The sandwich problem for odd-hole-free and even-hole-free graphs

TL;DR

This paper proves that the $\mathcal{P}$-Sandwich-Problem is NP-hard when $\mathcal{P}$ is the class of odd-hole-free or even-hole-free graphs. It builds on and adapts existing reductions, notably from $C_5$-free and 3-SAT constructions, using gadgetry based on five-cycles, six-cycles, and augmented chordal structures to encode truth assignments. For the odd-hole-free case, the reduction yields that any sandwich graph must avoid odd holes and certain antiholes, and, via a complement argument, establishes NP-hardness of the corresponding problem; for the even-hole-free case, the construction adds auxiliary vertices to enforce even-hole absence and shows a corresponding correspondence with satisfiability. The results advance understanding of Sandwich-Problem complexity for perfect-graph related classes and raise open questions about the relative hardness of Not-$\mathcal{C}$-Free vs $\mathcal{C}$-Free variants and related configurations.

Abstract

For a property $\mathcal{P}$ of graphs, the $\mathcal{P}$-\textsc{Sandwich-Problem}, introduced by Golumbic and Shamir (1993), is the following: Given a pair of graphs $(G_1, G_2)$ on the same vertex set $V$, does there exist a graph $G$ such that $V(G)=V$, $E(G_{1})\subseteq E(G) \subseteq E(G_{2})$, and $G$ satisfies $\mathcal{P}$? A {\em hole} in a graph is an induced subgraph which is a cycle of length at least four. An odd (respectively even) hole is a hole of odd (respectively even) length. Given a class of graphs $\mathcal{C}$ and a graph $G$ we say that $G$ is {\em $\mathcal{C}$-free} if it contains no induced subgraph isomorphic to a member of $\mathcal{C}$. In this paper we prove that if $\mathcal{P}$ is the property of being odd-hole-free or the property of being even-hole-free, then the $\mathcal{P}$-\textsc{Sandwich-Problem} is NP-hard.

Figures (5)

• Figure 1: This figure illustrates the construction that we discuss in the proof of \ref{['thm:odd']}. In the case which is illustrated the clause contains the literal $\overline{x}_{1}$. Edges represented with dashed line segments are optional edges, that is, edges of $E_{2} \setminus E_1$; the solid edges are forced, that is, edges of $E_{1}$. The omitted edges are forbidden, that is, edges of $E_{3}$.
• Figure 2: On the left-hand side the graph $P_{6}^{c}$ and on the right-hand side the graph $\{P_{4}+P_{1}\}^{c}$.
• Figure 3: A gadget which corresponds to the clause $(X+Y+Z)$ in the construction from bodlaender. Solid edges are forced, dashed edges are forbidden, and omitted edges are optional.
• Figure 4: Adding a four-vertex path in the construction from bodlaender. Both the vertices $W_{1}$ and $W_{2}$ have degree two in $G_{2}'$.
• Figure 5: If $i = (X + Y + Z)$ is a clause with three negative literals and $G$ the corresponding sandwich graph, then the four-cycle $\{H, K_{\overline{Z}}^i,F,K_{\overline{X}}^i\}$ which is formed by the optional edges which are assumed to be present in $G$ implies that the optional edge $K_{\overline{Z}}^iK_{\overline{X}}^i$ must be present in $G$ as well. Solid edges are forced, dashed edges are forbidden, and dashed-dotted edges are the optional edges known so far to be present in $G$.

Theorems & Definitions (6)

• Lemma 1.1
• proof : Proof of \ref{['lem:complements']}
• Theorem 2.1
• proof : Proof of \ref{['thm:odd']}
• Theorem 3.1
• proof : Proof of \ref{['thm:even']}