Sandwiches missing two ingredients of order four

Abstract

For a set $\mathcal{F}$ of graphs, an instance of the $\mathcal{F}$-free Sandwich Problem is a pair $(G_1,G_2)$ consisting of two graphs $G_1$ and $G_2$ with the same vertex set such that $G_1$ is a subgraph of $G_2$, and the task is to determine an $\mathcal{F}$-free graph G containing $G_1$ and contained in $G_2$, or to decide that such a graph does not exist. Initially motivated by the graph sandwich problem for trivially perfect graphs, which are the $\{ P_4,C_4\}$-free graphs, we study the complexity of the $\mathcal{F}$-free Sandwich Problem for sets $\mathcal{F}$ containing two non-isomorphic graphs of order four. We show that if $\mathcal{F}$ is one of the sets $\left\{ \mathrm{diamond},K_4\right\}$, $\left\{ \mathrm{diamond},C_4\right\}$, $\left\{ \mathrm{diamond},\mathrm{paw}\right\}$, $\left\{ K_4,\overline{K_4}\right\}$, $\left\{ P_4,C_4\right\}$, $\left\{ P_4,\overline{\mathrm{claw}}\right\}$, $\left\{ P_4,\overline{\mathrm{paw}}\right\}$, $\left\{ P_4,\overline{\mathrm{diamond}}\right\}$, $\left\{ \mathrm{paw},C_4\right\}$, $\left\{ \mathrm{paw},\mathrm{claw}\right\}$, $\left\{ \mathrm{paw},\overline{\mathrm{claw}}\right\}$, $\left\{ \mathrm{paw},\overline{\mathrm{paw}}\right\}$, $\left\{ C_4,\overline{C_4}\right\}$, $\left\{ \mathrm{claw},\overline{\mathrm{claw}}\right\}$, and $\left\{ \mathrm{claw},\overline{C_4}\right\}$, then the $\mathcal{F}$-free Sandwich Problem can be solved in polynomial time, and, if $\mathcal{F}$ is one of the sets $\left\{ C_4,K_4\right\}$, $\left\{ \mathrm{paw},K_4\right\}$, $\left\{ \mathrm{paw},\overline{K_4}\right\}$, $\left\{ \mathrm{paw},\overline{C_4}\right\}$, $\left\{ \mathrm{diamond},\overline{C_4}\right\}$, $\left\{ \mathrm{paw},\overline{\mathrm{diamond}}\right\}$, and $\left\{ \mathrm{diamond},\overline{\mathrm{diamond}}\right\}$, then the decision version of the $\mathcal{F}$-free Sandwich Problem is NP-complete.