Structure-Based Comparison of Biomolecules Benedikt Christoph - PowerPoint PPT Presentation

Arc-Preserving Common Subsequences Definition (Arc-Preserving Common Subsequence) Let S = ( s 1 s 2 ... s m , P s ) and T = ( t 1 t 2 ... t n , P t ) be two arc-annotated sequences over an alphabet Σ . A string is called an arc-preserving common subsequence of S and T if there exists a common subsequence w of s and t and a mapping ϕ consistent with w such that 1 s i = t j for all � i , j � ∈ ϕ , and 2 for all pairs of elements ( � i 1 , j 1 � , � i 2 , j 2 � ) from ϕ ( i 1 , i 2 ) ∈ P s ⇐ ⇒ ( j 1 , j 2 ) ∈ P t . 13 / 45

Example Σ = { A , G , U , C } ϕ = {� 1 , 4 � , � 5 , 5 � , � 6 , 6 � , � 9 , 8 � , � 10 , 9 � , � 11 , 11 � , � 12 , 12 �} – – – A U C D A G C G A U – C G S: G U A A – – – A G A – A U G C G T: 14 / 45

Longest Arc-Preserving Common Subsequence (LAPCS) Definition (LAPCS(L EVEL 1 ,L EVEL 2 )) By LAPCS(L EVEL 1 ,L EVEL 2 we denote the optimization problem for two arc-annotated strings S ∈ L EVEL 1 and T ∈ L EVEL 2 to find the longest common arc-annotated substring. 15 / 45

LAPCS(P LAIN ,P LAIN ) Theorem The optimization problem LAPCS(P LAIN ,P LAIN ) is computable in O ( m · n ) , where m and n are the length of the input strings. Proof. This problem is the same as the global alignment problem discussed in a previous talk. We can leverage dynamic programming and backtracking to solve this. 16 / 45

NP-Hardness of LAPCS(C ROSSING ,C ROSSING ) Theorem LAPCS(C ROSSING ,C ROSSING ) is an NP-hard optimization problem. Idea: Consider D EC LAPCS, the corresponding decision problem of LAPCS. Reduce input instance of C LIQUE to D EC LAPCS. 17 / 45

Recap: C LIQUE Problem Definition Let G = ( V , E ) be an undirected graph. A subset V ′ ⊆ V is called a clique , if every two vertices v i , v j ∈ V ′ , where v i � = v j are � � connected by an edge, i.e., v i , v j ∈ E . Definition (C LIQUE Decision Problem) Input: An undirected graph G = ( V , E ) and a positive integer k . Output: Y ES if G contains a clique V ′ of size k , N O , otherwise. Clique is a well-known NP-complete decision problem. 18 / 45

Example: C LIQUE Is there a clique for k = 3? v 2 v 1 v 3 v 5 v 4 19 / 45

Example: C LIQUE Is there a clique for k = 3? v 2 v 2 v 1 v 1 v 3 v 3 v 5 v 4 19 / 45

Arc-Annotated String Construction from Input-Graph v 2 v 1 v 3 v 5 v 4 S : b a a a a a b b a a a a a b b a a a a a b b a a a a a b b a a a a a b Block for v 1 Block for v 2 Block for v 3 Block for v 4 Block for v 5 20 / 45

Arc-Annotated String Construction from Input-Graph v 2 v 1 v 1 v 3 v 5 v 4 S : b a a a a a b b a a a a a b b a a a a a b b a a a a a b b a a a a a b Block for v 1 Block for v 2 Block for v 3 Block for v 4 Block for v 5 20 / 45

Arc-Annotated String Construction from Input-Graph v 2 v 1 v 3 v 5 v 4 S : b a a a a a b b a a a a a b b a a a a a b b a a a a a b b a a a a a b Block for v 1 Block for v 2 Block for v 3 Block for v 4 Block for v 5 20 / 45

Reduction construction formally Definition A undirected graph G = ( V , E ) , with | V | = n can be encoded as an arc-annotated string s = ( s , P s ) . � � n ba n b s = arcs encoding edges � �� P s = (( i − 1 )( n + 2 )+ j + 1 , ( j − 1 )( n + 2 )+ i + 1 ) |{ v i , v j } ∈ E ) ∪ { (( i − 1 )( n + 2 )+ 1 , i ( n + 2 )) | i ∈ { 1 ,..., n }} � �� arcs between two b ’s of a block 21 / 45

Analog: Construction of the Clique v 2 v 1 v 3 T : b a a a b b a a a b b a a a b Block for v i 1 Block for v i 2 Block for v i 3 Note that | T | = k · ( k + 2 ) , where k is the size of the clique. 22 / 45

Input for D EC LAPCS(C ROSSING ,C ROSSING ) Is there an arc-preserving common subsequence of size | T | ? S : b a a a a a b b a a a a a b b a a a a a b b a a a a a b b a a a a a b b a a a b a a a b b a a a b b T : 23 / 45

