At 04:18 AM 5/17/2006, Gary McGuire wrote:
If you then "mix" the columns, by taking the 1st, 4th, 7th, 2nd, 5th, 8th, and then 3rd, 6th, 9th
columns (BTW, "mix" is its own inverse), you get:
[ 1 4 7 2 5 8 3 6 9 ] [ 2 5 8 3 6 9 1 4 7 ] [ 3 6 9 1 4 7 2 5 8 ] [ 4 7 1 5 8 2 6 9 3 ] [ 5 8 2 6 9 3 4 7 1 ] [ 6 9 3 4 7 1 5 8 2 ] [ 7 1 4 8 2 5 9 3 6 ] [ 8 2 5 9 3 6 7 1 4 ] [ 9 3 6 7 1 4 8 2 5 ]
This matrix now satisfies the Sudoku criteria.
It also has a very simple structure.
I now wonder how many (what % of) sudoku squares can be generated from the direct product of two 3-element groups, followed by some sort of "mix"-ing operation, such as the one above.
Up to isomorphism there are two groups of order 9, Z3 x Z3 and Z9.
Don't you mean the only _commutative_ groups? Aren't there a lot more non-commutative groups of order 9? (I'm very rusty on my group theory...) Latin square operation tables only require "solvability" of equations, but not necessarily the associativity or identities of a group operation. But I'm willing to look at only group operation tables for the moment.
I assume "mix"-ing means doing the usual equivalences: -permuting rows/columns within a band -permuting bands -transposing
No, "mix"-ing does _not_ preserve sudoku-ness. This is a different operation that can create sudoku-ness out of a non-sudoku.
The group of these equivalences has order 2*6^8=3359232. Your Z3 x Z3 grid has an automorphism group of order 648 (the max possible) so you can get 3359232/648 = 5184 different grids. For the other group of order 9, the cyclic group, you also get a grid with 648 automorphisms, so another 5184 grids.
123 456 789 456 789 123 789 123 456
234 567 891 567 891 234 891 234 567
345 678 912 678 912 345 912 345 678
Gary McGuire