The Sackin Index of Simplex Networks

Abstract

A phylogenetic network is a simplex network if the child of every reticulation node is a network leaf and each tree node has at most one reticulation node as its child. Simplex networks are a superclass of phylogenetic trees and a subclass of tree-child networks. Generalizing the Sackin index to phylogenetic networks, we prove that the expected Sackin index of a random simplex network is asymptotically $\varTheta (n^{7/4})$ in the uniform model.

This work was supported by MOE Tier 1 grant R-146-000-318-114.

Appendices

Appendix A

Proposition A.1. Let $S_{k}= {n\atopwithdelims ()k} \frac{(2n-2)!}{2^{n-1}(k-1)!}$ and $k_0=\sqrt{n+1}-1$. Then,

• $S_{k}\le S_{k+1}$ for $k\in [1, k_0]$;

• $S_{k} > S_{k+1}$ if $k_0<k<2k_0$;

• $S_{k}>2S_{k+1}$ if $k\in [2k_0, \infty )$.

Proof

Note that

$$S_{k+1}=\frac{n-k}{k(k+1)}S_k.$$

If $k\le k_0$, $(k+1)^2\le (\sqrt{n+1}-1+1)^2=n+1$ and, equivalently, $k(1+k)\le n-k$ and thus ${S_{k+1}}=\frac{n-k}{k(k+1)}S_k\ge S_k$. Similarly, ${S_{k+1}}< S_k$ if $k>k_0$.

If $k\ge 2k_0$, $(k+2)^2\ge 4(n+1)$ and $k^2+k\ge 4(n-k)+k\ge 4(n-k)$ and therefore, ${S_{k+1}}\le \frac{1}{4} S_k< \frac{1}{2} S_k$.

Proposition A.2. $D_C(k, n-k)$ is a decreasing function of k on $[1, n-1]$.

Proof

By Eq. (1) and Eq. (12),

$$D_C(k, n-k)=\frac{n(2n-1)}{k}\left( \frac{4^kk!k!}{(2k)!}-1\right) . $$

Hence,

$$\begin{aligned}&n(2n-1)[D_C(k, n-k)-D_C(k+1, n-k-1)]\\= & {} \frac{2^{2k}(k-1)!k!}{(2k)!} - \frac{2^{2k}4k!(k+1)!}{(2k+2)!} - \frac{1}{k(k+1)}\\= & {} \frac{2^{2k}k!(k-1)!}{(2k+1)!} - \frac{1}{k(k+1)}\\= & {} \frac{1}{k(k+1)}\left[ \frac{2^{2k}k!(k+1)!}{(2k+1)!} - 1\right] \\= & {} \frac{1}{k(k+1)}\left[ \frac{ 4\times 6\times \cdots 2k\times (2k+2)}{ 3\times 5\times \cdots (2k+1)} -1\right] \\> & {} 0. \end{aligned}$$

Appendix B

Proposition B.1. Let P be a phylogenetic tree on n taxa. there exists a 1-to-1 mapping $\phi : \mathcal{T}(P)\cup \{\rho \}\rightarrow \mathcal{L}(P)$ such that u is an ancestor of $\phi (u)$ for each $u\in \mathcal{T}(P) \cup \{\rho \}$.

Proof

We prove the fact by mathematical induction on n. When $n=1$, we simply map the root $\rho$ to the only leaf.

Assume the fact is true for $n\le k$, where $k\ge 1$. For a phylogenetic tree P with $k+1$ leaves, we let the child of the root $\rho$ be u and the two grandchildren be v and w. We consider the subtree $P'$ induced by u, v and all the descendants of v and the subtree $P''$ induced by u, w and all the descendants of w.

Obviously, both $T'$ and $T''$ have less than k leaves. By induction, there is a 1-to-1 mapping $\phi ': \mathcal{T}(P')\cup \{u\} \rightarrow \mathcal{L}(P')$ satisfying the constraints on leaves, and there is a 1-to-1 mapping $\phi '': \mathcal{T}(P'')\cup \{u\} \rightarrow \mathcal{L}(P'')$ satisfying the constraints on leaves. Let $\phi '(u)=\ell$. Then, the function that maps $\rho$ to $\ell$, u to $\phi ''(u)$ and all the other tree nodes x to $\phi '(x)$ or $\phi ''(x)$ depending whether it is in $P'$ or $P''$. It is easy to verify that $\phi$ is a desired mapping.