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15. Dynamic Programming

15. Dynamic Programming. Heejin Park College of Information and Communications Hanyang University. Contents. Introduction Assembly-line scheduling Matrix-chain multiplication Elements of dynamic programming Longest common subsequence. Longest common subsequence. Definition Character

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15. Dynamic Programming

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  1. 15. Dynamic Programming Heejin Park College of Information and Communications Hanyang University

  2. Contents • Introduction • Assembly-line scheduling • Matrix-chain multiplication • Elements of dynamic programming • Longest common subsequence

  3. Longest common subsequence • Definition • Character • Alphabet: A set of characters • English alphabet: {A, B, …Z} • Korean alphabet: { ㄱ,ㄴ,…ㅎ, ㅏ,…ㅣ} • String (or sequence): A list of characters from an alphabet • ex> strings over {0,1}: Binary strings • ex> strings over {A,C,G,T}: DNA sequences

  4. Longest common subsequence • Substring • CBD is a substring of ABCBDAB • Subsequence • BCDB is a subsequence of ABCBDAB • Common subsequence • BCA is a common subsequence of X=ABCBDAB and Y=BDCABA

  5. Longest common subsequence • Longest common subsequence(LCS) • BCBA is the longest common subsequence of X and Y X = A B CB D A B Y = B D C A BA • LCS problem • Given two sequences X=<x1,x2,…,xm> and Y=<y1, y2,..,yn> to find an LCS of X and Y.

  6. Longest common subsequence • Brute force approach • Enumerate all subsequences of X and check each subsequence if it is also a subsequence of Y and find the longest one. • Infeasible! • The number of subsequences of X is 2m.

  7. Longest common subsequence • Dynamic programming • The ith prefixXi of X is Xi=<x1,x2,…,xi>. • If X = <A, B, C, B, D, A, B> • X4=< A, B, C, B> • X0=<>

  8. Longest common subsequence • Optimal substructure of an LCS • Let X = x1, x2, ..., xm and Y = y1, y2, ..., yn be sequences, and let Z = z1, z2, ..., zk be any LCS of X and Y. • If xm=yn, then zk=xm=yn and Zk-1 is an LCS of Xm-1 and Yn-1. • If xm≠yn, then zk≠xm implies Z is an LCS of Xm-1 and Y. • If xm≠yn, then zk≠yn implies Z is an LCS of X and Yn-1.

  9. Longest common subsequence • If xm=yn, then zk=xm=yn and Zk-1 is an LCS of Xm-1 and Yn-1. • Suppose zk ≠ xm. • Then, we could append xm = yn to Z to obtain a common subsequence of X and Y of length k + 1, Thus, zk= xm. X = A B C B D A B Y = B D C A B

  10. Longest common subsequence • If xm=yn, then zk=xm=yn and Zk-1 is an LCS of Xm-1 and Yn-1. • Suppose that there is a common subsequence W of Xm-1 and Yn-1 with length greater than k - 1. • Then, appending xm = yn to W produces a common subsequence of X and Y whose length is greater than k. X = A B C B D A B Y = B D C A B

  11. Longest common subsequence • If xm≠yn, then zk≠xm implies Z is an LCS of Xm-1 and Y. • Since zk≠ xm, Z is an LCS of Xm-1 and Y. X = A B C B D A B Y = B D C A A

  12. Longest common subsequence • If xm≠yn, then zk≠yn implies Z is an LCS of X and Yn-1. • symmetric to the proof of 2. X = A B C B D A B Y = B D C A A

  13. Longest common subsequence • If xm= yn, • Find an LCS of Xm-1 and Yn-1 and append xm= ynto this LCS. • If xm≠ yn • Find the longer between an LCS of Xm-1 and Y and an LCS of X and Yn-1.

  14. Longest common subsequence • c[i, j]: The length of an LCS of the sequences Xiand Yj. • If either i = 0 or j = 0, so the LCS has length = 0.

  15. Longest common subsequence j 0 1 2 3 4 5 6 i 0 1 2 3 4 5 6 7 ↑ 0 ↑ 0 ↑ 0 ↖ 1 ← 1 ↖ 1 0 ↖ 1 ← 1 ← 1 ↑ 1 ↖ 2 ← 2 0 ← 2 ↑ 1 ↑ 1 ↖ 2 ↑ 2 ↑ 2 0 ↖ 1 ↑ 2 ↑ 2 ↑ 1 ↖ 3 ← 3 0 ↑ 1 ↑ 2 ↑ 2 ↑ 3 ↖ 2 ↑ 3 0 ↖ 3 ↖ 4 ↑ 1 ↑ 2 ↑ 2 ↑ 3 0 ↖ 1 ↑ 2 ↑ 2 ↑ 3 ↖ 4 ↑ 4

  16. Longest common subsequence • Computation time: Θ(mn) • Space: Θ(mn) • Space reduction: Θ (min(m,n))

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