much better player, much faster than any other thing, including arenas and guides.  And the reason is, Evil NG isn’t just a bigger, higher-density board than the other preset game types - if you’ve played it, you’ll know that games in that mode are specifically crafted, such that somewhere in the board, there are what are called “high-rank” moves required.  The rank of a move is the quantity of number squares you have to consider together in order to identify a safe move from logic (it’s a term from matrix math).  In normal gameplay, nearly all moves are rank 1 or 2, and rarely 3.  At some point in every Evil NG game, though[3], you will only be able to make further progress, can only find the next safe square, if you can see 4, 5, or even 6+ “moves ahead”, in a chess sense.  Moves where you have to accurately make reasoning like, “so, these two squares have a mine, which covers half of this 2, so the other part of this 2-square has one mine, so these other 2 squares have a mine, therefore the spots behind that 1-square are open”, or even longer chains where you might go horizontally at first, draw an intermediate conclusion, and then go vertically, rely on that intermediate conclusion, and use it to prove that something is safe, or a mine. 

And if you don’t see it, can’t figure it out, the great thing about playing Evil on WoM is, you just click Hint and it tells you what you should’ve seen.  Every time.  For free.  Sure, there are some event quests that require that you solve Evil NG games without using any hints, and likewise in NG arenas, hints are not available.  But in general gameplay or in daily quests, you can get practice at these scenarios over and over again, training yourself for general high-ish-density logic play (for the vast majority of the board), and then on the advanced / rare techniques that Evil requires (for the high-rank part of the board).  Those high-rank moves occur pretty rarely in non-NG games, but when they do, recognizing them may be the difference between winning and losing.  But also, the way of thinking about the chained, multi-step logic required to find those moves, ends up being very similar to how to think about estimating probabilities when deciding what to guess in high-difficulty situations, or how to discover proof-by-contradiction solutions.  So it’s a very necessary part of your mental toolkit.

(at left: a Rank-7 move in Evil NG, requiring considering 7 cells at once in order to find the logic move.  7 might be the limit of the WoM hint solver.)

When you can sometimes solve Evil NG games without having to resort to a Hint, you’re then ready to get some value out of the rest of this guide.

There are too many techniques that Evil NG requires for us to recap them all.  But there are two in particular, whose way-of-thinking is essential to the more advanced topics in this guide.  They both revolve around being able to keep track of what the effective number of mines is for a given square, and then account for them in different ways.  We won’t practice these via puzzle sets - playing Evil is practice enough - but we will explain them, so we can refer to them later.

Neighbor Mismatch

The first technique involves spotting two cells which:

(A) share at least one adjacent unknown square, and

(B) have different effective numbers of mines remaining.

There’s a trivial example of this on linear fronts that we all know well, where we have 1-2 patterns which prove a mine’s location over on the effective-2 side, which then bounces back to a safe square on the effective-1 side.  At right, we have a 1 (in blue) next to a 2 (in yellow); this allows us to reason that there must be a mine in the two overlap squares between them, and thus a mine above (only touching the 2) and one safe square below (only touching the 1).  We have a mismatch of the effective numbers of neighbors, and as a result there is a conclusion that locates a mine on the higher-number side, and a conclusion of a safe square on the lower-number side.

Along linear fronts, this is a core tool in your toolbox: if two neighbors share some unknown squares, and are not the same effective number, we immediately recognize it as a spot for a 1-2 move (e.g. at right).

The same is true in more-complex situations too, though.  Situations where you have an effective-1 up against an effective-2, maybe with a buffer row or column between them, and more than one shared square.  And in those cases, the key, often-answerable question you should ask is: “How many mines can these shared squares hold?”  Generally you’ll be considering possible answers of zero, one or two.  But when there is a mismatch in total mines demanded on either side of that set of shared adjacencies, you can usually determine that there is only one possible answer to the question.  Consider the following:

 →

The middle 2, which is an effective-1, shares two cells with the corner 2 (in yellow).  Can those cells contain zero mines?  No, that would under-serve the corner 2 (it would need two mines from only one remaining square).  Can they contain two mines?  No, that would overflow the middle 2.  Therefore those shared squares contain exactly one mine, and even though we don’t know which square it is yet, we can use that information to make other moves, and proceed.

At left, we need to focus on the only effective-2 along the linear front on the left: that interior-left 4.  It has an effective-1 above it (the interior 3) and below it (the 2).  So we need to consider the area right around there, in particular the groupings at right: the yellow group is really our “pivot” here.  You might imagine that the yellow squares could contain 0, 1, or 2 mines.  But can it contain zero?  No: That would mean the 4 below it would need both its remaining mines from the blue squares, which would overload the lower 2.  Can the yellow group contain two mines?  Again, no: that would overload the 3 above it.  Therefore the yellow group has to contain exactly one mine.  This then leads to further logic: because the purple group overlaps the yellow, it means the square above that interior 3 is safe (with further moves above).  It also means that the blue group contains exactly one mine, meaning the square below-right of the 2 is safe, resulting in another safe square farther below, and several chances for progress.  Identifying this logic move is only possible by spotting the “neighbor mismatch” of the effective-2 amid a sea of effective-1s.

This next example is actually from a standard game, not NG, but this reasoning is typical of Evil NG-type logic.  The keys are the interior-3 at left, which is an effective-2, and the 4 just to its right, which is an effective-3 (!).  They share 3 squares between them (in the orange group), and thus they are a “neighbor mismatch”.  Considering from the right side, we know the orange group has to contain at least two mines, because the 4 needs three total from its four remaining adjacencies, and there’s only one square on its right side.  So orange has to have at least two of them, and maybe all three.  But from the left side’s perspective, orange could have zero, one or two, but it can’t have three, because that would overload that interior-3 next to it.  Therefore we have requirements of that orange group of “at least 2” and “at most 2”, meaning it must contain exactly 2.  This gets us a nice set of moves:

Or consider this simpler situation:

Here, we have an effective-1 (the left 3 towards bottom left), near an effective-2 (at far left).  Between them we have a shared square group, in blue.  From the 2’s perspective, the blue group could contain one or two mines; but from the 3’s perspective, since it only needs one more, that group could contain zero or one.  Therefore, it must contain exactly one mine.  And we don’t know which square of the blue group that mine is in, not yet, but we can draw conclusions from it anyway: there must be 1 mine below the 2 (the higher side) and 1 safe square below that 3 (the lower side).  Other moves then follow, as seen at right.

These situations can often be subtly hidden.  Simple 1-2 patterns on linear fronts will jump off the screen at you, if you’re even an intermediate-level player.  But things like this, below, look benign at a quick glance, and it’s easy to skip right past them:

 or

If you’re not on the lookout for effective-2s in the midst of effective-1s, you would probably zoom right on past this one.  Even if you see it, if you haven’t practiced asking yourself the questions, the idea that a logic move is there may not occur to you.  That’s where playing Evil a lot helps.

Neighbor-mismatch situations need not involve effective-1s vs effective-2s, either.  Sometimes you’ll have an eff-1 vs an eff-3, or an eff-2 vs an eff-3, in which case the situation ought to jump out at you all the more as being one that must offer a logic move.  E.g.:

 →

If normal logic moves fail you, but you have areas where two parts of your front share some unknown squares and nearly touch, then across those gaps, look for neighbor mismatches.

Projection

This technique involves considering the remaining adjacencies of an unsatisfied number square as a group, using them to partially account for the mines of another square (probably towards the interior of the board) which shares all of those same adjacencies, and then note the residual amount of mines demanded by the interior square, and see if those can satisfy something adjacent.  And then repeat the process, while we still have further squares we can “project” onto.  We are taking the remaining mine(s) from a number square, and “projecting” the group onto another square.  So I’ll call the technique “Projection”, although WoM refers to it as a “Dependency Chain” in the patterns guide. The key difference from basic logic moves is that we’re having to move groups of squares around in our mind, which collectively have a known number of mines - “these two squares account for one mine” - “...then its second mine is accounted for by these other three squares”, etc - instead of reasoning about individual squares that are specifically a mine or safe.  The specific square where the mine is, within that group of squares, remains unknown as of yet - but it’s also irrelevant for this technique.  It’s irrelevant because no matter which square the mine actually is on, if we can account for all mines around a given square using these ghostly “square groups”, and if there are other squares it also touches, then those other squares must be safe.  As a simple illustration, consider this position:

 →  →

An experienced player, seeing this, will immediately see the groups in the left image: the lower squares, in the yellow group, must contain one mine (per the 1 to its right).  So if we then consider that upper-left 3, it has one mine marked, another mine accounted for by the yellow group, so its third mine must be in the residual squares - marked in the blue group.  However, if there must be one mine in the blue group, then that fully satisfies the 1 to its left, so all three other squares around that 1 must be safe.  If you click the WoM Hint button, you get the image on right - the blue highlighting means “you have to consider these three squares together” in order to reach the conclusion that the green-highlighted cells are safe.  Sometimes it’s easier to visualize as square groups.  But either way, this is the basic thought process.  And this is just a rank-3 move (three squares together).  They can end up being a lot fancier than that.

Most of the common high-rank patterns seen in Evil NG fit into the “projection” category, so let’s illustrate the general concept using some specific example patterns:

Hole 2-3: An H2 hole pattern reveals three cells in the “hole”, the middle one of which is a 3, and one (or both) of the side ones is a 2.  You’ll see the situation at left:

 →  →

In this case, you start out by thinking about that wide 2: there’s one mine accounted-for by the effective-1 on the linear front (yellow group), and thus the residual area for that wide 2 has only 1 mine in it (blue).  We’re particularly interested, however, in the part of that blue group which overlaps with the central 3 (dotted-purple).  If you then switch your thinking to trying to account for all the mines on that central 3 (right image), you’ve got one mine coming from that hole 1 (teal group), and since the dotted purple is a subset of the blue group, the dotted purple area contains “at most one mine”.  So if we have two of our three mines for that 3 accounted for by those (orange arrows), and there’s only one square remaining, then that square must be a mine!  And then, based again on the teal group and dotted-purple, we can switch our definition of that purple group from “at most one mine” to “exactly one mine”, because otherwise the 3 is under-served.  And now that we know there’s one mine in that purple group, it means (yellow arrow) the residual squares in that blue group, based on the wide 2, must therefore be safe!

If you click the Hint button, what you see is this: it highlights in blue all the squares that must be “considered together” in order to identify the next move that results in safe squares.  This move requires considering four squares together, so you’d say it’s “rank 4”.

Sometimes, this pattern is available on both sides of the hole pattern’s 3 interior squares, if those hole squares are 2-3-2.  In that case, you can run this pattern on each side, marking the mine to the back of the wide 2s, clearing the wide squares away from the hole, and having another safe square emerge directly behind the central 3.

 →  →

Crossing Pattern: This pattern relies on using the interior of a hole pattern, doing projection from one direction, drawing an intermediate conclusion, then doing projection from the other side of the hole pattern cells, and finally overlaying the two to discover a safe cell.  Like this:

 →  →

From the left, we do projection: the yellow group has one mine.  Then based on the left-interior 2, the blue group has one mine.  Then using that, the central interior 2 says that the pink group must have one mine.  Hold that thought.  But then, consider it from the right instead: that 1 on the edge covers three cells (in purple).  But the pink group said there was one mine on just the rightmost 2 squares of that purple group!  Therefore, since both groupings are true, the pink “accounts for” the mine within the purple, and the residual square (bottom-left) must be safe.

In Evil NG, that safe cell will often turn up a 1, and then you can use projection from the top in order to show that the mine for that new 1 must be on (in this image) either the left or the right of it, making the cells along the bottom row to be all safe.  In this case, though, if it doesn’t, we have another projection option open to us too: from the top-left, the left-hole 2 now has one mine (blue group), which satisfies the lower-left 2, which proves all of the cells in the purple group safe.

Wheel 3-2-1: This is a very common endgame for an Evil NG game, and can sometimes be seen prior to the endgame as well.  On the interior of a hole pattern, you get a 3-2-1 set of number squares (or rarely, 3-2-2), and the 3 then abuts a side of the front.  The idea is, we will account for all three mines of that 3, one by one, using two-square groups, and then rely on one of those square groups to show that a specific cell must be safe.

 →  →

To recognize the situation, you’d see that 1-2-3 across the hole cells, at left, and the fact that we have a view onto the 3 from the side (here, the right side, with the 4 and the 3) as well as from the front that gave us the hole pattern.  Then (middle image) we reason that the yellow group has one mine, the blue group has a second mine, and therefore from the middle 3, the purple group must contain its third mine - winding its way around the 3 as if it were a wheel.  We then use that conclusion (right image) to go back to the top and project onto the 2 in the middle of the hole cells: the teal group has one mine, the purple group has one mine (as previously noted), and so therefore the one cell in the bottom left here must be safe.  In this case, that cell can turn up a 0 (in which case there are mines on the top-left and on the right square of the purple group), or it can turn up a 1 (in which case there’s a mine on the left square of the purple group, and the top-left is safe), and either way it solves the region, using either three or four mines.

Other times you encounter this, the safe square may not immediately solve the region like that, but it will at least enable follow-up logic moves.  Below, the safe square turns up a 1, but we can do a minor bit of projection to mark 2 more safe squares, and proceed from there.

 →

Don’t assume that every 1-2-3 Hole line is this Wheel pattern.  Sometimes it’s just simple linear projection, like this at right: the 2 to the right of the hole projects one mine for the 3, so its remaining squares (above and left of the 3) have two mines, satisfying the middle 2, and clearing to its left.  Most of these formations require some thought in order to consistently solve them correctly, like this one - it’s probably not going to be as automatic as normal logic-based sweeping can be.

While we’re at it, here’s two other slightly-less-common projection patterns from Evil NG, briefly:

Hole 2-4: This pattern is the cousin of the Hole 2-3, but slightly less common.  It requires a 4 to turn up on the side of a hole-pattern’s three interior squares.  Next to it is generally a 2.

Of the six adjacencies that the 4 has, we can account for one mine in two front squares (teal), and “at most one mine” in two other squares (dotted purple subset of the blue group), making the other two squares definitely mines.  Then, returning to the “at most one” blue group, we can conclude that it is “exactly one mine” in those two purple squares, and thereby prove a safe square on the other side of it (the third square of the blue group).

It is common in Evil NG play for that one safe square to turn up a 1, providing a path to a follow-up move, via another bit of simpler projection logic (on the right: yellow accounts for one mine, therefore blue has one mine, therefore the interior 1 must have safe squares to its left).

Pond Bridge: This involves a small enclosed space where the two sides almost touch, and some interior squares of the front are known (on at least one side, maybe both).  From there, you can project mines out as a “bridge”, and thereby account for some of the mines on the other side’s interior number square, hoping its remaining squares must either be all safe or all mines.

Above, the group above that lower-middle 3 contains two mines in three squares.  However, that group fully satisfies the interior-3 at top.  Even though we don’t know which of those three cells are mines, we know by projection that the other squares around that upper interior 3 are safe.

Above: In the high-rank portion of this game, we are faced with this narrow region that has interior numbers on each side.  By projection, the right-side 1 accounts for one mine (yellow), leaving the 3 with two mines from the three squares to its left (blue group).  The interior 5 has one mine marked, gets two mines from the blue group, proving its other two squares, above and below, as both mines - and therefore the squares above and below those new mines are safe.

These can be hard to spot, because unlike most logic, it’s not coming from something along your front - it’s coming “across the pond” from something in an entirely different direction.  But with enough practice, it’ll become part of the things you see when scanning a front, and situations like the one at left will have the logic-move solution immediately occur to you.

While these are all different patterns seen in the high-rank segments of Evil NG, they all depend on the same underlying idea: we don’t need to know exactly where a mine goes, if we can at least narrow it down to a small group of squares that can then be usefully “projected” out to account for another square’s mines, and create a downstream conclusion from that reasoning.  That principle, if borne in mind in high-difficulty play of all types (e.g. the example at right is from Standard mode), can find you moves (or favorable probabilities) in situations that might otherwise leave you guessing.  We’ll refer to it frequently while discussing more-advanced techniques.

These two techniques (neighbor mismatch and projection) are not even close to exhaustive for all the patterns and logic you may need to reliably solve Evil NG boards.  For example, starting moves, i.e. the first few moves of a game, are more varied in Evil (and NG arenas).  You’ll also need to be prepared to use minecounting.  But these two topics account for a sizable fraction of what you need to know, beyond the basic patterns.  If you’ve mastered the visualization required to identify and use these techniques, you will likely be well-equipped to reason any given logic situation out from first principles.  You may even come to notice that you’ll get to the end of an Evil NG board, and not even remember tackling the high-rank section of it.  Because you did it from muscle memory and pattern recognition - you never had to engage your deeper logical-reasoning muscles.  That’s when you’ll know you’re truly ready for high-difficulty play.

Oh, and there’s one more aspect of Evil NG (and No Guessing mode in general) worth mention:

Because-it’s-NG Logic: Evil (and indeed any NG game) has another tool you can employ, from time to time.  Because NG guarantees that there are never any forced guesses, you will sometimes see a situation that looks like it’s forming a 50-50, or nearly so.  Sometimes a thought along the following lines will occur to you: “Either X has to happen, or that’s a 50-50 / forced guess”.  Well, in NG mode, that means that X has to be true!  Because something that looks like it’s almost a forced guess, could maybe be one, cannot be one.

At left: if the left-hand 4’s remaining mine is on either of the unknown cells to its left, those cells are a 50-50.  Alternatively, its final mine could be above the right-hand 4.  Because this is an NG game (actually an NG arena), we know that 50-50s are impossible, and can thus conclude that the mine has to be above the right 4, and the cells at left are both safe.  We would have eventually reached that conclusion anyway, via other logic coming from other directions… but that would take a lot more effort.  Instead, we can take the cheesy way out, and just gain a shortcut from the mere knowledge that this game is no-guessing.

At right: which cell, if a mine, completes the 50-50?  Then, given that this is in NG, you can be assured it is a safe cell.  Maybe we would’ve had an alternate solution to this area later, and come back and solved it in a normal way.  But we can solve it now, if we want - because it’s NG!

This is sometimes termed “NG metalogic”, i.e. logic-about-logic available in the game.  It’s borderline cheating, because it’s not a move that the game wants you to see and make.  But in the end it’s just a shortcut, and in my opinion it’s a legitimate move to make, if you see it but don’t see anything better.

Screenshotting of Hints

On custom boards on WoM, you get a maximum of 5 Hints per game.  And there’s no worse feeling than being nearly done with a board, hitting the end of available moves from logic, and having no hints left.  Now, a top-tier High Diff player like Scar or Lovesick or Hao1501 might be able to reason out where the best-probability moves are, and make optimal guesses even without using hints from the server.  Indeed, some top players look down on anyone using hints at all - may even consider it “cheating”.  But for the rest of us mere mortals, our 5 hints per game are precious, and we don’t want to waste them if we want to finish big, hard boards.

The most obvious way to not waste hints, of course, is to be able to play with perfect logic and identify all of our available safe moves.  Then only use hints once we run out of those.  This is a lot more achievable than it might sound: an experienced player can usually spot all logic moves, with enough practice.  Prudence will dictate that before using a hint, at least take a second sweep around the board, carefully look at each part of the front, and see if there might not be a move that occurs to you (even if it’s very unlikely to result in meaningful progress) that you missed the first time around.  Sometimes I’ll use a hint and it’ll just say “A solution was found” and show me something I should’ve seen, and I’ll feel dumb.  Maybe it was just a bit of cleanup from a solved part of the board, where in my haste I forgot to open a square, but there was no real doubt about it.  Maybe it was something important that I missed, or some high-rank logic 4 or 5 steps deep that didn’t look promising at first glance.  In which case the hint wasn’t a waste, per se, because I gained some insight from it.  And of course, if you have a 50-50, there’s usually no need to ask the server - you know what you’re looking at, and should just guess it.  But generally, if we’re careful, we can use hints only when we’re truly out of inferences to draw, and then a Hint will deliver probabilities to us, which is what we want.

But then an annoying thing happens: it shows you the probabilities, but the instant you guess your next move, all those numbers disappear.  Not for no reason, of course: guessing a square (and not dying) means that at least some of the probabilities have shifted, so the numbers that were true previously may no longer apply.  But because so many regions of the board are isolated from each other, from marked mines interposed between areas of the front, it may also be true that most of those probability numbers are, in fact, still good.  And you might wish you could just pull up those old numbers, rather than have to pay a hint to see them again.

Well, that’s why I recommend, in this situation, screenshotting the probability display.  I’ll usually use the “part of screen” method of screenshotting, and then pull up the image in an editor.  So then if I need to refer to it again (to make another 13% guess or 15% guess or something), I have it available.  And when I’ve made enough progress that that image no longer provides any useful guide to me as to where to guess next, I delete it, and spend my next Hint to get a new probability display (when next needed), and screenshot that one before it goes away too.  But at that point, I’ve saved myself from spending a bunch more hints (which do cost honor points), and greatly reduced the chances that I run out before I finish the game.

Other Useful Guides

The following guides, and principles from those guides, are high-quality and helpful for high-difficulty play.  You should read them, and I will sometimes refer to them later on:

(1) What is a 50-50? - by Scar (a former #1-in-overall-trophies WoM player)

Scar covers stuff that adds a little more color to our section on 50-50s above, but describes the same basic patterns (line-and-1, box-and-2), though he doesn’t use those terms.  He emphasizes that corners are much likelier to hold 50-50s than anywhere else, and that you should wait until you’re certain that something is a 50-50 before guessing it - both of which are important enough to repeat. He also discusses things like “extended” versions of the line-and-1 (at right), and 33 / 67 situations.

(2) How to Play Difficult Boards Optimally - by Archeaic, f/k/a Anonymous2484838

This guide lays out broad strategic principles for high-difficulty play, all of them useful, including:

• Always start a high difficulty board by clicking all 4 corners, in order to take a large fraction of 50-50s off the table before you invest time into a board.
• Guess confirmed 50-50s right away, again to prevent wasting a lot of time
• Given the choice on a large board, proceed around the edges of a board, because the 50-50s and guessing regions are much likelier to be found there, so you want to find them as early as possible.
• Place all flags when playing high-difficulty, in case you come down to an endgame minecount.  It also prevents errors of “going too fast” that are common to no-flag play.
• In guessing situations, guess in favor of situations that use the fewest number of mines possible.  He gives a few good examples.

Many of these we cover in greater depth below, but it’s worth reading because his discussion might resonate with you better than mine.

(3) The Art of Guessing, by Koro (frequently top-10 in many things WoM)

This guide focuses on a set of techniques found most helpful by one of the best high-difficulty players in the WoM community.  He covers a lot of topics also covered in this guide (Assume Fewer, Probability via Lock-In, Helpful Guesses and Dead Cells, etc), but he does it in his own style, which makes it worth a read.  He also goes into some topics not fully covered here:

1. Gauging odds of a guess resulting in progress, based on the number of unknown cells around it and other considerations
2. Pseudo-50-50s (guesses which are either safe, or are part of a 50-50, and thus either way are appropriate to guess immediately upon discovery)
3. Brute-forcing of endgames, by identifying and mapping all possible mine configurations
4. For Mastery play, how to learn the proper opening lines that optimize beginning a game successfully and the odds of proceeding into regular logic play
5. How to use MSCoach’s Solver / Analyzer to learn from experience, when you lose a game on what might’ve been a sub-optimal guess

His discussions and examples are very good, and he himself is a world-class high-difficulty player, far better than I am.  So his discussion might “click” for you in a way that this doesn’t.

(4) Lastly, I recommend this video series by top mastery player GlennGould (on youtube: MineBuoy).  Each video presents a minesweeper reasoning puzzle, using some of the logic common to advanced situations.  You might prefer a spoken narrative explanation, and his videos are good at illustrating them.  It’s a 15-video playlist, and they map to the set of techniques and topics I cover in this guide, as follows:

1. 4-square-box logic
2. Evil NG logic (several kinds: neighbor mismatch + projection / crossing pattern)
3. Minecount (5) plus proof-by-contradiction
4. Helpful Guesses
5. Hypothesis Testing
6. Minecount (13) plus proof-by-contradiction
7. 4-square-box logic, plus some downstream logic
8. Helpful Guesses
9. Hypothesis Testing[4]
10. Proof by Contradiction
11. Minecount plus some minor conditional logic
12. Hypothesis Testing (helped by some Evil NG-style logic)
13. Minecount plus some 50-50 wisdom
14. Probability Estimation (and wisdom about optimizing for Win % over Safe Guess %)
15. Hypothesis Testing - an absolutely brilliant puzzle requiring deep scenario reasoning

Most other minesweeper puzzle videos on YouTube would be considered trivial by advanced players, IMO[5], but these are great.  GlennGould also has two interesting videos on opening (initial) moves, which he terms The Bongcloud and The Classical.  I recommend them, as openings are not a topic I cover in depth here.

Rule of Three

Here is an easy and fast (but non-obvious) way to determine if you’re facing a 50-50 or not: If there is some number square that borders three unopened cells, some of which also border other number squares, then there is (very likely) a logic pathway to solving the area.  If all your number squares on the front only border either two or four unopened cells, then you are very likely facing a 50-50, because no logical combination of considering multiple squares together can help you find a further move. (if they border just one unopened cell, then obviously that’s your next move, either to clear it or mark it - the key here is if you’ve got squares that are adjacent to specifically three).

This usually only applies in endgame type scenarios where you’re left with only a single high-density area on the board, and are trying to figure out if there are any logic moves left available to you, or if you’re left with only guesses or (gulp) even some 50-50s.  And yes, this is speaking about a matter of logic: a deterministic move you can make (or not).  But it’s a rule of thumb that I’ve found helpful in determining whether I can make further progress, or if I’m stuck and need to burn a hint (or worse).  Sometimes the situation is confusing, and this clarifies it.

(A)  (B)

Two very common applications of this rule: At left in (A), you’ve got what looks a bit like an “extended line-and-1 50-50 with a corner bend”.  But I know as soon as I look at it that it’s solvable, because that middle clump has 3 unknown squares, not 2 or 4.  By that mere fact, it’s solvable, and we just need to look for the (obvious) logic move, which I leave to you.  In (B), we have a likewise simple “projection” case: The 4 in the middle has three unknown squares around it, therefore by the Rule of Three, we are not facing a 50-50.  A little inspection shows us that the 2 above that 4 is an effective-1, and thus accounts for the 3rd mine around the middle 4 (via projection), therefore its 4th mine must be to its left, and logic follows which clears the area.

On the other hand, we will also see things like these:

(C)   (D)

In both of these cases, all number squares here are adjoined by either 2 or 4 unknown squares (or 6 squares, in the central 3’s case at left in C).  All pairs of cells are connected to each other, and no amount of logic will reveal which of the two mine layouts is correct here.  By the Rule of Three, both of these scenarios are confirmed as 50-50s.

What we’re adding with this rule is not some deep logical insight, but rather a handy heuristic so you can move quickly even in murky, dense endgame situations.  Speed of recognition is the goal.  Using this will also save you a hint, if you don’t need to confirm that the whole formation is a 50-50.  And all that separates examples A and B from examples C and D is a single square in the former which has 3 unknown squares around it; from that, a full solution is possible.  Lacking it (in C and D) means you’ve got a 50-50 forced guess, even if it’s an unusual formation.  (Sidenote: certain guesses in D are better (more “helpful”) than others - can you tell which?)

The Rule of Three is phrased very specifically: it can prove solvability, but if its condition is not satisfied, it doesn’t necessarily mean you’re facing a 50-50.  If a position fails the Rule of Three, and looks pretty doomed, but is nevertheless solvable, most of the time this happens because the space is mine-countable.  We’ll discuss minecounting in-depth shortly.  But for purposes here, if the Rule of Three says it could be a 50-50, double-check how many unknown squares are on the board: if it’s an odd number of squares, it’s solvable by mine-counting.  Often it will be even, and there’s nothing more you can do.  But sometimes you get something like this:

(E)  →

(F)  →

In those examples, the Rule of Three can’t prove solvability (all number squares face up to 2 or 4 squares only), but because there’s an odd number of remaining squares on the board, it is still mine-countable.  If it weren’t - if the situation were slightly different and had an even number of unknown squares - it would be a 50-50.

The example at right fails the Rule of Three test: all squares face up to either 2 or 4 squares.  However, it is obviously minecountable, too: a quick inspection will show that it must contain either 2 or 3 mines, and in either case is quickly solvable.  So we need not guess it immediately - we can (likely) wait till the endgame, and solve it then.

It might be possible that there are some exceptions to the Rule of Three, but as best I can tell, I believe they all fall into one category: The numbers that have three open adjacencies (for one or two mines) do not also have those adjacencies, those unopened squares, also bordering other numbers.  Or they only border numbers which have the exact same squares for their adjacencies.  An isolated 1-in-3 (one mine in three squares) does nothing for you.  But if you have a tight grid of confusion where some initial reasoning makes you worry you’re facing a 50-50, then look for numbers that have three adjacencies - those will likely be your way out of that mess, if one is to be found.

Here’s a bit from Scar and Koro’s world-record HD video (at 37:50): we have a formation that greatly resembles the circular 50-50 kind of scenario (the lower half is exactly that - the corner cells that might disambiguate are all mines).  He was worried for a second that it was a 50-50, but then noticed that the upper interior 1 (to the right of his mouse cursor) is a 1-in-2, while the 3 above it covers three cells (a 1-in-3) which include the former two.  Therefore we know just from logic that there’s a safe cell, which he quickly finds (under his mouse cursor).  This solves what would otherwise have been a 50-50.

Try this quiz using the Rule of Three: Simply decide which of the below are Solvable or 50-50s.  Pay no heed to whether they might become 50-50s after a next step; if there is a cell that can be marked safe right now, call it Solvable.  Maybe write down your answers somewhere as you go, then when done, check against the answer key below.

(A)   (B)   (C)

(D)   (E)   (F)

(G)   (H)   (I)

(J)   (K)   (L)

(M)   (N)  (O)

(P)  (Q)

(R)  (S)

Some of the above, even if they’re 50-50s, do have some Helpful Guesses that would be preferred to others.

Answer key:

1. Solvable (the left-interior 4, from 2 directions)
2. Solvable (the right lower 4 projects onto the 4 to the left, solving the box)
3. 50-50 (but there’s one cell you really don’t want to be a mine, so you’d guess that one)
4. 50-50 (but, there are 2 more-useful guesses: the square above the interior 4, and the upper square on the right front)
5. Solvable (projected from the right 3, the square above the left-hand 2 is a mine, satisfying the 5 above; or, projected from the 5 on down, next square down is a mine)
6. 50-50
7. 50-50 (an extended Swirly pattern)
8. Solvable (Project the 3 from mid-right; lower-right square of the box will tell you 1 or 2)
9. 50-50 (in a beginner game, crazily enough)
10. Solvable (project down from the middle 2; bottom-middle is a mine, solves from left)
11. Solvable (project left from the interior 3; the 5 tells you the lower cell at left is a mine)
12. Solvable (bottom is a 4-square-box 2 mine diagonal; project upwards and you see the cell to the left of the upper 3 is a mine)
13. 50-50 (if that upper 6 were a 5, though, it would be solvable)
14. Solvable (two cells on the middle column facing 3 squares; then it’s likely minecountable)
15. 50-50 (but there are 2 cells that are worse guesses than the rest, can you find them?)
16. Solvable (there are 2 safe cells you can identify right now)
17. 50-50 (it’s a trick question: this is mine-countable if we solve the rest of the board first.  But there’s no safe square as of right now.  Don’t believe it?  Check out the game.)
18. Solvable (logic move, from overlap: right box into upper extension)
19. Solvable (knowing it’s solvable can be seen through the Rule of Three, but finding the solution requires 4-square-box logic, covered later in this guide)

Isolated Linear Fronts

This set of logic rules might really be more of an Intermediate Technique kind of thing, but it’s worth discussing because they affect gameplay speed / pattern-recognition significantly, and the concepts behind it are a good foundation for some of the advanced techniques we’ll discuss.

Here, we’re talking about a front which is three things:

1. Isolated, meaning that the mine configuration within it cannot be determined by the immediately adjacent squares on the edges of the front, and furthermore
2. One in which we have an open number square facing up to the squares on the edge of the front (it’s not partially obscured by mines), for reasons that will become clear, and
3. A front that is Linear, i.e. all the squares of the front area are in a straight line.

At left, the green area is isolated and linear.  We could mark the mines below it (below the 3) and it would still meet the criteria.

In the second picture, the blue and pink areas - even though they’re separated by an intervening mine - are not isolated from each other, they share a square in between.  Marking that shared square as a mine, however, would make the two regions isolated from each other, leaving the pink front as a 3-square isolated linear front (and it’s a 1-2-1, a common logic pattern).

The third image above (bottom-right) isn’t linear: any logic we might apply to it wraps around that corner, and no portion of it is isolated into a linear fragment we can consider by itself.  Plenty of logic rules might still apply to it, but the shorthands here can’t help with it.

The red box at left looks isolated and linear, but the mine to the left of the 4 means that we only have one number touching the leftmost part of the red-boxed front.  Although this particular region still offers a logic move, it’s not an example of an ILF.

The orange region at left is an isolated linear front: it is “capped” at top and bottom, and we have number squares facing each end of the linear portion.  Although the full region isn’t linear (below the orange, it winds around and continues), the linear part in the orange box can’t be determined from the squares below it, so it still meets our definition.

Got it?

So the fun thing is, most isolated linear fronts (ILFs) are fully deterministic and solvable!  Let’s discuss two classes of them: Ones where the whole front consists of effective-1s, and ones where it doesn’t.  These solutions can be determined by reasoning from first-principles, because in all cases we have a 2-over-3-square logic rule available to us (WoM calls it a “1-1” pattern).  So I’m just summarizing the scenarios so you can apply them faster.

(a) ILFs with solely effective-1s on the front line.

These have solutions for most lengths of the front:

• 3 squares: The 1 mine is in the middle square (#2)
• 4 squares: The 2 mines are at the ends, i.e. top (#1) and bottom (#4)
• 5 squares: Indeterminate.  The middle square (#3) is safe, but the others form a 50-50; mines could be at #1&4, or at #2&5, and only additional information from outside can help determine which.
• 6 squares: The 2 mines are at positions #2 and #5, while the other 4 squares are safe.
• 7 squares: The 3 mines are at positions #1, #4 and #7.
• 8 squares: Indeterminate.  Squares #3 and #6 are safe, and the others have two possibilities (1/4/7 or 2/5/8).  It will be a 50-50 between those two configurations, unless you have help from above or below.[6]
• 9 squares: The 3 mines are at positions #2, #5 and #8, each in the middle of 3 squares.

3 squares: the mine is in the middle square in each of these.

 4 squares: the mines are on the ends (first and last squares)

 6 squares: Mines are at spots 2 and 5, as marked; others are safe.

 7 squares: Mines are at spots 1, 4 and 7, as marked.

These are all very handy shortcuts, in the instances where you encounter isolated linear fronts in the wild.  Bigger ones (6+) are a bit rarer, but small ones are extremely common - and being able to dispose of them quickly can add a lot of speed to your gameplay.

 At left: a 9-square All-1s ILF

(b) ILFs with number squares that are not solely effective-1s.  There are only a few distinct situations here, and they each have clear solutions.  While you can reason them all out using basic logic rules, it is actually a lot faster to just perceive what the effective numbers are for the adjacent front, and apply the standard rule for that pattern.  Each pattern given here applies in rotation, depending on what direction your front is going, and can be flipped top-to-bottom or left-to-right as needed too, of course.

3 squares:

• (1-1-1: The one mine is in the middle square, the outside ones are safe)
• 1-2-1: Mines are on the outside, the middle one is safe, same as the normal 1-2-1 pattern appearing anywhere.
• 2-2-1: The left square is full (2 squares for 2 mines), so the mines are there, and the third is safe.  Likewise if it’s 1-2-2.
• 2-3-2: All 3 squares are mines.  Obviously.

 You didn’t really need me to tell you to mark above the lower 4 here, on the 2-2-1 ILF.  But sometimes the obviousness of the move is disguised a bit by needing to first see the effective numbers.  This is especially obvious with 4-square ILFs.

4 squares:

• (1-1-1-1: Mines are on the 1st and 4th squares, 2nd and 3rd are safe, as above)
• 1-2-1-1: Mines are against the 1st and 3rd squares (like the normal 1-2-1 pattern), 2nd and 4th are safe.
• 1-2-2-1: Mines are in the middle 2 squares, #2 and #3; outside squares #1 and #4 are safe (just like the normal 1-2-2-1 pattern).
• 2-2-2-1: Mines are in squares #1, #2 and #4; square #3 is safe.
• 2-3-2-1: Mines are on the left 3 squares, only the 4th is safe.
• 2-3-3-2: All 4 squares are mines.  Obviously.

Some of these will always be obvious, because the edge of the linear front has a square that is “full”: all its adjacent squares are mines.  So in general, the 4-square situations you’ll encounter that aren’t obvious will be ones where you’re really just trying to discern which of three patterns you’re facing: 1-1-1-1 (mines on outside), 1-2-1-1 (mines on first and third spots), and 1-2-2-1 (mines in middle).  The trick is perceiving what the effective numbers are on your front.  That’s why this is useful shorthand - because otherwise, without these firm principles, it becomes a lot easier to make a mistake in trying to reason out the logic.

5 squares and up: It’s not worth setting out a general set of cases, because your fundamental play patterns will still apply to a subset of the front (e.g. 1-2-1, 1-2-2-1), and you should try to identify those spots and apply logic from there.

N.B.: sometimes you have an ILF where you have information from both sides, like in an endgame where you’ve gotten the unknown region down to just one column.  In these cases, you don’t necessarily need to have all the squares of the front adjoined by number squares on one side; it’s enough to have number squares on one side or the other for each square.  This can help a lot if your front is interspersed with mines, in determining which pattern you face.

(c) Here’s a quiz on the 4-square ILFs.  It’s mostly about fast recognition of what the effective number squares are.  Can you spot which patterns apply in each of the following ILF cases?

(A)   (B)   (C)

(D)   (E)   (F)

(G)   (H)  (I)

(J)   (K)   (L)

(M)   (N)

Answer key, all effective numbers listed from either left-to-right or top-to-bottom:

1. 1-1-1-1 (outside edges are mines)
2. 1-2-1-1 (1st and 3rd squares are mines)
3. 1-1-2-1 (2nd and 4th squares are mines)
4. 1-2-2-1 (middle squares are mines)
5. 1-1-1-1 (outside edges)
6. 1-2-2-1 (middle squares)
7. 1-1-1-1 (outside edges)
8. 1-1-2-1 (2nd and 4th squares)
9. 1-1-1-1 (outside edges)
10. 1-2-1-1 (1st and 3rd squares)
11. 1-1-2-1 (2nd and 4th squares)
12. 1-2-2-1 (middle squares)
13. 1-2-2-1 (middle squares)
14. 1-1-1-1 (outside edges)

The Eight Square Opening

When playing high difficulty boards, you’ll note that the game’s engine calculates a “win probability” that might seem awfully low.  But a great number of these would-be failures are actually preventable, through some smart choices for opening.  If you could turn a “Win rate: 1.36%” (the WoM-calculated rate for 100x100/2183) into, say, 5%, that would be pretty great, wouldn’t it?  To do so, all you need to contribute is a little patience at the very start.  To get a winnable board on higher-difficulty games more often (that boost in win rate), with a minimum of frustration, first note this important observation: 50-50s happen more often at the corners of the board than elsewhere.  Recall that in a board corner, it requires only 3 mines to be placed in just the wrong place to create an unsolvable dilemma, requiring a 50-50 guess:

At left is the dreaded “overhang” , the line-and-1 corner variant, with 2 mines shielding an area of 2 squares with 1 mine between them.  Either the corner, or the square to its right (in this example), is a mine, and nothing will tell you which, other than guessing.  The same is true of corner-located box-and-2 50-50s (example at right): If we’d approached this area from any other direction, it would have ended as a 4-square box-and-2 50-50, requiring guessing.

While these are the simplest and most common, there are lots of creative ways that a corner can create a 50-50 guessing situation, too.  (see e.g. at right).  Because you can only have information coming from 2 sides, instead of 4 sides in the middle of the board, it’s easier to run out of logic.

For this reason, it is common advice in other guides that when playing high difficulty, you should open all 4 corners first, and only if they are each safe should you proceed trying to play the rest of the board.  That’s the “4-corners opening”.  It’s recommended for 2 reasons:

1. If there are 50-50s right on the corners, you’ve just guessed them correctly already (or blasted, and moved on to the next game), so you’re taking a good fraction of 50-50s off the board before you start really playing it.
2. Openings (zero-mine squares) are more likely on the corners than anywhere else, because they only require 3 surrounding squares not be mines in order to yield an opening, as opposed to 5 squares on the edge and 8 squares in the interior.  So you’re more likely to start off with the information from an opening to continue from.  Efficiency players often do it for this reason alone.

This is good advice, and I don’t disagree with it.  However, I actually take it a step further: When playing high difficulty, I will open 8 squares to start, not just the 4 corners.  I suggest that at each corner area, open the corner square, and then also the square 2 rows and 2 columns “inland” from that corner.  So if starting from the top left corner, you click the corner, and then the square in the 3rd row / 3rd column.  Then likewise for the other 3 corners, and only if you survive all 8 squares do you proceed trying to play.  That’s the “8-square opening”.

Doing it can be annoying: If you’ve got a 22-25% mine density, it may take a while to get a board with all 8 of those particular squares safe.  At 25% density, the odds that you get a board where 8 randomly selected squares (which these effectively are, since you’re just guessing into space) are all safe, is (75% ^ 8) = 10%.  And so sometimes it could take as many as 30 or 40 board tries or so in order to have that 10% board come up, and let you begin playing it for real.  Whereas, for a board with 19% density, that equivalent number is (81% ^ 8) = 18.5% chance, or nearly twice as likely.  So the difference in that density can really affect how easy and quick it is to find a playable board.  But blasting in your first couple of moves and moving on to the next board takes only a couple seconds.  Whereas playing a whole board with dragons lurking near the corners (if you leave them alone and don’t open at the start) could cost you 15-20 minutes of work, only to then end with a 50-50 guess, often wasting all that time you spent.

So, what does doing the 8-square opening gain you, in addition to what you’re already getting from opening the 4 corners?  Some strategic benefits for completing big and/or hard boards:

3. While 50-50s are indeed most likely on the corners, the regions around the 3rd row in from the edge contain a much higher incidence rate of mines that are a part of 50-50s which aren’t exactly on the corners, but are near the corners.  So you end up avoiding a bunch more 50-50s by preemptively getting these inland squares too.  Of course, you’ll also blast a bunch on mines that were never 50-50s, but the same is true of the 4-corner opening,  too.  If you can save yourself some bigger heartache later, it’s worth taking a bit longer to find a playable board to begin with.
4. Even when there’s no 50-50 exactly on the corner, the corner-ish regions of the board are the likeliest areas in which to run into an area where you run out of logic moves and have to make a guess (even if your odds on it are far better than 50-50).  We’ll talk about such guessing later, but these tend to happen toward the corners because the walls, as they approach, shut down possible avenues to solve, mark and clear an area. It’s harder to clear around edges, and harder still to clear around corners, and you get stuck proportionally more.  So, by getting this second guess in each corner region, you’ll find that sometimes, that extra piece of information can help unlock a few additional moves, as you get close to closing out a corner region.
5. By having two known squares in a corner area, if you run out of logic moves on a board (even if you’re nowhere near the corner region), the spots right around that inland square you’ve opened may provide a lifeline of a good-percentage guess for you to try to continue from.  And you can chain some guesses together (see Guessing Rule #5 below) if you get going there, and think you can smell an opening in that direction.

To convince you of #5, consider the following percentages, based on Expert density (20.6%):

In the first (a 1-1 corner, i.e. 1 on the corner itself and 1 on the inland square), your interior square has 11%s around it for all squares except the one shared with the corner.  For a 2-1 opening (the middle image), it’s even better: 7%.  You might expect the 3 squares around the corner to each have 67%, but because we have a 1 sharing it, that “pulls away” some of the probability for the shared square, and in fact that shared square is only 52%.  But 52% on that 1 square is just fine, because it leaves only 48% odds to be distributed evenly among the other 7 squares!  And even on a 2-2 corner opening (the image at right), your odds are still at 18%.  So while you might not choose to guess around a (say) 2-4 opening, any of these low-number ones provide pretty good odds - and you only have them available if you’ve done the 8-square opening.  And that’s at Expert density (20%); your advantage, with the percentages relative to the background rate, increases with mine density.

To convince you of #3 above, allow me to provide just a few images from games where I blasted in the course of guessing an inland-from-the-corner square - and then could see, from the revealed mine pattern, that I’d run my ship aground on the rocks of what was a 50-50 anyway.  In each case, the corner itself was safe, but the inland square (3rd row / 3rd corner) was a mine, and so only by doing the 8-square opening did I spare myself dealing with this 50-50.  You’ll see all kinds here, but mostly line-and-1 50-50s where that inland square provides part of the “line”:

(Above, 3rd image, you’ve got a “half-full bucket”, on left edge)

(Above row are all “doubles”, e.g. at right, you have a line-and-1 just above the blast square, but you also have a box-and-2 to its right, for which the blast square is one of the framing corners)

         (Jeez)

And ugliest of all, a quadruple 50-50:

I trust that makes the point well enough.  With some of these, your odds of clearing it successfully would’ve been far less than 50%, because you’d end up with a guessing area requiring more than one guess, or a higher-than-50% mine density for the area.  Some of these are worst-case-scenarios which happen very rarely, of course, but this kind of thing is frequent enough that it’s worth a little more effort (in finding a playable starting board) to avoid the risk of it.  You’re trading a little time upfront to take some percentage of 50-50s off the board.  For large, high-difficulty boards where it can take 20-30 minutes or more to play through it, this is a tradeoff well worth making.

Now, does this opening guarantee there isn’t still a 50-50 in the vicinity of a corner?  Of course not - see at right.  It just makes it much less likely.

Lastly, in defining this 8-square-opening tactic, there are two implementation details worth noting:

First, perhaps obviously, you only open an interior square dictated by this plan if you don’t get an opening from its corner square.  If you get an opening, the 3rd row / 3rd column square may even be provably a mine based on the revealed pattern, and you don’t want to blast intentionally just on general principle.  So the real rule is “open the corner, and then if it’s not an opening, also open the 3rd row / 3rd column square, and if both are safe then move on”.

Second, part of the point of doing this 8-square opening is to get more information to work with to start the game.  So, if at any point I have what I consider 2 “useful” openings from the corners, ones in which I know there are logic moves that follow, I then revert from also opening that inland square, to just opening the remaining corners.  So for the 4th corner, and sometimes for the 3rd one too, I may not open that inland square.  This is in order to not throw away a very promising opening position, just for the (more minor) gains of the additional inland squares.  You might sometimes see a great big opening and you think “hmm, I’d like to play this!”, and then while trying to complete your 8-square opening, you blast.  That can be very frustrating!  On the other hand, if you avoid the corners too, you may end up playing that very tempting opening for a while, get to the end of the game, and discover a 50-50 on one of those corners that you could have avoided by clicking it at the start.  Far worse!  So my compromise position is, abandoning the “inland squares” part of the tactic, if and only if I get two useful openings from corners.  You may find yourself feeling more conservative (“do the 8-square regardless!”) or more liberal (“revert to 4-corners even with only 1 good-looking opening!”), based on your own tastes.

Again, you’ll have plenty of times where by doing this, you’ll blast on a mine that wasn’t part of a 50-50 or guessing region, and was perfectly playable.  That’s just a consequence worth accepting: the goal with this is to remove the highest fraction of 50-50s that we can, and leave ourselves to play boards that have the highest likelihood of being solvable.  This helps with that.

Rules for Guessing

Here are my own rules that I follow, when reaching a point in a game where I don’t have any further moves to make through logical reasoning.  These are mostly mid-game rules, not initial-moves strategy.  These apply to situations where guessing is likely required (there is a nonzero chance of losing), but the choice of guess can be tactically optimized.  In that sense they’re not really rules, they’re more “things to look for”.  I’ll elaborate on each category with examples shortly:

1. Check if it’s mine-count-able.  If so, don’t guess, come back to it at the endgame.
2. Find a spot where we can make the assumption of fewer mines; remember, mines are still less likely than not-mines (even at 25% density).  Look for spots where the possible configurations include some that would require more mines; those are less likely.
3. Think about which side of a guess would be more helpful if your guess is right.  Sometimes there’s a chain where if you guess a 50-50 you’re left with another 50-50, but if you guess a certain square and you’re right, then you’re not facing an additional 50-50.
4. Look at the general ratio of squares described by a peripheral square.  Sometimes you can get a 1 (or effective-1) against 5 or even 6 unknown squares, making the odds for any surrounding square a lot lower.
5. Chain your guesses diagonally, if you get 1s in space.  To try to find an opening, if you’ve guessed into terra incognita and gotten a 1, keep going in a diagonal direction from the 1 (and if you get a non-1, go back and try another direction).
6. Up against a wall alternative, skip a row and try the next wall-spot over, because if your guess comes through with a 1, then you get three (or more) squares for free.

For some of these, they are not my original thoughts: they have been explored, sometimes at length, by prominent players who came before me, and discovering their work reinforced or structured my thinking on the principle.  I’ll credit them when appropriate, and if I’m missing any contributory work, please write to me.  Anyway, let’s discuss each of those now, with examples.

(1) Check if it’s mine-countable

A mine-countable situation is one where, if you knew how many mines were contained within an area, you could find its unique solution.  Most commonly, it’s seen where an area could contain either one or two mines (but could be a two-or-three split, or more), but you won’t know which it is until you reach the endgame and solve all other parts of the board.  When you’re mid-game, it might still look like a forced guess.  But it’s not!  Now, sometimes it might make sense to guess it immediately, but you don’t have to, because minecount situations are nearly always solvable, once you reach the endgame and have a final mine count for the remaining area.

(a) Simple Mine Counting

These smaller mine-counting situations are collectively termed “N vs N+1” positions.  In each case, once you recognize it, you can leave it and come back at the end of the game when you know whether it holds N or N+1 mines.  Here are some examples of 1v2 minecounts:

Examples of 2v3 minecounts:

For the one at right (with the bend), note that there is still no number square that sees more than 2 numbers.  The key aspect that makes it obvious (as you’re cruising along solving) that this is a mine-countable situation is that it features an odd number of unknown squares[7].

How about a 3v4, while we’re at it?  In this case it became the 3-mine count, at endgame:

  →

There are limits as to how much solving this can provide, though.  For example, if you end up with more than one mine-counting spot on a board, there are some scenarios (endgame mine counts) that end up unsolvable after all.  Suppose you have two mine-countable regions on the board, both of which are 1v2s.  If you finish everything else and are left with a mine count of 2, then great!  Both of them can be solved with the 1-mine option, and you’re done.  Likewise if you happen to end with a mine count of 4 - then they’re both the 2-mine option, as unlikely as that is.  But if you end up with a mine count of 3, you don’t know which one is the 1 and which one is the 2.  You’re forced to guess, and it’ll be a 50-50 (but luckily, if you’re right on the first one, then the second one is determined).  Here’s an example, where we have a 1v2 and a 2v3 in the endgame.  We can solve it for 3 mines (or 5), but 4 would have been a split-region 50-50:

(b) Generalized / Advanced Mine Counting

The more you play advanced boards, the more you will encounter endgames where a larger region may be solvable with minecount, but it is not a simple (1-vs-2, 2-vs-3) either-or question.  To attempt to solve it using minecount, you will want to attempt to apply the following process:

1. Using number squares from your front that borders the area, identify groups of unknown squares (“square groups”) in which you know there is 1 mine (or more).  This is often done with test clicks (fake chording) to confirm which squares you’re looking at.  
2. Attempt to mentally select, for your set of square groups, only sets of squares which are mutually exclusive, i.e. where they do not overlap with each other.  Sometimes this is impossible - but see (e) below.  If you can’t keep it all in your head, see the note below.

1. It’s also OK to make into a square group an area that can have N or N+1 mines (most often, 1v2), and to run with that with your count.  So instead of counting “1, 2, 3…” you’ll be counting “1, 2 or 3, 3 or 4…” as you go.  This can still help you solve, so don’t be afraid to include a “conditional” group like this in your count if it’s the only way forward.

3. As you go through the available squares, keep a running count in your head of the number of mines accounted for by the square groups you’ve identified so far.
4. Your running count may reach the number of mines remaining on the board.  If so, and only if there are other unknown squares your square groups did not touch, you can safely clear those other squares, because no mine could be in them - all the mines have to be in the spots you’ve identified.  And then, likely, the new information from those squares will help you solve the remainder.

1. But remember that your candidate grouping of squares must satisfy all frontline number squares for it to be a valid candidate layout.  It doesn’t have to cover all unknown squares touching all frontline numbers - just has to meet the minimum number of remaining mines that those squares need.  But if your proposed grouping ignores an (unsatisfied) square entirely, then it can’t be valid.

5. In some situations, you may not be able to keep your square groups mutually exclusive, but you may also notice that you can only reach your full minecount if you were to allow square groups to overlap a bit (usually on just 1 square).  If so, this means that the mines for those groups have to be on those overlapping squares, or else you don’t have enough mines to satisfy all number squares available.  Given that mines are less common than not-mines, this happens often enough that it can be quite convenient.
6. In even-rarer situations, you’ll find that the only way you can imagine all the mines fitting is to assume that mines are not on the overlap squares between some overlapping square groups - you have a really high mine count remaining, and this forces a formation of them that maximizes density.  This is a “max count” or “mine-cramming” situation.

Let’s first consider a simple example of this thought process:

Suppose we have a minecount of 2 remaining.  It’s obvious that there’s one mine in the lower two squares, and one in the upper three.  Now if you were to look at the upper three squares in isolation, you can consider it from either direction (top or right) and isolate a set of two squares in which one mine must exist:

From the perspective of the right side (the 3, facing the orange group), the left square must be safe, because the orange group accounts for the one remaining mine.  But then again, from the perspective of the top side (the 2-5, facing the blue group), the bottom square of the three must be safe (in the orange but not in the blue), because the blue group accounts for the one remaining mine.  Which is correct?  Well, if the overlap square were safe, there would have to be mines on both the non-overlap squares, which would overflow our minecount.  Therefore the mine is in the shared corner (the top-right square), and the others are safe.  And then, from logic, the bottom square of the board here is safe and the cell above it is the remaining mine.  Trivial, perhaps, but the process extends to situations that are far from trivial.

Minecounting a larger area depends on being able to “account for” all remaining mines on the board, while using fewer than all remaining squares on the board.  Take the example at left: We don’t have a simple 1v2 or 2v3 scenario here to solve, but it is still easily mine-countable:
1 mine is accounted for by the interior 2 (1 mine in those 5 orange squares), and the other mine in the two squares at bottom (in blue).  The two areas are mutually exclusive (have no overlap), so together they must account for the 2 remaining mines.  Therefore the other 3 squares (highlighted in green) must be safe.

Tool Note: When identifying square groups, you can usually do it all in your head for simple cases, but as they get more complex, it might become hard to keep track of, or make mistakes more likely.  If you want some help in marking your square groups, many players on WoM will opt for using the “Open in the Editor” button, which takes a picture of the current game situation, and opens it in an MSPaint-like in-browser graphics editor.  You can click the “paint bucket” icon to then fill in each square with a color.  This then lets you color-code different square groups to test your grouping in a safe space, and also to quickly undo (via the undo button, or Ctrl-Z) so you can play around.  If you prefer, or if you aren’t on WoM, any graphics editor will do - just take a screenshot.

(c) All Groupings Are Correct

In minecount-able situations, it is fairly common to be able to find more than one valid square grouping (layout of the mines marked in square groups), if you play around enough with options.  If a grouping (A) can satisfy all squares, (B) has no overlaps, and (C) is sufficient (accounts for the full count), then it is valid.  So if you see multiple such possibilities, how would you choose between them?  The good news is, you don’t have to!

The squares that are never part of a square group in any of the possible groupings of a mine-countable region are, pretty obviously, safe.  But less obviously, it is also true that an unknown square that is excluded from the square groups in any one of the possible minecount groupings, is definitely safe.  If it’s possible to construct a square-grouping that accounts for all remaining mines, without using all available squares, then those residual squares are all safe - full stop. Any permutation of those square groupings just serves to narrow down the options as to where the mines go.  I call this the “all groupings are correct” rule[8].

For a simple example, consider the below endgame (from here), with 3 mines remaining:

 →   or:

Seeing that region, we have (in the middle image) a grouping that relies on the three effective-1s: the 4 on top (yellow group), the 2 at left (blue), and the 3 at right (pink).  Because the three mines are all accounted for, and all number squares can be satisfied from those groups, and the groups don’t overlap, it is a valid grouping, and we can conclude that the two other squares are both safe (in green).  However, if instead (right image) we start from the interior 5, which is an effective-2, and re-use the blue group, we get another valid grouping which shows that all three mines must be contained within purple + blue, so therefore a mostly-different set of squares must all be safe.

Which grouping is correct?  Both!  Combining the conclusions from the two groupings, we find that the left square of the yellow group is safe (so its right square is a mine), likewise the mine next to the 3 (pink group) is on its upper square, and this satisfies the 5.  The location of the final mine in the blue group will quickly be revealed after opening those safe squares.

Now for a less-intuitive example. Suppose at endgame we have a minecount of 5, and see this:

Firstly, as we get into more complex minecounting regions, it’s easy to forget that our square groupings still have to satisfy all number squares that still need a mine (or mines).  If we were naively playing around with just the perimeter front (the exterior of this region), notice what happens if we cram all five mines in that perimeter, by assuming the lower-left corner is safe:

→

The mine marked above the middle 3 accounts for the second mine in that “bridge” 2 (top-middle, orange-circled), therefore the other squares around it would be safe, and that “signal” propagates out to the right and finishes our marking.  But it’s marked our five mines, yet there’s a square that’s unsatisfied: the 2 in the center of the left formation, which got one mine from below but needs another from the top - and we don’t have any left to give it!  Oops.

We have, in essence, sort of proved that the arrangement above gets it exactly backwards. Instead, is there a way we can arrange the square-groups to account for all five mines, while satisfying all number squares?  The key insight we’d need is that the blue-box area above, accounting for one mine in three squares, is always going to contain a mine because of the hole pattern, coming from the 1 below it.  So we could lay out the square groups this way:

 or:

Each of these groups is defined by having a front-line number telling us it has one mine.  So this accounts for all our five mines, in either layout of groups, and therefore the unused squares here are safe.  But note that the pink group can be flipped (horizontal or vertical), while keeping the grouping valid.  What does this fact mean for our mine placement?  It means that the mine is actually in the only shared square between those two possible groupings, i.e. the lower-left corner.  Logic on which particular squares are safe or mines follows from there:

The whole board isn’t marked with this solution (the blue group in the “hole”), and that’s okay.  We’ll be able to figure out where the remaining mine (above that 2) is, specifically, once we clear things.  Our only goal with doing a minecount is to identify at least one safe square.  If we do that, logic usually can yield progress from there.

The fact that we can generalize about multiple groupings like this, is useful in several ways.  Although you may only need one valid grouping in order to make progress (if it yields a safe square), sometimes certain layouts are easier to visualize than others.  If you’re playing for efficiency, being able to spot multiple ones (and thus draw conclusions on mine placements) can help with saving clicks.  Consider this, with 4 mines remaining:

At first glance, the fact that we have that interior 3 pretty much clinches it for us, because by reduction (the effective-1 below it), there’s one mine in the 3 squares above that 3, untouched by anything else.  So there’s a pretty simple square-grouping we can do here to get to 4 mines:

As you can see at right, using just the 2 mines from that interior-3, plus one at the left and one at the right, gets us to our minecount of 4.  So we firstly know, right off the bat, that those 4 uncovered squares towards the right are all safe.  That’s probably enough to solve this region, most of the time.  But we can go further!  Because we also have this grouping available to us:

Just flipping things around, if we start by observing “hey that interior 4 is an effective-2, that’s half the mines I need!” and then fill in, we can get another, sufficient (covers all 4 mines) grouping, one that has no overlapping square groups.  By our rule, this grouping is also “correct”, i.e. true.

But now look at those two groupings together.  Notice that in order for them both to be correct, the only place for the 2nd mine to go for that interior-4 is to its top-left, i.e. the right square in the orange group (in the first grouping).  Likewise, the teal groups in both groupings are “correct”, so the mine for them must be the shared corner, the lower-left square right above that 4.  Which together, leads us to our complete solution:

(d) N-1 Groupings: The “all groupings are correct” rule works to demonstrate safe squares, but it can also work to demonstrate mines.  If you’ve got a board with N mines left, and you can find a square grouping that covers N-1 mines which uses all-but-1 square on the board, the rule lets you conclude that the final square must be a mine.  For example, take this, with 4 mines left:

 →

Looking at the scenario at left, there’s no grouping that covers all 4 mines, nor are there logic moves.  But there is a grouping that makes three one-mine square groups, and leaves out one square. So by this rule, that square (lower-right) has to be the 4th mine (image at right).  Logic moves then follow.

This isn’t limited to N-1 counts, either; they are just more common.  At right is an N-2 minecount, there are 5 mines remaining.  After three one-square groups, there are two mines needed from two residual squares, so they are both mines.  A little logic and smart guessing later, this becomes solvable.

(e) Partial-Overlap Minecounts: In our process steps above, “(e)” noted that sometimes, you can only reach your full minecount if you allow yourself to include square groups that overlap (usually by one square), and get comfortable with creating “1v2” or “2v3” type situations this way.  This shouldn’t be a first resort, but if you’re staring at something you just know is mine-countable, and yet you can’t make the square groups work, this is a further step you can take.  Consider the following region, with 9 mines remaining:

 →

Try as we might, we can’t get a fully mutually-exclusive set of square groups that amount to our 9 mines.  Excluding the pink group entirely, orange + yellow + blue + teal gets us to 8.  If we’re strict about groups being mutually-exclusive, we can’t include both the pink and blue in a single grouping.  But, do we always have to be strict?

Maybe not. Think about it this way: The pink and blue overlap by a single square, and there are 2 mines in the pink and 3 mines in the blue.  Either the overlap square is safe, in which case the combined area has 5 mines (2 in the left pink squares, 3 in the remaining blue squares), or the overlap is a mine, in which case the combined area only has 4 (the overlap, the 1 remaining in the pink, and the 2 remaining in the blue).  So the blue-pink area together is a “4v5”.  Aside from this area, we have 5 other mines that ARE in mutually-exclusive areas, and 9 mines total.  So our blue+pink region has to have no more than 4 total.  Therefore from our 4v5, we end up choosing the 4-mine option, making that overlap square a definite mine, and then we can make a fully valid grouping, with the other squares outside the groups being definitely safe.

A lot of real-world minecount situations in high-difficulty play involve handling partial-overlap situations and groupings.  Truly “clean” minecounts, where you can get an exhaustive square grouping, are rarer, because they often require a low background rate.  In higher-density situations, endgames that are mine-countable are much more likely to be N-1 “cramming” situations, or partial-overlap ones where you might have to keep track of one or even several overlap groups.  To help sort those situations out, the procedure in the next sub-section will be very helpful.

(f) Minimal vs Maximal provisional counting

As step “(f)” from the generalized process reminds us, sometimes when making overlapping groups, our count will tell us to resolve in favor of the more-mines option, with mines on the non-shared squares, and the shared ones safe.  This can be described as having to “cram” mines into a region.  So if you’re looking at an endgame that seems to have a lot of mines for the remaining squares, and it looks mine-count-ish, you’ll often mark “provisional” mines on it, as a way to assess the area.  This provisional minecount can be done in two ways:

1. “Minimal”, where you choose provisional mines with a view towards trying to mark the region with the fewest mines possible.  You assume untouched cells are safe.  And for partial-overlap grouping choices, you resolve them with the lower number (a mine on the shared square).  So e.g. with a 2v3 area, assume it’s the 2, and see what that leads to.
2. “Maximal”, where you do the opposite: choose a provisional marking where you’re opting for more mines over fewer mines at every opportunity, including resolving partial-overlap groups in favor of the higher possible number of mines they could have (their shared squares are assumed safe), and also assuming that untouched cells are all mines.

You’ll count how many mines you end up with in both scenarios[9], then look at your minecount: is it equal to one or the other?  If so, great!  You’ve just solved the board.  If the mine count is in between those two numbers, though, you haven’t won, but you’re also not lost, either: you’re actually pretty close.  If there are untouched cells, you can easily calculate the mine odds on each of them (minecount minus minimal count, divided by # of untouched cells), which may be very low.  You may be able to use the Assume-Fewer principle from the next chapter to guess a 1v2 or 2v3 partial-overlap group with favorable odds, and if your guess succeeds, you probably can solve the area.

Let’s look at a more-advanced minecount example.  I solved this via minecount, even though there was a logic move available (that I couldn’t find - can you?), so let’s ignore that for now and pretend there isn’t.  We’ve got a big 2-region endgame here, with 9 mines remaining:

Let’s start with the left region, as it’s simpler.  There’s four square groups, and then the top-left part is a 1v2 partial-overlap group (its middle square, above the 1, is either the area’s one mine, or it’s safe and there’s a mine on both sides of it):

Placing these groups is straightforward, though we have to favor some front-facing numbers and ignore some others to get there.  The teal would normally confound a beginner trying to do this, but we can be comfortable with a little ambiguity - it’s either one or two mines, but there are no untouched squares here.  So this region has “either 5 or 6 mines”, total.

Then we have the right region, which is even more ambiguous, but we can show, luckily, that there have to be at least 4 mines there.  Here’s two groupings:

(A)   (B)

Firstly, sidenote: That interior 3 on the right side, that is the key to this whole thing - without having previously found that safe square, we couldn’t solve this.  And it took a bit of vision: we had to use the 4 on the left (grouping A, in orange) to project through to the 4 in the middle (blue group), creating a 2-over-3 pattern between the purple group in grouping B and what the 2 next to the 4 is telling us, making interior-3 square safe.  Standard Evil NG logic, but non-obvious.

And since we did find that move, we can use that 3 to spot the teal group in both groupings via the Hole pattern, getting us up to “minimum four mines” in this area.  Notice then that with the orange groups in both groupings, we have a choice there - so the leftmost column here really is a 1v2 pattern (and likewise for the purple groups, another 1v2).  But that’s the least of our issues, because we’ve also got two untouched squares.  So in theory, this area could contain as many as 8 mines (opt for two mines with the orange areas, two via the purple areas, place two via our other groups, then make the two unknown squares mines), but no fewer than four are possible.  So if this region could have more than four mines, but isn’t asking for 8, we’re boned.

