Code:
A
/ \
B C(-13)
/ \
D(1) E
/ \
F G(-13)
/ \
H(2) I(-10)
The gote count of F is (2 + (-10)) / 2 = -4. F is a simple gote, so the gote count is its count.
Code:
A
/ \
B C(-13)
/ \
D(1) E(-8.5)
/ \
F(-4) G(-13)
/ \
H(2) I(-10)
The tentative gote count of E is (-4 + (-13)) / 2 = -8.5.
The tentative gote move value of E is 4.5. At the child F, the follow-up gote move value is 6.
The tentative gote move value of E is smaller than the follow-up move value: 4.5 < 6. This condition identifies a local sente.
Therefore, we discard the tentative values of E but determine the sente count of E as the count of the sente follower: -10.
Code:
A
/ \
B C(-13)
/ \
D(1) E(-10)
/ \
F(-4) G(-13)
/ \
H(2) I(-10)
The tentative gote count of B is (1 + (-10)) / 2 = -4.5.
Code:
A
/ \
B(-4.5) C(-13)
/ \
D(1) E(-10)
/ \
F(-4) G(-13)
/ \
H(2) I(-10)
The tentative gote move value of B is 5.5.
E is a local sente with the follow-up sente move value 3.
The tentative gote move value of B is larger than the follow-up sente move value at the child E, that is, 5.5 > 3. This condition identifies a local gote. (Besides, the follow-up gote move value at the child D is 0 for a pass. The tentative gote move value of B is larger than the follow-up gote move value at the child D, that is, 5.5 > 0. This condition also identifies a local gote.)
So B is a local gote and we keep its tentative gote move value as its gote move value.
We can check for traversal: We have B <= F <=> -4.5 <= -4 in a White-Black sequence so traversal applies, E-F may be pruned and we traverse to I. Hence we get the following equal diagram.
Code:
A'
/ \
B(-4.5) C(-13)
/ \
D(1) I(-10)
The line from B to I represents the sequence (move to E) - (move to F) - (move to I).
In the simplified game, the tentative gote count of A' is -8,75.
Code:
A'(-8.75)
/ \
B(-4.5) C(-13)
/ \
D(1) I(-10)
In the simplified game, the tentative gote move value of A' is 4.25. At the child B, the follow-up gote move value is 5.5.
The tentative gote move value of A' is smaller than the follow-up gote move value, that is 4.25 < 5.5. This condition identifies a local sente.
Therefore, we discard the tentative values of A' but determine the sente count of A' as the count of the sente follower I: -10.
Code:
A'(-10)
/ \
B(-4.5) C(-13)
/ \
D(1) I(-10)
The local endgame A' is a local sente with the already calculated sente count -10 and the sente move value 3.
Since A = A', we derive the values for original game A:
Code:
A(-10)
/ \
B(-4.5) C(-13)
/ \
D(1) E(-10)
/ \
F(-4) G(-13)
/ \
H(2) I(-10)
The local endgame A has the sente count -10 and the sente move value 3.
***
Let us go back to the analysis of A before we used traversal to simplify the game:
Code:
A
/ \
B(-4.5) C(-13)
/ \
D(1) E(-10)
/ \
F(-4) G(-13)
/ \
H(2) I(-10)
We have already identified B as a local gote with gote count -4.5 and the gote move value 5.5.
Now, we proceed to analyse A.
The local endgame A has the tentative gote count -8.75.
Code:
A(-8.75)
/ \
B(-4.5) C(-13)
/ \
D(1) E(-10)
/ \
F(-4) G(-13)
/ \
H(2) I(-10)
The tentative gote move value of A is 4.25. At the child B, the follow-up gote move value is 5.5.
The tentative gote move value of A is smaller than the follow-up gote move value, that is 4.25 < 5.5. This condition identifies a local sente.
Therefore, we discard the tentative values of A but determine the sente count of A as the count of the sente follower E: -10.
Code:
A(-10)
/ \
B(-4.5) C(-13)
/ \
D(1) E(-10)
/ \
F(-4) G(-13)
/ \
H(2) I(-10)
The local endgame A is a local sente with the sente count -10 and the sente move value 3.
Locally, without ko threat play, the sequence from A to I is traversed as Black's sente. The sente move value 3 at A and E and the follow-up move values 5.5 at B and 6 at F must be considered for the timing in the context of the global temperature.
***
For the local endgame A, traversal provides an optional technique with which type, count and move value of A can be determined. The type is a local sente. Traversal describes that, once Black has started the local sente sequence at a suitable global temperature, it may as well traverse to I.
Since the count of C equals the count of G, Black stopping at A incurs the same risk as stopping at E that White plays reverse sente. Since the follow-up move value of B is 5.5 and smaller than the follow-up move value 6 at F, if White should not stop at B allowing Black to achieve the count 1 at D, White should even less stop at F allowing Black to achieve the larger count 2 at H.
Ta-ta--ta---taaa!