How many different $2 \times 2$ Sudokus are there

combinatoricslatin-squarepermutationssudoku

PROBLEM
How many different $2 \times 2$ Sudokus are there?

APPROACH
This seems pretty easy to brute force. There are $576$ Latin squares of size $4$ (which are the sudokus without restriction on boxes), of which $288$ are valid sudokus.
One way two sudokus are not different is by permuting the symbols.

Example

1 2 | 3 4        2 1 | 3 4
3 4 | 1 2        3 4 | 2 1
----+----        ----+----
2 1 | 4 3        1 2 | 4 3
4 3 | 2 1        4 3 | 1 2

are equivalent because it simply renames $1\rightarrow2$ and $2 \rightarrow 1.$

This is also called Relabeling Symbols specifically.

An easy way to prevent relabeling is to fix the first row to 1 2 | 3 4. Now there are 24 Latin squares (which makes sense, since each one would represent $1\times2\times3\times4$ permutations of the numbers, and $24\times24=576$), and $12$ of these are valid sudokus $(12\times24=288)$.

Now they are grouped by Row Permutations (where exchanging the rows will lead them to being equivalent)

1 2 | 3 4      1 2 | 3 4        1 2 | 3 4      1 2 | 3 4
3 4 | 1 2      3 4 | 1 2        4 3 | 2 1      4 3 | 2 1
----+----      ----+----        ----+----      ----+----
2 1 | 4 3      4 3 | 2 1        2 1 | 4 3      3 4 | 1 2
4 3 | 2 1      2 1 | 4 3        3 4 | 1 2      2 1 | 4 3



1 2 | 3 4      1 2 | 3 4
3 4 | 1 2      3 4 | 1 2
----+----      ----+----
2 3 | 4 1      4 1 | 2 3
4 1 | 2 3      2 3 | 4 1



1 2 | 3 4      1 2 | 3 4        1 2 | 3 4      1 2 | 3 4
3 4 | 2 1      3 4 | 2 1        4 3 | 1 2      4 3 | 1 2
----+----      ----+----        ----+----      ----+----
2 1 | 4 3      4 3 | 1 2        2 1 | 4 3      3 4 | 2 1
4 3 | 1 2      2 1 | 4 3        3 4 | 2 1      2 1 | 4 3



1 2 | 3 4      1 2 | 3 4
4 3 | 2 1      4 3 | 2 1
----+----      ----+----
2 4 | 1 3      3 1 | 4 2
3 1 | 4 2      2 4 | 1 3

Does this mean that there are $4$ essentially different $2\times2$ sudokus? Are there any more transformations that lead to equivalence?

I've ruled out stack/column/band transformations, since the first row is fixed to 1 2 | 3 4. I am unsure if the row transformations are all essentially equivalent (in particular, swapping a row in the first band with a row in the second band), so there could be $6$ different puzzles. A post on this ancient forum claims that there are only $2$ essentially different puzzles, but I can't find any more transformations.

Here is a link to the brute force code I've used: https://wandbox.org/permlink/6vhSnSRPaqCX4cse

Also is there any slick reasoning why exactly half of all $4\times4$ Latin squares are valid sudokus?

Best Answer

The transformations you can apply to a sudoku are: permuting the numbers, rotating the sudoku, band permutation, line or column permutation inside a band, horizontal, vertical or diagonal symmetry... All of these transformations form a group, many of these transformations can be deduced from other, see wikipedia for a more detailed description of the group of transformations.

Your first sudoku is equivalent to the third one by the following transformations:

1 2 | 3 4 
3 4 | 1 2 
----+---- 
2 1 | 4 3 
4 3 | 2 1

Rotation to the right:

4 2 | 3 1 
3 1 | 4 2 
----+---- 
2 4 | 1 3 
1 3 | 2 4

Permutation of the two first columns:

2 4 | 3 1 
1 3 | 4 2 
----+---- 
4 2 | 1 3 
3 1 | 2 4

Taking permutation $\sigma = (4, 1, 3, 2)$

1 2 | 3 4 
4 3 | 2 1 
----+---- 
2 1 | 4 3 
3 4 | 1 2

By combining transformations, you should reduce the number of non equivalent tranformations to 2.

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