Table of Contents
Fetching ...

On hypercube statistics

Noga Alon, Maria Axenovich, John Goldwasser

Abstract

Let $d \geq 1$ and $s \leq 2^d$ be nonnegative integers. For a subset $A$ of vertices of the hypercube $Q_n$ and $n\geq d$, let $λ(n,d,s,A)$ denote the fraction of subcubes $Q_d$ of $Q_n$ that contain exactly $s$ vertices of $A$. Let $λ(n,d,s)$ denote the maximum possible value of $λ(n,d,s,A)$ as $A$ ranges over all subsets of vertices of $Q_n$, and let $λ(d,s)$ denote the limit of this quantity as $n$ tends to infinity. We prove several lower and upper bounds on $λ(d,s)$, showing that for all admissible values of $d$ and $s$ it is larger than $0.28$. We also show that the values of $s=s(d)$ such that $λ(d,s)=1$ are exactly $\{0,2^{d-1},2^d\}$. In addition we prove that if $0<s< d/8$, then $λ(d, s) \leq 1 - Ω(1/s)$, and that if $s$ is divisible by a power of $2$ which is $Ω(s)$ then $λ(d,s) \geq 1-O(1/s)$. We suspect that $λ(d,1)=(1+o(1))/e$ where the $o(1)$-term tends to $0$ as $d$ tends to infinity, but this remains open, as does the problem of obtaining tight bounds for essentially all other quantities $λ(d,s)$.

On hypercube statistics

Abstract

Let and be nonnegative integers. For a subset of vertices of the hypercube and , let denote the fraction of subcubes of that contain exactly vertices of . Let denote the maximum possible value of as ranges over all subsets of vertices of , and let denote the limit of this quantity as tends to infinity. We prove several lower and upper bounds on , showing that for all admissible values of and it is larger than . We also show that the values of such that are exactly . In addition we prove that if , then , and that if is divisible by a power of which is then . We suspect that where the -term tends to as tends to infinity, but this remains open, as does the problem of obtaining tight bounds for essentially all other quantities .

Paper Structure

This paper contains 5 sections, 5 theorems, 8 equations.

Key Result

Theorem 1

Let $s$ and $d$ be integers. Then $\lambda(d,s)=1$ if and only if $s\in \{0, 2^d, 2^{d-1}\}$. If $1< s <2^{d-1}$, then In particular, $\lambda(d+2, d,s)=1$ iff $d+2\leq \omega(s)$. When $s=1$, we have $\lambda(d,1) \leq \lambda(d+2, d, 1) = \pi(d+2, 3)$ for $d< 6$, and $\lambda(d+2, d,1)= 3/4$ otherwise.

Theorems & Definitions (10)

  • Theorem 1
  • Theorem 2
  • proof : Proof of Theorem \ref{['ub']}
  • proof : Proof of Theorem \ref{['lb']}
  • Proposition 3
  • proof
  • Theorem 4
  • proof
  • Theorem 5
  • proof