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.
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