[Paper Review] Minimal resolving sets for the hypercube

Ashwin Ganesan|arXiv (Cornell University)|Jun 18, 2011

Graph Labeling and Dimension Problems3 references3 citations

TL;DR

This paper proves that for the n-dimensional hypercube $Q^n$ with $n \geq 5$, a specific set of $n-1$ vertices—namely, all standard basis vectors except one—forms a minimal resolving set, improving upon a previously known resolving set of size $n$ from Erdös and Renyi. The proof uses symmetry and distance vector analysis without exhaustive computation, establishing that proper subsets of the original $n$-vertex resolving set can still resolve all vertices uniquely.

ABSTRACT

For a given undirected graph $G$, an \emph{ordered} subset $S = {s_1,s_2,...,s_k} \subseteq V$ of vertices is a resolving set for the graph if the vertices of the graph are distinguishable by their vector of distances to the vertices in $S$. While a superset of any resolving set is always a resolving set, a proper subset of a resolving set is not necessarily a resolving set, and we are interested in determining resolving sets that are minimal or that are minimum (of minimal cardinality). Let $Q^n$ denote the $n$-dimensional hypercube with vertex set ${0,1}^n$. In Erdös and Renyi (Erdos & Renyi, 1963) it was shown that a particular set of $n$ vertices forms a resolving set for the hypercube. The main purpose of this note is to prove that a proper subset of that set of size $n-1$ is also a resolving set for the hypercube for all $n \ge 5$ and that this proper subset is a minimal resolving set.

Motivation & Objective

• To determine whether a proper subset of the $n$-vertex resolving set for $Q^n$ described by Erdös and Renyi is itself a resolving set.
• To establish minimality of such a subset, showing no proper subset of it can resolve all vertices.
• To provide a non-computational, analytical proof that a minimal resolving set of size $n-1$ exists for $Q^n$ when $n \geq 5$, avoiding brute-force verification.

Proposed method

• Leverages the hypercube's vertex-transitive symmetry: if $S$ is a resolving set, then $S + x$ is also a resolving set for any $x \in V(Q^n)$, as shown in Lemma 1.
• Analyzes distance vectors $d(v, S)$ for vertices $v \in V(Q^n)$ to distinguish all pairs of vertices using only $n-1$ reference points.
• Considers the resolving set $S = \{e_2, e_3, \ldots, e_n\}$, where $e_i$ is the vertex with 1 in position $i$ and 0 elsewhere, and proves it resolves all vertices uniquely.
• Uses case analysis on pairs of vertices based on their intersection with the support of the resolving set elements, particularly focusing on elements in $\{1,2\}$ and $\{3,\ldots,n\}$.
• Applies symmetry to generalize the result to the set $\{e_1, e_2, \ldots, e_{n-1}\}$, which corresponds to the original $n$-vertex set from Erdös and Renyi with one element removed.
• Uses the product structure $Q^n = Q^{n-1} \times K_2$ and known results on resolving sets in graph products to construct resolving sets of size $n-1$ recursively.

Experimental results

Research questions

• RQ1 Can a proper subset of the $n$-vertex resolving set for $Q^n$ described by Erdös and Renyi still resolve all vertices of the hypercube for $n \geq 5$?
• RQ2 Is the resulting subset of size $n-1$ minimal, meaning no proper subset of it is a resolving set?
• RQ3 Does a non-computational, analytical proof exist for the minimality and resolving property of such a set, avoiding exhaustive distance vector checks?

Key findings

• For $n \geq 5$, the set $\{011\ldots1, 1011\ldots1, \ldots, 111\ldots101\}$—a proper subset of size $n-1$ of the $n$-vertex resolving set from Erdös and Renyi—is a resolving set for $Q^n$.
• This $n-1$-sized resolving set is minimal, as no proper subset of it can distinguish all pairs of vertices in $Q^n$.
• The proof avoids exhaustive computation by using symmetry and case analysis on vertex pairs based on their intersection with the resolving set elements.
• The result holds due to the existence of a resolving element $s$ for any pair of distinct vertices $x$ and $y$ such that $d(x,s) \neq d(y,s)$, which is guaranteed by the structure of the hypercube and the choice of $S$.
• The construction generalizes via automorphisms: any permutation of coordinates preserves the resolving property, so all such $n-1$-element sets formed by omitting one standard basis vector are minimal resolving sets.
• The metric dimension of $Q^n$ is strictly less than $n$ for $n \geq 5$, as demonstrated by the existence of a resolving set of size $n-1$.