Proof (I): Polynominal Time Reduction Lemma The input ( S , T , | T | ) to D EC LAPCS(C ROSSING ,C ROSSING ) from ( G , k ) can be performed in polynomial time. • S can be directly constructed from G and has quadratic length in the number of vertices. • A fully connected graph G T of size k can be constructed in polynomial-time. • Analogously to S , now also T and | T | can be constructed in polynomial time by constructing a fully connected graph G T . 24 / 45

Proof (II): Correctness “ ⇒ ” Lemma Existence of a clique of size k in G implies existence of an arc-preserving common subsequence of S and T of size | T | . • Let { v i 1 ,..., v i k } be a clique of size k in the input graph. • We can align k blocks of S to the k blocks of T . • In each block again k symbols are matched to symbols at positions i 1 ,..., i k in the block of S . • Arcs between two b ’s are matched since we always map complete blocks to complete blocks • v i 1 ,..., v i k are vertices of a clique, thus their corresponding arcs between a ’s are spanned by a arcs. 25 / 45

Proof (III): Correctness “ ⇐ ” Lemma Existence of an arc-preserving common subsequence of S and T of size | T | implies a clique of size k in G. • | T | = k · ( k + 2 ) . • Due to arcs over b framing a block only blocks can be mapped to blocks. • T represents a clique of size k and blocks are constructed the same way as in S . • Thus i 1 ,..., i k blocks that are matched from T to S � � ⇒ v i 1 ,..., v i k is a clique of size k . 26 / 45

NP-hardness of LAPCS(N ESTED ,N ESTED ) Theorem LAPCS(N ESTED ,N ESTED ) is an NP-hard optimization problem. • Proof [Lin et al., 2002] not presented here due to many preliminaries. • Idea: Reduction to variant of Maximum Independent Set (cubic planar graph) using several graph transformations with book embedding. 27 / 45

Complexity Results Overview for LAPCS Classes P LAIN C HAIN N ESTED C ROSSING U NLIMITED U NLIMITED NP-hard C ROSSING NP-hard O ( nm 3 ) N ESTED NP-hard C HAIN O ( nm ) P LAIN O ( nm ) Table: Complexity Results for LAPCS(L EVEL 1,L EVEL 2 ) Due to hardness results: LAPCS approximation algorithms. 28 / 45

2-Approximation Algorithm for LAPCS(C ROSSING ,C ROSSING ) Idea: Use Longest Common Subsequence without arcs as a starting point and remove arc-conflicting parts successively. 2-Approximation Algorithm for LAPCS(C ROSSING ,C ROSSING ) Input: Two arc-annotated strings S = ( s , P s ) and T = ( t , P t ) with S , T ∈ C ROSSING . 1 Determine longest common subsequence w of s and t . Let ϕ a mapping consistent to w . 2 Construct the conflict-graph G ϕ from ϕ . 3 For each connected component in G ϕ delete every second vertex. 4 From the resulting graph G ϕ ′ construct output string w ′ 29 / 45

Construction of the Conflict-Graph Definition (Conflict-Graph) Given a mapping ϕ that is consistent with by the longest common subsequence w of s and t . G ϕ = ( V , E ) • V = {� i , j �|� i , j � ∈ ϕ } • E = {{� i 1 , j 1 � , � i 2 , j 2 �}| either ( i 1 , i 2 ) ∈ P s or ( j 1 , j 2 ) ∈ P t } Note: G ϕ describes position pairs that are not arc-preserving. 30 / 45

Conflict-Graph – Example ϕ = {� 1 , 1 � , � 3 , 2 � , � 4 , 3 � , � 6 , 5 � , � 7 , 6 � , � 8 , 7 � , � 9 , 9 � , � 10 , 10 � , � 11 , 11 � , � 12 , 12 � , � 13 , 13 � , � 14 , 14 � , � 15 , 15 � , � 16 , 16 � , � 17 , 18 � , � 18 , 19 �} A A C G G U A C – G U A C G U A C – G U S: A – C G U U A C G G U A C G U A C C G U T: 31 / 45

Conflict-Graph – Example ϕ = {� 1 , 1 � , � 3 , 2 � , � 4 , 3 � , � 6 , 5 � , � 7 , 6 � , � 8 , 7 � , � 9 , 9 � , � 10 , 10 � , � 11 , 11 � , � 12 , 12 � , � 13 , 13 � , � 14 , 14 � , � 15 , 15 � , � 16 , 16 � , � 17 , 18 � , � 18 , 19 �} A A C G G U A C – G U A C G U A C – G U S: A – C G U U A C G G U A C G U A C C G U T: 1 , 1 3 , 2 4 , 3 6 , 5 7 , 6 8 , 7 9 , 9 10 , 10 11 , 11 G ϕ : 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 18 , 19 31 / 45

