Fredsweeper Patterns and Rules.

1. If a number is touching the same number of cells,
2. then these cells are all mines.

1. If a number is touching the same number of flags,
2. then all adjacent cells can be opened.

1. Look at the left 1.
2. It touches the two yellow cells, so they contain one mine.
3. Now look at the right 1.
4. It also touches the two yellow cells, so it already has a mine and the third cell can be opened.

1. Look at the left 1.
2. It touches the two yellow cells, so they contain one mine.
3. Now look at the right 1.
4. It also touches the two yellow cells, so it already has a mine and all the remaining cells can be opened.

1. Look at the 1.
2. It touches the two yellow cells, so they contain one mine.
3. Now look at the 2.
4. Its first mine is in the yellow cells, therefore, the second mine is in the third cell.

1. Look at the 1.
2. It touches the two yellow cells, so they contain one mine.
3. Now look at the 4.
4. It also touches the two yellow cells, so it already has one mine, therefore, as there are only three remaining cells, they must be mines.

Classic version of the 1–2 pattern:

1. Look at the purple cells, what can we say about them?
2. They can't both be mines, because that would overflow the 1. They contain a maximum of one mine.
3. Now look at the 2.
4. If the purple cells can't both be mines, then there must be a mine in the remaining third cell.

1. Look at the purple cells, what can we say about them?
2. They can't both be mines, because that would overflow the 1. They contain a maximum of one mine.
3. Now look at the 4.
4. If the purple cells can't both be mines, then there must be three mines in the three remaining cells.

The 1–2–1 pattern has one solution. Actually, it's just a combination of two 1–2 patterns:

1. Apply the 1–2 pattern from the left.
2. Apply the 1–2 from the right.
3. And here's the final result.

The 1–2–2–1 pattern has one solution. It's also just a combination of two 1–2 patterns:

1. Apply the 1–2 pattern from the left.
2. Apply the 2–1 from the right.
3. And here's the final result.

1. 2 turns into 1, since it already has one mine.
2. 4 also turns into 1, since it already has three mines.
3. Therefore, 2–4 turns into the 1–1 pattern and the third cell can be opened.

1. 2 turns into 1, since it already has one mine.
2. 3 also turns into 2, since it already has one mine.
3. Therefore, 2–3 turns into the 1–2 pattern and there must be a mine in the third cell.

1. 2–3–2 turns into the 1–2–1 pattern.

1. Look at the bottom 1.
2. It touches the two yellow cells, so they contain one mine.
3. Now look at the top 1.
4. It also touches the two yellow cells, so it already has a mine and all the remaining cells can be opened.

1. Look at the bottom 1.
2. It touches the two yellow cells, so they contain one mine.
3. Now look at the top 1.
4. It also touches the two yellow cells, so it already has a mine and all the remaining cells can be opened.

1. Look at the bottom 1.
2. It touches the two yellow cells, so they contain one mine.
3. Now look at the top 1.
4. It also touches the two yellow cells, so it already has a mine and all the remaining cells can be opened.

1. Look at the bottom 1.
2. It touches the three yellow cells, so they contain one mine.
3. Now look at the top 1.
4. It also touches the three yellow cells, so it already has a mine and all the remaining cells can be opened.

1. Look at the purple cells.
2. They contain a maximum of one mine, because they touch 1.
3. Now look at the 2.
4. If the purple cells contain a maximum of one mine, then there must be a mine in the remaining cell.
5. If you put the flag, you can see that there is one mine in the purple cells. So 1 already has a mine and all the remaining cells can be opened.

1. Look at the purple cells.
2. They contain a maximum of two mines, because they touch 2.
3. Now look at the 3.
4. If the purple cells contain a maximum of two mines, then there must be a mine in the remaining cell.
5. If you put the flag, you can see that there are two mines in the purple cells. So 2 already has mines and all the remaining cells can be opened.

1. Look at the purple cells.
2. They contain a maximum of two mines, because they touch 2.
3. Now look at the 4.
4. If the purple cells contain a maximum of two mines, then there must be two mines in the remaining cells.
5. If you put the flags, you can see that there are two mines in the purple cells. So 2 already has mines and all the remaining cells can be opened.

1. Look at the purple cells.
2. They contain a maximum of one mine, because they touch 1.
3. Now look at the 2.
4. If the purple cells contain a maximum of one mine, then there must be a mine in the remaining third cell.
5. If you put the flag, you can see that there is one mine in the purple cells. So 1 already has a mine and all the remaining cells can be opened.

1. Look at the purple cells.
2. They contain a maximum of one mine, because they touch 1.
3. Now look at the orange cells.
4. They also contain a maximum of one mine, because they also touch 1.
5. Now look at the 3.
6. The purple and orange cells contain a maximum of two mines, so the third mine must be in the last corner cell.
7. 3 has two mines left. One of them must be in the purple cells, the other must be in the orange cells. Therefore, 1–3–1 corner has one solution.

1. Look at the purple cells.
2. They can't both be empty, because that would overflow the 2. So they contain a minimum of one mine.
3. Now look at the orange cells.
4. They also can't both be empty, because that would overflow the 2. So they also contain a minimum of one mine.
5. Now look at the corner 2.
6. The purple and orange cells contain a minimum of two mines, so 2 already has mines and the last corner cell can be opened.
7. The first mine must be in the purple cells, and the second mine must be in the orange cells. Therefore, 2–2–2 corner has one solution.

1. Look at the purple cells.
2. They contain a maximum of one mine, because they touch 1.
3. Now look at the orange cells.
4. They also contain a maximum of one mine, because they also touch 1.
5. Now look at the 2.
6. The first mine must be in the purple cells, and the second mine must be in the orange cells. Therefore, all the remaining cells can be opened.

1. The left 1 touches the two yellow cells, so they contain one mine.
2. The right 1 touches the two purple cells, so they also contain one mine.
3. 2 touches the two purple cells, so the remaining orange cells also contain one mine.
4. Now look at the 3.
5. The yellow and orange cells contain a maximum of two mines, so the third mine must be in the last corner cell.
6. Look at the 3 again. The first mine is in the corner, the second mine is in the bottom cells, so the third mine is in the two remaining orange cells. Therefore, cells marked with green circles can be opened.

Usually long dependency chains can be resolved by parsing from both sides.

1. First, let's start parsing from the top.
2. There is 1 mine in the yellow cells.
3. There is 1 mine in the purple cells.
4. Now, let's start parsing from the bottom.
5. There is 1 mine in the orange cells.
6. There is 1 mine in the red cells.
7. There is 1 mine in the white cells.
8. There is 1 mine in the brown cells.
9. There is 1 mine in the cyan cells.
10. Look at the green 2.
11. The first mine must be in the purple cells, the second mine must be in the cyan cells. Therefore, cells marked with green circles can be opened.

When the game comes to an end, you can determine empty cells by the number of remaining mines:

1. There are 2 mines in the yellow cells.
2. There is 1 mine in the purple cells.
3. There is 1 mine in the orange cells.
4. There is 1 mine in the red cells.
5. There is 1 mine in the brown cells.
6. The colored cells contain 6 mines in total, just as much as the number of remaining mines. Therefore, cells marked with green circles can be opened.

In rare situations, it may be necessary to enumerate possible combinations:

1. Choose an arbitrary cell. For example, the yellow cell.
2. Let's check if there can be a mine.
3. If there is a mine in the yellow cell, then for the final solution we need 6 flags. This combination is not possible, because there are only 5 mines left.
4. Therefore, the yellow cell cannot be a mine and can be opened.