One basic strategy for any Sudoku game is to look at the rows and columns of the 9×9 Sudoku grid and to identify squares in which an integer must be placed. To do this, choose any number and look across the columns to see which columns already contain the particular numeral. For instance, you may choose to start with 1, and look from left to right for any columns with a 1 already in it. If you find any of the three sets of three columns in which two columns already possess a 1, you may be able to logically assume where to place the 1 in the third column.

To do this, find which 3×3 subsection in the set of three columns does not contain a 1. Since we have established that two columns already have the number, and the numbers cannot be in the same 3×3 square, only one should remain. In this 3×3 square, two columns have been already eliminated, leaving a maximum of three individual boxes to choose from, or, if luck is on your side, one empty box and two occupied boxes. Of course, if you find only one empty box, you will place the 1 in the empty box with certainty.

If there are two or three empty boxes in the column within the 3×3 square without the 1, you will need to look at the rows for clues to guide you to the correct box. Of your three rows in this set, look to see if any 1’s exist in the other sets. Hopefully, you will find one or two, which will narrow your options. If a 1 already exists on the same row as one of your empty boxes, that box will not receive another 1, since only one of each integer can be on the same row.

### Narrow Focus – Individual Boxes

While the method above looks first at the whole grid, then narrows the focus to just one square, this strategy chooses one square and then broadens the scope to include the rest of the grid.

On your puzzle, choose an empty square with a moderate amount of given numbers around it. First, look at the 3×3 subsection in which your empty square resides. You can eliminate any givens within that subsection as contenders for this particular square, since each subsection cannot have more than one number of the same value. For example, let us assume that the subsection of a particular empty square already contains a 2, 5, and 6. We can definitely rule these integers out of our consideration set for our chosen empty square.

Now, look at the row on which your empty square is found. You can safely rule out any numbers on that row. For example, our chosen square (which we already know cannot contain a 2, 5, or 6) has an 3, 7, and 8 on its row. Therefore, just from looking at the row and the subsection surrounding our chosen square, we have eliminated six of the nine possible numbers for our square.

Finally, look at the column above and below the empty square. Whatever givens exist on that column cannot exist in this particular square. For instance, if the same square we have mentioned has a 4 and a 9 in the same column, we can rule out those two possibilities. So, for our chosen square, 2, 5, and 6 were eliminated by looking at the 3×3 subsection, 3, 7, and 8 were eliminated by looking at the row, and 4 and 9 were eliminated by looking at the column. By the process of elimination, then, only one possibility remains for our square – 1.

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