Kernelization for Counting Problems on Graphs: Preserving the Number of Minimum Solutions

TL;DR

This paper develops polynomial-time preprocessing algorithms for certain counting problems on graphs, enabling size reduction while preserving the exact number of minimum solutions, thus advancing the theory of kernelization for counting problems.

Contribution

It introduces the first polynomial kernelization techniques for counting minimum feedback vertex sets and dominating sets in planar graphs, preserving the number of solutions.

Findings

Polynomial-time algorithms either compute the exact count or produce a polynomial-sized equivalent instance.

The approach leverages bounds on the number of solutions relative to input size and parameter k.

This work opens new avenues for kernelization in counting problems, previously thought infeasible.

Abstract

A kernelization for a parameterized decision problem $\mathcal{Q}$ is a polynomial-time preprocessing algorithm that reduces any parameterized instance $(x,k)$ into an instance $(x',k')$ whose size is bounded by a function of $k$ alone and which has the same yes/no answer for $\mathcal{Q}$. Such preprocessing algorithms cannot exist in the context of counting problems, when the answer to be preserved is the number of solutions, since this number can be arbitrarily large compared to $k$. However, we show that for counting minimum feedback vertex sets of size at most $k$, and for counting minimum dominating sets of size at most $k$ in a planar graph, there is a polynomial-time algorithm that either outputs the answer or reduces to an instance $(G',k')$ of size polynomial in $k$ with the same number of minimum solutions. This shows that a meaningful theory of kernelization for counting problems is possible and opens the door for future developments. Our algorithms exploit that if the number of solutions exceeds $2^{\mathsf{poly}(k)}$, the size of the input is exponential in terms of $k$ so that the running time of a parameterized counting algorithm can be bounded by $\mathsf{poly}(n)$. Otherwise, we can use gadgets that slightly increase $k$ to represent choices among $2^{O(k)}$ options by only $\mathsf{poly}(k)$ vertices.