John VanSickle <evi### [at] hotmailcom> wrote:
> Warp wrote:
> >   Inside a 3x3 square, where all the numbers between 1 and 9 are put, the
> > number of possible permutations of those numbers is 9! = 362880. Since
> > there are 9 such squares, the total number of combined permutations is
> > 9^362880. Sure, this is a huge number, but it's a fixed upper bound and
> > makes the solver O(1).

> Uh, shouldn't that be (9!)^9 instead of 9^(9!)?

  Yes. You are right.

  Btw, 9!^9 = 109110688415571316480344899355894085582848000000000, which
would be the total number of possible combinations of those numbers.
(Of course only a small percentage of all these combinations form valid
sudoku solutions, and if equal all symmetric positions, it gets even
smaller.)

-- 
                                                          - Warp

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