> If we call the three kos x,y,z from top to bottom, then a succesfull > White ladder amounts to > (x || y) && (y || z). Which is equivalent to y || (x && z). > So with y currently false, and White unable to flip it, White should > take the bottom ko to make z true. > Black can the make x false, but that allows White to make y true, > after which she can successfully escape > in a ladder.
I have attempted to reduce this y || (x && z) problem to the minimum number of stones at the bottom of my Go page http://tromp.github.io/go.html, which also contains an sgf link. Direct link to image: http://tromp.github.io/img/WO5lives.png regards, -John _______________________________________________ Computer-go mailing list Computer-go@computer-go.org http://computer-go.org/mailman/listinfo/computer-go