But luckily, it doesn’t!  Because we have a total minecount of 9 between the two regions.  Therefore it’s the “minimum” option for both: five mines on the left, and four on the right.  On the left, this means the 1v2 resolves in favor of “1” (on the middle square inside the teal), and logic moves follow which mark and clear the whole area.  On the right, because we only have four mines available, this means “both groupings are correct” and we can use the overlap of both the orange and the purple groups to mark 2 mines, and to clear the 2 untouched squares.  The remaining mines follow quickly from there, and we win.  Nice!

(g) Scenario Identification: So that’s how to execute solving via mine-count.  But how do you spot that something is mine-countable in the first place?  Or likely enough to be that you will skip guessing it for now, and try mine-counting at the end?  Two ways:

Small / Thin region: Look for an odd number of squares.  If it’s an obvious 3-square (1v2) or 5-square (2v3) isolated region, you know how to do that. Any odd number of squares will do, in fact, as long as none of them are untouched squares: that indicates a simple N vs N+1 scenario.  Return at the endgame with your minecount of either N or N+1, and it will then be solvable.  There is a huge, decisive difference between this…

  …and this:   

(both have minecount of 4)

At right: a (thin-region) 6v7 minecount!  The key is the single cell underneath the interior 5, making this an odd-number-of-cells region.  This is actually from a no-guessing Arena game, and given the minecount of 6, we’re able to fully determine all 6 mines’ placement.

Larger region: Look for minimal untouched squares.  The region, ideally, should have either a few interior squares that minecounting would reveal to be safe (a long linear front makes it more challenging), or to not be that “thick” in the first place (have a front coming from several directions, with only a few rows in between).  Basically, you want to be able to “See” nearly all the squares, from the perspective of the safe cleared squares on your front.  One or two untouched squares is not a huge barrier - maybe you can prove they’re safe! - but having a bunch of them, or worse a row or two in between the sides of your front, that you can’t see into from your front, is an obstacle to solving this way.  It is very unlikely that a region like that will be mine-countable, because it would require that every untouched square is either a mine or not a mine.  If that isn’t true, then you’re still in probability-land, and can’t use this technique[10].

At right: unlikely!  But the fewer untouched cells, the more possible this sort of thing becomes.

Not all larger regions with minimal “terra incognita” will be mine-countable, of course.  But the fewer (untouched) squares you have, where you know nothing about them, the likelier it is that all of them are safe (or mines), which is a precondition to being able to solve via minecount.

Practice Puzzles

Link to solution section.

These puzzles start with a few simple scenarios, but then get into ones that involve the more-advanced mine counting techniques from this chapter.  Don’t feel obligated to try all of them; find a difficulty that feels like the right degree of challenge for you.  The first half are pretty much just concept checks, making sure you understand and can apply the principles here.  Difficulty 6 / 10 and up are meant as progressive workouts for your brain muscles, to demonstrate true mastery of the conceptual category.

Puzzle 1-1 “Boat”: 2 mines remaining, 2 to mark, 3 to clear.  Difficulty 1 / 10.

1-1(b) Bonus: How would the answer change if it were 3 mines remaining?

Puzzle 1-2 “Trail”: 3 mines remaining, 1 to mark, 2 to clear.  Difficulty 2 / 10.

Puzzle 1-3 “Boxed Out”: 4 mines remaining, 3 to mark, 3 to clear.  Difficulty 2 / 10.

1-3(b) Bonus: Repeat while assuming there were 5 mines remaining.

Puzzle 1-4 “Empty”: 3 mines remaining, 2 to mark, 7 to clear. Difficulty: 3 / 10.

Puzzle 1-5 “Needle”: 4 mines remaining, 0 to mark, 1 square to clear.  Difficulty: 4 / 10.

Puzzle 1-6 “Zigzag”: 4 mines remaining, 4 to mark, 6 (the rest) to clear. Difficulty: 4 / 10.

Puzzle 1-7 “Journey”: 7 mines remaining, 6 to mark, 9 squares to clear.  Difficulty: 5 / 10.  N.B. this is also mostly-solvable using 4-square-box logic, but put that aside and try to do it starting with minecount.

Puzzle 1-8 “T-Bone”: 6 mines remaining, mark ‘em all, clear ‘em all.  Difficulty: 5 / 10.

Puzzle 1-9 “Projector”: 4 mines remaining, 1 to mark, 4 squares to clear.  Difficulty: 6 / 10.

Puzzle 1-10 “Trunk”: 8 mines remaining, mark ‘em all, clear ‘em all.  Difficulty: 6 / 10.

1-10(b) Bonus: If this minecount were unknown, what could you conclude about it?

Puzzle 1-11 “Maps”: 5 mines remaining, 4 to mark, 14 to clear (all but 2).  Difficulty: 6 / 10.

Puzzle 1-12 “Forgot About Dre”: 6 mines remaining, 2 to mark, 4 to clear.  Difficulty: 7 / 10.

Puzzle 1-13 “Fields”: 6 mines remaining, 4 to mark, 7 to clear.  Difficulty: 8 / 10.

Puzzle 1-14 “Donut”: 7 mines remaining, 6 to mark, all-but-2 cells to clear.  Difficulty: 9 / 10.

Puzzle 1-15 “Waltz”: 6 mines remaining, 2 to mark, 2 to clear.  Difficulty: 9 / 10.

Puzzle 1-16 “Constellation”: 19 mines remaining, 15 to mark, all-but-6 cells to clear.  Difficulty: 10 / 10.

(2) Assume fewer mines

Here’s an important “Minesweeper Theorem”, whose consequences we will explore here:

If there is an isolated part of your front which could contain either N mines or N+1 mines, each with equal numbers of permutations, and the background rate is B%, then the odds of that front containing N+1 mines is B%, and the odds of it containing N is (1 - B%).

Believe it or not, this is a somewhat profound conclusion.  One of WoM’s top players, Obelus, even wrote and published a mathematical proof of it[11], so it is often called Obelus’s Principle.

This has to do with guessing scenarios, not ones where logic can help you make progress.  We’ve already seen an example when discussing the 4-Square Box:

The front of that box is isolated from the rest of the board, and the 3 squares which border the front could either contain 1 mine (on the shared corner), or 2 mines (on the wide corners).  The one square not touching the front, of course, has a background-rate probability, by definition.  But note that the wide corners - the “more mines” scenario - also have odds equal to the background rate, B, while the “fewer mines” scenario, i.e. the mine being on the shared corner, equals the opposite of that (i.e., “1-B”).  And just about always, B will be <50%, usually in the 15-25% range, so the odds favor the “fewer mines” scenario, often by 3:1 or more.

Why is this principle true?  I think of it this way: For a mine configuration to contain more mines rather than fewer (given that both are possible), it has to “pull” one extra mine out of the “pool” contained within terra incognita.  And the odds of the configuration of this board just happening to be one that puts more mines in your front, when fewer would do, “costs” you, in terms of probability, the same as for a random square to be a mine.  Consider the universe of possible mine configurations that could exist on the still-unknown squares on your board: while mine density is under 50%, the more mines available to that pool, the more possible different arrangements that can be made.  Each time you reduce the number of mines in that pool, you reduce the number of possible configurations.  You’re on a row of Pascal’s Triangle and are moving one step toward the edge of your row, dropping the # of combinations that exist.  So if a guess would require more mines than another guess would, it’s relying on a smaller fraction of the universe of possibilities than the one with fewer mines would.  That means the odds of it being the real case, the real configuration, are lower.

Another way to think of it is, if you pick one random untouched square, what are the odds it’s a mine?  The answer is B, the background rate, by definition.  Now suppose we have 2 squares, across which we know there is one mine.  Absent no other information, what are the odds of each square?  50-50, of course.  But further suppose that we now have one additional square on the front, adjoining one of those two squares of ours, but not both - much like the example above, with two fronts of 2 squares each, with one shared square between them.  If there were no other information, each square would have odds = B, but because they’re connected, either outside square being a mine (odds B) means the shared square is not a mine (odds still B), and thus the odds of that shared square being a mine is 1-B.  In algebraic terms: The outside squares are independent, whereas the shared square and only the shared square is dependent.

It’s important to highlight some key parts of the principle and its definition:

• “Isolated”: As defined at the top of this guide, this rule only applies if the section of front we’re considering is isolated, i.e. its possible permutations are unaffected by the placement of mines outside of it.  There are times when we have a very large, singular front, facing up dozens of number squares, possibly wrapping around some corners, and this means that our rule above does not apply.  Because the different scenarios for mine placement are influenced by squares offering many possible mine permutations.
• “Could contain N or N+1”: This means that, for this section of the front, there are only two possible quantities of mines, one with N and the other N+1, and equal numbers of layouts for each (usually 1 each, sometimes 2 each).  If we have a more complex front (but still only two possible quantities), we likely have more than 2 possible scenarios, and this rule instead refines to: “the ratio of probabilities of each layout containing N mines to each layout containing N+1 mines is (1-B) : B, and so while together the N-mines layouts are still likelier and N+1 collectively still unlikelier, the final probabilities of each depend on the quantity of layouts in each group”.  And it’s hard to disaggregate those two groups of scenarios, usually requiring further information.  More on this extended version of the theorem later, but for now just note the constraint.
• There are also times where a complicated front could have N+2 mines (as well as N or N+1), or even more groups.  In these scenarios, the N+2 case is even less likely, in fact its probability is roughly B^2, on the same reasoning as above: in order to “pull out” even more mines from the background pool, the odds of that just happening to be the real configuration go down in proportion to the background rate.  Every time you need an additional mine for your scenario, that is only “B% as likely” as the incrementally less-demanding scenario.  At right, the region can have 2 or 3 mines just on the middle row, but if the top and bottom have mines, it needs 4 total.  At endgame, though, this region would be mine-countable (and was).

Let’s see some common-type examples where this can be useful:

(A)   (B)   (C)

In Example A, the front (at the top) can contain either one or two mines.  The “one mine” version, i.e. the mine on the shared square, is far likelier.

In Example B, we’ve got a mirrored situation here, which makes it a bit more fun.  Our background rate is down to 13%, but we’re not yet in mine-counting territory.  So for each of the two corners that we can see into, the Assume Fewer rule applies: the rate of the wide corners (= 2 mines around those 3 squares) is B, 13%, while the rate of the shared corner is 1-B (88%)[12].  This is by far the most common formation where the rule applies, too - a “corner 1v2”, where a corner on the front is a mine, but we can see from both sides, and it’s a “50-50 front” from each, forming a “1v2” that’s easily recognizable.

And note that in Example A, guessing those frontline 22% cells is also likelier to result in progress, and a full solution to this region, than guessing one of those two untouched cells.  Suppose you guess the lower-right cell, and it turns up a 3 (as would be likely).  What’s your next move?  Whatever it is, it’s another guess: either the cells shared with the corner 1 contain the mine, or they each have a mine directly above them - and you lack information to know which.  At least the former is likelier, but you’re still risking a guess.  Whereas, if you make one of the frontline 22% guesses, logic very likely follows that clears the entire area safely.

In example C, we have a framed “linear 1v2” along the top: either 1 mine is below that central 1 on the top front, satisfying the 3 to its left and the 2 to its right, or there are 2 mines, one below the left 3 and the other to the right of the interior 3.  Because of the mines framing it on the outside, this section is independent[13] of any adjacent areas, so its probabilities are only influenced by this one factor.  If we were to guess it, we’d want to guess the wide squares, thereby making an assumption of 1 mine rather than 2.  This is the second-most-frequent formation where you can make use of Assume Fewer Mines.

Here’s a more complicated situation where the background rate actually doesn’t come into play at all, but the logic for it illustrates the general “Assume Fewer Mines” principle.

This one is fun because the first thing you’ll notice is, there are no untouched squares[14].  Because we have that 2 in the middle, all squares in this isolated region are touched by a number.  Secondly, we have a firm minecount: looking at the region around the 2, plus the 2-square fronts at top and bottom, we can infer there are exactly three mines in here.  Third, notice that around both the corners we have on this front (at top, the 1-2 vs 3-2, and at bottom, the 1-3 vs the mine-3), we have a situation in which each could contain either one or two mines, so it’s sort of the scenario described by our rule.

However, the choice of mine for the 2 in the middle actually determines the entire configuration in the region!  Consider it for a second: mentally place a mine in one of the 5 spots around that central 2, and clear the other squares around it; the placement of all other mines are thereby determined.  Whereas, the reverse is not true: if you place a mine on either of those interior corners, it does not, on its own, determine where the mine is around the central 2.  So the 2 is “independent” and the rest are “dependent”, and therefore there are only five scenarios for mine placement: one each for each of the 5 unknown squares around that 2.  If and only if its mine is, say, the square directly above it, then the top corner has its “2-mine” scenario; in the other four possibilities, the top corner has its 1-mine scenario.  And likewise for the bottom corner.  Hence, the shared corners here will be mines in 4 of 5 (80%) scenarios, and the wide corners, only in 20%.  So we don’t even need the Hint button to know our probabilities.

What, then, should we guess here?  Well, the worry is, we’re somewhat close to having a box-and-2 50-50, on one half of this region or the other.  Unless the mine around the 2 is the one to its direct left, in any other case one of either the top or bottom 4-square box will contain 2 mines on a diagonal, and no way to discern where it is.  So we can hope for the 2’s mine to be to its left, and we shouldn’t guess it (because even if it’s not a mine, you’ll gain no additional useful information).  But if we guess literally any other one of the 20% squares, one of two things will happen: (a) it will show an effective-1, indicating that the mine in that half of the area is on the shared corner, and that the other half of the area is now a confirmed 50-50 (gulp), or (b) it will show an effective-2, indicating that the other half of the area has only one mine (in the shared corner), and further that this half was a 50-50, but we just guessed it successfully, or maybe it’s to the left of the 2, but either way, the whole area is now solved and determined by logic.  And if you run through the 5 possible configurations, outcome A (getting a 1 and thus a 50-50 that needs guessing) happens only in 2 of them (40%), which we survive half the time, while outcome B is 60%, which we clear every time.  So as long as we don’t guess that square to the left of the 2, we have an 80% chance of surviving this area.

So, this “theorem” may seem to you like a simple, even obvious result, but it has a couple of consequences worth mentioning:

1. Firstly, sometimes you’ll be in a position on WoM where you can’t get a Hint to show the probabilities (yet), not until you’ve exhausted all available moves knowable from logic alone - on the entire board.  Until then, hitting Hint will just show you the next available move via logic, even if you’re not interested in that part of the board yet (“no, just show me the hint for this area specifically!”, you might shout, uselessly).  So when you come to a situation where this rule applies, you’d have to either move on and then come back once the entire board is done - perhaps only to be handed a 50-50 anyway - or take your chances now on a guess where your odds of being wrong are equal to the background rate of mines, i.e. the board’s mine density.  Depends on how much you value your time.
2. Unlike just guessing into terra-incognita, if you can find a place on your frontline where you’re offered odds that are just as good as the background rate, it will be much more likely to result in progress than a random guess into space would be.  This is because the number of squares that have to be safe in order to have a logic move follow-up are many fewer than in a guess into space (except, perhaps, for a corner of the board), and also because there may be dependent squares, so your guess, if safe, opens several other cells too.  This is especially true on higher-density boards, where a guess into space, usually done hoping to discover an opening, will be mostly or entirely futile.
3. If one mine’s placement would determine another’s, the probability edge you have from this rule can be “compounded” into even better %s.  The more mines are “locked in” to their placement via one single assumption, the less likely that assumption is to be the case.  This is because out of the universe of possible mine configurations on your board, the more mines a scenario would put into definite locations, the lower share of all configurations will contain exactly those placements.  And it’s that compounding, that line of reasoning, that provides the basis for judging probabilities in advanced guessing scenarios, as we’ll discuss in Technique #10 below.
4. In the endgame, this rule might not apply, or rather, may not be as helpful as it would normally be.  You might discover that actually the fraction of mines in remaining squares is a lot higher than the original board density, maybe even over 50%!  When you reach those endgame situations, take notice of your mine count, because it’s possible this rule gets flipped on its head (i.e., the odds now say you should assume more mines), or becomes far less useful (a background-rate 30-50% guess is not great for your win %).
5. Simple mine-counting situations are usually “N vs N+1” dilemmas, and so if you’re not going to wait until the endgame to solve it and want to guess, you’ll want to guess as if it’s the N rather than the N+1.  E.g. this 3v4 minecount situation shows those odds, which will persist and as we approach the endgame, it will eventually converge on 0% / 100%, if and when it’s the only remaining unknown part of the board:
    
   (3v4 minecount before endgame, solvable at endgame)

For #4, it’s unusual, but in the endgame we can end up with a background rate that is super high, if we’ve cleared all the low-density areas of the board, and the only ones left, the hardest ones we’ve left till last, happen to be much higher-density.  For example, look at the gray numbers in these:

That’s a background rate of 79% and 82% respectively.  Not ideal!  But also very rarely seen.  In these situations, the presumption of “assume fewer mines” is flipped on its head, as here:

That’s a background rate of 62%, which of course is above 50%.  Ignore the green squares for a sec and focus on the corner on the far left side.  The shared corner - the choice where, if you guess and click it, you’re presuming more mines - actually has the lower percent chance of being a mine (38%), whereas the choice where you’d be assuming more mines, i.e. mines on the squares to the 38’s top and right sides, are the worse choices there (62%).  By guessing the corner square and presuming more mines, your odds of surviving are better.  But only because of an extreme background rate.  Normally the odds would favor the opposite of that.

In normal situations, the background rate is often a bet you’d rather take, over some front-adjacent options that might all be high-risk.  But seeing an endgame background rate (for a high-difficulty board) that gets up to 30, 35, even 40%+, is not all that unusual, and that can and should factor into your guessing choices.  Likewise, a guess into space, unless you have an available corner to try, is less likely to result in progress (which usually requires an opening) than solving some part of your frontline.

So for those reasons, it’s very useful to be able to spot the (non-minecountable) situations where we have a “N vs N+1” opportunity to get a favorable guess.  How do we look for those, when we’ve run out of logic moves to make?  Some tips:

• Shared corners.  If you have a front where there is a corner, and the corner itself has a mine on the corner of the front, but its adjacent ones are not, you could be looking at a 1v2 situation.  Note that implied 50-50 fronts that extend out into space coming from another corner around the front can count as a “shared corner” for purposes of making this situation.
• Dependency along the front.  If the frontline is single number squares interspersed with mines between them, so no squares are adjacent to each other (except perhaps around a corner), they are “independent” and that’s bad here.  We want number squares that are adjacent to each other, within reason, if we’re going to find a place to apply this rule.
• Lack of curvy, un-isolated frontline.  If you have a long frontline with curves and varying numbers and such, one which has all the numbers connected to each other via a lack of intervening mines, the number of possibilities that exist for mine placement along that front are probably many more than the simple 1v2 we’re looking for - and may make an assumption along these lines totally wrong.
• Play out scenarios in your head.  Verifying a guess is usually easier than identifying it by inspection.  If you mentally put a mine down on one square, and then draw logical conclusions from that along the front until no more can be made, how many mines did that place?  If you then flip that and assume that square is actually safe, and draw your logical conclusions, do you end up with the same number of mines?  If the quantities end up different, you might have found a good guessing spot (in favor of the Fewer Mines option, of course).

Three specific sub-category situations which rely on this principle are worth remembering and spotting in games:

(a) Flat Framed Front: A decent fraction of the opportunities to use this rule come from “corner 1v2” or “linear 1v2” assumption situations.  But there is another class of situation you should know and look for in guessing situations, which I’ll call “Flat, Framed Front”: On a linear front, there are two adjacent number squares, and mines on either side of the pair.

The front here goes “mine-3-3-mine”, with the 3s as effective-1s, and the mines on top and bottom as the “framing”.  I’ve labeled the front-facing squares.  And note that there are 3 mine configurations available for that front: either there’s one mine on one of the two As, or there’s a mine on both of the Bs.  We don’t know which.  But the Bs scenario requires 2 mines, while the As scenario requires only 1, so (per our theorem) it’s far likelier that the mine is on one of the As.  Also, helpfully, if one of the Bs is safe, then by logic, the other one is safe as well, so you get two squares for the price of one guess.  The mine odds of these particular squares, from top to bottom, are 15%-44%-44%-15%.  If you spot this sort of front in a guessing situation, you’re probably well under 20% chances if you guess one of the corners - and with a nontrivial chance of being able to make progress if you’re correct, too.

(A)   (B)   →

In A, we have the same 15-44-44-15 percentages as in the top example.  In B, those numbers become 13-43-43-13 (twice), due to a different background rate.  But the same conclusion holds, qualitatively: guess the outside squares of the flat framed front.

Under the right circumstances, using projection, flat framed fronts can also occur around a corner, such that it doesn’t even appear “framed” by mines at all.  In the example at right, the upper cells of the 4-square box, accounting for one of the mines of the corner 2, effectively provide the bottom “framing” for the remaining mine of that corner 2 and its partner above.  Because the front is otherwise isolated, and otherwise identical to our normal FFF situation, it does in fact behave the same way percentage-wise.  If you’re wondering in a game if this situation applies, just ask whether it’s presenting a 1v2 proposition, and if its probabilities are un-influenced by any outside squares.  If both are yes, as here, then it should behave the same way, and offer the same guessing advantage.

In the same way that line-and-1 50-50s can be extended by repeating every 3 squares, in the same way that ILFs have the same solution pattern if you extend it by 3 squares, you can also have an “extended FFF” situation.  In the basic FFF layout, we had two adjacent effective-1s on the front. Above-right, we have a row of 5 number squares, framed by mines on the edges, so it faces up against a 7-square front.  And instead of a 1v2 proposition, it presents a 2v3 proposition.  But the same logic applies: there are two different configurations on the front which use two mines, but only one using three mines, so we get better-than-background odds on the squares of the 3-mine scenario.  Likewise for the vertical example, at right.

The flat-framed front need not be fully isolated: it also “plays nice” with effective-1 adjacencies, if those adjacencies have a decent square coverage ratio.  See at left.  The wide squares of the flat-framed front are already unlikely, but it’s even-more unlikely now, since the wide-squares scenario would also satisfy the 3 to the left, clearing several squares and placing a mine above the left 4.

However, with adjacencies, you need to be careful that the same basic logic of “1v2” that an isolated flat-framed-front can provide, still applies.  For example, at left, the FFF on the right side interacts with the 4 up top, and so this is just a “2v2”, with 3 equally-probable layouts (33% each).  Or, at right, we have our usual FFF on the right side.  If the wide squares are mines, you need 2 mines for that column.  But if the wide squares are safe, that means there’s a mine to the left of the upper 2, against the wall, so you also need 2 mines for that layout.  We lose our “assume-fewer” edge for this case!  Here the wide squares are actually 61% to be mines.

The logic also doesn’t hold with adjacencies that are effective-2s, which pull enough mine probability into the wide squares that it overwhelms the 1v2 advantage you would’ve had in the absence of interfering adjacencies.  See the FFF, lower-down at left: instead of the wide squares being low odds, the 5 (effective-2) makes them worse guesses than going with the interior squares.  And the best guess is up on a shared corner, away from the effective-2s on the right and below.

Higher mine-to-square ratios on the cells adjacent to your FFF can even turn the odds from this pattern fully on their head, as seen at right and above-right, where the probabilities are flipped from the basic FFF layout.

Final caveat: The same pattern does not hold when the FFF’s front squares are effective-2s (and is thus a 2v3 decision) - it’s only useful when they’re effective-1s (and thus is a 1v2).  This is mainly because those eff-2 fronts have just one fewer-mines arrangement, using two mines, while there are two more-mines layouts using three mines, so the Ratio Rule wrecks it.  Note the (unhelpful) percentages:

(b) Split Background Rate: It’s common to have an isolated part of your front be an N-vs-N+1 option.  However, sometimes, the “N+1” option, i.e. the one with more mines, will also have one of its mine placements be “split” across two squares.  So if you were to guess that the N option is correct, then the squares making up the N+1 option would actually have half the mine probability on the squares splitting that 1 mine.  This can lead to finding the most-favorable guess on the board.  Consider this front, and let me note that the background rate is 24%:

 →  →

This is a simple 1v2 situation: either there’s 1 mine on the shared squares (above the two 3s), satisfying them both, or there isn’t, in which case you need a mine on either side.  Via labels, there’s either 1 mine on B/C, or 2 mines: one on A, and one on D/E.  However, note that the right side of this proposition, satisfying the right 3, shows that in the 2-mine scenario, the remaining mine would be in either D or E.  Now, we already know that the 2-mine scenario has probability = B, i.e. 24%, so therefore the 1-mine scenario has probability 1-B = 76%.  But in that 2-mine scenario, it’s still only one of D or E that’s a mine.  So each of them carry only 12% odds!  Getting half the background rate for a guess is about as good as you’re ever going to do.

(incidentally, you’d want to guess E, not D.  E gives higher chance of progress, because it only requires four squares to be safe in order to turn up a 3 and be chord-able, whereas guessing D would require seven squares safe for that.)

This “split background rate” pattern - sort of a flat-framed front with an extra square - is common enough to be worth remembering.  See another at right.

Another very-common variation is this 5-square front situation seen at right (it’s not an ILF!).  Here, there are 3 possible layouts for this front: one in the middle square, requiring one mine on square #3 on the front, or two others either on squares 1 & 4, or squares 2 & 5, along this front, requiring two mines.  By our principle, the 1-mine scenario has a high probability (1-B%), while the 2-mine scenarios each have probability <B% (see below for how the exact numbers come about), which here is a pretty attractive proposition.  You would guess the bottom square in this case, since it’s against a wall, so the odds of progress are better: only three other squares need to each be safe in order for this to turn up a 1 and be chord-able.

Another is seen below, in the second “flat framed front” example from above:

See the bit at bottom left?  The left-hand 4 has 1 mine coming from above, and thus 2 mines coming from the 3 squares to the left.  If its remaining safe square is on the shared square with the 3 below it, then the area requires 3 mines to solve; otherwise, it only requires 2.  Therefore the 3-mine scenario has probability = B (28% here), and the 2-mine scenario, probability 1-B (72%).  But in the 3-mine scenario, the placement of that third mine, next to the 3, could be on either square.  So those squares split the background rate, and are each 14%! (you’d guess the lower one for better progress odds, btw)

The SBR is a type of Assume-Fewer situation that pops up from time to time.  But the important lessons here are: (A) if you can find one of these, it will probably be your best guess.  And, (B) 1v2 corners and short linear fronts are not the only ways you can encounter an Assume-Fewer situation.  Some are wrapped around a little, like this one.  Don’t look past it just because it doesn’t fit one of the most common templates - instead, focus on envisioning the ways that a front can be solved, and whether they differ in the number of mines required.

(above: a fancier case of “split background rate”.  B = 10%; if any of the green cells are mines, it forces an additional mine onto the front.  It’s easiest to see with the lower ones, which together represent an N+1 scenario, carrying B% odds, so each of them have B/2, or 5%.)

(c) Illusory Corner: One category of situation where it is NOT an N-vs-N+1 scenario is what I’ll call a “dependent corner” - one where, despite the corner squares themselves offering an N-vs-N+1 option set, some of those choices are impacted by probabilities immediately off the corners, in adjacent spots on the front.  As a result, there is not a clean isolation of the corner area such that its choice is independent of the adjacent spots - it is actually (somewhat) determined by those nearby squares.  Here’s a game (courtesy of Swimispro) where looking only at the corner will lead you to the exact opposite and wrong conclusion:

Naively, someone looking for a guess might see the corner there at top right and think it’s a 1v2.  But if you then extend the conclusions of the “1” (corner mine) choice outwards, you find that a mine on that corner means 4 total mines on the perimeter of the region, while if the corner is safe, you only need 3 mines to cover that perimeter!  Turns out, that naive assumption gets the probabilities backwards: the corner has the background rate, and the wide squares next to it are likelier mines.

These situations occur fairly often.  At left (4 mines remaining), the dependency chain of the perimeter includes the lower-left and top-right squares, meaning the front consists of five squares, not just the corner three.  And it flips the N-vs-N+1 on its head: the upper-left corner being a mine requires three mines for that front, while it being safe means only two are needed.  This type of layout can easily lead to mistakes, so always count your squares before guessing it!

Sometimes to see the “illusory” corner situation, where the corner square has the N+1 mines scenario (and thus lower probability), you have to look a bit farther afield for the square groups, and even do a bit of Evil NG-style projection.  Take this example:

 labeled →

The first thing that most players would notice is the upper-right corner, which looks like a typical 1v2 opportunity for a background-rate guess.  However, note: The orange group (from the 5) has one mine, so the residual squares for the 3 next to it, the AE group (blue), has the 3’s other mine.  So this means, looking at the E-A-B corner, if E has a mine, then so does B, and if one is safe the other is too; they are dependent.  Extending the logic around the full front there, one scenario puts mines on E-B-D-G, while the other has mines on A-C-F.  Three mines is likelier than four, so the upper-right corner (D) is actually the background-rate square, not the adjacent squares (C / F).  Hence the odds at right.

If you have a front with only two possible layouts, to check if it’s N vs N+1, simply count the number of squares in it.  If even, it’s probably ~50-50, barring some weird interference.  If odd, it’s probably B vs (1-B), as above.  Consider this position and try to count the front squares:

 →

The left side has some projection that induces an N-v-N+1, and ensures only two front layouts here.  From the left 3, there’s one mine in yellow, which means by the 3 to its right, the blue group (AB) has one mine.  But by other front squares, so do each of BC, CD, DE, EF, FG, GH, and HJ.  That’s 9 squares. and if you have a mine on A then you do on A-C-E-G-J; if the mine is on B, then it’s on B-D-F-H.  That’s 4v5, so we have a background-rate guess on the A-C line:

As a sidenote, the probabilities around the lower-left corner are an interesting case of dependence.  Consider the four squares around that interior 2 that aren’t in the blue group (AB).  If the 2’s other mine is to its right, then there’s a mine on the left corner (left square of the yellow group); if it’s to the 2’s left (below the middle 3), then the corner is safe and there’s a mine at bottom-left, facing the 1.  So that’s a 2v2, there’s no Assume-Fewer edge to be had.  This means there are only four possible configurations here: the four options of the other mine’s location around that 2.  And the corner is fully dependent on that 2’s choice.  So each layout has 25% probability of occurring, with the corner itself being a mine in three, or 75%.  All the other squares are B% or (1-B)%.

Here’s a trickier example where it looks like it might be “N vs N+1”, but in fact it’s not.

 →

Here, on first glance, the lower-left part of the region has that mined corner that appears to create a 1v2 situation around that area (blue box).  The one-mine scenario would put the mine on the upper-left corner of the blue area, the two-mine scenario would put mines below and to the right of that corner.  However, the blue region is not isolated - it has an overlap square with the area below the middle teal-highlighted 3 (in orange).  We know, from that 3 and the contributions of the yellow region (which has one mine in two squares), that the orange area has one mine in its three squares.  If the overlap square is a mine, then there’s another mine at bottom left.  So the Blue+Orange region has exactly two mines, never more or fewer, and in fact the orange area is what’s independent (33% each).  Therefore the blue region is 67% to the corner and 33% on the wide squares - a lot worse than the B% bet we might have made.

As you can see, it’s the untouched squares here that are actually the best bets safety-wise.  Everything else has a higher chance of losing.  Are those the best choices to win the game, though?  It’s possible that the right-most 22% square offers some progress potential: if the mine under that teal 3 is actually to its lower-left, then that square can turn up a 1 (on a wall-skip basis), and allow clearing around it.  So that’s one reasonable play here.  Another is to prevent the potential box-and-2 50-50 that’s shaping up at top-right: we have a 4-square box with three framing corners.  If the square below the teal 3 is a mine, then we do have a confirmed 50-50.  If you plug the layout into the analyzer, it says the left or right untouched squares have 54% win rate, while “breaking” the 50-50 wins at ~48%, and all others are worse. (in the actual game, I guessed the wall-skip option and blasted 🙁).

The lesson here is, make sure your N-vs-N+1 candidate is actually independent of outside inputs before potentially relying on it.  If it’s partly dependent on adjacent squares, that will skew your percentages, and might turn the usual reasoning on its head.

At right: instead of that upper-left corner being the likelier mine spot, guessing its wide squares is actually a terrible choice: if the corner were a mine, that would lock in the 3 remaining mines for the 5 below it, and also determine the 3 below that, and the 2 to the upper right.  That corner is not a simple “corner 2v3”, because its placements are connected to adjacencies, on both ends in fact.  Not only is the usual mined-corner logic wrong, it’s way, way wrong.  Look - and think - before you leap.

Ratio Rule: We noted earlier that one of the constraints of Obelus’s Theorem requires that for it to apply, on a front that could have either N or N+1 mines, each such group of layouts needs to have equal numbers of possible layouts within them.  Usually that means one layout each for N and for N+1, sometimes two for each, rarely more than that.  But it is often the case that the numbers of possible layouts differ between the N and N+1 categories.  How does this affect the mine-probability math?

Although Obelus used that simplifying assumption for the sake of his proof, his class of situation is merely a specific case of a more-general principle, when comparing the odds of “More Mines” vs “Fewer Mines” on a front.  We alluded to the general principle early in the chapter, which I call the Ratio Rule, and it ties a lot of loose threads together about how to reason about mine probabilities.  The formal statement of it goes something like this:

If there is an isolated part of your front which could contain either N mines (with X layout permutations) or N+1 mines (with Y layout permutations), and the background rate is B%, then the odds of each permutation occurring for that front is given by a ratio, the denominator for which is ( (X*(1-B)) + (Y*B) ), and the numerator for which is (1-B) for each layout with N mines, and B for each layout with N+1 mines.

There is a much less-formal way to think about this that should help it make sense: The frequency of each more-mines layout, as a fraction of each fewer-mines layout, is the same as the ratio of untouched squares being mines to being safe (i.e., B / (1-B)).  Imagine you have a background rate of 20%, facing an isolated front which could have one or two mines, each in a single layout.  The two-mine layout is 20% likely and the one-mine layout is 80% likely, by Obelus’s original theorem; the ratio between the two probabilities is B : (1-B), i.e. 1:4.  For any given five random rolls of the dice, you expect one to yield a two-mine layout here, and the other four to yield a one-mine layout.  Now instead, suppose there are two potential one-mine layouts, vs still just one potential two-mine layout.  This time, the single two-mine layout “counts for 1” in the ratio, but each layout with one mine “counts for 4”.  So our frequency ratio of all three scenarios is 1:4:4, and combining the one-mine layouts together, the odds between the two groups is 1:8.  And overall, there’s 1/9th chance for the 2-mine layout (1 / (1+8)), and 4/9ths each for the 1-mine layouts.  So the chance of our two-mine layout being true is now 1/9th, or ~11%, well below the background rate!

The Ratio Rule is the underlying reason for why the Flat-Framed Front pattern works, and gives below-background-rate guesses: In an FFF situation, there can be either one mine directly facing the number squares, in either of two positions, or alternatively there are two mines, in a single fixed layout on the “wide” squares.  So we’ve got three possible layouts, and can apply this rule.  At a 25% background rate (so untouched squares are 75% safe, 25% mines, a 3:1 ratio), these layouts occur in a 3:3:1 ratio.  So the two-mine situation would be 1/7th (~14%), and the one-mine options would each be 3/7ths (~43%).  Recall the example from earlier in the FFF discussion, at left (with a B=24%, so not exact, but close enough).  Drop the background rate to 20%, meanwhile, and the ratio becomes 4:4:1, with the percentages being 11% (1/9th) and 44% (4/9ths).  As seen at right (with B=21%).  And notice that 88% up top: that square is a mine under both fewer-mines layouts, while the squares facing the 2s below only have one layout each, so the upper square gets both chunks of 44% probability, added together.

Conversely, sometimes you’ll have a front with a single one-mine layout and multiple two-mine layouts, and some easy arithmetic can tell you whether the resulting odds are favorable.  Recall this example, also from earlier: either there’s a single mine facing the middle 2, or there are two mines, in either of two pairs above and below that 2.  The background rate is roughly 20%, so our possibilities occur in ratios of 4:1:1, so each of the two-mine layouts has odds 1/6th (~17%), while the single one-mine layout is 4/6ths (67%).  17% is still a little better than the background rate, so there’s OK guesses here, but it’s not quite as favorable as the odds from a true FFF.

The Ratio Rule also explains the Split Background Rate pattern: there, you have two layouts each for the two possible counts: a single mine directly facing either one of the number squares (each of which “count” for 1-B in the ratio), or two mines, one of which has only a single position on a wide square, but the other mine can be on either of two squares on the other side.  Therefore the odds of the two-mine layouts collectively is still B%, but the odds of each individual two-mine layout (and thus the %s on the “split” wide squares) is merely B/2.

One final example to show how this works in tight, high-difficulty situations (via Hao1501).  We have this mid-game situation (i.e. no minecount available), in which we are not sure whether we are facing two mines or three mines remaining in the region:

We have a 5-square box-and-tail on the right side, with an extra cell to the left, and the 3 (an effective-1) in between.  If that left cell is a mine, then there are three total mines in the region, and furthermore we know exactly where they are: one top-left, one above the 5, and the third between the two 2s.  If that left cell is safe, however, then there are only two mines; one between the 4 and 5, and the other is on one of the two squares shared by the interior 2 and 3 - in other words, there are two possible layouts for two mines, and one layout for three mines.

Now, we know from analysis of the 2-mine, 5-square box (next chapter) that there are three possible layouts for it, and inspection can show which cells are the 33% vs 67%.  But because we have this ambiguity about how many mines there are total (from that one extra cell to the left), the percentages are controlled by the generalized Obelus Theorem, i.e. the Ratio Rule. With a background rate of 20%, the two 2-mine layouts and the one 3-mine layout occur in the ratios of 4:4:1, so the former have 44% odds each, while the latter (and thus the cells that are only mines under that layout) is only 11%, give or take.  And the MSCoach Analyzer confirms it:

Now let’s try some examples of spotting and using these “N vs N+1” situations.

Practice Puzzles

Link to solution section.

Note that instead of asking for mines to be marked or squares to be cleared, these puzzles are looking for the “optimal guess”, i.e. the square(s) where you have the best chance of not-dying on the next move.  In some cases, where a guess succeeding would determine one or more other squares’ status, we may specify “independent” moves, if more than one is requested.

Puzzle 2-1 “Depths”: Find the better move(s).  Difficulty: 1 / 10.

Puzzle 2-2 “Pairwise”: Find the better move(s).  Difficulty: 3 / 10.

Puzzle 2-3 “Minnesota”: Find at least two independent good moves.  Difficulty: 3 / 10.

Puzzle 2-4 “Box-Curious”: Find the better move(s).  Difficulty: 4 / 10.

Puzzle 2-5 “Induction”: Find the better move(s).  Difficulty: 4 / 10.

Puzzle 2-6 “Spillway”: Find a good move.  Difficulty: 4 / 10.  Bonus: Find a 2nd good move.

Puzzle 2-7 “Double Duty”: Find at least two independent good next moves.  Difficulty: 5 / 10.

Puzzle 2-8 “Whale”: Find three independent good next moves.  Difficulty 5 / 10.

Puzzle 2-9 “Swingline”: Find a set of good moves.  Difficulty: 5 / 10.

Puzzle 2-10 “Wild West”: Find at least two independent good next moves.  Difficulty: 5 / 10.

Puzzle 2-11 “Clockwork”: Find the better move(s).  Difficulty: 6 / 10.

Puzzle 2-12 “Staircase”: Find at least two independent good moves.  Difficulty: 6 / 10.

Puzzle 2-13 “Upside Down”: 12 mines remaining.  Find the better move(s).  Difficulty: 7 / 10.

Puzzle 2-14 “Rifle”: 14 mines remaining.  There are two squares that are below background rate; find at least one.  Difficulty 7 / 10.

Puzzle 2-15 “Onion”: Find three independent good moves.  Difficulty: 8 / 10.

(3) Guess the more-helpful result

This rule applies to an uncommon situation, but when it arrives, it will likely determine your endgame and whether you win or lose.  The situation is this: (a) You have an area left with more than one dilemma presented, (b) two (or more) of those dilemmas are connected somehow, and also, (c) there exists one or more squares on the remaining spots where, if they are not mines, then they provide information that makes the difference between other things being a 50-50 and not - i.e., they turn what might’ve been an unsolvable 50-50 into a solvable area.  In other words, such squares, while their odds of being a mine might be as poor as 50-50, would be “helpful” if they are not - because if so, it prevents something else from becoming a 50-50.  If all of that applies, then you should guess the helpful squares, because if you do and survive, you will optimize your odds of winning.  Sometimes the guess that gives you the best chance of winning the game[15] is not the same as the safest guess (lowest odds of being a mine).

There are a few particularly important and common categories of guesses with a “helpful result”, and we’ll cover these before giving a more general theory below.

(a) Double Line-and-1 50-50

Our first type of “helpful guess” involves two 50-50s that are connected.  Most often, this is two line-and-1 50-50s.  We’re down to two mines for four squares remaining.  What do you do here?  A naive guesser would choose randomly for one of the pairs (A/B or C/D), and then if successful, choose randomly for the other pair.  Sometimes that’s all you *can* do.  But we can actually do better than that here!

Recall the structural requirements for a line-and-1 50-50: you need a full line on either end of the two squares, three mines long.  This is because you need all sources of information about the two squares in the middle to only be able to see both squares, none of them can speak about just one square or the other.  Here, open square C provides the lack of information requirement for the A/B 50-50, and likewise square A provides it for the C/D 50-50.  But what if we knew one of them?  Well then the other group wouldn’t be a 50-50!  In fact, if you guess either A or C - which, to be fair, is still a 50-50 proposition - then if you’re right, the other group is no longer a 50-50!  If I guess C, and it’s not a mine, then I mark D.  And C will either be a 3 or a 4.  If it’s a 3, then A is not a mine, and B is; if it’s a 4, then A is a mine, and B isn’t.  I could do likewise by guessing A: if I’m right, then its number will decide the C/D split for me.  If I’m wrong, oh well, I lost, but it was a 50-50 anyway.

This is the better play for winning the game, because if you guess B or D, and then you’re right, the other pair of squares still is a 50-50 proposition to you.  That means you only have a 25% chance of winning the game by doing that.  But if you guess A or C, the interior parts here, you have a 50% chance not only of surviving your guess, but also of winning the game.  Still not ideal, but literally twice as good!

These happen from time to time.  Just remember to guess one of the interior squares, to “prevent the line” from completing the other 50-50!

(b) Box-and-tail (5-square box)

There is a category of situation you will see many times, where you have 5 squares remaining in a region, and 2 mines.  And those squares are arranged in a 4-square box, plus one square on the end, like a “tail”.  Here’s an example from an intermediate game:

  → Probabilities:

This might look simple: if that square that’s not in the 2x2 box is the “tail”, then the opposite square, the 67%-er, is the “head”.  Don’t guess the head, right?  But everything else is 33%, so what’s the difference?  Well, here’s the difference: If you guess the tail, you either lose (33% of the time), or in the other 67% of the time, all you’ve done is reduce the problem to a box-and-2 50-50.  So you took a 33% chance of losing, and then lose half of the time anyway - guessing the tail gives you only a 33% chance of clearing the area.  Same as the chances of guessing the “head” square, the red 67% one.  Furthermore, if you guessed the top-middle square, between the tail and the, uh, belly, you either lose (33% of the time), or you get an effective-2 result which doesn’t help you: you’ve just reduced the region to a different 50-50.  Boy wouldn’t you feel dumb!

However, the other two squares from among the five - the “belly” (at top-left here) and the “back” (bottom middle here), each have a 33% chance of losing, but if you don’t lose, then the rest of it can be determined conclusively.  Hence, those two squares have a 67% WP, while the other three have a 33% WP.  Kinda a significant difference.  So that’s what to remember: In a 5-square, 2-mine “box-and-tail” scenario, guess the belly or the back.

Some caveats:

Caveat 1: There appears to be some sort of “it depends on what sides of the 5-square formation are facing your front” conditionality to this.  Sometimes the 67% square is the “belly”, not the “head”.  The logic presented above (leading to “guess the belly or the back”) seems to apply most of the time, but if you have a view from “below” the box-and-tail but not “above” it (i.e. from the side facing the “head” and “back”), as with the image here at left, run your hint and get your percentages.  Some of the same reasoning still applies - don’t guess the tail - but the questions of “which cell is the 67%-er” and “which 33% cell in the box just creates another 50-50” do appear to change.  I haven’t tested out all the combinations of sides you can see the box from, and what each one does for the correct tactics.  If in doubt, use a hint, and don’t guess the 67%.

Caveat 2: Sometimes, you might have a view from three sides of the formation.  In the region at left, it looks very very similar to the 5-square situations we’ve looked at (there is local minecount), but is in fact directly solvable with logic.  Any advanced player would see it, if you had them focus on it.  But if you’re pattern-matching quickly, and this looks like any other 5-square box and you proceed to guess it, you might be going too quick to notice that it’s actually solvable.  The guessing tactics presented here really only apply when we either can’t see the 3-square side at all (it’s not on our front), or we have only a single square’s view of it (probably from the middle). If we have a near-complete view of it, like here, it’s probably solvable by logic.

Caveat 3: For a 5-square “box-and-tail” with three mines, mind you, the proper guess is less nuanced: you guess the “head”.  If you don’t blast, you’ve solved it.  Note the lone 33% square.  That’s for most shapes, at any rate (per Caveat 1 above) - at right you’ll see the “belly” be the lone 33% square.  The version at left is a lot more commonly arrived-at, though.

When in doubt, use a hint and check your percentages on the 5-square box-and-tail, because you should always be able to get a 67% WP on them.

(c) “Prevent” the 50-50

A corollary situation to this “guess the more helpful result” rule is, sometimes you’ll reach the end of what logic gives you and be faced with a guess, and at the same time, there will be a construct on the board that looks like it might be a 50-50.  It’s nearly there, a lot of the elements are there - but if one (or two) particular squares prove to be safe, then it isn’t a 50-50.  In those situations, if you’ve got to guess anyway, you might as well guess the square that would make sure that spot isn’t a 50-50.  Sure, you might be wrong and lose… but in that case, you’d have faced a 50-50 anyway, dropping your odds of winning the game by half.  So even if those squares aren’t the odds-on best guess from a probability perspective (which is only telling you the odds of dying on your next guess), they may actually be your best bets in terms of chance to win the game.  Because if it did prove to be a 50-50, then your odds of winning would be whatever it would have been by guessing the best guess and being able to logically proceed from there… divided by half.  So in that situation, better to just guess and hope that there’s a hole in the 50-50 - that the trap being set for you isn’t complete.

(A) Preventing box-and-2 50-50

 → Helpful:

(B) Preventing line-and-1 50-50 (edge variant)

 → Helpful:

(C) Preventing “overhang” line-and-1 50-50 corner variant

 →

N.B.: If yellow is a mine, we have a 50-50 to its right.  But also, all squares to its left can only be one number, if safe.  Whereas yellow could be a 4 or a 3, so it’s helpful in two ways.

Strategically, this guess choice doesn’t even have to yield immediate progress in order to add value.  If you can eliminate all forced guesses (especially 50-50s) from a board, it becomes solvable, by definition.  So even if your guess doesn’t immediately give you a way to solve that area (though it does in example (A) above, by chance - can you see why?), it can still mean that you can go off to solve elsewhere, and come back later with more information, more things cleared, you can be assured that there will be a solution you can determine through logic.

Other guides refer to this as “breaking” a 50-50, though that’s a bit of a misnomer because of course it isn’t a 50-50 yet.  To some extent this is just generalizing from the “double line-and-1” category above.  Although that one is visually distinctive, the same principle applies anytime you can see a potential for a double-guess coming up, and there’s a way you might reduce that to a single guess.  That’s the whole point about “helpful guesses” - reducing the risks you must take.

(d) Pseudo-50-50s: Among the situations we’re describing, one group of them that’s easier to recognize are what are called “Pseudo-50-50s”, or a “pseudo” for short.  A pseudo-50-50 is a situation where a cell must be either safe, or be part of a 50-50 - and therefore, either way, it is appropriate to guess immediately upon recognition.  Like actual 50-50s, the two most common categories either resemble a box-and-2 50-50, or a line-and-1 50-50.  But importantly, it is still a “pseudo” situation if you have not yet cleared around the box or the line enough to have confirmed it as a 50-50.  Consider the box-and-2 type (discussed more in Chapter 7):

At left, square A could be a mine, or not, we don’t know. If A is safe, however, then box BCDE is a box-and-2 50-50.  If A is a mine, then the 3 is satisfied, and solves the box: B, C and D are all safe, and E is a mine.  So in fact regardless of the status of A, it is always a good move to just guess that E is a mine, and open either C or D.  If A is a mine, we’re clear, and if A is safe, then we had to guess the 50-50 anyway (and we just did!).

Or take the situation at right, less-common but with similar reasoning: in the 4-square box, it’s either a box-and-2 50-50, or there are three mines in it (top-right is safe, other three are mines).  You could minecount it at endgame, but even before then, guessing that top-right square is the correct move: it’s either safe (and turns up a 4), or it was part of a 50-50 and you’d have had to guess it eventually anyway.

The same reasoning applies, in the same way, to situations that are nearly, or could become, line-and-1 50-50s.  At right, we’ve guessed the corner already, as an initial move.  But the wall of mines in the third column means that the second row here is now a pseudo: either the corner 1’s mine is to its right, and both squares in the second row are safe (33% chance), or the mine is below the 1, in that second row, and it’s a 50-50 (67% chance).  Guessing it right now thus carries a 67% chance of having no 50-50 to deal with there, and in either case, guessing it right away is the correct play (and it doesn’t matter which square you guess: there’s guaranteed logic to follow, if safe).

If you’re facing a region that requires a guess, sometimes identifying a pseudo shows the best guess to make, from a win-probability standpoint.  In the region at left (from here), take a close look at the left column.  We have repeated 50-50 fronts there going top to bottom, with the line of mines below; it looks very much like an extended line-and-1 50-50.  But there is one other possibility that still remains.  Can you see it?

The answer is at right.  That interior 4 has three adjacencies, not just two, and so in the (admittedly unlikely) chance that the teal-framed square is a mine, it would clear above and below the 4, determine the rest of that column, and reveal that the column was never a 50-50 to begin with!  As such, the yellow-framed cell is either safe, or part of a 50-50.  Therefore it is appropriate to guess it, under all circumstances.

Incidentally, the “cornered 5-square box-and-tail” that we discussed earlier, is a common type of pseudo: we know there are two mines in the region, and so either there’s one in the “tail” and one on the shared-corner of the box, in which case we should mark it and clear, or the box is a 50-50 and we have to guess it anyway.

There are many more situations that can invoke that reasoning (“this is either safe, or part of a 50-50”), beyond the ones having to do with a 4-square box or a line against a wall of mines.  You’ll see many other arrangements, many of them relating to the various types of exotic 50-50s that can crop up, and happening more often in high-difficulty play.  We mention it here[16] as a general principle, because recognizing pseudos and guessing them promptly is part of playing optimally.

(e) Avoid Dead Squares

The examples and situation-categories discussed so far in this chapter, and in the practice puzzles below, are instances of an even-more-general concept, one which has applicability in much more complex guessing situations.  That concept is the idea of “dead” squares, and it ties together all the specific cases and patterns covered in this chapter.  A dead cell is one that, if safe, can only be one number; as such it cannot add information to your game.  In other words, guessing a dead cell will either (A) blow you up, or (B) leave you in exactly the same position you were in before you guessed it.  Such squares are “dead” in the sense of “dead end” - they cannot yield progress.  Therefore, it is never a correct move to guess a dead square, unless you have no other possibilities to solve the region (which is the case when the dead square is part of a 50-50 forced guess).  Consider this trivialized example:

If you guess one of the wide squares here, then with equal 33% probability, you will either blast, turn up a 5 (meaning the middle square is safe and the other side is the mine), or turn up a 6 (meaning the middle square is the mine).  However, if you were to guess the middle square, then you will either blast (33% of the time), or the rest of the time, you will turn up a 5 and only a 5: there is a mine on one side or the other, but we still don’t know which side, and now we’re left with a 50-50 guess.  Therefore, the middle square is “dead”, and we should never guess it.  Put into the phrasing of this chapter, the wide squares are helpful guesses (and bring win probability 67%), and the middle square is an unhelpful guess (WP = 33%).

In the example here, a common three-mine edge formation, the square to the left of the interior 3 is dead: it can only turn up a 2, and if it does, we have merely reduced this region to an extended box-and-2 50-50. Likewise the cell above it, the 2nd one down on the left column: if safe, it can only be a 2, as the area’s 3rd mine must be on the bottom row. If clicked and safe, we are left with a “Swirly pattern” 50-50.  So despite being 33% like several other cells, those two cells are dead, and should not be guessed.  Whereas the 33%s on the top and bottom (and right edge, which is dependent on the other two) will either be a mine, or will yield information and progress, sufficient to solve the area.  They have WP 67%, the others WP 33%.

There is a dead cell in the region at left (which has 3 mines remaining), can you spot it?  Take a second and look.  Answer: Trick question, most of the cells there are dead.  But the cell to the left of the 4 is obviously so: it sees the exact same squares that the 4 does, so if it’s not a mine, it does not add any additional information that helps you solve the region.  If you click it and survive, you have just reduced the area to a 50-50.  The others, with a single exception, can take only one value.  That single exception is the untouched square, two down on the left: it can be a 2 (in which case the cell to the left of the 4 is a mine), or a 3 (in which case that cell is safe, and the 67%s are mines).  So that “living” cell has a 67% win probability, while the dead cells mostly have 33%.

Endgame guessing regions often have dead cells, and it’s important to look for them and exclude them from the guesses you consider.  In this region, we have four mines remaining:

 →

We can account for at least three of the four mines (top-left, around the 4 with ABC, and on the right corner at FGH).  However, note that cell A sees the exact same squares as the lower 4.  Therefore it’s a dead cell: it can only be either a mine or, if safe, will definitely turn up a 6.  So it’s out.  More subtly, cell G is also a dead cell: if G is safe, then both F and H are mines, which uses up all of our mines for the region (we only have four, and two others are accounted for), so if safe, G can only be a 5 (three mines marked, plus F and H forced to be mines then).  Not that you’d guess G anyway, because it’s a 1v2 corner (and 85% mine odds), but it’s still a dead cell.  There are many good choices to guess here (I went with J, and won; the solver recommends B, which also would have won), but dead cells should be the first ones crossed off the list.

Most of the categories we’ve been discussing in this chapter have really been about spotting common types of dead cells.  In the double-line-and-1 situation, the interior squares are “live”, and can take 2 values based on what they say about the other line, but the outside squares are dead, because if safe, they can take only one value.  In the 5-square box-and-tail discussion, the “tail” cell is a dead cell: it can only turn up one value, providing the same information as its neighbor (or blow you up).  It is quite common in high-density guessing situations that some cells are dead, and recognizing them can help you avoid mistakes. A big factor in guessing properly in minesweeper is to never guess dead cells (unless they’re a 50-50).  Dead cells cannot be good guesses.

If you use MSCoach’s Solver / Analyzer, which is considered the best-in-class minesweeper solver, it will recognize and mark dead cells (as dark gray) on any scenario you give it.  And it will never recommend that you guess a dead cell, even if it’s got high safety odds (low mine %), because doing so would not help you advance in the game.  Take the example at right (a fairly common isolated edge), with 4 mines remaining.  The square below that interior 3 is dead, because if it’s not a mine, it has to be a 2 (sees the same squares as the 3 above it, which is an effective-2).  And, if it’s not a mine, you’re just left with a 50-50 forced guess.  Whereas guessing one of the exterior 40% cells, one of the non-corners, leaves you with a 60% chance of clearing the area.

Sometimes complex 50-50s can have live squares.  Basic 50-50s are all-dead squares, but in bigger formations, there is still sometimes a way to smartly guess so that a 50-50 doesn’t become a double 50-50 - just like the double-line-and-1, above.  Those “smarter guesses” will be live cells.  Here’s a “triple” box-and-2 50-50 situation, from the appendix:

 →

When we plug it into the Analyzer, 10 of the 12 cells are dead, but two are living - the would-be corners framing the left-side box-and-2 50-50!  Those can be 2s or 3s, and if safe will reveal the layout of the left-side box here.  They each have 50% WP (along with the two squares diagonal from them in their own boxes, which are dependent).  All the others, if guessed, would then force one of the other boxes to be a second 50-50, so they each have only 25% WP.

Or, take example C from our Rule of Three Quiz.  If we put it into the Analyzer, it tells us:

“The solver has calculated tile (3,4) has a 50.00% chance to solve the isolated edge.  4 possible solutions remain.”

Out of the 10 unknown squares here, all are 50%, 8 are dead, but two of them are living!  Those squares can turn up more than one value, and that difference will tell you how to solve the rest of the region.  With either of those 2 living squares, you have a 50% win probability, whereas any other guess would force you into a second 50-50 guess later, and yield only a 25% win probability.

Many high-difficulty endgames with guessing regions will have a few dead squares; some will have lots. Here’s one, with local minecount known (7 mines):

 →

The left of the formation here makes total sense as dead cells (those are clearly a 4-square box with two mines), and the box next to it is a pseudo, but the ones to the right are non-obviously dead until you stare at it for a while.  If you provisionally mark some mines as test scenarios, though, you can get a feel for it, and realize that they can take only one value.

(f) Generalized Process

“Helpful result” situations, at their root, are about optimizing the chances to win an endgame that is forcing at least one guess.  One where there’s only a few squares remaining, and logic alone cannot solve it, but it’s also not a plain old 50-50.  In these situations, it may help for you to stop and consider the implications of each guess you could choose to make, and work out what the % chance to win the game is under each option.  That process looks roughly like this:

1. Calculate the odds of each square on the board being a mine based on your information right now.  You might get those odds via a hint, via brute-forcing the endgame, or by quick inspection if it’s simple enough.
2. Make a list (perhaps just in your head) of all candidate guesses you might pick as your next move.  Some guesses might be equivalent to others, some might be clearly worse and not worth considering.
3. For each candidate guess, consider what happens if you make that guess, survive, and face the rest of the board.  What possible numbers might it turn up?  What would each potential result tell you, and would that help you make further moves, or merely put you up to further guesses?
4. Repeating this, you’ll build a sort of logic tree in your head, where you’ll have logic like:

1. Ok so 50% of the time I lose on the guess, but then
2. With the result, 50% of the time if the mine is here I can solve it by logic, and
3. The other 50% of the time, the mine is there, and I end up with a 50-50.

5. Then you multiply the numbers through, like a weighted average, to calculate the chance of winning the game from that guess.  So taking the numbers above, you’d say 50% * 0% win rate (from A) + 25% * 100% win rate (from B) + 25% * 50% win rate (From C), which = 25% + 12.5% = 37.5%.  So that guess wins the game 37.5% of the time.
6. Repeat for each candidate guess.  There may be significant differences between them.  Make a guess (or one of the guesses) which has that best Win Probability.

A quick example: At right, we have 2 mines remaining, 4 squares, and just one window into them.  What’s the right guess?  Each square is a candidate, each has 50% chance of being a mine, so let’s go through our process for each:

1. If we survive, then we’ve just shown that B is a mine… but now we have to blindly guess between C and D, leaving us a second 50-50.  Winning percent: 50% * 50% = 25%.
2. If we survive, then we know A is a mine (from logic), but we still have an “overhang” style 50-50 between C and D remaining.  Winning percent: 50% * 50% = 25%.
3. This will turn up either a 3, implying that B and D are both mines, or a 2, implying that D is but B is not, so A is the other mine.  Either way, surviving the guess wins. WP = 50%.
4. This will likely turn up either a 5 (B and C are mines) or a 4 (C is a mine, B is safe, so A is a mine).  As with C, our WP is 50% * 100% = 50%.

Among those options, guessing C or D will have a 50% win probability, while guessing A or B has only a 25% win probability.  So while A or B might feel more comfortable since they’re connected to an area we already know, guessing into the unknown is actually the right play.

Or take this one, which is more fun.  We can see three potential 50-50s here: the line-and-1 at the upper row, a line-and-1 on the right column, and a box-and-2 at the bottom 4 squares.  We can’t prevent all of them.  All cells also have a 50% mine chance.  But if we guess, say, the top-left square (to prevent the box-and-2), it puts a mine to its right, which confirms the line-and-1 50-50 on the right column. WP = 25%.  Guessing top-middle will yield logic half the time, and a 50-50 half the time, so its WP = 37.5%.  However, guessing the upper square on the right column - and only that square - yields a 50% win rate: if it’s not a mine, the whole area becomes solvable.  (All other squares are dead)

It’s rare that you’ll have to fully reason out a situation like this, and calculate out the odds for several options.  The specific scenarios we covered above are more common, with optimal moves known.  But they’re just applications of this process, special cases.  For anything more unusual, you may need to apply it in full to optimize your chance of winning the game.

Practice Puzzles

Link to solution section.

Notes:

• Instead of asking for mines to be marked or squares to be cleared, these puzzles are looking for the “optimal guess”, i.e. the square(s) where you have the best chance of winning the game (taking into account the chance you blast on the guess itself).  In the solutions, we’ll notate that metric as Win Probability or WP%.
• Several of these puzzles offer “hints” next to the initial situation image, in the form of showing you the probability distribution (if it’s not obvious, like 50-50 everywhere).  They don’t solve the puzzle for you (in fact, in at least one case the probabilities are misleading), but they may make some mental math a little easier.
• Puzzles 3-7 and 3-8 together provide a thorough explanation of situations you’ll see fairly often in high-diff play, so they are especially useful to understand and remember.

Puzzle 3-1 “U-Turn”: Find an optimal guess.  Difficulty: 1 / 10.

Puzzle 3-2 “Hockey Stick”: Find an optimal guess.  Difficulty: 2 / 10.

Puzzle 3-3 “Combo”: Find an optimal guess.  Difficulty: 2 / 10.

Puzzle 3-4 “Snuggle”: Find an optimal guess.  Difficulty: 3 / 10.

Puzzle 3-5 “Screen”: Find an optimal guess.  Difficulty: 3 / 10.

Puzzle 3-6 “Clown Car”: Find an optimal guess.  Difficulty: 4 / 10.

Puzzle 3-7 “Box and Tail”: 3 mines remaining, find an optimal guess. Difficulty: 4 / 10.

Puzzle 3-8 “Mirror on the Wall”: 3 mines remaining, find an optimal guess.  Mirrors the previous situation, with a twist.  Difficulty: 5 / 10.

Puzzle 3-9 “Extended Box”: Find an optimal guess.  Difficulty: 5 / 10.

Puzzle 3-10 “Ounce of Prevention”: Find a helpful guess.  Difficulty: 5 / 10.

Puzzle 3-11 “Daily Double”: Find an optimal guess.  Difficulty: 6 / 10).

 → Probabilities:

Puzzle 3-12 “Gauntlet”: 2 mines remaining, find an optimal guess.  Difficulty: 6 / 10.

Puzzle 3-13 “Spiderweb”: Find an optimal guess by identifying the dead cells. Difficulty: 7 / 10.

Puzzle 3-14 “Slow Down”: Find at least one helpful guess.  Difficulty: 7 / 10.

Puzzle 3-15 “Overlays”: Find an optimal guess.  Difficulty: 8 / 10.

  →

(4) Square Coverage Ratio

This rule goes: “if forced to guess, look for an effective-1 number square that has the greatest number of unknown cells around it, and make your guess around it and away from your front”.  Often, we have a guessing front that bends around a corner, and maybe has one square we uncovered from logic that goes around that corner or into space behind a marked mine, but gives no further moves.  Such a square might have 5 or 6 or even 7 unknown cells around it, only one of which is a mine.  1/5 is 20%, 1/6 is 16.7%, so those can end up being pretty favorable guesses (especially if we have to guess more than once!).

Here’s a commonly-seen example, where a square extending out into space is an effective-1, and has many adjacencies for that one mine to be hiding in, giving us favorable guess odds:

  All share the same %s →  

We call the ratio of “# mines” to “# cells those mines might be in” our “Square Coverage Ratio”, and the average probability of its adjacencies we call the cell’s “Local Density”.  In high-density endgame situations, you’ll often be considering e.g. “do I guess this 1-mine-in-4-squares, or this 1-in-3 that will unlock more logic”.  You’re considering the square coverage ratio then, too.  But we’re extending that idea: even when you have to guess out into space, with large amounts of the board remaining, the same principles apply (and may guide you to a good choice).

A few nuances of this tactic to emphasize:

• Effective-1: This means you’re not just looking for a number 1; the square you want might be showing a 2 or a 3, but it has only one unmarked mine.  The key is to focus on the # of available cells around it first, usually right around / near the corners of your front.
• “Away from your front”: once you’ve found a cell with a high square coverage ratio, you’ll note that some of those surrounding squares are shared with another piece of your front.  Usually two, but sometimes one or three, of your open squares will be adjacent to other squares on your front.  Those will, almost invariably, have higher fractions of the mine probability, and the squares that don’t touch the rest of your front will have lower probabilities.  So your lowest-risk guess will be the ones that don’t touch the front.
• Background rate is a factor.  If you’re playing Expert (background rate: 20.6%), a 1-in-5 coverage-ratio guess really isn’t that great.  If there are unguessed corners, you might prefer those.  Maybe some open edge space, even.  But if you’re up to a 24-25%+ background rate, all of a sudden more SCR guesses may look attractive enough to try.

Because this technique usually finds a better-than-background-rate guess, it’s often a preferable choice over an “Assume Fewer Mines” type guess (which generally gets you a background-rate guess, just one that’s likelier to result in progress if you survive it).  However, note that guesses of this “SCR” type are more “opening-hunting” than they are “progress-hunting”.  It’s unlikely a move here will cascade into a chain of logic moves that let you continue through big chunks of the board, but it’s possible that hiding behind this low-ratio cell is an opening, or one not far beyond it, and if you find it, that opening will offer you progress.  So if you’re deciding between this and an assume-fewer kind of situation, you’ll want to really consider the likelihood that you can make progress off of the assume-fewer guess.  Because this technique, even though it’s happening “near the front”, is essentially closer to a “guess into space”.  Better odds of surviving a guess, as we know, does not always translate into “better odds of winning the game”.

On the other hand, this technique’s effectiveness will be usually inferior to a guess suggested by Rule #10 - Guessing via Lock-in.  Those are harder to spot and take more thinking, however, and require a lot of practice to get right.  So a Square Coverage Ratio option may be the easier one to see and choose, if it’s there when you need to guess.

It’s pretty common to see 1/5 and 1/6 coverage ratio options.  Rarely, you can even get all 7 adjacent squares involved (which can only happen if certain other things align).  If so, that’s an even-better 14.3% average probability (and even more favorable for the squares away from the front).  You won’t see this often, but if you do, good things are usually coming to you.

