1. Speaker: Chuang-Chieh Lin National Chung Cheng University
Constructing Phylogenies from Quartets: Elucidation of Eutherian Superordinal Relationships
A. Ben-Dor, B. Chor, D. Graur, R. Ophir, D. Pelleg
Journal of Computational Biology, Vol. 5, 1998, pp. 377390.
Elucidation: 說明、解釋
Speaker: Chuang-Chieh Lin
National Chung Cheng University
2019/1/14

2. Computation Theory Lab, CSIE, CCU, Taiwan
Outline
Introduction and preliminaries
Problem description
The dynamic programming algorithm
The space complexity and the time complexity
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3. Computation Theory Lab, CSIE, CCU, Taiwan
Evolutionary trees
Let S be a set of taxa and | S | = n.
An evolutionary tree T on S is an unrooted, leaf-labeled tree such that the leaves of T are bijectively labeled by the taxa in S, and each internal node of T has degree 3.
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4. Computation Theory Lab, CSIE, CCU, Taiwan
Evolutionary trees
For 4 taxa a, b, c, d, we have 3 possible topologies: a c a b a c b d c d d b
[ad|bc]
[ab|cd]
[ac|bd]
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5. Evolutionary trees (contd.)
For 5 taxa a, b, c, d, e, how many possible evolutionary trees can we derive?
The answer is: 5  3 = 15. a c
There are 5 possible positions for e to be inserted. b d
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6. Evolutionary trees (contd.)
For n taxa, how many possible evolutionary trees can we derive?
The answer is (2n  5)!!
This observation can be verified by induction on n.
For an odd positive integer n, it is defined that n!! = n (n  2)  (n  4)  …  3  1.
If n = 15, (2n  5)!! is approximately 8  1012.
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7. Computation Theory Lab, CSIE, CCU, Taiwan
Let us analyze n!! in another way.
For a nonnegative integer m  0, let n = 2m + 1.
Then we have ( 2 m + 1 ) ! = P . S o ( 2 m + 1 ) ! = O .
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8. Computation Theory Lab, CSIE, CCU, Taiwan
(2n−5)!! = O(nn−2)
For n taxa, we have (2n  5)!! = O((n  3)n2)
= O(nn−2) possible evolutionary trees.
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9. Computation Theory Lab, CSIE, CCU, Taiwan
Quartet topologies
A set of four taxa is called a quartet.
Given an evolutionary tree T and a quartet {a, b, c, d}, the quartet topology of {a, b, c, d} induced by T is obtained by the following procedure.
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10. Computation Theory Lab, CSIE, CCU, Taiwan
Step 1: All leaves but a, b, c and d are deleted from the tree. Edges adjacent to these leaves are also deleted. a b c d f e g T a b c d
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11. Computation Theory Lab, CSIE, CCU, Taiwan
Step 2: Internal nodes with degree two are contracted and deleted, so their two adjacent nodes become connected. This process is repeated until no internal nodes of degree two are left. a b c d a b c d
For simplicity, we denote the quartet topology above by [bc|ad], which is a kind of bipartition of {a, b, c, d}.
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12. Computation Theory Lab, CSIE, CCU, Taiwand
For simplicity, we denote the quartet topology above by [bc|ad], which is a kind of bipartition of {a, b, c, d}.
Note that each input quartet topology t is accom-panied by a positive weight Ct .
2019/1/14
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13. Computation Theory Lab, CSIE, CCU, Taiwan
Problem description
Input:
A list of weighted quartet topologies over n taxa.
Output:
A binary tree with n leaves such that the total weight of the satisfied quartet topologies is maximized.
This problem was shown to be NP-hard.
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14. Computation Theory Lab, CSIE, CCU, Taiwan
Quartet method
The fact that small phylogenies are easier to infer than large ones leads to another approach – the quartet method.
First, consider subsets of 4 taxa, one at a time, and infer the phylogenies (i.e., quartet topologies) for these subsets.
The next stage combines the multiple quartet topologies into a single phylogeny.
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15. Computation Theory Lab, CSIE, CCU, Taiwan
Given a set of quartet topologies Q, how to determine whether an evolutionary tree T is “good” or “bad”?
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16. Computation Theory Lab, CSIE, CCU, Taiwan
Given an evolutionary tree T and a set of quartet topologies Q.
We say that T satisfies a quartet topology tq of a quartet q if the induced quartet topology of q by T is exactly tq. a b c d f e g T
For example, T satisfies [ab|dg], [ce|fg], [ad|bc], etc.
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17. Computation Theory Lab, CSIE, CCU, Taiwan
Score
We denote by S, where S  Q, the set of quartet topologies that are satisfied by T, and let U = Q  S.
We define the score of the evolutionary tree T as follows. P s 2 S C + 1 3 u U .
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18. Computation Theory Lab, CSIE, CCU, Taiwan
Score (contd.)
The latter term was chosen because there are three possible topologies for every quartet.
Therefore this term equals the expected increase.
In a variant of the same method, the latter term is zeroed, so the quartet topologies which are not satisfied by T do not contribute to the score. P s 2 S C + 1 3 u U .
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19. Computation Theory Lab, CSIE, CCU, Taiwan
Score (contd.)
It can be easily derived that is an upper bound on the score of any evolutionary tree T. P q 2 Q C
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20. Preliminaries for the dynamic programming algorithm
For technical reasons, the following discussion deals with rooted evolutionary trees.
For a node v, its left and right children are denoted by vl and vr respectively. v vl vr
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21. Preliminaries for the dynamic programming algorithm (contd.)
Given a rooted evolutionary tree T and a node v in it we denote by T(v) the subtree of T rooted at v. u v w
T(v)
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22. Preliminaries for the dynamic programming algorithm (contd.)
We denote by L(T) the set of leaves (i.e., taxa) of the tree T. u v w
L(Tv) …
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23. Preliminaries for the dynamic programming algorithm (contd.)
For a pair of nodes u, v, the least common ancestor of u and v, lca(u, v), is defined as an ancestor p of both u and v such that no node in T(p) other than p is an ancestor of both u and v.
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24. Preliminaries for the dynamic programming algorithm (contd.)
The lca of a and c. a b c d a b c d
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25. Preliminaries for the dynamic programming algorithm (contd.)
Definition: Given a quartet topology t = [ab|cd] and an evolutionary tree T, the quartet least common ancestor of t, qlca(t) is defined as a node p that is the lca of two or more pairs of elements from {a, b, c, d}, and no node in T(p) except p is the lca of two or more pairs of elements from {a, b, c, d}.
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Computation Theory Lab, CSIE, CCU, Taiwan

