[Paper Review] Equivalence Classes of Full-Dimensional 0/1-Polytopes with Many Vertices

William Y.C. Chen, Peter L. Guo|arXiv (Cornell University)|Jan 2, 2011

graph theory and CDMA systems10 references4 citations

TL;DR

This paper presents a computational method to enumerate equivalence classes of full-dimensional 0/1-polytopes in the 6-dimensional hypercube $Q_6$ with more than 12 vertices, leveraging symmetry reductions via the hyperoctahedral group and cycle index techniques. The key contribution is the complete enumeration of $F_6(k)$ for $k = 13$ to $16$, yielding exact counts: 290,159,817, 1,051,410,747, 3,491,461,629, and 10,665,920,350, respectively.

ABSTRACT

Let $Q_n$ denote the $n$-dimensional hypercube with the vertex set $V_n=\{0,1}^n$. A 0/1-polytope of $Q_n$ is a convex hull of a subset of $V_n$. This paper is concerned with the enumeration of equivalence classes of full-dimensional 0/1-polytopes under the symmetries of the hypercube. With the aid of a computer program, Aichholzer completed the enumeration of equivalence classes of full-dimensional 0/1-polytopes for $Q_4$, $Q_5$, and those of $Q_6$ up to 12 vertices. In this paper, we present a method to compute the number of equivalence classes of full-dimensional 0/1-polytopes of $Q_n$ with more than $2^{n-3}$ vertices. As an application, we finish the counting of equivalence classes of full-dimensional 0/1-polytopes of $Q_6$ with more than 12 vertices.

Motivation & Objective

• To resolve the open problem of counting full-dimensional 0/1-polytope equivalence classes in $Q_6$ beyond 12 vertices.
• To develop a systematic method for computing $F_n(k)$, the number of equivalence classes of full-dimensional 0/1-polytopes in $Q_n$ with $k$ vertices, for $k > 2^{n-3}$.
• To extend prior computational results by Aichholzer, who enumerated classes up to 12 vertices in $Q_6$, by completing the count for larger $k$.
• To utilize the hyperoctahedral group's action and Pólya theory to reduce the computational complexity of equivalence class enumeration.

Proposed method

• Use the cycle index of the hyperoctahedral group $B_n$ to compute $A_n(k)$, the total number of 0/1-equivalence classes with $k$ vertices.
• Apply inclusion-exclusion to isolate $H_n(k)$, the number of non-full-dimensional classes, by analyzing intersections of $Q_n$ with its spanned hyperplanes.
• Reduce the computation of $H_n(k)$ for $2^{n-3} < k ≤ 2^{n-2}$ to pairwise intersections of hyperplanes, exploiting dimensionality constraints.
• Leverage symmetry to identify representative configurations via $w$-transforms, reducing the number of distinct cases to evaluate.
• Use generating functions $C_H(u_1, u_2)$ to count vertex colorings on hyperplane intersections, with coefficients extracted via series expansion.
• Apply the formula $F_n(k) = A_n(k) - H_n(k)$ to derive the final count of full-dimensional equivalence classes.

Experimental results

Research questions

• RQ1 What is the number of equivalence classes of full-dimensional 0/1-polytopes in $Q_6$ with 13 vertices?
• RQ2 How many full-dimensional 0/1-polytope equivalence classes exist in $Q_6$ with 14 vertices?
• RQ3 What is the exact count of full-dimensional 0/1-polytope equivalence classes in $Q_6$ with 15 vertices?
• RQ4 What is the number of full-dimensional 0/1-polytope equivalence classes in $Q_6$ with 16 vertices?
• RQ5 Can the enumeration of full-dimensional 0/1-polytope equivalence classes in $Q_6$ be completed beyond the 12-vertex threshold?

Key findings

• The number of full-dimensional 0/1-polytope equivalence classes in $Q_6$ with 13 vertices is 290,159,817.
• For 14 vertices, the count is 1,051,410,747 equivalence classes.
• The number of equivalence classes for 15 vertices is 3,491,461,629.
• For 16 vertices, the count reaches 10,665,920,350 full-dimensional 0/1-polytope equivalence classes.
• The method successfully completes the enumeration of full-dimensional 0/1-polytope equivalence classes in $Q_6$ for all $k > 12$, resolving a long-standing open problem.