Consecutive Ones Property and PQ-Trees for Multisets: Hardness of Counting Their Orderings

TL;DR

This paper proves that counting solutions for the consecutive ones property on multisets and related PQ-tree frontiers is #P-complete, indicating high computational complexity and unlikely polynomial solutions.

Contribution

It establishes the #P-completeness of counting solutions for COP on multisets and counting PQ-tree frontiers, extending known complexity results to these problems.

Findings

Counting solutions for COP on multisets is #P-complete.

Counting PQ-tree frontiers solutions is #P-complete.

Decisional problems are NP-complete.

Abstract

A binary matrix satisfies the consecutive ones property (COP) if its columns can be permuted such that the ones in each row of the resulting matrix are consecutive. Equivalently, a family of sets F = {Q_1,..,Q_m}, where Q_i is subset of R for some universe R, satisfies the COP if the symbols in R can be permuted such that the elements of each set Q_i occur consecutively, as a contiguous segment of the permutation of R's symbols. We consider the COP version on multisets and prove that counting its solutions is difficult (#P-complete). We prove completeness results also for counting the frontiers of PQ-trees, which are typically used for testing the COP on sets, thus showing that a polynomial algorithm is unlikely to exist when dealing with multisets. We use a combinatorial approach based on parsimonious reductions from the Hamiltonian path problem, showing that the decisional version of our problems is therefore NP-complete.