26. Preliminaries for the dynamic programming algorithm (contd.)
The qlca for [ab|cd]. a b c d a b c d
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27. Another equivalent definition for the quartet least common ancestor
Definition: Given a quartet topology t = [ab|cd] and an evolutionary tree T, the qlca of t is a node p such that
|L(T(p)){a, b, c, d}|  3.
For any child s of p, |L(T(s)){a, b, c, d}|  2.
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28. Computation Theory Lab, CSIE, CCU, Taiwan
Some observations
Every quartet topology t has a unique qlca(t).
Given a tree T and a quartet topology t, the subtree rooted at qlca(t) determines whether t is satisfied in the evolutionary tree T.
Let t = [ab|cd] and v = qlca(t). We look at vl , vr , T(vl) and T(vr).
At least one of these subtrees contains exactly two taxa e, f from {a, b, c, d}.
Then t is satisfied iff the pair {e, f} is either {a, b} or {c, d}.
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Computation Theory Lab, CSIE, CCU, Taiwan

29. Some observations (contd.)
Given a quartet topology t = [ab|cd] and an evolutionary tree T, let v = qlca(t). Then T satisfies t if and only if at least one of the following holds:
{a, b}  L(T(s)).
{c, d}  L(T(s)).
where s = vl or s = vr.
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30. Computation Theory Lab, CSIE, CCU, Taiwan
The algorithm
We denote by SATQ(T(v)) the set of quartet topologies t  Q such that t is satisfied by T, and qlca(t) is a node in T(v).
Let TOPQ(T(v)) SATQ(T(v)) be the set of quartet topologies in Q that have v as their qlca and are satisfied by T.
最底下等式(recursive formula)右邊三項為disjoint，所以等一下的score加總沒問題。 W e t h n a v S A T Q ( ) = O P [ l r :
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31. Computation Theory Lab, CSIE, CCU, Taiwan
The algorithm (contd.)
For a set A  Q of quartet topologies, let denote the sum of their weights.
The score of the subtree T(v) (with respect to Q) is defined as s u m ( A ) = P t 2 C s c o r e Q ( T v ) = u m S A :
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32. Computation Theory Lab, CSIE, CCU, Taiwan
The algorithm (contd.)
By the above equation, we have s c o r e Q ( T v ) = u m O P + l :
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33. Computation Theory Lab, CSIE, CCU, Taiwan
The algorithm (contd.)
Let S be a set of three or more taxa.
Denote by opt_scoreQ(S) the maximum score with respect to Q among all trees that have S as their set of leaves.
We denote by opt_treeQ(S) a tree which attains the maximum score.
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34. Computation Theory Lab, CSIE, CCU, Taiwan
The algorithm (contd.)
For every proper partition of S into two subsets S1 and S2, let T(S1, S2) denote a tree whose left subtree equals opt_treeQ(S1) and its right subtree equals opt_treeQ(S2).
We then have s c o r e Q ( T S 1 ; 2 ) = u m O P + p t l :
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35. Computation Theory Lab, CSIE, CCU, Taiwan
The algorithm (contd.)
This implies that
By employing the dynamic programming paradigm, we can avoid wasteful repetitions.
To do this, we scan the subsets S {1 ,2 …, n} by increasing size of S. o p t s c r e Q ( S ) = m a x 1 [ 2 u T O P ; + l :
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36. Computation Theory Lab, CSIE, CCU, Taiwan
The algorithm (contd.)
For simplicity, the details of implementing the dynamic programming algorithm are omitted.
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37. The space complexity and the time complexity= 2 k O ( 3 ) ; w h e r s t z o f p u q a l g .
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Computation Theory Lab, CSIE, CCU, Taiwan

38. Thank you.

39. Computation Theory Lab, CSIE, CCU, Taiwan
References
[S92] M. Steel: The complexity of reconstructing trees from qualitative characters and subtrees. Journal of Classification, 9 (1992), pp. 91−116.
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