On Extracting Common Random Bits From Correlated Sources on Large Alphabets

Abstract

Suppose Alice and Bob receive strings X=(X <sub xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">1</sub> ,...,X <sub xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">n</sub> ) and Y=(Y <sub xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">1</sub> ,...,Y <sub xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">n</sub> ) each uniformly random in [s] <sup xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">n</sup> , but so that X and Y are correlated. For each symbol i, we have that Y <sub xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">i</sub> =X <sub xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">i</sub> with probability 1-ε and otherwise Y <sub xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">i</sub> is chosen independently and uniformly from [s]. Alice and Bob wish to use their respective strings to extract a uniformly chosen common sequence from [s] <sup xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">k</sup> , but without communicating. How well can they do? The trivial strategy of outputting the first k symbols yields an agreement probability of (1-ε+ε/s) <sup xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">k</sup> . In a recent work by Bogdanov and Mossel, it was shown that in the binary case where s=2 and k=k(ε) is large enough then it is possible to extract k bits with a better agreement probability rate. In particular, it is possible to achieve agreement probability (kε) <sup xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">-1/2</sup> ·2 <sup xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">-kε/(2(1-ε/2))</sup> using a random construction based on Hamming balls, and this is optimal up to lower order terms. In this paper, we consider the same problem over larger alphabet sizes s and we show that the agreement probability rate changes dramatically as the alphabet grows. In particular, we show no strategy can achieve agreement probability better than (1-ε) <sup xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">k</sup> (1+δ(s)) <sup xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink">k</sup> where δ(s)→ 0 as s→∞. We also show that Hamming ball-based constructions have much lower agreement probability rate than the trivial algorithm as s→∞. Our proofs and results are intimately related to subtle properties of hypercontractive inequalities.