Here’s an all-7-squares scenario, very favorable for a few reasons.  We’ve got that interior 1 with a 7 SCR, so we’re starting with a 14% average probability.  But then, the adjacencies shared the front (three of them, here) will tend to “pull” probability towards them and away from the squares that only touch the interior 1 (see below for more on this).  Although the squares below it only end up with 13% each, the square to its left has 35%, leaving the other 4 squares with 10% each.  And then secondly, that 10% cascades up the left-side front, creating 90/10 splits for the next several pairs along that linear portion.  So that one guess yields at least three squares’ worth of progress, and likely more.  And even if it doesn’t, you still have those interior squares to try, and maybe get an opening.

 →

This all-7-squares SCR option is even better than you’d expect.  Partly because of the front-adjacent square above it that’s pulling probability towards it is a 50-50 front (so the ones out into space are just 10%, as they split the residual odds).  But also because the highlighted 7% squares are an “assume fewer mines” situation all on their own, due to the 1v2 corner at the right - if the highlighted squares are mines, it forces an extra mine from the “reservoir”, which is only B% (15%) as likely as the reverse.

Combining with Assume-Fewer

Sometimes in mid-game guessing situations, a good option that you spot on the basis of Square Coverage Ratio will be nearby an 1v2-style Assume-Fewer guessing spot, often around corners and such.  Generally, you might assume, since SCR gets you below-background-rate guesses (But with lower chances of progress), and Assume Fewer guesses get you a background-rate guess (but with higher chances of progress), you’re going to prefer guessing into space in line with this chapter’s tactic, before moving to the other ones that bring more blasting risk.  But if the two options share some squares (even via dependent squares), you can get the best of both worlds, some below-background-rate guesses which also have very high odds of progress.  Here’s an example, can you see any good guesses?

There are two 2s sitting out in space there, and while the lower one is still an effective-2, the upper one is an effective-1.  You would expect good odds on all the squares except the one above-left of it, which basically shares a 50-50 front with the upper-left corner area.  But what does that mean for the corner area just below that 2?  That corner area is a 1v2 corner, if it were fully isolated you would expect ~20% odds for the wide squares and ~80% mine odds for the corner square, based on Assume Fewer.  But note what happens to the wide-square option: not only does it force more mines into the area, it also would satisfy the interior 2 above it!  That means that all the other options for that 2 would be removed, limiting the possible configurations for the area.  So because the Assume-Fewer corner here has its usual logic reinforced by the Square Coverage Ratio option adjacent to it, you actually get the best of both worlds: good safety, and good odds of progress.  See the odds at right: You’d much rather guess the 14% cells, giving you guaranteed progress, than the 11% cells which carry very low progress odds.  That’s the best guess here, taking into consideration all factors.

Square Coverage Ratio can help you find a decent guess out into space (away from your front), but it’s a much more powerful guide, much likelier to give you progress, when you can use it to identify a better guess that’s on your front like this.  A few more examples of this situation:

In this 100k attempt (at 1:07:00), middle-left, we have a 1v2 corner helped by a favorable SCR:

 →

First we have to spot the interior 2 in the middle there, which needs one mine from 6 squares.  The corner to the right of the interior 2 isn’t a 1v2 corner, because the 2 on the right edge is not a 50-50 front.  But the corner above the interior 2 is a 1v2, and so the fact that it shares an adjacency with that interior 2, means that the corner’s wide squares are close to (actually a bit below) the odds of the residual probabilities for the interior 2.  Most of the 2’s probability is pulled toward that front to its right.  But the corner above is the best choice for a guess, because you have good progress odds, at well-below-background rate risk.

 →

In this game, we’ve got those 2-1-2 squares at left, so we know there’s at least one mine on that top row above the 2.  There could be two mines in them, though, if and only if the left 1’s mine is below-left of it. But that would force an extra mine into the region (a 4th one - you can get only three mines if you have a mine on the right corner, on the 61% to the left of the 1, and then one more above the 2 somewhere).  The interaction with the 1v2 corner on the right side, though, means that the corner’s wide squares are below background rate, because if the mines were on those wide squares, it satisfies that 1 in space, forcing a 4th mine below-left of the lower front 2.  So because we’ve got that high-SCR 1 in space, it gives a safety “boost” to the 1v2 corner.

Minecount can come into play, when finding advantageous relationships with high-SCR tiles:

 →  →

Here we have a minecount of three (which is important); as such, we know from square groups that once we take away the single mines in teal (per the interior 2) and in yellow (from the right 3), the final mine must be in the two squares to the left, in pink.  Likewise, from the upper-left 1 on the front, there’s one mine on the purple.  So we have a 1v2 corner: either the groups share a mine at upper-left, and there’s one in any of the other five squares around the 2, or they don’t share, and the mines are at lower-left corner and to the left of the 2.  Because of that 2’s high square coverage ratio, the more-mines option here is much less likely than having one on the pink-purple overlap square at upper left.

The interaction with Assume Fewer need not involve a corner - a linear 1v2 will do just as well:

 →

The interior 2 has a local density of 20%, one mine in five squares, but while it has a 50-50 (-ish) front to its right, the front to its left is a 1v2: a mine above the 3 forces an extra mine in the region, to the top-left of the middle 4.  Because of this interaction, the four adjacencies to that 2 that aren’t shared with the 50-50 front to the right, all split the residual probability, each of which is below background rate (15% against a B=27%) because of that Assume Fewer situation.

Pulling Probability

In our discussion of Square Coverage Ratio, we have spoken a bit about how, among the cells surrounding a high-ratio number square, the ones adjacent to our front will tend to have higher-than-proportional odds of a mine, and the other ones, out in space - where the only information comes from our effective-1 with the high ratio - will tend to have lower-than-proportional odds.  This is the first example of a certain effect on the probability odds of mines, which I think of as a sort of gravity: if there is a high-density area that interacts with (is partially adjacent to) an area that is lower-density, then the squares adjacent to the high density area will “pull” some of the probability odds towards it, and away from the low-density squares that are not adjacent to the high-density area.

At right, consider the left-side front there.  In blue, we have a linear front which, absent information from the 5 above, would be just an ordinary 50-50 front the way the bottom front is.  However, because we DO have the 5, which needs three mines from four squares (a high coverage ratio), the 5 “pulls” probability away from the lower of each square pair in blue, and towards the upper square pair, making the bottoms a good guess.  Why does this happen here?  Two reasons:

1. In the combined yellow+blue region, a mine on the overlap square means it’s satisfied with four mines, while that square being safe makes it require five mines.  So we have Assume Fewer working in our favor.  But also:
2. There are four possible arrangements for the remaining mines around the 5, but only one of the four has that overlap square being safe.  So even if the mine requirements were equal, that square is only rarely safe.

Another example at left.  In the blue linear front, we have a 1v2 situation: one mine in the middle square (above the central 1), or two mines, one in each pair to the left and right.  And that interior 3, at right, needs two mines from six squares, so the average probability around it is 33%.  Absent the “pull” of that 3, we might assume the blue region resembles the “split background rate” front from the Assume Fewer chapter.  However, the left squares and right squares of the pairs in blue are not equal!  The 3, which has a higher local density (coverage ratio), “pulls” some of the mine probability between those pairs toward it, making the left squares of the pairs the better guess.

If you’re trying to gauge probability odds by inspection, this can often lead you to one of the optimal guesses, or at least a pretty good guess.  And they can be found without a ton of deep thought.  It’s also an effect that we’ll see impacting a whole bunch of guessing situations, and will refer to throughout our probability-estimation discussions.

This “Pulling” effect can be seen in one of the most common corner-opening positions:

You get a 1-2-1 corner opening, you click your safe square off the corner, and it turns up a 1.  So now, technically, we don’t have any guaranteed-safe squares.  However, even beyond the 1/7 square coverage ratio, we should notice something about the two front groups (orange and yellow).  They are not truly independent or 50-50, because it’s impossible for both groups to have their mine on the interior square.  That alone “pushes” probability towards the edge squares, landing about 64-36 in favor of the mines being on the outside (edge) of a given pair.  But consider what that does to the interior-1: if those numbers are correct, 72% of its mine probability is accounted for with the two squares adjacent to the front.  It has 5 other adjacent cells that account for the other 28%, making each of them a little under 6% odds!  As such, the green square there is almost a free guess, it’s only a mine once in a blue moon, and fairly often it will yield an opening (24% of the time in expert) that probably lets you make progress.  So I pretty much guess it as a habit when I see it.  All of which comes from the fact that the front-facing squares are “pulling probability” away from the rest of the interior square’s adjacencies.

If you’ve got a square with a high coverage ratio, it’s worth asking whether its “pulling” effect has some Assume Fewer implications, too - basically, would certain choices result in the area requiring more mines?  If so, then the fewer-mines option will pull probability towards it.

In the example at left, the interior 4 (an effective-3) has a local density of 50% (3 mines in 6 squares).  But the square directly to its left is not 50%, it’s much higher than that, because if that square were safe, then the whole front there would need an extra mine: three around the 4, two on the extended line to the left, and the one on the lower 15% square.  So the 4 is pulling a lot of probability towards it on that front to its left.  Those 15% squares also benefit from a Split Equivalence effect, where either of them being a mine puts a mine on those 30% squares going left, so each of them individually are half of those squares’ odds.

Pseudo-Hole Pattern: One practical use of this pulling effect can be found in a pattern that’s like a probabilistic extension of the Hole pattern.  Suppose we have a linear front, we have a number square in the interior which is an effective-1, but it is in front of a mine.  And we have 50-50 fronts both above and below it on that linear front (or both left and right, if horizontal).  So without a square immediately behind it, we can’t use the Hole Pattern logic rule to be absolutely sure that the squares one row off from the front are safe.  However, because we have a 50-50 front above and below, two things are true: the squares just above and below the interior square can’t both be mines, and likewise, the odds that neither is a mine (which would mean the one remaining mine is in the “hole”, two rows off the main front line) are, while not zero, fairly low probability.  What’s more, the three squares behind that frontline, in the second row, all divide that residual probability (of “neither”) equally between them.  So the result is that some pretty good guesses can be hiding behind that type of layout.

At left is a standard hole-pattern situation.  We exposed the interior 1 with a 1-1 pattern, and then can reason that because the 1 mine for that interior 1 is accounted for by the “backing” 1, on the front to its right (via the two yellow squares - we don’t know which one is a mine yet, but we know it’s one or the other), therefore the squares behind it - the green ones - are all safe.

Well, similar things are true, or at least true-ish, with the “pseudo-hole” at right.  You have that interior 2 that’s an effective-1, but it’s backed by a mine, not a number square, so no hole-pattern inference is possible.  But still, it starts out with decent odds on a square coverage ratio basis: one mine in 5 cells, so even under a naive assumption, those squares are a 20% mine probability - not very exciting, still likely better than the background rate, if only just.

However, due to the “pulling” effect, we can see it’s much better than that.  Note that it has a 50-50 front both above it (the 2-3), and below it (the 2-2).  We can reason that for the squares shared with the interior 2 on those two pairs, it’s unlikely that they’re both safe.  The odds of that happening (and thus the hole squares containing a mine) are well less than 50%, i.e. the odds that those two squares have the interior-2’s mine are higher than 50%.  In other words, the shared front (adjacencies above and below which had higher mine probability) has “pulled” some of that probability towards its shared squares (increasing their mine odds above 20%), and away from the non-shared “hole” ones (reducing their mine odds below 20%).  For the hole squares, you’re now dividing <50% worth of mine odds, equally across three squares.  So each one is probably a pretty nice guess, on its own.

Here’s another example of a pseudo-Hole:

 →

The interior-2 here is backed by a mine on the front, so there’s no hole pattern available.  But we have 50-50 fronts above and below it (sorta - the lower one is complicated by the lower 4 sharing one of its squares).  So we reason that the chances of the squares above and below that interior 2 both being safe, and thus its mine being in the left column, are probably pretty low.  In fact, here, it’s 36% - not all that low - but when you split those odds across the 3 interior squares, each one is pretty attractive!  12% sure beats our background rate of 22% here.

And then furthermore, if you guess a hole square, it has a chance to turn up as a 1.  Which would mean that its single mine is shared with the front squares (or the interior-2, if guessing the middle), so we can use a hole pattern on it (H-3 in WoM terminology) to clear its other adjacencies.  Wouldn’t that be nice!  At right, the 5 other adjacent cells to that guessed 1 are safe.

A few more pseudo-hole examples (and in D, it worked!):

(A)

(B)   (C)   (D)

Notice that in (B), we can see the pattern works even when the interior cell is shielded from the rest of the front!  It shares only two adjacencies with other front squares (one of which is a one-mine-in-three, the other a 50-50 front with some probability pulled away), but together that’s enough to pull nearly 2/3s of the mine probability towards those two, and away from the cells below.  The other three adjacencies split the residual mine odds, giving us 12%, good chances.  And in fact, the square below that 4 yielded a 1, offering a hole pattern and progress from there.

This pattern works to find good guesses, only when you can make an assumption that the fronts on either side of the interior cell are at least somewhat close to 50-50.  If they’re not even close, if the squares shared by your interior cell have low mine odds themselves, then its residual probability can be high, and then the “dividing it in 3” effect here still yields bad guesses.  The outside 5s here are pulling a lot of probability towards them, so the pseudo-hole squares are actually higher than background rate.

When you have this pattern, which “hole” square do you then guess, the middle one (directly behind the interior-2), or the wide squares?  That’s a matter of taste.  They both have the same safety chances.  The wide squares, however, require five other squares to be safe in order to turn up a 1 (the ones not shared with the front), while the middle square requires only three other squares to turn up that 1 (the ones not shared with the interior-2).  As a result, at expert density, the wide squares have chances of progress (which is mostly the odds of yielding a 1) of about 33%, while the middle square has ~46% odds of progress.  The former is less good, in terms of progress chances, than a wall-skip move, but the latter is better than a wall-skip (see the next chapter for odds of an opening).  On the other hand, if you do get that 1, you get to clear more squares (five) with guessing the wide square, than you do with the middle square (three).  So both are defensible choices, and both of them have roughly equal chances of having a zero (opening) behind them somewhere.  But the whole tactic has one main thing going for it: Because of the pulling-probability effect discussed here, the odds of blasting on the guess itself are usually well below the background rate (whereas for a wall-skip, it’s equal to the background rate).  So trying this is closer to a “free roll of the dice”, more than it is a big risk.  And if you find progress with it, the payoff can be pretty great.

Interaction with Background Rate: This effect makes a certain amount of sense intuitively: you can think of mines like electrons, existing not definitively in one spot, but rather as a probability distribution, where some places have more odds of it, some have less (or basically none), and then think of number squares as having electrical charge, bending that probability distribution towards certain places, or pushing it away.  But alongside our number squares, the background rate serves as a sort of “atmospheric density”.  As #1-ranked PvP player Oreh1337 puts it, the strength of the “pull” depends on how your square coverage ratio compares to that background rate B: if a square’s local density is higher than B, it pulls toward itself, if it’s lower than B it pushes probability away, and if it’s equal then there’s no pulling effect at all, no “net charge”.  This is just a mental concept for what is, under the hood, a representation of the odds of all possible and valid mine configurations within each region, and how frequently among them a mine appears in each square.  But it’s still a useful model as you scan a front for opportunities.

(A)  or (B)  vs (C)

In (A), the interior 3 has a square coverage of 20% (1 mine / 5 squares), lower than B=22%, so the square it shares with the 2 (which is otherwise a 50-50 front) has some of its probability pushed away.  It is (slightly) less likely to be a mine than the square to its left, because a mine on the square under the 2 effectively would increase the background rate, forcing more mines into fewer remaining untouched squares (the same underlying effect that was the basis for Obelus’s Principle).  For (B), the interior 3 has a local density for its adjacencies of 20% (1 in 5), which happens to be higher than the background rate (only 17%), so with greater local density comes a pulling towards it, resulting in 54-46 odds on the adjoining two-square front.  And then you’ve got (C), where the interior 3 has only four adjacencies (25% local density), and the background rate happens to be 25% too, so the adjacent 2-square front is exactly 50-50: the interior 3 neither pulls nor pushes.

We will see other situations illustrating this pulling-probability effect; these are just the first.

Examples

We don’t have practice puzzles for this chapter, because there’s not a lot of subtlety here if you know what to look for.  However, instead of puzzles, we do at least have a few examples where this rule nicely applies, and you can just look at them and practice spotting them.  In all cases, the best guesses will be the ones that do not share any adjacent squares with number cells other than the effective-1 that goes “out into space”.

Try to quickly spot the cells with a high square coverage ratio, and which guesses around them will be favorable.  When you see it, check yourself in the Solution Section.  Most don’t need much discussion, so the “solutions” are usually presented without comment, but any example whose sequence letter has an * next to it will have some comments in the solution, on other aspects of the example that I think are interesting to note.

(A)   (B)

(C*)   (D)

(E*)  (F)

(G)   (H)

(I*)  (J*)

(K*)  (L*)

(M*)

(5) Chain the guesses for 1s-in-space

This is an early-game tactic, that amounts to hunting for openings.  The idea is this: if on the corner of your front, you get an effective-1 that has 4+ unknown cells around it, guess diagonally away from your front, and if you get a 1 on that, keep guessing diagonally away from the front until you either hit an opening or get a number higher than 1.  And if/when you get a high number, then backtrack, and go diagonally away in a different direction from one of those 1s.

 →

Above: at left, we see that the corner of our front has an effective-1 with 6 unknown squares.  We know from probabilities that the remaining mine is likelier to be on one of the two adjacent cells that is shared (below with the other 2s, or top-right with the 3).  So going diagonally away, while not guaranteed safe, is the likeliest to be safe.  We get a 1 on that corner, so we decide to keep going in that diagonally-away direction.  And on the very next guess (at right), we get an opening, and can proceed with logic from there.

Why is this a good approach?  Because of the square coverage ratio.  The first effective-1 you have on a corner of your front might have 5 or 6 squares around it for that one mine, so your chances start at 20% (1/5) or 16.7% (1/6).  But they’re actually better than that, for the front-adjacent 1, because the adjacent squares that are shared with other parts of the front will “pull some of the probability” away from the others, and end up higher-risk than the ones that are truly out into space.  So you guess out into space, “chaining your 1s”.  And then, guessing diagonally away means you’re landing a 1 (if you get it) that has 7 unknown squares around it (5 of them new to you and not shared with any other known square), and the only adjacent number square is a 1 anyway so there’s not a lot of risk lurking around.  And even without the pulling-probability effect, 1/7 is even better odds (14.3%).  On a board that might be 20%, 22% or even 25% background rate, that sounds like a good proposition!

Why go diagonally?  If you were to chain guesses going horizontally or vertically, each new guess would share four unknown cells with the previous guess, while adding information for three new cells that you’re accessing.  But by going diagonally, your new guess shares only two cells with the previous guess, while adding 5 previously-unknown cells to the front.  Given that each guess carries with it a ~15% chance of dying, give or take, the more efficiently you can gather information about the board, the better you can probe it for promising spots.

And what do you do when you stop getting 1s in your “chained” guesses going diagonally?  You backtrack, and go in a different diagonal direction from your chain of 1s:

 →

Basically, you want to avoid anything that’s shared with the 2 you departed from on your front, and also away from the 2 (or higher) you just uncovered deeper into the board.  Typically, the squares with the lowest mine odds may be the ones shared by two of your 1s (dark blue).  But in my experience, your best chances of finding an opening, which is our goal after all, are to proceed diagonally away - the light blue squares.  For example, if you were to guess the diagonally-outwards light blue squares away from that topmost 1 in the chain, a lot of the probability of where that 1 mine is pulled towards that interior 2, so by avoiding the cells shared with it, your odds stay good.  If you get stymied on a chain and get spooked by a higher number appearing, maybe you try the dark blues instead, to see if there are any small openings there.  Or, if you’re mid-game and want to play it as safe as possible and hope that a few moves later, logic becomes possible again, you might opt for the dark blues.  But my tactic here suggests the light blues, to best hunt for an opening.

Some more examples requiring a backtrack:

I use this tactic in two situations:

1. When it’s very early-game, and I want to either get a foothold in the game and start making progress, or blow up and move on to the next one, but either way I want it to happen quickly.
2. When it’s early-to-mid-game and I’ve got a promising initial position, but I got stuck and can’t spot any good guesses (nevermind logic moves) on my remaining front.

#1 only applies if you’ve already guessed the corners, by the way.  The corners are still the best spots for openings.  But if it’s a high-diff game, you’ve probably opened them all already as part of your opening moves.  So if instead you’re (say) playing Expert for mastery, and you get a situation like what’s described here, definitely go to the corners before anything else.

And, of course, sometimes you won’t have a corner of the front that provides this sort of attractive-looking jumping-off point.  Sometimes you’ll really have to think and visualize percentages around your front.  But that’s a lot harder and slower.  If you’ve got one of these available to you, it’s simple to spot and quick to use.

In one sense it’s a bid to just “get lucky”, but in a broader sense you’re playing the percentages and putting yourself in the best position to get lucky.  Gotta take some risks anyway, so might as well do it as intelligently as you can.  Bending the percentages a little bit in your favor is the name of the game in advanced play.

But of course, it can still fail.  No probability-based rule works all of the time.  At right: “you gotta be kidding me!”

Opening Chances

Recall a very common piece of advice for early-game strategy: “Guess the corners first”.  Why does everyone nearly always start with a corner guess, and then if they still lack safe moves, they go and guess another corner?  Because you’re a lot more likely to get an opening there, i.e. a “0-number” square that then chords itself and reveals a bunch more squares and information on them, giving you (we hope) a starting point from which to start reasoning and clearing the board.  If you can’t get that, then your odds of just guessing your way square-by-square into a position where you can start doing logic are quite low.  If you try to start without getting openings, just guessing adjacencies to your previous guesses, you will lose an expert game 99%+ of the time, but if you start by guessing corners, you can get as high as a ~38% win rate.  Smart early-game guessing is an important part of tilting the odds in your favor.

To get an opening on a corner, you only need 4 squares to not have mines (the one you’re guessing, plus the 3 adjacent to it).  An opening on an edge requires 6 squares to be clear (your guess, the two edges adjacent to it, and the 3 spots one square “inland” from each of those 3).  But to get an opening in the middle of the board, not on an edge, you need 9 squares to be not-mines.  Wherever you guess, if you’re hoping for an opening, it requires two things to go right: you need your guess to not be a mine in the first place (odds = the background rate), and then you need the squares surrounding that guessed square to each not contain a mine, such that your guessed square is a “0” and the game can auto-clear around it.  So if we take the first risk (dying on the guess itself) out of it, what are the odds that a guess onto a safe square will result in an opening?  With some simplifying assumptions,[17] that’s a pretty straightforward probability calculation:

As we can see, even for Expert, your odds of a safe interior square being an opening (15.8%) are lower than your odds of dying just in making the guess (20.6%), whereas the chances of getting an opening on a corner are pretty great: you lose 20.6% of the time, get an opening 40% of the time (50% of safe guesses), and the other ~40% of the time, you’ll turn up a 1, 2 or 3 and at least get to guess some more.  Those odds go down in higher mine densities, but even for the highest density that WoM considers playable (25%), you still have a better chance of a corner being an opening (~32%, = 75% safe guess * 42.2% corner opening) than being a mine.  That’s true for any density up to ~27.5%.  Whereas the same is not true if you guess an edge, on any density higher than Expert.

These probabilities have been well-known for a while, e.g. on this site, and come alongside another consideration: the size of the opening, in # of cells opened.  Bigger openings give better odds for better quantities of logic moves available after they’re done clearing.  And so it’s also true that openings towards the middle of the board tend to be larger, on average, than ones on the edges, and ones from the corners are even smaller.  But in terms of your chances of winning the game, opening size has less of an impact than the chances of getting an opening at all in the first place.  If you’re guessing like this, you’re stuck and need something to go right to resume making progress.

So, the above tactic (chain-the-1s), the next one (wall-skip), and to a certain extent the previous one (square-coverage ratio) and some others, all are predicated on hopes for an opening, which then offers us logic moves and further progress.  That’s a lot more helpful for winning a game than trying to brute-force our way through a bunch of front-line dilemmas through a combination of luck and a slight edge from probabilities.  But we do need to bear in mind what kind of chances we’re taking here, because in early- and mid-game scenarios, hoping for an opening might actually be a fool’s errand, or might be our best move, and we need to know our numbers in order to know which it is.

Examples

No practice puzzles for this chapter, but I’ll include some examples of good jumping-off points for the technique, so you can begin to recognize the pattern and use it when it’s available.

(A)   (B)

(C)   (D)

(E)

(the interior 2, between the two 4s in the middle)

(F)   (G)

(H)  →

(I)

(the 1 sticking out into space at far right)

(J)  (two for the price of one!)

(6) Skip a wall-spot and hope for a 1

So we just covered odds of finding an opening, and some tactics to hunt for them, particularly in early-game.  How can we apply that to mid-game, or when we lack options on the board that fit with those tactics?  We can do so in the following situation, which is quite common:

• You have no safe moves left to make through logic
• You’ve opened all four corners
• There exists on your front line some numbers that are on the very edge of the board, which present a two-square split (i.e., one mine on two squares, one of which is an edge square and the other is directly inland from it)

If so, then guess the edge square that is two squares past your last known number, two rows or columns past the front, “skipping” the two-square split along the wall.  And, simply, you hope to get a “1” there, and if so, you can clear some squares for free.  More on that in a moment, but first here’s an example.  It meets the conditions: All corners open, no more logic moves to make, and there are two good spots to do your wall-skip guess (i.e. where one mine is either in the edge square or the square immediately inland from it).  Can you spot them?

Well, here they are:

The pairs of yellow squares each contain one mine between them, so we “skip over” them and guess the next square along the wall, away from our front.

The one at the top results in a 2, which we have to leave aside - one of the other 3 surrounding squares is a mine, and we don’t know which yet.  We’re not dead, but we can’t make progress using this.

The one at the bottom, though, results in the 1 we’re hoping for.  Which means its one mine is contained within the two squares it shares with our front-line 3.  We don’t know which one has the mine, but it doesn’t matter, we can use the hole pattern to clear the other three safe squares immediately, and can open them “for free”.  So we basically got four squares for the price of one guess.

In this particular case, the other three surrounding squares were 1s as well, which isn’t uncommon.  And from here, we see that because each pair of yellow squares here contains a 1 (per the wall-adjacent cell), it means the ones on top of them - the four green squares - are therefore all safe and can be cleared.  And this leads on to other logic, so the one guess basically allowed us to resume making steady progress, the way we normally hope to.

If you get a 2 or a 3 on your wall-skip guess, you move on and look for another thing to guess, because there are some mines there that you can’t make assumptions about.  If (rarely) you get a 4, you can mark those other 3 squares as mines, though it still doesn’t help you make further progress.  But your odds of getting back into the flow of making meaningful progress by doing this - skipping over a wall and getting a 1 - are generally better than other options you might have when reaching this point.

Here’s another example.  Assume the corners are guessed; spot the two wall-skip opportunities.

Big hint was in the name: they’re along the wall.  In this case, they’re at (1) skipping past the 3 at top-left, jumping two columns to the left there, and (2) center-right, down below the 2 on the wall (the guess would be two rows below that 2).

In this case, both of the guesses yield 1s, meaning we clear around them, and we can see that further logic proceeds from there.  At top-left, we got a (minimal) opening, and the 1 on the corner means the interior cells are safe.  At bottom-right, our clearing gets us two mines (on the 2s below) and then clearing the 4 cells to their left.  Which puts a mine in between the 4 and the 3 on the front, which lets us clear from there, etc.

Now, of course, doing this technique will blow you up at a rate equal to the background rate of the board, so it’s only something to resort to after you’ve run out of logic.  This makes it a good early-game move, in situations where you’ve gotten part of the way through the board (and have guessed the corners), but not so far that you’re inclined to use a hint and try to optimize your guess by identifying some low-probability square.  You’re willing to take a guess at the background rate, because if it pays off then your effective chance of winning the board goes way up, but you’re not so committed to this game that you’ll be upset in the ~20% chance that you die trying this move.

Also note that this requires your edge square to be faced up to only two unknown squares.  If your front has a mine on the edge, and a number square one space inland, that’s facing up to three squares on the front, it’s not a 50-50 front.  If it’s facing 3 squares, then even if we make our wall-skip guess and turn up a 1, we can’t be certain that the non-shared squares are safe, because the mine could be on the 3rd square down.  So make sure that either you have a number square on the edge of the board, or a formation that limits the mine to the 2 squares closest to the edge

What are the odds that it “works”?  Similar to the chance-for-an-opening calculations earlier, there are two risks: dying on the guess itself (= background rate), and then not turning up a 1 and being able to make progress.  Here’s why it works, though: The odds of a wall-skip guess turning up a 1 are equal to the odds that a freshly-clicked corner is an opening.  This is because getting a 1 on an edge like this merely requires that three other untouched cells (the ones not shared with your frontline) not be mines.  Same as a corner, which only has three bordering cells.  So if you like your odds of a corner opening, such as on Expert (=50%), this technique effectively gives you an additional corner guess, and if it pays off you get at least three more safe cells to work with (some of which might become openings, or lead you to openings shortly after through logic).

Is it an optimal choice?  That just depends on whether there are better-odds guesses available.  Sometimes, the entire frontline is just a high-percentage mess, and a wall-skip (or guess to a corner, or somewhere in space) is actually the best move.  For example:

At left, the lowest frontline percentage is 25%, whereas the background rate is 16%.  So I would do the wall-skip at the top here, every time.  Or a corner, if available.

Whereas, in the first example above, there are actually 3 cells with only an 8% mine chance. See those 4x interior front-line 4s at bottom left?  The second from lowest-left has its remaining mine being “pulled” by the two-mines-in-three-squares 4 from below, so the odds on the three squares to its right are actually really low.  If you use a guess here, then you probably guess the upper one of those, hoping that you get a 3.  In the second example above, looking in the middle of the board, that lower-left 1 dangling in space there looks juicy: for the squares to its left and bottom to be a mine, a lot of other things would have to be true first.  If you really know how to look for probabilities, you can sometimes find better-odds guesses than doing a wall-skip.  But if you find yourself stuck, needing to guess, and no great choice recommends itself to you, then doing the wall-skip guess can prove helpful, and is a good tool to have in your bag.

Rear Attack

There’s a corollary technique to the wall-skip, one that gets employed in endgames, rather than mid-game.  It can solve a very particular and annoying category of situation often encountered in advanced play (and sometimes in not-advanced play as well).  The situation is this:

• You’re down to a few remaining guessing regions in the endgame
• One such region is exactly 2 rows deep of unknown squares (usually up against a wall)
• That region has mines or wall on the front-facing corners, but has few mines in between
• The front has high-ish mine percentages, but the background rate is lower / normal

In this case, it is my practice - and I believe the correct play - to guess behind the front line onto the back line.  Jump over that front line, and see what comes.  Because often, what comes is that with that guess and some projection logic, we can usually clear straight along that back line, marking a few mines, getting past them, and maybe clearing the whole area via logic (or at least making our odds better).

To do this, you skip past the front row and guess a back-row square where two things are true:

1. The cell you’re guessing, if it’s not a mine, will lead immediately to a next step via logic (because you can use projection from the front row)
2. The cell you’re guessing is at worst a background-rate guess (maybe against a wall or a wall of flags, or maybe it’s wrapped around a corner giving you assume-fewer odds)

At right is an example, from this beginner game.  We have 4 mines left, and the front accounts for at least 3, meaning the rear line has at most 1 mine and maybe even zero.  So we are on pretty sound footing to guess onto the rear.  But where do we guess?

If we pick the right-side wall, the 3 can’t help us, because its projection covers three columns, so even if we turn up a 1, we can’t be sure the square to its left is safe.  If we opt for the second-from-right column in order to solve this, then what if we turn up a 2?  We don’t know if the other mine is on the right or the left.  The only choice that satisfies condition #1 is the left-most column.

From there, it turns up a 1, so the cell to its right is safe. It also turns up a 1, so we have two safe squares to its right. This leads to a 1-1 pattern that opens the cell above the right-most 2, which gives us a hole pattern, revealing a back-row 2 in the middle column.  That tells us where the (lone) back-row mine must be, and the game is solved a few moves later.

These opportunities come up pretty often, in fact.  Take this one, from this intermediate game:

Similar situation - no logic, no minecount.  And no obvious probability-based guesses.  So we rear-attack into the corner, where we have the right-column 2 to give us logic from there.  We get a 1, which gives us the cell to its left (a 1), which gives us two safe squares to the left again, and then it leaves two squares for two mines to satisfy the 4.  The rest follows.

Here’s a trickier case (from this game, at 2:35), with five mines remaining.

At the start we have a 2-row endgame here, with no logic moves.  Our choice to “rear attack” here leads us to jump over the 2-square front at right and guess next to that right-interior 2.  Why that one?  Because unless we blast, we will have logic available to us going left from there.  And because we have the right-interior 2, the corner there is a 1v2, and gives us a background rate on it.

We turn up a 2 (an effective-1), which by projection lets us conclude that the square to its left is safe.  If that square turns up a 1, we can open the one to its left, etc.  If it turns up a 2, then the one to its left is a mine, but the interior 2 then would give us another safe square to the left of that.

This proceeds along until we identify there’s a mine under the interior 3.  This clears the 3 and lets us now go back and clear the top line, solving the board.

Much like the general wall-skip approach, Rear Attack is best employed when either you can’t spot a good guess, or you are sure (through a Hint) that there aren’t any.  E.g. here:

You can get 21% odds at lower-right, but the chances of progress from there look slim, to me.  I would be considering two options of a guess into space: Either the 19% at top-left, where getting a 2 or a 4 would lead to logic, or along the wall in the big open area, where in theory I might luck into an opening.  Top-left is a rear attack, of sorts: not quite as guaranteed-progress as the previous examples, but high enough odds of progress to be a good choice here.

“Rear Attack” is like an endgame special-case of “Wall Skip”, and it works for a similar reason: because you’re up against a wall, a guess to the back will be able to use projection from the “front row” and yield a follow-up move, almost guaranteed (if you don’t blast).  You don’t need to know or guess at which of the front-line square group has the mine - at least, not at first.  But if you use this to clear around the back row, eventually you can reach logic that will let you solve the front row, too.  Though as with any guessing tactic, it doesn’t always work - e.g. see at left.

No practice puzzles or examples for this chapter - just keep an eye out for opportunities to use this technique, when you lack better options.

Deep Tactics

Each of the chapters here describes a class of situations you may find yourself in a game, particularly a very hard game.  Like before, each chapter contains some principles, examples, and then a “problem set” to practice on.  The goal of this guide is to train you to spot and then solve these situations, to optimize your chances of winning hard boards.  Even if you’re equipped with perfect logic, these tools will not magically turn unsolvable, guess-dense boards into fully solvable ones.  But they can help around the margins, by reducing the number of guesses you’ll have to make, and raising your win rate from the games you encounter them in.

These last few techniques build on each other.  Particularly, #s 8 and 9 are building blocks for understanding how to spot situations where you can apply #10.  #10 is sort of the master-key of guessing situations, but it won’t feel intuitive unless you’ve first practiced the earlier techniques.

(7) The 4-Square Box

How to handle a framed 2-by-2 grid of unknown squares, which I will call a “4-square box”, is important to some advanced techniques and reasoning.  Some of those situations yield safe squares and their patterns are often called “Box Logic”, while others offer ways to make optimal guesses.  The general category involving these boxes occurs very often in gameplay (and especially often in high-difficulty), and so a proper, deep understanding of it is essential to making good decisions.  We’re speaking here of situations where the following applies:

• There is a 2-by-2 grid of unknown squares (a 4-square box)
• The corners surrounding the 4-box (its “frame”) are either mines, or are also unknown (if 2+ corners are unknown, though, you’re probably facing a different logic situation).
• You have a front which faces up to edges of this box, on at least two sides, and
• For each edge of the box that you can see, the front suggests there is one mine in the two adjoining squares of the box (a “50-50 front”).  If there were zero or two mines in those squares, of course, you’d just clear them or mark them and it’s uninteresting.

Examples just to illustrate what these situations look like:

The relevant “4-square box” is marked in orange.  In each case, there is some “framing”, with mines on two or three corners, and we’ve got a “50-50 front” on a few sides.  When first seeing one, we often don’t yet know how many mines are in the box.  But because of these “framing” mines on the corners, these boxes can often be hard to work with just from logic moves.

Now, a few rules and patterns apply for 4-square boxes.  Firstly, you could be facing a 50-50 “box-and-2” type forced-guess situation, so let’s get those out of the way:

• If you have all four edges of the box known (as 1-mine-in-2-squares, or 50-50 fronts), and all four corners are mines (or wall), then and only then do you have a confirmed 50-50, of the “box-and-2” type.
• If the box is on the board edge, and the other three edges are 50-50 fronts, and the two inland corners are mines, then you have a box-and-2 50-50, edge-variant.
• If the box is in the very corner of the board, with the two available edges known to be 50-50 fronts and the one inland corner is a mine, then you may have the corner variant of the box-and-2 50-50.  But only if you are positive that you have two mines in the box, which usually requires that it be the final space on the board, with a minecount of 2.
• Just because there are two mines in the 4-square box, arranged diagonally (“2-mine-diagonal”), doesn’t mean it’s a 50-50 forced guess; that requires having mines / wall in all four corners, too.  If you have a safe corner, you can almost always[18] solve it.

Where it gets interesting is where we’re not yet sure whether it’s a 50-50.  So here are the useful rules about 4-square boxes, with examples to follow:

1. If you have a 50-50 front on three sides of the box, then it is 50-50 when viewed from the 4th side as well (and furthermore, there are definitely two mines in the box, arranged diagonally).  And you can only determine which diagonal arrangement is present if one currently-unknown corner can be revealed later (i.e. it isn’t a mine).

1. In particular, this means that you can use this fact to make assumptions about mine coverage over on the unknown side.  We’ll show some examples shortly.
2. It’s important to remember that you need the 50-50 front confirmed from three sides before you decide it’s a 50-50 box with 2 diagonal mines.  During gameplay, once you get some corners of a box marked as mines and a 50-50 front on two sides, it can be easy to assume that you’re facing a 50-50 box, cause it kind of starts to look like one.  But it might not be!  There might just be one mine on the shared corner, or even three mines in the box.  You can’t be sure yet, and sometimes assuming so too early will result in avoidable mistakes.

2. If you have a 50-50 front on two opposite sides of the box (and the other two sides are unknown, or mines, or wall), then there are definitely two mines in it, but you can’t yet be sure if they’re arranged diagonally, or are in-line with each other horizontally or vertically (depending on which two sides you have).  This may sound like an obvious point, but I mention it to distinguish from the following, more-useful scenario:
3. If you have a 50-50 front on two adjacent sides of the box, with the other two sides being unknown /  mines / wall, then there could be one, two or three mines in the box, and which count you have remains unknown.  Sometimes further clearing can reveal a third side to you, or the endgame minecount can tell you which count situation you’re at, and then if it’s one or three mines, you can solve it.

1. Because mines are less likely than safe squares, the one-mine possibility is a lot more likely than the others, and the three-mine possibility is quite rare compared to the others.  This derives from Guessing Rule #2 above (Assume Fewer Mines), but is a very common real-world application of that rule.
2. If the box is merely a corner along a larger front, and you have some known adjacent squares around the box, then ask yourself: can this box contain one mine? Two? Three? See if any of those scenarios would lead to a contradiction involving those adjacencies.  Or, can those answers prove the box is a two-mine diagonal - in which case you can use Rule 1B.
3. If you end up having to guess in this box, without benefit of knowing the final minecount, you will want to guess anything other than the corner shared by the known sides.  If you do, then across all scenarios (one, two and three mines in the box) your odds of dying are less than the board’s background rate!

Let’s illustrate these scenarios, particularly those from rules 1A and 3B.

1A: 3 sides confirms the 4th.  Occasionally, using this “three sides confirms the 4th” rule can be used to directly solve some game moves, through a chain of logical reasoning.  Top player Gramana has an explainer to walk through an example of this. These occur often enough to be important.  Here’s a simple, if frustrating, example:

 →

Looking at the left, we have a 50-50 front from three sides: top, left, and bottom.  So even though the right side has some effective-1s with three unknown squares around them, we can conclude that the 4-square box is a “2-mine-diagonal”, that there’s one mine facing the 4th side of the box (in the blue group), and therefore the safe squares and mines at right follow logically.  However, it’s still a forced-guess 50-50 (mines on all 4 framing corners), so we’re screwed… but at least we had a nice moment of logic before then :).

Often when this appears, we aren’t also screwed at the same time.  Here’s a happier example:

 →  →

We have a 4-square box, framed by mines on the top two corners (and the lower-left is safe but we have no information from that side, so it can’t yet be used to solve the box).  The middle picture shows that, from the perspective of the top (orange), the bottom (blue - consider the lefthand 3), and from the left (yellow), it’s a 50-50 front.  If that’s the case from three sides, then by our rule, we know it’s the case from the 4th side as well (teal, dotted line).  So with respect to that upper-right 4, regardless of which diagonal the mines in the box are on, the 2 squares facing that 4 from inside the box (from the teal group) account for 1 mine.  Therefore that 4’s 4th mine has to be below it, and conclusions flow downward from there (the picture at right).  We can then make progress from there, solving the box, too - the upper green safe square will turn up either a 2 or a 3 and determine the box.  Then the box solution will let us solve the box’s lower-left corner 4, mark as appropriate, solve the 4 to its left, and continue on.

Going forward, in our analysis notation, we will use a transparent square with yellow outline on a 4-square box’s interior squares, to indicate that there is a box with a 2-mine diagonal arrangement in it, because such boxes often provide square groups via projection to support other logic, or even probability-based guessing.  Here’s an example, from this game:

The 4-square box at top is a 50-50 front from three sides (top, left, right), and is thus a two-mine diagonal. The 4 on its bottom side (pink frame) therefore gets one and only one mine from that box above. (we also know the box will eventually be solvable, because its lower-right corner isn’t a mine, it’s that other 4).  So, the pink 4’s remaining mine must be in the squares below (blue group).  This satisfies the 4 below it, resulting in two safe squares to its left.

Sometimes by being very clever, we can construct a logic path, or projection, that gets us the 50-50 front from a third side, in order to confirm that two-mine-diagonal arrangement.  The projection might not prove a safe square directly, like it can in Evil NG, but if it proves that a 4-square box is a two-mine diagonal, it can still be useful. Consider this position (from here):

 →  →

From that bottom-edge 1, we have the teal group, which means the interior 2 above it must have its remaining mine in the squares above (blue group).  Combined with 50-50 fronts from the left and top, that means the blue group is part of a 2-mine-diagonal box.  So at right, the box provides the only mine for that interior 1, clearing four safe squares.  The keys here were (A) seeing the interior 1 which offered logic if the box was a 2-mine diagonal, and (B) finding the third 50-50 front, via projection, to prove that it was one.  They get fancier sometimes:

 →  →

Looking at that central interior 4, it has one mine marked.  It can also get one from below (pink group, from bottom 3), and one from above/left (teal group, from that 3).  That leaves its last mine that must be to its right, in the yellow group.  This yellow group provides the third 50-50 front side to the box at the right, along with its top and right sides.  As a two-mine diagonal box, then, its bottom side forces a mine above-left of the bottom 4, and other conclusions follow.

More broadly, there’s other logic about this “3 sides confirms the 4th” rule, which intertwines with minecount.  In the scenario at right, you might see the 4-square box, framed by two mines on the interior corners, and quickly think “hey that’s gotta be a 50-50 2-mine-diagonal box”.  Therefore, the 3 in the middle has its 3rd mine coming from above, therefore we can clear below.  But DOES it mean that?  Our rule 1A above says we’re only confirmed to have that “2-mine-diagonal” layout if we have a 50-50 front from a 3rd side - and we have only two.  It’s not proven.  And in fact, if we don’t know the minecount, there are three possible mine layouts here, and only in two of them do we have a 2-mine-diagonal arrangement.

1B: Only 2 sides, minecount-dependent.  The scenario above at right occurs in the wild fairly often.  Scan it for a second and think how you might proceed.  The first thing that may occur to you is that minecount matters a lot.  If we start from a labeled version, to aid discussion…

Firstly, notice that if square E is a mine, then it satisfies the 3 adjoining the box from the bottom (below D), and so C and D are not mines.  Therefore B is a mine (based on the right-side front), and A is not.  So that is the only 2-mine scenario: if there’s 2 mines, then they’re at B and E, and the others are safe.

On the other hand, if E is *not* a mine, then the A/B/C/D box is a 50-50, and furthermore F is a mine and we have three total mines.  Put another way, if there are three mines, we then know that E is safe and F is a mine.  How do we know the box is a 50-50, though?  Go back to rule 1: the box looks 50-50 from two adjacent sides already.  If E is safe, then the 3 (south of D) sees the box from a 3rd side, and needs one mine: the box would be a 50-50 front from its perspective.  And by rule, if it looks 50-50 from three sides, then it’s 50-50 from the 4th (and we thus have two mines, on one diagonal or the other).  So you can reason this out in either direction.

On the third hand, and more usefully: what if we don’t know the minecount?  What if this is just the start of a big board, and we’re not going to delay solving this until the end?  It makes a certain amount of sense to deal with it now, because one strategic objective in high-difficulty play is to identify your 50-50s as early as possible, and then guess them immediately upon identification, so that you waste a minimum of time on losing efforts.

The area here has either two or three mines, so the odds between those scenarios is equal to the background rate[19].  If that’s (say) 20%, then the odds of a two-mine scenario are 80% and the odds of a three-mine scenario are only 20%.  That’s super helpful!  So, 80% of the time, the mines are on B and E, and the other 20% of the time, there’s a mine on F, and mines are 50-50 between A+D or B+C.  So all you need to do is guess one of A, C or D (which are 90% safe), and avoid B (which is 90% a mine), and logic will proceed based on your results.  If you (say) guess C and it’s safe, then if it shows a 2, then you’re in the two-mine scenario (mark B, clear A and D), and if it shows a 3, then you’re in the three-mine scenario (mark A and D, clear B) - and also, by the way, you just survived a 50-50.

Meanwhile, guessing F would be safe 80% of the time (i.e. in the two-mine scenario).  But in the 20% of the time when you’re wrong, you always die (and also, the box was a 50-50).  Whereas if you had guessed randomly inside that box, you’d have at least survived half of the time.

These are fairly common, i.e. framed 4-square boxes which are either 50-50 forced guesses (two mines), or have a single, definite solution.  They are one type of what some players call a “pseudo-50-50”.  In the example at left, we have a framed box (DEFG), with minecount unclear.  Either the region has just two mines - one in the box (upper right, on F) and the other on either B/C - or the region has three mines, two in the box (which is then a 50-50), and a mine on A.  We can wait for the endgame to tell us which (and did), but even before then, we actually have a probability advantage!  There are, again, two possible arrangements with two mines (F-B, or F-C), and two possible arrangements with three mines (A-D-F and A-E-G).  Obelus’s principle tells us that the two-mine scenarios are ~80% likely at expert density, and the three-mine ones are ~20%... but both two-mine scenarios have F as a mine.  F can only be safe in the three-mine scenarios (20% odds), and then only half of the time.  So it’s only safe 10% of the time, it’s 90% a mine, and thus E/G are only 10% mines - twice as safe as the background rate!  That makes it worth guessing early, usually.

1B: Only two sides, Box has a “tail”.  In this scenario (from here), we have what looks a lot like a box 50-50 on the edge, but is not yet confirmed to be one (a pseudo).  And note: here, to the box’s right, the 3 is an effective-2.

In that 4-square box on the left side, we do have the two framing corners as mines, and we have a 50-50 front from two sides.  But, importantly, not from all three sides.  So, could it not be a 50-50?  Let’s consider whether we could have something other than two mines, arranged diagonally:

One mine: We get a contradiction here, if we play it out.  If this 4-square box only has one mine, it has to be in the upper left corner, satisfying the two fronts.  Then the other three squares are safe, and then to satisfy the 3 on the right front of the box, we need both the squares below it to be mines.  But this would overload the 2 that’s next to it.  And we have no other way to arrange the mines.  So that doesn’t work.

Three mines: If the upper left of the box isn’t a mine, then the two known fronts dictate that there have to be mines in the upper right and lower left.  If we further assume the bottom right is a mine too, then those mines satisfy the 3 on the right front of the box, all on their own.  This then means that the two other squares below that 3 are safe (green), and therefore that the one to its right is a mine, to satisfy that 2.  So that’s a valid possibility.  So this COULD be a 3-mine configuration in the 4-square box!  It’s just unlikely, because that requires more mines.

If we look at the probabilities, we see the box is estimated to nearly be a 50-50.  But the probabilities of those 4 squares add up to more than 200%, in fact, 210%.  There is a small possibility of that 3-mine configuration.  In that situation, and only in that situation, that square towards the right (in the teal box) ends up being a mine, and that’s the one possible layout that gives it its “10%” here.  In all other situations, it’s safe.  You might think it’s a good choice to guess: not only is it a low chance of blowing up, it’s likely to enable further progress (it determines the purple square, by 1-1 or 1-2 logic).  That’s the good news.  The bad news is, if we’re guessing that blue square, then either we’re dead (10% chance), or we’re alive but we now know for sure that the 4-square box at left is a 50-50 (90% chance, but only a 45% contingent chance of surviving it).  

So then, what’s the best play?  Look at those probabilities again, and it’ll tell you.  The upper-left square of the 4-square box is 45% a mine, while the others are 55%.  If you click it, in the 90% scenario where the box has two mines, then 50% of the time you lose, but 10% of the time there will be three mines in the box, and your click will be safe and yield a 4 on the square. You’d then clear below the 3, mark the blue square as a mine, and expect some further logic available.  So with that square, your odds of surviving are a small bit better than 50-50 (here 55%).

In reality, that upper-left square was safe, it showed a 3 (indicating we’d just survived a 50-50 box), and a bit of reasoning later, the blue square was a 2, the purple square was also a 2, and by the Hole pattern, the 3 squares behind it were safe too - we’re off making progress again.  That whole area is solvable as a result of smartly guessing what turned out to be a 50-50.  But if we hadn’t considered the possibility of it not being a 50-50 at all, we might not have known to guess the upper left, and we’d have been incrementally likelier to lose.

3B: “Check Arrangements”: Corner of a longer front, 50-50 from 2 Adjacent Sides, with other Adjacencies: Along with Rule 1B situations, this is a category that will sometimes yield logic moves (100%-safe squares) that enable further progress.  These cases occur when the box is not “cornered” along a wall or row of mines, but itself serves as the corner for a longer front.  It’s 50-50 from two adjacent sides, we have mines on three corners, with the interior one still unknown.  And we have another key element: some number squares next to those adjacent sides, which might limit the possible mine arrangements in the 2x2 box (or make use of them).  The rule tells us to “check the possible arrangements”, i.e. to ask whether the box can have one, two, or three mines in it.

If one of your neighboring adjacencies is an effective-2, the first thing to check is, can both of its needed mines be in the box?  Usually the answer is “No”.  At right, we have a framed box (CDEF), with an effective-2 to its left: the teal-ringed 3.  It also has an effective-1 to its top (in pink).  For the teal 3 to get both its mines from the box, that would mean C and F, so D would be safe, so E would have to also be a mine.  But mines on E and F would overload the pink 3, so that’s impossible.  Therefore teal can only get, at most, one mine from the box.  But because it only has one other adjacency (B), it must also get at least one mine from the box.  Meaning it gets exactly one mine from the box (the C/F front), providing our third 50-50 front and meaning the box is a two-mine diagonal.  As such, the pink 3 definitely gets its third mine from below, and thus G and H are both safe!  And since B is a mine, A is safe too.  So this is a “the box can’t have three mines” type of logic.

This situation crops up a fair bit, i.e., one neighbor to a corner box that’s an effective-2, and another that’s an effective-1, and so proving that the box must be a 2-mine-diagonal.  MS Discord #puzzles frequenter Mythers45 offers this example:

Here, the corner box (at upper left) has an effective-2 to its right (the 4), and an effective-1 below it (the interior 2).  The box having one mine would reach a contradiction to its right (if the 4’s mines are below and below-right, it overflows the 3 next to it), so the 4 needs at least one mine from its left, in the box.  The box having three mines, meanwhile (all but the top-left shared corner), would obviously overflow the interior 2 below it.  Therefore the box has exactly two mines, in a two-mine diagonal, and both neighbor adjacencies take this and yield safe squares.

You may also have to check whether any of the other usual possible arrangements of mines within the box would break something, downstream.

 →

Our effective-3 in the interior there, from a wall-skip guess, gives us an interesting situation.  The would-be 4-square box, ABDE, has an effective-1 to its left (the 3 to the left of A), and of course that 3 on the bottom edge.  Can the box have one mine (on B)?  If A, D and E are all safe, it means mines are on C-F-G, and that works.  It can’t have three mines in the box, though: that requires two sides that are effective-2s, and the 3 to the left is just an effective-1, so mines on A-D-E would overload it.  For two mines, if they’re on B-D, then A-C-E are safe, G is a mine, so F must be the bottom 3’s other mine - that works.  But if they’re on the other diagonal, i.e. mines on A-E, then B, C, D and G all have to be safe, and that leaves the bottom 3 needing two mines from one square (F).  So in this case, B has to be a mine, A & E have to be safe, therefore F and G are both mines as well, and the last one is either on C (67%) or D (33%).

3C: Cornered Box with 2 Adjacent Sides.  In each of the examples below, we have the scenario described in 4-Square Rule 3C above, a 4-square box with a 50-50 front from two adjacent sides, which is up against a wall (or a wall of mines, which functions the same).

Only in the middle one are we actually on a corner.  But in all cases, the same guessing rule applies: either we can wait until we have a final minecount (impractical for very large boards with bigger fish to fry), or if we’re going to guess it now, then the odds of survival are best if we guess either a wide corner (not the corner shared by the known sides), or the rear square.

Why?  Because if you work out the odds of 1-mine, 2-mine and 3-mine scenarios, and the chances of each, that’s the best play.  If you just hit the Hint button, it looks like this (the left example, from this game): the background rate B is 23%, so they forecast the odds of the near (shared) corner at (1-B) = 77%, the two wide corners at B = 23%, and the far corner, an untouched square, at B as well.  But if we were to calculate out the probabilities of 1-, 2- and 3-mine scenarios, and add up our weighted chance of survival in each, we get:

 (Expert density)

 (23% density)

(it’s calculable from the Bernoulli Distribution for n=4).  Notice that your odds of survival are in fact higher than the background rate would suggest, e.g. for a background rate of 23% (= guesses into space are 77% safe), your odds of survival are roughly 79.6%.  This is because we do know something about these four squares: they contain neither zero nor four mines.

The 3-mine scenario is somewhat rare, even at very high background rates (e.g. at 30% density, it’s still only 10%).  So you’ll want to guess as if it’s not the case, unless you’re near an endgame minecount.  If it’s the 1-mine scenario, you win with the wide squares or the rear one.  If it’s the 2-mine scenario, you have a 50-50 anyway, so any guess is as good as another.

Cornered boxes are not quite a Pseudo, but they’re a cousin of pseudos.  The reason Rule 3C holds - the reason you should guess them early - is similar to the logic of pseudos: either there’s one mine in the box, or there’s two and you’d have to guess it as a 50-50 sooner or later. Unlike true Pseudos, there’s the third possibility (three mines) which we can’t entirely discount, and so minecount can sometimes be a help, if the endgame is near.  But if you’re playing big boards, or PvP, or speed, or really for anything other than mastery, the strategic play is likely to guess them as if they’re a pseudo.

(e) 6-Square Box: One variation on this theme, which is uncommon but not rare, looks like this:

Instead of a 4-square box, we have a 6-square box.  But it, too, is framed with mines on the corners, and also mines above/below the middle column, preventing logic from determining placement.  So this is actually a similar mine-counting exercise.  The most common arrangement here is a 2-mine scenario (on the top-left / top-right corners).  Very rarely, this can hold a 4-mine scenario (on the other 4 squares).  But if it’s a 3-mine count, you have a 50-50 between this:

  vs  

What’s the best approach?  What I do is, unless it’s endgame and I know the minecount, I guess the interior middle square (bottom-center here).  It’s analogous to guessing the remote corner in the 4-square box.  If it’s a 2-mine arrangement, it will show me a 2 and I can mark that.  If it’s a 3-mine scenario, 50% of the time I’m already dead, and 50% it’ll show me a 3 and I can mark that (the left-hand 3-mine picture above, or real example at right).  If it’s 4 mines, I’m dead, but that almost never happens.  Because I largely discount that 4-mine scenario, this means the minecount is almost irrelevant to me: if it’s actually 3 mines in the area, it was a 50-50 anyway, and any guess is as good as any other.  So for that reason, I don’t wait until the endgame - I guess that interior square straightaway, and either it’s a trivial 2-mine situation, or I just guessed a 50-50.

← N.B at left: it also depends a bit as to your vantage points into the box.  This formation, seen occasionally, has all 33% cells as dead cells.  Guessing one and being correct just reduces the other four cells to a box-and-2 50-50.  Every cell has a 33% win prob.

(f) “Finned Box”: There’s a rarer type of box formation, also worth mention. Back in Chapter 3, we covered the 5-square, 2-mine box-and-tail, when it’s up against a wall or mines and thus forces a 67% / 33% guess.  There, some squares were dead, and some were optimal guesses with a 67% WP.  But sometimes in a game, there’s a 5-square box that’s “open-ended”: one side with a three-square front is not facing up against a wall or mines.  In those cases, there can be a logic move available (analogous to Rule 1A above, “3 sides confirms the 4th”) that basically yields a square group with one mine, projecting out toward the 4th side of the box.  Archeaic calls this a “Finned Box”, which is as good a term as any (Tail?  Fin?  Whatever).

The move derives from the same principle that 4-square box logic does: with a 2x2 box with 3x 50-50 fronts, there are only two mine configurations, and both put one mine facing the 4th side.  Here there are three mine configurations that exist, and each one might or might not be equally likely, but whichever one is true, there is one and only one mine on the open-ended side.

Take an example from this game (in NG!) by steph7892, shared by Koro:

 →  →

In this isolated edge, there’s two mines in that 5-square “finned box”, to the left of the interior 3.  One around the upper-left 2 (yellow), one from below (blue), and then of course the interior 3 is an effective-1 (purple).  The three possible mine configurations for the box portion are:

(A)   (B)   (C)

The upper-left 2 can have a mine in any of the three spots around it, and for each choice, the rest of the box is determined as well.  But notice that each time, the right column of squares here (the “front side facing three squares”) contains one and only one mine.  From this we get the solution in the top-right image: there is a one-mine square group, in purple, and it provides the third mine for the interior 3, thus clearing the two squares above it (and guaranteeing a solution for the box, too).  This is a pattern worth remembering, by high-difficulty players.

Zonal Math: The functioning of this logic pattern can also be understood in a different way, as a special case of a technique we will discuss much later called Zone Reduction.  Consider the isolated edge at right (from here).  Take a second and look at it, before reading further.  The first thing to notice is that there is a definite “local minecount”: from the top and bottom, there are exactly two mines in the region.  But then, note that the 5 at right needs both of them.  So if we think of it as “subtracting” the 5’s adjacencies from the full area, the two left-edge squares must be safe!

Similarly, In the Finned-Box pattern, we have five squares which have a local minecount: the squares contain exactly two mines, and we know this because we can cover the 5 squares with two one-mine square groups, just like we do with advanced minecounting.  But then, it turns out that from a different direction, we can “take away” two of the squares with an overlapping one-mine square group, and that leaves us the remaining three squares, which must collectively have one and only one mine.  It sounds like random voodoo, but I promise it works!  See here:

We have a box that’s almost a typical 4-square box, at bottom here.  We have a 50-50 front from the left and bottom, but the right side is just a 3-square group.  And, crucially, we have that interior-2 that’s an effective-1.  So firstly, observe that the 5-square box region has exactly two mines, as shown by the blue groups: from the left and right sides, the square groups of their adjacencies compose that blue “zone”.  Now, pivot to thinking about coming up from the bottom, from that 4, which gives the red zone at bottom.  If the blue area collectively has two mines in those five squares, and one mine must be the red group at bottom, then the other mine must be in the dotted-red group above that, in the blue zone’s residual squares!  And in this case, that square group is enough to satisfy the interior 2, and prove two safe squares above it.  In effect, our ability to move square groups around has created, or re-shaped, a new square group from the old ones.

We’ll generalize and explore more complex cases of this “Zone Reduction” concept later, but this pattern is the first, and probably most common, opportunity to use it.

(g) Check Yourself: Lastly, and maybe most importantly, before any other consideration of a 4-square box, make sure your known sides are actually a 50-50 front.  Nothing’s worse than guessing into something where you missed a logic move, just because of a false visual resemblance to a 50-50 pattern.  Don’t jump to conclusions, check first.

At right: “Hmm, this 4-square box looks like a 50-50
or a pseudo, I’ll just guess over here and…” NO.
Stop, take a second, look again.  Count your mines,
be sure of your effective-mine numbers.

Practice Puzzles

Link to solution section.

These puzzles mostly deal with “box logic”, from either Rule 1A (“3 sides confirms the 4th”, including clever projection), Rule 3B (“Check the arrangements”), or Finned Box patterns.  They won’t focus on the related guessing rules involving probabilities around 4-square boxes, or box pseudos.  Sometimes you will need to use an understanding of what the possible arrangements of mines in a box or box-and-tail can be, in order to show something to be certain or impossible.

Also, two additional puzzles involving 4-square boxes can be found on GlennGould’s youtube playlist of minesweeper puzzle videos - in particular, this one (#1) and this one (#7).

Puzzle 7-1 “Tail Help”: Mark 2, clear 2.  Difficulty: 2 / 10.

Puzzle 7-2 “Hollow Chamber”: Mark 2, clear 4.  Difficulty: 3 / 10.

Puzzle 7-3 “Tetris”: Mark 1, clear 3.  Difficulty: 3 / 10.

Puzzle 7-4 “Unboxed”: Mark 0, clear 2.  Difficulty: 4 / 10.

Puzzle 7-5 “Corner Store”: Mark 2, clear 7.  Difficulty: 4 / 10.

Puzzle 7-6 “Intruder”: Mark 1, clear 4.  Difficulty: 5 / 10.

Puzzle 7-7 “Squeeze”: Mark 0, clear 1.  Difficulty: 6 / 10.

Puzzle 7-8 “Platform”: Mark 0, clear 3.  Difficulty: 6 / 10.

Puzzle 7-9 “Glass House”: Mark 2, clear 4.  Difficulty: 6 / 10.

Puzzle 7-10 “Backup”: Mark 2, clear 4.  Difficulty: 6 / 10.

Puzzle 7-11 “Stamp Press”: Mark 0, clear 3.  Difficulty: 7 / 10.

Puzzle 7-12 “Salmon Run”: Mark 3, clear 5.  Difficulty: 7 / 10.

Puzzle 7-13 “Moving Day”: Mark 2, clear 3. Difficulty: 8 / 10.

Puzzle 7-14 “Great White”: Mark 0, clear 2.  Difficulty: 8 / 10.

Puzzle 7-15 “Isthmus”: Mark 4, clear 9.  Difficulty: 9 / 10.

(8) Proof by Contradiction

In high-diff play, there will almost always be some residual areas of a board where you’ve run out of logic - regions which aren’t 50-50s, but are high local density.  Usually this means a guess is coming, even if it’s a good-odds guess.  But rarely, in smaller highly-constrained areas, you can see that there are really only two possible scenarios for where the mines are - there are lots of two-squares-one-mine pairings, and they’re chained, such that if you’ve got pairs of squares (A,B), (C,D), (E,F) etc, then if A is a mine (so B is not), it must mean C is a mine and D is not, and E is a mine but F is not, and so on.  Sometimes this sort of dilemma is unresolvable.  But sometimes, rarely, you can follow the chain of logic around and you’ll come to discover that one of those scenarios, one of those possibilities, is actually impossible.  That following it to its logical conclusion would eventually lead you to contradict the information in one of the known squares.  Therefore, you can conclude, that scenario is wrong, the other one is right, and you can mark mines and clear cells accordingly.  We call this “proof by contradiction” (“PBC”).[20]

When this happens, it usually involves chains of logical reasoning that exceed the number of moves-ahead that the WoM solver (i.e. Hints) can handle.  So if you’re facing one and you use a Hint, it will tell you “no solution found, calculating probabilities…” but then show you that a few squares are 0s (safe) and few are 100s (definite mines).  You’d expect those could have been deduced from the normal pattern inference process in minesweeper, but they can’t!  These solutions can only be determined by following a long logic chain and discovering a contradiction embedded in it, sometimes as many as 10+ moves deep.  And if you want to save a hint, or go without hints, it helps to learn to spot the situations where this might apply, and reason it out.

For a simple example (from here), consider the top-middle cell, to the right of the 2.  What happens if that one is a mine?  Well, let’s mark it and imagine how things play out downstream.  If it’s a mine (red, with teal frame to show that it’s the assumption), then the two squares below it are safe (green), so the lower-left shared cell between the 3s is a mine (red), so the other two cells from that lower-left 3 are safe (green), so the mine at bottom would be above the bottom 3 (red).

But then a funny thing happens: per the blue-framed 3, we then need to clear the cell above it, as that 3 has all of its needed mines.  But the cell to its left, in orange frame, still needs one more mine - and we don’t have any more squares where it could be coming from!  Therefore, we can conclude that this is actually an impossibility, and that the first mine we marked, our assumption, must be wrong - it cannot be a mine.  And indeed, the hint shows us that for certain:

This is a trivial example, partly because that safe square gives us no additional information, and you end up having to guess anyway.  But it’s the process that I want to illustrate: make an assumption, and see what comes of it, until you either run out of downstream conclusions that your assumption would determine, OR, ideally, you come to a contradiction like the above: your assumption cannot be correct, and therefore it’s the other way.

The best way I’ve found to actually attempt to DO this proof-by-contradiction, is by marking “provisional mines” on the board itself.  In other words, I try to bear in mind where the definite mines are that I’ve already marked (it’s usually pretty obvious), and then I choose one scenario to test, and go about placing mine flags which, in that scenario, would necessarily be mines, and go on until I have no more conclusions I could draw from my scenario, and see if I’ve reached any sort of contradiction.  But because I don’t want to actually guess the safe squares yet (I’m just testing it out!), the only way I can really do this and keep track of my provisional logic is to mark the provisional mines.  And then I get to the end and that seems like it’s a safe scenario, so I unmark the provisional mines, and try again under the other competing scenario, and see if that leads me anywhere.

Use the Editor: Instead of provisionally marking on the board itself, though, many high-difficulty players on WoM will opt for using the “Open in the Editor” button, which takes a picture of the current game situation, and opens it in an MSPaint-like in-browser graphics editor.  This then lets you color-code different cells to do the provisional marking in a safe space, and also to quickly undo one scenario (via the undo button, or Ctrl-Z) so you can try the other one.  For highly complex situations, this is probably better than marking on-board, not least because you can open multiple editor windows and compare scenarios side-by-side.  And also because you’re less likely to make a mistake and leave a provisionally-marked mine flag in, or accidentally open a provisionally-safe square.  But for simpler situations, on-board is faster.

