The Math Behind Sudoku

4. The 4x4 case

Let us look at the Sudoku of rank 2, or the 4×4 Sudoku grid, in order to get a

better understanding of the ideas we have discussed so far.

In order to count the number of 4×4 Sudoku grids, we start - as in the 9×9 case

- by filling in the top left block in the standard way. This accounts for a factor of

4! in the number of distinct grids, since there are 4! ways to relabel four digits.

After doing this, 1 and 2 occur in the first two entries of the first row, leaving 3

and 4 to go in the second two entries. For each of the 4! labelings, there are two
ways to insert 3 and 4 into two cells, and this contributes to the count of distinct

grids. However, for a count of essentially different grids, the order in which we

enter 3 and 4 here does not matter, since swapping the third and fourth columns

preserves the validity of the grid. So we may assume that 3 goes in the third cell

and 4 in the fourth cell of the first row.

Since 1 and 3 occur in the first column of the grid already, 2 and 4 must be

placed in the third and fourth cells of that column. Since there are two ways to

do this, our count of distinct grids is now up to 4!×2×2. For counting essentially

different grids, we note that swapping the third and fourth rows is a symmetry of

the grid, so we may assume that 2 goes in the third cell and 4 in the fourth cell

of the first column. Since there is a 2 in the third row and a 3 in the third column,

the only possibilities for the cell in the (3,3) position are 1 and 4.