anthonyberet wrote:
> Tim Chase wrote:
> >> I want to work on a sudoku brute-forcer, just for fun.
> >
> >
> > Well, as everybody seems to be doing these (self included...), the
> > sudoku solver may become the "hello world" of the new world :)
> >
> >> What is the equivalent way to store data in python? - It isn't obvious
> >> to me how to do it with lists.
> >
> >
> > Several other answers have crossed the list. I've done it using a
> > dictionary of tuples:
> >
> > grid = {}
> > for row in range(1,10):
> > for col in range(1,10):
> > grid[(row,col)] = value
> >
> > item = grid[(3,2)]
> >
> > etc.
> >
> > Seemed fairly quick and worked for me.
> >
> 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.
Here's another way:
def region_map():
for row in range(9):
for col in range(9):
region = (row/3,col/3)
print region,
print
def identify_region(cell):
return (cell[0]/3,cell[1]/3)
def create_regions():
regions = {}
for row in range(9):
for col in range(9):
rowcol = (row,col)
reg = (row/3,col/3)
if regions.has_key(reg):
regions[reg].append(rowcol)
else:
regions[reg] = [rowcol]
return regions
grid = [[0,0,6,0,7,0,0,0,0], \
[0,3,5,4,0,0,0,9,0], \
[0,0,0,0,0,6,1,2,0], \
[0,0,0,0,0,3,2,0,8], \
[0,0,0,0,6,0,0,0,0], \
[4,0,7,2,0,0,0,0,0], \
[0,5,2,1,0,0,0,0,0], \
[0,7,0,0,0,5,9,1,0], \
[0,0,0,0,4,0,3,0,0]]
print 'the grid:'
for g in grid: print g
print
print 'the region ids:'
region_map()
print
# create the region dictionary
regions = create_regions()
# pick an arbitrary cell
cell = (5,5)
reg_id = identify_region(cell)
print 'cell:',cell,'is in region:',reg_id
print
print 'region',reg_id,'contains:'
reg_data = regions[reg_id]
for c in reg_data:
print grid[c[0]][c[1]],
"""
the grid:
[0, 0, 6, 0, 7, 0, 0, 0, 0]
[0, 3, 5, 4, 0, 0, 0, 9, 0]
[0, 0, 0, 0, 0, 6, 1, 2, 0]
[0, 0, 0, 0, 0, 3, 2, 0, 8]
[0, 0, 0, 0, 6, 0, 0, 0, 0]
[4, 0, 7, 2, 0, 0, 0, 0, 0]
[0, 5, 2, 1, 0, 0, 0, 0, 0]
[0, 7, 0, 0, 0, 5, 9, 1, 0]
[0, 0, 0, 0, 4, 0, 3, 0, 0]
the region ids:
(0, 0) (0, 0) (0, 0) (0, 1) (0, 1) (0, 1) (0, 2) (0, 2) (0, 2)
(0, 0) (0, 0) (0, 0) (0, 1) (0, 1) (0, 1) (0, 2) (0, 2) (0, 2)
(0, 0) (0, 0) (0, 0) (0, 1) (0, 1) (0, 1) (0, 2) (0, 2) (0, 2)
(1, 0) (1, 0) (1, 0) (1, 1) (1, 1) (1, 1) (1, 2) (1, 2) (1, 2)
(1, 0) (1, 0) (1, 0) (1, 1) (1, 1) (1, 1) (1, 2) (1, 2) (1, 2)
(1, 0) (1, 0) (1, 0) (1, 1) (1, 1) (1, 1) (1, 2) (1, 2) (1, 2)
(2, 0) (2, 0) (2, 0) (2, 1) (2, 1) (2, 1) (2, 2) (2, 2) (2, 2)
(2, 0) (2, 0) (2, 0) (2, 1) (2, 1) (2, 1) (2, 2) (2, 2) (2, 2)
(2, 0) (2, 0) (2, 0) (2, 1) (2, 1) (2, 1) (2, 2) (2, 2) (2, 2)
cell: (5, 5) is in region: (1, 1)
region (1, 1) contains:
0 0 3 0 6 0 2 0 0
"""
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