On Sep 28, 10:27 pm, AndyB <[EMAIL PROTECTED]> wrote:
> ...
> This is in a program that generates random numbers to do a brute force
> solve on a sudoku-like puzzle. Once a certain level of difficulty in
> the puzzle is reached, performance goes off a cliff because the
> duplicate checking code, although fast, is executed so many times.
For a sudoku solver, you may be better dodging the problem, and
maintaining a set per row, column and box saying which numbers have
been placed already - and thus avoiding adding duplicates in the first
place. It may be better to use a bitfield instead (ie use bits 1..9 of
an int to represent these sets). You should also check out Donald
Knuth's 'Dancing Links' algorithm as a clever implementation of a
brute-force search that works perfectly for sudoku solving. (It uses
linked lists, but the idea still works in python).
But to answer your actual question:
> I have found a lot of material on removing duplicates from a list, but I
> am trying to find the most efficient way to just check for the existence
> of duplicates in a list. Here is the best I have come up with so far:
>
> CheckList = [x[ValIndex] for x in self.__XRList[z]]
> FilteredList = filter((lambda x:x != 0),CheckList)
> if len(FilteredList) > len(sets.Set(FilteredList)): return
In general, the natural way to test for duplicates is as you have it,
except that set is now built-in, so you can write
def has_no_duplicates(x):
return len(x) == len(set(x))
But in your case, you're having to construct the list and filter, so
it's better just to loop and do everything at once:
found = set()
for x in self.__XRList[z]:
cell = x[ValIndex]
if cell != 0 and cell in found:
return False
found.add(cell)
Since your elements are probably numbers 1 to 9 (or 0), you might use
bitfields. I'd expect this to be faster, but you should check:
found = 0
for x in self.__XRList[z]:
cellbit = 1 << x[ValIndex]
if cellbit != 1 and (cellbit & found):
return False
found |= cellbit
--
Paul Hankin
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