Capturing a phylogenetic tree when the number of character states varies with the number of leaves

We show that for any two values α, β ∈ (0, 1) for which α+β > 1 then there exists a value N so that for all n ≥ N , and any binary phylogenetic tree T on n leaves there exists a set of at most n characters that capture T , and for which each character has at most n states. Here ‘capture’ means that T is the only phylogenetic tree that is compatible with the characters. Our short proof of this combinatorial result is based on the probabilistic method. Suppose that k characters on n taxa captures some phylogenetic tree T (i.e. the characters are compatible with T and no other tree; thus T is a binary unrooted tree with the n taxa as its leaves). If each of the k characters has at most r states, then a simple and known inequality states that k must be at least (n − 3)/(r − 1) (Proposition 4.2 of [5]). Remarkably, this lower bound was recently shown [2] to be sharp for every value of r ≥ 2, provided that n ≥ nr, where nr is some (increasing) function of r. In this note, we consider how small k can be when r is allowed to depend on n. From [1, 3] it is known that k = 4 holds for a certain series rn for which n/rn = O(1), so we focus on the setting where both rn and n/rn tend to infinity with increasing n. We consider what happens when rn grows as a sub-linear function of n, by constraining rn to be less or equal to a n for α ∈ (0, 1), in which case the inequality k ≥ (n− 3)/(r− 1) implies that k must exceed n for β = 1−α, for n sufficiently large. The following result is independent of the result from [2] mentioned above, in the sense that neither result directly implies the other. Our short proof involves a simple application of the probabilistic method, the Chernoff bound, and a property of the random cluster model on trees established in [4]. Theorem 1. For any two values α, β ∈ (0, 1) for which α+β > 1 there exists a value N so that for all n ≥ N , and any unrooted binary phylogenetic tree T on n leaves there exists a set of at most ⌊n⌋ characters that capture T , and for which each character has at most rn = ⌊n ⌋ states. Proof. Let X denote the leaf set of T . Consider the random cluster model on T in which each edge of T is independently cut with probability pn = rn/4n, or left intact with probability 1−pn. This leads to a partition ofX corresponding to the equivalence relation that two leaves are related if and only if they lie in the same connected component of the resulting graph. We will regard such a partition as equivalent to a character (with the number of ‘states’ of the character being the number of blocks of the partition). Notice that limn→∞ pn = 0. Let Y denote the random number of edges of T that are cut. Then Y has a binomial distribution Y ∼ Bin(2n − 3, pn), which has mean μn = (2n − 3)pn = ( 1 2 − o(1))n . By a multiplicative form of the ‘Chernoff bound’ in probability theory, P(Y ≥ 2μn) ≤ exp(−4μn/3) and since rn > 2μn we obtain: (1) P(Y ≥ rn) ≤ exp(−4μn/3). The number of blocks of the partition of X induced by randomly cutting edges of T in this way is at most Y + 1. Thus, the probability that a character, generated by the random cluster model with pn value as specified, has strictly more than rn states is at most P(Y + 1 > rn) = P(Y ≥ rn) ≤ exp(−4μn/3), by (1). Thus if we generate a set Sn of ⌊n ⌋ such characters the probability that at least one of these characters has more than rn states is, by Boole’s inequality,