The largest crossing number of tanglegrams

TL;DR

This paper establishes improved bounds on the maximum crossing number of tanglegrams, showing it is close to half the total possible crossings, with bounds that differ by a term involving n and logarithmic factors.

Contribution

It provides tighter upper and lower bounds on the maximum crossing number of size n tanglegrams, refining previous estimates.

Findings

Maximum crossing number is at most (1/2) * binomial(n, 2) - n/4.

For infinitely many n, the crossing number is at least (1/2) * binomial(n, 2) - (n log n)/4.

Bounds relate to the Unbalancing Lights Problem of Gale and Berlekamp.

Abstract

A tanglegram $\cal T$ consists of two rooted binary trees with the same number of leaves, and a perfect matching between the two leaf sets. In a layout, the tanglegrams is drawn with the leaves on two parallel lines, the trees on either side of the strip created by these lines are drawn as plane trees, and the perfect matching is drawn in straight line segments inside the strip. The tanglegram crossing number ${\rm cr}({\cal T})$ of $\cal T$ is the smallest number of crossings of pairs of matching edges, over all possible layouts of $\cal T$. The size of the tanglegram is the number of matching edges, say $n$. An earlier paper showed that the maximum of the tanglegram crossing number of size $n$ tanglegrams is $<\frac{1}{2}\binom{n}{2}$; but is at least $\frac{1}{2}\binom{n}{2}-\frac{n^{3/2}-n}{2}$ for infinitely many $n$. Now we make better bounds: the maximum crossing number of a size $n$ tanglegram is at most $ \frac{1}{2}\binom{n}{2}-\frac{n}{4}$, but for infinitely many $n$, at least $\frac{1}{2}\binom{n}{2}-\frac{n\log_2 n}{4}$. The problem shows analogy with the Unbalancing Lights Problem of Gale and Berlekamp.