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On Slepian-Wolf Theorem with Interaction

机译:论互动的睡莲定理

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In this paper we study interactive "one-shot" analogues of the classical Slepian-Wolf theorem. Alice receives a value of a random variable X, Bob receives a value of another random variable Y that is jointly distributed with X. Alice's goal is to transmit X to Bob (with some error probability ε). Instead of one-way transmission we allow them to interact. They may also use shared randomness. We show, that for every natural r Alice can transmit X to Bob using (1+1/r) H(X|Y) + r + O(log_2 (1/ε)) bits on average in 2H(X|Y)/r+2 rounds on average. Setting r = [{the square root of}(H(X|Y))] and using a result of [2] we conclude that every one-round protocol π with information complexity I can be compressed to a (many-round) protocol with expected communication about I + 2{the square root of}I bits. This improves a result by Braverman and Rao [3], where they had I + 5{the square root of}I. Further, we show (by setting r = [H(X|Y)]) how to solve this problem (transmitting X) using 2H(X|Y)+O(log_2(1/ε)) bits and 4 rounds on average. This improves a result of [4], where they had 4H(X|Y) + O(log 1/ε) bits and 10 rounds on average. In the end of the paper we discuss how many bits Alice and Bob may need to communicate on average besides H(X|Y). The main question is whether the upper bounds mentioned above are tight. We provide an example of (X, Y), such that transmission of X from Alice to Bob with error probability ε requires H(X|Y)+ Ω (log_2 (1/ε)) bits on average.
机译:在本文中,我们研究了古典思维狼定理的互动“单次”模数。 Alice接收一个随机变量x的值,Bob接收另一个随机变量y的值,该值与x共同分布.Alice的目标是将X传输到Bob(具有一些错误概率ε)。我们允许他们互动的单向传输。他们也可以使用共享的随机性。我们展示,对于每个天然的R Alice,可以使用(1 + 1 / r)H(x | y)+ r + o(log_2(1 /ε))位平均在2h(x ​​| y)中传输x到鲍勃。 / r + 2轮平均。设置r = [}(h(x | y))]和使用[2]的结果,我们得出结论,每个单次协议π具有信息复杂性的,我可以压缩到(多轮)关于I + 2的预期通信的协议{I位的平方根。这通过Braverman和Rao [3]改善了结果,在那里他们拥有I + 5 {Square Root。此外,我们显示(通过设置r = [h(x | y)])如何使用2h(x ​​| y)+ o(log_2(1 /ε))位和4轮平均地解决这个问题(发送x) 。这改善了[4]的结果,其中它们具有4h(x | y)+ O(log 1 /ε)位,平均10轮。在本文的最后,我们讨论了除H(x | y)之外,鲍勃可能需要平均通信的比特和鲍勃。主要问题是上面提到的上限是紧张的。我们提供了一个(x,y)的示例,使得通过误差概率ε从Alice传输X到鲍勃,因此平均需要H(x | y)+ω(log_2(1 /ε))比特。

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