On Thu, 24 Jan 2008 21:09:42 +0100
Thomas Thiel <[EMAIL PROTECTED]> wrote:
> Neither fast nor user friendly, but very concise:
This is a bit faster:
options = set([str(i) for i in range(1, 10)])
def allow(puzzle,i):
exclude = set(x if i//9 == j//9 or i%9 == j%9
or i//27 == j//27 and (i%9)//3 == (j%9)//3
else '0' for j,x in enumerate(puzzle))
return options-exclude
def solve(puzzle):
zeros = [i for i,x in enumerate(puzzle) if x == '0']
if not zeros:
print puzzle
else:
i,R = min(((j,allow(puzzle,j)) for j in zeros),
key=lambda x: len(x[1]))
for option in R:
solve(puzzle[:i] + option + puzzle[i+1:])
P.
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