Let’s explore a few categories of proof-by-contradiction opportunities:

(a) Either-Or: To seek out a proof-by-contradiction, we generally need some sort of binary choice, picking a “key” spot and coming up with two possibilities, which are exhaustive between them (i.e. there are no other possibilities besides those two), which each create downstream conclusions.  The most typical way we find a good spot for this reasoning is when we can cleanly divide the needed mines for a square, and say “either this has one mine to one side and the other one below, or both to that side and not the one below”, something like that.  Here’s a simple example (lower left in this game) with an unusual-looking hole pattern formation:

 →

Because the BCD group has two mines (which is unusual, but not rare), and because we have that upper interior 1 (in purple frame), it suggests we check two possibilities: either the BCD group has two mines above (on C-D, with B safe), or it doesn’t (one mine on C/D; B is a mine).  Why would we check this?  Well, we also see that CDEF is a 4-square box, with two 50-50 fronts known, from the top and right.  So that interior 1 suggests there might be logic available.

If both C and D are mines, then E is safe and F is a mine too.  But this would overload that interior 1: both C and F would be mines, and it can take only one.  Therefore, by contradiction, that cannot be true, and the other possibility must be true: C and D can have only one mine between them, and therefore B is a mine, and A is safe.  However, taking it farther, since we now know that C-D is a 50-50 front, the CDEF box is a two-mine diagonal (mines on either C-E or D-F), and that interior 1 is always getting one mine from its C-F front.  Therefore the 1 is satisfied from that direction, and its other adjacencies (in that blue box) are all safe.

Splitting scenarios for an effective-2 with three adjacencies is a common way to try this.  Consider the below (from here), where we can see that the interesting part is the right-hand portion:

 →

The upper-left 4 is an effective-1, while the lower-right 4 is an effective-2 (as is the bottom 4).  As a “neighbor mismatch” type scenario (from Evil NG), we should create two options to test that upper-left 4.  Either its remaining mine is to its right (A/D, making ABDE a 2-mine-diagonal box, because it would provide its third 50-50 front), or it’s not, in which case the mine is below, on C.  Now we ask: is the latter possible?  If C is a mine, it means that A and D are both safe.  If A is safe, then B is a mine (per the top 4), which means that E is safe.  If E is safe, and D is previously known safe, then the lower-right 4 needs two mines, but only has one remaining square (G), so therefore this is impossible.  Therefore C cannot be a mine, and has to be safe (so F is a mine).  By similar reasoning, D and E cannot both be mines (it leads to A and D both being mines, as B must be safe, and that would overload the upper-left 4).  Therefore ABDE is a 2-mine-diagonal box, and so G is a mine too.  This leads to a fuller solution (below)... and the unfortunate fact that we’re now screwed, as we have two box-and-2 50-50s to guess.  Sigh.

The point here, though, is the thought process: we see a spot on the board where we can say “one of two things must be true”, and, importantly, can see that assuming either of those two things would then each force downstream moves.  If there’s a cascade of conclusions from a potential assumption, then it’s maybe worth trying out to see if a contradiction follows.  If found to be impossible, then the other side of the “one of two things” is the correct one.

(b) Split-Front 2s: Another common pathway for “proof-by-contradiction” reasoning is when you’re looking at some effective-2s, among a bunch of effective-1s, which have their squares split between different parts of the front.  Here, the type of question you want to ask yourself is, “Can both remaining mines be on that side?  Or can only one be over there?”  Maybe you have two pairs of two squares on either side of an effective-2, and so the question is “on this side, can we have two mines?  One mine?  Zero mines?”.  So you test each possibility, run through the consequences and see if it leads to a contradiction.  This sort of logic is commonly required in Evil NG games, e.g.:

For an experienced player, the 2 in space will immediately jump out at them, especially since they can connect it with the 1 below.  The question then becomes, OK, there’s shared space between an effective-2 and an effective-1 (the yellow group).  We know it can’t contain two mines, because of the 1 below.  Can it contain 0 mines?  If so, then the 2 needs both of its mines above the yellow area, which would overload the 1 that’s above that 2.  So therefore, it can’t have two mines and thus must have exactly one mine.  Based on that assumption we can clear below to the right of that lower 1, and above-right of the upper 1, from logic.  This is the “Neighbor Mismatch” pattern from before.

But that’s Evil NG, which at most asks you to consider together four or five squares in order to show something is safe or a mine.  In standard-mode high-diff situations, we can end up with solutions that the Hints Engine can’t directly find with pattern logic - it can only point them out after running all the permutations and giving you probabilities.  For example (from this game, at bottom), not even Evil NG will come up with something like this:

The way this region is mirrored makes us worry a bit about a 50-50.  If both bottom corners are mines, we’d have a box-and-2!  But let’s think about this, because the 4s here are effective-2s.  Each one has two squares to the inside, to the middle box, and a square shared with the cell below.  Can the 4s provide just one mine to the inside?  Yes, clearly - that would make the remaining mines be below them, and then create a 2-mine diagonal on the inside, but that works fine.  OK, but can the 4s provide both their remaining mines to the inside?  Think about what happens if we try: then the 2s on top clear the other top square.  But we also have a safe square below the 4, leading to a mine along the bottom row, leading to that interior 2 on that bottom row being satisfied, meaning both squares on the other side of that 4-square box would have to be safe.  But that can’t be the case, because the 4 on the other side needs to put its mines somewhere!  Therefore, the 4s can’t provide two mines to the inside, they have to provide only one mine to the inside and have one mine below it.  This leads us to this solution, from which the rest of the area is solved immediately (and sparing us a 50-50!).

One more example, involving some projection logic, from this 100k attempt (left side, at 45:00).  A lucky guess (the interior-2) has given us a hole pattern, deep into this high-density region.  The first thing we should see here is that the cells are in a nice little row there with our hole pattern, starting from the 5 on the right - and that 5 is an effective-2.  So we have two possibilities: either both the 5’s mines are to its left, or only one mine is left and the other is below it.

If we assume both the 5’s mines are to the left, then the 3 immediately to its left is fully satisfied by a wall of three mines to its right.  This would imply that all of three squares in the yellow group are safe, however.  If that were true, then the middle 3 here, which also touches the purple group, would have only two squares (in purple) and need three mines.  So that can’t be the case!  That middle 3 needs at least one mine from the yellow group.  And so if we go back and assume conversely that the 5 has a mine below it, and thus only provides one mine to the left, then the yellow group has that one mine, and both squares in the purple group are mines.  Since our two scenarios are exhaustive, the latter one has to be correct.  See right, and note the 100%s:

Although we didn’t get any safe squares out of this reasoning, we did mark three mines, which makes guessing easier, so that’s something.  But the principle - the idea of seeing a situation where there can be only two possibilities, and then reasoning through them to look for patterns or contradictions - that’s the point, the main technique we’re focused on here.

(c) Minecount PBCs: Another type of contradiction you can reach is one from minecount.  A potential solution might be possible, based on the board layout and number squares, but in some cases in order to be true, it would necessarily require more (or fewer) mines than we have left to place.  Our usual mine-counting approach, with square groups and N-v-N+1s, may not be able to reach that conclusion directly.  But by making an assumption and playing it out, we can determine that the assumption would necessitate breaking our minecount.

For a simple example, consider this endgame (from Cryogen), with four mines remaining:

 (A):  (B):

Anyone who’s read Chapter 2 would see that the perimeter front here is an illusory corner, where a mine on the corner (Layout B) actually requires more mines from the perimeter (here, three) than assuming it safe (Layout A: only two), and might guess that directly. But notice the interior 4 here: in A, both those perimeter mines touch the 4, so it requires only one more mine from the interior squares.  But in B, only the corner mine touches the 4, so it needs both interior squares to be mines.  So the front options here really are “three mines or five mines, but never four”.  And wait, we only have four mines remaining here!  Therefore, by minecount PBC, the five-mine scenario cannot be the case, and the three-mine scenario must be true.[21]

While that’s an almost-trivial example, the approach is useful in much more complex situations.  In the example at right, our rightmost column (plus bottom edge) of the front looks like a two-scenario option: the front line is fully connected and there’s only two ways mines could be placed along that right edge.  Let’s test out how those two scenarios look.

If we assume that the bottom-right corner is safe (teal frame), then it marks two mines, clears one more square, and otherwise doesn’t determine too much.  But look what happens if we assume that square is a mine, in the second layout: it forces two mines above the lower-left 3, it satisfies the upper interior 2, cascading to other mine placements at top and left.  And then we turn around and need a 4th mine for the interior 4, and there’s only one place left for it to go (the purple square)... but we’ve placed six mines already, and we only have a minecount of six.  We can’t place a 7th!  So therefore this scenario is impossible, via minecount.  Minecount alone can’t find a solution (there’s no exhaustive square grouping), but we can use it to prove-by-contradiction that the left-hand layout is the only one that’s possible.

(d) Identification: We’ve covered a few classes of situation where a proof-by-contradiction approach yields a definite solution.  But how do you identify places to try it, when playing a real game?  The closest description I can give is “when it looks like it’s very close to a known pattern, but is just shy of the necessary certainty to use it”.  I.e., if you see a layout that looks like it would be one of the high-complexity patterns that experienced players know, except it is (usually) one square shy: a square group we need has a third square, there’s an extra adjacency for a number square, that kind of thing.  So for example:

1. With our “Either-Or” technique, notice in the example that the situation at upper-right is very close to one where you could use 4-square-box logic, except for the presence of the extra cell from the bottom (G), preventing D-E being a 50-50 front.  If G were safe, we’d have that logic pattern to use.  So this leads to asking “well let’s assume G is a mine, does that lead to a contradiction?”
2. With our “Split-Front 2s” technique, this is an extension of the neighbor mismatch from Evil NG: You’ll see something that looks close to a spot you could use it, but you have an extra cell that adds a little ambiguity.  If that cell were safe, it would be a straightforward logic move.  Well then, what happens if we assume it’s not safe?  Sometimes you can find a contradiction that way.
3. With Minecount PBCs, you’d only try it if you were in an endgame where it looked possibly mine-countable.  So you’re trying to get your square groups exhaustive, and you can’t find a full minecount-based solution… but you notice that you run out of mines readily under some circumstances.  So try to reach a minecount PBC!  Make some assumption that forces more mines into the front region, and see if minecount can prove it impossible.
4. Other high-complexity patterns can end up being “very close, but just shy” of the situation, and so recognition of a PBC opportunity depends on recognizing “hey, this is really close to <pattern>”.  For example, the Hole 2-3 pattern can very-nearly appear, except you have an extra cell that makes it a split-front.  If that extra cell were safe, you’d have a definite move, so what happens if you assume it was a mine?

For an example of (D), consider this formation (from here):

What does this remind us of?  The Hole 2-4 pattern from Evil NG.  We can get a one-mine group to the interior 4’s left, but the fact that the middle 4 is an effective-2 means we can’t apply the usual rule for that pattern.  But are we sure we can’t?  It looks pretty close!  If the square shared by the two 4s were a mine (at right, in teal), then there would be exactly one mine to the right of that middle 4 (yellow group), so one mine in the blue group, and therefore by the Hole 2-4 pattern, that 4 implies two mines below it, and thus a safe square above-right (by orange arrows).  So we turn to PBC and ask, well, can the teal square be safe?  That would mean yellow contains two mines (on both its squares), satisfying the pink group, but also implying the blue group has zero mines in it.  This under-flows the interior 4, and is therefore impossible.  So by contradiction, we can show that the teal square must be a mine, and the Hole 2-4 pattern applies here.

In this case, the visual resemblance to the pattern we know from Evil NG leads us to consider “well, could we maybe use that?  What’s stopping us?”, and then trying out that scenario.  The same trigger, that pattern-matching instinct, is useful to help identify most situations where PBC could work.  So you should think of PBC as an extension onto the patterns you already know, a higher-complexity version of them.  

PBC doesn’t always work, of course.  Sometimes, there are two possible scenarios, both of which work to explain all information we’ve got from open squares, and we don’t reach a useful conclusion that one is impossible (and thus the other must be true).  So you end up looking for a favorable guess along the front, or guessing into space somewhere promising.  Sometimes, testing these options only serves to confirm that some big extended areas are, in fact, merely exotic 50-50s.  But at least then, you can be assured there was never any alternative to flipping a coin.  The hope is that for some percentage of these situations, you’re able to discover a safe move, when you might otherwise have guessed at your peril, or at least used a hint.

Practice Puzzles

Link to solution section.

The thing to emphasize with these puzzles is that they are not “stare at it until the solution comes to you” kind of things.  They are “put it in an editor and play around with it, start by assuming a scenario, and see what comes” puzzles.  It’s very hard to discover solutions here, especially the further you go in this problem set, if you’re not “just trying stuff” - that’s how you acquire a familiarity for the relationships and contingencies that make the puzzle tick.

Puzzle 8-1 “Last Chance”: Mark 2, Clear 2.  Difficulty: 3 / 10.

Puzzle 8-2 “Into the Heart”: Mark 3, Clear 4.  Difficulty: 3 / 10.

Puzzle 8-3 “Assembly Line”: Mark 3, clear 3.  Difficulty: 4 / 10.

Puzzle 8-4 “3’s A Crowd”: Mark 1, clear 2.  Difficulty: 4 / 10.

Puzzle 8-5 “Vortex”: Mark 4, clear 4.  Difficulty: 5 / 10.

Puzzle 8-6 “Shadows”: Mark 1, clear 3.  Difficulty: 5 / 10.

Puzzle 8-7 “Uninvited”: Solve the whole region.  Difficulty: 5 / 10.

Puzzle 8-8 “Two-Step”: Mark 3, clear 4.  Difficulty: 6 / 10.

Puzzle 8-9 “Bubbles”: Mark 3, clear 5.  Difficulty: 6 / 10.

Puzzle 8-10 “Wiggle Room”: 7 mines remaining.  Mark 1, clear 3.  Difficulty: 7 / 10.

Puzzle 8-11 “Swim Lanes”: Mark 2, clear 3.  Difficulty: 7 / 10.

Puzzle 8-12 “Bon Voyage”: Mark 2, clear 3.  Difficulty: 7 / 10.

Puzzle 8-13 “Hat Rack”: Mark 6, clear 7.  Difficulty: 7 / 10.

Puzzle 8-14 “Lagoon”: Mark 5, clear 5.  Difficulty: 8 / 10.

Puzzle 8-15 “Bookends”: Mark 2, clear 9 (clearing 8 is acceptable).  Difficulty: 9 / 10.

(9) Hypothesis Testing

Last chapter, the “proof by contradiction” technique we used involved imagining a few scenarios, and then following each around as its assumptions locked into place certain mines and safe squares.  That technique gave definitive results only when it could prove an entire scenario impossible.  Other times, when faced with a situation with no logical moves remaining, you’ll have to just guess, and the only question is, which guesses are better than others - probability estimation, which we’ll cover next.  But there’s another class of situation where, rarely, if you were to play out some possibilities, test some hypotheses, you will slowly discover that certain squares will always be mines or never be mines, even if there is no straightforward logical path to prove that, or to disprove some alternative.  You can only show it, by playing out the implications of a few scenarios, and then seeing if there are useful trends in the results.  I call this “hypothesis testing”, but we can also think of it as “guess and check”.

The difference from the previous technique, Proof by Contradiction, is that one was trying to demonstrate that a certain scenario or choice cannot be a mine.  This one, by contrast, isn’t trying to prove that, but rather trying to show that under all scenarios, certain squares are not mines, in fact are never (or always) mines.  We could call it “proof-by-exhaustion”, but either way it’s a different (and tougher) thing to do.

The basic idea of hypothesis testing is, you use your imagination to start testing out just a few scenarios, comparing the downstream implications of those scenarios, and hoping to notice a trend - one that can lead you to deducing that certain squares are mines and (especially) that some are safe, because no scenario you’ve run through has involved them being a mine (or failed to involve them as a mine).  The full technique goes as follows:

1. Pick a square whose status, based on looking at its surroundings, will tend to determine a lot of other squares’ status as well, in a sort of chain reaction (similar to the proof-by-contradiction above).  We’ll call this your “pivot square”.
2. Assume that square is a mine, mark it as such, then reason out every other calculation of mine / not-mine that must necessarily derive from that assumption, marking the mines.

1. You may hit an impossibility, like being forced to over-mark or under-mark a square.  If so, congrats, you’re in proof-by-contradiction land - see above.
2. You may be at endgame, with minecount.  If so, pay attention to that minecount - sometimes, playing out a scenario may require placing more mines than you have.  If so, congrats, you’ve disproven this scenario!  But you’re not done, so continue with the process, noting that this branch of the logic tree requires impossible assumptions.  More on that in a second.
3. More likely, neither of the above apply - you just reach the end of what logic dictates about this scenario.  That’s fine for our purposes here.

3. Once you’ve fully played out that assumption, take a screenshot.  If it’s really an elaborate chain of reasoning, or you really care about this, perhaps edit the screenshot to remind yourself which mines were provisionally (hypothetically) marked, and which squares were safe based on the assumption.
4. Go back, unmark your provisional mines, and then make the opposite assumption - that the pivot square is not a mine.  Play out the logical possibilities from there, marking the mines that would need to result from that assumption.  Take a screenshot, perhaps indicating the (hypothesized) mines and safe-squares under this scenario.
5. Compare the two screenshots.  Are there any mines that are marked under both scenarios?  Are there any safe squares under the assumptions of both scenarios?  If yes, great!  Mark those mines, clear those squares, and then see if the new information gives you enough to resume making progress on the board.
6. If there are no overlaps in the consequences of that scenario testing, you have two options to continue the process if you think it might bear fruit:

1. Try a different pivot square, one which might cause a greater or more-useful combination of chain reactions that determine mines / safe squares downstream.
2. Make further assumptions (additional pivot squares), under a logic tree, where you assume a combination of choices for multiple pivots and then see what results from it, then repeat for each combination of assumptions on those pivot squares.  This can get laborious, but if it’s maybe the difference between completing a really high-diff board and not, you might do it.  And then compare the results of those scenarios to see if there are universal trends.

Simpler Example: In this endgame, we’ve got five mines left, and while it takes a minute of thought, this leads us to a fairly straightforward minecount solution (at right).

But what if it were a minecount of six?  We notice in that five-mine solution, that it is forced to pick a specific square for the bottom-facing 5’s remaining mine.  That is the only choice if we want to solve this using five mines.  But if we choose either of the other squares for the bottom 5’s mine, we end up with a six-mine layout.  Two layouts, in fact:

The first (A, below at left) assumes a mine in the middle square above the 5 - a teal frame will be our indication for “we assume a mine here” - and this leads to two other mines, two safe squares, and a two-square group with one mine (in blue).  The second layout (B, below at right) assumes the mine in the lower-left square, which leads to three other definitively-marked mines, since we need to satisfy the upper 3. (this would also lead to a definite solution for the upper part of the region, but let’s leave that aside for now - we know there’s two mines there).

(A)   (B)

So if we have six mines left in this area, these are our two possible layouts - they’re fully determined by the choice of mine for the lower 5, and the third choice, a mine on the lower-right square, leads to the five-mine solution that we don’t want.  What do we see when we look at these layouts together?  They are opposite in some respects, but in two squares they are the same: the right side of the front.  There is a mine facing the 5, and a safe square facing the 4, in both cases.  So those overlaps of the two layouts are our hypothesis-testing solution!  If we were to clear the cell left of that 4, it might turn up a 4 in both cases.  But there’s a chance it turns up a 5, in which case we know we’re in Layout A, and furthermore, the mine for the blue group is on the lower (central) square - which would solve the region without a 50-50 guess.

What made us choose the lower 5 as our pivot square to test around?  A few things: It touches several other parts of the front, it’s an effective-1 (the upper 3 would’ve been tough to work with) and it offers an ability to exhaust all possible configurations by testing it.  Any other choice (including that upper 3) would’ve required a further assumption, a second pivot square, in order to detail out the options.  The main thing I focused on was the extent to which that 5 determines the mines on both its sides.  If a square has that property, of not just touching but determining parts of the front in multiple directions, then those downstream squares really are dependent, and it’s very useful to posit independence for that square and work with it.

Generally, most useful results (guaranteed safe squares) will reveal themselves with just using one pivot square, and I’ve seen a small, two-pivot-square logic tree successfully find results once or twice only.  It’s highly unlikely that going any further than that will yield fruit.  However… it’s important to note that this sort of extended hypothesis-testing process with a logic tree is exactly how the probability calculator works, when you do a Hint against a board containing no further logical solution.  The hint algorithm tries to compute all of the possibilities for all the different permutations of where the mines could be placed along your front, recognizing that some assumptions force other conclusions.  Some squares will have mines on a higher percentage of those scenarios than other squares will, which is how it arrives at the percentages that it displays to you.  It tries to brute-force its way through every conceivable permutation, and then just shows which cells are mines in more of the permutations than others.  A human brain won’t be able to run through all of those scenarios, but a computer can.  If you’ve used the Hint function to show probabilities a lot, there may have been some times where it didn’t show a logically-provable next move, but nevertheless showed a 0% square that can be cleared.  That’s a result you would’ve gotten to with (and maybe only with) hypothesis testing!

Medium Example: In this game by Bushtimy, at 3:45, he’s down to this, with five mines left:

Try as we might, there is no logic or minecount available to us here.  Trying proof-by-contradiction doesn’t solve it either, but it does show us that for the perimeter (the top/left front), there are exactly two layouts for the mines.  So let’s see what those layouts look like when assuming one or the other.

At left (call it Layout A), we assume that top middle square is a mine (teal frame).  This implies mines and safe squares to the right and down the left column.  But what we should then note is, our assumption has put two mines around that interior 3, so it is satisfied.  This clears the squares below it (bigger orange arrow).  So the bottom-right square must be a mine, and by minecount, the bottom middle square must be, too.

How about the opposite assumption?  If we assume that top middle square is safe, instead (Layout B, at left), then we get the exact opposite square status on the perimeter.  That interior 3 isn’t yet satisfied.  However, because we only have five total mines, we’re down to one remaining, for the squares in that yellow box.  This means that one remaining mine must be shared by the 3 and the 2, and the other three squares must be safe.

We don’t yet know which of those two layouts is correct; they are both equally plausible.  But they’re also exhaustive: one of them has to be the case.  And if you stare at the two layout images taken together, is there anything that has to be true under both scenarios?  In fact, there is!  The middle square of the left area, directly below the 3, is safe under both layouts.  For that square to be a mine, you’d have needed six mines remaining, in which case it would look a lot like Layout B except the yellow box would have two mines (top-left and bottom-right) instead of one.  We don’t have six mines, so that one square must be safe.  And as we can see from the two layouts, in Layout A that cell would turn up a 4, and in Layout B that cell would turn up a 3, so one way or another that square will solve the region for us and win the game.

Was this just a Minecount Proof-by-Contradiction, though?  Well, it could have been, if we had known upfront to test on that specific cell.  But there was nothing really calling to us about that middle cell.  Once our attention is focused on it, we can quickly verify that it must be safe, because assuming it’s a mine leads to a contradiction based on minecount.  But that’s very different than being able to spot it in the first place.  The thing we could easily spot was that the top-and-left perimeter had only two possible layouts, so we might as well try them and see what happens.  And then we overlaid the two of them, and saw that they shared a (single) conclusion.

Fancier Example: consider the following absolute classic Beginner game, from WoM server admin FracturedAnvil.  The odds of this happening in Beginner mode have to be infinitesimal, and yet, here we are.

If you play around with this a bit, you’ll see that the situation is “precariously balanced” - pretty much any one thing being confirmed as a mine would determine most of the rest of the board.  And yet, no logic moves are available to us, and there is no proof-by-contradiction either.  So to start out with the hypothesis testing, we first pick a pivot square.  Most of them would be good choices, but I think the board is more cleanly determined by making a choice on ones at the top.  Then we play out the scenario, showing what must result from this assumption.

Making that first assumption determines all but two of the mines on the board, and opens up 8 safe squares (in green) via straightforward logic.  What if we then make the opposite choice?

Note that the assumption on the pivot square here (marked with a *) forces a mine to the square below-right of it, which forces B to be a safe square.  Then, because there must be one mine across the two A squares (per the #1 right below the * square), the cell to the left of B and below the As must be a mine (in yellow box).  Therefore the square below it is safe (marked with a C).

So compare our two scenario results.  Is there an overlap?  Yes!  Specifically, the yellow-box mine and the square marked C in the second scenario are the same in the first scenario.  So under all possible scenarios, those two things are true.  So we’re able to mark and clear them (at right), and then other conclusions follow from logic alone, and on we go to glory.

An unusual situation?  Yes, undoubtedly.  And especially in a Beginner game.  I’ve seen thousands of 50-50s in Beginner games (which are themselves pretty unlikely, roughly 3.5% of games), but I’ve never seen something like this!  But at higher difficulties, these situations do pop up, rarely - particularly in endgame scenarios in big boards, where the only regions remaining once logic is exhausted tend to be high-density areas with lots of mines on the perimeter.  Those where logic is impossible, and you’re forced to assess the probabilities and make a guess (sometimes with help from a Hint).  But in some fraction of those situations, hypothesis testing can instead reveal a next-step that either solves the whole area, or gives you an ability to make some progress, enough to go on that your eventual number of guesses becomes fewer, and so your odds of winning go up.

Identification: How do you find situations where this technique works?  Or where it’s even likely enough to be worth trying?  They’re impossible to conclusively identify, but some trends include:

1. Even though it’s a high-density front, a lot of the squares share some overlap such that they’re not independent.  i.e. They’re not “isolated”: There aren’t a bunch of two- or three-square bits that are segregated from each other by intervening mines, such that they don’t interact with each other or determine each other.  If it looks like one mine assumption would be enough to start drawing multiple downstream conclusions, it might be a good candidate.
2. If you look like you have two plausible scenarios based on one (or even two) pivot squares, and you’re going to check it for any proof-by-contradiction possibilities.  Trying and failing a proof-by-contradiction is a great way to do the legwork for hypothesis testing!  They involve the same steps, just a different type of result at the end (and a need to keep track of your intermediate conclusions, rather than throw them out if they haven’t borne fruit).
3. You have “depth” to your knowledge of the region, i.e. it’s not just a flat or two-edge front, you have some squares on the interior that are known, or better still, several sides of a rectangular area.  Much like the kind of situation that results in employing logic in Evil NG, where you need to be able to reason about “these two squares contain one mine, therefore those three have a one mine, therefore (etc)”.  If you know some squares on the inside, behind the front, like in the example above (where we have some open squares in the left-most two columns, in rows 5, 6 and 7), it might be a good candidate.  In most of our example puzzles here, your front faces multiple sides of the region.
4. If there is some chain-of-conclusions, based on some pivot square, which would then proceed in multiple directions.  A key factor in my choice of pivot square above is that choosing it immediately creates downstream conclusions to its bottom-right, to its bottom-left, and also above (trivially).  This is common for the scenarios where we can get something out of hypothesis testing: In order to get lucky enough that some scenarios have overlapping implications, we need those scenarios to “send their conclusions out” in multiple directions, so that they can affect squares 3-4+ moves away from that pivot.  Only if things go that far, and then have some conclusions determined by those “waves of information” coming back together, is there a realistic chance that they reach a harmony like this.

Choosing a Pivot: OK, so once you’ve found a situation that looks plausible for hypothesis testing, how do you choose your pivot square?  The heuristic description above said, “one that determines a lot of downstream conclusions”. That takes practice to see, but is a good filter.  If you’re looking back and forth across your front, and one number square’s status seems to be the linchpin of the whole thing, then try that.  But, importantly, while it’s nice if we can find a useful binary (2-scenario) pivot, don’t be afraid of a ternary (3-scenario) pivot!  Sometimes comprehensively covering options requires testing three of them.  Making use of a secondary pivot square is rarely needed or helpful, but a single pivot with three options can get you there fairly often.  Particularly if you’ve already tried a couple binary pivots and gotten nowhere.

Another pivot-choosing technique is what I’ll call “near-minecount”, as proposed by top efficiency expert Llama.  Once you’ve chosen a situation to try this in, one of your first steps might be to try and find a square grouping for a minecount, because the conditions to try hypothesis testing overlap a bunch with those where minecount might help.  If you’re unable to find a grouping that fully determines your minecount, but you can get very close (within one, or maybe within two), while using all but a handful of squares (within-1 leaving 2-3, or within-2 leaving 3-4), then the residual squares of your near-minecount are good spots to try a hypothesis test.  Take the hypothesis-testing example from the WoM Patterns guide:

​​There’s no full minecount that solves, but you can get a near-minecount (at right; credit Llama).  The squares in his groups account for four of the five needed mines, leaving one mine across two residual squares.  Those two leftover squares are the natural spots to try your hypothesis test.  You’d want to test them both, to see if they could both be mines or both be safe, or if only one can be.  In this example, the contradiction is reached via minecount - an assumption of the top-left square being a mine ends up requiring six mines from that scenario, when we have only five.

Nor is that the only grouping that helps: at right is another grouping, by top player Koro, which has four mines in the red groups, meaning the residual blue squares must contain one mine among them.  This simplifies the set of possible choices to test around: you should pick one of the residual squares as your pivot, assume it’s a mine, and see how it goes.  The same applies when trying to reach a minecount contradiction, any other proof-by-contradiction, or an overlay-style hypothesis test as here.

For Proof-by-Contradiction puzzles in the last chapter, we could make our assumptions in one spot, play out the scenario, and then usually discover that something else was impossible (perhaps the entire scenario is impossible).  In the situations for which Hypothesis Testing applies, if you were to know in advance which cells were safe, you could disprove the opposite: assume the specific safe cell is a mine, and then play it out and usually there’s a way to show that it leads to a contradiction.  But that is just validating the answer; it doesn’t really help you discover it, unless you’re going to do that for every cell on your frontline, which isn’t practical.  Instead, the Hypothesis Testing process here is a fairly elegant way to test dozens of cells at once, by looking for the overlap of conclusions in a Mutually-Exclusive-Collectively-Exhaustive (MECE) set of scenarios.  It’s a little labor-intensive, but it’s still somewhat practical to do when you’re facing a situation you think should or at least could have some available moves, except that they’re invisible to standard MS logic.

Practice Puzzles

Link to solution section.

As with the previous chapter’s puzzles, the key here is to take the image, put it in an editor, and start playing around with it to see what pops out.  There are better and worse choices of where to explore, as we’ve discussed - but basically nobody can solve the more-advanced situations here just by staring at them until a Eureka moment arrives.  Instead, you’ll learn by doing.

Puzzle 9-1 “Boxed Out”: Mark 2, clear 2.  Difficulty: 4 / 10.  N.B. this puzzle can also be solved with other techniques; please try to use the ones from this chapter instead.

Puzzle 9-2 “Single Step”: Clear 3 cells.  Difficulty: 5 / 10.

Puzzle 9-3 “Gradient”: Clear 2 cells.  Difficulty: 5 / 10.

Puzzle 9-4 “Keyhole”: Clear 2 cells.  Difficulty: 6 / 10.

Puzzle 9-5 “Stowaway”: 5 mines remaining, 1 to mark, 1 to clear.  Difficulty: 6 / 10.

Puzzle 9-6 “Merry-go-round”: Mark 4, clear 3.  Difficulty: 6 / 10.

Puzzle 9-7 “Bad Wheel”: 1 cell to clear.  Difficulty: 7 / 10.

Puzzle 9-8 “Vise Grip”: 5 mines remaining, 2 cells to clear.  Difficulty: 7 / 10.

Puzzle 9-9 “Trident”: 6 mines remaining, 3 cells to clear.  Difficulty: 8 / 10.

Puzzle 9-10 “Shell”: 1 cell to clear.  Difficulty: 8 / 10.

Puzzle 9-11 “Bridge”: 8 mines remaining; Mark 3, clear 3.  Difficulty: 8 / 10.

Puzzle 9-12 “Oasis”: 7 mines remaining, 2 to mark, 4 to clear.  Difficulty: 9 / 10.

Puzzle 9-13 “Watchtower”: 6 mines to mark, 7 cells to clear.  Difficulty: 9 / 10.

Puzzle 9-14 “Saddle”: mark 1, clear 2.  Difficulty: 10 / 10.

(10) Guessing via Lock-In

Believe it or not, the process of hypothesis testing (last chapter) teaches an essential skill for making optimal guesses, even though the former is searching for certainty, and the latter is about managing uncertainty.  If you’re able to successfully solve some of the puzzles in the two previous chapters (Proof by Contradiction and Hypothesis Testing), then you’re ready for what I think is the last step to being an expert solver: Being able to make reasonably accurate estimations as to which cells along your front are higher or lower percentage guesses, and therefore not really need to resort to the hint engine that much.  Along with that skill comes an instinct on the quantification of each cell’s odds, one that’s good enough to know if one move is better than another, or whether anything is better than the background rate.  Developing this instinct is really really hard, and honestly I’m still “just OK at it”, compared with the true masters.  But if you listen to this video with Scar and Koro, even for a few minutes (after they get into the gameplay / play-by-play commentary), you’ll hear a thought process that is as refined as it gets:

New Minesweeper High Difficulty World Record Set By Mufasa - Full Game Review

(and here is the game they’re reviewing)

Until you’re tackled the previous topics in this guide, until you can look at the board and not even have to think to see some of the underlying building blocks (#-of-squares ratios, # of forced mines, etc), their discussion about situations and strategy will sound like absolute gobbledygook.  But for high-level players, that’s what optimal reasoning sounds like.

More importantly, those guys are actually better than a probability calculator, because they can take into account things a calculator can’t, or would need some deep research to incorporate.  The most important of these considerations is, what moves are likelier to result in progress.  Listen at around 8:00 and you’ll hear them talk about how the corner squares (around a front) are often the lowest probability, but they yield progress a lot less often (because they won’t be a 1 very often, etc).  A probabilistic calculator can tell you where the best chances are of not blowing up in your very next move.  It can’t[22], however, tell you which guesses optimize for the chance that you end up winning the game; that requires wisdom gained from playing many many hours of high difficulty, encountering a lot of these situations and developing some habits.

Optimal guessing incorporates an understanding of “likelihood that this guess would result in progress”, which involves a few difficult-to-quantify heuristics:

1. Suppose you guess a square; how many remaining adjacent unknown cells does it have available to it?  If it’s four or five, then the odds that those are all safe (and thus the square you guessed can be chorded) are pretty low indeed.  If you guess and it’s an effective-1, you’re still kinda stuck.  But if it’s only two or three new cells, odds are better that the guess result is immediately chordable.
2. How much information a guess will give you, based on it being adjacent to a fairly well-determined set of squares, such that adding one new fact will be a “tipping point” that lets you draw multiple downstream conclusions?
3. Is this guess on a long linear front of effective-1s, such that the cells are dependent and a guess on one will let you clear every 3rd cell, giving you several opportunities to find a new path forward?
4. Can I see from nearby front areas that it looks like a low-density area is over there, and so I’m thinking about hunting for openings in that direction?
5. Is there an angle here that might override other considerations, such as preventing a 50-50, or guessing a pseudo?

At this level of play, the key guessing decisions amount to an art, not a science.  The best we can do is show what factors led experts to choose one thing over another, and teach through enough examples.  But there are trends that can consistently lead you to the best choices.

One essential concept we can use to address these kinds of situations is what I call “Lock-In”: When evaluating candidate guesses, focus your attention on squares where, if it were a mine, that would then determine a high number of downstream conclusions forced by this single assumption.  In other words, if this given square were to be a mine, how many other squares does that then imply are also mines?  The higher the number in answer to that question, the more simultaneous coincidences are required in order for that assumption square to be a mine.  Which lowers the odds of that exact scenario being the true case, and improves the odds for that given square to not be a mine!  So therefore, you want to guess the square that, if it were a mine, would “lock in” the highest number of other squares into having a mine.  That makes it the least likely to actually be a mine!

This is the primary heuristic Scar is using to evaluate his options, on the numerous occasions where he was forced to guess.  And using it is better than a probabilistic solver, for the reason that he (and all of us) can take into account meta considerations like progress likelihood, degree of dependence / independence, preventing 50-50s, and so on.  Probabilistic solvers use a brute-force method on the front regions, counting all possible mine configurations and then giving you the fraction of those in which each square was a mine.  None of us can do that in our head, beyond trivial situations.  With this technique, though, all we need to keep track of is downstream dependency: how many mines would this guess “fix in place” (“fix” in the sense of “placing it definitively”).  As that number rises, the number of possible mine configurations where all of those mines are in those exact positions becomes fewer and fewer, and consequently, odds of that square being a mine get lower too.

Let’s give some examples of what we mean, so the video above will stop sounding like a foreign language and actually let you follow along.  Firstly, take this one (Video: 15:35, game: 2:00).  Scar has a long and very connected front:

He zeroes in on the square to the right of the 4 at upper-middle, noticing that if it were safe (or equivalently, if the square to its right were a mine), it locks in a whole lot of other mine positions.  If we were to assume it (in teal frame) to be safe, then it forces these downstream conclusions:

That’s FOURTEEN mines that would need to be in those specific locations, in order for the teal-framed square to be safe.  Now, not all of those mine locations are fully dependent (if they were, you’d have expected the odds of this square to be 99%+), but enough of them are, particularly above the 4 itself and down the right-hand side, to drive the percentages for this particular case quite low:

Sidenote: See how the 4-square box towards the bottom does NOT provide what we might otherwise expect (a 2v3 around that corner, with a mine assumed on the upper-left share corner of it).  In fact, that square is extremely likely to not be a mine - it’s the worst guess we could make here!  That’s because if it is, it locks in all three unknown mines from the 5 below it, the 3 below that, and two on the 3-3-2 wall to the right. Same effect, different use.

The teal (assumption) square is 95% to be a mine, the one to its right only 5%, because of this lock-in effect.  There are many possible mine layouts along this front (probably thousands), but very few with all of those mines in those specific places.

Another good example comes at video 46:35, game time 30:45.  Scar has this isolated front:

The immediately tempting spot, for some of us, would be the lower corner, assuming a mine on the corner itself.  And it’s not bad: it’s N-vs-N+1, with the N+1 running into a high-SCR 2, such that your odds are a bit better than background rate.  But there’s also a far better choice.

One general idea to keep in mind is that when you’re looking at a front, two adjacent numbers that differ by one effective mine (neighbor mismatch) is usually a good place to find logic, and even if no logic is available, it’s a good place to hunt for solid guesses.  If you have adjacent numbers that differ by two effective mines, you almost certainly have logic or an incredibly good guess available.  One such place is offered here at the top-left of the front, with the row of 1s extending up to that 3 that pokes upwards.  Focus on that for a second, see if you can reason about what the lock-in rule says about it.

As Scar remarks, there is a certain square (and dependent squares lower down) where, if it’s a mine, then you’d clear the other squares around the top-left 1, which then forces all three squares above the interior 3 to be mines.  It would also fix the location of other mines below, making the square hugely unlikely to be a mine:

I would certainly consider 1% to be a pretty solid bet, when guessing in minesweeper!  That’s how dramatic the lock-in effect can be when it compounds across a number of different spaces, locking in a bunch of different mines that would otherwise have some degree of independence.  And even better, here, that guess gives you two other dependent squares down the linear front, giving Scar here three chances to break through and resume making progress - which he does.

Identification

How do you identify good spots to guess based on lock-in?  It’s the same basic process as what you did for Proof-by-Contradiction or Hypothesis testing:

1. Pick a spot you think might be a decent candidate to guess - one you think likely safe.
2. Assume it’s a mine instead.  Mark it (provisionally).[23]
3. Think in your head about what squares would therefore be safe if that one were a mine.  Don’t actually clear the imagined safe squares, obviously, but play it out mentally.
4. This will lead you to other squares that would need to be mines.  Provisionally mark ‘em.
5. Go in both directions along your front, if possible / relevant.
6. When you’ve hit the limit of any further squares that would necessarily be safe or mines, stop and count how many mines you’ve provisionally marked.
7. Un-mark your provisional mines, and pick an adjacent square to check the same way.  Repeat.  Chances are, this will leave you with a different number of forced mine placements.  Sometimes there’s a third candidate right around an area too.
8. You can go to another part of the front and try again if you want to explore more options.
9. When you’ve exhausted the reasonable candidates for being a good guess, whichever square that (if a mine) would have forced the most number of mine placements - that’s almost certainly your safest guess.
10. If choosing between several equally-attractive options, consider whether one of them would reveal more board or more angles on the front, and so provide likelier progress.

The hard part, mentally, is the dance of considering what’s locked in by assuming a square is a mine, and if substantial, then turning around and deciding that square is likely safe.  That dance can be a bit counterintuitive the first few times, but a little practice can make it feel natural.

There are a few rules-of-thumb for good spots to look for and then analyze this way:

• Longer fronts without many interspersed mines.  A few is OK, but if you can find a higher concentration of number squares, making a longer chunk of non-isolated front, it’s likelier to find a good guess around there.
• Guesses with a lot of dependent squares, such that one guess implies a few other safe squares - this gives you more chances to get progress out of your guess.
• “Neighbor Mismatch”: Sudden differentials in the number of effective mines remaining on adjacent squares, such as the 1-to-3 corner situation in the second example above.

This process does not require a big mostly-linear front in order to be usable.  The locked-in mines determined by a guess can all be around even one square, if it locks in enough to move the probabilities.  Consider the example at right, which was a counterexample from the “flat-framed front” section, showing that the presence of adjacencies can undo the favorability of guessing the wide squares from an FFF (the upper 2s here).  But look below that, to the left: facing the lower part of the front, the 6 requires five mines from six adjacencies.  This means that if the square the 6 shares with the 2 to its above-left is not a mine, it locks five mines into place into its other five adjacencies.  That’s a big number, if using the lock-in principle!  This means that square is highly likely to be a mine, and you can see the square next to it, representing the opposite assumption, is only 5%, probably the best guess on the board.  So you don’t have to look very far, sometimes, in order to spot a good guess from lock-in.  You just have to be able to ask the question “if this is [or isn’t] a mine, how many other mines does that lock into place?”

Adding lock-in considerations to a situation that is close to, but not exactly matching, one of our previous guessing situations, can make it an even-more-favorable guess. It can even flip something from “does not fit our heuristics” over to “actually, there’s a good guess here”.  Recall from our Assume Fewer chapter, that if a flat-framed front’s wide squares share adjacencies with effective-1s, it “plays nice” (still has good odds), but not when it shares with effective-2s.  Well, we can add an exception-to-the-exception now:

At left, we have a three-square framed front, and if the 4 at bottom were just a mine, not a number square, then the wide squares here would be favorable (roughly 17% each, vs ~65% in the middle one).  But the 4 below does exist, and is an effective-2.  So instead, consider firstly the total number of mines of different layouts: If the 2 and 4 share a mine, we need three mines total, but if they don’t share, we can still have three mines if the other is next to the 1 (in the middle), or need four mines if the one to the right of the lower 2 is a mine.  But then, notice that if the latter layout is true (a mine next to the lower 2), it not only requires four total mines, it also locks in the mines around the 4, and a dependent one above it too.  If the 2 and 4 share a mine, meanwhile, there is still a degree of freedom - the 4’s other mine can be in either of two spots.  So the square to the right of the lower 2 (the 11% at right) is made very favorable by two effects working together: lock-in, and also assume-fewer (via the Ratio Rule).  The result is the best guess on a big board.

Why only count mines?  Why not consider lock-in of safe squares, too?

Good question.  First, mechanically, mines are fewer in number than safe squares, and we can provisionally mark them on the board as we play out a scenario and count what’s locked-in by that assumption, so it’s easier to think in these terms.  But second, if a certain square locks in a meaningfully greater number of safe squares, and roughly the same number of mines, that square might in fact be a good choice, for the same underlying reason we count mines: any potential guess that, if a mine, would meaningfully limit the universe of overall mine configurations that the board could have, is less likely to be the true case.  In high difficulty as mine density approaches 25%, the ratio of safe squares to mines is about 3:1 (and at 20%, or “slightly lower than expert”, it’s 4:1).  So we can add some nuance to our rule-of-thumb to account for this: When considering what’s locked-in by a potential guess, if one option locks in more safe squares, count the additional safe squares as if they were locked-in mines in a ratio of “three safe squares counts as one mine”.  As mine density decreases, the value of locking in safe squares is lower, with respect to a guess’s odds, whereas at very high densities like in an endgame, safe squares might be equally valuable or even more so.

Here’s an example of that coming into play (from this game):

As we can see, we’ve already guessed to prevent the 50-50 at lower-left, and it’s left us with two interior 3s that are effective-1s.  Now, a naive first thought might be, look at that top with the two 4s - they share a square, so maybe we can just Assume Fewer and guess the shared square.  The problem with that is, the corner on the right means that it’s not 1v2, it’s actually 2v2, so we’re not saving a mine in either case.  See it at right: assuming a mine in the upper blue square forces one in the lower-right blue square, so Assume Fewer doesn’t make blue more likely - blue means the same number of mines.

But this initial dead-end is actually the start of a better idea: Notice that in the blue-square scenario only, all the other squares surrounding that righthand interior 3 would be safe.  So the yellow scenario sets 2 mines and 2 safe squares (the blue ones), but the blue scenario sets 2 mines and gets us 6 safe squares: the yellow ones, plus the other four around that right 3.  That’s worth a mine saved - even a little bit more than a mine!

Likewise the lower-left area.  If you see the image at right, A and B might normally be a 50-50 front, but the fact that the 3 shares a square changes that dramatically.  If A is a mine, then all 5 squares around that interior 3 (the green area) are all safe.  Whereas if B is a mine, we don’t necessarily know anything else.  Those 5 extra safe squares count as like a mine and a half of extra lock-in, and make square A here (and the blue-square scenario at upper-right) the safest choices to guess, as we can see[24] from the solver:

Is that necessarily the end of the story?  Well, not fully: It gives us one of our goals (safety), but says nothing about the other goal (progress).  Suppose we guess cell A at lower-left.  To get immediate progress, we’d need both cells to the right to be safe (or both mines), which is unlikely.  Instead, it probably shows a 4 (effective-1), and we’re right back to guessing.  Whereas, up top, guessing the 18% blue squares let us mark and clear 2 additional squares, and thereby see additional angles on several more squares.  So it’s likelier to result in progress.  The blast odds are close enough (13% to 18%) that I would tend to opt for the upper guess.

So, yes, you can count safe squares as well as counting mines.  But in most situations, it’s counting the number of mines forced by an assumption that will give you the clearest indication of safety.  You’ll also have to keep fewer things in your head, and have less chance to get confused.  You should only bother counting safe squares if it helps to break a tie, or if the difference between two possible guesses is a huge gulf in safe squares, as above.

That’s the general theory of how to think about “probability via lock-in” - how different guess choices would limit the universe of possible mine configurations to a greater (good guess!) or lesser (bad guess!) degree.  You can use it to reason out any situation from those principles, if you need to.  But there are also a few common and recognizable formations where it applies, beyond the ones discussed previously, which are worth a brief mention, because pattern matching is easier to pick up on and identify in-game.  These include:

(a) Shared Corner - an eff-2 and eff-1 on a bend: You have an effective-1 and an effective-2 which have both shared squares and non-shared squares, on a front around a corner, such that no “1-2” logic patterns apply.  In this case, assuming the mines are on the non-shared squares, and the shared squares are safe, would lock in three mines (and also require more mines than the two required by the opposite assumption, so Assume Fewer compounds the advantage), so one of the wide squares next to the effective-1 ends up being a great guess.

(A)   (B)   (C)

Examples A and B here are basically the same formation, usually isolated and easy to spot: a 4 that’s an effective-2 next to a 3 that’s an effective-1; you Assume Fewer, and so the wide square (which is only a mine if neither shared square is a mine, which locks three mines into place) is a good guess.  Example C is more subtle: here the lower 3 is an effective-1 and the upper 3 is an effective-2, it’s still a slight neighbor mismatch with two shared squares, but getting wrapped around the 4-square box at top can draw your attention away.  Same reasoning applies, though.  And the other good-guess there is because of / dependent on this one: if the 8% cell at lower-left is safe, so is the other one at right, via 4-square-box logic.

(D)  →

In D, your eyes might first go to the upper-right 3 in the corner, as mine did in the real game, thinking that if its upper-left square is safe, logic will follow (which is true).  But that upper 3 is an effective-2, and there are three squares above it not shared with the (effective-1) 3 below it.  So there is no lock-in effect available here, because there are many configurations which have both mines on top, not just one.  Instead, I should have looked around a bit more and seen the shared-corner 3-4 down below, where this exact formation was available: eff-2 with two wide squares, eff-1 with one wide square, which makes the latter a great guess, because they are very likely to have a mine on the shared squares.  And you get a free second square with it!

This pattern is available all the time[25] in high-difficulty guessing situations - it’s an important one to practice and internalize.  All you have to look for is an eff-2 next to an eff-1, where the eff-2 has two non-shared adjacencies.  If it only had one non-shared adjacency, of course, you’d have a 1-2 logic pattern.  And if it has three non-shared squares, you’re in that situation above in (D), the misleading one.  But with two non-shared squares, there is a wide adjacency by the eff-1 which can only be a mine if both of the eff-2’s non-shared squares are mines.  That’s the guess you want, usually coming in at around 25% of the background rate.

Incidentally, this pattern works because of the Ratio Rule: There are multiple fewer-mines layouts, and only one more-mines layout, so the latter is even less likely than usual.  But that actually gets it backwards.  The Ratio Rule and our Lock-In rule in this chapter both derive from the same underlying truth about Minesweeper: the squares which, if a mine, would most substantially reduce the number of possible mine configurations for the board, are the least likely to actually be mines.  All of our guessing rules and patterns are just illuminating this fact.

(b) Split Equivalence - dependency around a corner, divided in 3: You have an effective-1 on a corner, and if any of the three squares on one side of the corner (including the corner square) were to be a mine, then so would be the square one off along the front in the other direction.  Therefore the probability of that lone equivalent square equals the sum of the probabilities of all three squares on the other side - so each “split” square has low mine odds.

(A to D, left to right)

1. The cell above-left of the left 3 (the 22%) is equivalent to the 3x 7%ers down below: a mine above implies a mine among the bottom ones, and safety above implies safety for the three below.  Thus the probability of the upper one is equivalent to that of the group below, and the odds of the ones below collectively add up to the odds of the upper one.
2. The cell directly to the left of the 3 on the left is equivalent to the three green squares at top, the 6% great guesses: if that 18% square is safe, then by a 1-1 pattern going up from there, so are the 3 on the top.  The odds of that square here are lower than usual because of an Assume Fewer situation, but even if it were, say, 33%, the top squares each being (33% / 3) = 11% would still look pretty good.  
3. The square to the left of the upper-left 3 is 20%, and those odds are split amongst the three squares at the bottom of the front, around the corner 2 (they’re not the same %s because of a lock-in effect, but they’re all low).  The same does not work in the other direction because of the interspersing mine meaning that the mine assignment there would still be ambiguous even if the mine were located to the left of the corner 2.
4. We have an Assume Fewer corner, 2v3 around the upper-left there.  However, the odds on the “3 mines” squares are even better than background-rate, because of split equivalence at bottom-right: a mine at the bottom-right square implies one at one of the two 10% squares on the 2 just above there.  So the “3 mines” corner scenario would also lock-in another mine at bottom right, which reduces its odds to a tasty 10%.

The corner has to be an effective-1 to work, however it does not necessarily have to be an exact and neat corner, as long as the “split equivalence” effect can be seen (if the off-corner square is safe, then so are the three off-corner squares, and vice-versa).  Taking the example at left, note that the 40% cell, second from right, is equivalent to the 3x 13% cells at far left; they are dependent, the former being a mine puts a mine among the latter, and the former being safe means all three at left are safe too.  This pattern is defined by that dependence relationship.

(c) Side Hole - the wide squares around a hole pattern: You have a hole-pattern mushroom going into the unknown squares, and all of the squares on the back row of the hole are the same number, ideally 2s.  In this case, it is often true that the wide squares, on either side of that back row but not directly behind it, are lower-than-background guesses, because mines there would tend to significantly limit the possible configurations for the rest of the formation.

(A)   (B)   (C)

These formations can provide good-probability guesses, safety-wise, but it is often also true that they offer low chances of progress.  That can be acceptable if you have nothing else to try, or if it’s lower-density and you might turn up a 1 or an opening.  But where this situation can really help is when the “side-hole” squares are up against a wall (or row of flags), limiting the number of adjacencies you need safe in order to find follow-up moves.  In those cases, side-hole moves can be great choices even in higher-density play, e.g.:

(D)  (E)  (F)  (G)

Those side-hole guesses against the wall (D/E/F) are pretty good choices.  I would usually guess the one closest to the front; you can turn up a 1 if the other mine on the neighboring 2 is against the middle of the hole squares, which only requires two other squares to be safe.  If you get a 2, though, you can guess the other wide square next to it as a good-odds followup.  In that first group of examples, meanwhile, getting progress would require a lot more squares to be safe (except for C, where the top one is tucked into a wall of flags).

Also, as with example G, sometimes you don’t have a clean front, with the hole squares separated from the front by exactly two unknown squares (making the interior squares collectively holding one mine), but the formation offers good odds anyway.  This side-hole pattern works based on the wide squares of the hole having probability “pulled” toward the middle of the hole formation, so make sure that’s the case before guessing it.

There are some caveats / exceptions:

(H)   (i)   (J)

Firstly, again, you need the hole squares to be effective-1s (after accounting for the squares they share with the front).  Nobody would think example H above would be a good place to guess.  Second, sometimes you get different numbers along those hole squares.  If the middle one is an eff-1 but the wide ones are not (example I), you mostly have bad odds all around.  But if one wide square is a higher number, but the other side isn’t, it can still be good (as in J).

Hole-pattern “Gradients” and the Back Row: That last example above, J, is a generalize-able and useful pattern to learn, in and of itself.  Sometimes you get nice neat little hole patterns like A through G above and the side-hole works, but many other hole-pattern situations don’t end up that way.  But there’s a second type of hole-pattern formation that always has one good guess attached, and it occurs often enough to be worth remembering.  It goes like this: if the hole squares, viewed from one side to the other, are increasing or decreasing in (effective) number, i.e. if there’s a “gradient” to the squares’ values, then there’s a good guess behind the lower side square, on the “back row”.  We see it in J with that 5% square.  Here are a few others:

(K)   (L)   (M)

K has a 2-2-3 gradient, so the square “behind” the wide 2 is the good guess.  L has a 1-1-2 where the inside 1 doesn’t share much with the front, but there’s still a good guess behind it.  M is a 2-2-3 where you have some interference from a 1v2 corner below it (by the 4), so in this case you have some good guesses on an Assume Fewer basis, helped by Side Hole - but the best (or, well, safest) guess is still the square behind the wide lower number on the Gradient.  Progress odds will be low, but not zero, and it gives you a safe chance to maybe get lucky.

Extreme Lock-In: Below are a handful of “extreme probability” situations, where squares have <3% probability (but still not 0%, not guaranteed-safe with logic or even hypothesis testing), each of them created by this lock-in effect.  In order for the low-% squares to be mines, the consequences of that state would end up locking in a very large number of mines downstream, which we will illustrate.  Compare the given probability here with the presented “consequence map”, the flow of what happens if a good-choice square were to be a mine, to see that cascading effect when it grows large (and thus the odds of it being the case, small).  The idea here is to get you thinking about seeing these lock-in chains, showing some dramatic examples, so you can learn to recognize more-typical, less-extreme examples in the wild.

(A) From this game:

 →

If the teal-framed square (or any of the ones in the dotted-blue group, really) are a mine, then so is the one at top (which adds an extra, going against Assume Fewer), and so is the one below-left of the 3, clearing the square below the 3, and forcing two mines to the left of the 4 in specific places.  This means a total of 4 other mines locked in, plus 8 safe squares, so a total of 6 2/3s mine equivalents.

(also note how the #s under the 3 add up to 101!  Getting rounding error on WoM is very rare, it requires that the probabilities here be exactly 8.5%, rounding to 9, and 91.5%, rounding to 92)

(B) From here at 1:03:

 →

The assumption mine (in teal) would push two mines to the right of the 5, clear around the 1 and 2 to its left, which pushes three mines to the left of the 4 (note the Neighbor Mismatch between that 2 and 4), and the cell below that upper-left 4 being safe then locks in the mines to the left of the middle 4, and clears the cells below the lower 4.  All told, that’s 7 mines locked in plus 8 safe squares, for a 9 2/3s mine-equivalent effect.

(C) From this innocuous-looking situation:

 →

The assumption mind puts a safe square below it, and clears the corner 3 above it, putting a mine at top-left.  But that safe square means the right-hand 4 below it needs three mines from three squares, so they all get marked.  This clears the to the right, puts a mine to the right on that front, then the 2 on the corner clears two squares to its right, putting a mine at top-right.

(D) This game at 4:12:

 →

The assumption mine flows conclusions upwards, clearing the 1 at top-right, and forcing an additional mine into the purple group (odds supposed one mine there; now there’s two) and resolving the top-left.  It also flows to the right, clearing to its right and putting two mines on the column to the right of the assumption square.  This leaves only two squares left for the left-interior 3, so it forces mines on both, and a safe square to the right.  It also clears the 3 down that column, giving a safe square around the mined corner, putting two mines toward the bottom-right.  Finally there’s a 1-2 pattern that puts another mine and safe square at very bottom-right.  That is 9 additional mines locked in, plus 10 safe squares, so 12 ⅓ mine-equivs.

(E) From this 1k HD board, at 8:25:

→

The assumption mine implies a mine to the right, which clears the 1 next to it and then down to the corner next to that 3.  That clears the 2 around the corner.  Then we have a 1-2-1 pattern above, which puts a mine at the top-right (next to the top-right 1), clearing that 1, and the other two cells below that.  Going downwards from there, we have two more mines along that long linear front, which clears the interior 3, putting a mine on the lower-left corner, and another two to the right.  In total that one assumption mine would lock into place 9 other mines, and clear 20 (!) safe squares, so that is at least 15 2/3s mine equivalents (!!), and given that the board has a 29% background rate, it’s probably even higher.

What about progress?  Lock-in can show you safety for guessing moves, but it can’t tell you which is the best move to make, in terms of winning the game.  There are several other considerations for that, but the most important of them of course is progress - the chance that this move will result in there being follow-up moves available from logic afterwards.  If you’re picking between several good guesses, of roughly equivalent safety, here are the main things that should make you choose between them:

1. # of Dependent squares.  More than anything else, the number of additional dependent squares you get “for free” with your guess is huge in terms of finding progress.  Each extra cell that your guess allows you to clear if safe, gives you another roll of the dice to find a low-density area you can get into and work with logic.  A long linear front, where you find a good guess and get 3-4 cells along the front with it, can easily turn up a 1 on one of them, giving you a hole pattern and plenty of chances for further logic.  Sometimes a guess will enable some projection logic around a corner or box - those are dependent squares too.  e.g.:
   
2. Odds of surrounding squares.  All else equal, if you have two or more independent good guesses in the same neighborhood, it suggests good things about there being a “soft spot" in the mine formation behind them.  “Loner” good-guess cells are fine, but if you have neighboring ones that are all good guesses (perhaps from split-equivalence or pseudo-holes or square coverage ratio), there’s better chances of serendipity.
   
   At right: The best guess here is to the right of the 3 in yellow (the 6%), not the 5%s, because you have a ~40% chance of getting a 4 that is immediately chordable.  Even if you get a 5, one of the guesses above could yield a 2, giving you a hole pattern.  The blue group all being together (and independent) also suggests that a hole pattern, or even an opening, might be lurking.
   
3. # of Adjacencies.  Like Square Coverage Ratio and our Wall-Skip tactic, it matters how many other squares are next to a potential guess, because all of them would have to be safe in order to get an immediately-chordable result.  The more squares that have to be safe, the less chance that has of happening (and likewise, the higher % those squares are - see (B) above - the less chance of that, too).  When a guess is up against a wall, or a wall of mines, and only has three other adjacencies after your guess, it’s likelier to be hiding an opening - and so you might take it over a lower-percentage guess that just leads out into a sea of untouched squares.  At right: the 15%s are against a wall, but the left 13% has the same three adjacencies as a wall guess, plus a bit more logic potential, so it’s optimal.
   

And a few other less-common things also worth weighing, if they are relevant to your guessing options:

• Proximity to a wall.  If a guess would put you right up next to a wall, it’s probably better, because the odds of coming up with a chord or opening afterwards are better.  You may not have fewer adjacencies now, but you would then.  You might also get follow-up moves from 1-1 patterns.  At left: the wall guess is better because, while both pairs of 17%s offer a second dependent square, the left (wall) guess would then only need three other squares to be safe to offer progress, while both squares above the 2 have an equal probability of offering a hole pattern, so that’s a push.  It’s maybe harder to see that the rightmost 17% is less-favorable for progress, but it’s pretty easy to see that the left (wall) one is more-favorable.
  
• Guaranteed logic.  Sometimes the only things that a guess could turn up are going to lead you to one conclusion or another.  You don’t know what that conclusion is yet, but you know it’ll give you one either way.  That’s worth paying a price (in higher mine odds) for.  At right, the 5% is an even better guess than it looks, because it will only have two adjacencies (so possible 5 and chord), and if it turns up a 6, then you have logic with the 4 (above-left of the 4 must be safe), and so will have follow-up moves guaranteed either way.
• Helpful-guess considerations, such as preventing a 50-50, or guessing a pseudo.

Putting it together: It’s worth mentioning that using this technique in a game situation, reasoning from first principles, can be really labor-intensive and kind of a pain in the butt.  Top players will very rarely take a guessing situation and assess the lock-in value of each and every square on their front before making their move.  But our goal here with this chapter has been to strengthen your skillset in a few ways:

1. Teach you specific, recurring special-cases of the lock-in principle, so you can recognize them in-game as a sort of pattern.
2. Learn to recognize how the lock-in consequence chains can work, so you can spot when a situation can tip into a flood of downstream conclusions.  This is about training your instincts to recognize “there’s probably a great guess right around here”, making mental assessment of this lock-in a bit faster, even if you don’t diagram it all out.  Maybe you focus in on a few potential guesses that look like good candidates (which is what Scar frequently does in that video), and then evaluate them to whatever level of depth feels appropriate.  But the more you try to reason this out, instead of reaching for the Hint button, the better your instincts will become.
3. And when absolutely necessary, you can fire up the Editor or start marking provisional mines, and actually reason things out fully.  Maybe you’re out of hints on a 100k board, or want to seek an achievement without using hints at all, or maybe you are at endgame on a hard board and it’s worth investing more time for a slight edge in your chance of winning.  But the more you use this technique, the easier it will be to spot optimal guesses quickly, prove them to your own satisfaction quickly, and apply judgment (often about progress) that the hint engine can’t offer.

In that World Record Game, Scar makes 28 successful guesses (!).  Of those, he wins 4x 50-50s, and has 6x guesses in the first minute of play where he is being somewhat cavalier, before he gets emotionally invested in playing it well.  But he also faces 18 other guessing situations, after the initial moves and which aren’t 50-50s: in 14 of the 18, he makes the best available guess (percentage-wise), and in the other 4, he chooses a square that isn’t the lowest percentage, but which has a much higher chance of leading to progress, making it a better choice.  How did he recognize those guesses were the best moves?  Through the principles explained here.  Practice enough, get your thought process to line up with how he’s thinking in the video, and your game can start looking more like that, too.

Nor do you need to be as professorial and computational as Scar in order to be a great guesser and great high-diff player.  Mario Pro Gamer, currently #3 in trophies on WoM, says he never computes guessing situations - he plays “by instinct”.  Here is a community-famous game by him, where he was about to run out of time on a L8E High Diff arena, and ended up finishing it on his final game with 1 second left in overtime (!).  Without hints available, and with only a few seconds’ thought, he makes two guesses at 4:20 and 4:26, on a rather large front, which were both the safest moves on the board (5% and 8%) and offered great progress odds: the first via Lock-In, the second a Flat-Framed Front guess.  He has trained himself over time to spot these same classes of situation we’ve discussed, no math involved, to the point where they’ve become instinctual.  And so can you!

Practice Puzzles

Link to solution section.

This final set of puzzles asks for “the best guess[es]”, taking into account both safety (probability of not-dying on your guess) and progress (probability of meaningful advancement toward winning the game).  Right answers will not come in the form of corner 1v2s or other assume-fewer or square-coverage-ratio type guesses, but instead will require that you use this chapter’s techniques to find some option(s) that improve upon what our other guessing heuristics would have recommended.

We’re including more puzzles than usual here, partly because of how crucial it is for advanced players to get good at this reasoning, but also partly because the simpler examples (difficulty 7 and below, for me anyway) can mostly be solved by visual inspection, and the explanations are shorter.  So we’ll give you the opportunity for a bit more practice.

Puzzle 10-1 “Parapets”: Find the best guess (there are multiple).  Difficulty: 2 / 10.

Puzzle 10-2 “Smoke Stacks”: Find the best guess.  Difficulty: 3 / 10.

Puzzle 10-3 “Scepter”: Find the best guess.  Difficulty: 3 / 10.

Puzzle 10-4 “Balcony”: Find at least one good guess.  Difficulty: 3 / 10.

Puzzle 10-5 “Axolotl”: Find the best guess.  Difficulty: 4 / 10.

Puzzle 10-6 “Slipknot”: Find 2 independent good guesses.  Difficulty: 4 / 10.

Puzzle 10-7 “Monument”: Find the best guess.  Difficulty: 4 / 10.

Puzzle 10-8 “Monkey Bars”: Find two (semi-)independent good guesses. Difficulty: 5 / 10.

Puzzle 10-9 “Slinky”: Find two independent good guesses.  Difficulty: 5 / 10.

Puzzle 10-10 “Propeller”: Find two independent good guesses.  Difficulty: 5 / 10.

Puzzle 10-11 “Doorbell”: Find the best guess.  Difficulty: 6 / 10.

Puzzle 10-12 “Bento Box”: Find two independent good guesses.  Difficulty: 6 / 10.

Puzzle 10-13 “Pride Rock”: Find the best guess.  Difficulty: 6 / 10.

Puzzle 10-14 “Toggle”: Find two independent good guesses.  Difficulty 7 / 10.

Puzzle 10-15 “Waterfall”: Find two independent good guesses.  Difficulty: 7 / 10.

Puzzle 10-16 “Pillars of Sand”: Find 3 independent good guesses.  Difficulty: 7 / 10.

Puzzle 10-17 “Sword and Shield”: Find two independent good guesses.  Difficulty: 7 / 10.

Puzzle 10-18 “Thunderstorm”: Find 3 independent good guesses.  Difficulty: 8 / 10.

Puzzle 10-19 “Rhydon”: Find 3 independent good guesses.  Difficulty: 8 / 10.

