OK. Given a game, H = {d|e||f|g}, with d > e > f > g, we have a simple test for sente and gote. Let x = (d+e+f+g)/4. If x > e, then the game is a Black sente. If f > x, then the game is a White sente. If e > x > f, then the game is a gote. If d-g >> e-f, the game is unlikely to be gote. That would be threading the needle.
Gérard TAILLE wrote:
For the time being I tested several functions against couples of local endgames according to the kind of tree above.
OK. Let there also be another game, J = {p|q||r|s}, with p > q > r > s. And let there be an ideal environment with temperature t. Since H and J are considered to be double sente, let (d-e), (f-g), (p-q), and (r-s) all be greater than 2t. Thus, each will be played before a play in the environment, and the same goes for H and J.
It doesn't matter which side is to play, the lesson is basically the same. So let Black play first.
1) Let Black play in H first.
1a) Let White reply in H. Then the result after an even number of plays will be
e + q.
1b) Let White reply in J. Then we are left with {d|e} + {r|s}. Depending on which play is larger, our result will be
d + s, or
e + r.
2) Let Black play in J first.
2a) Let White reply in J. The result will be
e + q.
2b) Let White reply in H. Then the result will be either
p + g, or
q + f.
The environment does not play a role in this decision, as the reader may verify. Also, the result will be the same, whichever position Black plays first in, if White replies in it. So we treat the double sente as gote, to decide between them.
The question then becomes that of finding
max(max(d+s,e+r), max(p+g,q+f))
Black plays in H first if
(
d + s ≥ p + g, i.e., if
d - g ≥ p - s,
and
d + s ≥ q + f, i.e.,
d - f ≥ q - s
)
(Back around 50 years ago, when I solved this question, I called d - f the
forward value for Black of playing in H. (I was still in the swing value camp back then.) And I called q - s the
backward value for Black of playing in J.)
or if
(
e - g ≥ p - r
and
e - f ≥ p - q
)
OC, e - f and p - q were traditionally called the value of double sente for each play. Note that choosing which to play based upon that value alone is not correct. However, you could start with that comparison.
Without loss of generality, let e - f ≥ p - q . Then if the backward value for H is greater than or equal to the forward value for J, play in H.