The worst case complexity of Maximum Parsimony

Transcription

1 he worst case complexity of Maximum Parsimony mir armel Noa Musa-Lempel Dekel sur Michal Ziv-Ukelson Ben-urion University June 2, 20 / 2

2 What s a phylogeny Phylogenies: raph-like structures whose topology describes the inferred evolutionary history among a set of species. Modeled as either rooted or unrooted labeled binary trees, where the input entities are assigned to the leaf vertices. 2 / 2

3 haracter based methods for phylogenetic reconstruction Each specie is characterized by a sequence of letters. We are given a subsitution scoring matrix over the letters. Position independence is assumed. : 2: : : 5: / 2

4 rooted/unrooted phylogeny he decision whether to model phylogenies as rooted versus unooted depends on the substitution scoring matrix. Modeling phylogenies as unrooted trees requires the assumption of symmetric scoring matrices. oday, many applications apply asymmetric scoring matrices. / 2

5 Parsimony Maximization classical approach for phylogenetic reconstruction. he Parsimony Maximization approach seeks the phylogenetic tree that supposes the least amount of evolutionary change explaining the observed data. here are two classical problems inferred from phylogenetic parsimony maximization: Small Parsimony (SP) and Maxmimum Parsimony (MP). 5 / 2

6 Small Parsimony Problem (SP) Input: multiple alignment, tree topology on n leaves. : 2: : : 5: 5 oal: ssignment to internal vertices that minimizes the scoring function. 5 Score = 5 Score = 2 6 / 2

7 Small Parsimony Problem (SP) Input: multiple alignment, tree topology on n leaves. : 2: : : 5: 5 oal: ssignment to internal vertices that minimizes the scoring function. 5 Score = 5 Score = 2 We note that known algorithms for Small Parsimony traverse the tree in a bottom up manner. 6 / 2

8 Maximum Parsimony Problem (MP) Input: multiple alignment : 2: : : 5: oal: topology and assignments to internal vertices, that minimizes the SP score. 5 Score = 5 Score = Score = 2 7 / 2

9 Maximum Parsimony Problem (MP) Input: multiple alignment : 2: : : 5: oal: topology and assignments to internal vertices, that minimizes the SP score. 5 Score = 5 Score = Score = 2 he Maximum Parsimony (MP) problem is NP-hard [L. R. Foulds and R. L. raham (982)]. 7 / 2

10 Measuring SP and MP complexity in terms of assignment operations ssignment operation - time to compute the assignment for a single vertex. his depends on the scoring scheme employed, for example: Fitch s algorithm (Hamming distance) O(m), Sankoff s algorithm (weighted edit distance) O(mΣ 2 ). 8 / 2

11 Our contribution Previous results: avalli-sforza and Edwards (967) - (n ) (2n )!! assignment operations. Hendy and Penny (982) - branch&bound algorithm for MP. Where (2n )!! = 5... (2n ). 9 / 2

12 Our contribution New results: avalli-sforza and Edwards (967) - (n ) (2n )!! assignment operations. Hendy and Penny (982) - branch&bound algorithm for MP. Worst case running time: Θ( n (2n )!!) assignment operations. new, faster algorithm which executes Θ((2n )!!) assignment operations. Where (2n )!! = 5... (2n ) 9 / 2

13 he algorithm of avalli-sforza and Edwards 0 / 2

14 he algorithm of avalli-sforza and Edwards avalli-sforza and Edwards showed that the number of rooted phylogenies with n leaves is (2n )!!. 0 / 2

15 he algorithm of avalli-sforza and Edwards avalli-sforza and Edwards showed that the number of rooted phylogenies with n leaves is (2n )!!. he algorithm enumerates all phylogenies with n leaves, and then solves the Small Parsimony (SP) problem on each tree. 0 / 2

