BeginnersEndgameExercise9/Attempts

I will count only territory inside the middle region, letting that be the score (more-positive score therefore being in White's favour). Not sure if this matches the usual conventions.

There are two first moves to consider: a and b. (Anything else by White is obviously selling herself short, and Black cannot jump in as he gets cut off easily.)

For White, a is better because it creates a corridor, while b allows Black to threaten the territory on the next move. That is, after either a or b, White can move to 3, but with b, Black can move to (2 | 0), while with a Black can only move to (2 | (1 | 0)).

For Black, with either a or b, he can move to 0 next with c. With a, White blocks at c, moving to (2 | 1). With b, White can only make a corridor again - i.e., moving to (2 | (1 | 0)). Thus b is better for Black, speaking strictly from a points perspective. Of course, playing b and then c leaves a 3-point ko threat for White. I will ignore this.

Thus: White can move at a to (3 | (2 | (1 | 0))). Black can move at b to ((2 | (1 | 0)) | 0). The game tree is thus ((3 | (2 | (1 | 0))) | ((2 | (1 | 0)) | 0)).

(2 | (1 | 0)) reduces to 1.25 at a temperature of 0.75. Assuming the current temperature is at least 0.75, we get ((3 | 1.25) | (1.25 | 0)). However, in that case, the current score would be constrained to equal 1.25, and the temperature would be 0, with a sente play for each side (basically, a ko threat). This is a contradiction: therefore there is a reversal, and we can't make that simplification. :( How exactly do I proceed from here? An actual algorithm would be appreciated.

Bill: Hi, Karl. :) Since you are counting White as positive, let's draw the game tree you came up with.

                    A
                   / \
                  B   C
                 / \ / \
                3   D   0
                   / \
                  2   E
                     / \
                    1   0

It should be fairly obvious that A = D. OK? :)

Karl: I am still confused about how temperature interacts with the sente-gote evaluation, apparently. Or rather: do we evaluate these trees bottom-up or top-down? Or is there some other way?

Bill: Well, you can always evaluate bottom up, but you have to check for sente and reversals, as you did.

Let's fill in the tree with counts.

                    A (1.25)
                   / \
          (2.125) B   C (0.5)
                 / \ / \
                3   D   0  (D = 1.25)
                   / \
                  2   E (0.5)
                     / \
                    1   0

Note that the counts for A and C are not the averages of the counts of their immediate followers. That is because of reversals.

Working graphically there is no problem, because you just have to watch for intersecting lines. But here is how we might work this out with the tree bottom up.

                    A
                   / \
          (2.125) B   C (0.625)
                 / \ / \
                3   D   0  (D = 1.25)
                   / \
                  2   E (0.5)
                     / \
                    1   0

When we reach this point, taking the arithmetic averages of the immediate followers, we notice that the value for C is greater than that for E. That means that the play from C to D is sente or it reverses. Let's assume first that it reverses.

                    A
                   / \
          (2.125) B   --- C (0.5)
                 / \     / \
                3   D   1   0  (D = 1.25)
                   / \
                  2   E (0.5)
                     / \
                    1   0

Then we can rewrite the tree like this, and now the count for C is the same as that for E. (If it were less, it would be sente, and we would just correct it to be the same. No reversal.)

Continuing:

                    A (1.3125)
                   / \
          (2.125) B   --- C (0.5)
                 / \     / \
                3   D   1   0  (D = 1.25)
                   / \
                  2   E (0.5)
                     / \
                    1   0

Now the value for A is greater than that for D. First, let's assume a reversal, and rewrite the tree like this.

                    A (1.25)
                   / \
                  2   C (0.5)
                     / \
                    1   0

That checks, since the value for A is the same as that for D, and we are done. :)

Working top down, we could have saved ourselves some trouble by noting that A = D, because White's best play followed by Black's best play (in the new position) leads to the same or equivalent position as Black's best play followed by White's best play. :)

Ko threats excepted, this position is equal to the original.

Here we have a difference game with the and stones indicating the competing plays. Does either player have an advantage?

White to play can tie.

But Black to play can win.

The difference game favors Black, so Black's play is dominant.

Now, because of the reversal, it is unlikely to matter. However, if White is looking for a ko threat in the original position, we know which one to play. :)