by frazer » Mon Jul 18, 2005 9:43 am
In reply to skeptic, your argument looks good at first. There are indeed N=9!x(2612736)^2 ways to fill in B2, B4, B5, B6 and B8, for exactly the reasons you observe (B2 and B8 are independent of B4 and B6). However, different arrangements for B2, B4, B5, B6, and B8 have a different number of ways to complete to a full grid, which gets rid of the square factor. For example, if all but one of these N ways to fill in B2, B4, B5, B6 and B8 can be completed to a full grid in, say, 1000 ways, and the other one can be completed in 1001 ways, then the total number of full grids will be 1000*(N-1)+1001=1000N+1, no longer divisible by N.
As you observe, given B2 and B4, the number of ways of filling B1 is a complex calculation, and, indeed, it is here that the variation starts happening. There may be as few as 384 ways to complete B1, or as many as 448, if I remember correctly. Both B1 and B9 have this variation (and I see no reason why they should have the same number of possibilities, given B2, B4-B6, B8, although it's possible that it is true). You then point out, "the filling of B3 and B7 is deterministic and unique", but this is not quite right -- it is deterministic in that at most one solution exists, but it is actually much more likely that there are no solutions at this step. So again, you can have either 0 or 1 way to complete to a full grid given your first 7 blocks.
Frazer