Computing Quartet Distance is Equivalent to Counting 4-Cycles
November 15, 2018 Β· Declared Dead Β· π Symposium on the Theory of Computing
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Authors
BartΕomiej Dudek, PaweΕ Gawrychowski
arXiv ID
1811.06244
Category
cs.DS: Data Structures & Algorithms
Citations
12
Venue
Symposium on the Theory of Computing
Last Checked
4 months ago
Abstract
The quartet distance is a measure of similarity used to compare two unrooted phylogenetic trees on the same set of $n$ leaves, defined as the number of subsets of four leaves related by a different topology in both trees. After a series of previous results, Brodal et al. [SODA 2013] presented an algorithm that computes this number in $\mathcal{O}(nd\log n)$ time, where $d$ is the maximum degree of a node. Our main contribution is a two-way reduction establishing that the complexity of computing the quartet distance between two trees on $n$ leaves is the same, up to polylogarithmic factors, as the complexity of counting 4-cycles in an undirected simple graph with $m$ edges. The latter problem has been extensively studied, and the fastest known algorithm by Vassilevska Williams [SODA 2015] works in $\mathcal{O}(m^{1.48})$ time. In fact, even for the seemingly simpler problem of detecting a 4-cycle, the best known algorithm works in $\mathcal{O}(m^{4/3})$ time, and a conjecture of Yuster and Zwick implies that this might be optimal. In particular, an almost-linear time for computing the quartet distance would imply a surprisingly efficient algorithm for counting 4-cycles. In the other direction, by plugging in the state-of-the-art algorithms for counting 4-cycles, our reduction allows us to significantly decrease the complexity of computing the quartet distance. For trees with unbounded degrees we obtain an $\mathcal{O}(n^{1.48})$ time algorithm, which is a substantial improvement on the previous bound of $\mathcal{O}(n^{2}\log n)$. For trees with degrees bounded by $d$, by analysing the reduction more carefully, we are able to obtain an $\mathcal{\tilde O}(nd^{0.77})$ time algorithm, which is again a nontrivial improvement on the previous bound of $\mathcal{O}(nd\log n)$.
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