Self-referential instances of the dominating set problem are irreducible

TL;DR

The paper analyzes the domination number in Erdős-Rényi graphs $G(n,p)$ and proves an irreducibility result: for carefully chosen $p(n)$ and target $k=\ln n$, no induced subgraph of size at most $n^c$ (0< c<1) suffices to decide whether a dominating set of size $k$ exists. It establishes this via two probabilistic regimes: with positive probability there is a unique DS of size $k$, and with high probability there exist near-dominating $k$-sets when no DS exists; both rely on first and second moment calculations for appropriate $p(n)$. The irreducibility is then shown by a symmetry mapping that, using only vertices outside any given $H$ with $|V(H)|\le n^c$, can flip the global existence of a DS without altering $H$, thereby preventing sublinear subgraphs from solving the problem. The results suggest that the extreme hardness of DS in random graphs stems from self-referential and near-independence structure of the solution space, rather than local graph structure, with broader implications for understanding computational hardness in combinatorial problems.

Abstract

We study the algorithmic decidability of the domination number in the Erdos-Renyi random graph model $G(n,p)$. We show that for a carefully chosen edge probability $p=p(n)$, the domination problem exhibits a strong irreducible property. Specifically, for any constant $0<c<1$, no algorithm that inspects only an induced subgraph of order at most $n^c$ can determine whether $G(n,p)$ contains a dominating set of size $k=\ln n$. We demonstrate that the existence of such a dominating set can be flipped by a local symmetry mapping that alters only a constant number of edges, thereby producing indistinguishable random graph instances which require exhaustive search. These results demonstrate that the extreme hardness of the dominating set problem in random graphs cannot be attributed to local structure, but instead arises from the self-referential nature and near-independence structure of the entire solution space.

Figures (1)

• Figure 1: A symmetry mapping between two classes of instances. (a) → (b) Initially, the vertex $v$ has exactly one neighbor $u\in S$. The symmetry mapping transforms $G$ which has a unique dominating set of size $k$ into a graph with no dominating set of size $k$. (b) → (a) Initially, $v$ is the only vertex not dominated by $S$. The symmetry mapping transforms $G$ which has no dominating set of size $k$ into a graph with a unique dominating set of size $k$.

Theorems & Definitions (8)

• Definition 2.1
• Theorem 2.1
• Lemma 2.2
• proof
• Corollary 2.3
• proof
• Lemma 2.4
• proof