What is the "move value" of a simple ko position A
Code:
A A
/ \ / \
/ \ / \
B C B c
/ /
/ /
D d
let's call c and d the counts of leaves C and D.
How to justify that the "move value" va of position A is equal to the value m = (d - c) / 3 ?
1) let's take first a temperature t > m
Let's take a rich environment Rich(t, N)
In this environment the score of the game if black decides to gain the ko is
score1 = d - g(Nt/N) - g((N-1)t/N) + g((N-2)t/N) - ...
while the score of the game when black plays in the environment and white finishes the ko is
score 2 = c + g(Nt/N) + g((N-1)t/N) - g((N-2)t/N) - ...
=> score 2 - score1 = c - d + 2(Nt/N) + 2((N-1)t/N) - 2((N-2)t/N) + …
=> score 2 - score1 = c - d + 2t + 2t/N ((N-1) - (N-2) + (N-3) – (N-4) …)
If N is odd then ((N-1) - (N-2) + (N-3) – (N-4) …) = (N-1)/2
And if N is even then ((N-1) - (N-2) + (N-3) – (N-4) …) = N/2
in any case we have ((N-1) - (N-2) + (N-3) – (N-4) …) ≤ N/2
=> score 2 - score1 ≤ c - d +2t + 2t/N (N/2) = c - d + 3t < c – d + 3m = 0
Eventually t > m => score2 > score1 => black plays in the environment to reach the best score.
2) With the same kind of reasoning you can also prove that
t < m => black must play in the ko to reach the best score
3) and 4) same result with white to play
Finally the “move value” of position A is va = m = (d - c) / 3
OC such proof is quite boring but I am very convinced it works.