45 lines
2.5 KiB
Markdown
45 lines
2.5 KiB
Markdown
# Spec
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## Best first guess
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### All different
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[C1, D2, E3]: 4.27
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[A1, D2, G3]: 4.25
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### Same notes
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[D1, D2, D3]: 4.80
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### Same octaves
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[C2, D2, E2]: 4.34
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## Tips to improve `nextGuess`
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The best results can be had by carefully choosing each guess so that it is most
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likely to leave a small remaining list of possible targets. You can do this by
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computing for each remaining possible target the maximum number of possible
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targets it will leave if you guess it. This you can do by computing, for each
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remaining possible target, the answer you will receive if it is the actual
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target, and then compute how many of the remaining possible targets would yield
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the same output, and take the maximum of all of these. Alternatively, you can
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take a more probabilistic approach, and compute the average number of possible
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targets that will remain after each guess, giving the expected number of
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remaining possible targets for each guess, and choose the guess with the
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smallest expected number of remaining possible targets.
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Unfortunately, this is much more expensive to compute, and you will need to be
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careful to make it efficient enough to use. One thing you can do to speed it up
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is to laboriously (somehow) find the best first guess and hard code that into
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your program. After the first guess, there are much fewer possible targets
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remaining, and your implementation may be fast enough then.
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You can also remove symmetry in the problem space. The key insight needed for
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this is that given any guess and an answer returned for it, the set of
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remaining possibilities after receiving that answer for that guess will be the
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same regardless of which target yielded that answer. This suggests collecting
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all the distinct answers for a given guess and for each answer, counting the
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number of targets that would give that answer. Since there are much fewer
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answers than possible targets, this can save significant work.
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For example, suppose there are ten remaining candidate targets, and one guess
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gives the answer (3,0,0), three others give (1,0,2), and the remaining six give
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the answer (2,0,1). In this case, if you make that guess, there is a 1 in 10
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chance of that being the right answer (so you are left with that as the only
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remaining candidate), 3 in 10 of being left with three candidates, and a 6 in
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10 chance of being left with six candidates. This means on average you would
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expect this answer to leave you with 1/10 * 1 + 3/10 * 3 + 6/10 * 6 = 4.6
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remaining candidates. You just need to select a guess that gives the minimum
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expected number of remaining candidates.
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