Sandwiches Missing Two Ingredients of Order Four

For a set ${\cal F}$ of graphs, an instance of the ${\cal F}$-{\sc 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 ${\cal F}$-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 ${\cal F}$-{\sc free Sandwich Problem} for sets ${\cal F}$ containing two non-isomorphic graphs of order four. We show that if ${\cal F}$ is one of the sets $\left{ {\rm diamond},K_4\right}$, $\left{ {\rm diamond},C_4\right}$, $\left{ {\rm diamond},{\rm paw}\right}$, $\left{ K_4,\overline{K_4}\right}$, $\left{ P_4,C_4\right}$, $\left{ P_4,\overline{\rm claw}\right}$, $\left{ P_4,\overline{\rm paw}\right}$, $\left{ P_4,\overline{\rm diamond}\right}$, $\left{ {\rm paw},C_4\right}$, $\left{ {\rm paw},{\rm claw}\right}$, $\left{ {\rm paw},\overline{{\rm claw}}\right}$, $\left{ {\rm paw},\overline{\rm paw}\right}$, $\left{ C_4,\overline{C_4}\right}$, $\left{ {\rm claw},\overline{{\rm claw}}\right}$, and $\left{ {\rm claw},\overline{C_4}\right}$, then the ${\cal F}$-{\sc free Sandwich Problem} can be solved in polynomial time, and, if ${\cal F}$ is one of the sets $\left{ C_4,K_4\right}$, $\left{ {\rm paw},K_4\right}$, $\left{ {\rm paw},\overline{K_4}\right}$, $\left{ {\rm paw},\overline{C_4}\right}$, $\left{ {\rm diamond},\overline{C_4}\right}$, $\left{ {\rm paw},\overline{\rm diamond}\right}$, and $\left{ {\rm diamond},\overline{\rm diamond}\right}$, then the decision version of the ${\cal F}$-{\sc free Sandwich Problem} is NP-complete.

💡 Research Summary

The paper investigates the graph sandwich problem for families F consisting of two non‑isomorphic graphs on four vertices. An instance of the F‑free Sandwich Problem is a pair (G₁, G₂) with the same vertex set, G₁ ⊆ G₂, and the task is to find a graph G such that G₁ ⊆ G ⊆ G₂ and G contains no induced subgraph from F, or to decide that no such G exists. The authors first enumerate all possible unordered pairs of 4‑vertex graphs up to complement, obtaining 30 distinct families F. They then classify the computational complexity of the decision version for each family.

Key technical tools include:

• Observation 1.1, which gives four reduction rules: symmetry of the sandwich set, component‑preserving reductions when all forbidden graphs are connected, removal of universal vertices when no forbidden graph has a universal vertex, and a constructive algorithm when each forbidden graph has a unique F‑free supergraph on the same vertex set.
• Ramsey theory (R(4,4)=18) to bound the size of instances for certain families.
• Known polynomial‑time algorithms for special graph classes such as P₄‑free (trivially perfect) graphs, complete bipartite graphs, and paw‑free graphs (via Olariu’s characterization).
• Structural decompositions that split the problem along connected components of G₂ when G₂ is disconnected.

The authors prove polynomial‑time solvability for fifteen families, notably:

• {diamond, K₄}, {diamond, C₄}, {diamond, paw} (via Observation 1.1(iv));
• {P₄, C₄} (trivially perfect graphs, using the Ramsey bound);
• {P₄, K₁∪F₂} with F₂∈{K₃, P₃} (component‑wise reduction);
• The complete bipartite case (Lemma 2.4), which reduces to a simple merging procedure akin to 2‑SAT;
• Several families involving paw, claw, and their complements, where Lemma 2.6 (Olariu) reduces the problem to checking triangle‑free or P₃‑free graphs, both of which are already known to be tractable.

Conversely, seven families are shown to be NP‑complete: {C₄, K₄}, {paw, K₄}, {paw, \overline{K₄}}, {paw, \overline{C₄}}, {diamond, \overline{C₄}}, {paw, \overline{diamond}}, {diamond, \overline{diamond}}. For these, the authors construct reductions from known hard problems (e.g., 3‑colorability, SAT) by encoding logical constraints into the presence or absence of the two forbidden subgraphs. The reductions exploit the fact that each pair simultaneously forbids a clique and a cycle (or their complements), which forces the sandwich instance to simulate a coloring or satisfiability condition.

The paper concludes with a complete complexity map for all 30 families, leaving eight cases unresolved as open problems. The work demonstrates how fine‑grained structural properties of small forbidden subgraphs can be leveraged to either design efficient algorithms or prove hardness, thereby enriching the theory of graph sandwich problems and suggesting directions for future research on larger forbidden families or related parameterized versions.