16 he algorithm of avalli-sforza and Edwards avalli-sforza and Edwards showed that the number of rooted phylogenies with n leaves is (2n )!!. he algorithm enumerates all phylogenies with n leaves, and then solves the Small Parsimony (SP) problem on each tree. Each phylogeny has exactly n internal vertices, therefore the algorithm has a running time of (n ) (2n )!! assignment operations. 0 / 2

17 he algorithm of Hendy and Penny Preliminaries: 2 2 / 2

18 he algorithm of Hendy and Penny Enumeration space: 2 2 / 2

19 he algorithm of Hendy and Penny Enumeration space: 2 2 / 2

20 he algorithm of Hendy and Penny Enumeration space: 2 2 / 2

21 he algorithm of Hendy and Penny ssignment operations: 2 2 / 2

22 he algorithm of Hendy and Penny ssignment operations: 2 2 / 2

23 he algorithm of Hendy and Penny ssignment operations: / 2

24 he algorithm of Hendy and Penny ssignment operations: / 2

25 he algorithm of Hendy and Penny ssignment operations: / 2

26 he algorithm of Hendy and Penny ssignment operations: / 2

27 he algorithm of Hendy and Penny ssignment operations: / 2

28 he algorithm of Hendy and Penny ssignment operations: / 2

29 he algorithm of Hendy and Penny ssignment operations: / 2

30 he algorithm of Hendy and Penny ssignment operations: / 2

31 he algorithm of Hendy and Penny ssignment operations: / 2

32 he algorithm of Hendy and Penny ssignment operations: / 2

33 he algorithm of Hendy and Penny ssignment operations: he search space tree is developed in top-down order, while the recalculations of assignments is done in a bottom-up order. / 2

34 he algorithm of Hendy and Penny ssignment operations: he complexity of the algorithm equals to the number of assignment operations. / 2

35 he algorithm of Hendy and Penny heir algorithm was originally proposed for the purpose of branch and bound and its worst case bound was not previously properly analyzed. Using combinatorial methods we managed to achieve an exact bound. 2 / 2

36 he number of assignment operations Let Numnc(v) denote the number of ancestors of x in F v. / 2

37 he number of assignment operations Let Numnc(v) denote the number of ancestors of x in F v x 5 / 2

38 he number of assignment operations Let Numnc(v) denote the number of ancestors of x in F v x 5 he number of ancestors of x in F v is equal to the number of assignment operations executes in node v. / 2

39 he number of assignment operations Let Numnc(v) denote the number of ancestors of x in F v x 5 he number of ancestors of x in F v is equal to the number of assignment operations executes in node v. Let H i be the sum of Numnc(v) for all nodes v in level i +. / 2

40 he number of assignment operations Let Numnc(v) denote the number of ancestors of x in F v x 5 he number of ancestors of x in F v is equal to the number of assignment operations executes in node v. Let H i be the sum of Numnc(v) for all nodes v in level i +. By definition, Numnc(v) = n i= H i. / 2

41 he number of assignment operations Lemma H i = (2i)!! (2i )!!. heorem 2 he assignment operations complexity of the algorithm of Hendy and Penny is Θ( n(2n )!!). / 2

42 new, more efficient search space tree 5 / 2

43 new, more efficient search space tree 5 / 2

44 new, more efficient search space tree 2 5 / 2

45 new, more efficient search space tree 2 5 / 2

46 new, more efficient search space tree 2 5 / 2

47 new, more efficient search space tree 2 5 / 2

48 new, more efficient search space tree 2 5 / 2

49 new, more efficient search space tree 2 First challenge: he same node can be obtained from two different sequences of merge operations. 5 / 2

50 new, more efficient search space tree 2 First challenge: he same node can be obtained from two different sequences of merge operations. 5 / 2

51 new, more efficient search space tree Second challenge: We need to count the number of nodes in the search space tree. 5 / 2

52 new, more efficient search space tree he new algorithm develops the search space tree bottom up, which flows along with the natural bottom up direction of Small Parsimony. 6 / 2

