# Quaret Compatibility: New Results and Surprising Counterexamples

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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

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• 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).

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

• 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.

• 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

• 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

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