Re: Paper “Complexity of Go” b y Robson

> 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
[email protected]
http://computer-go.org/mailman/listinfo/computer-go