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Method of Multiples
Path: Endgame · Prev: Count · Next: MiaiValue
Difficulty: Beginner
Keywords: EndGame
Under Construction The method of multiples, while seldom a practical way of determining the count of a position, provides a good explanation of it, I think. :-) If there are no kos or future kos in a position, and some combination of multiples of that position has a definite score, the count of that position is that score divided by the number of multiples. (The combination has a definite score if the net score after correct play is the same, regardless of who plays first. IOW, the combination is miai.) Example 1:
(Per convention, stones next to empty space are immortal.)
The score of this position is +4 (for Black). If White plays first the result is the same (e. g.,
Note that Example 2:
Regardless of who plays first, the score is +5. Since there are 4 copies, each has a count of 1 1/4.[1][2] Note that each move is gote. Also that the number of copies is a power of 2. Example 3:
The final result is +11.
The score is +11 regardless of who plays first.
There are 8 copies, so each one has a count of 1 3/8.
Note that the first play is gote, that the number of copies is a power of 2, and that the second player plays last. (You can also see what I meant about being impractical. ;-)) -- Bill Spight [1] Note on fractional counts: Some people object to fractional counts because the final scores of go games are integers. There are two answers to that. First is that, if the game is not over, a fractional count can serve as an estimate of the local score. The second is that, as with these examples, you can have a whole board position that is miai, and therefore with a definite score, that includes regions with fractional counts.
Hyppy: Just throwing a twig in the fan here on the example above.
Bill: Yes, to use the method of multiples you have to find correct play. :-) Path: Endgame · Prev: Count · Next: MiaiValue This is a copy of the living page "Method of Multiples" at Sensei's Library. ![]() |