53 new, more efficient search space tree he new algorithm develops the search space tree bottom up, which flows along with the natural bottom up direction of Small Parsimony. We first define a graph n, and then, using canonical representation for each node, we transform n into a tree n by removing redundant edges. 6 / 2

54 new, more efficient search space tree he new algorithm develops the search space tree bottom up, which flows along with the natural bottom up direction of Small Parsimony. We first define a graph n, and then, using canonical representation for each node, we transform n into a tree n by removing redundant edges. Using combinatorial analysis we showed that the size of n is approximately e (2n )!!. 6 / 2

55 new, more efficient search space tree he new algorithm develops the search space tree bottom up, which flows along with the natural bottom up direction of Small Parsimony. We first define a graph n, and then, using canonical representation for each node, we transform n into a tree n by removing redundant edges. Using combinatorial analysis we showed that the size of n is approximately e (2n )!!. raversing the search space tree requires performing exactly one assignment operation per node, thus the complexity of our new MP algorithm is Θ((2n )!!) 6 / 2

56 From n to n For a node u in the search space, we denote by F u the corresponding forest. For a tree in a forest define the label of to be label( ) = min{label(v) : v is a leaf in }. he label of a forest F is label(f ) = min{label( ) : is a non-singleton tree in F }. For a node u. let u denote the tree in F u for which label(f u ) = label( u ). Definition he search space tree n is a tree whose nodes are the nodes of n. node v is the parent of a node u in n if F v is obtained from F u by deleting the root of u. 7 / 2

57 From n to n 2 8 / 2

58 From n to n 2 8 / 2

59 From n to n he definition of n gives a characterization for the parent of a node in the tree. In order to perform a top-down traversal of the search space tree, we need a characterization for the children of a node. Such characterization is given in the following lemma. Lemma 2 node u is a child of a node v in n if and only if F u is obtained from F v by merging two trees and 2 from F v with label( ) < label( 2 ) and either. = v (and in particular, is not a singleton), or 2. is a singleton and label( ) < label( v ). 9 / 2

60 From n to n 2. = v (and in particular, is not a singleton), or 2. is a singleton and label( ) < label( v ). 20 / 2

61 From n to n 2. = v (and in particular, is not a singleton), or 2. is a singleton and label( ) < label( v ). 20 / 2

62 omplexity of the new search space Let n i denote the number of nodes in level i of n. Let L n i,k denote the number of nodes v in level i for which label(f v ) = k. For example, n 0 =, n = ( n 2), and n n = (2n )!!. Let µ n denote the size of n. i.e. µ n = n i=0 n i. Lemma L n i+,k = (n i k) n i l=k Ln i,l. Lemma L n i,k = (2i )!!( ) n k+i 2i. Lemma 5 n i = (2i )!! ( ) n+i 2i. 2 / 2

63 omplexity of the new search space Let n i denote the number of nodes in level i of n. Let L n i,k denote the number of nodes v in level i for which label(f v ) = k. For example, n 0 =, n = ( n 2), and n n = (2n )!!. Let µ n denote the size of n. i.e. µ n = n i=0 n i. Lemma L n i+,k = (n i k) n i l=k Ln i,l. Lemma L n i,k = (2i )!!( ) n k+i 2i. Lemma 5 n i = (2i )!! ( ) n+i 2i. heorem he assignment operations complexity for solving MP using the new search space is ( + o()) e (2n )!!, i.e. µ n e (2n )!!. 2 / 2

64 Summary We studied the classical problem of exact Maximum Parsimony, focusing on the running time complexity of various algorithms for the problem. he first approach proposed by avalli-sforza and Edwards yields an assignment operations complexity of (n ) (2n )!!. he second approach we analyzed was proposed by Hendy and Penny. Its theoretical running time complexity has not been previously analyzed. We showed that the assignment operations complexity of this approach is smaller by a factor of Θ( n). We proposed a new, faster MP approach, whose assignment operations complexity is smaller by a factor of Θ( n) than the complexity of the Hendy and Penny approach. 22 / 2

65 hank You 2 / 2