mitsun wrote:
Here are the graphs for the two positions Bill provided as Go diagrams earlier. The node counts and branch move values are for the gote assumption, with probability 50% for every branch.
Code:
o 5.25 o 2.75? really 3
/ \ / \
/ \ -2.75 / \ gote -1.25,
/ \ / \ really sente
/ \ / \
8 o 2.5 4 o 1.5
/ \ / \
/ \ -2.5 / \ -1.5
/ \ / \
5 0 3 0
Bill demonstrated that the W branch is local gote in the first graph and local sente in the second graph. Is there an easy way to see this from the graphs? I guess the contradiction in the second graph is that the gote count of the starting position (2.75) is less than a terminal position (3) which B can always guarantee reaching, by treating the W move as sente.
Is there an easy way to see this from the graphs? Yes, indeed, as you have actually shown.

Let me repeat the graphs without the calculations.
Code:
o A o C
/ \ / \
/ \ / \
/ \ / \
/ \ / \
8 o B 4 o D
/ \ / \
/ \ / \
/ \ / \
5 0 3 0
When we play the gote strategy for A in the four copies, we get these four results: 2*8 + 5 + 0 = 21. The sente strategy for Black yields these four result: 4*5 = 20. Black does better with the gote strategy, so we can use those results to find its territorial count: 21/4 = 5¼.
Using the graph we would first find the count of B, which is (5+0)/2 = 2½, and then find the count of A, assuming it to be gote, which is (8+2½)/2 = 5¼. OC, we can see at a glance that the count of A, assuming it to be sente, is 5, which is less than 5¼, so we conclude that A is gote. Using the graph gives us more information, and may be quicker.
When we play the gote strategy for C in the four copies, we get these four results: 2*4 + 3 + 0 = 11. The sente strategy for Black yields these four result: 4*3 = 12. Black does better with the sente strategy, so we can use those results to find its territorial count: 3.
Using the graph we would first find the count of D, which is (3+0)/2 = 1½, and then find the count of C, assuming it to be gote, which is (4+1½)/2 = 2¾. OC, we can see at a glance that the count of C, assuming it to be sente, is 3, which is greater than 2¾, so we conclude that C is White sente. Using the graph gives us more information, and may be quicker.

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Suppose we wish to find a general rule to find out whether such a position is sente or gote. Let's relabel the graph with letters.
Code:
o A
/ \
/ \
/ \
/ \
a o B
/ \
/ \
/ \
b c
The result with the gote strategy is
2*a + b + c
The result with the sente strategy is
4b
Taking gote as the default, we get this rule:
Code:
o A
/ \
/ \
/ \
/ \
a o B
/ \
/ \
/ \
b c
A is White sente if and only if 4*b > 2*a + b + c.
Noting that we can subtract b from both sides of the inequality, we get
Code:
o A
/ \
/ \
/ \
/ \
a o B
/ \
/ \
/ \
b c
A is White sente if and only if 3*b > 2*a + c.
Now tell that rule to a regular go player.
Archibald MacLeish wrote:
A poem should be wordless
As the flight of birds. . . .
A poem should be motionless in time
As the moon climbs. . . .
A poem should not mean
But be.
