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Method of Multiples
Path: Endgame · Prev: Count · Next: MiaiValue
Difficulty: Intermediate
Keywords: EndGame
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. ;-)) Now let's look at a different kind of play. Example 4.
If Black plays first the score is 5.
If White plays first the score is 4.
One instance does not give us an answer --We did not think it would--, but it tells us that the count lies between 4 and 5.
When Black plays first the score is 9, for an average of 4.5.
When White plays first the score is 8, for an average of 4.
Still no answer, but we now know that the count lies between 4 and 4.5. Let's try 4 copies.
When Black plays first the score is 17, for an average of 4.25.
When White plays first the score is 16, for an average of 4.
Enough! The trend seems clear. If we have N copies, the score when White plays first is 4N, for an average of 4, but the score when Black plays first is 4N + 1, for an average of 4 + 1/N. We are never going to have a miai, no matter how many copies we have. The reason is that this is White's sente. When White plays first she plays with sente in every copy, but when Black plays first he gets one reverse sente, leaving the rest for White to play with sente. (In fact, you can take the fact that there is never a miai, no matter how many copies we have, as a definition of sente.) So what is the count? We have to take 4 as the count. If that is not clear, suppose that we decided to take the average of 4 and 5, or 4.5. Then our count for ten copies would be 45, but we know that the score will be either 40 or 41. 4 is the only count where the error is bounded. We will never be off by more than 1. This is why, when we are evaluating positions, we assume that any sente plays will be played with sente. While it is possible that the reverse sente will be played, we do not, ahem, count on it. (We can also see from this that there is something fishy about the idea of Double Sente. Whose sente do we use for the count?) STILL UNDER CONSTRUCTION Kos next. -- 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. [2]
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. ![]() |