anthonyberet wrote:
> I want to work on a sudoku brute-forcer, just for fun.
>...
> Thanks for the advice (to everyone in the thread).
> I think I will go with nested lists.
> However, I am running into a conceptual problem.
> My approach will be firstly to remove all the impossible digits for a
> square by searching the row and column for other occurances.
>
> However, I wondering how to approach the search of the nine regions of
> the grid. I am thinking of producing another nested list, again 9x9 to
> store the contents of each region, and to update this after each pass
> through -and update of- the main grid (row and column).
>
> I am not sure how to most efficiently identify which region any given
> square on the grid is actually in - any thoughts, for those that have
> done this? - I don't want a massive list of IF conditionals if I can
> avoid it.
Some 'UselessPython' :
import math
def SudokuOrder( length ):
block_length = int(math.sqrt(length))
for block in range(length):
row_offset = block_length * ( block // block_length )
col_offset = block_length * ( block % block_length )
for i in range( block_length ):
for j in range( block_length ):
yield i+row_offset, j+col_offset
grid = list(SudokuOrder(9))
BLOCK1 = grid[:9]
BLOCK2 = grid[9:18]
BLOCK9 = grid[72:81]
print
print 'BLOCK1 ->', BLOCK1
print
print 'BLOCK2 ->', BLOCK2
print
print 'BLOCK9 ->', BLOCK9
BLOCK1 -> [(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2,
1), (2, 2)]
BLOCK2 -> [(0, 3), (0, 4), (0, 5), (1, 3), (1, 4), (1, 5), (2, 3), (2,
4), (2, 5)]
BLOCK9 -> [(6, 6), (6, 7), (6, 8), (7, 6), (7, 7), (7, 8), (8, 6), (8,
7), (8, 8)]
Gerard
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