Puzzle 10-20 “Freefall”: Find at least 4 independent good guesses.  Difficulty: 9 / 10.

Epilogue: That’s it!  You’ve reached the end of the guide.  What follows from here is just an appendix about 50-50s, and the solution walk-throughs of all of the practice puzzles.  If you were able to solve even half of the puzzles in the problem sets, particularly these last few sets, you are likely an expert player now.  If you were already an advanced player, I hope this guide provided you with a few new techniques or ways of thinking that helped you get even better.

Exotic 50-50 Graveyard

Below is an appendix showing some unusual 50-50s that I’ve come across, much like the samples at the bottom of Scar’s 50-50 guide.  If our 50-50s section in Part 1 was like the “101” level intro to 50-50s, this is the advanced course.  These are arrangements that extend beyond the typical formats of the “basic” 50-50s (box-and-2, line-and-1), and their common variations.  There are some trends and styles to how a 50-50 can get extended to include additional pairs of squares, while still keeping to an A-or-B, 50-50 scenario.  Such as how they can be made to bend around corners, change which row they’re on (it requires a few strategically placed mines!), or even to form circles.  I’ve included it in this guide because you’ll need to be able to recognize what is actually a 50-50, vs something that is mine-countable, vs something that might have a proof-by-contradiction, vs something that is solvable just with basic logic patterns.  And it’s nice not to have to use a Hint just to confirm your instincts.

To start, there are a few principles that apply to how and where 50-50s can exist, beyond the two basic types, and they form additional categories for us:

1. Line-and-1 50-50s can often form extended patterns when you have a repeating sequence of 3 cells in a row, where 2 are unknown and contain 1 mine between them, and the third is known (and could be a mine or safe, it doesn’t matter).  This is why, for a line-and-1 edge variant, the line of mines has to be 3 rows off the edge of the board (so the 2 squares / 1 mine can be between the line and the edge).  And to extend it by one “repetition”, the line has to be six squares off the edge, so the 50-50 pattern takes up a 5-square sequence: two pairs of unknown squares, with a spacer in between.
2. A line-and-1 50-50 sequence can be bent 90 degrees, with no loss of indeterminacy (i.e. while remaining unsolvable), if there is one pivot square around which it “bends”, and there are mines located at at least 2 specific squares around that bend, preventing discernment of whether the nearest squares in the pattern are clear or not.  This is like the corners of a box-and-2 50-50 being required to be mines, so that you can’t use them to get the information to solve the area.
3. An extended 50-50 line-and-1 sequence can have its row or column adjusted by one position, if some other cells right around the adjustment spot are mines.  This unusual - there are a lot of squares that could solve the 50-50 if they’re not mines too - but it does happen.  The Rule of Three can nicely prove or disprove a 50-50 here.
4. Box-and-2 50-50s can be extended into sequenced or conjoined 4-square boxes, if (A) the exterior corners of the sequence are mines, or wall, or shared in close alignment so as not to violate the Rule of Three, and also (B) the alternating-diagonal pattern of mine placement that exists inside them is repeated (not mirrored) across the boxes.  One correct guess usually still solves the whole region.
5. Sometimes an extended sequence can combine a box-and-2 50-50 and a line-and-1 50-50, if the combination has a 1-square spacer in between, and the corners of each formation are mines to prevent determination.
6. Double Trouble: Sometimes you’ll get a big formation of 50-50 stuff that is actually two separate 50-50s.  i.e., you could correctly guess one, and still be left with more 50-50 to solve, squares that weren’t connected to the one you solved.  For example, this.  Much like the “helpful guesses” chapter above, there will sometimes be certain guesses that will solve all parts of a 50-50, while other guesses would, even if you’re luckily correct, still leave you some further guessing to do.  Other times, the 50-50s are independent.
7. There is a certain arrangement of 4 remaining squares with 2 mines in an “L” formation, almost always found in a corner of the board, which isn’t quite a line-and-1, and is frustratingly common.
8. There is a particular “swirly” pattern of 6-7 squares / 3 mines, with appropriately-placed mines to prevent discernment, which will be encountered sooner or later in high-diff play.
9. Certain regions with some untouched squares are 50-50s because of minecount, i.e. the front-line squares are 50-50 and the untouched squares offer no advantage due to the number of mines remaining.
10. Then there’s a few things you’ll run into that are just off-the-wall, truly wacky and defying easy description.  The ones you complain in chat about.

We’ll now show some examples, in each of these categories.

And if you’re thinking “hey, I don’t need to know this, by definition I can’t solve these… this section is only here to mourn the many ways the game can force a loss!”... well, you might be a little right.  But not 100% right.  Some of the value of understanding this is to be able to recognize quickly whether something is or isn’t a 50-50, and either solve it, or guess it without having to burn a hint.  Half of playing the game is pattern recognition, after all, and these are patterns.  That, plus I needed some spacing between the puzzle sets and their solutions. 🙂

(1) Line-and-1 extended patterns

(1b) Extended half-full-bucket variants:

And these next ones look a bit weird, because they’re not really staying in a line, but they don’t really look like the “bent” or “displaced” categories below either.  I guess it makes more sense if you imagine the non-extended variant, in the corner, where it’s just a boxed-in choice between two squares.

(2) Line-and-1 extensions with bending at “joints”

Simple:

Fancier:

(3) Line-and-1 extensions with row/column displacement

Twins, sorta:  

(4) Box-and-2 extended patterns

There are some sub-types to these.  4A, “Split Box”:

4B, “Double Box”:

(on these, note that even having the interior corner between the boxes available to you, isn’t enough to solve it, because the diagonal mine patterns are repeated in each box).

4C, “Conjoined Box”:

4D, “Box-and-a-Half”:

Someday, I am sure, I will encounter the Unholy Grail: a triple box-and-2.  [June 2025 Edit]: …We’ve done it, guys!  Note that there are two not-dead cells here, making them best guesses.  Can you find them?

(5) Combo “box-and-2 plus line-and-1” sequences

In these, at least solving one will solve the other, they’re not “double” 50-50s.  But still.

(6) Double Trouble: multiple coin flips, faced sequentially

(For the 2nd one: Guessing correctly on the line-and-1 at the left does not help you solve the one at right, and vice-versa.  But if the spot above the middle 4 hadn’t been a mine, we would’ve had a connection between the two that likely would’ve reduced it to a single 50-50.)

Pretty brutal here, at left: Extended line-and-1, with a bend, then combo’d into a box-and-2 guess too.  Even worse, the middle (horizontal) portion does not offer a Helpful Guess optimization: if the left-hand one (above the lower-left 4) is a mine, then you have a confirmed box-and-2 from all directions.  However, if it’s not a mine, then the one to its right has to be a mine - and that becomes the last nail in the coffin to make a line-and-1 50-50 out of the bottom-right portion (it “completes the line”, you could say).  So you can’t avoid having to make two guesses on this one.

Above: Not bad, got 2 out of the 4 sequential guesses before the 3rd one got me!  

(7) Corner L

This is a 4-square, 2-mine endgame, almost always found in the corner of a board.  There’s not a lot of variety to how these are laid out, but they thoughtfully form an “L” shape, to inform you of the game result you’re about to be handed.

Basic:    

They can also kinda be extended, in ways that can blur the distinction between this and Category 2, “line-and-1s with a bend at a joint”.  Here’s a few:

(8) Swirly Pattern

These involve an 8-square region (4 by 2), in which 6 of the squares are unknown, containing 3 mines, and the other 2 squares are either safe or mines but either way are known.  Much like the box-and-2 pattern, it requires either a wall, or framing around the corners of the region.

Note how these are framed on the corners by mines / wall, in a manner similar to that of simple box-and-2 50-50s.  We know they are 50-50s, in part because they fail the Rule of Three.

They can also be “extended” by having one of the corner mines replaced by another 50-50-ish bit: e.g. below left, it replaces the upper-right corner’s mine with a pivot to a line-and-1 50-50.

(9) Minecount-based 50-50s

These have some untouched squares, which minecount says are half mines.  With a different minecount, some might be guaranteed safe or guaranteed mines.  But, as-is: 50-50.

(10) Even-crazier stuff I can’t adequately put into words

The sheer variety here should convince you that there is no general solution, no algorithm, that can look at a mine pattern and identify 100% of the 50-50 situations.  We can spot and categorize most of them, but there will always be more-exotic stuff cooked up by the server than we can possibly anticipate.  Especially when playing high-difficulty.  Life’s tough, get a helmet.

Practice Puzzle Solutions

Below are solutions for each of the scenario / puzzle sets given in the chapters above.  At the start of each are some links, to allow jumping from a problem set to its solutions.  Puzzles have been presented in order of increasing difficulty, as best as I could assess it.  YMMV.

I will usually try to link to the original / source game for a given puzzle in its solution, especially for harder ones.  If you want, click over to it and start your own version (“Continue Game”), and you can play it from that point, including provisional mine-marking and test-chording and using the built-in “Editor” graphics tool.  Ideally, once you’ve done a bunch of these puzzles, you’d be able to solve it just by staring at the image… but frankly I’m not at that level myself, beyond a 5 or 6 / 10 difficulty or so.  Sometimes you just need to interact with the game scenario to get a feel for the situation.  So don’t hesitate to go to a source game and continue it, if it helps you (as long as you resist the temptation to click Hint).

(1) Advanced Mine Counting

Link to discussion section.  Link to puzzle section.

Puzzle 1-1 “Boat”: 2 mines remaining, 2 to mark, 3 to clear.  Difficulty 1 / 10.

Solution: This looks a little funny because it’s wrapped around a bit, but that 5 is still the determinant: a mine below it means a 3-mine solution, a mine to its right means a 2-mine solution.

Think of the square to the right of the 5 as a “pivot”: either it’s a mine, satisfying the 2 and 1 above it, or it’s not, requiring the rightmost square to be a mine and the one below it to be too.

1-1(b) Bonus: Make a 3-mine solution for this same region.  As discussed, just flip the marks:

Puzzle 1-2 “Trail”: 3 mines remaining, 1 to mark, 2 to clear.  Difficulty 2 / 10.

Analysis: For our mine-counting strategy, we need to identify groups of squares in which certain mines are accounted-for, and see if we can reach our total from mutually-exclusive (non-overlapping) groups, without using all of the squares available.

The easiest way to do this is to make use of the 4 in the middle, which is an effective-2, has two mines unmarked.  So that orange area has two mines in the three unknown squares.  And then down below, the yellow area has one.  That accounts for our three mines.  Therefore anything else beyond that must be open.

So we clear the top two squares here, and then can draw a conclusion about the orange area - the top-left has to be a mine.

Solution:

The remaining two mines are 50-50 between the remaining four squares, in two groups.  We’re not told what number those safe squares will show, but the region can be solved either way.

Puzzle 1-3 “Boxed Out”: 4 mines remaining, 3 to mark, 3 to clear.  Difficulty 2 / 10.

Source game is here.

Analysis: As the Part-B to the puzzle suggests, this is a 4v5 minecount, not that much bigger than the simple cases discussed above.  The only complication is that there isn’t an odd number of unknown squares (there are 8).  But we can reach our minecount by carving off areas, pretty easily, in several different ways.  Here’s one:

There are two mines in the yellow (per the 5) and two mines in the orange (per the right-hand 4), which is four total and accounted-for just from those squares.  Therefore the other two squares are safe (top-left and bottom-right), and the other conclusions follow from there.

Solution:

1-3(b): To make a 5-mine scenario out of this, observe that one key to making the four mines work was that one of the lower-left 5’s remaining two mines was put on the shared square with the 3 above it.  So, to cram more mines into this area, one approach is to just flip that, i.e.:

  Or:  

So if we don’t know the minecount, there are three scenarios (the four-mine solution, and the 2x 5-mine solutions above).  So in 2 / 3 cases, the upper-left and the lower-right squares are safe, even leaving minecount aside.  If you were forced to guess this via probability estimation, those are your candidates, and you’d guess upper-left because it offers guaranteed progress.

Puzzle 1-4 “Empty”: 3 mines remaining, 2 to mark, 7 to clear. Difficulty: 3 / 10.  Source here.

Analysis: This puzzle is about the “all square groupings are correct” rule.  A little inspection will show you why.  The first thing you’ll see if you look at this is, three mines is actually really few here!  We have that 3 showing in the interior, which is an effective-2, and then we have a 50-50 front at the bottom for the 3rd mine.  So the first possible square grouping is, almost-trivially:

Pretty simple, right?  Well, there’s actually another layout here, both for the two mines at the top and for the one at the bottom (and they could be mixed-and-matched, as well):

That is an equally-valid square grouping.  So which is correct?  Both!  By the “all groupings are correct” rule, any square that is safe under any valid, sufficient grouping is safe under all such groupings.  So this leads us very quickly to pick out two cells that are mines, and others safe.

Solution: There remains one two-square group (in blue), the overlap squares of the blue groups in the two layouts above.

The most common mistake here would be to not see the second square grouping, particularly the upper groups, or not realize that both of them are “correct”.  In that case, you’d still get that the upper-right square is clear, so the one below it is a mine, etc, and you would probably work out a few other clear squares just from logic.  If you were playing a real game, the numbers from the safe squares would quickly reveal for you the rest of the mine locations.  That would only be a 1- or 2-difficulty puzzle.  But to know nearly the whole thing upfront, as here, requires a little more knowledge and thought.

Puzzle 1-5 “Needle”: 4 mines remaining, 0 to mark, 1 to clear.  Difficulty: 4 / 10. Source here.

Analysis: There’s a right way and a wrong way to look at this, and the wrong way is a fair bit easier to come up with, probably the first thing you see when you try to visualize the minecount.

Wrong:

The easy approach here is to start from that central 3, use that as a square group, and add in the others from there.  You would correctly conclude that the region contains four mines.  But doing so, you would also (wrongly) conclude that there are no definitely-safe squares.

Instead, you want to use different groupings…

Solution: By changing the direction from which we’re gathering the square groups, we can do it differently and find the one safe square.  Note that the orange group comes from the 3 at left.

Puzzle 1-6 “Zigzag”: 4 mines remaining, 4 to mark, 6 to clear (clear ‘em all!). Difficulty: 4 / 10.

Source: this game.

Analysis: This is one where the minecount really helps.  There are five-mine configurations, but they are uncertain.  But there’s only one with four.  Instead of starting with square groupings, let’s actually show this with some counterexamples first.  Let’s make our first assumption, that the square shared by the two 5s is a mine:

If we start with that one, then it makes the two squares to its left safe, the square at top-left a mine, complete the 4 below, clear the bottom-left square for the 2, and then using the projection rule (orange grouping) which proves the square at the end of that arrow is a mine, then clear what’s left, and all we have is that orange pair that has one mine, which thus has to be on the lower square… except, we’re out of mines!  We need a 5th and we only have four.  So this square cannot be a mine.

Alternatively, what if we start with the bottom-left square being a mine? (at right)  Then, we clear the square above it, mark the square next to the 4, which satisfies the upper 5 and clears the other two.  Then by the same projection logic (from the orange group), the square bottom the bottom-middle 2 is safe, the square above that is a mine… and we’ve used our four mines, without completing the orange group (which would have a mine on the upper square).  So this reaches a contradiction too!  In both cases due to minecount.

If instead we decided to just start with a square grouping, it’s not too hard to find a four-mine grouping.  Since the middle and right-hand 5s have a bit of uncertainty around them, let’s avoid them to start and focus on the other squares.  Starting from the bottom-right, we quickly get:

 or, alternatively:

The teal group follows from the orange one, serving the bottom-middle 2.  Your choice of direction for the yellow one depends on whether you’re focusing on satisfying the bottom-left 3 (yielding the right-hand layout), or the upper-left 4 (yielding the left-hand one).

Of particular importance here is that, while not all safe number squares have all their unknown squares touched by our square groups, they have enough squares touched.  That was the big failure of the attempts above.  In particular, the bottom-right 5 needs two more mines, and in our first (left-hand) layout, it is touched by exactly two squares.  By our “all groupings are correct” rule, that means those are in fact where the mines are, and the others fall into place easily.

Solution:

Puzzle 1-7 “Journey”: 7 mines remaining, 6 to mark, 9 squares to clear.  Difficulty: 5 / 10.

Comes from this game.

Analysis: So firstly, you’ve got the 4-square box at lower right, which is 50-50 from three sides (top, right and bottom).  In combination with the 2 on its left, there’s some logic that can flow from that.  The puzzle asks us to not use that to solve.  But we can still use the fact that there are (provably) two mines inside that box, out of our 7 mines remaining.

Secondly, because we have that interior 2 on the left side, backed (at its left) by an effective-1, we should remember the Hole pattern and consider the three squares to its right to be one mine, in a group.  We’ll choose this because the 4 on the other side doesn’t really lend itself well to a minecount.  We then have a choice of three ways to group the other squares, basically either horizontally or vertically from the upper-left square (is how I think about it):

The differences between the 1st and 2nd are in flipping the yellow, and in moving the orange to split that interior-2, rather than below.  The difference between the 2nd and 3rd is solely in the maroon group at lower-left.  You could also blend 1st and 3rd by having the orange shift back down from the 3rd image to match the 1st.  In all cases, we easily account for all 5 mines.

What does this mean for our mine placement?  By the “all groupings are correct” rule, we’ve actually identified five of our 9 desired squares to clear already, and logic follows from there:

 →

So that’s 8 squares cleared and five mines marked.  As a final step, we project that hole-pattern region (blue), which contains one mine, onto the 4 on the right.  The 4 has two mines marked, one safe square below, and one mine in the blue to its left - so its 4th mine has to be above.

Solution:

Puzzle 1-8 “T-Bone”: 6 mines remaining, mark ‘em all, clear ‘em all.  Difficulty: 5 / 10.

Source is this game.

Analysis: Once you’re sure there’s no logic moves remaining on a endgame situation like this, the next thing you do is start thinking about square groups and whether you can get to the minecount with known cells.  There no untouched cells here, so it looks pretty plausible.  The first thing to note is that in-line with that 5, it’s a 1v2: either there’s a mine on the shared square between the 5 and the 2, or there’s two: to the left of the 5, and below that 4 near the vertical line of mines.  We can show this with two (equally-valid) square groupings for the whole board:

(A)   (B)

These are two perfectly-valid square groupings for all 6 remaining mines.  The bottom two are trivial, as is the pink in top-right.  But the other three are flexible, and can slide between either option.  So remember our rule: All (sufficiently-marked) square groupings are correct.

Which yellow box is correct?  Both!  So the mine is on the shared square there.

Which blue box is correct?  Both!  So the mine is on the square shared by those two 3s.

Which orange box is correct?  Both!  So the mine is on the overlapping square, the top-left of the orange box in (A) and the right-hand square in the orange box in (B).

What consequence does this have for the overlap between the orange box in (A) and the blue box in (B)?  Because those overlap too, don’t they.  Why couldn’t they have a mine?  Well, all the rule really guarantees is that they have the same fate.  The squares shared there (above the central 2) either have one mine or they’re both safe.  In theory, there could be a mine at the upper-left of the blue box in (A) (to the left of the righthand 4), and not have one on the overlap between the two blue boxes, and instead it’s on one of the squares below that central column of mines, in that middle overlap area.  However, this would then demand 7 mines from us…

…and we only have 6 remaining.  So the blue-B and orange-A overlap squares (shown as a blue box above) can’t contain a mine, they have to both be safe.  This leads us to our answer.

Solution:

Puzzle 1-9 “Projector”: 4 mines remaining, 1 to mark, 4 squares to clear.  Difficulty: 6 / 10.

This comes from this classic Beginner game by FracturedAnvil.

Analysis: If we start out mine counting, it’s tempting to use that big 2 in the middle of the front to account for two mines.  It’d be easy to think,

“Well, I’ve got two mines in the yellow region, and the teal region could have either one or two.  So I can’t get to my four mines fully.  But if the teal region has two, then all four of the other squares are open, and if there’s only one in teal, then there’s one mine in those four squares, so I could just guess one of those and have only a 25% risk.”

The same could be done if we take the pairs of squares along the left side, fronted by 1s, and say “well there’s two mines in those four squares, and I don’t know how many are in range for that central 2”.  But in fact, we DO know that!

The right way to think through this is to see the similarities between that left-side front and the 2-mine extended version of the line-and-1 50-50.  The key is the “1” above that 2.  Only one of those two squares can be a mine!  So the possibilities are that on that row, there are mines in columns 1 and 4, or in columns 2 and 5.  This means there is one mine in the three squares below the 2.  In other words, either the two yellow squares or the two blue squares are mines, but no other combinations of those squares are possible.  Then the three squares below (the orange ones) contain one mine.  So those 7 squares contain three mines between them.  Therefore because we have a minecount of four, the teal region has one mine (in the middle) not two (on the edges of the teal box).  And the other unmarked squares are free.

Solution:

Puzzle 1-10 “Trunk”: 8 mines remaining, mark ‘em all, clear ‘em all.  Difficulty: 6 / 10.

Analysis: As you start poking around this one, the first thing you realize is that this is the opposite of most minecount puzzles, where you’ve got few mines and lots of places to put them - here, you have “too many” mines, and need to find a way to cram them all in.  The other problem is that the mines breaking up the linear front at the top, making some semi-isolated sections, really prevent an idealized choice of square-grouping it.  Instead, we kinda have to brute-force it a bit.  But first, let’s prove to ourselves that a minimal grouping (a provisional marking while minimizing mines) is not the way to go here.

The minimal-mines approach can start from the left - if the square shared by the 5 and the 3 is a mine, that probably minimizes our mine usage.  Then we have one to the left on the 5, on a half-full-bucket 50/50 (blue box), the other two squares above the 3 are clear, the two squares shared by the 4s are mines, and so on.  On the right, we could have one or two mines there (either the square below the 1 is a mine, or both the top and bottom are mines).  So with that 1v2, we have either 6 or 7 mines in this configuration.  But we have to place 8!  So this is clearly not going to work: we need a maximal-mines setup.

So, going maximal, our first observation is, the right side (orange box above) can contain one or two mines.  Either the square to the right of the rightmost 3 is a mine (clearing above and below), or it’s not, in which case there’s 2 mines (below, to the right of the 5, and then one of the top 2).  Since we’ve already ruled out minimal mines, we’re going to need the bigger option.  So let’s start by assuming it’s the 2-mine configuration.  Moreover, putting the upper mine at the top-right corner forces mines towards the right, giving us more space to the left, so that’s probably the maximal choice.  Then we can proceed leftwards:

 →

After we get to the leftmost 3 on our front, we no longer have a determined layout.  Instead, we have another choice.  The 3 there has one mine for two squares (blue), and the 5 to its left has two mines for three squares (yellow), and they overlap so it’s not directly determined.  However, what this choice does for us is create a 2v3 situation out of the residual area: either there’s a mine on the shared square (same as the minimal approach above), for two mines in these four squares, or the shared square is safe, and there’s one directly above the 3, and then two mines above and above-left of the 5, for three total mines in that area.  So we look back at our previous choices: we’ve marked five mines with our attempt at a maximal configuration, and we have a 2v3 choice.  But we need 8 mines, so the 2v3 is the “3” mine option.  Done.

Solution:

Puzzle 1-10(b) Bonus:

What’s really fun about this puzzle is, if we don’t know the count, it’s still (mostly) solvable!  If it were a 6-mine count, we’d have our minimal approach, from above, and just choose the “1” in the 1v2.  If it were 7 mines, however, our only two layout choices would be (A) our initial, minimal-mine approach, with the 1v2 resolving in favor of the “2”, or (B) our later, maximal-mine approach, with the 2v3 resolving in favor of the two mines.  However, in both scenarios, there are a few shared conclusions - in particular, the two mines above the 4s, safe squares to their sides, and the mine above that middle 3.  So we could just mark those mines, and then use the numbers from the revealed safe squares to determine whether we were in situation A or B.  In all minecounts, this puzzle has progress available!  There’s just one hitch: If you end up being in situation A, with minecount 7, then you end up with a 50-50 guess required for the last mine, at left.  But it’s still pretty cool that you can get that far.

Puzzle 1-11 “Maps”: 5 mines remaining, 4 to mark, 14 to clear (all but 2).  Difficulty: 6 / 10.

Source is this game. In the game, I guessed using wall-skip, but I didn’t need to (as we’ll see).

Analysis: Give it a try first, just placing hypothetical mines.  You’ll notice that the hard part of this is just finding a combination of mines that even works at all.  For example, just with the right half of the board (the two parts are isolated), I can construct a layout that uses all five mines just on that right half, plus three more on the left half:

Clearly, that isn’t going to work.  Of course, that contrived layout was chosen for maximum mines, not minimum.  But if we’re going to get to a minimum, it will require satisfying squares through the highest use of mines on shared squares as possible.  Notice which squares are the ones which satisfy multiple front-line numbers at once:

If a mine is in any one of these five teal squares, it is meeting the needs of at least two numbers at once.  These are the “pivot points” where we can either decrease our mine usage by putting mines on them, or increase our mine usage by having them be safe.  But notice that on the left side, we can’t use all three of them at once - using the middle one (with a * on it) as a mine precludes the use of the other two on that side.  So using it means it doesn’t actually save us a mine, because while we save one (on the shared square between the 3 and the 5), doing so also costs us two extra mines (by forcing the other teal squares on the left side to be safe), netting out +1.  And here, with such a low minecount, we can’t afford that.

So that leaves us with using the other four pivot squares and using them as mines.  We have a total of five mines, which leaves us with just one mine whose location is unknown - the other two squares above the central 4.  So in this case, we didn’t even need to use square groups.  But we could have!

This choice of grouping shows it pretty plainly: the blue squares need one mine, the yellow needs one, the orange needs two, and that leaves us with one between the teal and the purple.  So clearly, that one is in the shared square.  Then the choice of blue is obvious (has to satisfy the 5 at right), and same with the yellow (meet the 4 below) and half of the orange (has to help the 4s to its right).  We don’t know where the last mine is, but this lets us mark four of the five while satisfying every number that’s showing, and that’s good enough to solve the puzzle.

Solution:

Puzzle 1-12 “Forgot About Dre”: 6 mines remaining, 2 to mark, 4 to clear.  Difficulty: 7 / 10.

Source is this game.

Analysis: I love this one in particular, because it seems to break the typically-iron logic about a 4-square box that’s framed by mines/wall (i.e. that it’s a box-and-2 50-50).  Which is what we have on the lower parts of the lefthand region.

So the first and most obvious thing, besides the fact that we have two isolated regions, is that the righthand region is a 1v2 minecount.  We either have one mine on the shared square between the two 4s, or that’s one safe and the other two are mines.

Now let’s consider the lefthand region.  The key to all of this is the interior 4, which is an effective-3.  In combination with the two pairs at left, that’s minimum five mines in this region.

(how many more could we get, though?  As many as 7 →
Because we have two points of flexibility: the lower-right square here is untouched, could be a mine or not and affects nothing else, while the interior-4 and the 3 below-left of it either share one mine, or the shared square is safe and they have two mines split out, one below the 3 and the other to the lower-right of that 4.)

“Minimum five mines” for the left region sounds pretty good, because we only have a total of six mines anyway.  We can only do six if we have five on the left and one on the right.  So let’s mark the righthand region as the “one mine” option, clear those two squares, and then focus on the left region again.  We got to five mines on this option because that interior 4 counted for three of them.  But luckily for us, relying on that for our square-groups leaves us with some cells that must be safe:

The other mine we need to mark follows immediately from the adjoining 3.

Solution:

Note that because of how the lefthand region can flex to as many as 7 mines, and the righthand region is a 1v2, this board is only solvable by minecount if the count is 6 (as here) or 9 (7+2), which is a pretty big spread.  If you punch this into the analyzer, you’ll also see that with 7 mines, there’s an optimal guess with a 60% WP, and with 8 mines, there’s one with a 50% WP.

Puzzle 1-13 “Fields”: 6 mines remaining, 4 to mark, 7 to clear.  Difficulty: 8 / 10.

Source is this game.  I won the original game but didn’t see it, had to use a hint.

Analysis: So, you can twist yourself in knots trying to see logic steps here, or trying to do hypothesis testing to reach conclusions.  Far easier is a pure mine-counting approach: use the two interior 2s here as the basis for two-mine square groups each, and then the 3rd row becomes simple: the left pair has one mine, and the next group over must have the last mine on it (so there’s only one place it can be).

I included the orange and teal overlap so you can see how the 6th mine must be placed - there’s only one possibility, since “all groupings are correct”.  A few moves follow logically from there:

Next, consider the blue group.  Note that we have, in the 3s below it, a logic move available to us: the right 3 (orange) is an effective-1, and the one next to it is an effective-2, while the square to the left of that has already been proven safe.  So even without uncovering it, we know that the lower-left square in the blue square group must be one of the two mines.  We mark that, and then turn our attention to where the other mine in the blue group must be.  There’s only one mine, and the its fronts from the lower-right 3 (orange) and the rightmost 2 (purple) must both have it, therefore it’s on the shared corner.  So we can mark that too, and clear the other three squares in the blue square group, and we’re done.

Solution:

Puzzle 1-14 “Donut”: 7 mines remaining, 6 to mark, all-but-2 to clear.  Difficulty: 9 / 10.

Source game is here, from an old account of Jessi - it was posted in the WoM chatroom.  Some wise person suggested it was mine-countable, there was debate, and I don’t know if she won the game, but at least it was fun to discuss and solve.

Analysis: This is a mess to do square-groupings on.  We have to get to 7 mines, somehow, so let’s just start by taking a first stab at mutually-exclusive groups.  Some are pretty easily chosen (the 4 groups on the left side here, in particular: brown, yellow, teal via projection, and dark blue).  After a little tinkering, and some creativity on projecting a hole pattern (in black), we arrive at this:

So while we get to the black group by pulling one mine from the 2 that the arrow starts at, we don’t end up using the square group under that 2, because we can more-usefully divide that top row of cells into two mines (shown in blue and orange) rather than just one.  But it helps us get the 7th mine included, which means we’ve counted the whole thing.  This only gets us two safe squares (the two cells below the black group), but it’s a start.

For the next step, I’m sure there’s a better way than what I’m about to do - so if you know it, please comment! - but when all else fails I always go with “guess and check”, and we can show that certain configurations are impossible (we run out of mines) and then reason from there.  

The most straightforward of these is to focus on the rightmost column of unknown cells, where between the 4 and the 3, we have sort of a 1v2: we can put a mine on the shared square, or we can mark that safe and put in two, one in the upper-right corner and one left of the 3.  Because we feel a little mine-constrained, let’s imagine it’s “1”, that the shared square is a mine.  What happens then?

 →

At left, the mine choice allows us to clear the 4 and the 3, shrinking the orange group from three cells below the top-right 2, to two cells.  But it still has a mine!  And, crucially, as we see from the arrows in the second picture, that central 2 now has its mines accounted for by (1) our assumption, and (2) the orange group.  Which implies the two cells below it are safe as well.  But that’s a contradiction, because we have a 1 below that on the bottom row, and from our logic, now all five of its surrounding cells would have to be safe - but it needs a mine!  So that configuration is impossible.  Therefore, the 1v2 is the “2” option: the mines are immediately to the left of the right-side 4 and 3.  That has several consequences that follow:

 →

At left: The determination of those two mines at right clears the two other cells below the top-right 2 (the former orange group), and it also satisfies the bottom-right 1 (pink arrow), enabling clearing the other two squares down there.  (BTW: Why does the mine next to the lower-right 3 need to be to the 3’s left, rather than lower-left?  Because otherwise we have nowhere for the rightmost 2 along the interior line of numbers to get its 2nd mine from.)

Then, at right, we need a mine in the top-center to satisfy the left 2 along the top row (above the central 3), which clears the cell below the 4.  But also, that left 2 on the middle row needs one more mine, and only has one more option for it, so the cell below the central 3 is a mine as well (curved arrow).

Next, we note that the central 3 needs a 3rd mine, and only has one option: the top-right of the teal group.  Which in turn satisfies the central 4, forcing a mine onto the left side of the yellow group (below the 5).  Which brings us to our solution.

Solution:

Being able to draw the logical conclusions here from our choice of scenario is pretty easy.  Spotting a contradiction once you’re playing out a scenario is doable for most advanced players.  But knowing which cell to hypothesize about - that is more art than science, and is what is actually hard here.  The more reps you get at that, the easier not just mine-counting situations will be for you, but more-advanced midgame situations and guessing choices will be easier too.

Puzzle 1-15 “Waltz”: 6 mines remaining, 2 to mark, 2 to clear.  Difficulty: 9 / 10.

Source is this expert game.  I… did not see the solution, and guessed poorly.

Analysis: This one is real tough, because you can stare at it forever trying to come up with a valid square grouping for six mines, and come up empty.  It’s solvable using minecount, but not just minecount.  As you stare at it, you’ll come up with a few candidate square groupings, none of which fully mark out the area without overlapping.  Some of the groups you’ll see include:

(A)    (B)

Neither is a fully valid grouping, because in (A) we’ve only called out 4 mines, and in (B) we have this overlap of the yellow and teal, where we don’t know (yet) if their shared square is a mine or not.  But this is the beginnings of a solution, believe it or not, if we combine elements of both of these groupings.  To start with, note in the (B) candidate grouping that the yellow and the teal, together, form a 2v3 minecount split: either the shared square is a mine (satisfying the 2 that abuts the yellow, and leaving needing 1 mine to satisfy the 5 from the lower-2 squares in the teal), or the shared share is not a mine and the other 3 squares there are all mines.  So let’s look at it this way, isolating that righthand column from the others:

Amazingly, this grouping actually accounts for all six mines in a mutually-exclusive manner!  It requires that we opt for the “2” out of the 2v3 split.  (it also requires that we just ignore that interior 3, despite it kinda calling to us).  But having done so, we now have our conclusions.  Firstly, the lower-left square is safe, therefore the one to its right is a mine.  And secondly, because the 2v3 is actually two mines, the square shared by the 2 and the 5 has to be a mine; and while we don’t know which one below it is the other mine for the 5 (yet), we do know the one above it, the other half of our original yellow group, is safe.  So that’s our answer.

Solution:

Puzzle 1-16 “Constellation”: 19 mines remaining, 15 to mark, all-but-6 cells to clear.  Difficulty: 10 / 10.

This is the minecount endgame from Scar’s world-record 4-million-difficulty game (at 1:08:00).

Analysis: Given the context of the game and moment that it’s from, this board is possibly the most famous and significant minecount situation in the history of Minesweeper Online.  It would be hard to top it for importance, but frankly it’s very hard to top in terms of sheer scale, too (19 mines is A LOT for mine-counting).  Let’s start with a few remarks, things to take note of:

1. We clearly have to deal with the two isolated regions separately.
2. We don’t need to mark all 19 mines (just a mere 15 of them!).
3. This is a very thick minecounting region, 4 columns wide.  So because we’re told it’s mine-countable, and we don’t have nearly enough mines to stuff the interior there full of mines, we have to assume a minimal count, i.e. that every untouched square is safe - one at the top, six at the bottom.  We’ll use a circular green dot on those.
4. There are four mines on the bottom region that are fully isolated from the rest: the interior 4 at mid-left, which has three, and the one to its left up against the wall.  That area will always have four mines, and nothing else can push an extra into it, rely on any mine from it (no shared squares), etc.  So that part’s in the bank for four mines.  Because of its isolation, we can conclude that those are the mines we can’t determine at this mine-counting step, and it’s all the rest that make up the 15 we need to find.

OK so let’s start with the upper region.  We don’t have many degrees of freedom here, but we do have two: the lower-left corner (which is sort of a 1v2), and the interior 5, which has three unknown squares around it - rather than the usual two for this sort of chain.  Let’s make that 5 be our pivot point, and play out the 3 layouts based on that:

The teal-ringed square is our assumption square; in each case it fixes the corner to the left of it, and out to the right as well.  But notice that only in the left layout (layout A) do we have safe squares on our outer edges and four mines total in the region; in layouts B and C, we have mines on the outermost squares (top and right sides), and five total mines.  So if we need four mines from this area we’re good, if we need five mines that’s OK too - we can distinguish layout B from layout C by using the free square, which will turn up either a 6 or 7 depending on the true layout.  But we have either four or five coming from the region.

The bottom region is big, but we have a few simplifying things we can do to help: first is that four-mine isolated area that we don’t want to deal with (in purple).  Second are our free untouched squares, marked by the green circles.  Third is that, although we have to pick two pivot squares - and I’m going to suggest we come in from the top and from the mid-left - but based on those, we can play out the assumptions and fill the whole area and see what we got.

Let’s start with an attempt at a minimalist count layout.  This means, at top, marking the two corner mines for the 4 on the edges, and propagating down from there to fill the row of 3s.  And from the left, we mark the shared square on the 5 and 2, which then propagates right and forces a mine at the end of that orange arrow:

(A)

Then a cool thing happens: the rest of the area follows directly.  That mine at the end of the lower orange arrow is the 3rd mine for the 3 above-right, which clears below it (blue arrow), which forces a mine on the right, a safe cell below the right 4, which then forces the other three mines for the interior 5 there to be on the shared squares with the interior 4 opposite it (yellow arrow).  This then fills in the bottom part.  And excluding our purple area, this means we’ve marked exactly 11 mines in this region.  Awesome!  Combined with the low-count four mines in the upper region, that makes 15, which is what we have to solve.  So this is a valid solution.  But is it the only solution?

To find out, let’s next attempt a maximalist count layout, reversing our assumptions and trying to cram as many mines in here as possible.  We’d start by flipping the assumed mine at mid-left, and also making an adjustment up top.  There’s two switches we could make to the top-right 4: either the top-left assumed mine flips to the top-left square, and we put another mine in the open square below it (which would keep the same flow down the column of 3s), or we flip the lower assumed mine inside, make the column of 3s change, and see where it leads us.  The second one sounds more chaotic, so let’s do that. 🙂

As you can see from the resulting layout, we end up with two square groups (in the red outlines), neither of which are forced.  But we’re doing a maximal approach here, which means keeping mines leaning away from our working area.  So let’s further assume that the mines are on the upper and left-side squares of these groups (mines framed in yellow), and continue.

Our next choice is how to populate this area.  My first instinct would be, we’ve got that 4-square box at bottom with a 50-50 front from only two sides; the right side could be mines and the top side could be anything.  In theory, I bet we could make that box carry three mines!  What happens when we do that?  (assumed mines from this step here framed in orange)

Well, what happens is, we quickly see that our next move is that the (remaining) shared squares between the interior 4 and the 5 need to contain one mine between them.  The 4 is now an effective-1 and the 5 an effective-2, so those shared squares cannot contain two or zero mines, they have to contain one.  This means (orange arrows) that the square above the 4 is safe, the square above the 5 is a mine, so the square above that is safe.  And then we reach a contradiction: the blue-framed 3 needs its third mine from one of the two squares below it, but the 3 to its right wants to place it below-right, and the yellow-framed 2 below still needs its 2nd mine, which can only be directly below the blue-framed 3, and that blue-framed 3 can’t have both.  So, contradiction, this layout doesn’t work.

OK let’s back up, maybe we can’t cram three mines in that 4-square box.  What if we put a mine to the right of the yellow-framed 2?

If so, we need a mine on the shared square of the two 3s on top here, which puts a mine above the 5, which means the 4 and 5 share two mines out of those three squares, which then implies the square below the 4 is safe, and we fill in the rest.  This works!  And the total minecount we’ve placed is then… 13.  Call this Layout B.

OK.  We’ve got two possibilities left to explore.  Firstly, what happens if the yellow-framed 2 and the blue-framed 3 have their needed mine on the shared (rightward) square between them?  Well, then the layout for this lower part of the region ends up looking just like Layout A above, except for a mine on that one square.  This then has a total minecount of 13 too.

(C) (D)

Layout C has the same lower-right area as Layout A, except we were able to cram two more mines into the lower area based on our upper area choices.  Layout D comes from taking that corner shared by the yellow-framed 2 and the blue-framed 3 and opting for two mines instead of one; it implies a mine above the 5, meaning the 5 needs two more from its shared squares with the 4, which is fine but that satisfies the 4, so the square below it is safe.  So D, too, totals 13.

OK so we have multiple ways to make 13 mines.  Can we make 11 a different way, though?  The only remaining plausible variation to try is to revert to Layout A’s upper assumptions, but keep the flipped assumption at left from Layout B.  Let’s see what happens:

(E)

We have to keep the mid-left teal-framed mine, which then implies a square group on the row of 2s to its right.  We’re now trying to keep mines minimal, so we will assume one on the right-side square - framed here in yellow.  Based on that, it satisfies the yellow-framed 2 to its lower-right (orange arrow), which clears the other squares around it, which forces a mine to the lower-right of the blue-framed 3 (blue arrow), which ultimately forces two mines on the shared squares between the 4 and the 5.  Then the 4th mine that the 4 needs has to be below, so that corner has two mines.  This is a total minecount of 12, for Layout E (above).

I think at this point we’ve tried everything, so we’re forced to conclude that Layout A is the only choice that has 11 mines.

Solution: We need 15 mines total, so we have to take the (singular) 4-mine layout for the upper region, and the (singular) 11-mine Layout A for the lower region together.  And incredibly, unbelievably, they uniquely solve this puzzle, leaving only the four mines in the purple area to be solved post-minecount.

We can also see here that because of our safe squares, the remaining 4-mine area that can’t be mine-counted will nevertheless be solvable, because all potential configurations will still fix whether there’s a mine above the interior-4 on that area, and then the squares to its left can be determined by the numbers.  Giving Scar his win on a 4-million-difficulty board.

(2) Assume Fewer Mines

Link to discussion section.  Link to puzzle section.

Puzzle 2-1 “Depths”: Find the better move(s).  Difficulty: 1 / 10.

Analysis: This is our very basic “1v2 around a corner” situation.  What’s down below is not something we can draw strong conclusions about.  But up top, there’s either a mine on the shared square on the corner, or if that’s safe, then there’s two mines to its left and its bottom.  One mine is more likely than two mines, so that’s our answer.

It’s worth noting, however, that the two-flag separation on the right side of the front means there’s no information shared with what’s below.  If that wasn’t the case, then these odds might not be like this, they could be worse, or there could be logic moves available to us.  It’s that two-mine wall that provides the independence for our reasoning here.

Solution:

Puzzle 2-2 “Pairwise”: Find the better move(s).  Difficulty: 3 / 10.

Analysis: We have three effects here in this little region, two of them directly from our discussion above, and all of them helpfully compounding in their effect on the probabilities.  Let’s consider this region as being three sections: the one at left (first 4 columns), the middle section below the 2-3 along the top, and then the one at right with the corner and the 4s there.

Firstly and most clearly, we have a “flat framed front” in the middle section along the top, the 2-3 squares (Both effective-1s).  Our rule says that there is likelier to be one mine in the shared squares between them, and less likely to be two mines on the respective outsides.

Secondly, if you look at the left section a bit, you’ll notice that it’s a 1v2: either the mine is above the 2 (satisfying both 4s above it), or it’s not, in which case we need a mine above the 1 and above-right of the 3.  The one-mine scenario is likelier.  So that’s two separate reasons to think that it’s likely that the square shared between the 4 and the 2 is safe.  In combination, these two regions are really a “2v3”, as follows: two mines in yellow, three in blue.

Thirdly, although we’ll generalize this later in the guide, notice that if we have our “more mines” scenario, it also fixes (mandates) the location of the mines at the right section.  Because if there’s a mine to the lower-right of that 3 along the top, then the 4 to its right is satisfied, which also requires the mine from the far right to be lower, next to the 2.  Because such an assumption would require that several other mines be placed just-so, as a consequence, it is less likely than other assumptions that would preserve more possibilities for mine locations.  So that adds to the probability that the “3-mine” scenario above (in blue) is not the case.

Solution: Make the assumption of fewer mines among the left two sections.

Puzzle 2-3 “Minnesota”: Find at least 2 independent good next moves.  Difficulty: 3 / 10.

Source here (see hint at 5:06, or stop it right at 5:30 to see).

Analysis: We only have three independent parts of the front here, and the one at the bottom is an ordinary 50-50 front, so that’s not it.

The middle portion with the two 3s (effective 1s) is a “flat framed front”.  Either there’s one mine directly to the left of one of the 3s, or there are two mines on the wide spots above-left and below-left of them.  One mine is likelier, so the wide squares there are good bets.

At top, it’s more subtle, but only slightly.  Instead of an ordinary corner where a 1v2 is obvious, that corner continues to the left and it might not be obvious at first glance that there isn’t conflicting information coming from that 4.  But because the 4-2 bit implies one mine below the 2, and one mine left of the 4, those two placements are independent of each other.  Therefore the extended area around the corner is a 2v3, and the “2” option implies a mine on the corner.

Solution: Good guesses are as-marked below.  The ones around the flat framed front are to be preferred here, partly for the lower odds but partly to “prevent” the 50-50 at bottom.

Puzzle 2-4 “Box-Curious”: Find the better move(s).  Difficulty: 4 / 10.  Source game.

 → Labeled:

Analysis: So first let’s label our frontline squares (above), to make this make sense better.

The first important observation here is that ABDE is a (partially-framed) 4-square box with two mines on a diagonal.  How do we know this?  Start from the right side: the 4 (facing F/G) takes one of the two mines from its neighboring 3, meaning that from the perspective of that middle 3, D/E has one mine - it’s a 50-50 front.  Then from above (A/D), and from left on that 4 (A/B), we have two more fronts with one mine on its two adjacencies.  So a mine on A implies a mine on E, and a mine on B implies a mine on D.

Secondly, while the choice of mine for F/G is independent of the rest, that’s not the case for the lower-left: C is a mine if A is, and is safe if A is safe. A-C-E are all one dependency chain.

Therefore, for those five squares at left, there are two configurations, and they aren’t equal: we can have mines on A-C-E, or mines on B-D.  For there to be a mine on A, C, or E, it would force one extra mine out of the “pool” and fix its location.  Instead, we should “assume fewer”: the likelier configuration is that the mines are on B and D.

Solution: The top-left or bottom-left squares are the ones to guess.

Puzzle 2-5 “Induction”: Find the better move(s).  Difficulty: 4 / 10.  Source game.

Analysis: This puzzle asks two things of the viewer.  Firstly, to realize why the obvious framed corner there is not a good “corner 1v2” guess.  And then secondly, to spot a common and favorable layout, which is disguised a bit.

For that first observation, it should be obvious, at least, that that upper-left corner is not isolated.  Against both of its wide squares is an interior 3 that is an effective-2, implying 40% local density.  As such, the wide squares start from a presumption of being 40% odds, not background rate (here B=17%).  But more importantly, because the choice of mine location in that upper-left box affects those adjacent 3s, it is not a cleanly isolated corner, nor does it have linear extensions that just have ordinary two-way dependence.  There are not merely two layouts for that box, one with a mine in the corner and one with mines on its wide squares; in fact there are many, for both those choices, and it is difficult to see any clean ratio between the two groups.  So Obelus’s Principle does not apply there.  (And in fact, the ratios in that box are closer to 50-50 than to B vs (1-B), because those adjacencies are eff-2s.  If they were both 2s that were eff-1s, it might well be different.)

Our eyes might next go to the lower left portion, where we have what could, in theory, be a linear 1v2: either one mine above-right of the 4, or that square safe and two mines placed: below-right of the 4 and below the left interior 3.  But that suffers from the same problem: That adjacent left-interior 3 needs two mines from five squares.  If you fix one safe square (via the supposed one-mine layout on the linear stretch below it), it then needs two mines from four squares, for which there are six possible arrangements, locally.  If you fix one mine (via the supposed two-mines layout below), then it needs one more mine from four squares, for four possible layouts.  So while the fewer-mines choice below is still more likely (66%) than the more-mines one (34%), they are not favorable guessing odds, and again it’s because the requirements of Obelus’s Principle - having the same number of possible layouts which use N mines and which use N+1 - are not met.  So there is no background-rate guess here, either.

Instead, we need to look at the upper-right portion and see that it is inducing a flat-framed front (hence the problem name, induction).  Recall that the definition requirements for the FFF are that it needs a two-square front which is open on both ends, i.e. two number squares facing four front squares, generally with mines on each side of its front.  But a square group (mine location as-yet unknown) can serve to provide the same effect on the front there as a mine, without being a mine.  We just need to be left with a four-square front that can contain either one or two mines.  See it labeled here:

The group AB here provides one mine for the corner 2, which then needs one more mine from CDE.  Likewise, the 3 above it needs one mine from DEF.  On that full CDEF front, there are two possible one-mine layouts (a mine on either D or E), and one possible two-mine layout (mines on both C and F).  So it actually meets all the requirements for a flat-framed front!  Moreover, the interior 3 next to A does not present an obstacle like it does for the other discarded candidate solutions here, because regardless of how many mines are in that vicinity or where they are arranged, there is still one and only one mine in the group AB.  So the front CDEF is truly isolated from squares aside from it; the AB group does not influence square C at all.  And as an FFF, it enjoys the usual below-background-rate odds, making it the best guess.

Solution:

Puzzle 2-6 “Spillway”: Find a good move.  Difficulty: 4 / 10.  Bonus: Find a 2nd good move.

Source game - an intermediate game, surprisingly.  Big guessing regions in int are fairly rare.

Analysis: Our first observation is, this whole front is connected, nothing is independent.  As a result, the upper-left corner here does not provide the usual joy, because a mine on the corner leaves a mine required for the 4 below it, whereas the corner itself being safe means a mine on the shared square between the two 4s, satisfying that lower 4.  So the two configurations for this area both involve 2 mines, there’s no 1v2 reasoning we can apply.

Layout A:   Layout B:

Layout B definitely contains 2 mines, while Layout A has at least 2; we’d have to then go look at the areas outside this sub-region to see if it influences mine placement elsewhere.  It gets complicated.  But one thing’s for sure: a mine on the corner does not reduce the number of mines needed on the front.  No easy corner 1v2 for us!

Instead, let’s consider the top-right area.  Just taking the top row of numbers, we can quickly see that the 2-1-1 front there (above the interior-3) can be satisfied with either 1 mine (below the middle 1) or 2 mines (below-left of the 2 and the other on the edge).  So this is a classic linear 1v2.  But wait, you might say - it’s not fully independent!  And yes, that’s true - to a point.  But it’s an answerable objection.  In this case, if we have our 2-mine scenario (mine on the edge, mine below-left of the top 2), it propagates out around the front and determines that there’s a mine in the upper-left corner, which we’ve already determined has no fewer mines than the alternative.  So making our “2-mine” assumption of just the top-right corner does not result in fewer mines elsewhere (which would reduce the probability edge that our rule gives us).  Instead, it leaves us with the usual advantage: guess the 1-mine scenario and you get a background-rate guess.

Now for the Bonus… it’s all about the lower-left front here.  Firstly, there is one square where, if it’s a mine, it determines (fixes in place) the location of two other mines.  Can you find it?  No other square would determine more than one downstream mine (And some wouldn’t determine any!).  Let’s use the 2 below the 4 as our pivot square, and consider the implications of each potential option:

(A)   (B)   (C)

Making the assumption (teal-framed mine) leads us to mark 0, 2 and 1 further mines, respectively.  Also, option C provides the 4th mine for the 4 above this area, leading to the 1-mine option for the upper-left corner (Layout A, above).  Therefore option B there is much less likely than the others, because it being a mine requires definite positions for at least two other mines, which eliminates a lot of possible configurations.  So the teal-marked square in option B here is therefore the least likely square to be a mine on this board, and the best answer.  We cover this technique in a later chapter, but if you can spot it here now, big kudos to you.

There is a further answer available, too.  Note that among these 3 configurations above, in two of them, we end up with a 1v2 at the very bottom (implying that the upper-left corner of that box is likelier to be a mine), and in the third one, the upper-left corner of the box is definitely a mine, and the square below it definitely safe.  Even without option B here, the lower-left square would be a background-rate guess; including option B as a possibility, even if only a remote one, reduces the chances for that bottom-left square even lower (a bit below the background rate).

Solution: The most obvious answer is the 1v2 at top-right, but two bonus options are available in the lower-left, as marked here in green.

As it happens, the good guess at top-right did not lead to much further progress, but the one at lower-left did, and eventually let the whole region be solved.

Puzzle 2-7 “Double Duty”: Find at least 2 independent good next moves.  Difficulty: 5 / 10.

Source game.

Analysis: Scanning the perimeter here, there’s basically four regions, all independent of each other: the upper left, the lower right, the upper right, and the middle region (the 4-3-4) along the lower-left curve.  None of them share squares, so we can discuss them entirely separately.

Starting upper left, the thing we’ll notice is that the very edge is nearly a line-and-1 50-50.  For that reason alone, guessing the square on the 3rd row (at the end of the row of marked mines) might be a worthy guess, as we’ll discuss in the “Helpful Guesses” chapter.  But there’s a good reason to prefer that square anyway: right below it is a “Flat Framed Front”, the two 3s there being effective-1s with mines at top and bottom.  So it’s far likelier that their third mine is on one of the two shared (middle) squares, and not at the edges where it would require two mines.  So above and below those 3s, that’s a pair of good guesses (one of them dependent on the other).

Next, at lower right, we have an obvious 1v2 situation, on either side of that interior 4.  Either the mine is to the right of the 4, or there’s mines on the right wall and to the left of the 4.  The “1” scenario is far likelier, so the wide squares are good (dependent) guesses.

At top-right, we have a trickier scenario to spot.  Although we won’t cover “lock-in” until Chapter 10, the situation here intertwines that with “assume fewer”.  Consider what a mine in either of these two situations would mean for the middle 4 in this formation, initial assumption marked with a teal framing:

(A)   (B)

In (A), if we assume the teal-framed square is a mine, then the one to its right is safe (as the 3rd mine is above in the yellow group), and therefore the 4 needs both of its other squares below to be mines.  In (B), assuming the teal-framed square is a mine, implies the two squares to the left of the 4 are mines.  In both cases, one assumption “locks in” 2 other mines, fixes their location.  This makes those squares both way less likely to be mines (in fact, lower than the background rate).  The two are not linked, in the sense that one being safe implies the other is safe, necessarily.  But if one of them is safe, then it does create a 1v2 “assume fewer” situation with the other one:

(C)  (D)

If the teal-framed square is safe, then the adjacent is a mine, and the 4 creates a residual 1v2.  In C, we would have a linear 1v2 (like the bottom-right area in the full region) below the 4, and in D, we would have a corner 1v2 (with the 1-mine scenario being the interior corner, just to the left of the 4).  In both cases, the “1 mine” option is in blue and the “2 mine” option in pink.  So either one of these two wide squares being safe makes the other one very likely to be safe, though not certain.  But either one is a good guess.

Finally, the lower-left-ish curve-y part of the region, in theory has one good-guess square, but in practice that one is very hard to find (I couldn’t tell you how to spot it, frankly) and also unlikely to lead to progress even if it’s safe.  So I wouldn’t consider anything there to be a good guess.

Solution: The best moves are around the flat-framed front, and the two fronts along the right wall.  Primary guesses are in green, and dependent guesses are in yellow.  My choices would be firstly the ones at top-left, and then after that the ones along the wall at right, as both are likelier to lead to progress than the interior ones.

In the actual source game, if you make all of these guesses (including the residual 1v2 that occurs after guessing at top-right), the entire region becomes fully solvable.  Which is nice.

Puzzle 2-8 “Whale”: Find 3 independent good next moves.  Difficulty 5 / 10.  Source game.

Analysis: This one is chock full of 1v2 corner-ish situations.  Perhaps easiest if I just show the triangles that illustrate the 1v2 nature:

The yellow group, I trust is obvious: one mine on the corner, or two on the wide squares.

That the orange group is independent from the blue group, however, is less obvious.  Start by considering the two 4s below the blue group.  The right 4 says there’s one mine in the two squares above it (horizontal blue); the left 4 says, residually, that there is one mine in the two squares to its left (vertical orange).  But the choice of one would not fix the placement of the other!  They can exist in any combination.  However, in combination with the 3-2 front (the horizontal orange), it, too, is a 1v2.

The upper right front (dotted-line blue) is dependent on the main blue 1v2 corner below it.  The choice of the latter determines the former, so that isn’t part of the answer.  The answer is, assume the corners of these triangles above are the mines, and guess wide squares around them.  In this case, you’d be right 2 out of 3 times!

Solution: The best guesses are the squares in green, as-marked.

Careful observers will see the 9% square above that middle protrusion.  That’s due to the lock-in effect discussed in Technique #10 (if that square were a mine, then it would force 2 mines to the left of the interior 3).  It’s also highly unlikely to lead to progress, because we don’t have an ability to see that front-line 2 below the 9% square from the bottom, so even if it turns up a 1 we likely can’t clear anything else.  This illustrates one of the reasons why using the assume-fewer technique is better than guessing an edge among untouched squares: you have a high likelihood of immediately having an available next move, because you already have information from the squares behind where you’re guessing.

Puzzle 2-9 “Swingline”: Find a set of good moves.  Difficulty: 5 / 10.  Source game.

 → Labeled:

Analysis: So firstly, observe that using projection from below, the blue group above the 2 is independent: its squares will be 33% each (for now), because they are only touched by that 2.  It’s the other six squares that are more interesting.

On those squares, we would first presume that ABC is a 1v2 corner: A and C have equal status, if B is safe then they’re mines, if B is a mine then they’re safe.  But a mine on C will also clear D and E, leaving a mine on F.  Whereas a mine on B will leave A and C safe, and therefore also make F safe, again by projection: there must be one mine on D or E (per the middle 1), therefore whichever it is, F is safe.

This means that we have three possible mine layouts: A-C-F, B-D, or B-E.  The two-mine layouts are more likely than the three-mine layout, in B : (1-B) ratios, but there are also two layouts for the fewer-mines option.  By the Ratio Rule, these configurations will occur in a ratio of B : (1-B) : (1-B), and so if B is ~20% then our (single) more-mines layout has odds of only 1 / (1+4+4) = ~11%.  So the squares in that lone three-mines layout must be great guesses!

Solution: …and indeed they are:

Puzzle 2-10 “Wild West”: Find at least 2 independent good next moves.  Difficulty: 5 / 10.

Source game, of which this is only part of it.

Analysis: Opportunity abounds here across the front, if you’re looking at it right.  There are four, arguably five, independent(-ish) areas here we can focus on.  Clockwise from top right:

(A) The upper pair of 3s at top-right is a flat framed front.  So the wide squares are a favorable guess.  Furthermore, the lower square is useful, too: if it’s safe, then the 3s below it aren’t a 50-50 forced guess.

(B) The lower pair of 3s, recessed in a little bit, are a 50-50 front with not much to do there, except hope that the square to the left of the upper row of mines is safe, as noted above.

(C) The area above the two middle 4s is a 50-50 front for the moment, although it’s (slightly) connected to the part to the left via that interior 3.

(D) The 5-square linear front above that 3-1-2, just to the left of that interior 3, is an interesting case - the very reason this was selected as a puzzle.  In that five-square front, there are three possibilities for mine arrangement:

Mines could be on cells A and D, on cells B and E, or on cell C alone.  If this sounds like a 1v2 - you’re right!  The choice of “there’s a mine on cell C” is likelier than the others, for just that reason.  However, because there are more possible mine configurations for the 2-mine scenario, the odds of a 2-mine option are slightly higher than the background rate (here, 30% rather than 22%).  But that’s across both options (A/D and B/E).  How can we choose between them?  Well, notice that for the B/E option, the mine on E would satisfy that interior 3.  And if that 3 is satisfied, then it fixes a mine location above the 4s to the right (above the right-most 4).  That “lock-in” effect is enough to make the odds of a B/E configuration a little less likely than the A/D configuration, and therefore B/E is a good guess.

(E) At the lower-left is a long linear front, one which is not obviously an Assume Fewer type situation.  But consider the 2 just below the upper 3 (adjacent to D/E/F, labeled below).

That square offers 3 possibilities: D, E and F.  First, if there’s a mine on E or F, then it satisfies the 3 above it, clearing G and H in both cases.  A mine on F clears D and E and puts a mine on C and then A.  A mine on E clears D, F and also C, putting a mine on B and clearing A.  And a mine on D clears C, E and F, putting the mine in on B again.  So our 3 possibilities here for this front are mines on:

(A-C-F)

(B-E)

(B-D-G/H) (One of either G or H; they’re both safe on the other two)

This means there’s a fewer-mines scenario: The B-E option.  But notice in particular that B is a mine not just in the fewer-mines scenario, (B-E), but also in two of the more-mines scenarios, (B-D-G) and (B-D-H).  This makes B very likely to be a mine, and dependent squares A and C correspondingly likely to be safe, in fact lower than the background rate.

Solution: The best guesses are the very lower-left, the shared square to the left of the interior 3, or the wide squares on the flat framed front at top-right.  Any of these are good choices:

(note the probabilities above that interior 3, too: 14%s.  The shared square with that 50-50 front above the 4s means that it takes an extra share of the probability from the other squares around that 3, pulling odds towards it, and making the whole area a good candidate for progress).

Puzzle 2-11 “Clockwork”: Find the better move(s).  Difficulty: 6 / 10.  Source is here.

 → Labeled:

Analysis: There’s only three untouched squares here, but the action really isn’t on those left-hand squares with the interior 3.  One of the 3’s mines is on the two squares to its right (on D/E, per that 4 to its right), and the other is among the four other squares, with no other sources of information (so they’re 25% each).  That’s nice to know, but not helpful to making progress.  We’ve labeled the others, pretty much going clockwise, since those are our consideration set.

So one temptation here is to look at D/E/F and say “well this group can solve for the two 4s at right either through one mine (on E) or two mines (on D/F), therefore we should prefer the E assumption (and guess D/F).  But this region is very connected to the areas at top and bottom!  We need to dig further before we can judge that properly.

For one thing, we can quickly disprove that D/E/F is a 1v2 situation.  Consider the 3-4 squares just below it.  Based on the 4, one of H/J is a mine, and therefore the remaining mine for the 3 above it is within G/F.  So if F is a mine, G is safe, and vice-versa.  So D/E/F is really “D/E/F/G”, and the mines are either on D/F or E/G, with neither being any likelier than the other based on this logic alone.

But we’re not done.  Note that a mine on D forces C to be safe, and a mine on C forces D to be safe.  That section is really an extension of the A/B/C corner at top!  So let’s look at our 1v2 corner up there: if A is a mine, then B is safe, C is a mine, D is safe, as is F, and E/G are mines.  Call that the A/C/E/G mine option.  Alternatively, if A is safe, then we have mines on B/D/F, and the other 4 are safe.  So this whole thing is a 3v4!  But the likelier (fewer-mines) option is that the mines are actually on D/F, the opposite of the logic you’d come to if you just were considering the D/E/F frontline.

Likewise, because of the logic on that lower 3-4, the squares above the 3 (F/G) are independent of the ones below it (H/J/K).  Whether the mine is on J, or on H/K, the 3 is still getting only one mine from its left and the remaining one is above it on G/F.  So despite having shared squares and no mines interspersed, those sections are independent.  And H/J/K, if isolated, is pretty obviously a 1v2 situation, with the likelier result being that the mine is on J.

Solution: The interior corners at top and bottom (B & J) are best assumed to be mines, and therefore the squares around them (A / C / H / K) are the best guesses.

Guessing E would be equivalent, of course, but is less likely to result in progress than clearing A and C, which you’d just do immediately after.

Puzzle 2-12 “Staircase”: Find at least 2 independent good moves.  Difficulty: 6 / 10.  Source.

Analysis: As with “Double Duty” above, what we have here is a front with four already-isolated, independent areas.  Top left, bottom left, top right, and middle-right (the 3-4).  We’re asked to find two good moves, although three of these four areas yield decent options.

Firstly (and I hope obviously, at this point), the top-right is a flat framed front.  So that’s an easy one: the wide squares are good, low-probability guesses, while the middle ones directly facing the 2s there are much likelier to be a mine.

Secondly, the top-left doesn’t yield anything useful.  The interior 3, suggesting there’s 3 mines in the surrounding 7 squares, has a local density of 3/7 = 43%.  Although this tends to pull some of the probability of the neighboring 50-50 fronts (from the corner) towards the 3, towards the interior, it doesn’t reduce the odds to a useful point.  If this interior corner were a 1, of course, we’d know what to do.

Third, in bottom left, that interior 3 is an effective-1, and it neighbors a 50-50 front to its left.  It may not look obvious, but this is actually a 1v2: either the mine is in between the 3s, satisfying all nearby numbers and opening a bunch of safe squares, or it’s above the 1 on the wall, which then requires another mine in one of the three squares above the interior 3.  So any of those squares above the 3 are good choices, though the middle one has the best odds of progress.

Finally, there’s that 3-4 section at middle-right, and this is more subtle so let’s label that one:

Pause and think for a second about what possible mine layouts there are for these five front squares.  The labeling might help; for example, one valid layout is mines on (A,C).  One thing that will quickly become apparent is that one certain square is almost never a mine.

There’s another (non-brute-force) way to think about it, though.  Think about the squares above the 4, A and B.  Clearly, between them, they can’t contain zero mines; that would force the 4’s other 2 mines into C and D, which would overload the 3.  They can contain 2 mines; this means C and D are both safe, and E is the 3rd mine for the 3.  Or they can contain 1 mine between them, which means one other mine is on C/D and E is safe.

So for E to be a mine, it requires two things that decrease the odds of that happening: both A and B have to be mines (fixing it in place; see the “Lock-In” chapter).  But also, that just increases the number of mines we need: 3 (via the A-B-E combo) rather than 2.  The combination of the two effects means cell E is the least likely cell on the board to be a mine.

Solution: So your best bets are the wide squares around the top-right FFF, the cell below-left of the interior 3 on the right side, and the cells above the interior 3 at lower-left.

Puzzle 2-13 “Upside Down”: 12 mines remaining.  Find the better move(s).  Difficulty: 7 / 10.

From this game.

Analysis: There’s a hint in the puzzle name, because the background rate here (which applies only to the five untouched cells) is 66%!  How can you see this?  An experienced player, faced with this board, might first ask “what can I do with mine counting?”, just to get a rough idea of density, and maybe get lucky with it being mine-countable.  And you’d get 5-6 mines on the frontline of the top region, and 2-4 mines on the frontline of the bottom region, so 7-10 mines there across all fronts, and therefore between 2-5 mines on the five untouched squares (three at top, two at bottom).  So even a rough approximation (3.5 mines on fiveb squares?) would tell you this is a 60-70% background rate.  That means our usual “assume fewer” logic is backwards - here is the rare situation where we can’t assume fewer, we need to “assume more”!

Let’s start with the lower-right region.  Although we don’t know how many mines are here, we do happen to have two 1v2 situations:

The orange squares are a standard “1v2 around a corner” type of situation.  And then, because the left-hand two orange squares account for one of corner 3’s remaining mines, the other must be in either the righthand yellow or the blue square.  But, per the squares to its left, this means we have a 1v2 here as well: either the yellow squares are mines, or the blue one is.  Normally, this would mean, assume the mines are on the blue square and on the corner of the orange squares - so guess one of the others.  But here, because of the super-high background rate, our favorable guesses are actually the opposite: guess the blue square, and guess the corner orange square.  Those guesses at least get us only a 30-40% chance of losing.

