On Nov 13, 2007 3:30 PM, Jason House <[EMAIL PROTECTED]> wrote: > > > > On Nov 13, 2007 3:13 PM, Imran Hendley <[EMAIL PROTECTED]> wrote: > > Looking at my code I first check if the number of pseudoliberties is > > less than or equal to 2 (this is necessary but not sufficent for a > > string to be in atari given the way I compute pseudoliberties), which > > is very fast (it just involves a lookup because I have already > > computed pseudoliberties for all strings when the last move was made). > > If this is true I then do a slower check to see if the string in > > question is actually in atari. This gets the best of both worlds > > because most strings (ones with more than two pseudoliberties) are > > ruled out very quickly and only a few are looked at with the slower > > routine that uses real liberties to check for atari. > > > I'm curious how you do that. The simpler problems can fit in 3x3 patterns > such as 5-7 stones forming a "C", or 8 stones forming an "O". There can be > larger, more pathological cases such as an E or butterfly shapes that can > have disparate stones counting liberties more than two times.
I was afraid someone would ask that! I forgot to emphasize that I do not calculate pseudoliberties in the standard way (I use something a little more complicated to address the kinds of problems you mentioned), and that the "2" in if pseudoliberties <= 2 would most likely be a different number for someone else, but now I guess I will have to go back over my code and see if what I'm doing is actually correct. I remember thinking about this problem and spending a lot of time on it, but it was so long ago that I can't remember how it ended up working in the end, so I'll try to get back to you soon. _______________________________________________ computer-go mailing list computer-go@computer-go.org http://www.computer-go.org/mailman/listinfo/computer-go/
