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Quaret Compatibility: New Results and Surprising Counterexamples. Stefan Gr ünewald. Compatibility Problem. Given a set of phylogenetic trees with overlapping taxa sets, does there exist a single phylogenetic tree on the union of the taxa sets that contains the information of all input trees?. - PowerPoint PPT Presentation

Quaret Compatibility: New Results and Surprising CounterexamplesStefan Grnewald

Compatibility ProblemGiven a set of phylogenetic trees with overlapping taxa sets, does there exist a single phylogenetic tree on the union of the taxa sets that contains the information of all input trees?

RestrictionsA restriction of a phylogenetic tree T to a subset S of L(T) is the tree obtained from the smallest subtree of T containing S by suppressing all vertices of degree 2. bghcefad

RestrictionsA restriction of a phylogenetic tree T to a subset S of L(T) is the tree obtained from the smallest subtree of T containing S by suppressing all vertices of degree 2. ghced

RestrictionsA restriction of a phylogenetic tree T to a subset S of L(T) is the tree obtained from the smallest subtree of T containing S by suppressing all vertices of degree 2. ghced

DisplayingT displays a binary tree T if and T is the restriction of T to L(T). bghcefad

DisplayingT displays a binary tree T if and T is the restriction of T to L(T).

bghcefadDisplayedcdah

DisplayingT displays a binary tree T if and T is the restriction of T to L(T).

bghcefadNot displayedafbh

QuartetsA quartet is a binary phylogenetic tree with 4 taxa. The quartet separating a and b from c and d is denoted by ab|cd.

ab|cdac|bd ad|bcadbcacbdabcd

CompatibilityFor a set P of binary phylogenetic trees, let L(P) be the union of the taxa sets of all trees in P. P is compatible if there is a tree T with L(T)=L(P) such that T displays every tree in P. If T is unique then P defines T.

CompatibilityIt is NP-complete to decide if a given set of binary phylogenetic trees is compatible, even if all trees are quartets (Steel 1992). It can be decided in polynomial time if a set Q of quartets with |Q|=|L(Q)|-3 defines a phylogenetic tree (Bcker, Dress, Steel 1999). Given a set Q of quartets and a tree T that displays Q. The complexity of deciding if Q defines T is unknown.

Thin Quartet SetsA set Q of quartets is thin if every subset of the taxa set with k4 elements contains at most k-3 quartets.

Theorem 1: If Q is thin then Q is compatible.

Maximal HierarchiesTheorem 2 (Bcker, Dress, Steel, 1999): Every minimum defining quartet set contains a maximal hierarchy of excess-free subsets.

Dezulian, Steel (2004): one of the most mysterious and apparently difficult results in phylogeny.

Maximal Hierarchiesbg|afah|bfcd|ghcf|dg

Maximal Hierarchiesfghbabg|afah|bfcd|ghcf|dg

Maximal Hierarchiesfghbcfghdabg|afah|bfcd|ghcf|dg

Maximal Hierarchiescfghbafghbcfghdadbg|afah|bfcd|ghcf|dg

Freely compatible quadruple setsA set of quadruples is freely compatible if, for every choice of one quartet for each quadruple, the obtained quartet set is compatible. A set of quadruples is thin if every subset of the taxa set with k4 elements contains at most k-3 quadruples.

Freely compatible quadruple setsBy Theorem 1, every thin set of quadruples is freely compatible.

Question: Is every freely compatible set of quadruples thin?

Freely compatible quadruple setsBy Theorem 1, every thin set of quadruples is freely compatible.

Question: Is every freely compatible set of quadruples thin?

Answer: No! A set of 13 quadruples on 13 taxa such that every two quadruples intersect in exactly two taxa is a counterexample (SG, Li Junrui).

The Quartet Graphabcdefab|de, bc|ef, cd|fa

The Quartet Graphabcdefab|de, bc|ef, cd|faabcdef

The Quartet Grapha,bcdefab|de, bc|ef, cd|faabcdef

The Quartet Grapha,bcde,fab|de, bc|ef, cd|faabcdef

The Quartet Grapha,bc,de,fab|de, bc|ef, cd|faabcdef

Closure Rulesabcdeab|cd, bc|de infer ab|ce, ab|de, ac|de.

A quartet set is called closed if no closure rule can be applied to any subset.

An ExampleThe quartet set corresponding to the picture is strongly closed but not compatible (SG, Steel, Swenson, 2007).

An incompatible quartet set with little overlap

In1992, Mike Steel used a counting argument to show that are incompatible quartet sets such that two quartets in the set have at most one taxon in common.

An Example

An Example