Can we do better than that on the top region?  Let’s first notice that the 6 is an effective-3, i.e. three mines in its remaining four adjacencies.  That’s tough, but here it’s actually somewhat helpful.  Absent any external inputs, that means a ~75% mine chance for each square around it, including the one to the lower-left, shared with the neighboring 2.  But there are also two neighbor effects we need to account for.  Firstly, the 5 to its right is a 50-50 front, so absent the “pull” from the 6, those would be 50-50.  Because the default rate for the squares around the 6 is 75%, the rate for the shared square is going to be somewhere in between 50% and 75% (in fact it’s 64%) - this is the “pulling probability” effect discussed as part of Technique #4 - which slightly lowers the odds for that square and thus raises them proportionally on the other squares (to 79%, in fact).  But secondly, notice that the remaining front to its left is actually a 2v3:

Either the three yellow cells are mines, or the two blue cells.  Normally we would then make the assumption that the fewer-mines option (blue) is likelier.  But here, our odds are upside-down, and it’s actually the yellow cells that are likelier to be mines.  This fits in nicely with the 79% rate for the cells adjoining the 6, and that rate propagates out to the frontline cells.

Solution: Guess the lower-left corner of the upper region.  If forced to guess the lower region, guess the interior corner of its upper-right box.

Puzzle 2-14 “Rifle”: 14 mines remaining.  There are two squares that are below background rate; find at least one.  Difficulty 7 / 10.  Source: this game.

Analysis: I gave the minecount because we can account for roughly 13 mines on the front here, out of 14 total, and there are 20 untouched squares.  So the background rate is already 5%.  Thus a guess into space (on that right wall) is likely the best play to make progress and finish the board.  But still, we want to find two front squares with even-better odds, via assume-fewer.

The middle parts of the region (from the 3-4-3 corner and down through the front with the interior 2) don’t really fit what we’re looking for.  Yes, you can get either 4 or 5 mines on the 3-4-3 corner, but they blend into the adjacent fronts anyway, so you end up with the same number of mines in total under most scenarios.

The bottom section is isolated, and leaving aside the corner 1, it’s interesting on its own:

Since the upper-left 4 is an effective-2, we already know that C/D/E has one mine across the 3 squares, and nothing else faces those squares, so those are all 33%ers no matter what.  It’s A/B/F/G that’s the interesting part.  Suppose that upper 4’s remaining mines are above it, on A and B. Then F is safe, and we require a mine on G.  However, while A/B can’t have zero mines, they could also have one mine.  This would mean a mine on F, G is safe, and the other mine is on A or B.  So, together, this is actually a 2v3!  If F is a mine, we have only two mines in this area, whereas if it’s not, then we have three.  So F being safe would pull another mine out of the “pool”.  Therefore, F is very likely a mine, and G is very likely safe.  Which is nice, because otherwise we’d probably have a 50-50 down below around that corner-1.

And then there’s the top section, where we only have three number squares, all effective-1s:

Let’s consider the possibilities based on that central 3 as the pivot square.  B and D are, of course, our overlap squares: a mine on either one will satisfy the numbers on both sides.  If B is a mine, it means A/C/D are all safe (2 mines for the sub-region), and if D is a mine then so is A, and B/C/E/F/G are safe.  So those are the two-mine options.  Then there’s square C.  If that’s a mine, then B and D are safe, meaning A is a mine, and there’s a third mine somewhere within E/F/G.  So that’s a three-mine scenario, meaning square C, specifically, is below background rate (3%, actually).  So that’s the other part of the answer, although if safe it really doesn’t help you make progress, because there’s still nothing to help choose between B and D.  In fact, C is a “dead square” - it has to turn up a 2.

Solution: You want the square to the right of the bottom 4.  And if that doesn’t turn up a 3, then guess into the big glut of untouched squares, where you’ll likely find some joy.

Puzzle 2-15 “Onion”: Find 3 independent good moves.  Difficulty: 8 / 10.  Source.

Analysis: This one is hard to see the answer, but (thankfully) easy to explain.  We’re asked for three independent moves, so let’s peel this “onion” back layer by layer, as we uncover each one.

Firstly and most obviously, there is the right side, from the top down to the bottom corner.  This is transparently a 2v3 option: either the three blue squares are mines, or the two yellow squares are mines, but those are the only two layout options for this part.  Two mines is likelier, so we would guess the blues.

The second one is more subtle: we need to do some projection reasoning from the right, through to those middle 3s.  As seen on the left, as per the 2 on the right, the yellow group contains one mine.  Therefore, residually, as per the middle 3, the orange group contains one mine.  Therefore, as per the leftward 3 (which has two of its three mines marked from the yellow and orange groups), its residual mine is in the dark blue group.

However, per the 1 to the upper-left in the corner, there is also one mine in the two squares facing it (in teal).  Between the blue group and the teal group, we have a (subtle) 1v2!  So the corner of it (below the interior 2) is likelier to be a mine, and the wide corners (below the 1, and below-left of the 3, where the arrow is pointing) are good guesses.

Finding the third one depends on first pulling the already- described areas out of consideration.  Let’s mark the outer (right-side, yellow) 2v3 and the inner (left-side, blue) 1v2, and consider the two remaining bits of information we have.  We have that middle pair (in orange) from last time… but we also have the 1 at top, facing two squares (in purple).  Those two can be jointly satisfied by one mine (the overlap) or by two mines (the non-overlap squares), and one mine is likelier.  So that’s the third option we can guess here.

Solution: All cells that would be safe in the fewer-mines scenarios are good guesses here.  Because it offers multiple dependent squares, though, you would probably guess the right-most column first, because for your 19% risk, you get three safe squares if you don’t blast.

(3) Guess the more-helpful result

Link to discussion section.  Link to puzzle section.

Puzzle 3-1 “U-Turn”: Find an optimal guess.  Difficulty: 1 / 10.

Analysis: This is a slight twist on the “double line-and-1” situation.  The left two-thirds of this 50-50 structure is a typical sight: an extended line-and-1 with a bend.  But the right two-thirds of it has a characteristic we look for: if either of the “inside” squares (The ones that touch between the right two pairs of squares) are mines, then the other side of it is another required guess.

(A)   (B)

Either of these represent a double-guess situation: If in (A) the square above the 6 is a mine, then we have a forced-guess line-and-1 above it (the yellow).  In (B), if the square to the right of the right-hand yellow group is a mine, then “the trap is complete”, we’ll have to guess the yellow groups’ 50-50.  So if either of these is the case (because we guessed the other squares of their respective pairs), we’re facing a double guess.  Therefore, in keeping with our principles, the right choice is to guess in a manner so as to prevent having to make another guess.

Solution: Either of the interior squares on the right side are good guesses (in teal, below).  Being right on that guess lets you solve the rest of the region, and thus have a 50% win prob.  Guessing anywhere else ends up necessitating a double guess and thus has a 25% WP.

 (yellow is equivalent to the lower teal square)

Puzzle 3-2 “Hockey Stick”: Find an optimal guess.  Difficulty: 2 / 10.

Analysis: Similar to the previous puzzle, this is an extension / variation on the “double line-and-1” situation.  We have a very-extended line-and-1 run from that interior-5 on downward, but that’s all connected, and should be thought of as a single unit.  What’s distinct is the top bit, below the 4, “around the bend”.  That pair, below the top 4, has shared squares with the bit next to the interior-5 just to its lower right.  Imagine what happens if either of the shared squares are a mine: then the other pair becomes a forced-guess 50-50.  Therefore, we need to prefer a guess that, if survived, will resolve the other 50-50 through logic.

Solution: Either of the shared squares (in teal) have a 50% win rate, because surviving that one guess solves the whole region.  The yellow squares are equivalent to the teal one that’s in-line with them, so they work too, albeit indirectly.  Any other guess either loses (50%) or results in a 50-50 for the other part of the region, so thus has a 25% win rate.

Puzzle 3-3 “Combo”: Find an optimal guess.  Difficulty: 2 / 10.

Analysis: We are 1 square away from having a locked-in 50-50 forced guess, in 2 spots.  We’re close to a box-and-2 (wall variant) at lower-left, where if the cell above the 3 is a mine, we’re hosed.  And likewise, we’re close to an extended line-and-1 (wall variant) 50-50 to the right of that 4-square box, with the “line” being only 2/3s complete (between the two 4s): if the cell below that column of 2 mines is also a mine, the trap is complete and we’ve got a 50-50 there too.  A mine on either one means a guess in the other region; however, because the two spots touch each other, knowing one will tell you the other.  This leads us to…

Solution: The teal squares below (which are the crucial ones we just described) are good guesses; if you survive them, the rest of the region can be determined, and therefore you’ve got a 50% win probability with them.  Same with the yellow squares, as those are functionally equivalent to knowing the teal ones (But they work only because they determine the teal ones, are dependent on/with them).  Any other guess will end up implying a full 50-50 in the adjacent area, forcing a second coin-flip guess and leading to a 25% win rate.  Don’t do those.

Puzzle 3-4 “Snuggle”: Find an optimal guess.  Difficulty: 3 / 10.

Analysis: This area is a (compressed) example of the “prevent the 50-50” discussion.  Visualize this being broken into two pairs of squares:

It’s fairly obvious that each pair here contains 1 mine (the upper 4 is an effective-1 facing only those two squares, and the lower 4 is an effective-2 that shares both blue squares, so the orange squares contain its residual 4th mine).  So let’s ask ourselves two questions:

(1) If we know the blue squares, do we automatically know the orange?  Answer: No.  The orange squares, with their framing mines to the left, are a classic line-and-1 50-50 corner variant (the dreaded “overhang”).  Given where things are placed, no additional information can prevent a guess on the orange squares.

(2) If we know the orange squares, do we automatically know the blue?  Answer: Yes!  That would give us a view onto one of the blue squares but not the other, which would tell us whether the lower blue square was a mine or safe.  So the blue section is solvable from below, if we know the orange.  Therefore,

Solution: Guessing either of the lower squares (orange in the above diagram) is correct and prevents a double-guess, 50% win rate.  Guessing either of the blue squares would force a second guess, so a 25% win rate.  Put in the more-formal terms of the chapter: while all squares are 50% mine vs 50% safe, the blue squares are “dead”, whereas the orange squares are “live”.

Supplement: This process applies whether or not we have that lower 4 (in this puzzle) available.  The exact same analysis holds for this layout:

In this case, we’re told the minecount (2), or else it might be a trivial matter of guessing the 50-50 above the 5.  But here, with 2 mines left, we have a confirmed 50-50 (the overhang: line-and-1 corner variant) on the right side.  Even if the middle square (above-right of the 5) is safe, it can’t tell us which of the two right-hand squares is safe.  But the reverse is not true: knowing those two rightward squares, whichever one is safe will tell us if the middle square is safe.  So either of the right 2 squares (away from the front line) is a good guess, win prob 50%, while guessing the others gives you a second 50-50 to guess (win prob 25%).

Puzzle 3-5 “Screen”: Find an optimal guess.  Difficulty: 3 / 10.

  → Let’s label it →

Analysis: Working from the labeled version, let’s start by noting that each pair that faces a number square (A/B and C/D) is a 50-50 independently, i.e. we’ll have to guess one of them.  But suppose we guess A, and we’re right.  Then B is a mine.  We’re left with having to guess C/D on a prayer - it’s a double guess.  Likewise if we guess D to start; then C is a mine, but we have no information about A/B.

But suppose instead we guess B (and we’re right, and don’t die).  Then A is a mine, but B itself will have either turned up a 3 (in which case C is safe) or a 4 (in which case C is a mine).  Likewise for guessing C, which determines B.  B and C have information about each other; but A and D cannot help you.  This is a sort of corollary situation to sec 3(a), “double line-and-1 50-50”.  The exterior guesses force you into a second guess if you’re right, but the interior squares, assuming you don’t die, let you solve the rest of the puzzle.

Solution: B and C are correct: they have win rate of 50%, while A and D have win rate = 25%.

Puzzle 3-6 “Clown Car”: 3 mines remaining, find an optimal guess.  Difficulty: 4 / 10.

Analysis: This is the flip situation of some minecount situations we sometimes find ourselves in; every single square that’s interior to our front is a mine.  But note that the lower-left pair and the right-hand pair are connected, via the shared 3 between them.  Whereas the lower-left and the upper-left pairs are connected only via their unknown squares, not their number squares.  That creates an analogous situation to our “double line-and-1” guideline.  Let’s label them:

So let’s move to some questions to assess determinism and direction of this setup:

1. If we know the yellow pair, do we then thereby know the orange pair?  Yes.  Because of the shared 3, the lower yellow square will either provide its 3rd mine, or be safe and the 3rd mine for that shared 3 will be the right-hand orange square.
2. If we know the blue pair, do we then know the orange pair?  No.  Because of how the orange pair is up against that wall to its left, even having the lower blue square be safe (which would turn up a 6) does not tell us any more than the 5 below the orange group would.  It’s the same information from the other side, not new information.
3. If we know the orange pair, do we then know the blue pair?  Yes!  Solving the orange squares, one of which is a mine but both of which can see the bottom blue square, will tell us whether that bottom blue square is a mine or not, and thus solve the blue pair.

So the orange and yellow pairs are connected, knowing one determines the other.  But the orange and blue are connected in only one “direction”: orange can solve blue, but blue cannot solve orange.  Therefore:

Solution: Any guess in the orange or yellow squares (as labeled above) has a 50% win rate, and are correct.  A guess on the blue squares necessitates still making a guess in the orange-yellow area, making it a double-guess and 25% win rate, so that’s wrong.

Puzzle 3-7 “Box and Tail”: 3 mines remaining, find the optimal guess.  Difficulty: 4 / 10.

Analysis: So this is the “box-and-tail” situation we discussed in the “Assume Fewer” chapter, but with some variation.  Given that there are 3 mines remaining, we can see that there are 2 independent (semi-isolated) areas here: the 2 squares under the 3, and the 4-square box to its left.  We can tell that the 4-square box is a box-and-2 50-50, because it’s fully framed, and there’s nowhere else to push the mines off to (there’s no 1-mine or 3-mine possibility), so therefore there’s exactly 2 mines in it.  So let’s ask ourselves a simple pair of questions:

1. If we know the layout of the 4-square box, do we then know the tail?  Yes.  Either of the squares under the 4 will determine the square to the lower-left of the rightmost 3.  E.g., if the one just under the 4 turns up a 5, then that square to its lower-right is a mine, and if it’s a 4 then that square is safe.
2. If we know the layout of the tail, are we able to solve the 4-square box?  No.  If the square under the 3 is a mine, then all the other square does is provide the 3rd front to our box-and-2 50-50: it’s still a remaining guess.  If the square under the 3 is safe and the one to its left is a mine, then with a known 2 mines remaining there’s still no possibilities eliminated, it’s a 2-mine-diagonal 50-50.  So we just guessed only to have to make a second guess.

Therefore,

Solution: Guessing any of the leftward 4 squares (in the 4-square box) has a 50% game win rate, and are “correct”, while guessing either of the 2 rightward squares (in the “tail”) has a 25% game win rate (because you have to survive two 50-50s), and so are worse (i.e. wrong) choices.

This is actually a not-too-uncommon situation, so it’s an important result to remember: Don’t guess the tail!

(that 3rd one actually only had a minecount of 2, thankfully, so it was solvable directly.)

Puzzle 3-8 “Mirror on the Wall”: 3 mines remaining, find an optimal guess.  Mirrors the previous situation, with a twist.  Difficulty: 5 / 10.  Source: this game.

 → Label it →

Analysis: Continuing our 6-square box-and-tail discussion, this presents a slightly different scenario.  We have 3 fully independent pairs of squares (A/B, C/D and E/F), each of which has 1 mine.  

Firstly, note that the rule derived from the previous puzzle - “don’t guess the tail!” - still applies. If we were to guess D and be right, then C is a mine, and we are left with two more 50-50 coinflips to guess the remaining pairs.  Win rate: 12.5%.  Big yikes.  If we were to guess C and be right, at least we can mark D, but then the remaining squares with 2 mines leave us with a box-and-2 50-50 (corner variant), so we have only a 25% win rate.  Still a yikes.

Can we do better than that?  Let’s next consider the E/F pair.  If we guess E (thus marking F), we will turn up an effective-1, split between the A/B pair.  If that pair has a mine on B, then we’re back to our 3 coin flips (12.5% WP), and if it has a mine on A, and we click B, at least it determines C for us (25% WP).  But that’s no better than guessing C, then.

And in fact, neither is guessing B outright: with A thus a mine and one mine coming from E/F, we could turn up a 3 (meaning C is safe) or a 4 (meaning C is a mine).  So 50% of the time, then, we get a safe C square that will determine E/F for us, and 50% of the time C is a mine and E/F is a line-and-1 50-50.  Let’s do that conditional probability out: We take a 50-50 guess on B, and then if we’re alive, then half the time it’s fully solvable (C is safe), and half the time it’s a 50-50 again (C is a mine, have to guess E/F).  That works out to a 37.5% win probability.

And what of guessing F?  This is a clever move, actually, because assuming we survive it, we will be fronting up to 3 squares, A/B/C.  And we already know A/B has one mine between it.  So if we guess F and mark E, then F will turn up either an effective-1 (that mine accounted-for by A/B, so C is safe), or an effective-2 (one mine on A/B, C is a mine, will need to guess A/B).  So it’s similar to guessing B: half the time we lose on the guess itself, but if we don’t, then half the time remaining, we win, and the other half of the time we face another 50-50.  This, too, works out to a 37.5% win probability.  So…

Solution: Guessing squares B or F results in a 37.5% win probability; all others are worse.

N.B. this reasoning and solution is identical to this example too (from here):

Puzzle 3-9 “Extended Box”: 4 mines remaining, find an optimal guess.  Difficulty: 5 / 10.

Analysis: This is a variation on the “5-square box-and-tail” that we discussed, but with one twist.  That twist is that the 5-square box here has 3 mines in it, as you can tell from quick inspection (the interior-3 is an effective-2, covering the top 3 squares of the box, and then the bottom 2 squares have 1 between them).  So the third caveat to our analysis applies: instead of “guess the belly or back”, the rule for 5-square-3-mines is “guess the head”.  That’s a shortcut here, a correct answer with 67% win probability.  But we can look at this a bit more deeply to understand how we get there, and how else we might get there if we don’t remember the rule-of-thumb.

To begin with, note that it is the 4 to the right of the 5-square box that is the independent variable here: It has 2 mines to its 3 squares to the left, all 3 are valid possible configurations, and each one fully determines the rest of the board:

A:   B:   C:   D:

Those 3 possibilities (A, B and C) determine our square-by-square probabilities in (D), if we were to hit the Hint button.  In each layout option, the mines / safe square of the independent square (the 4 to the right of the box) are in bright colors, while the downstream / dependent squares are the duller red/green colors.

But there are still, clearly, some wrong answers.  For example, suppose we guess the 33% square that’s to the left of the upper 4, i.e. the one in the “extended tail”.  So that’s a 33% chance of losing right then and there.  But if we survive it, then the square above is a mine, and so is the square below it.  And now we’re either in layout option A, or layout option C.  In fact, we have set ourselves up such that we have a box-and-2 50-50!  So we have to guess between those options, A and C, with no help to choose between them!  Our 33% guess forced us into a 50-50, meaning we have only a 33% chance of winning if we guess that.

Solution: There’s only 1 other “33%” square on the board, and it’s the one that’s the “head” of the box-and-tail, the one our rule of thumb tells us to guess.  It’s safe in layout options B and C, but in either case, it will either turn up a 4 (for layout B) or a 3 (for layout C).  In both cases, it will fully specify the rest of the board, letting us mark and clear it.  All other squares within the box have a 67% mine probability, as you can tell from the layouts.  So that lower-left square is the right answer, the only one that gives you a 67% win probability.

Puzzle 3-10 “Ounce of Prevention”: Find a helpful guess.  Difficulty: 5 / 10.  Source game.

Analysis: We’re given the probability chart upfront, partly because calculating out the permutations would’ve been near impossible.  But even with it, this probability chart looks like a dog’s breakfast, just a mess of nasty-looking stuff everywhere.  We could take that 23% up top.  But before we do, let’s consider what else is ugly around here.  Up top we have a 50-50 front at left, but that might work out from below.  Likewise at lower-left, a 50-50 front, but mostly for want of information from above or below.  In the middle we have some high-density area above that interior 5, with nothing looking much more attractive than anything else.

But at the bottom, we see that we’ve got a 4-square box, and it’s framed on 3 corners (top-right to the right of the 5, and bottom-left and bottom-right).  So we’re 1 corner away from having a definite 50-50.  In other words, if we don’t guess that 4th corner, then our win probability for any other approach vector would be cut in half.  So let’s guess it.

Solution: So you guess that 4th corner.  Jump scare!  You just guessed into a 7.

But not only does guessing it prevent us having a 50-50, it’s useful in several other ways: it solves our 4-square box directly, and it determines a bunch of mines that reduce our puzzle’s residual region to something approaching “manageable”.

A follow-on Part B to this question might ask what the right play is from here, because we’re pretty close to having some line-and-1 50-50s.  Feel free to go through the options in your head, but unfortunately, the square that “prevents” the 50-50s - the one just below the upper 4 - also has a mine % of 67%.  And if it’s a mine, then we have a line-and-1 50-50 in the upper left.  Either way, best we can do is a 33% win probability, and that’s no fun.

Puzzle 3-11 “Daily Double”: 3 mines remaining, find an optimal guess.  Difficulty: 6 / 10.

Source is this game.

 → Probabilities:

Analysis: This is another one where you have to consider each of the scenarios in turn, in order to find what the “helpful guess” is, the one that prevents or minimizes further guessing.  Firstly, let’s note that the choice of mine around the lower-right 3 determines the rest of that column: if the mine next to that 3 is the high square, then the pair above pushes the mine to its top; otherwise, the mine is on the lower square (above the interior 3).  So we can describe layout options relative to the choice of mine next to that 3, either High, Middle or Low.

You would think that guessing the High square, the one that breaks the line of mines on that 3rd row from the bottom, would be the right choice, because if that square is a mine then we have a double 50-50 below it (yikes!), two line-and-1 guesses which are independent of each other.  But a successful guess on the High square does not, actually, lead to much joy.  It does fix the mine above it, but it can turn up any of 3 things:

25%: shows a 2.  The two mines are along the bottom row.  We can chord it.

50%: shows a 3.  This means we’re facing a box-and-2 50-50 below us now, mines diagonal.

25%: shows a 4.  The two mines are both on their higher-up square options, and we can mark those and clear.

So we take a 33% risk initially, and then what thanks do we get?  50% of the time we can clear the board, but the other 50% of the time, we have a 50-50!  This works out to exactly a 50% win probability, if we were to make this guess.

What about guessing the middle square?  Well, we already know it’ll turn up a 5, with one mine either above or below it, and one mine to its left, which we’re already split on.  The results of that guess will not enable any sort of progress or deduction.  So let’s put that aside.

That leaves guessing the Low square (the one touching the wall).  It has one significantly helpful aspect: because it has one mine to its right, and one mine accounted for in the 2 squares to its left, the square above it (its only other adjacency) is flexibly determined by this one: we’ll either turn up a 3 that makes that square a mine, or a 2 that makes that square, the middle square left of the lower-right 3, safe.  So if we get a 3, it means the mine was in the middle square, we can clear the upper square, and mark the top bit accordingly.  If we get a 2, it means the mine is in the High square, solving the top bit… but leaving us with a line-and-1 50-50 to our left.  So we take a 33% chance, and then half the time we solve the region, but the other half the time we’re left to a residual 50-50 again.  That works out to a 50% win probability, same as guessing the High square.

Finally, one other possibility is to guess the lefthand column.  It’s a 50-50 front right now, not necessarily a forced guess, but if we’re guessing anyway, let’s consider it.  If we guess the upper option on that left column, then success just means we have a mirror of the 3 on the right side - no new information.  However, if we guess the lower-left square, then one of two things results: it shows an effective-1, meaning the mine for the right-hand 3 is in the lower 2 squares and the HIgh square must be safe.  Or, it shows a face 1, effective-0, and we can chord it and mark the High square as a mine.  In both cases, the region is fully solved!  So, if we’re going to guess that 50-50 front anyway, guessing the lower-left square gives us a 50% win rate.

Solution: Any of the 3 teal squares are good guesses, with a 50% win probability.  The yellow is equivalent to guessing the upper teal square.

Puzzle 3-12 “Gauntlet”: 2 mines remaining, find an optimal guess.  Difficulty: 6 / 10.

 → Labels:

Analysis: This one is deceptive!  If you don’t know the minecount coming in, you might well “Assume fewer”, guess D, and hope that it’s 1 mine (you’d clear it at a rate of 1-B, which is way better than taking the 50/50 by guessing A/B).  But here we do know the minecount, there’s 2 mines.  So let’s consider what happens if we guess each option.

Guess A: we mark B, and then have a 50-50 random guess between C/D.  WP: 25%

Guess B: we mark A, and the number on B will tell us if C is a mine or not.  WP: 50%

Guess C: this could turn up a 5 (= B and D are the mines), or a 4 (= only one of B/D are mines, but because we have a minecount of 2, it means they’re on A and D).  So it’s fully solvable.  WP: 50%.

Guess D: This will turn up a 5, we mark C, and then have a 50-50 guess on A/B.  WP: 25%.

Solution: As with others where we have more known squares, this scenario follows the same rule: Guess an interior square, they’re the most helpful guesses.  In this case, B or C solve the region for you if you don’t blast on the guess itself, while A or D lead to a second coinflip guess.

Puzzle 3-13 “Spiderweb”: Find an optimal guess by identifying the dead cells. Difficulty: 7 / 10.

 → Labels:

Analysis: So right off the bat, we can see that we have a local minecount.  There are five mines in these 9 squares: one each in AB, CD, GH, and two in EFJ.  The fact that the lower-right 4 there, facing EFJ, has three possible permutations of its mines, suggests that there might be some dependency that could enumerate our possible layouts.

Secondly, the only box-and-2 50-50 that we’re worried about here is ABEF.  It is framed by mines on 3 sides, so if H is a mine, it’s fully framed.  However, it’s also possible that J is safe and E-F are both mines (in which case A is as well), which would make it *not* a 50-50.  So H is something to keep our eye on.

But the problem asks us to identify the dead cells - something we haven’t done exhaustively in the problems up until now, but which is important here.  Let’s do this by trying to enumerate our layouts (based on which of EFJ is the safe cell), but instead of labeling safe cells with a green square overlay, we’re going to put their face number on them under that layout.  This should help us see whether they are dead or not.

X: Y: Z:

(not all logic-forcing arrows shown)

As we can see from these layouts, six squares are safe under only one of the three scenarios, namely E (layout X), A and F (layout Y), and D, G and J (layout Z).  Those are 67% mines.  They’re also dead cells, but trivially so.  Now, of the other three squares, note that while C is safe in layouts X and Y, it is a 4 in both of those.  Therefore it is a dead square too!  If you click it and it’s safe, you won’t know whether you’re in X or Y.  However, you can see by comparing these layouts that cells B (5 or 6) and H (4 or 5) can take more than one value, and that value will reveal which layout we’re in, and thereby solve the rest of the region.

Solution:  Our two non-dead cells are our optimal guesses.  Both B and H have a 67% WP.

(Sidenote: astute observers might point out that cell C being safe implies cell H is also safe, i.e. they are dependent.  And so while cell C is dead by definition, it is not, in this case, a “bad guess”, because it leads to cell H which solves the area.  It’s true, and as a result it has the same 67% WP as B and H.  But seeing the dead cells is what we’re here to practice.)

Puzzle 3-14 “Slow Down”: Find at least one helpful guess.  Difficulty: 7 / 10.

Analysis: We’re given the probabilities, and we see a background rate of 24%, with no squares improving on that for their face probability.  What we need to do is think forward from a potential guess to the impact it might have on win probability.  Let’s consider 3 options that most advanced players would gravitate towards when looking at this layout:

(A) Prevent the 50-50.  You can see there’s a 50-50 forming at top-left.  If the square below that 3 is a mine, then the two to its left are a forced guess.  Therefore, our Guessing Rule #3 here would tell us that the square below the 3 (above the 5) is a Helpful Guess, because if safe, it prevents a later 50-50 guess.  If it’s safe, we can solve the top-left, and also mark the square to its right as a mine, open the one to its right, maybe more.  In short, progress feels fairly likely.

(B) Guess the upper-right corner.  This is a guess into terra incognita, but our odds of an opening are not that hopeless, they’re estimated around 45% (as we’ll discuss), but are actually a bit lower than that, about 39%.  So while it’s a shot in the dark, there’s a method to it: the odds we can use what we find there to make progress feel decent.  Even if we turn up, say, a 2, that alone meaningfully lowers the background rate for the rest of the board, because we have so little of the board left.  But it’s also a solid bet to make progress directly.  So 24% of the time we lose, and (39% * 76% = 29%) of the time we get an opening that probably gives some progress, and some fraction of the rest of the time we have helpful info that can steer our next guess.

(C) Guess the lower-right corner.  Why that one?  Because (1) there’s a real risk of a 50-50 forming around it, and guessing that square would break it, and (2) even if there isn’t one there, we have enough number squares above us here that the odds of progress are actually quite good.  So what about the 50-50?  We’d be looking at a half-full bucket, if the 74% there (to the right of the 2) is a mine, and the one to its right is also a mine.  Then the square to the right of the 3, which is thus safe, would likely turn up a forced guess between the bottom-right corner or the square just above it.  So if that’s the formation we’re facing, we might as well guess it now.  And if it’s not, that square could very well turn up a 3 (effective-0, chordable) and let us probably clear the whole area above it; it would only require a 26% and a 24% square to both be safe, which is ~56% combined.

So those are 3 pretty good options.  I don’t love the 26% next to the 3 at lower-right because even though it’s on a front, it wouldn’t offer progress to above, and progress to the right would depend on those 3 squares on the wall either all being mines or all being safe, which is low probability.  I also don’t love guessing around that interior-3 in the middle of our formation, it’s very unlikely to lead to progress.  If I’m going to do that, I’d rather guess top-corner.

Solution: How do we choose?  Firstly, consider this: while we have a general guideline here of “prevent the 50-50”, it is not a universal priority, i.e. it does not override other, better options.  And if either of the other two guesses gave us some momentum (24% is not so high a background rate that that’s dubious), we might well be able to come through and solve that area on the left directly, either preventing the 50-50 or proving that it’s there and we have to guess it anyway.  I’d rather have that done for me at a 24% risk than take a 43% chance of losing right now in order to definitely stop it (or lose).  So for me, (A) is less-preferred.  The choice between B and C is a matter of taste.  To me, 56% chance of progress (getting a chordable 3 at the corner, and then likely being able to clear above) is meaningfully better than the 39% odds of an opening at top-right (only some of which lead to progress), so I would probably go C, then B if it doesn’t work.  Only if neither of those got me going again would I try A.

But I should also note, in fairness, that the MSCoach Analyzer prefers the 50-50 breaker, A.

Puzzle 3-15 “Overlays”: Find an optimal guess.  Difficulty: 8 / 10.  From here.

  →  → Labels:

Analysis: The challenge here starts with there being, not one, not two, but three possibilities for a 50-50 forced guess.  In the below, the orange represents “if there is a mine here…”, yellow means “...in conjunction with these other mine placements…”, and the red represents “...we would have a 50-50 guess between these squares”.

(1)   (2)   (3)

As you can see, these are a box-and-2 basic formation, a box-and-2 edge variant, and a line-and-1 edge variant, respectively.  Those would be bad, but guessing at our peril and then having one of these be the result, would be even worse.

With that context, let’s consider some potential guess paths.

• Suppose we guess the upper-left box, and pick a 33% square (B or C), leaving A as a mine.  This directly stops 50-50 scenarios 1 and 3, but it could create 50-50 scenario 2 above, meaning we would have a 33% WP.  Is that necessarily the case, though?  More on this in a minute.
• Suppose we guess the square below the right-hand 2, square E.  We either lose (33%) or we turn up a 2, and then have another guess required.  It’s a dead cell.  This is similar to the “5-square box-with-tail” logic we gave above.
• For the same reasoning, square D is a bad guess as well: if safe it is a 4, and it doesn’t generate progress.  We’re supposed to guess “the belly or the back”, which were squares C or B respectively, discussed above.  So D and E are out.
• That’s it for 33% squares.  But what about the lower line, guessing F or G?  Suppose we guess the left one, square F.  It will turn up a 4, with 1 mine across the 3 squares above it (C/D/E).  There’s not much good that can do us, other than forcing a guess into the resultant 5-square box.  That means (if we then guess B or C) we have a 67% win probability after taking a 50% initial risk, so we end up with the same 33% WP.
• But how about square G, the lower-right?  If we don’t blow up on the guess, it will either turn up a 1 (effective-0, chordable), implying there’s a mine on B, or a 2 (effective-1), implying there’s a mine in the two squares above it and, from the interior-2, square B is safe.  Either way, the rest of the area will be cleared!  So that works out very nicely: It’s one way to solve the area deterministically, without running into those 50-50 scenarios.  And we end up with a win probability of 50%, which is optimal here.

Now let’s return to the matter of guessing B or C.  It turns out that the 50-50 scenario #2 above is actually… kinda a lie.  It’s true that such a 50-50 could be the case, if we guess B+C.  Cell B is a dead cell: if safe, then C is safe too, and B has to be a 2.  But Cell C is not dead!  The two remaining mines (other than A) could be both on the left (D+F), or both on the right (E+G), in which case cell C will turn up a 5 or a 3 respectively.  If the mines are diagonal (D+G or E+F), then C will turn up a 4, and we have a 50-50 to guess.  But given that there are four possible mine configurations (in the cases where B/C are safe), two are immediately solvable, while the other two result in that 50-50.  So guessing C (which has a 33% mine probability itself), if safe, will yield a win 100% of the time under 50% of the layouts, and a win 50% of the time under the other 50% of the layouts.  Adding that up yields a 50% win probability for guessing cell C.

Solution: Guess the lower-right square, “G”, for a WP of 50%.  Or guess cell C, the wide squares of the upper box, which also has a 50% WP, but with the possibility of another guess to follow.  All others have a WP of 33%.  Here’s how the solver views it:

(the solver sees the two not-dead cells, and given that they both have 50% win probability, recommends the one with greater initial safety - not that it matters which one you pick, really)

One way we could have foreseen this a little faster, without all the brute-force, is to look again at the overlay of the three 50-50 scenarios we’re worried about.  There is no square that is shared by all 3 of the red areas. However, there is one and only one square that is shared by two of the red areas (#2 and #3), and also would provide the missing piece for the third (#1), i.e. the framing.  That square is the lower-right one, Square G.  Although we don’t cover this in depth in this guide, this consideration is one way to help choose a guess in a high-density environment: squares where they would solve or prevent multiple guess scenarios.  Sometimes that leads you to a square that might have a higher mine % itself, but if you survive it, it provides a higher WP% for the game.

(4) Square Coverage Ratio

Link to discussion section.  Link to examples section.

Answer images here are presented mostly without comment, because, well, this is a pretty simple thing and we’re just trying to train you to spot it, as a reflex.

Solutions indicated by the Square Coverage Ratio technique are in green.  Other good guesses (or fallbacks if those guesses don’t work out) are shown in yellow, where applicable.

(A)  

(B)

(C)

The square below-left of the interior 2 is shared with the 2 below the mine, so that’s not part of what this technique recommends.  However, notice that the square to the left of that 2, in yellow, if it were a mine, would fix the location of both remaining mines for the 4 below it.  Which makes it less likely to be a mine itself.  So that’s where I would go next if the green guesses looked too treacherous to keep poking around there.

(D)

The background rate of this region is actually 17% (see top row); the 13%s seen here are actually an even split of the mine equity on that interior 1, where 25% is to its right, and the remaining 75% is split over 6 squares (so 12.50% each, rounded to 13%).  Same is true right of the 4.  So I would either guess above-left of that interior 1, per the Square Coverage Ratio, or I would go for the 1v2 corner at lower-left, which is influenced by the interior 1’s odds.

(E)

(source) The interior 2 below the green boxes is an effective-1 that has six squares around it, so while four of those six squares are shared with some other adjacency, the two ones marked here are not.  If the one directly above this 2, for example, turns up a 1, then we know that mine is shared with the 2 on the upper front, and we can clear three other squares.  So progress is decently likely from this.  However, failing these, the yellow-square series is probably our best bet.  If they were mines, then first of all it would clear all squares around that interior 2, but it would also clear the 2 to the left of these yellows, further forcing a mine above-left of the 3 next to it.  So these are below background rate (actually 14%).

Also, pay no mind to the logic move available at top (the 4 is fully marked). 🙂

(F*)

The guesses that SCR would recommend are illustrated in green (and, less-favorably, yellow).  In particular, the square in green just left of the 3 is a good guess, it adds a lot of information.  But worth noting that there is an Assume Fewer-based guess on the left side of the front, which is actually the best guess on the board at lower minecounts (can you spot it?), so in fact this is a bad example of when to use SCR to pick a guess.  But if you saw the green and/or yellow regions and those registered as being SCR-type guesses, consider yourself “correct” here.

(G)  (source game)

(H)

The 10% square below-right of that interior 2 we’re highlighting here is not what the Square Coverage Ratio technique recommends; instead, it derives from Chapter 10’s Mine Lock-In rule (A mine there fixes the two mines by the right 3, fixing in place a few other things downstream).  But guessing the squares in green (same reasoning as with Pseudo-Hole, but with four other adjacencies rather than three) is still a good move, and far easier to see.

(I*)

The area around that interior 1 (green framing) is your best bet.  Failing that, you have a “split background rate” situation lower down (yellow framing), with similarly good odds.  The square up against the mine wall is particularly good - high likelihood of progress.

(J*)

The two squares above the 1 as marked are your best bets; that there’s another one that also has 10% odds is nice, but is shared with the 1 below it, so it can’t be an opening.  But then, consider the yellow areas.  The righthand one is a 1v2 corner box situation with a little extra juice: mines in the wide squares determine the 1 we’re focused on for this question, and greatly limit possibilities around the upper interior 2, as well.  Over at left, is a simpler 1v2 corner “Assume fewer” situation, and in fact I believe it’s that one which leads a player to the full solution for this board.

(K*)

There are several pretty plausible options here with good square coverage ratios.  But the best one is at top-right: that interior-2 has 1 mine in 6 squares, better than the others.  Of those squares, 2 are shared with the 3 to the right, and the square above also shared there, so it’s the other 3 that are the best bets (in green).  Of those, the top-right square offers the best chance for progress, because it will turn up a 5 unless the square below it has a mine (which would clear the rest of the 2), and either way you can make further moves.  But if that leads to a dead end, the yellow ones are not bad: the righthand yellow has 4 adjacencies that aren’t shared with other frontline squares, and the lower square in it would only require a very few cells to be safe in order for it to be chordable.  At left, most of that interior 2’s adjacencies are shared, but that above-right one isn’t, and could lead to a set of chained guesses into space (see Technique #5 for that).

(L*)

This one is mildly tricky because the 7 will scare people away.  But if you look at that interior 3 to its lower left, which is an effective-1, you’ll notice that most of its mine equity is pulled toward that 7, since the 7 needs two mines from three unknown squares.  So the 3’s remaining squares look pretty good, in fact they’re only 6-7%.  Below-left of the 3 (the 6%) is the best one: it offers better progress odds than the green box, because it will often turn up a 2 (chordable), and if it turns up a 3 (eff-1), it makes the upper cells in the green box into even more favorable guesses.

(M*)

You have two fine examples here of the split-background-rate pattern.  There’s two good pseudo-hole opportunities here, too: the top-left green box has its top-right square offering one, while the bottom-right green box has its bottom-left square offering another.  The best guess is probably below the interior 3 in the top-left box, however, as it has the fewest adjacencies needed to generate progress.

(7) The 4-Square Box

Link to discussion section.  Link to puzzle section.

Puzzle 7-1 “Tail Help”: Mark 2, clear 2.  Difficulty: 2 / 10.

Source is this 100k attempt, dial the clock to 46:00 and look about halfway down on the left side.

Analysis: On first glance, this looks very similar to the example from the discussion of Rule 1A (“3 sides of a 50-50 front confirms the 4th side”).  But there is a crucial difference: here, the 4th corner of the 4-square box is not  a mine.  It’s an unknown square.

The important thing to notice is that we can prove the 4-square box is a 2-mine-diagonal, because we have a 50-50 front from 3 sides:

Sorry for the awkward overlapping boxes there, but I’m trying to illustrate which number squares are providing the 50-50 front for which sides.  Point is, we have it from the left, top, and right.  So, by Rule 1A, it’s 50-50 from the 4th side, too.

And once we know that we have a 2-mine-diagonal 4-square-box, there’s one little insight we can find that makes all the difference: that 3, below the box, is an effective-1, and its 3rd mine is accounted for by the box (even though we don’t yet know which square!)

Solution:

Puzzle 7-2 “Hollow Chamber”: Mark 2, clear 4.  Difficulty: 3 / 10.  Source game here.

Analysis: Although there are a few candidates to choose from, it should be visually obvious that we have two 4-square boxes at lower-right, particularly the one at far right.  That box is a 50-50 front from 3 sides: the top, right, and bottom.  Therefore it’s a 2-mine-diagonal.

If we take the 1 mine from the 4th side, that solves the middle 3 between it and the box to its left, allowing us to clear and proceed from there:

 →

Once we’ve used the box logic to clear that 3, it now implies there’s a mine above the lower 2 (orange arrow), which means we can clear the 3 just above that, which then implies a mine farther to the left.  This is the whole of what we need to clear here.

Solution: Turns out, that middle 4-square box was “hollow”, just 1 mine in it, but we needed to use our 4-square-box rules in order to see that.

Puzzle 7-3 “Tetris”: Mark 1, clear 3.  Difficulty 3 / 10.  Source here, at bottom-center.

Analysis: There are two 4-square boxes here, abutting each other without a boundary between them.  So the choice of which box to focus on may not be obvious.  But they work together to provide a solution, and the key insight is to think about the effective numbers here, particularly that 4 at middle-right, which is an effective-1.  Also, let’s recognize that in neither case are we initially sure, just from inspection, that it’s a 4-square box with a confirmed “2 mines on a diagonal” situation.  That could be true on the upper box, or on the lower box, and if true on one box then it’s not true on the other (because of that right-side 4).  But we only have information showing the 50-50 front from 2 adjacent sides, on each of the upper and lower one - insufficient to prove its status.  However, there is a quick bit of insight that can tell us which one is the 2-mine-diagonal one.

The key bit we need is to figure out that the upper 4-square box is actually a “2 mine diagonal” box, which we can prove by finding a third 50-50 front, coming from below:

Within the cyan box, the lower 3 is an effective-1 with two unknown squares (in orange), but those orange squares account for one of the two unknown mines for the upper 3, so therefore the other two squares (yellow) contain one mine.  Therefore the upper box has a third side from which it is a 50-50 front, and therefore (by Rule 1B) it’s true from the 4th side as well - the one facing the 4.  So the 4 gets its 4th mine from the upper 4-square box, meaning the two squares below it are safe.  Which leads to showing the lower box just has 1 mine, on the shared corner of the two adjacent sides, and the other 3 squares in it are safe:

Solution:

It’s worth noting that if that “pivot” 4 there on the right had instead been a 6, then the lower 4-square box would have been a three-mine box, with the lower-left corner being safe and the other 3 squares (which here are safe) being mines.  And if it had been a 5, then the logic we applied here would just have proven that the lower 4-square box was 50-50 from a 3rd side too, and therefore both of them were box-and-2 50-50s, in a sort of double, conjoined fashion - and you’d have to guess it.  But as a 4 (or a 6), there’s enough information to solve it.

Puzzle 7-4 “Unboxed”: Mark 0, clear 2.  Difficulty: 4 / 10.  From this game.

Analysis: This one’s a bit subtle, and if you don’t see the Finned Box pattern immediately, it will test how well you understand mine arrangements in a framed box.  The box we’re talking about is at upper-left, so let’s focus on that area.  We have a 50-50 front from the top and the left, but not quite from either the right or the bottom.  In fact, what we have is a 5-square “Finned Box”, if we include the cell below it, i.e. the other one faced by the interior 4.  The box is not isolated, so the odds of each possible mine placement are not equal (as with our isolated usual box-and-tail from Technique #3).  But the possible mine placements are still the same: there’s only three configurations, one for each adjacency on that interior-4:

(A)  (B)  (C)

Again, unlike when this box is isolated, these layouts are not all equally likely.  But they are still the only layouts possible within these five squares, which must contain exactly two mines.

And then we need to have a bit of insight: in those three layouts, they each put a mine in a different square on the right column of this 5-square box, one for each option.  All three of which face that 3 just to the right of the box.  So collectively, they account for one and only one mine that that 3 needs.  It’s actually a “Finned box”, covered in the chapter.

So, let’s just treat that as being a “one mine in three squares” group (blue frame at left).  Then the 3rd mine for that neighboring 3 (the effective-2) has to be in the 2 squares below it (yellow group).  Which leads to a provable safe square, per the next 3 over (an effective-1).  Then another logic move follows to the right, which completes our answer.

Solution: Our Finned-Box logic leads us to clear two cells to the right of our “5-square box”.

Puzzle 7-5 “Corner Store”: Mark 2, Clear 7.  Difficulty: 4 / 10.

Source is this game.  Which I eventually lost, but at least we got a good puzzle out of it.

Analysis: That interior-1 that we got on the corner, which was clearable from direct logic, means some good things for us.  If we can figure out where its one mine for it has to be, we can probably clear a lot off of it.  We strongly suspect that it’s got to be coming from above, that the 4-square box above it is a 2-mine diagonal.  But maybe it’s the 4-square box to its left that’s a 2-mine diagonal, or neither.  Let’s see what we can prove, first.

So looking at the right-hand 4-square box, we can prove it’s a 2-mine-diagonal box pretty simply.  That’s a good first step.  It’s a 50-50 front from the top, the right, and the bottom, so that decides it.  But then, as a second step, note that the alternation inherent in the right-hand box gets “mirrored” across the gap, that 2-3 down the top middle, and into the upper-left box.  How can we see this?  Consider the alternatives:

The orange squares share a mine (from the 1 on top), and therefore the blue squares share 1 mine too (from the 2 above them, which was an effective-1 given the orange squares).  So among those 4 squares bridging that 2-3 interior line, the orange squares have 1 and the blue squares have one.  But is it the case, too, that the two mines here have to be on opposite sides of the gap, 1 mine per side?  If that were not the case, then both mines would have to be on one side - and one of the 4-square boxes would have to have a 3-mine scenario:

…both of which are impossible, as they reach a contradiction quickly.  At left, putting both mines for the interior 2 on the left means a 3-mine configuration there, which overloads the interior 1 (in the yellow).  At right, putting both mines on the right means you overload the bottom-right 2, with 3 mines (also in yellow).  To say nothing of the impossible downstream implications, too.  Either way, it’s a firm conclusion that the two mines for that interior 2 need to be on alternating sides, left or right: one on each side / column, and one on each row.

But this is just another way of saying it’s providing a 50-50 front on the 4-square box.  And since we already have that coming from the right-hand box, we can now infer that the middle (upper-left?) 4-square box is also a 2-mine diagonal, mirroring that of the upper-right box: We have a 50-50 front from the left, from the top, and now from the right as well.  3 sides is enough.

Therefore that middle box is a 2-mine diagonal box, and so it’s 50-50 from the 4th side as well - the bottom.  But if it’s 50-50 from the bottom, then the mine for that interior 1 is accounted for with just the squares above it.  Therefore, the other five squares surrounding it - the sides and bottom - are all safe (in fact, below the 1 is an opening).  This leads to marking two other mines, and from there two other safe squares, which follow from that clearance of the 1.

Solution:

Puzzle 7-6 “Intruder”: Mark 1, clear 4.  Difficulty: 5 / 10.  Source is this 100k replay at 20:00.

Analysis: What should jump out at you first is the two interior squares in the middle, the 4-3: one of them an effective-2 (the 4) and the other an effective-1 (the 3).  That’s the key to this problem, which becomes a “Check your arrangements” situation.  Because the 4 and neighboring 3 share two squares, we can first reduce this to asking ourselves about the distribution options of that 4.  What are the allowable options for how the 4’s remaining mines can be distributed, between the squares shared with the 3, and the squares in the 4-square box above?  We clearly can’t have both of them in the shared squares, as that would overload the 3.  We can probably have 1 of them shared.  But could we have 0?

If there’s 0 mines in those two shared squares, then they’re both in the 4-square box up top.  We already know that box is a 50-50 front from both the top and the right.  What would that look like?

At left we can see it.  The assumed mines (in the teal frame) go into the box.  This would make the two shared squares safe, which is fine for now.  But then, clearing around the sides of the 4-square box, we would end up clearing both the top two squares: the right one from the right side, the left one from the interior 3 at upper-left.  This would leave the top side numbers (in yellow frame) without an option for their remaining mine!  Therefore we’ve hit a contradiction: this can’t be the case, there can’t be both the 4’s needed mines in the squares in the box.

This realization has several conclusions.  Firstly (And obviously), this means that one of the 4’s mines is in the two shared squares with the 3, so both squares below that 3 are safe.  Which means you can mark a mine below that, too.

But secondly, if the 4 is providing one mine upwards to that box, it now means we have a 50-50 front from a third side of the box (the bottom).  Therefore it’s a confirmed 2-mine diagonal, and we can conclude that there’s 1 mine on the left side of it, as well.  This would be the third mine for that interior 3, which means we can determine that the two squares below it are safe.

Solution:

Puzzle 7-7 “Squeeze”: Mark 0, clear 1.  Difficulty: 6 / 10.  Source is this 1k attempt.

Analysis: This is a “clever projection” type 4-square box situation, where the box is not obvious or even mostly-framed, and it emerges only by projecting square groups from different areas.  It also misleads you a bit, because the 6 draws the eye, and there are several other possibilities for 4-square boxes here.  You might have to stare at it a bit to see this:

 +  =

At left, we project up from the bottom.  The bottom 2 says one mine in the blue, and the lower interior 3 says its other mine has to be in the yellow.  In the middle image, we get our 50-50 front from the left: there must be one mine each in the blue and purple, so going from the left, from the interior 2 to the left-interior 3, the 3’s remaining mine has to be in that yellow.  Combine that with the 50-50 front from the right side’s 3, and we have our three sides for our box.  With that 2-mine-diagonal box, the 3 just above it is satisfied, which proves a safe square to its left.

Solution: Isn’t it fun how that works out!  Most 4-square boxes are not this hard to visualize how the 50-50 fronts come together, but sometimes you have to work a bit, like this, to see them.

Puzzle 7-8 “Platform”: Mark 0, clear 3.  Difficulty: 6 / 10.  Source is here.

Analysis: This one is a “Finned Box” puzzle, but in disguise, and relying on logic similar to how we reason about 4-square boxes.  It doesn’t concern the box toward the top of the front. Instead, we’re talking about that little peninsula going out to the left, which gave this puzzle its name.  And there’s two ways to get to the solution: reasoning from basic box principles, or direct use of the Finned Box pattern.

Firstly, from the basics: If you stare at it for a little while, you’ll see that the 4-square box on that peninsula is a 50-50 front from two sides, but not necessarily from three.  In the left image below, the yellow box has a mine, so the remaining mine for the lower 3 has to be in one of the right squares (orange).  But up top, the 2 above the box could have a mine in any of three squares.  So we can’t definitely conclude that it’s a 2-mine diagonal.  But if we adopt a little flexibility, a little comfort with ambiguity, we can see that there are two possibilities…

Either there’s a mine on the square shared between the upper 4 and the 2, or there’s not.  If there is a mine there (middle image), then the region is determined: there’s a mine to the right of the lower 3, and one that is 3 squares to the right of our initial assumption (on, let’s call it the “neck” of this peninsula).  If there isn’t a mine there on the shared square, then we have our 50-50 front from the third side - the top - as per the 2 above the box!  See above-right.

That third 50-50 front would mean the box is a 2-mine diagonal, and thus 50-50 from the right side as well.  So under all possible configurations, there is a mine in one of these three squares:

This process just reinvents or re-proves the logic of the Finned Box pattern, though, so let’s try that route.  Consider instead the 5-square area, built from the middle 1 on top, and the 1 below: the blue groups show that the finned box has exactly two mines in it.  So if we then project from the left, that orange group has one mine, so we can “subtract” that from the blue region, and what’s left from it is the pink box with one mine again.  Which proves the safe squares at right.

 →

Solution: The above reasoning leads us to conclude that there’s a mine on the neck of the peninsula or just to the left of it, meaning all cells to the right of the 1 below that neck are safe.

Note the 8% square in the box above the area we were focusing on.  This will be covered in Chapter 10, but: imagine that square were a mine.  If so, then it forces two mine placements for the 4 above the box (the other squares to its right), and it clears the 3 below it, forcing a mine to the right of the lower 2, clearing three cells below that, and forcing a mine below the 2 on the peninsula.  A lot cascades from that decision!  Which thus makes it very unlikely to be a mine.

Puzzle 7-9 “Glass House”: Mark 2, clear 4.  Difficulty: 6 / 10.  Source game here.

Analysis: So by now it should be obvious where the 4-square box is, at upper-left.  And we can quickly confirm that it is a 2-mine-diagonal, because we have 50-50 fronts at top, left and bottom.  But we can’t make direct use of that fact, because there’s no square to the right of the box that is an effective-1.  All we have is that interior 3!  And you might then search in vain for other logic moves around the region.

But if we look a bit closer, we can use something akin to Evil NG “Wheel 3-2” logic to mark out the mines for that interior 3.

We start with the 1 mine we must be getting from the 4-square box (blue box), and then add in the mine coming from the top 4 (yellow box).  So the remaining mine for the interior 3 must be in one of the two squares in the teal box.  We’ve identified a hidden square group!  Then:

Next, we realize that the square below that interior 3, the 2, is actually an effective-1, and the teal group solves it.  From which we can clear squares below-right and above-left.  The latter is part of the 4-square box, so it means we can fully solve that box, marking and clearing it.  Then, lastly, we can discover one other safe cell below the lower already-cleared cell.

Solution: As described above, we use the 4-square box to narrow down our mine locations for the interior 3, eventually solving the box itself, plus a little bit else.

Puzzle 7-10 “Backup”: Mark 2, clear 4.  Difficulty: 6 / 10.  Source here.

Analysis: This is broadly similar to Puzzle 7-6 “Intruder”, in that it’s about checking the possible arrangements in the box, their agreement with adjacencies on the open ends, and finding what is possible and what isn’t.  Ignoring the hole-pattern area at top and the area by the 5 at bottom, let’s focus on our middle area, with the box framed from the left.

The key thing to see here is that interior 2, which has no flags around it and remains an effective-2.  With the 1s to its right, we might imagine there is something like a 1-2 pattern available, which gives us optimism to look more closely at the box logic.  And indeed, something like that does eventually work out.  Consider these groups:

Teal comes from below, purple comes from the front on the left; we have 50-50 fronts on two adjacent sides.  We also want to split the interior 2’s adjacencies between the blue group (in the box), and the pink group (outside the box).  We don’t know how many mines each contains yet, but we need to ask the questions about those areas, to check our arrangements (per Rule 3C in the chapter).  If the blue box has one mine, then the 4-square box is a two-mine-diagonal, and we can do stuff with it.  So let’s ask: how many mines can be in blue?

Can blue hold zero mines?  Then the one mine in the box is on the shared corner (between purple and teal).  Then pink needs to hold both of the 2’s mines, and it overlaps with a 1 to its right, which would be overloaded.  So, no, it can’t.

Can blue hold two mines?  Then the 1 to the interior 2’s right would get its mine from its top-right, so that scenario is OK from that direction.  But if blue holds two mines, recall that the box logic means that the lower-right box cell (the right side of teal) would imply a mine at the top-left of the box (top of the purple group).  There would thus be three mines in the box.  That might be fine in other cases, but here we have that 2 above the box, which is an effective-1.  If blue has two mines, and there’s one on top of purple, that means it’s getting two mines from below, which overloads it.  So, no, it can’t.

Therefore the blue group has exactly one mine, and therefore the 4-square box is a two-mine-diagonal.  Because 3 sides proves the 4th, this offers logic:

The box projects a square group facing up to the 2 above it, providing the 2’s needed mine, which clears the 2’s other adjacencies.  This also puts a mine above that (per the 4), and also further conclusions going upwards.  Also, going right, since the interior 2 is getting one mine from the box, the pink group contains one mine, which shows a safe square by a 1-1 pattern.

Solution:

Puzzle 7-11 “Stamp Press”: Mark 0, clear 3.  Difficulty: 7 / 10.

Source is this high-density game (at 0:50) by Cryogen / TheMemeLocomotive.

Analysis: The first thing that should jump out at us is the neighbor mismatch here: the front on the right side has an effective-2 above it (the interior 4) and an effective-1 below it (the edge 2), with an effective-2 projected group coming from the 3 at left, for good measure.  That projection is an important first step: the 3 gets one mine from above/below, per the 2 to its left, so the three-square group on its right must have two mines.

 (pink group, projected from left, has two mines, others one)

If we focus on the 4-square box facing the framed front on the right side, what we should feel a need to do is to check our arrangements: ask ourselves what possible layout of mines could exist inside that box.  Can there be only one mine?  No, we’ve got the yellow group from the right, and the pink group from the left has two mines, so even if we push one to the top, the 4-square box has at least two mines.  Let’s vary our choice of safe square for the pink, and see what layouts that creates:

A:  B:  C:

So Layout A is our first attempt: assume the safe square in the middle, next to the 3.  This means a mine below it, which clears the edge 2, which puts another mine above it on the right front.  OK, what if we flip that, let’s have the safe square at the bottom of the pink group.  This puts both of pink’s mines facing that interior 4, which clears below it, forcing a mine to the lower part of the right front, satisfying the edge 2 again.  Both of those put two mines in that 4-square box.  Can the box hold three mines?  That would mean assuming two mines in the box from the pink group, so the safe square on top (Layout C).  That clears the edge 2 again, putting a mine next to the right-side 3, just like in Layout A.  So we can get three mines in the box, under that layout.  To no great surprise, it means shading the safe square towards the eff-1 adjacency, and the double mine sides of the box toward the eff-2 adjacencies.

OK, those are our possible layouts for the box.  What can we say about them?  Well, perhaps obviously, the square to the left of the lower edge 2 is safe under all assumptions.  Just too many mines coming from the top and around the box via box logic.  Less obvious is that under all three layouts, the interior 4 above the box is satisfied from those squares below it.  In fact, in Layout B, we used that fact to continue the box logic; if that 4 was a 5 instead, the further logic would not have been implied here, and those conclusions wouldn’t be certain.

If you’re thinking that this resembles a Finned Box pattern, except with three mines in the box rather than two, you’d be correct!  In the original grouping image, yellow plus pink equals three mines in that 5-square finned box area (including the upper adjacency from pink).  So if we go to subtract the teal group (from the edge 2), the residual part of the box has “zone math” like this:

 →

Three mines in the yellow + pink zone (the finned box) means that if we subtract out the teal, then either teal accounts for one mine in the box, or the brown square is the mine.  If the brown square is the mine, though, then the residual area (purple) has three mines, not just two.  That would overload the 4, which is only an effective-2.  So brown can’t have the mine; it’s the other squares in teal that must have the mine, and thus purple has two mines in it.  The purple group satisfies the 4, clearing its two other adjacencies above it.

Solution: Either way you do it, you get the same answer.  And a lot of 67% cells.

Puzzle 7-12 “Salmon Run”: Mark 3, clear 5.  Difficulty: 7 / 10.

Analysis: It turns out this is just a classic finned-box puzzle, at its root, but there are a whole lot of steps involved to get there.  The square groups are jumping from lower-left to upper-right, kinda like salmon jumping upstream… if you’re whimsical.  Anyway, let’s focus on that lower-left region, since that’s where all the action is.  The interior 2 we see looks quite juicy, since it’s an effective-1 and would clear a bunch, if we can figure out where its other mine is coming from.  The box to its right can draw more attention than it needs, but we’re not going to start there.

What probably draws your attention is the 4-square box to the 2’s left, above the interior 4.  There’s a 50-50 front from the left, and we can get a 50-50 front from the bottom with some clever projection.  But from the top, working with the 3s above it, we end up only with a three-square group.  They come together like this:

  Also:

As seen in the left image, from the bottom, the blue and pink groups have one mine each, leaving the interior 4 with one mine needed from the teal group.  From the top, the purple group has one mine, so the corner 3 has one mine left from the yellow group below it.

Also important is the right image: because we have 50-50 fronts on this box from the bottom (teal) and left (brown), the box’s top-left and bottom-right squares are equivalent.  They’re either both mines or both safe, one determines the other.

Now comes the zoning math!  Because of the adjacent 50-50 fronts here, with the square equivalence, the 5-square box defined by the yellow and teal groups is a Finned Box, and our Finned Box pattern works with it.  Yellow plus Teal equals two mines, with three possible layouts.  Therefore if we subtract out the brown box (one mine) from left, what remains of the finned box is another three-square group with exactly one mine:

(1)  → (2)

Step (1) is our zone reduction: Take the brown group’s one mine out from the finned box, and the pink group must therefore contain the other mine.  This satisfies the 2 and clears its other adjacencies: two to its right, and also one to the bottom-left of the 2.  Doesn’t matter all the assumptions that were used to build up to the finned-box pattern (and the fact that the original blue group had one mine); this logic proves that lower safe square, and solves the blue group.  So in step (2), we have further conclusions that follow: a mine on the lower-right corner of the right-hand box, a safe square above that, and on the bottom row here, a mine left of the 3 (original blue box), which also solves to the left.  A lot of immediate results from this!

Solution:

Puzzle 7-13 “Moving Day”: Mark 2, clear 3. Difficulty: 8 / 10.  Source is this 1k endgame.

Analysis: This region’s got more boxes than a loaded moving truck.  Annoyingly, this puzzle doesn’t tell us whether the region has 7 or 8 mines remaining (which would make it a fairly straightforward minecount problem).

Let’s start with what’s obvious: We have a 2-mine-diagonal box on the right side, so the interior 5 (an effective-3) gets one mine from its right. Its other two mines are as-yet unknown. The middle area below the 5 has a 50-50 front from the right (by projection) and below (from the corner 3), but not yet from the left or top, so we don’t yet know it’s a 2-mine-diagonal. Projecting that interior 2 from below gets us nowhere.  But we do have two separate one-mine square groups from the left.  It looks like:

Now let’s consider that interior 2, because it’s the crux of the puzzle.  Suppose it gets both of its mines from the left, i.e. suppose the mine in orange is on the right side.  That would fully satisfy the 2, clearing its other three adjacencies above and to the right.  This forces a mine on the shared-corner of the middle box (bottom right), which clears above it, which means a mine to the left of the 5.  But hey now, because the right box can only provide one mine from the 5’s right, that means the 5 is left with needing one mine and having nowhere to get it from.

 →

So that’s not it.  The bottom-left (orange group) therefore has to have its mine on the left, above the 4, and the other is safe.  Similar reasoning applies to the square above the 2; it forces two safe squares to its right, leaving the 5 under-served.  That has to be safe too.

In fact, the only way that the 5 could get two mines and not have just one of them below it, would be if that middle box had three mines in it, on the squares other than the shared corner.  Like this:

But if you think about it, while this satisfies the other squares around it, that pesky interior 2 causes trouble again: no matter what other choices you might make, it’s getting one mine from the blue group on the left.  The three-mine arrangement in the box below the 5 would give it two mines from its right, and it needs one from the left, so that overloads the 2.  There can’t be three mines in that middle box, and we’ve already shown there can’t be one.  Therefore there are two mines in the middle box, it has a 50-50 front from the left, and is a 2-mine-diagonal box too!

Solution: This conclusion about the middle box forces a mine to the left of the 5, which puts a safe square below the top 4, and combined with our other results, gives us our answer.

Puzzle 7-14 “Great White”: Mark 0, clear 2. Difficulty: 8 / 10.  From this global-quest endgame.

Analysis: This is a pretty simple for an 8/10 difficulty puzzle, but there are a few things that make it tougher: you have to think beyond our usual patterns, really understand and imagine the mine configurations within a box, and ignore several other confounding spots competing for your attention (like the plausible boxes at left, top, and right, or trying to do projections with box logic from the left side).  The real box of interest is at the bottom. And while it’s ultimately similar to a Finned Box, the fin isn’t that of some ordinary fish… it’s, like, the fin of something much bigger.

That bottom box has a 50-50 front from the bottom and the left, two adjacent sides, but the two other sides are those 4s, which are an effective-3 (at top) and an effective-2 (at right), it really looks like you can’t prove a 2-mine-diagonal in the box.  But it turns out, that’s OK.  Let’s first reduce the problem a bit, with some clever projection from up top.  The central 2 in the hole pattern has one mine from above (teal), so it must have exactly one mine below it (yellow), leaving the middle 4 with two mines from its right and below (pink group):

In addition to those square groups, it’s important to note that from our basic box logic, the two squares that orange arc is pointing to are dependent and equivalent: either they’re both safe or both mines.  One square implies the other, because of how boxes work when you have 50-50 fronts from two adjacent sides.

So now, let’s consider the 5-square finned box, defined by the pink group and the rest of the framed box below it.  That pink group plus orange arc end up demonstrating that there are only three (equally-likely) possibilities for mine layouts in the box.  

Depending on which square you choose for the pink group’s safe square, you get these:

From a zone-reduction perspective, this is even easier to visualize, and you don’t even need to consider the different permutations.  The pink box (two mines) plus the square group below it (bottom of the box) has three mines total.  Then you “subtract” the group from the left, like this:

 →

Because pink plus blue has three mines, and both blue and brown groups have one mine, we know that yellow must have exactly two mines in it.  The layouts and logic above prove that, but if you trust that zonal math works, it’s more straightforward to see.  No matter how you visualize it, that yellow group has two mines in it.  Which means the 4 to the right (an eff-2) is satisfied by that yellow group, and its other two adjacencies are the safe squares we’re looking for.

Solution: It’s a Finned Box pattern, but instead of each group having one mine, this time some of them have two mines.  That it works anyway takes some effort to understand, but it does!

Puzzle 7-15 “Isthmus”: Mark 4, clear 9.  Difficulty: 9 / 10.  From this game.

Analysis: So part of the difficulty here is the sheer size of the region, the decision of where to focus.  When hunting for logic moves, whether relating to a 4-square box or anything else subtle (like Evil NG patterns), it usually helps a lot to have a lot of interior safe squares, which give information from several sides of an area.  So we should focus initially on the narrower area between the left and right regions, which has many interspersed number cells.

