1. Dynamic Programming Kun-Mao Chao ( 趙坤茂 ) Department of Computer Science and Information Engineering National Taiwan University, Taiwan E-mail: kmchao@csie.ntu.edu.tw WWW: http://www.csie.ntu.edu.tw/~kmchao

2. 2 Dynamic Programming Dynamic programming is a class of solution methods for solving sequential decision problems with a compositional cost structure. Richard Bellman was one of the principal founders of this approach.

3. 3 Two key ingredients Two key ingredients for an optimization problem to be suitable for a dynamic- programming solution: Each substructure is optimal. (Principle of optimality) 1. optimal substructures 2. overlapping subproblems Subproblems are dependent. (otherwise, a divide-and- conquer approach is the choice.)

4. 4 Three basic components The development of a dynamic- programming algorithm has three basic components: –The recurrence relation (for defining the value of an optimal solution); –The tabular computation (for computing the value of an optimal solution); –The traceback (for delivering an optimal solution).

5. 5 Fibonacci numbers.for 21 1 1 0 0       i>1 i F i F i F F F The Fibonacci numbers are defined by the following recurrence:

6. 6 How to compute F 10 ？ F 10 F9F9 F8F8 F8F8 F7F7 F7F7 F6F6 ……

7. 7 Tabular computation The tabular computation can avoid recompuation. F0F0 F1F1 F2F2 F3F3 F4F4 F5F5 F6F6 F7F7 F8F8 F9F9 F 10 011235813213455

8. 8 Longest increasing subsequence(LIS) The longest increasing subsequence is to find a longest increasing subsequence of a given sequence of distinct integers a 1 a 2 …a n. e.g. 9 2 5 3 7 11 8 10 13 6 23 7 57 10 13 97 11 3 5 11 13 are increasing subsequences. are not increasing subsequences. We want to find a longest one.

9. 9 A naive approach for LIS Let L[i] be the length of a longest increasing subsequence ending at position i. L[i] = 1 + max j = 0..i-1 {L[j] | a j < a i } (use a dummy a 0 = minimum, and L[0]=0) 9 2 5 3 7 11 8 10 13 6 L[i] 1 1 2 2 3 4 ?

10. 10 A naive approach for LIS 9 2 5 3 7 11 8 10 13 6 L[i] 1 1 2 2 3 4 4 5 6 3 L[i] = 1 + max j = 0..i-1 {L[j] | a j < a i } The maximum length The subsequence 2, 3, 7, 8, 10, 13 is a longest increasing subsequence. This method runs in O(n 2 ) time.

11. 11 Binary search Given an ordered sequence x 1 x 2... x n, where x 1 <x 2 <... <x n, and a number y, a binary search finds the largest x i such that x i < y in O(log n) time. n... n/2 n/4

12. 12 Binary search How many steps would a binary search reduce the problem size to 1? n n/2 n/4 n/8 n/16... 1 How many steps? O(log n) steps.

13. 13 An O(n log n) method for LIS Define BestEnd[k] to be the smallest number of an increasing subsequence of length k. 9 2 5 3 7 11 8 10 13 6 922 5 2 3 2 3 7 2 3 7 11 2 3 7 8 2 3 7 8 10 2 3 7 8 13 BestEnd[1] BestEnd[2] BestEnd[3] BestEnd[4] BestEnd[5] BestEnd[6]

14. 14 An O(n log n) method for LIS Define BestEnd[k] to be the smallest number of an increasing subsequence of length k. 9 2 5 3 7 11 8 10 13 6 922 5 2 3 2 3 7 2 3 7 11 2 3 7 8 2 3 7 8 10 2 3 7 8 13 2 3 6 8 10 13 BestEnd[1] BestEnd[2] BestEnd[3] BestEnd[4] BestEnd[5] BestEnd[6] For each position, we perform a binary search to update BestEnd. Therefore, the running time is O(n log n).

15. 15 Longest Common Subsequence (LCS) A subsequence of a sequence S is obtained by deleting zero or more symbols from S. For example, the following are all subsequences of “president”: pred, sdn, predent. The longest common subsequence problem is to find a maximum-length common subsequence between two sequences.

16. 16 LCS For instance, Sequence 1: president Sequence 2: providence Its LCS is priden. president providence

17. 17 LCS Another example: Sequence 1: algorithm Sequence 2: alignment One of its LCS is algm. a l g o r i t h m a l i g n m e n t

18. 18 How to compute LCS? Let A=a 1 a 2 …a m and B=b 1 b 2 …b n. len(i, j): the length of an LCS between a 1 a 2 …a i and b 1 b 2 …b j With proper initializations, len(i, j)can be computed as follows.

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23. 23 Longest Common Increasing Subsequence Proposed by Yang, Huang and Chao –IPL 2005 Improvement for some special case: –Katriel and Kutz (March 2005) –Chan, Zhang, Fung, Ye and Zhu (ISAAC 2005)