An improved bound on the zero-error list-decoding capacity of the 4/3 channel

Abstract

We prove a new, improved upper bound on the size of codes C ⊆ {1, 2, 3, 4} n with the property that every four distinct codewords in C have a coordinate where they all differ. Specifically, we show that such a code has size at most 2 6n/19+o(n) , or equivalently has rate bounded by 6/19 ≤ 0.3158 (measured in bits). This improves the previous best upper bound of 0.3512 due to (Arikan 1994), which in turn improved the 0.375 bound that followed from general bounds for perfect hashing due to (Fredman and Komlós, 1984) and (Körner and Marton, 1988). The context for this problem is two-fold: zero-error list decoding capacity, where such codes give a way to communicate with no error on the "4/3 channel" when list-of-3 decoding is employed, and perfect hashing, where such codes give a perfect hash family of size n mapping C to {1, 2, 3, 4}.

On Colorful Edge Triples in Edge-Colored Complete Graphs

“…We remark that the upper bound is implicit in Elias [11], and that for c 4;4 some improvements were found recently by Dalai, Guruswami, and Radhakrishnan [10]. From now on we are focusing here only on the b ¼ k ¼ 3 case.…”

Beating the probabilistic lower bound on perfect hashing

“…q q−1 for all q 2. Arikan [2] improved this bound for q = 4, and then Dalai, Guruswami and Radhakrishnan [7] further improved this upper bound. Recently, Guruswami and Riazanov [11] discovered a stronger upper bound for every q ≥ 4.…”

“…, q − 1}, and (v, w) ∈ E if and only if v = w. If we want to ensure that the receiver can identify a subset of at most q − 1 sequences that is guaranteed to contain the transmitted sequence, one can communicate via n repeated uses of the channel using the perfect q-hash code. See [9,7] for more details.…”

On the trifference problem for linear codes

“…By the pigeonhole principle there is a hyperplaneH 2 ⊂ F d 3 such that H 2 ∩ H 0 = H 1 and |H 2 ∩ (X \ Y )| ≥ |X \ Y |/3. We conclude that m ≥ |H 2 ∩ X| = |H 2 ∩ Y | + |H 2 ∩ (X \ Y )| ≥ |Y | + 2n − |Y | 3 ≥ m ′ + 2n − m 3 = 2n + 3m ′ − mNow we study(5). After plugging in n ′ = (m − 1)/2 and removing floors we get m/c3 m/2−d ,…”

“…Over the years, there have been multiple interesting refinements of the Fredman-Komlós method which led to various slight improvements of the inequality from (2) for k ≥ 4. See for example [5] and [4] for some recent progress on the cases k = 4 and k ≥ 5, respectively. We also encourage the interested reader to consult [7], which is the paper that [4] builds upon and which also contains a wonderful comprehensive discussion of the history of the problem and of the Fredman-Komlós approach.…”