In an earlier post, I foolishly ventured that

Red Ed wrote:In restricted sudoku, there seem to be very few symmetries to exploit, so we would appear to be in the unfortunate situation of needing much more computing power to calculate a much smaller number ...

Well, not so

The number of restricted sudokus, such that no digit appears at the same offset in any two 3x3 boxes, appears to be 201105135151764480. Here's how ...

1. without loss of generality, we can assume the top left box is 1...9 in order. We can also assume that the 1s in boxes 2,3 (top middle/right) are in rows 2,3 respectively and the 1s in boxes 4,7 are in columns 2,3. So we solve the problem with those constraints and multiply the answer by 9! x 4.

2. naively, we can just exhaust over the 244 subproblems corresponding to each of the possible arrangements of 1s. Someone said earlier that there are 8784 arrangements without the constraints above, so that's 8784/9 = 976 with 1 in the top left cell and 976/4 = 244 with the additional row/col constraints. But we can do better.

3. following AFJ's and Bertram's lead, we can look for equivalence classes. The following operations leave the number of solutions intact: (a) rotate the whole grid; (b) transpose it; (c) switch the order of the row bands [by band, I mean rows 1,2,3 or 4,5,6 or 7,8,9]; (d) switch the order of the rows within each band according to the *same* permutation for each band. Applying these ops, we find that the 244 layouts of the 1s breaks down into 11 equivalence classes of sizes 1, 2, 4, 9, 12, 18, 18, 36, 36, 36, 72. The number of solutions for each of the 11 representative subproblems is different, suggesting this is the best we can do.

I could easily have messed up my sums here -- I've done it before. Anyone want to check it?