# 09 - Algorithms – Sequence Alignment Sequence Alignment...

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Unformatted text preview: Algorithms – Sequence Alignment Sequence Alignment II Design and Analysis of Algorithms Andrei Bulatov Algorithms – Sequence Alignment 9-2 Alignments Let and be two strings A matching is a set of ordered pairs, such that an element of each set occurs at most once. A matching is an alignment if there no crossing pairs: (i,j) and (i’,j’) are in the matching and i < i’ then j < j’ m x x x X , , , 2 1 K = n y y y Y , , , 2 1 K = if (i,j) and (i’,j’) are in the matching and i < i’ then j < j’ o c u r r a n c e o c u r r e n c e c o c u r r a n c e o c u r r e n c e c Algorithms – Sequence Alignment 9-3 The Problem Let M be an alignment between X and Y. Each position of X or Y that is not matched in M is called a gap . Each pair (i,j) ∈ M such that is called a mismatch The cost of M is given as follows: here is 0, a p penalty For each gap in M we incur a cost j i y x ≠- There is δ > 0, a gap penalty . For each gap in M we incur a cost of δ- For each pair of letters p,q in the alphabet, there is a mismatch cost For each (i,j) ∈ M we pay the mismatch cost Usually,- The cost of M is the sum its gap penalties and mismatch costs . pq α . j i y x α . = pp α Algorithms – Sequence Alignment 9-4 The Problem (cntd) The Sequence Alignment Problem Instance : Sequences X and Y Objective : ind an alignment between X and Y of minimal cost. Find an alignment between X and Y of minimal cost. Algorithms – Sequence Alignment 9-5 Graph Based Approach Having and construct a square grid-like graph m x x x X , , , 2 1 K = n y y y Y , , , 2 1 K = 3 x XY G Lemma Let f(i,j) denote the minimum weight of a path from (0,0) to (i,j) in Then for all i,j, we have f(i,j) = OPT(i,j) 1 x 2 x 1 y 2 y 3 y 4 y Weights: δ on each horizontal or vertical arc on the diagonal arc from (i,j) to (i + 1, j + 1) j i y x α . XY G Algorithms – Sequence Alignment 9-6 Backward Search We introduce another function related to OPT Let g(i,j) denote the length of a shortest path from (i,j) to (m,n) 2 x 3 x Lemma Then for all i,j, we have 1 x 1 y 2 y 3 y 4 y )} 1 , ( ), , 1 ( ), 1 , 1 ( min{ ) , ( 1 1 + + + + + + + = + + j i g j i g j i g j i g j i y x δ δ α Algorithms – Sequence Alignment 9-7 Backward Search (cntd) Lemma The length of the shortest corner-corner path in that passes through (i,j) is f(i,j) + g(i,j) Proof t k denote the length of a shortest corner rner path that passes XY G Let k denote the length of a shortest corner-to-corner path that passes through (i,j) It splits into to parts: from (0,0) to (i,j), and from (i,j) to (m,n) The length of the first part is ≥ f(i,j), the length of the second...
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## This note was uploaded on 11/11/2009 for the course CS 405/705 taught by Professor Bulatov during the Fall '09 term at Simon Fraser.

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09 - Algorithms – Sequence Alignment Sequence Alignment...

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