Scrabble tile checking

Question

For tile checking in scrabble, you make four 5x5 grids of letters totalling 100 tiles. I would like to make one where all 40 horizontal and vertical words are valid. The set of

Answer 1

I would approach the problem (naively, to be sure) by taking a pessimistic view. I'd try to prove there was no 5x5 solution, and therefore certainly not four 5x5 solutions. To prove there was no 5x5 solution I'd try to construct one from all possibilities. If my conjecture failed and I was able to construct a 5x5 solution, well, then, I'd have a way to construct 5x5 solutions and I would try to construct all of the (independent) 5x5 solutions. If there were at least 4, then I would determine if some combination satisfied the letter count restrictions.

[Edit] Null Set has determined that there are "4,430,974 5x5 solutions". Are these valid? I mean that we have a limitation on the number of letters we can use. This limitation can be expressed as a boundary vector BV = [9, 2, 2, 4, ...] corresponding to the limits on A, B, C, etc. (You see this vector in Null Set's code). A 5x5 solution is valid if each term of its letter count vector is less than the corresponding term in BV. It would be easy to check if a 5x5 solution is valid as it was created. Perhaps the 4,430,974 number can be reduced, say to N.

Regardless, we can state the problem as: find four letter count vectors among the N whose sum is equal to BV. There are (N, 4) possible sums ("N choose 4"). With N equal to 4 million this is still on the order of 10^25---not an encouraging number. Perhaps you could search for four whose first terms sum to 9, and if so checking that their second terms sum to 2, etc.

I'd remark that after choosing 4 from N the computations are independent, so if you have a multi-core machine you can make this go faster with a parallel solution.

[Edit2] Parallelizing probably wouldn't make much difference, though. At this point I might take an optimistic view: there are certainly more 5x5 solutions than I expected, so there may be more final solutions than expected, too. Perhaps you might not have to get far into the 10^25 to hit one.