BrainSnack® Sudoku
How to solve Sudoku?
To solve a Sudoku you don’t need any mathematical insight, logical reasoning suffices. There are three logical strategies that allow you to solve practically every Sudoku - from beginners to experts.

1. Excluding
Which number belongs in box A? We can exclude 1, 4 and 9 because they already appear in the frame that A is in. We can also exclude 7 and 8 because they appear in the row that A is in. And 3, 5 and 6 appear in the column that A is in. So box A can only contain the number 2. Following the same logic box B can only contain the number 5. Box C can contain a 4 or a 5 and box D can also only contain a 4 or a 5. But because we just placed number 5 in box B, box C can only contain a 4 and box D can only contain a 5.

2. Situating
'Excluding' involves finding a box where only 1 number is possible. Here we’re looking for every number that can be used in a box (the candidates) and we look for a unique candidate in a row, column or frame. The first box, upper left corner, can only contain a 5 or a 6 because the other numbers already appear in the frame, row and column of this box. We call 5 and 6 the candidates for this box. So for the first row we’ve filled in all the candidates for every empty box. In box A we see that there is a unique candidate for this row. The number 3 doesn’t appear amongst the other candidates. In other words, box A must contain the number 3. You can apply this strategy to all the other boxes. In this way we discover that box B must also contain a 3.

3. Securing
You’ll be able to solve most Sudokus with 'Excluding' and 'Situating'. For the most difficult puzzles you need an extra strategy. This strategy looks at certain patterns in the candidates. Even if the same candidates appear in different boxes, we can use that logic to secure another box. In this situation you can only solve box D using 'Situating'. But you can’t solve the upper right frame with 'Excluding' or with 'Situating'. We see that box B and C both have the same candidates (2 and 6). In other words if B is 6 then C is equal to 2 or the other way around. This means that the boxes that are in the extension of B and C (box D) or those that are in the frame of boxes B and C (A, E and F) can’t contain a 2 or a 6. So box D becomes an 8 and box A becomes a 1.

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