The first thing we’ll see in focusing there is that there’s a logic move available, which we need:

And unlike most situations like this, we actually need the left-hand mine here, more than we need the safe square.  We might pick around a bit trying to find a contradiction on the right side, but the action is actually now leftwards.  The left side of this high-information region is attractive particularly because there’s a bunch of effective-1s surrounded by a single effective-2: the lower 2 in the column at left.  That’s why we should focus specifically on that spot.

There is a box available to us on that side, but it looks very unlike the 4-square boxes we’re used to reasoning about here.  It’s a 5-square “finned box”, below the front that has the 5-3-3 on top.  Everything facing it is an effective-1, except that 2 on its lower-left corner.  So, what possible mine configurations could it have?  See labeled at right:

If we assume A is a mine, then B is safe, and the 2 next to B needs a mine from D/E. So G is safe, and C is therefore safe too.

If we assume B is a mine, then A, D and E are all safe, forcing mines on G and C.

Either way, there is a mine on one of A/C, and not the other, so A/C is a square group!  This projects to the left of the ABCDE box, satisfying the 2 to its left, and marking all of H/J/K as safe.  Which also has downstream conclusions: The cell below-left of the left 5 here is a mine, which satisfies the interior 3 to its left.  That’s 7 safe squares and 1 mine known, from this box-logic move, plus the earlier logic move (1 safe, 2 mines), so 8 safe and 3 mines.  Where are the rest?  Well, let’s look at what we’ve got now:

There’s a simple logic move available here again.  Can you see it?  Look closely.

It relies on that 2 down below, between C/L/M.  It has the 1 next to it, on C/D/M, a neighbor mismatch.  By the 1-2 pattern, then, going left, L must be a mine, and going back right, D must be safe.  This is the rest of what we’ve been asked to solve for, the last piece of the solution.

How did we spot this box, when it’s not a normal shape, and know we could work with it?  Well, part of what might draw your attention here is the effective-2 that’s kind of floating in space.  But the main thing is, once we make our initial logic move, and have that 3 (between B/F) be an effective-1, it makes all the fronts on that box effective-1 fronts, i.e. A/B, B/D/E, and C/D/E all visibly have one mine between them.  So it suggests to us, visually, that a “3 sides confirms the 4th” type rule could help us out – and after some work, we can indeed show just that.

Solution:

Note that if we hadn’t seen the box, there is a proof-by-contradiction reasoning available by poking around that 5-square box: if you assume a mine to the left here (in the H/J/K group), it eventually overflows the 2 over at B/D/E, or similarly if you assume a mine on D, it forces a mine on A, leaving that effective-2 below with only 1 square to get its mines from.  That can be a separate exercise, if you like: prove this solution by contradiction, rather than box logic.  But that’s way harder!  And it’s the subject of the next chapter, not this one.

(8) Proof by Contradiction

Link to discussion section.  Link to puzzle section.

Puzzle 8-1 “Last Chance”: Mark 2, Clear 2.  Difficulty 3 / 10.

Source is this 100k attempt replay, at 2:46:20 (in top-center).

Analysis: This is pretty simple, there’s only 9 squares here, and 4 mines.  There is a fully-framed 4-square box at bottom, so one of two things is true: either (1) it’s a 50-50 box-and-2 forced guess, or (2) the only mine in the box is at lower-left, and another mine is to the right of the interior 3.  Let’s hope it’s the latter.  But for now, the important observation here is sort of a variation on the Rule of Three: we have a lot of 50-50 fronts on this region, but only two places where a number square faces more than 2 cells: the 3 at left, and the 4 at lower-right.  Those are good places for us to start.

Really savvy experts will be able to spot the conclusion here by inspection, but most of us will have to guess and check.  So let’s start guessing and checking on the 4 at mid-right, playing out what the conclusions are just within the bottom box:

Those look pretty straightforward, nothing unusual there.  But notice that the only thing all 3 have in common is that they all clear the 3rd mine for that 3 at middle-left, which clears the cell above-right of it, into the upper box.

This should make us think “hmm, is that a necessity?  What happens if there’s a mine there?”  I’ll tell you what happens! (at right)  It would clear the middle 3, forcing a mine to bottom-left corner, which clears the bottom-right.  This leaves the middle 4 (yellow frame) without an option for its last mine.  Therefore, this configuration is impossible!  We have proved, by contradiction, that that square cannot be a mine.  So the top box’s solution follows immediately from that.

Solution:  Worth noting that the first set of possible configurations explored, the set of all things satisfying that 4, constitutes a sort of “proof-by-exhaustion” that the upper cell had to be safe.  We can show that the reverse assumption leads to a contradiction, and generally if we spot that reasoning pathway, doing so will be faster.  But showing that “under all possible options, this has to be true” is a valid way to draw your conclusions too.

Even nicer than this solution is that, if you go look at the source game itself, the cell to the above-right of the 3 turns up a 5, meaning the 3’s last mine is to its right, which clears that 3 and avoids there being a 50-50 in the box below.  Who doesn’t love happy endings?

Puzzle 8-2 “Into the Heart”: Mark 3, Clear 4.  Difficulty: 3 / 10.  Source is this 100k replay.

Analysis: Firstly, since there are no untouched squares, it’s useful for us to show that this region has an exact minecount of five.  Square grouping is below.

Next, we notice that the interior 4 has one mine on its upper squares, and two mines below it.  This is a useful dichotomy.  Let’s try marking scenarios from its upper squares.  If we start by assuming a mine directly to that 4’s left, we get:

 →

The first image is straightforward, the logic proceeds around to that 3.  But then, we have a 4-square box situation in the middle: we have a 50-50 front from the 3 at right, in addition to what we previously had from the bottom and the left (the left is because the lower-left 2 accounts for one mine above it, so the remaining squares for the 3 next to it must have the other mine).  So therefore, it’s a 2-mine diagonal box, providing one mine to the top.  So based on that interior 4, which already has two mines marked and is now getting its third from the box, its fourth and final mine must be below-left, above the lower-left 3.  That’s Layout A.

The alternative scenario, Layout B, would put a mine to that interior 4’s right.  Let’s see:

 →

In the first image, the logic proceeds very nicely around the region: it goes around the corner, satisfies the lower-right 3, which clears to its left and forces a mine to the bottom-left of the 4-square box.  But then a funny thing happens: the upper 4 only has two mines marked, needs two more from two squares (blue box group)… meanwhile, the 3 from below has two mines marked and so it’s saying the three squares above it have one mine between them (yellow group).  This is a contradiction, they can’t both be satisfied.  Therefore the whole of Layout B is impossible, and Layout A must be true.

Solution: …and we’re screwed, because what’s left is a box-and-2 50-50.  So it goes.

Puzzle 8-3 “Assembly Line”: Mark 3, clear 3.  Difficulty: 4 / 10.  Source is here, at 1:55.

Analysis: This puzzle employs - twice! - the “split-front 2s” assessment technique discussed in the chapter.  All the action is on the top: The left 2 is sharing two of its three squares with the middle 2, which also shares two of its five squares with the right 3.

When we see this sort of multiple-square sharing, our mind should jump to the question “can these shared cells hold N mines?  How about N+1 or N-1?”.

Start with the left and middle 2s.  Can they share 0 mines, in the cells between them?  Clearly not, because the left 2 needs two mines and that would leave it only 1 square.  Can the left and middle 2s share two mines, in the shared squares between them?  Again no: that would satisfy the middle 2, clearing the other 3 cells around it, and leaving the right 3 only one adjacent square for two needed mines.  So they must share one mine (and therefore the cell below the left 2 is a mine).

Having decided that, consider the sharing between the middle 2 (which is now an effective-1) and the right 3.  Can they share 0 mines?  No, because that leaves the 3 with only 1 square for 2 mines.  So they must share 1 mine, and so the cell below this middle 2 is safe.

And if the middle 2 and the right 3 share one mine, then it means the square below that 3 holds its other mine.  This results in a few downstream logic moves along the right front, which solve our puzzle.

Solution:

Puzzle 8-4 “3’s A Crowd”: Mark 1, clear 2.  Difficulty: 4 / 10.  Source is here at 1:10.

Analysis: This one depends firstly on 4-square-box logic, and then using some proof-by-contradiction.  The same was true of Puzzles 7-6 and 7-12, which I put in the 4-square-box problem set.  This one also could’ve gone in either.  But I put it here, because I felt the box part was the easy part, and the PBC part was the harder part, and that wasn’t true with the other two.

So firstly, it’s trivial to show that the upper-left 4-square box here is a 2-mine diagonal.  We have 50-50 fronts from the top and left, and then by using the lower 5 and 3 in combination, from the bottom as well.  So the right side of that box, too, has 1 mine across those two squares.

Then all the action is with the 5, 3 and 1 on the right side.  With the information from the box, we know that one of the 3’s mines is accounted for on the left.  That leaves it three squares for which it needs two mines.  Could the 3’s safe spot be upper-right?  If it were, then the other two mines would be on the lower two squares, which are both shared with the 1.  That would overload the 1, which is a contradiction.  Therefore that can’t be the case - there has to be one mine on the upper-right (right of the 5), and the 3’s other mine has to be below, on one of those two shared squares.  This satisfies the 1, meaning that the two squares below it are both safe.

Solution: It’s basically just a 1-2 pattern, except one mine is split across two cells.

Puzzle 8-5 “Vortex”: Mark 4, clear 4.  Difficulty: 5 / 10.  Source game.

Disclaimer: Firstly, this can be solved directly by some very well-chosen projection / Evil logic and some 4-square-box logic, proving the cell between the upper 4 and 2 is safe.  But that requires insight about as deep as this PBC process requires anyway, so we might as well do it this way.  We can leave that insight as a further, bonus exercise to the reader from this puzzle.

Analysis: A little thinking should suggest that the entire outer perimeter of the region is fully connected - one choice determines the rest of the outside, and there are only two options.  Suppose we decide that the upper-left square is a mine.  That plays out as follows:

 →

First we’d mark the outside perimeter.  Then, we’d notice that the interior 3 at upper right is satisfied, so we can clear it (orange arrow).  This then forces a mine above-left of the 4, to satisfy that 4 (yellow arrow), which in turn would imply the pink square is safe.  But then consider: the 3 above that pink square needs a 3rd mine, and both of its options to get it have now been marked safe.  So this is impossible!  Instead, the top-left square on the perimeter needs to be safe, which results in that upper-right 3 only getting 1 mine from the perimeter instead of 2, and leaving us a coherent area behind.

Solution: Marked as above.  And we’re screwed (again!), because it’s a box-and-2 50-50 🙁

Puzzle 8-6 “Shadows”: Mark 1, clear 3.  Difficulty: 5 / 10.

Source is this game by nqm.

Analysis: This one depends on making a good choice of pivot on which to split our scenarios out.  The two numbers at top are both effective-1s, but the interior numbers below it are effective-2s, so that looks like a good place to find something in the friction between them.

Let’s make a simple observation about the top area, the corner 1 and the 2 at top-right: either they share their mine (on the two squares between them), or they don’t.

If they do share the mine, then it’s pretty straightforward: The squares below each of them are safe, implying a mine below-left of the 3, and a further safe square below.  The lower square of the yellow group and the square to the left of the interior 1 are linked, meaning they have the same status, but otherwise there’s not much going on here.

So what happens if they don’t share the mine?  Mayhem: mines below each of the 1 and 2 and safe squares in between, mean that we clear the interior 2.  The lower 2 gets one mine from its lower-right (orange arrow), but that mine also clears the interior 1 (in yellow), meaning the middle interior 2 (in blue) has nowhere to get its second mine from.  So, this is impossible, the top mine has to be on the shared squares.

Solution: The “mine on the shared squares” scenario above is the complete answer.

Puzzle 8-7 “Uninvited”: Solve the whole region.  Difficulty: 5 / 10.

Source is this 100k replay.  And while the minecount is demonstrable with a clever choice of square grouping, if I told it to you upfront, it would make the puzzle too easy.

Analysis: The first and most obvious tool to use here is that interior 2, which is an effective-1 that touches nearly half the squares of the puzzle region.  In theory we could make each assumption around it, five in total.  But we don’t need to be that thorough.  We can instead simply pair it with either of the number squares to its right that overlap partially with it - in the case of the middle 5 or the right 4, two of their three scenarios would satisfy the 2.  In the interests of starting from a central position (from which the “signal” of the scenario-testing can proceed in multiple directions), let’s pick that middle 5.

Going counterclockwise, here’s Layout A:

We first mark the assumption mine above-left, which clears the other two squares there.  But then, what this creates is a 4-square box with 2 mines on the diagonal, because we have a 50-50 front on it from 3 sides now.  This means it’s a 50-50 front from the 4th side on bottom, which faces up to that 6.  Because the 6 is an effective-2, this would imply the other unknown square to its right must be a mine.  But then, with that marked as a mine (which satisfies the central 2), it implies safe squares below, which imply mines on both squares of the bottom… which is a contradiction.  So this doesn’t work!  Let’s try Layout B:

 →

We quickly get to the same problem here as in Layout A: with the 2 satisfied from above, and the right-side 4 being otherwise cleared, this scenario would require mines on both the squares of that bottom row, which can’t be true.  So this layout isn’t right either!  How about Layout C:

 →

Here, our selection of the assumption square just right of the middle 2 first clears around that 2 (and the square shared by the 5s), and then there’s a bunch of logic that follows: the right 4 clears below-left of it, which forces a mine to the lower-left of the bottom box.  The upper 5 must have two mines to its left, and the 6 needs one more mine in its remaining square above it, so this fully specifies the entire area.  But we’ve also just proven that it’s the only possible logically-consistent layout for the area, too.  The same reasoning would’ve held true if we used some other squares (like that right-hand 4) as our pivot.

Solution: So it’s solved as discussed above.  

Sidenote: I had a caveat about the minecount above, but just to show: the count is directly provable by an exhaustive square grouping.  And that count is 5:

And the reason this would have made the puzzle too easy, is that there is an alternative square grouping, which takes advantage of knowing there are only 5 mines, and can show that two squares must be safe, on the basis of the “all groupings are correct” rule.

This re-grouping shows the two residual squares (in green) are safe.  Which means the upper 5 has two mines from two squares, so they’re both mines, and logic proceeds around to solve the area for us.  But that’s less fun!  Not to mention that when in mid-game, it’s pretty rare for an area to have an exhaustive square grouping like this (to show minecount), and very rare for that minecount to enable a solution of the isolated region - and rarer still to be able to see it as you sweep.  So the better, likelier approach with this region is to try out the scenarios, like we did.

Puzzle 8-8 “Two-Step”: Mark 3, clear 4.  Difficulty: 6 / 10.  From this game, which somehow had two proof-by-contradiction scenarios operating at the same time.

Analysis: The middle row from the hole pattern looks very much like the sort of situation you might do the Evil NG “hole 2-3” pattern on, using a 2’s back-side mine to wrap around and squeeze the 3’s third mine into a single cell.  But we can’t do that here, not directly, because facing the 2 on the left here is another 2 (under that upper-left 3), with split fronts.

An effective-2 with split fronts?  That’s the “split-front 2s” concept we covered in the discussion.  Let’s focus on that upper-left 2.  So either both its mines are on the right, or just 1 is, and the other is on the left.  Let’s start by imagining both the mines are on the right.

If both mines are on the right of the 2 (teal framed here), then it immediately satisfies the 2 at right, clearing the other 4 squares around it.  It also clears down to the 1.  This, however, leaves the middle 3 (in the pink frame) with only one more square around it, and it needs two mines!  So this is a contradiction, both mines for that upper 2 can’t be on the right; one has to be left, the other on the right.

What does that scenario look like?  Well, in that case we can actually do our Evil NG “Hole 2-3” pattern:

Assuming a mine to the left of the upper-left 2, the other is on its right (blue group).  This means there’s 1 mine in the other 4 squares facing the upper-right 2 (purple group).  Using purple plus the lower (yellow) group, means that the third mine on that interior middle 3 has to be below-right of it (orange arrow).  A simple step beyond this also shows that the top and top-right squares in the purple group must then be safe.

What else can we do?  Well, with the newly marked mine at middle-right, there’s a bit of logic available.  We can start with the 3 below that mine (now an effective-2), and notice that it has below it a straight 3.  So by our usual patterns, the cell below-right of that middle 3 is a mine, and then going back up again, the cell to the right of the marked mine must also be safe.

Solution:

Puzzle 8-9 “Bubbles”: Mark 3, clear 5.  Difficulty: 6 / 10.  Source game (at 5:00).

Analysis: This one is super fun, because firstly it looks like a standard Evil NG “projection” or “dependency chain” kind of situation, and kinda is and kinda isn’t.  But also because you can do the correct logic to find the solution coming from either the top or the bottom.  It turns out, the top is a bit more convoluted to do so, so we’re going to start from the bottom.

The bottom of what?  Well, the bottom of the three central columns here which represent a very “projection”-y sort of stack, starting with the 4 on the bottom.  That 4, which is an effective-2, has three adjacencies, so it doesn’t start out clean.  Let’s use our dilemma-busting Either-Or questions to clean it up: so the 4 has one remaining safe square around it.  Either that safe square is to its left, implying two mines above, or it’s one of the two squares above and there’s a mine to its left.  So, firstly, can the latter be true?  Can the 4 provide only one mine above it?

  →

If we test this out, our hypothesis is that the teal square group has one mine.  Therefore by the 3 above it, the yellow group has two mines.  This would satisfy the 2 immediately above the yellow, clearing those three squares around it, which then in the second image forces two mines to the sides of the upper 2 on our column.  Then based on that upper 2, the tiles above it (blue group) would have to be all safe.  But this would under-supply both the 3s there - the purple needs one mine, and the pink needs two mines and has only one other square.  So this whole test provides our answer: the bottom 4 cannot send just one mine above it, it has to send both its remaining mines up, and have a safe square to its left.

Incidentally, this implies a safe square to the right of that 4, above the 1, which becomes part of our solution.  Along with the 2 mines we just determined were above the 4.

Anyway, what becomes of our little tree here, now that we’ve determined that 4?

(1)  (2)  (3)

In Step 1, the same process applies, except now the yellow group has one mine instead of two.  Therefore the blue group has one mine (instead of all-safe), so the middle 2 in our stack implies that the purple group has one mine in it.  So far so good.  But now consider that upper 2 in the stack, in Step 2.  One of its mines is in the purple, which leaves four squares the other one could be in: the three above it, and the one below-right of it.  But if it were in the below-right square, we’d have the exact same problem with the 3s on top of the stack as we did in the previous attempt - they’d both be under-served.  So that can’t be true either, the square below-right of that upper 2 (to the right of the middle 2) must be safe.  Ok, so Step 3: now that we know the remaining mine of that upper 2 (inside the purple group) must be above it, in yellow, this means we can prove a mine to the right of that upper 3, and a safe square to its right.  It also means that, because of the upper-left 3 needing a mine in the blue group there, that the residual square of the yellow group, the one on the right of it, must also be safe.

Solution: So that’s the full story of it.

And again, you could pretty much do the same starting from the top if you wanted to, beginning with that top-middle 3 (pink, in our first scenario test).  Either that 3 has both of its mines supplied from its squares below it, leading to a contradiction, or it has only one mine from below it and the square to its right is a mine.  And that all proceeds in much the same way, with conclusions bubbling down instead of bubbling up.

Puzzle 8-10 “Wiggle Room”: 7 mines remaining.  Mark 1, clear 3.  Difficulty: 7 / 10.  Source.

Analysis: It surely caught your attention that this is the only puzzle in the set which offers a minecount in the problem statement, and that is for good reason.  It’s not solvable without it.  However, neither is it a pure minecount puzzle: minecount is used to arrive at a contradiction.  This somewhat blurs the distinction between a minecount proof-by-contradiction and the subject of the next chapter (Hypothesis Testing), in its solution process, but it fits well enough here.

Because we’re given minecount, let’s start with some square groupings.  There is no exhaustive grouping or even partial-overlap grouping available, but some are more useful than others:

        …Nope…                        …Nope…                                …Aha!

                Good:        

The first two attempts (pink indicating two mines, yellow indicating one mine) only get us six mines, with the 7th one floating about among four or five scattered squares, including the two untouched squares.  But the third one, using a grouping between the upper interior 3 and 5, can work: that pink group must contain either one or two mines.  We can’t be sure it contains two mines, so the grouping isn’t exhaustive.  But it could contain two, which means we’ve got some scenarios we can test with it.

A) Pink contains two: This is nice and simple.

 →

Because assuming two mines on the shared squares at top satisfies both the interior 3 and the 5, the other adjacencies are thus safe.  This implies mines in the other parts of the front there, plus a safe square at bottom-left.  We notice that this step 1 has assigned five mines, so we have two left.  There is only one configuration of remaining squares which satisfies all number squares with only two mines: the one in step 2 at right, with the mines placed on the shared squares.  So if pink contains two mines, this is our only layout.

B) Pink contains one, left column contains two: If we assume the pink group contains one mine, we have two choices for how we can resolve the upper-left interior 3: either its remaining mine is above it, which puts two mines on the left column, or the remaining mine is below it, which puts three mines on that left column.  Let’s start with the first assumption (assumption mine in teal):

 → (1)  or (2)

With one mine in the pink, the remaining mine for the upper-right 5 must be above it either way.  But the remaining mine for the upper-left 3… let’s assume it’s above it.  Then we end up with two mines on the left column, as seen in Step 1, and we’ve marked four in total so far.  Since we know from the first scenario that there is a two-mine option for the remainder, and one of those mines would imply the other, there are two possible layouts for the remaining three mines: either the opposite of the previous assumption (bottom row has the mine on the right instead of the left), which uses 3 mines to satisfy the known number squares.  Or, we do the same layout as the two-in-pink scenario above, leaving one free mine remaining, which then must be in one of the two untouched squares (in yellow outline, above-right).  Both are valid potential layouts.

C) Pink contains one, left column contains three: Let’s instead flip the assumption square at left.

 →

Assuming one mine in the pink and that the square above the upper-left interior 3 is safe, this gives us 3 mines on the left column, and five mines marked in total.  But the mine at lower-left satisfies the lower-left 4, so in Step 2 it clears to its right, puts a mine at lower-right, and this implies a mine next to the middle-right 3 above it.  But then the lower-right interior 3 has only two mines marked and needs one more from its only remaining adjacency (blue outline)… but we don’t have another mine to give it!  This Layout C would require 8 mines, and we only have 7, so it is impossible.  We can rule it out, based on minecount.

Solution: Putting it together, the remaining valid scenarios have something obvious in common: the lower portion of the left column.  There is a mine in the middle of that 4-4-5 stretch, and safe squares above and below it.  Furthermore, a bit of comparison will show that the square above-left of the lower-right interior 3 (the contradiction square in the last scenario, in the dark blue outline), must also be safe under all scenarios.  So that’s our answer.

Note that the two untouched squares are not definitively safe, either.  They are safe in layout A and B1, but not in layout B2.  If we had a minecount of 6, we would’ve been able to show that layout B2 was the only possible one, with both untouched squares safe, and this would’ve been a far simpler puzzle.  With a minecount of 8 or 9, this wouldn’t have been solvable at all.  But because of the interplay - particularly that lower-left 4 and the interior lower-right 3 - the 7-mine scenario here has a definite solution, too.

Puzzle 8-11 “Swim Lanes”: Mark 2, clear 3.  Difficulty 7 / 10.  Source here (requires some further probing).

Analysis: This one is tougher because of how many different spots are futile to try, or how many moves you need to play out after the initial assumption, in order to prove what you need to prove.  But it’s still very connected, and can cascade to a large extent.

So let’s start with what’s not going to help:

• The lower-left isolated 4
• The lower-right 4-2 on the wall, which is a 50-50 front that’s isolated (for now)
• The 4 untouched squares, in the middle of the lower-right block

Also, there’s one of our 3 cells to clear that’s available just through normal (NG-type) logic.  See at right.  You probably spotted this right away.  Its presence will help us, but if we’d known the actual tile value, it would have given away too much that made the puzzle interesting.

OK, now, what else can we make a supposition about that seems likely to have cascading effects?  Three candidates recommend themselves, all of them interlinked: both of the interior 5s toward the top and middle have 2 remaining mines from 3 squares, and the leftmost column is a flat framed front: either there’s mines on both the wide corners, or there’s one in the middle and the corners are safe.

Suppose the mine on the left-most column is in the middle squares, and the corners there are safe.  What can we conclude?  Not much: there’s a mine on the shared square between the two 5s below, and between the 2 and 5 at top, and then pretty much everything is reduced to an effective-1 in that neighborhood.  But what happens if that’s not the case, and there are mines on the corners?  In short: Mayhem.

The corners force safe squares next to those interior 5s, which means that each one forces 2 mines to be placed in the remaining squares around those 5s.  This then forces clearances for the 3s to their right, forcing mines among the 2s.  And then we hit the right side, where the 2 (in pink) needs 1 more mine… but the 2 to its left is ready to be cleared, so we’re saying those squares would have to be all safe! There’s our contradiction.  Therefore, those wide squares at far left can’t be mines, they have to be safe.  That leads to our solution.

Solution: Straightforward, it just required looking about 12 moves ahead to demonstrate it!

Puzzle 8-12 “Bon Voyage”: Mark 2, clear 3.  Difficulty: 7 / 10.  From this game (at 20:20).

Analysis: Sometimes, like here, half of the puzzle is figuring out what area to focus on.  In this case, there’s an area that looks like an Evil NG “Hole 2-3” sort of setup, basically in the middle of the region.  And the key we need is the “split-front 2s” technique.

No Evil logic is possible, not directly (as seen at left), nor can we use the Evil “projection” pattern from the bottom, or in the middle (from the left), but it does suggest that something might be available right around there.  Also suggestive is the fact that we have depth in our probing of the region right there - we have lots of interior squares available, so logic can go in multiple directions.  We do have depth in the lower portion of the region too, but that is so much higher-density, and we have so few interior numbers to work with, that drawing any conclusions there seems very remote.

Instead, if you were trying to figure out how to apply an Evil “Hole 2-3” pattern to the middle area, you’d notice one thing: you could get there in both directions, in each case if it weren’t for one pesky square.

(A)   (B)

In A, if we make a couple assumptions about extra safe squares (light blue), then our Evil NG projection logic from below means that there’s one mine in the yellow area.  Then there has to be just one in the blue area, and one in the purple group as well, and in combination that would prove 1 mine in the top-left there (and mean two safe squares on the right side of the blue group).

Likewise in B, if we assume the light blue square is a mine, then the yellow group only contains one mine (as per the 2 below-right of it).  So then, the blue group has one mine (per the 2), and we do our little hole-3 pattern on that middle 3 and could conclude there’s a mine below-right of that middle 3 (which would also mean two safe squares on the top part of that blue group).

In both cases, our ability to make definite logic moves is somewhat stymied by those 2s that are right around the yellow group.  That should lead us to notice that it fits into our “effective-2 with split fronts” case from the discussion.  So let’s focus on that, and ask: can we have both mines for those 2s be on their right, or does one of them have to be on the left?

Let’s try out the scenario where both mines for those 2s are on the right, and see where it goes:

If both of those mines around those 2s are on the right side, then it clears the 1 below it, forcing a mine onto the lower corner (orange arrow) and then a safe square to its right.  But crucially, it also fully satisfies the 2 to the right of the assumed squares (yellow frame), clearing all 4 other cells around it.  In combination with the safe squares around that lower corner, that means the 3s in the middle (blue frame, plus the one above it) only have one more square to get a mine from - and they both need 2 more!  So therefore, this can’t be true - both mines for that 2 can’t be on their right.  One of them has to be on the left (on the upper-left safe cell we’ve marked here), and one mine on the right.

This conclusion firstly means that we can clear the cell farther to the left of the mine we just marked.  But it also means that our hole-3 pattern B from above is actually true!  Now that we know there’s a mine on the light blue square, the rest of the logic follows: the mine it designates is in fact a mine, and then the rest of that pattern implies there’s 2 safe squares on the wide (top) part of the blue group.

Solution: And this is exactly how it goes.  There are some other very good guesses available, too, but these are definite moves that can be made, and they progress things quite a bit.

The key insight here is very similar to that of the “Two-Step” puzzle above - checking the split-front effective-2 - but the applicability of it is just more obscured by all the surrounding stuff.

Puzzle 8-13 “Hat Rack”: Mark 6, clear 7.  Difficulty: 7 / 10.

From this game by Cicada1.

Analysis: The first thing we have to do is see the projection logic that lets us mark 4 of the 5 mines around that 5.

The yellow group leaves 2 mines on the 4, and then the 3 implies 1 mine in the 2 squares to its left.  Given that the 5 has only 6 squares around it, this means that the other 4 squares around that 5, the ones not in the purple group, are all mines.  So let’s re-set our view on this area and look at what’s available then.

 → Labeled:

The key to this whole thing is the middle 3.  It’s got four squares around it and needs two mines.  One has to be on the left pair and one on the right pair.  But what about top vs bottom?  We can see easily that both its mines can’t be on the bottom (B and C), because that would overload the 1s on the bottom.  So it’s either one mine on top of the 3 (making A safe), or two mines on top (making A a mine and clearing B and C).

Can we have two mines on top of the 3?  I.e., mines on both F and J?  What that does is, it satisfies both of the middle 4s at the same time, so they would each clear around them.  And this would clear cells G and H above that row of mines.  Leaving no spot for the remaining mine for the 2s above that (yellow box).  So this can’t be the case - both mines can’t be above the 3 either.  Can only be one above the 3, and one of them below it - meaning A is safe.

OK, so between F and J, one but not both has to be a mine, and the other safe.  If J is the mine, then the 4 above is satisfied, and H, K and L are cleared.  If F is the mine, then the 4 above-right is satisfied, it clears G, which means H is a mine, clearing J, K and L.  Either way, K and L are cleared!  This is the insight that lets us solve the rest of the puzzle, because if K and (especially) L are safe, it means that M is a mine.  We can then clear to the left along the top front of the original puzzle.

Solution: After clearing A at bottom, and to the left of the middle-interior 4, we can proceed joyously around the top for the rest.

Puzzle 8-14 “Lagoon”: Mark 5, clear 5.  Difficulty: 8 / 10.  From this game.

Analysis: The real trick here is figuring out which square to use as a pivot, where we can ask the question “can this have 2 mines on the shared squares with ___, or does it need 1, or 0?”  Which means we are going to need something that’s at least an effective-2.  And for all the intriguing options around the ring here, going around that middle opening, they’re all pretty much effective-1s (or where we can very clearly delineate the square groups that satisfy them).  The only place that’s not true is on the right side, where we’ve got an effective-1 at top, followed by 2s and 3s below it.  That seems like the best place to look for a contradiction.

Since the main ring here is all connected and interdependent, let’s start by outlining the two possible layouts for that, and we can use them to make our analysis of the right side easier.  It almost doesn’t matter which square we choose as a pivot square, so let’s go with this:

(A)   (B)

In Layout A, we assume a mine on the teal-framed square to the right of the middle-left 3.  It propagates out to mines on the two “right corners” of this “ring”, let’s call it - the ones above the upper-right 3, and below the lower-right 3.  In Layout B, we make the opposite assumption, that the teal-framed square is safe, and then the end result is that these right corners of the ring would be safe squares.  The only other meaningful difference is that we need a mine from two squares between the lower 2s (the yellow group), and don’t know which, but otherwise it’s a mirror image.

So now we can look at the right-column squares here, and one thing jumps out as we think about it.  If we use Layout B for the ring, the two 3s on the right side both need 3 mines from 3 squares remaining (all of them on the right column).  That would put 5 mines in a row going down that right column.  But there’s a 2 in between them - it needs a safe square somewhere in there!  There’s no possible arrangement of mines, if the ring has Layout B.  Therefore Layout B is impossible, and the only possibility is Layout A.

Solution: There are multiple ways to get here, but they all arrive at the same place.

Puzzle 8-15 “Bookends”: Mark 2, clear 9 (clearing 8 is acceptable).  Difficulty: 9 / 10.

Source is this beginner game by Oreh1337, which confounded the entire MS Community Discord, who all spent a while arguing over whether it was a minecount problem or not. (it’s not)

Analysis: This puzzle really underscores the importance of actually picking a scenario and testing it, playing it out to see what happens and see if it reaches any contradiction that makes the assumption impossible.  It also might be the most complicated situation I’ve ever seen in a real-life Beginner game which still has a definite solution.

Before we get into it, I first want to emphasize one particular part of the board:

These squares here - a pair of 2s, the puzzle’s namesake “bookends”, which have 2 shared squares marked in orange - are the linchpin of the puzzle.  As we’ll see, much of the reasoning ultimately comes down to them, and we have to make a conclusion there before we can reach any part of the solution.  So if we refer to the “shared squares” of the bookends, that’s what I’ll be meaning.

Right, so let’s get to it.  Try as we might, no direct pattern logic is available here, and minecount can’t help us… but looking at this, it feels like it ought to offer something.  So our first step is to open it in an editor and pick something to assume as a mine, and see where that leads.  Let’s suppose we pick the following (teal frame) and play it out:

In Step 1, we assume the teal square as a mine, which by the 1 to its left, clears around that.  Some minor conclusions flow above.  In Step 2, we see this forces a mine to the left, which then clears around the 1 in the middle of the lower hole pattern, which leaves only 2 squares remaining for the 2 to its right - so those (the Shared Squares) would have to both be mines.  In Step 3, the shared squares clear the other bookend 2, which then leaves only one square for the 2 to its right (pink frame) - that’s a mine, but there’s no place for it to get its other needed mine.  So this whole thing is a contradiction!

Note that the proximate cause of our contradiction was assuming the Shared Squares were both mines.  Any scenario that results in needing two mines there, will create the contradiction for that pink-frame 2 in the same way.  So we can back up and conclude that the shared squares can’t both be mines.

That’s a useful realization.  But hang on, can we have a scenario where neither of those shared squares are mines?  Let’s test that:

(1)   (2)

In Step 1, we assume those two shared squares (teal frame) are not mines.  Then using the 1-2 to their left, a standard 1-2 pattern lets us conclude the square above-right of that bookend 2 is a mine.  This clears around the 1 above it.  Then in Step 2, that clearance forces a mine to the left of the gap there.  That mine would satisfy the 1 that we relied on in the previous step, meaning its other adjacencies are safe.  However, this leaves nowhere for the bookend 2 to get its second mine from!  Therefore, that’s impossible as well: the shared squares can’t both be safe, either.  Thus the shared squares between the bookend 2s must contain 1 and only 1 mine between them.

That’s a very useful intermediate conclusion, which we can now use to begin building the solution.  If the righthand bookend 2 is getting 1 mine from its left (yellow group), then it’s an effective-1, and this means we can use the 1-2 pattern to mark a mine above the far-right 2, and then three safe squares follow in that lower-right region:

(1)   (2)

Next, the logic we used initially about the square to the right of that interior-1 in the upper hole pattern - turns out, that same reasoning applies for all three squares to its right (but not the one below it, since that supplies the bookend 2 with its remaining mine).  And then, if those squares are all safe, we can locate a mine above that little column, and some further logic proves two more safe squares.

 →

That’s 2 mines marked and 8 safe squares cleared.  But the puzzle asks for 9 safe squares!  If you got this far, you should consider yourself as having solved the puzzle.  The final one requires the Hypothesis Testing technique from Chapter 9, or a lucky guess-and-check.  If you’re going to use Hypothesis Testing, you’d pick as a pivot the middle-left split-front 1s, with 3 scenarios to play out.  I will leave it as an exercise to flesh out fully, but the short version is:

• If the mine is on either of the two lower squares, then the 1 below it clears the 3 squares below that.
• If the mine is on the upper square, then it clears the 1 to its right.  Since the bookend squares contain exactly one mine, this leaves only two squares remaining for the middle 2 (to the left of the bookend squares) to find its second mine.  But the 1 next to it says that the three squares underneath it must contain one mine as well; this means that via a 1-1 pattern, the third square, the one to the left, must be safe in that scenario.
• Thus under all three scenarios (which are exhaustive), one particular square is safe: the one directly under the left-most 2.

Solution: Including that bonus safe square from hypothesis testing, the complete answer is:

(9) Hypothesis Testing

Link to discussion section.  Link to puzzle section.

Puzzle 9-1 “Boxed Out”: Mark 2, clear 2.  Difficulty: 4 / 10.  N.B. this puzzle can also be solved with other techniques; please try to use the ones from this chapter instead.

  Source is this game (at 42:20).

Analysis: This one is small (only 9 squares) but cleverly shows how the hypothesis testing approach can sometimes work.  The key for this is that every bit of the front is an effective-1 with 2 front squares… except for the interior square (which serves as a junction for conclusions coming from multiple sides), and for the 5 at lower-left, which sort of “flips” the sequence of mines and safe squares that otherwise proceeds around the left-hand 4-square box.  We can pick any square on the left side and declare it safe or a mine, and things play out in a similar fashion.  Let’s alternate our assumptions using the left-hand front, and see where it goes:

(A)   (B)

In Layout A, we assume a safe square at lower left.  This forces the square below the upper 4 to be safe, but also forces the other 2 squares above the 5 to both be mines.  In Layout B, we assume that same square to be a mine (teal frame), which forces the square below the upper 4 to be a mine, and forces the other two squares above the 5 to have one mine between them (yellow group).

And then here’s the key: those are the only two scenarios, they both follow directly from making one assumption (or not).  And in both cases, the central interior 3 is fully satisfied!  It either gets its needed two mines from below (A), or one from left and one from below (B).  This means that in all cases, the two squares to its right are safe, which is our answer.

Solution:  This is more of a warmup puzzle, and it can be solved using minecount

The simplicity of this puzzle makes it a great choice to illustrate one of the sort of general principles of identifying where these sorts of solutions can be found (whether by proof-by-contradiction or hypothesis testing).  The key is that the arrangement of squares we’re facing is such that, if we make an assumption in that left box, “information” (the assumption and its consequences downstream) proceeds in two directions from there.  An assumption anywhere in that box will go around it and determine the status of both the square to the left of the 3 as well as the ones below it.  So not only does this information in an assumption go in two directions, it goes out and then comes back together - it “rejoins” at the decision point at that central 3.  That characteristic was very common to the puzzles in the last chapter, and will be common to the analysis of ones in this chapter too - in fact, in the discussion section about how to identify situations of this type, it’s heuristic (D).  And it’s an aspect that you can’t use for easy pattern-recognition, but it’s understandable enough that a human can still look for it.

Also of note, the 4-square box at left that remains in this puzzle is not necessarily a 50-50 box-and-2.  That box cannot contain just 1 mine, obviously, but it could still contain 3!  If the lower safe square here turns up a 4 instead of a 5, that means that the square below the interior 3 is safe, which would mean that both squares to its left are mines and only the upper-left square is safe.  That is only 1 of the 3 possible permutations of mines in the remaining area, but its existence is enough to tilt our odds of clearing the area from 50% to 67%.

Puzzle 9-2 “Single Step”: Clear 3 cells.  Difficulty: 5 / 10.  From this wheel-event 1k board.

“A journey of a thousand miles begins with a single step.”

Analysis: You may have seen this puzzle layout, with just 8 number squares and zero mines marked, and thought “you must be joking”.  And yet we are not!  This is a beautifully-simple illustration of the technique that hypothesis testing asks for.  There might be a direct way to find the answer here, but in my view, hypothesis testing is the easiest path to the solution.

The obvious, no-brainer choice here for a pivot square is the top-right corner 1.  Any choice of assumed mine around that corner immediately creates a ton of downstream safe squares.  So let’s use that as our pivot, and try each of the three scenarios we can assume from it.

Layout A:  →  →

Layout B:  →

Layout C:  →

In Layout A, the assumption forces safe squares to the left, and then the central 1 has only one option left for a mine, so that’s a mine, which implies (from the 1 to its right) the two squares at right are safe.  Layout B is simple: the assumption square clears everything in the vicinity, leaving the 3 barely enough squares to mark all its mines (which satisfies the lower-left 2, clearing below it).  And in Layout C, the clearing of safe squares goes down first, then that central 1 forces a mine above-left, which marks the top-left as safe.  But we can’t say anything firm about the lower/left side, as the 2 and the 3 could share either one or two mines.

OK, so those are our 3 possibilities here.  Hypothesis Testing says we should look at the overlap from those, and see if there are commonalities between all of the layouts.  So compare the final image from each of the 3 layouts - are there some squares that are always mines or always safe?  Yes, in fact: two squares along the right edge are always safe, as is the top-left square on the edge too.  So we’ve just created 100%-safe moves even when no direct logic was available to find them.  Cool!

Solution:

You can also see, from the scenarios above, why the cells below the lower-left 2 are so low-percentage - they are all safe in Layouts A and B, and in Layout C, they can only be mines if there’s a mine above-left of the 3 (implying just one mine on the two squares shared between the 2 and 3), which is an unlikely coincidence to begin with.

As a sidenote, MS puzzle enthusiast LukewarmToasterOven believes this one can be directly reasoned from some (very fancy) projection logic.  I don’t quite follow his reasoning here, but this is how he did it, and he got the right answer, so maybe it’ll make sense to you:

Puzzle 9-3 “Gradient”: Clear 2 cells.  Difficulty: 5 / 10.  Source is this intermediate(!) game.

Analysis: The rising numbers in the interior hole, 1-2-3, strongly suggests some sort of logic is available (but alas, the Evil NG “Wheel 1-2-3” pattern is not possible).  The other parts of the front are semi-isolated.  So let’s focus on the top area.

The top is very connected, and in fact creates two distinct scenarios for what goes around that central 3: either there’s a mine at top-left corner, or there’s not.

(A)   (B)

In A, if we assume top-left is a mine, then a mine flows to the top-right corner, and by the 1 underneath that top-left corner, the two cells below it are safe.  That only leaves 2 spots for the other mines of the 3, so those are mines too.  Also the mid-right 2 is satisfied, so that implies the one below-left of it is safe.

Meanwhile in B, if we conversely assume that top-left is safe, then a mine ends up on the shared square between the right-hand 4 and 2, clearing down.  If we then use a 2-over-3 pattern (from the left) to mark the square directly under the 3 safe, then the one below-left of the 3 is the only one left and has to be a mine (yellow arrow).

Both of these scenarios are valid, and they are each exhaustive of the squares around the top region.  But do they have an overlap?  Yes, in two places: the left-hand column below the 1 is safe in both, and the square to the left of the 1 on the right side is also safe in both.

If we were to make one of these (probably the left one) a mine, and play it out, we’d eventually reach a contradiction.  But the easiest way to see it prospectively, rather than retrospectively, is to just run the scenarios and look at the overlap.

Solution: …and what’s more, clearing these squares end up letting you solve the board.

Puzzle 9-4 “Keyhole”: Clear 2 cells.  Difficulty: 6 / 10.  From this 1k game.

Analysis: So with this one, there’s a proof-by-contradiction approach you can take with it, or a hypothesis-testing approach, and the two are sorta linked.  But the PBC approach requires a very clever choice of bifurcating the scenarios, whereas the hypothesis testing can kind of start from multiple places, so we’ll approach it that way.

To begin with, you might try to assess this with minecount, and will find you can get a grouping with 7 mines many different ways, each of which leaving 1, 2 or even 3 squares behind, but there isn’t anything that groups every tile, and we don’t have a total count anyway.  We might also try to pick a pivot spot that determines a bunch of things downstream, but there’s really not a good one that provides binary scenarios to test (they all peter out very quickly with no conclusions we can use).

Instead, however, we have five squares we could choose which each provide ternary scenarios (they’re surrounded by 3 unknown squares), all on the left half of the formation: the lower-left 4, the middle interior 5, the middle 2 that’s below the interior 2, or the interior squares from either the interior 3 or the interior 2, by projection.  Some of those look a bit tricky to try out, so let’s suppose we do the lower-left 4.  Start with assumptions going clockwise:

The assumption cascades to the right and above and around.  And it quickly reaches a situation where we’ve marked everything on the left half of the region, except for the three squares above the interior 2.  Those three squares are spoken to by three different number squares: the 3 above-left, the 5 above-right, and the 2 below it, each of them now effective-1s.  So from logic, we know that the only solution there is one mine in the middle square (right image).  Next up:

This time for Layout B, we try assuming a mine on the interior square (the upper right corner of the 4).  This cascades to the right and down as it did in Layout A, and cascades (with opposite assignment) going up and around the top.  However, a funny thing then happens: the interior-3 is satisfied, so it wants to clear the remaining middle squares below it, and the 5 is satisfied as well.  But clearing below those would leave the interior 2 (in pink) under-served!  So that’s a contradiction: this cell can’t be a mine.  That’s one of our solutions!  Let’s do Layout C just for completeness.

If we now assume the square to the right of the 4 is a mine, then the conclusions at the top are the same as Layout B (mine top-left, mine to the left of the 5), but the conclusions to the bottom are different: with a safe square to the right of our assumption square, we now have a 2-mine diagonal in the 4-square box there (50-50 fronts from left, bottom and right sides).  And while we can’t locate additional mines, our markings thus far add some additional conclusions: there are two squares left from which the interior 3 can get its last mine (blue group), and between that and the 4-square box (in brown), the interior 2 is satisfied.  This clears the square above-right of that 2.

So looking at an overlay, we see that two squares are safe in both Layouts A and C, while B is impossible (and its assumption square is one of our two commonalities).  The other one, we could have done logic to infer from the first one: as we see in layout C, if the square above-right of the lower-left 4 is safe, then by projection, we infer that the square directly below the 5 is safe too.  This is our solution.

Solution:

Now, the more direct proof-by-contradiction approach here would have been to observe the projections of the interior 3 and interior 2, and note that they both ask for one mine from 3 partially-overlapping squares:

The teal group means that the interior-3 requires one mine from the blue group, and the purple group means that the interior-2 requires one mine from the yellow group.  Thus, we might have reasoned, either the blue and yellow groups have the mine on their shared squares, or they don’t.  That would have been the either-or choice we could have created that would lead us to the answer (because a test of the “maybe they don’t, maybe there’s mines on the wide squares” scenario would prove impossible for the same reasons Layout B did here).  But seeing that and making that choice is something that is asking a lot of a player facing the position naively.

Puzzle 9-5 “Stowaway”: 5 mines remaining, 1 to mark, 1 to clear.  Difficulty: 6 / 10.  Source.

Analysis: This one gets a difficulty of 6 rather than a 3 or 4, because it requires a preliminary step.  And that step is mine counting.  Mine-counting here will prove a mine, but not any safe squares.  Here’s how it goes (at right): We can get square groups from the bottom, top, and two interior groups of three, such that we can account for four of the five mines with only one square remaining.  This is an “N-1 grouping” or “N-1 minecount”, and can be used to prove that that residual square is a mine.  However, it doesn’t get us progress, i.e. a safe square.  But it does provide the groundwork to enable that progress.

To move forward, let’s call that mine marked, and try some scenarios to see if there’s commonality or contradiction.  Neither of the two-square groupings can be comprehensive in dictating the rest of the region, so we’re going to choose one of the three-square groupings as our pivot - let’s say the yellow one, since that touches just about everything else.  I’ll spare you the exact steps, but let’s mark each of the yellow-group squares as a mine (in teal) in turn, and play out what that does to the region:

(A) (B) (C)

In Layout A, we assume a mine on the rightmost square, and logic proceeds counterclockwise.  The left-hand 3 clears below it, which determines the lower row.  Layout B is nearly identical, just a different square shared by the 4s.  But Layout C, where we assume the square to the left of that lower 4, requires a sort of opposite layout: the other 4 pushes a mine to its left, and the middle 3 is now satisfied so it gets safe squares to its left, which forces a conclusion from the left-hand 3 and the opposite conclusions down the bottom.

So if we look at these 3 layouts, are there any commonalities (besides the mine we already proved with minecount)?  Yes there is, but just one, right beneath our mine from minecount.

Solution:

Note, also, that the safe square is not a “dead cell”: it can take different values.  Count the mines around it: in Layout A, it will turn up a 2, but in the others, it will turn up a 3.  And then the rightmost unknown square will thus be safe, and it will either turn up a 4 (Layout B) or a 5 (Layout C).  So even without knowing what this square will show, we already know - from our hypothesis testing - that the entire area is solvable!

Puzzle 9-6 “Merry-go-round”: Mark 4, clear 3.  Difficulty: 6 / 10.  Source: this 100k attempt.

Analysis: At first glance, this doesn’t look like a promising situation - there’s only one effective-2 in the operative part of the front (the top 4), all the rest are effective-1s which are semi-isolated by all those interspersed mines on the edge.  But they’re connected enough that we can try to run scenarios on them and see what we find.  It basically doesn’t matter where we start, so let’s just pick, say, the lower-left squares on the upper region.  We’ll put the mine in the corner first, and then not in the corner.  Layout A:

 →

Logic proceeds from that corner both upwards and to the right, putting a mine in the lower-right corner of the upper region.  This clears around that upper bridge 2 (and the lower bridge 2), cascading to the lower region, but also going upwards around the upper region, and filling back in along the right and upper edges of the upper region.

Now how about the opposite assumption?  Layout B:

 →

In this assumption, having a safe square at lower-left results in the opposite conclusions, at least initially.  We can make more progress going clockwise along the upper region, however, because the upper 4 is an effective-2, so a safe square in the upper-left corner fixes its two mines and allows more markings.  We end there with that upper bridge 2 having a mine to its upper-left (and a safe square directly-left).  Then logic can proceed into the lower region, where we get the opposite results along the left edge, but the same results (!) along the top and right edge, because either way, that upper bridge 2 is cleared.

This overlap in conclusions provides our solution.  Well, that plus the fact that there’s always a mine directly under the upper 4.

Solution: As we saw, the key to this was the two squares in between those interior mines, to the left of the upper “bridge” 2.  One of those two squares has to be a mine under all scenarios, and that fact clears the rest of the 2’s adjacencies and gives us the key part of our solution.

Puzzle 9-7 “Bad Wheel”: 1 cell to clear.  Difficulty: 7 / 10.  From this 1k game.

Analysis: This at first looks similar to a lot of Evil NG projection patterns.  But while we’ve got Hole 2-3, Wheel 1-2-3, and even Hole 2-4, they all require some elements we don’t have.  For example, although the hole cells fit the “x-2-4” needed for that pattern, we would also need an effective-1 on the top of the front here, and it’s obscured by a mine. So Evil NG logic won’t help.

Next we might try projection, and an Either-Or.  Two spots look likely: The 5-3 neighbor mismatch at left, and the 4-1 neighbor mismatch at lower-right (which is a big difference!).  The 5 and 3 can either share one or two mines, but neither results in all that much, and they don’t overlap or reach a contradiction.  Maybe we’ll come back to that.  The 4 and 1 can either share a mine or not share a mine, and the “they don’t share” scenario results in a nice cascade around the region, but the “they share” scenario doesn’t yield enough downstream conclusions.  But while trying both of these, we should get the impression that if we pair our reasoning with an assumption about the top row, we might get somewhere.  So let’s make the top-left our initial pivot square, and if we have to get weird with it, then we’ll get weird.  Layout A:

We assume the top-left is safe and the square next to it is a mine.  This makes the square above the 4 safe, and by projection to the central 2, we know the yellow group contains one mine.  What we realize next is that the 4 is now constrained: it has six squares around it, one is safe, and two others have at most one mine.  So the other three (to the 4’s right) must all be mines, and the remaining two squares (blue group) have a mine.  This implies the left square of the yellow group must be safe.  Then the mines forced to the right column satisfy the lower-right corner 1, putting safe squares to its left, and fixing the location of the yellow-group mine as the middle square, below the 2.  Then by projection, the 3’s remaining mine must be in the purple group, so the 5 needs its free square below to be a mine.  Pretty thorough!  Let’s try Layout B:

 →

Initially, we have two mines along the top, and the same projection down to the yellow group.  But this time the 4 is not (fully) constrained, so we look elsewhere.  The 5 is now an effective-1, but its neighbor the 3 is an effective-2.  So the blue group can’t have two mines (overloads the 5), nor can it have zero (the 3 would need two mines from the yellow group, which has only one), so it must have one mine.  This puts a safe square below the mine, and it also means that the 3 has its third mine below it, so the yellow group can’t have a mine on its right square - that one has to be safe.

So if we look at the two layouts, their conclusions do overlap on one square: the one below the 4, which thus has to be safe in all circumstances.  Great!

But that feels like a weird conclusion, with a lot left unsaid or undone.  Can we go a bit further, to help our understanding?  Well, that would require a further pivot, but in this case it’s easily done.  The 4 is nearly constrained, so we could say that the square under the central 2 is either a mine or it’s not - that should fix other things into place.  Let’s look at how that plays out:

(C)   (D)

Both are valid / possible.  In Layout C, we get the same right column of mines, which satisfies the 1, which pushes the mine out to the left of the yellow group, under the 3.  We can’t resolve the blue group between the 5 and the 3, but they’re both effective-1s now and everyone else is happy.  On the opposite assumption, we end up with the same blue group, but we now can’t draw any conclusions from the 4 (it has two mines from the three squares to its right) nor the corner 1 (one mine, two squares to choose from).  Certainly Layout C is a lot likelier, as Layout D could require 8 mines from the region instead of 7.  But we don’t find anything else that is always a mine or always safe.  So what we got from the first two layouts is all we’ll get.

Solution:

Puzzle 9-8 “Vise Grip”: 5 mines remaining, 2 cells to clear.  Difficulty: 7 / 10.

This is the endgame from this Evil NG game by Skywalker2022.  Despite him being a top player, he had to ask the WoM chat for help on it.  What made this such an unusual puzzle?  Read on!

Analysis: So with a square-shaped region and lots of interior squares like this, the first thing I’d try is minecount.  It’s pretty easy to get to 4 mines, in lots of different ways, but there’s no direct way to get a square grouping that gets you all 5.  Which makes this a “near-minecount” situation, susceptible to Llama’s technique, and thus worth trying some hypothesis testing.  We’ll do that first, and then we’ll show another way to get the solution as well.

Although it’s pretty flexible as to what pivot square we pick, I’ll choose the top-left of the formation, as it pretty directly cascades to the rest of the region.  Let’s assume it’s a mine first.

When we do this, we get a mine lower-left as well.  Then from the safe square to the right of our pivot square, we can infer a mine in the top-right two squares (blue), a safe square below them per the 2 at right, and a mine lower-right too (yellow).  That’s 4 mines, though.  We only have one mine left, and we need to satisfy the interior-left 3 and the middle 2.  So our only choice is the square shared by them.  With this as our 5th mine, all the five other squares must be safe.

So that’s Layout A.  In the converse assumption, call it Layout B, we get:

Our assumption of a safe square in upper-left cascades down the left column, but forces a mine to its right which also marks the upper-right squares as safe.  Then what follows is two mines marked down the right column, and the interior-3 at left clearing its two other interior squares.  Finally, the middle mine at right satisfies the interior-2, clearing one square below it.  The final (5th) mine must be in one of the remaining 3 untouched squares (blue region).

Compare Layout A and Layout B.  Are there overlaps?  Yes, there are!  Two squares, in fact, are safe under both assumptions: the one under the interior 2 (final one marked in Layout B), and the one below-right of the 3 at left.  So under both hypotheses, which are exhaustive (one of them must be true), in either case those two squares are safe, and can be cleared here.  Crucial to the solution was the minecount, because in Layout A that’s what forced the final mine to be shared by the interior 2 and 3 squares; if we had more mines, we would not have been able to assume that, and so the two layouts would not have had certain overlap.

Alternative: Now, that hypothesis-testing is a fairly straightforward path to finding the solution here, and might be the first one you try - might be fastest, actually.  But there is a direct logic path to get there as well, a slightly-fancier minecount proof-by-contradiction.

Let’s mark some square groups, in which we know a mine has to exist.  From left, the 2 gives us the blue group, from which then the interior 3 implies the purple group (all 3 squares to its right).  From right, we have the yellow group implied by the upper-right 2, and below that, the 3 has a mine in the teal.  That’s four of our five mines.  Now consider the perspective of the interior 2.  It’s definitely getting 1 mine from the right, but its second mine can either come from the purple group (if it’s on either of the upper two squares of that purple group), or if the purple group’s mine is not (i.e. if the purple mine is in the dotted-brown square at bottom), then it’s coming from the square below it - in pink.  So the pink square and the dotted-brown square are dependent, or linked - they have to either both be mines, or both be safe.  One’s status implies the other’s.

However, if both of them are mines, then it means the upper-left squares still need a mine.  And they can’t be getting it from the top square of the purple group, so instead it has to be coming from the top square of the blue group.  Which means we need a mine at the bottom-left square (dotted-gray).  But that would be a 6th mine!  So if the two linked squares are mines, it means we need at least 6 mines to solve the area.  If this weren’t an endgame, we couldn’t know which was true.  But because we know there are only 5 mines, we can conclude that this scenario is impossible - there can’t be 6 mines, so the two linked cells can’t be mines.  Therefore the two linked cells are both safe.

Solution:

The fact that the interior-2 is “squeezed from both sides” here, gives the puzzle its name - and the path to its answer.  Furthermore, because this is an NG game, experienced players will spot that subsequent moves must show that Layout B is the only possibility - because Layout A actually has an extended line-and-1 50-50 on the right column.  Good thing we’re in NG-land, where there are only happy endings!

Puzzle 9-9 “Trident”: 6 mines remaining, 3 cells to clear.  Difficulty: 8 / 10. From an int game!

Analysis: The 1 up the left edge there was a lucky guess, which (if I’d used a hint at that moment) I wouldn’t have made, and if I had used a hint, it would have recommended the 3rd cell along the bottom, which would have led to a solution from logic alone - much less fun!

Instead, faced with the exact situation here, we would probably try to approach via minecount first.  That would not yield a definitive solution, but would at least get us to an N-1 minecount:

So we can account for 5 of the 6 mines this way, with two squares left over.  Take careful note of them, above and below the set of hole-pattern 2s.  That grouping doesn’t prove a safe square or a mine, but it does two useful things:

1. By Llama’s “near-minecount” principle, the two residual squares here are a good choice for pivot squares for a hypothesis test; and,
2. Because we can apply this partial minecount even if one of those two squares were to be known, it means if one is safe the other is a mine, and vice-versa.  They can’t both be mines (would overflow the minecount), can’t both be safe (would underflow it), so they have to have opposite status.

So let’s try a hypothesis test starting from the upper square of these pivots.  The first layout has a lot that flows logically:

 →

The assumption square as a mine clears around that interior 4, which forces two mines down our right column.  Then, the upper interior 2 is satisfied, so it clears to its left, forcing a mine below those squares, and thereby clearing around the bottom.  We’re left with two of our six mines, shown in groups.

How about the opposite assumption?  Layout B:

In Step 1, our pivot square is assumed safe, so its “twin” (from the N-1 minecount) at bottom is therefore a mine.  Based on the pink group at lower right, the squares opposite that lower-interior 2 must be all safe, via projection.  This means the square above the lower-left corner 1 is a mine.  Next, if we project from the right side a bit higher up (brown group), across that upper 2, there is a mine in the blue group, meaning the squares above and below the mid-left 1 are safe.  Finally, if we return to the right column and start from that middle spot, and use the middle interior 2 (in purple), the yellow group has one mine so that purple 2’s other mine must be above-left of it.  This forces a safe square above-right of that mid-left 1.  That’s three mines marked, and the remaining three are in groups the same as in our partial minecount.

Are those scenarios enough to prove anything?  Yes, in fact there are 3 overlaps:

Solution:

…and those squares lead to a full solution through logic moves.  But getting there was hard!

Puzzle 9-10 “Shell”: 1 cell to clear.  Difficulty: 8 / 10.  Comes from this game.

Analysis: Looking at this, we’ve got a “double-box” at lower right, where we have some interior squares but we also have some framing of 4-square boxes, which themselves partially overlap.  That looks like a good place to try things, not least because the upper-left area has a bunch of untouched squares that make this technique largely useless (without a low minecount).

The first thing that jumps out at me is that the interior 4 and interior 5 of the lower-right area are both effective-2s, while the rest are effective-1s.  That’s usually a good sign.  Let’s try some scenarios where they share 2 mines in the two shared squares between them, and where they share only 1 (they can’t share zero, the 5 only has one other square).  Assuming 2 shared:

Looks pretty nice!  The cleared squares from our other adjacencies on that 4 and 5 solve the whole lower region and also the 3 to its left, letting us mark and clear half of the upper region too.

But what about if those two squares only share one mine?  Let’s mark that with a teal area (for “one mine in these two squares”) and see how it plays out.

So in this case, the assumption (teal) forces a mine to the right of the 5, instead of a safe square.  This means we don’t solve the lower-right 3.  However, it does mean the 3 is now facing a 50-50 front above it.  In combination with the 50-50 front to the right and top of the 4-square box, this means it’s a 2-mine-diagonal box.  Which means it provides a mine to that interior 4 from the left side of the box, which means the one remaining adjacency, to the left, is safe.

Solution: Looking at the pictures of these two scenarios, there is in fact one overlap: the square to the left of that middle interior 4.  So that is our answer.

If the safe square turns up a 5, then we’re looking at the 2-shared-mines scenario, and we have happy days.  However, if it turns up a 4, then we’re looking at the 1-shared-mine scenario, and not only do we immediately know a lot less, but it also means the 4-square-box at the right is a forced-guess 50-50, since its lower-left corner will then be a mine.

Puzzle 9-11 “Bridge”: 8 mines remaining; Mark 3, clear 3.  Difficulty: 8 / 10.

Source is this game, at 7:00.

Analysis: On the one hand, this looks a bit like a normal (if fancy) mine-counting puzzle, since there’s only one untouched square, and we give the count.  On the other hand, the key insights leading to the solution come from hypothesis-testing only.

The bottom region has three mines in its five squares, and the two squares that “bridge” the gap between the regions, that central 3 and 2, are thus determined by our choices over there.  So let’s pick one of these - let’s say, the central 2 - for which we will assume each of its three adjacencies in turn are safe (thus the other two are mines), and play those out.  Layout A:

Our assumption of a mine in the upper-right corner of the lower region puts two other mines in that box, and also a mine across on the upper region, on the 2’s other adjacency.  It then requires a mine above the 3 above it, and a safe square to the right.  So now, note that we’ve used 5 mines already, and we only have 8 total.  I’ve given a residual square grouping for the groups we know we have (to say nothing of the 1 untouched square, which clearly would be safe here).  Try as we might, there is no combination of mines here that produce a valid layout using only the three mines we have remaining.  So Layout A must be impossible.

Let’s try putting the mine in the upper-region square, above-right of that 2, for Layout B:

The assumption forces two mines to the left of our central 2, which determines the rest of the lower region.  Then those placements force two mines, above and to the right of our assumption square (middle image).  But this does a big thing for us, because the one to the right satisfies the lower 4, clearing the other two squares above it and forcing a mine to the upper of the rightmost squares.  This leaves us having used six mines to get to this point, with two left.  If we look at the groupings, this layout is only possible if the mine is on the shared square between blue and purple, i.e. the square where the orange arrow is pointing.  But it IS possible!  We don’t care where the yellow one is, but for this assumption to be valid, it forces a 7th mine, onto that square above-left of the upper 4.  So that’s one option, how about the third choice?  Layout C:

This assumption square forces mines around the lower region, as-before.  Note that there are commonalities there with Layout B (two mines and one safe square), which must therefore be part of the solution.  In the middle step, our assumptions have enabled both squares above-right of those 3s to be safe, which means we’ve only used four mines when getting to the square-groupings part.  The yellow, disjoint from the rest, accounts for a fifth.  We then have more possibilities forking out from there, which we can enumerate without diagramming them out using a secondary pivot:

• Suppose the left square in the teal group is a mine.  Then so is the upper square in the pink group, which reduces the rest of the blue group to having only one mine… but we only have one mine left, so it has to be on the shared square between blue and purple.
• Suppose alternatively that the right square in the teal group is a mine, satisfying the pink too.  We have two mines remaining, and the blue group demands both.  They can’t both be on the bottom square, then - one needs to be in the square shared with the purple.  But the final one could be on either square below the 4 in the blue, it doesn’t matter.

So that’s a total of three configurations possible here from our assumption in Layout C - complicated, sure, but still fairly limited.  In all of those, however, there is a mine in the square shared by the blue and purple groups.  And looking back at Layout B, the one valid configuration for that assumption also had a mine there.  That’s another overlap, so that’s a part of the solution - the mine there, which implies a safe square just above it.

Add to that that none of these configurations had room for a spare mine, therefore the untouched square (below the yellow groups) must also be safe, and we’re done.  Some squares are mines in 3 of the 4 configurations, some in 2 of the 4, others only in 1/4.

Solution:

Puzzle 9-12 “Oasis”: 7 mines remaining, 2 to mark, 4 squares to clear. Difficulty: 9 / 10.

Source is this game by Plural (f/k/a MC_Jacks), with hat tip to FracturedAnvil.

Analysis: We can choose just about any frontline square as our pivot square, but first we ought to observe two things:

1. The 5 at bottom needs 3 mines from 4 unknown squares, so it has only one free spot for a safe square; therefore, the interior 4 above it can’t have both its two safe squares on the bottom.
2. Although there’s no simple square-grouping for minecount, a quick scan tells us that minecount might come into play.  One hint to that effect is from the problem statement itself: Most hypothesis-testing gains yield just 1 new safe square, but here we’re told to clear 4.  So we should be open to a solution that ultimately derives from minecount.

Let’s start testing some scenarios.  Choice of pivot square is not that critical, but suppose we opt for the lower-right corner.  We’ll start by assuming that it’s a mine, and play that out:

The middle image just plays out the conclusions, particularly that with the square to the right of the 5 being safe, the three above it must all be mines.  Then for the right image, firstly the interior 3 at right is fully marked, so the 2 squares to its left are safe.  But also, the interior 4 has one mine remaining from two squares (in orange)... and now we’ve used up all 7 of our mines remaining, so the top two squares must be safe, from minecount.

Then let’s see what happens if our pivot square is instead safe:

Hmmm, we seem to have run out of further downstream conclusions we can draw (we don’t know how to mark the 5).  But we won’t give up yet!  We’re going to choose a second pivot square here, and see if our scenarios become limited somehow that way.

Let’s choose the bottom-left square, shared between the 5 and the two 4s.  We’ll start by assuming it’s safe, and play that out:

For the middle image, we note that our choice of pivot has meant that the two remaining squares above the 5 must both be mines.  That leaves one remaining mine from two squares for the interior 4 at left (marked in orange).  But now with our minecount, we’ve assumed six mines, plus one from the orange, and that’s our 7.  So the remaining four squares must be safe - the same ones as before.

Then finally, let’s make that bottom-left square a mine, for our third scenario in the logic tree:

We weren’t able to mark everything, but our two pivot-square assumptions leave us with two unmarked mines from three squares for the left-hand 4 (orange), and one mine from three squares for the right-hand 3 (yellow).  So this is five marked mines, plus three from groups… but that’s at least 8 mines, and we only have 7.  So therefore this scenario is impossible!  And so in fact, the two pivots are thus just one: if the lower-right square is safe, then the lower-left square must be safe too (and the two squares above the 5, both mines).

