Polynomial Kernel and Incompressibility for Prison-Free Edge Deletion and Completion

TL;DR

This paper investigates the kernelization complexity of the H-free Edge Deletion problem, demonstrating that for a specific graph called the prison, the problem admits a polynomial kernel, thus challenging existing conjectures in the field.

Contribution

The paper provides the first polynomial kernel for Prison-Free Edge Deletion, countering the conjecture that such problems are generally incompressible for graphs with at least five vertices.

Findings

Prison-Free Edge Deletion admits a polynomial kernel.

The problem for the complement of the prison is incompressible.

Refutes the conjecture on the universal incompressibility of H-free Edge Deletion.

Abstract

Given a graph $G$ and an integer $k$, the $H$-free Edge Deletion problem asks whether there exists a set of at most $k$ edges of $G$ whose deletion makes $G$ free of induced copies of $H$. Significant attention has been given to the kernelizability aspects of this problem -- i.e., for which graphs $H$ does the problem admit an "efficient preprocessing" procedure, known as a polynomial kernelization, where an instance $I$ of the problem with parameter $k$ is reduced to an equivalent instance $I'$ whose size and parameter value are bounded polynomially in $k$? Although such routines are known for many graphs $H$ where the class of $H$-free graphs has significant restricted structure, it is also clear that for most graphs $H$ the problem is incompressible, i.e., admits no polynomial kernelization parameterized by $k$ unless the polynomial hierarchy collapses. These results led Marx and Sandeep to the conjecture that $H$-free Edge Deletion is incompressible for any graph $H$ with at least five vertices, unless $H$ is complete or has at most one edge (JCSS 2022). This conjecture was reduced to the incompressibility of $H$-free Edge Deletion for a finite list of graphs $H$. We consider one of these graphs, which we dub the prison, and show that Prison-Free Edge Deletion has a polynomial kernel, refuting the conjecture. On the other hand, the same problem for the complement of the prison is incompressible.