On Mappings on the Hypercube with Small Average Stretch

May 27, 2019 · The Ethereal · 🏛 Combinatorics, probability & computing

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Authors Lucas Boczkowski, Igor Shinkar arXiv ID 1905.11350 Category math.CO: Combinatorics Cross-listed cs.DS Citations 1 Venue Combinatorics, probability & computing Last Checked 3 months ago
Abstract
Let $A \subseteq \{0,1\}^n$ be a set of size $2^{n-1}$, and let $φ\colon \{0,1\}^{n-1} \to A$ be a bijection. We define the average stretch of $φ$ as ${\sf avgStretch}(φ) = {\mathbb E}[{\sf dist}(φ(x),φ(x'))]$, where the expectation is taken over uniformly random $x,x' \in \{0,1\}^{n-1}$ that differ in exactly one coordinate. In this paper we continue the line of research studying mappings on the discrete hypercube with small average stretch. We prove the following results. (1) For any set $A \subseteq \{0,1\}^n$ of density $1/2$ there exists a bijection $φ_A \colon \{0,1\}^{n-1} \to A$ such that ${\sf avgstretch}(φ_A) = O(\sqrt{n})$. (2) For $n = 3^k$ let $A_{{\sf rec\text{-}maj}} = \{x \in \{0,1\}^n : {\sf rec\text{-}maj}(x) = 1\}$, where ${\sf rec\text{-}maj} : \{0,1\}^n \to \{0,1\}$ is the function recursive majority of 3's. There exists a bijection $φ_{{\sf rec\text{-}maj}} \colon \{0,1\}^{n-1} \to A_{\sf rec\text{-}maj}$ such that ${\sf avgstretch}(φ_{\sf rec\text{-}maj}) = O(1)$. (3) Let $A_{\sf tribes} = \{x \in \{0,1\}^n : {\sf tribes}(x) = 1\}$. There exists a bijection $φ_{\sf tribes} \colon \{0,1\}^{n-1} \to A_{\sf tribes}$ such that ${\sf avgstretch}(φ_{\sf tribes}) = O(\log(n))$. These results answer the questions raised by Benjamini et al.\ (FOCS 2014).
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