dicmentndnsnehffesolution.pptbbbbbbbbbbbb

dicmentndnsnehffesolution.pptbbbbbbbbbbbb

• 1.
  1 Longest Common Subsequence AndreasKlappenecker
• 2.
  Subsequences Suppose you havea sequence X = < x1,x2,…,xm> of elements over a finite set S. A sequence Z = < z1,z2,…,zk> over S is called a subsequence of X if and only if it can be obtained from X by deleting elements. Put differently, there exist indices i1<i2 <…<ik such that za = xia for all a in the range 1<= a <= k. 2
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  Common Subsequences Suppose thatX and Y are two sequences over a set S. We say that Z is a common subsequence of X and Y if and only if • Z is a subsequence of X • Z is a subsequence of Y 3
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  The Longest Common SubsequenceProblem Given two sequences X and Y over a set S, the longest common subsequence problem asks to find a common subsequence of X and Y that is of maximal length. 4
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  5 Naïve Solution Let Xbe a sequence of length m, and Y a sequence of length n. Check for every subsequence of X whether it is a subsequence of Y, and return the longest common subsequence found. There are 2m subsequences of X. Testing a sequences whether or not it is a subsequence of Y takes O(n) time. Thus, the naïve algorithm would take O(n2m ) time.
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  Divide and Conquer Canwe use divide-and-conquer to solve this problem? 6
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  Dynamic Programming Let ustry to develop a dynamic programming solution to the LCS problem. 7
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  Prefix Let X =< x1,x2,…,xm> be a sequence. We denote by Xi the sequence Xi = < x1,x2,…,xi> and call it the ith prefix of X. 8
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  LCS Notation Let Xand Y be sequences. We denote by LCS(X, Y) the set of longest common subsequences of X and Y. 9
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  Optimal Substructure Let X= < x1,x2,…,xm> and Y = < y1,y2,…,yn> be two sequences. Let Z = < z1,z2,…,zk> is any LCS of X and Y. a) If xm = yn then certainly xm = yn = zk and Zk-1 is in LCS(Xm-1 , Yn-1) 10
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  Optimal Substructure (2) LetX = < x1,x2,…,xm> and Y = < y1,y2,…,yn> be two sequences. Let Z = < z1,z2,…,zk> is any LCS of X and Y. b) If xm <> yn then xm <> zk implies that Z is in LCS(Xm-1 , Y) c)If xm <> yn then yn <> zk implies that Z is in LCS(X, Yn-1) 11
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  Overlapping Subproblems If xm= yn then we solve the subproblem to find an element in LCS(Xm-1 , Yn-1 ) and append xm If xm <> yn then we solve the two subproblems of finding elements in LCS(Xm-1 , Yn-1 ) and LCS(Xm-1 , Yn-1 ) and choose the longer one. 12
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  Recursive Solution Let Xand Y be sequences. Let c[i,j] be the length of an element in LCS(Xi, Yj). c[i,j] = 13
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  Dynamic Programming Solution Tocompute length of an element in LCS(X,Y) with X of length m and Y of length n, we do the following: •Initialize first row and first column of the array c with 0. •Calculate c[1,j] for 1 <= j <= n, c[2,j] for 1 <= j <= n … •Return c[m,n] •Complexity O(mn). 14
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  Dynamic Programming Solution(2) How can we get an actual longest common subsequence? Store in addition to the array c an array b pointing to the optimal subproblem chosen when computing c[i,j]. 15
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  16 Example yj B DC A xj 0 0 0 0 0 A 0 0 0 0 1 B 0 1 1 1 1 C 0 1 1 2 2 B 0 1 1 2 2 Start at b[m,n]. Follow the arrows. Each diagonal array gives one element of the LCS.
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  Animation http://wordaligned.org/articles/longest-c ommon-subsequence 17
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  18 LCS(X,Y) m  length[X] n length[Y] for i  1 to m do c[i,0]  0 for j  1 to n do c[0,j]  0
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  19 LCS(X,Y) for i 1 to m do for j  1 to n do if xi = yj c[i, j]  c[i-1, j-1]+1 b[i, j]  “D” else if c[i-1, j]  c[i, j-1] c[i, j]  c[i-1, j] b[i, j]  “U” else c[i, j]  c[i, j-1] b[i, j]  “L” return c and b
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  Greedy Algorithms There existsa greedy solution to this problem that can be advantageous when the size of the alphabet S is small. 20