Conflict-Graph – Example ϕ = {� 1 , 1 � , � 3 , 2 � , � 4 , 3 � , � 6 , 5 � , � 7 , 6 � , � 8 , 7 � , � 9 , 9 � , � 10 , 10 � , � 11 , 11 � , � 12 , 12 � , � 13 , 13 � , � 14 , 14 � , � 15 , 15 � , � 16 , 16 � , � 17 , 18 � , � 18 , 19 �} A A C G G U A C – G U A C G U A C – G U S: A – C G U U A C G G U A C G U A C C G U T: 1 , 1 1 , 1 3 , 2 4 , 3 6 , 5 7 , 6 8 , 7 9 , 9 10 , 10 11 , 11 G ϕ : 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 18 , 19 31 / 45

Conflict-Graph – Example ϕ = {� 1 , 1 � , � 3 , 2 � , � 4 , 3 � , � 6 , 5 � , � 7 , 6 � , � 8 , 7 � , � 9 , 9 � , � 10 , 10 � , � 11 , 11 � , � 12 , 12 � , � 13 , 13 � , � 14 , 14 � , � 15 , 15 � , � 16 , 16 � , � 17 , 18 � , � 18 , 19 �} A A C G G U A C – G U A C G U A C – G U S: A – C G U U A C G G U A C G U A C C G U T: 1 , 1 3 , 2 4 , 3 6 , 5 7 , 6 8 , 7 9 , 9 10 , 10 11 , 11 G ϕ : 12 , 12 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 18 , 19 31 / 45

Conflict-Graph Observation . . . A . . . B . . . C . . . ... ... i , j i , j S: . . . A . . . B . . . C . . . T: Lemma G ϕ has at most node degree two for two arc-annotated strings T , S ∈ C ROSSING . Proof. • Since T , S ∈ C ROSSING no two arcs share a common start/endpoint. • Incoming edge: w.l.o.g. at most one arc-mismatch for incoming edges • Outgoing edge: analogous. 32 / 45

Approximation Algorithm – Step 3 For each connected component in G ϕ delete every second vertex. 1 , 1 3 , 2 4 , 3 6 , 5 7 , 6 8 , 7 9 , 9 10 , 10 11 , 11 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 18 , 19 33 / 45

Approximation Algorithm – Step 3 For each connected component in G ϕ delete every second vertex. 1 , 1 1 , 1 3 , 2 3 , 2 4 , 3 4 , 3 6 , 5 6 , 5 7 , 6 8 , 7 9 , 9 10 , 10 11 , 11 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 18 , 19 33 / 45

Approximation Algorithm – Step 3 For each connected component in G ϕ delete every second vertex. 1 , 1 4 , 3 7 , 6 7 , 6 8 , 7 9 , 9 10 , 10 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 18 , 19 33 / 45

Approximation Algorithm – Step 3 For each connected component in G ϕ delete every second vertex. 1 , 1 4 , 3 7 , 6 8 , 7 9 , 9 10 , 10 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 18 , 19 17 , 18 18 , 19 33 / 45

Approximation Algorithm – Step 3 For each connected component in G ϕ delete every second vertex. 1 , 1 4 , 3 7 , 6 8 , 7 9 , 9 10 , 10 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 17 , 18 33 / 45

Approximation Algorithm – Step 3 For each connected component in G ϕ delete every second vertex. 1 , 1 4 , 3 7 , 6 8 , 7 9 , 9 10 , 10 G ′ ϕ : 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 33 / 45

Approximation Algorithm – Final Step Reconstruct corresponding arc-preserving common subsequence w ′ . 1 , 1 4 , 3 7 , 6 8 , 7 9 , 9 10 , 10 G ′ ϕ : 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 34 / 45

Approximation Algorithm – Final Step Reconstruct corresponding arc-preserving common subsequence w ′ . 1 , 1 4 , 3 7 , 6 8 , 7 9 , 9 10 , 10 G ′ ϕ : 12 , 12 13 , 13 14 , 14 15 , 15 16 , 16 17 , 18 A A C G G U A C – G U A C G U A C – G U S: A – C G U U A C G G U A C G U A C C G U T: 34 / 45

Correctness Proof (I) Theorem The Approximation algorithm computes a feasible solution for LAPCS(C ROSSING ,C ROSSING ) . Proof. • The string w ′ results from removing some symbols in w and thus is still a common subsequence. • Also, w ′ is arc-preserving: • Connected vertices in the conflict-graph G ϕ denoted violating position pairs. • The algorithm removes all edges from the conflict graph. 35 / 45

Structure-Based Comparison of Biomolecules Benedikt Christoph - PowerPoint PPT Presentation

Structure-Based Comparison of Biomolecules Benedikt Christoph Wolters Seminar Bioinformatics Algorithms RWTH AACHEN 07/17/2015 Outline 1 Introduction and Motivation Protein Structure Hierarchy Protein Data Bases 2 Arc-Annotated Sequences

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