How I redefine thermography in terms of play in ideal environments
Let there be a game, G
0, with possible subgames, G
1, G
2, .... Let the temperature of G
i be t(G
i). Let there be an environment consisting of an odd number of simple gote of the form {t
j|-t
j}, t
j > 0, for each t
j, such that t
j - t
j+1 = ∆, and such that the net score by alternating play for Black playing first for each t(G
i) = t(G
i)/2, and the net score by alternating play for White playing first for each t(G
i) = -t(G
i)/2; and furthermore that there are sufficient gote in the environment at each t(G
i) for all ko and superko fights at that temperature to be resolved at that temperature.
I hope that is clear and correct.
Edit: I see that I need to say that there needs to be an ideal environment for each game or subgame, G
i,
with t(Gi) > 0, and to specify that for each t
j, all games in the environment of the form, {t
j|-t
j}, are played before continuing to t
j+1.
The problem with the net final score of G
0 plus an ideal environment, E, is that it does not tell us the mast value of G
0. But the Lord giveth and the Lord taketh away. To get the walls of the thermograph of G
0 for each player we take the net results of G
0 + E and subtract the net results of E.
Let me illustrate with G
0 = {a||0|-2b}, a,b > 0.
Black first plays to
a. The net score with alternating play is s = a - t(G
0)/2. To find the Black scaffold we subtract t(G
0)/2 from that to get
s = a - t(G
0).
Next, White first plays to {0|-2b}.
1) Black replies to 0. The net score with alternating play is s = -t(G
0)/2. To find the White scaffold we subtract -t(G
0)/2 from that to get
s = 0.
The two scaffolds intersect at (s,t) = (0,a).
Because {0|-2b} is a simple gote, we know that t({0|-2b}) = b. If a < b then {a||0|-2b} is a White sente, and we are done. The mast rises vertically from (0,a). We may color it red up to t = b.
But what if a > b? Then
2) Black replies in the environment. Since {0|-2b} is a simple gote, we may add it to the simple gote of the form, {t
j|-t
j}, with t
j = b, without affecting the final score in E with alternating play. {0|-2b} + {b|-b} = -b. So Black plays to -b, and the final score with alternating play is s = -b + t(G
0)/2. To find the White scaffold we subtract -t(G
0)/2 from that to get
s = -b + t(G
0).
The two scaffolds intersect at (s,t) = ((a-b)/2,(a+b)/2).
G
0 is a gote, and the mast rises vertically from ((a-b)/2,(a+b)/2).
