BobC wrote:
I don't stand a chance of working this out but an interesting question from the point of view of mathematical simplicity is the following:
In the above go board a white piece is placed at "a" and its "value" determined. It is then placed at "b" and a further value determined. The test white stone follows a closed path ending at "n". All the values for each position are then added up.
The question is. Do all the values for white add up to zero??????
This isn't trivial.. those with a mathematical background will see this as a test for a conservative field. If the sum is zero then the maths becomes "easier"... actually still stink hard but all the same... Ant takers???
No, the values do not add to zero. Not by a long shot.
Edit: Your post has some ambiguity. Are we to evaluate the position, or the play of the White stone?
My answer was based on thinking that you meant to evaluate the play of the stone, but perhaps you meant to evaluate the position with one Black stone and one White stone. In that case the sum of the values is still not zero, but it is closer to zero. If they were zero, then either in all cases the values of the White play and Black play are equal, or some of the White values are better than the value of the Black stone. In that case, Black had a better play.
But, in fact, the Black play is considered better than most of the White plays, and roughly equal to some of them. I. e., the sum of the positional values should be positive, from the standpoint of Black.
