Gérard TAILLE wrote:your answer
"Without the hane at a the local temperature is the same as the temperature of the whole board" upset me a little. I understood that, in thermography, the local temperature correspond to the temperature of the low part of the mast going vertically to infinity.
$$B
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[go]$$B
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If I try to draw the termograph of this position I will draw the line -5 + 2t starting at the bottom from the point -5 at t= 0 till the point +8 at t = 6½ and then, from that point (+8, 6½) I will draw a vertical mast to intinity.
Such thermograpph is not quite usual but by definition the local temperature seems to me t = 6½ isn't it?
BTW you use the value -5 + t where I use the value -5 + 2t. How do you draw the thermograph Bill?
For the Black wall, let's find out the territorial count,
s, when t > 1.
$$B Black first
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[go]$$B Black first
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$$ | X 2 X O X . .
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,

= t
The final local score is -5.
s = -5 + 2t
Now for the White wall.
$$W White wall
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[go]$$W White wall
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,

= t
The result is the same. Big surprise.
But to find the vertical mast, we have to find out where the two non-vertical walls diverge. That can happen when White is indifferent to Black's threat to kill the corner.
$$B Black first
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[go]$$B Black first
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$$ | . . . , . . .[/go]

,

= t
s = 21 - 2t
The walls meet when
-5 + 2t = 21 - 2t , that is, when
t = 6½ , and
s = 8.
As advertised, because we know that we do not have a 5 move iterated ko where each move gains 5.4 points.
So when t ≥ 6½, the mast rises from (s,t) = (8,6½). When 6½ ≥ t ≥ 1, the mast is inclined along the line, s = -5 + 2t, and when 1 ≥ t ≥ 0, the mast is a vertical line at s = -3.
Let's confirm the value when 1 ≥ t ≥ 0.
$$B Black first
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[go]$$B Black first
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$$ | X 2 X O X . .
$$ | 4 X O O X . .
$$ | X 6 O X X . .
$$ | 5 O O X . . .
$$ | 1 O X X . . .
$$ | 3 X . . . . .
$$ | . X . . . . .
$$ | X X . . . . .
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$$ | . . . , . . .[/go]
Final local score, s = -3.
Ditto when White plays first.
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I probably gave the impression that the thermograph was inclined along the line, s = -5 + t when 6½ ≥ t ≥ 1, because that was the Black wall in that range for the other thermograph. I apologize for that.
But the thing is, the mast of this thermograph is inclined at, say. t = 4. The mast does not determine the local temperature, it's the other way around. The temperature determines the mast, when the two walls coincide. The Black and White walls are the same at every temperature.

So, even though the mast rises vertically at s = 8, that is not the average territorial value at t = 4. That value is -5 + 2t = 3.
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$$B
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[go]$$B
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The thermograph for this position, after White has captured one Black stone has the inclined mast when 6½ ≥ t ≥ 1, along the line, s = -5 + t. At t = 4, s = -1, which is 4 points better for White than s = +3. That is why capturing the stone gains, on average,
t.