So the shared (overlapping) conclusions of our two possible scenarios are the answer.

Solution:

Now, it turns out that there is a simpler path to this solution, using partial-overlap minecount.  Courtesy MineBuoy’s video on the same problem (!), consider the following square grouping:

We have four mutually-exclusive single-mine groups (in yellow), the residual interior group on the 4 (two mines, in pink), and the group around the 5, which has three mines (in teal).  If we regard the pink and teal as one partial-overlap group, the pink-teal group has either four or five mines together, depending on whether that one overlap square is safe or not.  However, there are three mines worth of square groups that are entirely outside this pink-teal group, and we only have 7 mines total.  Therefore the combined group cannot have five total mines, as that would overflow our minecount.  Therefore the overlap square is a mine, the combined group has four mines, and the residual two squares at top are safe.  Furthermore, because we weren’t even counting the central yellow group (to the left of the 3) in our count, the only place for that group’s mine is where it overlaps with the teal group - that overlap square has to be a mine too.  Therefore its other squares must also be safe.  That leads us to the same solution as above - just with less fuss over it.

If you didn’t see that grouping, though, our technique here (hypothesis testing, until we reach a minecount-based contradiction) would get us to the same place eventually.

Puzzle 9-13 “Watchtower”: 6 mines to mark, 7 cells to clear.  Difficulty: 9 / 10.

Source is this 100k attempt, at 31:45.

Analysis: This one is not the hardest puzzle, when it comes to making decisions on where to start your scenarios, or in their quantity (just 2), or in playing them out.  But it has one of the crazier sets of coincidences leading to the solution than just about anything I’ve ever seen.

Also, while there’s still a connection via the isthmus going towards the bottom of the image, that is just going to be derivative of whatever we conclude inside the upper region.  So we’re going to focus our attention on everything above that isthmus, and at the end if we can do anything on it, we’ll come back and mark it then.

To begin with, that interior 1, next to the interior 2 that’s an effective-2, looks very suitable for attempting proof-by-contradiction or hypothesis testing.  The left side of the region looks higher-density, we would probably need multiple tiers of assumptions, so let’s instead start at the bottom row of the island region (which is a 1v2) and hope it proves to be simple enough.

(1)   (2)

With our first scenario, Layout A, we start (Step 1) by assuming a mine in the lower-middle of the island.  This forces logic to the left and up, and also the 2 to the right of the assumption then can clear above it.  In Step 2, this then forces a mine above that interior 3 to the bottom-left, which then forces conclusions above it, all the way up.  Meanwhile, the interior 3 at lower-right also now has only one spot for its remaining mine.  Then, finally: (3)

In this Step 3, the 2 that’s at the lower-left of what is now our front is satisfied by the row of conclusions from the last step, and so it clears above it.  This forces 2 mines to the right of the interior-4, which had been an effective-2.  This leaves only two possible locations for the mine of that lonely 1 on the inside, and in fact, it’s now a 50-50 front facing that gray box from three sides: left, bottom and right (so it would be a 2-mine diagonal box, in this scenario, for whatever that’s worth).  But that’s as far as we can go.

OK, that seems like a lot of conclusion we get from that assumption.  What happens if we make the opposite assumption, that the teal-framed square is safe?  Here comes Layout B:

(1)   (2)

In Step 1, we assume the square above the 3 as safe, and this marks to the left and then up.  Then the square above that interior 3 is safe (it was a mine in Layout A).  In Step 2, we then have to use some Evil NG-style logic: we have 1 mine in the yellow group (from the 3s to its left), therefore we have 1 mine in the blue group (per the blue-framed 2), therefore the interior 1 is satisfied from the blue group, so the two squares to its right must be safe.  Cool!  Continuing:

(3)   (4)

Here’s where it kinda goes off the rails.  The key insight is in Step 3: from the 2 below that purple group, the purple group contains one mine.  So the 2 that’s within the purple group only has one place that its second mine could go: above-right, next to that 3.  With that marked, things proceed like an explosion in Step 4: starting with the orange arrows, we have two safe cells next to that rightmost 3, which then forces a mine above-left of that 2 near our assumption square (orange arrow).  Switching to yellow arrows, that new mine clears the remaining cells above the interior 1, and also the lower square on our left column, which then proceeds to mark things going upwards from there.  Lastly, all this clearing means that the interior 4 has its remaining two mines marked to the right (blue arrow).  Whew!

So let’s compare our final-conclusion images from Layouts A and B:

(A)   (B)

Quite a lot of overlap!  Some parts are mirror images, but the squares above and above-left of the interior 1, the square two spots to the left of that interior 1, and the conclusions above it, are all the same in both scenarios.  As is the square towards the lower-right of the island, below-left of the framed 3 on the right - and from that, conclusions flow downwards, including past even where we’ve marked.  Bottom line, our assumption area doesn’t end up with a conclusion (yet), so we never proved or disproved an entire scenario - but a whole bunch of things end up being definitely-safe or definitely-mines under both scenarios.

Solution:

Puzzle 9-14 “Saddle”: mark 1, clear 2.  Difficulty: 10 / 10.  Source: this 100k attempt.

Analysis: The first question you’re probably asking is, how can this scenario possibly fit our hypothesis-testing filters?  The upper portion of the region pretty clearly doesn’t - it has a few interior number squares, but it doesn’t quite connect bottom to top, or cross back over itself.  However, the lower part of the region does: there are no untouched squares, it’s kinda concave, and logic can pretty clearly proceed around the 4-square box at left.  We’ve also got that pair of 4s (being effective-2s) which only share two squares, so that’s a pretty nice source of some signal.  So we’ll start down below.  And we’ll chop down our screenshots to the operative area, too, to minimize clutter.

What do we choose as a pivot square?  This is where it gets funky.  Try as we might, there is no clean single pivot where we can have binary scenarios and have them play out to the degree we need them to.  Instead, we need to start from the only spot we really could, which is that central corner 1, lower part of the formation, with the two interior-2s around it.  And that is ternary - it has three adjacent squares, and thus three scenarios we’ll need to try.  But it’s the only thing that will do what we need.  You can kind of see that its choice controls the whole area around it, and so offers the possibility of “information” going out, going around, and coming back again.

Let’s start with a Layout A that assumes a mine on the lower-left square of our pivot:

Here, the assumption (teal frame) goes down and around the lower box.  Then, because that satisfies the interior-3 to its left, it clears all 4 squares above that, forcing a mine to the upper-left of the left-side box, and a safe square to its right.  That leaves just enough squares for the lower of the two upper 4s to place its mines, with one last safe square at the top.

Next let’s assume a mine above-right of our pivot.  This makes Layout B a bit more challenging:

By assuming the mine top-right, it clears four squares and puts a mine at lower-right of the lower box.  Then we need to focus on the left-facing interior-2, and the lower interior-3, which are now in opposition: the 2 needs two mines from the three squares to its left (blue box), but the 3 can only provide one remaining mine from its adjacencies (yellow box).  Therefore, in a classic Evil NG-style “neighbor mismatch” move, we can conclude there is a mine in the overlap squares, a mine in the non-overlap blue square, and the non-overlap yellow squares are both safe (middle step).  This conclusion leads us to mark around the left-side box, forcing mines up to the upper 4s and a safe square at top.  Pretty good so far!

Now let’s see what happens when we assume a mine upper-left of our pivot, as at right.  Initially, the marking proceeds around the lower box as normal.  And it might seem, to most eyes, like this is where we’re stuck.  That’s all we can directly conclude from this scenario, and while it has some overlaps with one of the other scenarios, it doesn’t overlap with both - there’s no shared conclusion among all three layouts.

However - and this is the key insight that makes a solution possible, and also a top difficulty rating - one thing we can do is bring back those overlays from Layout B and give it another think.  They still have the same overlap, except this time the 2 only needs one mine from the blue region, while the 3 can still only provide one.  There’s no direct logic conclusion available to us this time.  But, in a step similar to our “split-front 2s” logic, we do know that in this scenario, one of two things must be true: either the overlap squares have the mine needed by both the 2 and the 3, or they don’t.  So we can use that overlap region, the pair of overlap squares, as a second pivot point!  Let’s start by assuming the overlap squares have a mine, and play that out as Layout C:

 →

So here, if we assume there’s a mine in the overlap squares (teal group), then the 2 and 3 clear their other adjacencies.  This leads, as with previous layouts, to a mine in the upper-left of the left box, and mines to the right of the 4 above, in the same places.  But now, how about if we do Layout D, and conversely assume the overlap squares don’t have a mine?

So our assumption that the shared squares are safe means the 2 next to them must have a mine above-left of it.  We don’t yet know where the mine for the interior-3 is, though it’s down to two possibilities.  In the middle step, though, we make a move that’s the crucial insight of the whole puzzle: because the lower 4 now is an effective-1, we can use the 1-2 pattern to prove a mine at the top of the upper 4 (orange arrow), and therefore a safe square below-left of the lower 4 (yellow arrow).  This leads, in our final step, to a mine at top-left of the left box, a safe square below it, and therefore a mine above the interior 3.

Because we were able to get decently-expansive layouts under what is a comprehensive set of scenarios (one of the four scenarios must be true), we can now compare.  The squares next to the 4 are the same in layouts A-C, but not in Layout D.  The lower box is largely the same in layouts B, C and D, but not in Layout A.  However, the left box has 3 squares that, miraculously, entirely overlap in all four scenarios: the upper-left corner is a mine, and the two squares adjacent to it are thus always safe!  So that’s our answer.

Solution: We only get guaranteed safe moves in that left-hand box.  But as it turns out, that proves to be enough to solve the entire region!  This is one of only a very few scenarios I’ve ever encountered where a double-pivot logic tree was necessary, and a solution came of it.

(10) Guessing via Lock-in

Link to discussion section.  Link to puzzle section.

Puzzle 10-1 “Parapets”: Find the best guess (there are multiple).  Difficulty: 2 / 10.  Source.

Analysis: We intentionally omitted the mine count here (which is 4), because it directs attention away from the main focus, which is to practice visualizing lock-in chains.  But actually, the process of seeing how you fit different numbers of mines on the front here, is not too dissimilar from attempting a mine-count solution.  In particular, the front squares where assuming a mine would require four for that front line are the good guesses, and the ones that would only require three mines would be the higher mine probabilities.  Let’s firstly label the front:

Because we have an effective-2 (the middle 4) nestled in between three effective-1s, there are some chain reactions that occur with an assumption.  Let’s start from the right: If H is a mine, then F and G are safe, making D and E mines, making C and B safe, making A a mine.  So you need 4 mines here (A/D/E/H).  Same is true if G is a mine.  Whereas if F is a mine, G and H are safe, but you can say nothing definitive about the cells to the left; merely that, if D is a mine then so is A, while if E is a mine then D is safe and there’s either a mine on B, or on both of A and C.  An inability to draw lots of downstream conclusions here is a tell, given the topic of this chapter, that this cell is not particularly unlikely to be a mine; in fact F is 90% a mine.

Continuing this process will reveal that while G and H are good guesses (requiring four mines for the front), most of the others are not (they usually only require three mines).  But there is an exception, which a keen eye will observe: if C is a mine, then it clears both B and D, forcing mines on A, E and F - and a mine on F means that G and H are again safe.  So this means mines on A/C/E/F, four required mines.  That makes C the other good guess.

Solution: While C (the left 5%) is a decent guess, the best guess here is H (at far right) as it offers the best chance of progress.

Puzzle 10-2 “Smoke Stacks”: Find the best guess.  Difficulty: 3 / 10.  From endgame here.

Analysis: The first thing that should jump out at you is that the entire front consists of effective-1s (leaving aside the interior areas at top), with one exception: the lower 4 is an effective-2.  That means any of its three adjacencies being cleared would fix mines into the other two spots.  The second thing to notice is that certain guesses along the right-hand column would fix more mines into place than other guesses would.

Focusing on the first aspect, an initial thought might be to consider that if the square left of the 4 (between it and the lower-left 3) were safe, then it would require more mines from the area: three to satisfy the lower front, rather than two if the square above-left of the 4 were safe.  However, this assumption has one shortcoming, which is that it doesn’t tell us very much about the right column - just one additional safe square.

Other guesses will either fully force a layout on the right column, or on the lower front, but not both.  One guess connects both insights and areas, however: the bottom-right corner.

If that’s a mine (with teal border), then we get a safe square to the left, putting two mines left of the 4 and one safe square at bottom-left.  But then going up, we get two safe squares above, another mine, clearing two more safe squares, and then the very top bit as well.

Solution: The bottom-right corner would fix in place five other mines, while other guesses would only fix at most three.  Therefore it’s the best guess (along with its dependent square above it).

Puzzle 10-3 “Scepter”: Find the best guess.  Difficulty: 3 / 10.  Source is here at 1:35:00.

Analysis: This one combines our lock-in principle with an Assume Fewer insight.  Firstly, observe that all front-facing squares are effective-1s except for the bottom two (the 5 and 4), which are effective-2s, with three shared squares.  If they share both their mines on the shared squares, the region requires fewer mines, whereas if they only share one mine and each have one in the non-shared squares below them, it uses an additional mine.  Also, there is a certain arrangement that makes the region use 8 mines, rather than 7 or 6: if they only share one mine, and that shared mine is on the lower of the shared squares, so all three squares on the bottom of our region are mines.  Let’s look at this from a square-grouping perspective:

From the lower-most 2, the pink group has one mine, but then the yellow group (around the 5) and the teal group (around the lower 4) both have two.  The only way that the pink group’s mine isn’t accounted-for by the overlap squares between yellow and teal, is if its mine is in its top square, and there’s a mine in the lowermost shared square (along all three squares on the bottom).  In that sole case, the region has 8 mines.  Since 7 is likelier, and 6 is likelier still, that 8-mine case (putting a mine on the upper square of the pink group) is the least likely.  But then, also note that that case would fix the location of all mines in the whole region: it implies a mine on the top square of the blue, orange and purple groups, and lower square of the brown group at top left, plus three mines along the bottom row.  An assumption fixing 7 other mine locations - that is a very favorable guess, as that is very unlikely!  And indeed it is so.

Solution: The squares which would be mines only if the region contains 8 mines, are the best guesses here, partly due to Assume Fewer but partly due to lock-in.

Puzzle 10-4 “Balcony”: Find at least one good guess.  Difficulty: 3 / 10.  Source here at 3:00.

Analysis: This offers classic examples of two situations we have discussed, each yielding a good guess.

Firstly is a Split Background Rate example at the very top: note that if the square shared between the 3 and the 4 is a mine, the area (the five front squares they face) contains two mines, but if it’s safe, the area contains three mines, which is less likely.  So the shared square has rate (1-B) = 76%.  But then the two remaining adjacencies of the 3 split the residual probability, making each one have (24% / 2) = 12% odds.  So that’s a good guess, though it derives from Assume Fewer and isn’t our best answer here.

But then we have a Neighbor Mismatch type lock-in scenario, around the corner towards the upper right - see the 4 on the corner square.  That’s an effective-3 (the 4) up against an effective-1 (the 2 above it).  So we know the three squares below that 4 have at least two mines.  Could those squares below the 4 have three mines?  Yes, but that then demands 4 mines from that isolated corner area, whereas if there’s a mine to the left of the middle 2, then only three are required.  Anyway, if all three squares below the 4 are mines, then and only then is the square to the left of the upper-left 3 a mine; that square being a mine would “lock in” three mines below the 4 (plus a safe square below-right), making it a solid choice.  Indeed, that’s our best guess.

Solution: As noted.  The other 12% square (in the middle) is sort of dependent on, and downstream from, the Neighbor Mismatch lock-in insight.  But it’s a good guess too, since it offers reasonable chances of progress.  The 16% square at bottom might offer the best chance of progress, but it’s hard to find.

Puzzle 10-5 “Axolotl”: Find the best guess.  Difficulty: 4 / 10.  Source game here.

 → Key spot:

Analysis: We start out here by scanning for effective-2s that might provide an easier opportunity for lock-in.  (we would be looking for that purposes of searching for logic moves anyway, so this goes with the same process).  And we find two of them, middle-low-ish: the lower-left 3, and the 4 opposite it on the right.  There is no Neighbor Mismatch Evil NG move available, however this situation does offer some probability-based insight.

Those cells, the 3 and the 4, share two squares (teal group, above).  The shared squares can’t contain zero mines (it would underflow the 3), but they can contain either one or two.  However, observe what happens when they contain two mines, at right.  Our 4 of interest then has two safe squares above it, forcing a mine to the upper square of the two-square front at right, it forces a mine to the bottom of the region, and our 3 next to our shared squares has a safe square above it, putting a mine just above that.  That’s three other mines locked-in by this choice, and also requires more mines from the region (having only one mine on the shared squares could very easily use only four total mines here instead of five).  That’s a lot!

OK, so that particular scenario is thus unlikely, due to lock-in.  So let’s say the shared squares are very likely to have one mine between them; what does this tell us?  The 4 at the right then has one mine in the squares above it, so the corner around the 3s at right becomes a 1v2 corner, and so the corner mine (lower square next to the right-hand 3s) is likelier than two mines on the other squares.  So that’s a decent guess.

But there’s one that’s even better, because in the one-mine-on-teal scenario, it forces a mine above our lower-left 3, making the square above that one safe.  And unlike the 1v2 corner at right, that square is always safe in the one-mine scenario, whereas the others can still be mines in the one-mine scenario.  So that square is our lowest-odds choice, and the answer.

Solution: The 4% is the best guess, but it’s the 13% at right that offers the best chance of progress, because the squares in between it (outside of teal) have the lowest odds among themselves.  In fact, that 13% square turns up a 3, which is chordable, and solves the region.

Note that minecount can be helpful to see how demanding the two-mines-in-teal scenario is here, because the region can contain as few as 7 mines (if the two untouched squares are safe), while the two-mine scenario demands at least 8 and probably 9 mines here.

Puzzle 10-6 “Slipknot”: Find 2 independent good guesses.  Difficulty: 4 / 10.  Source game.

Analysis: We’ve got two classic examples of lock-in here.

Firstly, that extended line at top-left would normally be ignored on first glance, but in this case we run into an effective-2 in that 5 just below it.  That leads us to some interesting conclusions there: either there’s a mine above-left (next to the 3), or both squares below the 5 are mines.  If so, lots of conclusions flow downwards from there: the 4 below the 5 clears to its right, forcing a mine below-right of the 3, clearing the 2 around the lower-left box and putting a mine in the lower-left corner, plus some other safe squares.  Whereas, the opposite assumption (teal square is a mine) forces opposite conclusions going upwards on the line, but with the 5 reduced to an effective-1, it forces almost nothing below that 5, just a single safe square below-right of the 3.  Therefore this assumption (teal square safe) is extremely unlikely - in fact, it’s 91% to be a mine.  So the square above it is 9%, and that square (and its dependent square above it) are one answer.

This experiment showed us how precariously balanced that left column is.  What if we made some other assumptions around it?  Trying a few things, we see that an assumption of a mine in the lower-left corner doesn’t force much other conclusions, but a mine on the other squares of the lower-left box seems like it might.  Let’s try it out:

A mine assumed on the top-left or bottom-right corners of the lower-left box (they are equivalent, of course) forces five safe squares along both fronts (left and bottom), forcing two mines to the right and another three mines up top (once you account for the square group with the 4 at left, which can’t be determined yet, but with it we can fix a mine above-right of the 5).  That’s 5 downstream mine conclusions, clearly unlikely to be the case.  So the assumption squares are very likely safe (in fact, they’re 14%).

Solution: The two guesses discussed are the best ones here, principally because of their dependent squares, which give you many chances to make progress.  There are also the 16% squares down below, which are a bit harder to find but have many of the same characteristics of the ones mentioned.  The 13% at top-right is what Scar would call a “solver move”, something you’d never find on your own (and frankly doesn’t have great odds of progress, either).

Puzzle 10-7 “Monument”: Find the best guess.  Difficulty: 4 / 10.  Source: Corner here.

Analysis: You could brute-force this one if you had to (i.e. enumerate every single possible layout and then calculate probabilities manually), as there’s only one untouched square and only 4-5 mines required from the number squares.  Luckily we have the techniques to not have to do that, but just to say, this is a fairly simple formation here - deceptively so.

Firstly, if you were hunting for logic moves, you’d notice that there were some square groups of interest, even if they didn’t yield logic or a firm minecount:

(1)   (2)

For grouping 1, it’s from projection from below; the central 3 implies the yellow group, while the blue group is implied directly by the upper 3.  On the second grouping, the 6 from right has the pink group with two mines in three squares, which projects to the central 3 and implies the teal group contains only one mine.

So let’s reason this out: either the yellow and blue groups have their mine on a shared square (B or A), or they don’t.  If they have a mine on the shared squares, then the disjoint squares (the Cs) are safe; then the corner 1 needs a mine, the lower interior 3 needs two more, and the lower row of our front could have either one or two mines.  That looks like a 4v5 scenario.  Let’s examine them more closely.

If A is a mine, then it clears B and the Cs.  The pink group has one mine left, one of the two lower squares in the teal group has a mine; if it’s the lower square, then we can end up with only 4 mines from the region, and the corner 1 has two options for its placement.  So that’s a pretty likely scenario.

If B is a mine, then it clears A and the Cs, forcing two mines on the lower squares of the pink group.  The teal group is already accounted for (by the mine in B), so there’s consequently a mine in our lower-left square, and then another by the corner 1, so there are still two layout options for this scenario, but it definitely requires 5 mines from our front.

On the other hand, if the Cs are mines, then we have a fully specified mine layout:

The Cs being mines (they each imply the other) clear the cells below the upper 3, which forces two mines left and below-left of the 6, which clears the lower-interior 3 to its left, which forces a mine to the bottom-left.  Meanwhile, the corner 1 has its mine already, so the two other squares are safe.  So we have five mines from the region (already more than the minimum we could have), and we have no remaining flexibility, there is only one layout possible (pending the untouched square, of course).  While if B is a mine, it still has two layouts available (the corner 1 having two options), making B twice as likely to be a mine.

Solution: …therefore, the squares marked C are our best guess, as they fix the most number of squares into place.  And because there is a fewer-mines scenario (if A is a mine), the rest (B or C a mine) has total probability equal to the background rate, and the Cs are only mines in one of the three layouts in that case.  So this one ends up with very attractive numbers.  The upper square will also offer guaranteed logic as to where the mine near the corner is.

Puzzle 10-8 “Monkey Bars”: Find two (semi-)independent good guesses. Difficulty: 5 / 10.

Source game here.

 → Labels:

Analysis: There are at least 4 good options here, although the question just asks for two.  One of them, however, is more obvious and significant than the others, so a right answer would definitely need to include it.

The first thing your eye may have went to with this puzzle is the interior 2 at upper right.  It’s an effective-1 in a great spot on the formation, one mine from 6 squares and connecting a bunch of parts of the front.  The first thought might be, “if it shares its mine with the 3 to its right (J), or the 3 above-left (H), then the other 4 squares around it would have to be safe, so that’s probably a good start.”  But while that’s not a wrong hypothesis, playing around with potential mine placements around it yields an interesting discovery: an assumption of a mine to its right (on J) creates a cascade of downstream conclusions:

Assuming a mine on J clears H, which (going left) puts a mine on G and clears E and F.  Then going right it makes K (the corner) safe, and going down from there, puts a mine on L and N, which clears the 3 below N and the three cells to its left, which forces two mines below that 4, on R and S.  Which also then makes T safe.  So that’s 5 downstream mines, and 13 safe squares, all from that one assumption.  That makes it pretty unlikely!  In fact, it is the lowest mine odds in the whole region, it’s only 6% despite a 25% background rate.  And comes with two dependent safe squares (L and N), so better odds of progress.

Now, what else is out there?  A close observer would notice that the upper-left 4-square box, ABCD, is a pseudo: either there’s one mine on the corner, B, and A/C/D are safe (and a mine is on one of the other adjacencies to that interior 2), or there’s two mines in that box, it’s a 50-50, and the three cells below that interior 2 are all safe.  If B’s a mine then A and C are safe and should be guessed right away; if the box is a 50-50, then it should also be guessed right away.  So regardless, it is a correct move to guess that B is a mine, and A and C are safe squares.  The implied safe squares below that interior 2, then, make that guess even better than a background-rate guess, in fact A and C have probability 18% against that 25% background rate.

Those are the obvious ones.  There are also two more less-obvious ones:

• The assumption of J as a mine, above, cascaded down to that interior 3 and 4 (the 4 is an effective-2), marking N and clearing O, P and Q, forcing mines onto R and S.  But an assumption of O as a mine would do likewise, forcing mines onto R and S.  Basically, either R and S are both mines (unlikely), or N and O are both safe.  So O is a pretty decent guess too, if need be.
• If the interior-2 at upper left shares a mine with the 3 to its lower-right (cell E), then it forces a mine onto B, makes A, C, D all safe, as well as F and G, making H a mine, and thus mines on K and M as well.  That’s four forced mines from that assumption (B, H, K, M), plus 15 (!) safe squares.  Which is not much less than guessing J, in terms of downstream conclusions.  And indeed, it is the second-lowest mine odds, at 9%.

What makes this puzzle a 5/10 difficulty, implying the answers are a little easier to spot?  Well, the answers center around those interior 2s that are effective-1s, sticking out into the formation, threatening to clear a bunch of their adjacencies and cascade one mine assumption down across the region.  Those are often going to be the source of good guesses, is the lesson here.

Solution: Any two of the four good guesses constitute a correct answer here, but preferably you’d see the 6% ones, because that’s a great guess, giving you 3 cells to resume progress.  And while the two lowest-probability guesses (J, 6%, and E, 9%) are not fully independent, they are not fully dependent either, so either one is a very good choice.

Puzzle 10-9 “Slinky”: Find two independent good guesses.  Difficulty: 5 / 10.

Source here (at 1:40).

→

Analysis: This puzzle has a fairly-obvious but less-helpful guess, paired with a less-obvious but more-helpful guess.  Let’s examine from top-left, first.

The two 4s at top-left are an example of the Shared Corner type situation described in the chapter.  Although those mostly commonly involve a 3 and a 4 (with the 3 being an effective-1 and the 4 being an effective-2), here one 4 is an effective-1 and the other an effective-2, and they still share two squares and have three non-shared ones.  So, either there’s a 2-mine scenario (one mine on B/C, the other on D/E), or a 3-mine scenario (mines on A, D and E).  The 3-mine scenario is not only less likely to begin with, but also, because there is only one configuration that works for 3 mines (A/D/E), while there are 4 possible configurations with 2 mines, it makes that scenario even less likely.  This makes A very unlikely to be a mine.  However, it’s not that helpful: it will solve for B and C, but we will be left with a dilemma on D/E and unable to make further progress.

More subtly, try imagining a mine on each square as we go, and seeing if something can push “information” (downstream conclusions) very far, or outward in both directions.  The 4s at lower-right are effective-2s, so any given mine assumption seems unlikely to tilt them usefully.  However, there is that little “hop” between the upper 3s (below F and G), and the 2-3 in the middle (below K).  Those resemble a sort of pivot point, where conclusions on one transfer to the other side.  What happens if a mine lands on J?  Then it clears K-L-M-N and forces a mine onto O… and also clears F-G-H and forces a mine onto E (and likely clearing D as well, if A is safe).  That’s enough consequence to make J a very good guess.  In this case it’s 13%, but that’s against a background rate of 31% (!).  And because it opens up more possibilities, it’s actually a better guess than A.

Solution: The two squares A and J provide the best guesses here.  K and L actually provide nearly-as-good guesses as J does (16% to 13%), and make good followups if J doesn’t offer progress.  The rest of the board is very hard to make progress on, given that high background rate, but (if you go check out the actual game) there are further good guesses available after those, which give you a fighting chance.  For this scenario though, it’s mainly those two.

Puzzle 10-10 “Propeller”: Find two independent good guesses.  Difficulty: 5 / 10.  Source.

 → Labels:

Analysis: So one of the good guesses here ought to be obvious, if you’ve read the chapter.  We have a classic 2-2-2 hole pattern here, which is situated right up against the edge of the board.  This means that the side-hole squares, above and below those 2s, are going to have below-background-rate odds - and in particular the ones above it, being against the wall, provide good enough odds of progress that they are good guesses.

For the other good guess, let’s look at the wrap-around near those 4s, from C to H.  For the corner 3 next to B, one mine is marked, another is provided by A/B, so C/D provides the third.  So mines are either on C, E and G, or they’re on D & F, and then one of H/J/K.  Those are the fully dependent chains.

But wait - a mine on G will clear H, J and K and put a mine on L, clearing M and N.  That’s pretty good lock-in, enough to give below-background-rate odds.  If G is safe, then a mine on H or J will clear the others plus the 5 interior squares (via the 2 between H & J), while also implying mines on D and F, but that’s not locking in quite as many mines.

And, notice the split-equivalence!  If there’s a mine on any one of H, J or K, then by the corner 2 (between H/J), G would have to be safe, and thus F would be a mine.  So the odds of F have to equal those of the other three: F = H + J + K.  While sometimes we would use this to argue that H/J/K must have low odds, in this case we can’t quite be sure about that, but it likely means F (and thus D) have very high mine odds.

Why are we sure about this?  Because we already have an assume-fewer case here: the two scenarios above are C-E-G vs D-F (the latter doesn’t lock in a mine between H/J/K).  So absent any other effects, we would expect background rates for C-E-G, as the N+1 option.  But because a mine on G implies other things downstream that the others don’t - specifically, clearing H-J-K and M-N and putting a mine on L - the odds of those three have to thus be below background rate.  And by split-equivalence, we also think the odds of D/F are higher than default.  So we would reason that the C-E-G chain is a good set of guesses.

Solution:  The upper 14%s above the hole pattern (against the wall, so better odds of progress), and the chain of dependent 17%s as noted above (C-E-G), are the best guesses here.  The latter even more than the former, because of how many additional dependent squares you get “for free”, and the greater chances for progress from it.

Puzzle 10-11 “Doorbell”: Find the best guess.  Difficulty: 6 / 10.  Source game.

 → Labels and groups:

Analysis: We will use both square labels and some square groups here, as that will be the most helpful way to refer to these spots.  Examining the front, two things should jump out at us: the middle interior 3 (between E and F) is an effective-1 that offers some potential pseudo-hole possibilities, and the upper interior 3 (around which is the blue group) which is an effective-2, which complicates the A-B-A corner.

Taking the upper corner first, that upper interior 3 requires 2 mines from 4 adjacencies; if it gets one from A, then blue + ABA contains 3 mines (the two As plus one residual in the blue); if it doesn’t get one from A, then B is a mine, but it still has 2 more mines in the blue.  Either way that’s 3 mines, so there is no Assume Fewer reasoning available to us there.

But look further right, and we can see that there is a Split-Equivalence around that corner 3, as it’s an effective-1.  Namely, if A or C are mines (and can’t be both), then D is safe and E is a mine; put differently, E = A + C.  This could be very helpful if E were to be background-rate likely or less, but we can’t prove that yet.  We’ll come back to this.

Now let’s look around that yellow box.  It feels like there should be a pretty strong possibility that the remaining mine for that middle interior 3 is either on E or F, making the pseudo-hole behind it (yellow) split the residual probability 3 ways.  But first we need to assess E and F.  If E is a mine, and F safe, then it forces mines onto G and H both.  That makes K and J safe, and thus L a mine as well.  In short, the assumption of E forces three downstream mine placements plus two safe squares.  The reverse isn’t true: if F is a mine, we only know one of G/H and one of J/K are mines, with L dependent on J so K is thus the likelier.  But there are no firm downstream conclusions.  Therefore F is fairly likely to be a mine, and E a lot less likely.  But in combination, this means that the residual probability for that interior 3 - the yellow group - is itself low, and then gets split 3 ways.  So that has to be our best option!

Solution:  Our best guesses are the pseudo-holes to the right of the interior 3… but there’s also an epilogue to this story.  In the real game, we tried the upper pseudo-hole option, and guess what happened?  Exactly what you hope for with a guess like this:

 —> Behold: progress!  

Puzzle 10-12 “Bento Box”: Find two independent good guesses.  Difficulty: 6 / 10.  Source is this game at 2:00.

Analysis: There’s a fairly obvious lock-in situation at the 4-square box at lower right.  Consider:

The two purple-framed 2s there, both effective-1s, make the situation around that box precarious: if they get their remaining mine from the 4-square box (in yellow), then a lot more is determined from there.  So the question is, does the box contain one mine or two?  If it contains one mine, it’s in the upper-right, on the corner.  If it contains two, then they have to be on a diagonal, and either way it will satisfy both purple-framed 2s and cause some cascading effects.  So let’s look at how significant those are:

OK, pretty significant.  The assumption of the mines on the other parts of the 4-square box (not the top-right) clears everything around those pink-framed 2s, which forces one mine to the bottom, but it also fixes 3 mines around the long front at left, and - importantly - clears all 3 squares above that corner 1 at the top of the long front, which puts a mine into the lower-right corner of the upper area, forcing 3 other mines into the region.  That’s 9 mines fixed by this one assumption!  Plus an extra mine required from the area, since the counter-assumption only requires 1 in the 4-square box.  That makes this scenario very unlikely, and makes it very likely that the upper-right square of the 4-square box is a mine.

Now, what of another good guess here?  Well, while analyzing the consequences of the first guess, there are two effects that should stand out, which are not fully dependent on that guess:

1. On the long front, the lower-corner 1 is satisfied by many choices of mine along the lower front; there is only one square on that lower front which doesn’t fix the corner’s mine location.
2. Also on that long front, the upper-corner 1, if satisfied from its left, clears the squares above it, forcing the mine into the corner toward its right, and also other conclusions above it, just from that one assumption.

Those are the kind of consequence chains that good guesses are made from.  Let’s consider:

So if the upper corner 1 (next to the yellow group) gets its mine from the left (teal or yellow), it clears the two squares above it, forcing the mine to the right of the 2, and all the other conclusions in the dense region above (4 mines, 5 total safe squares).  The only square on the left side that doesn’t do that (if it’s a mine) is the one in purple, next to the middle 2 in that vertical line.  So that square is the likeliest of the bunch (in fact it’s 63%), because of all the conclusions above that it doesn’t force.  And if that’s a mine, then so is the one farther below.

Then considering below, the only square that doesn’t force the lower corner 1 to clear the green group, if it’s a mine, is the pink square.  That one leaves three possibilities for the corner, the cells in the green group, and in fact it is split-equivalence: all three cells to the left of the corner 1 (in green) sum to the odds of the pink square.  A mine in the blue group also forces mines to be on the teals, going upwards, and in turn force the conclusions to the top of our region as noted before.  That pink square is thus highly likely (in fact it’s 80%).

Lastly, in mirror fashion, the only cell that doesn’t force that lower corner 1 to clear the blue group, is the teal cell right above.  It’s the split-equivalence spot for the blue group.  It has a dependent square above it (shown with orange arrows), which would be a mine as well, but would force the upper-area conclusions, so it’s a bit less likely than pink.  But still likelier than the others, because a mine in teal wouldn’t force the corner 1 to clear the blue group.

So if purple, teal and pink are our “high likelihood” cells, what’s left?  Low likelihoods!  Basically every cell on that area (in yellow, in blue, the two in green that aren’t the purple one, etc) are good odds: mostly 9%, one is 12%.  But they’re all good guesses!  And we can find them by just noting the corner 1s, noting the conclusions at top which might-or-might-not be forced by the mine choices around that left front, and being careful about our split equivalence.

Solution: The 6%s around the 4-square box are best, but any of the 9%s here make good guesses too, with several dependent cells offering better odds of progress.  And the bottom-left edge here turns out to be an opening!

Puzzle 10-13 “Pride Rock”: Find the best guess.  Difficulty: 6 / 10.

As the name alludes to, this is from Scar’s world-record HD game (at 56:30).  Guess #26!

Analysis: Scar lays it out pretty well in the video (at 73:00), but this area really doesn’t offer that much in the way of diverse choices - there’s many untouched squares, there’s the semi-isolated 3 at bottom-right, next to a 5 that’s an effective-2, and so we’re left with only three real options to be a good guess.

First is the 4-3-2 at top-left.  Around that mine next to the 4, this section is a 1v2 Assume Fewer candidate - either the mine is next to the 3, or there’s mines above-right of the 4 and next to the 2, the latter of which also satisfies the 3 meaning it clears 4 other cells.  So the 2-mine scenario is not only less likely due to requiring more mines, it’s also less likely than the background rate too, due to the clearance it provides at bottom.  Oddly, these are 21% guesses (vs B = 23%, hardly better), when I’d have expected 15% or so.  It offers 2 safe squares if correct, which have 4 and 3 unknown squares respectively - decent odds of progress, but not wonderful.

Second, going down a bit we have the two interior 3s, which are both effective-1s.  You’d expect that given B = 23%, the odds that their one mine is on one of the two shared squares (clearing the other 7 around them) is pretty low; it would also lock in 2 other mines, above (making the 4-3-2 into the 1-mine scenario, mine to the right of the 3) and below (the third mine for that 3 just below it has to be below-right, next to the 4.  This comes with some potential for progress, but the guess does not lock in the mine (though if it’s a 5 it solves it, and if it turns up a 4 you do still have good guesses).  The square right between the 3s is actually 18%.

Third, at bottom-left, is the little corner around that L of 4 mines.  And there’s a few things to say about it.  First, from an assume-fewer perspective, if that lower-left corner is a mine, then both squares to the right of the 5 have to be mines, and it’s three total for the area, vs two mines needed if it’s safe.  So we’re already at background rate.  Furthermore, that layout would also solve the bottom-right 3, locking in that mine too.  So our odds are a bit better than background - actually 16%.  But the real beauty of this choice becomes clear for the reasons Scar lays out: guaranteed progress.  Assuming our guess is safe, there are two things that can happen.

(A) The lower-left cell turns up a 5.  We mark the mine to the right, and clear above-right.  Then in turn, that forces a mine to the right of the 3 above us, which satisfies the 3 above that, clearing 4 more squares (and solving the mine group above the lower-right 3, and so on).

 →

(B) The lower-left cell turns up a 6.  We mark the mine to the right, and also mark above-right.  This clears the cell next to the 3 above, but also fully satisfies the 3 to the right of the guessed cell, which allows its 2 other adjacencies to be cleared, likely leading to further progress.

 →

Either way, we’re in business here, in the 84% of the time that we don’t die.  Getting guaranteed progress off a guess is pretty rare, and in this situation it’s truly an oasis in the desert.

Solution: Guess the lower-left-most cell.  The other candidates are good guesses, but that’s a great guess.  And you’d have to say it worked out pretty well for Scar in the end.

Puzzle 10-14 “Toggle”: Find two independent good guesses.  Difficulty 7 / 10.  Source game is this 1k, at 13:15.

Analysis: We’ve now graduated into the hard part of the problem set, and one characteristic common to high-difficulty guessing situations is that you often end up with large, only semi-connected fronts for your guessing region.  This is the first situation with this large a front, but ideally the next few puzzles will make them feel less daunting to analyze in a game.

As with other puzzles in this set, there’s one obvious good guess here, but it’s unlikely to (and indeed doesn’t) lead to progress.  But to get it out of the way, focus on the upper-left of the formation.  There you have a classic Shared Corner, with the 4 (effective-2) and the 3 (effective-1) sharing two squares.  Unless both squares above the 4 are mines, the one below-right of the 3 cannot be a mine, and doing so would require three mines from those five squares rather than just two.  So below-right of that 3 is a very safe guess.

Which leads us to the more interesting part of the puzzle, for which the action is lower-right (in fact we’ll clip our screenshots to the area).  Although that hole formation has effective-2s on its right (the 4 and the 2), they are not equal: the 4 is asking for two mines from three squares while the 2 is asking for two mines from four squares.  And the 2 is at least slightly connected to the 5 below-left.

So let’s consider that left side.  How many mines can the 2 and the 4 share?  One possibility is “both”; in that case, the squares above and below them on the front are all safe (A).  This includes the one shared by the 2 and the 5, forcing a mine above the 5, and a mine to the right.

(A)   (B)   (C)

The other possibility is that they share only one mine, in which case the other one, below the 2, can be either on the left (B), or directly below it (C).  Case A uses 5 mines (including yellow group on the right), but fixes the location of 4 of them, making it less likely.  Case B also uses 5 mines, but two of them have flexible locations (in yellow), so that’s more likely - more configurations could use them.  But Case C uses 6 mines (!), while fixing the location of 4 of them.  So that has to be the least likely case of all, even though two of the mines have location flexibility.  And then notice the overlap in conclusions between cases A and B: specifically, the lower-right square and its dependent to the left are safe in both cases.  Only in the least-likely case C are those mines.  So that looks like a pretty good guess, based on a combination of assume-fewer and lock-in reasoning.

Solution: Two good guesses as marked:

The reason the second guess is so good to find is that you get almost guaranteed progress out of it.  In the 67% of the time that the mine along the right column there is shared with the 2 - which will be revealed by our 14% guess - then we get to chord the 2 and see what’s above it (and in the game, it was an opening).

Puzzle 10-15 “Waterfall”: Find 2 independent good guesses.  Difficulty: 7 / 10.  Source.

 → Labels:

Analysis: Looking at this fresh, without the benefit of seeing the solution, my first instinct goes to the middle 4 around the central corner, because of the 3 to the right, it looks like it should have some consequences.  I also notice that the 3 two cells below that 4 is an effective-2, which suggests some neighbor-mismatch potential to me; the only other effective-2 that I see is the bottom 4.  So those are the places I’m thinking are good spots to poke around.

Going back to the middle corner: if M is safe and L-N are mines, then the yellow group is safe (and K too), J is a mine, and F/G/H all safe.  That seems like an interesting set of consequences to shoot for.  Conversely, if something inside the yellow group is a mine - if they’re not all safe - then K and L are safe, putting mines on J and M, and thus O and P as well.  It seems like a lot of things can make K be safe, put a mine on J, and clear F/G/H up top, so there’s a split equivalence: the odds of K = F + G + H. my initial guess is that G and H, at least, have low odds.  F less so, because it’s shared with the 3 above it, so in some situations where J is not a mine, it’s likelier that F is a mine because that will tend to use fewer mines through the top part of the front.  And so G and H do turn out to be a good guess (10%).

But let’s consider the bottom 4, as well.  It needs two mines from three squares.  That means that if R is made safe, it forces mines on S and T.  The only thing that would do that is if Q were a mine.  What else happens then?  Mayhem:

In particular, that assumption (a mine on Q) happens to lock in the mines for both of our effective-2s on the front (the 3 above it and the 4 below it).  Going downwards, it makes R safe, so S and T are mines.  Going upwards, it makes P safe, leaving only two cells for two mines for that 3 above: they must be on N and O.  And with a mine on N, you get a mine on L, clearing the yellow group and K, mine on J, and F/G/H are safe again.  That’s 6 mines and 10 safe squares locked in by that single assumption.  That’s a lot!  And it makes Q our best guess on the map by far.

Similar logic to our initial reasoning will also show that the yellow group in our labeling (which only touches the 3 between K and L and so all three must have the same odds) must be low-probability too.  Because they are essentially splitting the residual probability that K and L leave behind for that 3, it’s a bit of a hidden split-equivalence.  So all three squares in the yellow group have low odds individually.  And of them, and you’d probably want to guess the corner square (below-right of that 3), so that you’re not sharing any adjacencies with any higher-odds squares like K or L.

So that’s one great guess, Q, and two sets of very good guesses (G/H and the yellow group).  Any two of which would suffice for an answer to this puzzle.

Solution: N.B., best odds of progress are the upper 10%s, not the lower 5%.

Puzzle 10-16 “Pillars of Sand”: Find 3 independent good guesses.  Difficulty: 7 / 10.

Source game (at 2:40).

Analysis: So this one is kinda intended to distract you with the pre-existing guess on that corner.  Our previous guessing rules would’ve said “just chain some guesses diagonally down from there”, but we now know we can do better than that.  The most obvious place to start, though, is right around that interior 1.  We know from square coverage ratio that the average percentage around it is 14%, which isn’t bad, but can we do better than that?  The obvious places to check first would be the two squares shared with the rest of the front.

We’ll go up first, because absent that interior 1, the obvious place to check would be that semi-framed box at top middle.  Those are usually good places to find good guesses, because there is usually some downstream dependency going off in one direction or another.

(A) An assumption of a mine directly above that interior 1, going to the left, fixes two mines around that corner. Going upwards, it puts one on the shared square of the 3 and 4 around the box, clearing that pink 2, and forcing a fifth fixed mine to its right.  All told, that’s 5 mines and 16 (!) safe squares forced.

Note that the 2 in the pink was the eye-grabber for me to begin with.  Even without that interior corner 1, I might have started with an assumption of a mine to its left, because that would clearly lock in a bunch of chained consequences.  And indeed, the assumption mine here and the one next to that pink (plus the one in between), are fully dependent: each implies the other two.  They therefore have the same odds, while the other forced mines under this scenario do not.

OK, so 5 mines and 16 safe squares sounds like a lot… but then again, that interior 1 implies really low density around there, maybe that’s just how things go.  What happens if we assume one to its left, instead?

(B) We get the mirror-image conclusions going upwards, with the 3/4 shared square being safe, and thus the top corner is a mine, not safe - so the conclusions stop short of scenario A.  Going left, however, we get a mine on the left corner, going down and providing the 3rd mine for the 3 at lower left here, clearing those two additional squares.  So we end up with 4 mines and 14 safe squares fixed by this assumption.

Again, sounds like a lot!  So how do we really tell if it’s “a lot”, in terms of lowering our probabilities below the average for that interior 1?  By looking at the third case: what happens if that 1’s mine is on a non-shared square.  Then we get:

(C) The immediately adjacent cells beyond that interior 1’s surroundings can’t have conclusions made about them yet.  We do have some 1-1 patterns (2-over-3 squares) to prove safe squares three spots away, and here those happen to force one mine in each direction.  All told, though, we get 2 mines and 10 safe squares from our assumption.

So what does this imply for our guess quality?  Well firstly, understand that this third case, Scenario C, is something like 5 times as likely as the others, because without evidence that cases A or B are more likely than average, case C happens for 5 of the 7 possible mine placements around that interior 1.  So the safe squares we see in Scenario C - which happen to be the dependent guesses of Scenario A (going upwards) and Scenario B (going leftwards) - are therefore likely to be safe.

But returning to the question of what the lock-in determines, let’s think of C as a baseline, and consider how much more is determined by A or B.  Relative to C (2 mines, 10 safe), A adds 3 mines and 6 safe squares, while B adds 2 mines and 4 safe squares.  That’s still a pretty solid amount!  This makes the dependent squares of A very unlikely - in fact the best guesses on the board - and also makes B pretty unlikely too (marginally better than each given guess out of the five non-shared squares in C, but with much higher chances of progress).  So those are two of our three independent good guesses!

Where else can we look for our third?  Well, there’s a part of the front we haven’t touched yet, the lower left.  Let’s try there.  Naively to start, we’d think we might have a 1v2 corner at bottom left: either a mine on the corner (shared by the leftmost 3 and 5), or two wide mines (bottom-left on edge, and below-right of the 5).  But wait, mines on those wide squares would also clear the middle 3, to the right of the corner… see Scenario D at right.  If fixing one additional mine is enough (via Assume Fewer) to get a background-rate square, then since this fixes two (plus a few extra safe cells), we’d guess this assumption is meaningfully below the background rate.

There’s one more alternative worth considering: we can see now that at the bottom-right of scenario D above, that rightmost 2 shares one square with the 3 above it, and so a mine there might have some lock-in potential.  Let’s check that flow:

(E) So an assumption of that shared square clears a bunch of cells below that 2, and above that 3.  Going up, that forces a mine above the 2, a safe square on the corner, and then we have a 1-1 pattern to get a safe square to the right.  Going left, we get a mine below the middle-left 3, putting a mine on the left corner and safe squares in between.  All told, 3 mines and 10 safe squares locked in by this assumption.  Sounds like a lot, again.

HOWEVER, there are a few things working against scenario E here, because even though on its face it would appear to lock in a bunch of stuff, the context actually argues otherwise:

1. If we limit ourselves to just the area considered in scenario D, both D and E use the same number of mines to mark the area.  There is also a two-mine scenario (mine on corner and three squares to its right), so while D and E are less likely than that, E isn’t any less likely than D.
2. Moreover, the conclusions going upwards from the assumption contradict what we concluded from Scenario B.  E would lock in safe squares on the upper-left corner and its dependent squares (to the right and below), which would normally predict that the assumption square is unlikely to be a mine, and those would-be lock-ins would be therefore more likely to be mines.  But we are already told by Scenario B (which is a much more extensive degree of lock-in) that its assumption squares are unlikely to be mines.  So the exact scenario above here, with those mines and safe squares, is a lot MORE likely than it might otherwise appear.

That doesn’t make the assumption in E a terrible guess, but it does make it worse than others we’ve considered - in particular, worse than B (far less lock-in, meaning B “wins” in terms of probabilities), and worse than D (since D and B can co-exist).  So despite superficially looking like a good guess, the assumption square in E is not awful but also not great: it’s actually 28%.  Plus you only get one cell with E, and it’s not particularly likely to yield progress, while you get three cells each (via dependent squares) with A, B and D.

Solution: Scenarios A, B and D are our good guesses, as marked below.  Reasoning out that scenario E is actually a not-great guess is what makes this problem somewhat hard.

Puzzle 10-17 “Sword and Shield”: Find two independent good guesses.  Difficulty: 7 / 10.

Source game (at 18:30)

Analysis: This one is a two-parter, with two very different approaches needed for each.

Let’s start with the easy one, along the top front.  The first thing that probably grabs your eye is the flat framed front at top-middle, under the pair of 3s.  It isn’t fully isolated, but its shared adjacencies are effective-1s, so it’s worth exploring.  If we look at it, an assumption on either of the squares shared by the two 3s doesn’t cascade at all.  However, an assumption on the wide squares does:

The assumption square(s) decide the entire top front, and then also clear below it, forcing a mine onto the lower corner and a safe square left, via a 1-2 pattern.  But more importantly, if you look at that top front, every other possible configuration of mines on it involves 3 mines; this is the only one using 4.  So not only does our flat-framed front logic apply (Assume Fewer), but it’s amplified by the fact that only one configuration remains possible there.  And amplified further by the downstream conclusions here.  This makes the assumption (and its two dependent squares along the top) very unlikely indeed.  So that’s one good guess.

The other one is more subtle, along the left side.  Here’s what we’re working with, labeled:

Recognition starts with seeing the split-equivalence: We have a corner(-ish) 1, in pink, between H/R/Q.  If H is a mine, then R/Q are safe, and there must then be a mine among T/U/V.  But if H is safe, then the pink’s mine is between R/Q, and so the 1 below it tells us that all of T/U/V are safe.  In other words, H = T + U + V.

(There’s no split-equivalence in the upper-right corner of the protrusion, by the L, because the corner square by L is a 2, not a 1)

It’s nice to see a split-equivalence, but if H is, say, a 75% probability, then proving that each of T/U/V have 25% mine odds doesn’t really do much for us.  But let’s at least poke around with it.  We would next notice that the continuation of the G-H front goes up around the D-E-F corner, and despite the intervening mine above D, continues left.  Our choices for mine coverage along those edges (A through H) include:

• If a mine is on the corner (E), then D/F are safe, and so is A (with a mine on one of B/C), and there’s a mine on G while H is safe.  That’s 3 mines, E-G-(B/C).
• If a mine is not on the corner (E), then there are mines on D-F-H, B/C are safe, and there’s a mine on A.  That’s 4 mines: A-D-F-H.

Based on Assume Fewer, we can conclude that A-D-F-H have background-rate odds (here 24%), while E, G, and (B+C) all equal (1-B), which is about 76%.  B and C split those odds based on other considerations beyond what we can address here, but it’s not necessary to sort that out, because the important thing is that we estimate H at about a 24% mine probability.

Because H is low (background-rate) odds, and we have the split-equivalence, that means each of T/U/V are ~8%, making them great guesses.  So that’s our answer.  Among them, you would probably guess V, because in the not-unlikely case that you get a 1, you have immediate progress, whereas the same isn’t true of the others (in fact, U is a dead cell: it’s either a mine or a 1).

There’s a third decent answer, though it’s subtle and not as likely to lead to progress.  You may have noticed that the lower-right corner, around O/P/Q, is a 2, but it’s surrounded by 1s.  Mentally we might just put its two mines into two pairs (one on N/O, one on P/Q), and certainly that is likelier.  But it’s also at least possible that both mines are on P & Q.  What happens if that’s the case?

If P and Q are both mines (in teal), then the 1 to its left clears R/T/U/V, so a mine on S.  The middle (pink) one clears H too, putting mines on G and E, while making F, D and A safe.  Going above the assumption squares, N/O are safe, therefore M is a mine.  And going around the interior rectangle again, there’s a mine on J/K, so L is safe.  That’s 4 mines and 11 safe squares locked in by this assumption, which feels like a lot, but then again we’re making two mines’ worth of assumptions.  Surely this means that this exact configuration is pretty unlikely, though.  And if it’s not the case - if there’s one mine on N/O and one on P/Q - then square M has to be safe.  The double mine on P-Q is the only scenario in which M is a mine.  This means M has very low odds!  But that’s hard to find and has low progress chances.

Solution: The good guesses are T/U/V (especially V) on bottom-left, and the FFF along the top.

Puzzle 10-18 “Thunderstorm”: Find 3 independent good guesses.  Difficulty: 8 / 10. Source.

Analysis: This puzzle is practically baiting you, with the hole pattern that puts the side-hole squares up on the wall… and yet also puts the stray 3 above it, probably cancelling out a lot of the usual favorable odds there.  We also have a nice open vertical front in the middle, but interrupted somewhat by the presence of the middle 4 (an effective-2), and the upper interior-2 at the top of that column.  And there’s an oddly-formed pseudo-hole possibility above that interior-2 at right.  These make for some interesting effects.

Firstly, let’s focus on the hole pattern.  That 3 in space makes it very likely that one mine is on the squares shared by the 3 and the upper-left hole 2 (there can’t be two).  If there were zero mines on those shared squares (in yellow), one of these two things would have to be true:

(A)   (B)

If yellow contains zero mines, then (in A) the side-hole square to the left could be a mine.  If so, the 3’s mines are all on the disjoint squares, it clears above the central hole 2, forcing a mine above the right hole 2, and two safe squares to its right.  Alternatively (in B), that side-hole square could be safe, but the one above the central 2 could be a mine, which would put 3 safe squares around the right hole 2.  Given how many mines this fixes in place, both of those are very unlikely - but A uses an extra mine to satisfy this front, so it is especially unlikely (<1%!).  Yellow is very likely to have one mine, making the assumption squares in both A and B safe.

But notice the pink group in those images: it illustrates the neighbor-mismatch situation we have with the corner 2.  We have no direct logic on it, but the existence of the mismatch helps influence the probabilities.  In particular, suppose yellow contains no mine: then the pink squares could have both mines, in which case purple is a mine.  But that would fix a lot of cells’ status; far likelier, pink contains one mine, and purple is safe.  Even if we assume (as is likelier) that yellow contains a mine, the corner 2 will be pulling probability towards it - from the blue groups (right squares favored), the square shared with the right hole 2, and above it (pink), too.

Let’s shift our focus over to the middle column, and label it.

Firstly, note that around that middle 4, its remaining mines are in pairs: it gets one from L/M and one from J/K.  Furthermore, some of these can’t coexist: if K is safe, then so is L (mines on J-L don’t work, only J-M), and if L is a mine, then so is K (because L clears both J and M).

And then there are the logic chains that help us see lock-in:

1. If B or F are a mine, it clears G-H, mine on J, clears L, mines on M-O-Q.  That’s 4 mines locked, plus 6 safe squares.
2. If G is a mine, it clears B-F-H-J, putting mines on L-K, clearing M, and thus mines on N-P.  4 mines, 7 safe squares.
3. If H is a mine, it clears B-F-G and J-L, mines on K-M-O-Q.  The mine on K also satisfies the interior 2, forcing another mine right.

Lastly, split equivalence due to the 1 next to G: A = F + G + H.

Let’s examine that 3rd scenario on the vertical column:

An assumption on square H, clears above it, forcing two mines to the left.  It clears below it (J/L), fixing mines for the middle 4, which clears the interior 2 above it and puts a mine to the right.  The mine on M also determines the bottom of the column.  Scenarios 1 and 2 would also determine the upper-left part of the front, but there are two differences that make this scenario less likely: Firstly, it forces two mines into the pink group (C-D-E), rather than just one; this increases the number of mines required for the full front and makes it less likely.  Second, it fixes a mine on K, which determines that right area around the 2.  All these things combine to make this guess (H) very unlikely to be a mine.

Now, what else fixes the mines for the middle 4?  A mine on L.  If L has a mine, then J and M are safe, so K is also a mine - it gets that lock-in to the right.  This is our scenario 2:

The assumption (G, or equivalently L) clears below on the column, forcing a mine above the 4, and the conclusions on the right.  It clears around the upper part (B-F-H), fixing a mine on A (and left), but it only leaves one mine for the pink group, not two.  And it also fixes the mines going down the vertical column, albeit in the other order from scenario 3.

By contrast, scenario 1 doesn’t fix the right hand side, doesn’t force a second mine into the pink group, and doesn’t even need to pick between B and F.  It is thus much likelier.

Lastly, although it’s harder to quantify, we need to observe the pulling-probability effect that the corner 2 has here: to avoid needing a second mine in the pink group, it pulls a lot of % rightward among the blue groups, i.e. specifically onto square B and away from A.  This is part of why scenario 1 is likelier, because at worst, square B starts with 50% probability, and in fact ends up much higher than that.

Solution: So our conclusions are (1) the side-hole square on the left wall, (2) the square above the central hole 2, and (3) square H on the central column, are all very unlikely to be mines.  Additionally, the chain of G-L-N-P (in yellow) is much less likely than its alternatives, and due to how many dependent squares it determines, is probably the best guess for progress here.

Puzzle 10-19 “Rhydon”: Find 3 independent good guesses.  Difficulty: 8 / 10.  Source game.

Analysis: As with other big complex fronts, the most useful thing to do first is to just go around the whole thing, and look for where the effective-2s are.  There are several - basically every 4 that’s on the perimeter - and one of them is an effective-3.

That effective-3, the corner 4 towards top-left, should draw our attention first, because it’s immediately adjacent to an effective-1 below it.  In the image below at left, basically if the 4 doesn’t share a mine with the 3 in pink, then all three of the cells in the blue group need to be mines.  Which is rather unlikely!  So how do we create that scenario?  Putting a mine below-left of the pink 3.  If that’s done, then it clears above and below, fixes those 3 mines, and clears another cell above besides.  And while it doesn’t fix a mine for the 4 below it, it does clear one of the cells, turning a relative density around that 4 from 40% (two mines in five squares) to 50%.  All told, that amount of lock-in is enough to make a good guess.

 → Scenario:

Next we look at the top area.  Again, the 4 needs two mines from three squares, so how can we fix those mines around it?  Force one safe square onto it: either to the right, or the one above.

Assuming a safe square above that 4 (or the opposite assumption - assuming a mine on either of the yellow-ringed squares), results in a big cascade going in both directions.  7 other mines are fixed by this!  This makes the assumption square likely to be a mine, and the yellow-ringed squares likely to be safe.  But not super likely, as we’ll see.

What if we instead force a safe cell to the right of the 4?  Well, that can be done by having a mine on either of the other squares around that 3 below it - on A or B inside the yellow.  That fixes the two mines above the 4, putting another one to its left along with two safe squares.  But notice that while a mine on B merely forces C to be safe, and no further conclusions to the right, a mine on A would force C to be a mine, which would push onto it all the right-side downstream conclusions of the previous scenario!  So A has to be a very unlikely square, even less likely than the previous ones.

That upper-right portion of the field seems to have a lot of potential to cascade some conclusions.  How else can we find high lock-in around there?  Let’s try our last effective-2, the 4 at far right.  It too needs two mines from three squares, so any safe square will lock it in.  Assuming a safe square right or above-right of it doesn’t seem to get us anywhere.  But assuming a safe square directly above it gets us the dramatic flow we’re looking for:

 Isn’t that pretty?

Our assumption fixes the two mines for the 4 at lower-right, but it also clears the 2 above the box, putting a mine by that upper-right corner 1, forcing mines to the left, which ultimately clears the 3 below the top 4, fixing the mines around the top 4, and pushing another mine and safe squares to its left.  Whew!  That’s 8 fixed mines and 12 safe squares from our assumption.  Clearly this is the least likely of all (in fact it is the safest guess on the board, at 5%).

And notice also that the square to the right of our cyan mine, while it wouldn’t fix the conclusions below by the 4, would still create the conditions for the other conclusions going up and to the left.  And it would also offer guaranteed progress, because the cyan mine above would be safe, and would indicate where the last mine for the lower-right 4 would be placed.  So that case is nearly as unlikely as the previous.

Solution: All told, that is at least 5 independent guesses available here.  My preference would generally be for the 14%s up top because you get two squares with it, but with that 3 on the front, the odds of progress are low.  So I’d probably make the guesses at far right, since taking a 5% and a 6% get me a split-equivalence 11% farther up the front, and 3 cells for 11% sounds better than 2 cells for 14%.  The 6% at top seems very unlikely to lead to progress, but the 6% on the left isn’t hopeless, either - it only needs three safe squares to turn up a 2 and yield a Hole pattern.

Puzzle 10-20 “Freefall”: Find at least 4 independent good guesses.  Difficulty: 9 / 10.

Source is this 1M difficulty near-miss by 324yeew.  The upper-left region of it, anyway.

Analysis: This puzzle is like a final exam, because it’s got a little bit of everything.  Also, it’s intentionally cropped so you don’t bother yourself with the stuff at the top-right edge.  Let’s start at the top left, and proceed around it discussing each sub-area and what it offers for good odds.

The top-left 4-square box there offers a 1v2, but that’s not exciting to us anymore.  But the top-right of that formation has that 1 sticking into space, which previously might suggest a Square Coverage Ratio-based guess.  But we’re better than that now, we don’t need to be that cavalier.  The three cells above the 1 are equivalent to the one two rows below that end 1 on the right… and you know what, let’s look at what happens if that’s a mine in the first place:

The yellow group can only contain a mine if the cyan (hypothesized) square is a mine.  And if so, then conclusions flow downwards along the front… and keep flowing.  In fact, that one assumption would lock in SIX other mines along this little edge.  So the cyan cell is itself very unlikely to be a mine, and since the yellow group’s odds in total equals that of cyan, each yellow cell has to be VERY favorable (in fact 2%s).

While we’re at it, notice that the downstream conclusions from that cyan hypothesis would also flow from the purple squares being mines, too.  Only the cell just below the cyan one being the mine around there wouldn’t force those conclusions below it, as it wouldn’t clear the 2 that’s above the 4-square box.  So the other cells around there (purple, and even below-right of that corner 2) are going to be good guesses as well - 7-9% odds.

Also in that same general area, the possibility of a pseudo-hole formation ought to suggest itself to you.  Labeled at right: the yellow group can only contain a mine if both G and H are safe.  And what are the odds that both G and H are safe?  Recall from the cascade in the image above that a mine on A or B forces mines onto all of D-G-J-M.  A mine on E likewise forces mines onto C-G-J-M (while clearing A and B and putting mines going upwards from there, too).  The degree of lock-in on these suggests that F is the likeliest mine between the D-E-F trio, and F being a mine doesn’t force one onto H.  But notice lock-in coming from the other direction, too: a mine on K or L clears J and forces a mine on H (making G safe and F a mine).  So G and H can only both be safe if you have mines on J-M and on F-C (and above).  That is not hugely unlikely, but it’s collectively somewhat unlikely.  Which makes the yellow group pseudo-hole guesses into pretty good guesses!  (9%s)

Next we need to move to the messy bottom of this front, which I will label:

It’s messy because we have a bunch of effective-2s (the 4 at upper-left, the middle 4 by J/K/L, and the lower-right 5 by M-N-O), and even an effective-3 in the 6 that’s there.  But with some effective-1s surrounding them, we know from our discussion that there has to be some decent guesses among the pile, too.  The blue group is useless to us (it only touches the hole 2, and is backed by an effective-1 below it, so the odds of each square are exactly 33%), but the pink group is a pseudo-hole formation, so we might like that option.

The most obvious place to attack here is that 6, which needs three mines from four adjacencies.  So if any one of them were to be safe, the other three would be mines, and that would fix a bunch of other conclusions too.  The two shared squares with the 4 above (D/E) seem tough to work with, so let’s ask what happens if G is the safe square:

What happens is, the whole bottom area of the board is determined.  G safe and H a mine, puts mines on D-E-F, clearing B-C and putting a mine on A.  Then going right, it marks J as safe, forcing mines onto K+L, clearing M, putting mines on N-O and making P safe.  That’s 8 mines and 5 safe squares!  That’s quite a lot.  So while the 6 alone wouldn’t have made H an incredible guess, the other downstream conclusions locked-in by it do make it an incredible guess (4%).  It has no dependent follow-ups (just mines on G and J), and low odds of progress, but it’s worth a shot.

But what of the right side of this area, and the pink group that we brought up?  Re-label it:

So we already know that pink has to be safe if either G or F are a mine, can only contain a mine if both G and F are both safe - that’s the pseudo-hole pattern.  But while the cyan group above likewise only touches one number square (the interior 3), J/K/L has to contain at least two mines, and in fact theoretically could contain three!  Why do we say “theoretically”?  Well, that would require that O be a mine, in which case H-M-N are safe, making J-K-L all mines.  It also puts a mine on G.  Hey wait, that sounds pretty unlikely!  Normally the O/P front next to that 5 at top right would be 50-50 or close to it, but in this case, a mine on O seems very unlikely - it forces 4 mines and 4 other safe squares.  So O has to be a good guess all on its own (6%).

And we’re not even done, because we’re still wondering about the pink group.  If G is a mine, then H is safe but there’s nothing else going upwards, and F is safe, forcing a mine on E, but nothing else going down.  So that’s fairly likely, via (lack of) lock-in.  But what of F?  

Well if F is a mine, then (going up) G is safe and H a mine, clearing all of M-N-O and putting a mine on P.  And going down, it clears E, forcing mines on C and D (from the bottom 5), making B safe, and putting a mine on A and also the square to A’s left.  That’s 6 mines and 9 safe squares from that one assumption.  Well, uh, F looks very unlikely to be a mine.  So, there’s bad news - the pink group isn’t that unlikely to contain a mine, because the odds F and G are both safe is non-trivial - but there’s also good news, because F itself is a good guess!

Solution: That’s 6 independent good guesses we explored, and if you can find at least 4 without help, you’re pretty awesome at guessing.