Twenty (simple) questions

Y. Dagan, Yuval Filmus, Ariel Gabizon, S. Moran
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引用次数: 11

Abstract

A basic combinatorial interpretation of Shannon's entropy function is via the "20 questions" game. This cooperative game is played by two players, Alice and Bob: Alice picks a distribution Π over the numbers {1,…,n}, and announces it to Bob. She then chooses a number x according to Π, and Bob attempts to identify x using as few Yes/No queries as possible, on average. An optimal strategy for the "20 questions" game is given by a Huffman code for Π: Bob's questions reveal the codeword for x bit by bit. This strategy finds x using fewer than H(Π)+1 questions on average. However, the questions asked by Bob could be arbitrary. In this paper, we investigate the following question: *Are there restricted sets of questions that match the performance of Huffman codes, either exactly or approximately? Our first main result shows that for every distribution Π, Bob has a strategy that uses only questions of the form "x < c?" and "x = c?", and uncovers x using at most H(Π)+1 questions on average, matching the performance of Huffman codes in this sense. We also give a natural set of O(rn1/r) questions that achieve a performance of at most H(Π)+r, and show that Ωrn1/r) questions are required to achieve such a guarantee. Our second main result gives a set Q of 1.25n+o(n) questions such that for every distribution Π, Bob can implement an optimal strategy for Π using only questions from Q. We also show that 1.25n-o(n) questions are needed, for infinitely many n. If we allow a small slack of r over the optimal strategy, then roughly (rn)Θ(1/r) questions are necessary and sufficient.
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20个(简单)问题
香农熵函数的一个基本组合解释是通过“20个问题”游戏。这个合作博弈是由两个玩家Alice和Bob进行的:Alice在数字{1,…,n}上选择一个分布Π,并将其宣布给Bob。然后,她根据Π选择一个数字x, Bob尝试平均使用尽可能少的Yes/No查询来识别x。“20个问题”游戏的最佳策略是由Π的霍夫曼代码给出的:鲍勃的问题一点一点地揭示x的码字。这个策略发现x平均使用少于H(Π)+1个问题。然而,Bob提出的问题可能是任意的。在本文中,我们研究了以下问题:*是否存在与霍夫曼码的性能完全匹配或近似匹配的问题集?我们的第一个主要结果表明,对于每个分布Π, Bob的策略只使用形式为“x < c?”和“x = c?”的问题,并且平均最多使用H(Π)+1个问题来发现x,在这个意义上与霍夫曼代码的性能相匹配。我们还给出了O(rn1/r)个问题的自然集,这些问题的性能最多达到H(Π)+r,并表明需要Ωrn1/r)个问题才能达到这样的保证。我们的第二个主要结果给出了一个包含1.25n+o(n)个问题的集合Q,这样对于每个分布Π, Bob可以仅使用Q中的问题来实现Π的最优策略。我们还表明,对于无限多个n,需要1.25n-o(n)个问题。如果我们允许r在最优策略上有一个小的松弛,那么大约(rn)Θ(1/r)个问题是必要和充分的。
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