On Fri, 05 Jul 2013 15:45:25 +0100, Oscar Benjamin wrote:
> Presumably then you're now down to the innermost loop as a bottle-neck:
>
> Possibilities= 0
> for d in range(1,10) :
> if Row_Digits[r,d] or Col_Digits[c,d] or Sqr_Digits[Sq_No,d] :
> continue
> Possibilities+= 1
>
> If you make it so that e.g. Row_Digits[r] is a set of indices rather
> than a list of bools then you can do this with something like
>
> Possibilities = len(Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No])
>
> or perhaps
>
> Possibilities = len(set.union(Row_Digits[r], Col_Digits[c],
> Sqr_Digits[Sq_No]))
>
> which I would expect to be a little faster than looping over range
> since the loop is then performed under the hood by the builtin
> set-type.
>
> It just takes practice.
indeed
> It's a little less obvious in Python than in
> low-level languages where the bottlenecks will be and which operations
> are faster/slower but optimisation always involves a certain amount of
> trial and error anyway.
>
>
> Oscar
I've tried the following version
def find_good_cell() :
Best= None
minPoss= 10
for r,c in Grid :
if Grid[(r,c)] > 0 : continue
Sq_No= (r//3)*3+c//3
Possibilities= 9-len(Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No])
if ( Possibilities < minPoss ) :
minPoss= Possibilities
Best= (r,c)
if minPoss == 0 : Best=(-1,-1)
return Best
All_digits= set((1,2,3,4,5,6,7,8,9))
def Solve(R_Cells) :
if R_Cells == 0 :
print("\n\n++++++++++ S o l u t i o n ++++++++++\n")
Print_Grid()
return True
r,c= find_good_cell()
if r < 0 : return False
Sq_No= (r//3)*3+c//3
for d in All_digits - (Row_Digits[r] | Col_Digits[c] | Sqr_Digits[Sq_No]) :
# put d into Grid
Grid[(r,c)]= d
Row_Digits[r].add(d)
Col_Digits[c].add(d)
Sqr_Digits[Sq_No].add(d)
Success= Solve(R_Cells-1)
# remove d again
Grid[(r,c)]= 0
Row_Digits[r].remove(d)
Col_Digits[c].remove(d)
Sqr_Digits[Sq_No].remove(d)
if Success :
Zuege.append((d,r,c))
return True
return False
which turns out to be as fast as the previous "dictionary only version".
Probably, set.remove is a bit slow
Thanks,
Helmut
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