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CS 575 Design and Analysis of Computer Algorithms Professor Michal Cutler Lecture 12 March 7, 2005

CS 575 Design and Analysis of Computer Algorithms Professor Michal Cutler Lecture 12 March 7, 2005. This class. Longest common subsequence Edit distance. Principle of optimality - shortest path problem. Problem: Given a graph G and vertices s and t, find a shortest path in G from s to t

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CS 575 Design and Analysis of Computer Algorithms Professor Michal Cutler Lecture 12 March 7, 2005

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  1. CS 575Design and Analysis ofComputer AlgorithmsProfessor Michal CutlerLecture 12March 7, 2005

  2. This class • Longest common subsequence • Edit distance

  3. Principle of optimality - shortest path problem • Problem: Given a graph G and vertices s and t, find a shortest path in G from s to t • Theorem: A subpath from s’ to t’ of a shortest path P is also a shortest path. • Proof: If there was a a shorter path from s’ to t’, we can cut the existing subpath from s’ to t’ in P and paste into P the shorter one. • The result would be a better solution than the optimal one.

  4. A problem that does not satisfy the Principle of Optimality • Problem: What is the longest simple route between City A and B? • Simple = never visit the same spot twice. • The longest simple route (solid line consisting of edges (A,C), (C, D) and (D, B)) has city C as an intermediate city. • It does not consist of the longest simple route from A to C plus the longest simple route from C to B. D Longest A to C B A C Longest A to B

  5. Biological Applications • Compare the DNA of two or more organisms • How similar are the two strands? • Is one a substring of the other? • Find a new longest strand in which the bases (A, C, G, T) appear in the same order as in the original 2 strands?

  6. Longest Common Subsequence (LCS) • Problem: Given sequences x[1..m] and y[1..n], find a longest common subsequence of both. • Example: x=ABCBDAB and y=BDCABA, • BCA is a common subsequence and • BCBA and BDAB are two LCSs

  7. LCS • Brute force solution • Writing a recurrence equation • The dynamic programming solution • Application of algorithm

  8. Brute force solution • Solution: For every subsequence of x, check if it is a subsequence of y. • Analysis : • There are 2m subsequences of x. • Each check takes O(n) time, since we scan y for first element, and then scan for second element, etc. • The worst case running time is O(n2m) or (2m).

  9. Writing the recurrence equation • Let Xi denote the ithprefix x[1,..i] of x[1..,m], and • X0 denotes an empty prefix • We will first compute the length of an LCS of Xm and Yn, LenLCS(m, n), and then use information saved during the computation for finding the actual subsequence • We need a recursive formula for computing LenLCS(i, j).

  10. Writing the recurrence equation • If Xi and Yjend with the same character xi=yj, the LCS must include the character. If it did not we could get a longer LCS by adding the common character. • If Xi and Yjdo not end with the same character there are two possibilities: • either the LCS does not end with xi, • or it does not end with yj • Let Zk denote an LCS of Xi and Yj

  11. x1 x2 … xi-1xi Xi Yj y1 y2 … yj-1yj=xi Zk z1 z2…zk-1zk=yj=xi Zk is Zk -1 followed by zk = yj = xi where Zk-1 is an LCS of Xi-1 and Yj -1 and LenLCS(i, j)=LenLCS(i-1, j-1)+1 Xiand Yjend with xi=yj

  12. x1 x2 … xi-1 xi x1 x2 … xi-1 x i Xi Xi Yj Yj y1 y2 … yj-1 yj yj y1 y2 …yj-1 yj Zk Zk z1 z2…zk-1 zk ¹yj z1 z2…zk-1 zk ¹xi Xiand Yjend with xi¹ yj Zk is an LCS of Xi and Yj -1 Zk is an LCS of Xi -1 and Yj LenLCS(i, j)=max{LenLCS(i, j-1), LenLCS(i-1, j)}

  13. The recurrence equation

  14. The dynamic programming solution • Initialize the first row and the first column of the matrix LenLCS to 0 • Calculate LenLCS (1, j) for j = 1,…, n • Then the LenLCS (2, j) for j = 1,…, n, etc. • Store also in a table an arrow pointing to the array element that was used in the computation. • It is easy to see that the computation is O(mn)

  15. LCS-Length(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

  16. LCS-Length(X, Y) cont. 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

  17. Example To find an LCS follow the arrows, each diagonal one denotes a member of the LCS

  18. Edit distance • Given two strings s and t • Edit distance = the minimum number of basic operations to covert one to the other • Basic operations are typically character-level • Insert • Delete • Replace • Often include also transposition • http://www.merriampark.com/ld.htm

  19. Dynamic programming for edit distance • Let s[1, 2, ..., m] and t[1, 2, ..., n] be the two strings. The recurrence equation is: • r(i, j) =0 when s[i] = t[j], otherwise 1

  20. The Greedy Method • Greedy algorithms make goodlocalchoices in the hope that they result in an optimal solution. • They result in feasible solutions. • Not necessarily optimal ones. • A proof is needed to show that the algorithm finds an optimal solution. • A counterexample shows that the greedy algorithm does not provide an optimal solution.

  21. Pseudo-code for Greedy Algorithm set Greedy (Set Candidate){ solution= new Set( ); while (Candidate.isNotEmpty()) { next = Candidate.select(); //use selection criteria, //remove from Candidate and return value if (solution.isFeasible( next)) //constraints satisfied solution.union( next); if (solution.solves()) return solution} //No more candidates and no solution return null }

  22. Pseudo code for greedy cont. • select() chooses a candidate based on a local selection criteria, removes it from Candidate, and returns its value. • isFeasible() checks whether adding the selected value to the current solution can result in a feasible solution (no constraints are violated). • solves() checks whether the problem is solved.

  23. Elements of the Greedy Strategy Cast problem as one in which we make a greedy choice and are left with one subproblem to solve. To show optimality: • Prove there is always an optimal solution to original problem that makes the greedy choice.

  24. Elements of the Greedy Strategy 2. Demonstrate that what remains is a subproblem with property: If we combine the optimal solution of the subproblem with the greedy choice we have an optimal solution to original problem.

  25. Activity Selection • Given a set S of nactivities with start time si and finish time fi of activity i • Find a maximum size subset A of compatible activities (maximum number of activities). • Activities are compatible if they do not overlap • Can you suggest a greedy choice?

  26. Example Activities 1 2 3 4 5 6 7 11 13 2 12 3 10 11 15 3 7 1 4 0 2 Time 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

  27. Counter Example 1 • Select by start time Activities 1 2 3 11 15 1 4 0 15 Time 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

  28. Counter Example 2 • Select by minimum duration Activities 1 2 3 8 15 1 8 7 9 Time 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

  29. Select by finishing time Activities 1 2 3 4 5 6 7 11 13 2 12 3 10 11 15 3 7 1 4 0 2 Time 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

  30. Next class Greedy algorithms

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