Improved Lower Bounds on the Compatibility of Multi-State Characters

TL;DR

This paper establishes new lower bounds on the size of incompatible sets of multi-state characters, advancing understanding of their compatibility conditions and providing constructions related to quartet and triplet compatibility.

Contribution

It improves the lower bounds on the function f(r) for multi-state character compatibility and introduces new incompatible sets with specific properties.

Findings

f(r) ge bfloor(r/2)ce c8 ce bce b7 ce b7 ce b7 + 1 for all r ge 2

Existence of incompatible quartet sets with specific properties for all n ge 4

Upper bound for incompatible triplet sets is tight, with constructions for all n ge 3

Abstract

We study a long standing conjecture on the necessary and sufficient conditions for the compatibility of multi-state characters: There exists a function $f(r)$ such that, for any set $C$ of $r$-state characters, $C$ is compatible if and only if every subset of $f(r)$ characters of $C$ is compatible. We show that for every $r \ge 2$, there exists an incompatible set $C$ of $\lfloor\frac{r}{2}\rfloor\cdot\lceil\frac{r}{2}\rceil + 1$ $r$-state characters such that every proper subset of $C$ is compatible. Thus, $f(r) \ge \lfloor\frac{r}{2}\rfloor\cdot\lceil\frac{r}{2}\rceil + 1$ for every $r \ge 2$. This improves the previous lower bound of $f(r) \ge r$ given by Meacham (1983), and generalizes the construction showing that $f(4) \ge 5$ given by Habib and To (2011). We prove our result via a result on quartet compatibility that may be of independent interest: For every integer $n \ge 4$, there exists an incompatible set $Q$ of $\lfloor\frac{n-2}{2}\rfloor\cdot\lceil\frac{n-2}{2}\rceil + 1$ quartets over $n$ labels such that every proper subset of $Q$ is compatible. We contrast this with a result on the compatibility of triplets: For every $n \ge 3$, if $R$ is an incompatible set of more than $n-1$ triplets over $n$ labels, then some proper subset of $R$ is incompatible. We show this upper bound is tight by exhibiting, for every $n \ge 3$, a set of $n-1$ triplets over $n$ taxa such that $R$ is incompatible, but every proper subset of $R$ is compatible.