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On Extracting Common Random Bits From Correlated Sources on Large Alphabets

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3 Author(s)
Siu On Chan ; Microsoft Res., Cambridge, MA, USA ; Elchanan Mossel ; Joe Neeman

Suppose Alice and Bob receive strings X=(X1,...,Xn) and Y=(Y1,...,Yn) each uniformly random in [s]n, but so that X and Y are correlated. For each symbol i, we have that Yi=Xi with probability 1-ε and otherwise Yi 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]k, 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)k. 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)-1/2·2-kε/(2(1-ε/2)) 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)k(1+δ(s))k 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.

Published in:

IEEE Transactions on Information Theory  (Volume:60 ,  Issue: 3 )