Forward values and backward values
Actually, I think I was on to something all those years ago.
I was still in the swing value camp, where I took the value of a play in this simple gote, {a|b}, given a > b, as a - b. I also was aware that the best play was not always the largest play as traditionally calculated. I also became aware of the concept of the environment, which, as we know, I modeled as a set of simple gote. I took the value of the largest such gote as my standard for evaluation.
From that perspective, how to evaluate {d|e||f|g}, with d > e > f > g? Well we compare it with a simple gote, {a|b}, given a > b, in an environment of temperature, t, to modernize my thinking slightly. It does not affect the argument. OC, a-b > 2t, but also d-e > 2t and f-g > 2t. Otherwise it is not an environment. And it doesn't hurt to assign a number to a couple of variables. So let's evaluate D = {d|e||0|-g}, given g > 0, with A = {a|0}, with Black to play.
1) Black starts in A. The result is
a + 0 - t/2 = a - t/2.
2) Black starts in D, with sente. The result is
e + a - t/2, which is better for Black than 1), OC. But White does not have to answer in D. Let White play in A.
3) Black starts in D, White answers in A. The result is
d + 0 - t/2 = d - t/2.
So we compare a with d. a is the swing value of A, our standard of comparison. And d is the
forward value of D.
OC, if d - e > a, then d > a, so the case of sente is covered. And for the case of gote, the forward value of D is easier to calculate than {d + e + f + g}/4. In fact, we do not have to worry about how to classify D.
In hindsight, we can see that I was actually anticipating the calculation of the sides of the thermograph at different temperatures in terms of the play in an ideal environment.
Which is how I redefined thermography in 1998.
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Whether to answer a double sente, given two of them, of the form {j|k||m|n}.
Let A = {a|0||-b|-c} and D = {d|0||-e|-f}, and let Black play in A to {a|0}. Should White answer?
Well, we already know the answer. We compare a with f, the swing value of Black's threat with the backward value of D (from Black's point of view).
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OC, I haven't worked anything out in detail, at least not recently, but let's explore Gérard's hypothetical with several such plays where the temperature of the environment is low enough.
First, let's compare threats and find the largest one. Next, compare its swing value with the backward values of each of the others. If it is greater than all of them, it's a good bet that Black should make that threat.
And, as things get more complicated, let's not forget that traditional evaluation arose for a reason. The more double sente there are, the more likely it is that they act as a rich environment for each other, and the traditional evaluation becomes more relevant. And by traditional I do not mean the traditional double sente value, since with even two of them we cannot assume that one of them will be sente against the other, much less double sente.
Edit: Difference games.
Also, as things get more complicated, since difference games allow us to draw conclusions about play in every non-ko environment, they can be useful. OC, working them out on the table can be daunting, but I have done some of the work for us.
Let A = {a|0||-b|-c} and D = {d|0||-e|-f}, and let Black play in A to {a|0}. Should White answer?
Well, when there are only two of them with a sufficiently low ambient temperature, White should answer if a is at least as big as the backward value of D (for Black, that is). But what about non-ko environments? When is answering in A at least as good as playing in D, in every non-ko environment? When a is greater than or equal to both the backward value of D and the forward value of D.
Vive la difference game!
_________________
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At some point, doesn't thinking have to go on?— Winona Adkins
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My mother and my wife.
Everything with love. Stay safe.