Chapter 6. Dynamic Programming. Modified from slides by Kevin Wayne. Copyright 2005 Pearson-Addison Wesley. All rights reserved.

Transcription

1 Chapter 6 Dynamic Programming Modified from slides by Kevin Wayne. Copyright 2005 Pearson-Addison Wesley. All rights reserved. 1

2 Think recursively (this week)!!! Divide & conquer and Dynamic programming are Recursive Q: What is recursion? How to think recursively? A: We have seen this in Divide & conquer: Size of the tree is O(n) T(n) Same problem different size, or parameter!! T(n/2) T(n/2) T(n/4) T(n/4) T(n/4) T(n/4) T(2) T(2) T(2) T(2) T(2) T(2) T(2) T(2) 2

3 reedy. Build up a solution incrementally, myopically optimizing some local criterion. Divide-and-conquer. Break up a problem into two sub-problems, solve each sub-problem independently, and combine solution to sub-problems to form solution to original problem. Dynamic programming. Break up a problem into a series of overlapping sub-problems, and build up solutions to larger and larger sub-problems. Network flow. Different story! 3

4 Dynamic Programming History Bellman. Pioneered the systematic study of dynamic programming in the 1950s. Kantorovich and Koopmans. Nobel prize in economics in 1975 Applications: Areas. Bioinformatics, Control theory, Information theory, Operations research, Computer science: theory, graphics, AI, systems,. Some famous dynamic programming algorithms. Viterbi for hidden Markov models. Smith-Waterman for sequence alignment. Bellman-Ford for shortest path routing in networks. Cocke-Kasami-Younger for parsing context free grammars. Optimal resource allocation.

5 6.1 Weighted Interval Scheduling (Introduction to Dynamic Programming) Problem Recursive formula Recursive algorithm: Top-down approach Brute force Memoizing Finding solution Running time Bottom-up approach. Define sub-problems Building up solutions Running time

6 Weighted Interval Scheduling Weighted interval scheduling problem. Job j starts at s j, finishes at f j, and has weight or value v j. Two jobs compatible if they don't overlap. oal: find maximum weight subset of mutually compatible jobs. a b c d e f g h Time 6

7 Unweighted Interval Scheduling Review Recall. reedy algorithm works if all weights are 1. Consider jobs in ascending order of finish time. Add job to subset if it is compatible with previously chosen jobs. Observation. reedy algorithm can fail spectacularly if arbitrary weights are allowed. weight = 999 b weight = 1 a Time 7

8 Weighted interval scheduling How to get a recursive formula? Start from an optimal solution, and try to learn about its structure. In dynamic programming we are trying reduce to an optimal solution of smaller problems.. 8

9 Weighted Interval Scheduling Notation. Label jobs by finishing time: f 1 f 2... f n. Def 1. OPT(j) = value of optimal solution to the problem consisting of job requests 1, 2,..., j. Def 2. p(j) = largest index i < j such that job i is compatible with j. Ex: p(8) = 5, p(7) = 3, p(2) = Time 9

10 Dynamic Programming: Binary Choice Case 1: OPT selects job j. can't use incompatible jobs { p(j) + 1, p(j) + 2,..., j - 1 } must include optimal solution to problem consisting of remaining compatible jobs 1, 2,..., p(j) optimal substructure Case 2: OPT does not select job j. must include optimal solution to problem consisting of remaining compatible jobs 1, 2,..., j-1 0 if j = 0 OPT( j) = max v j + OPT( p( j)), OPT( j 1) { } otherwise 10

11 Weighted Interval Scheduling: Brute Force Brute force algorithm. Input: n, s 1,,s n, f 1,,f n, v 1,,v n Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) Compute-Opt(j) { if (j = 0) return 0 else return max(v j } + Compute-Opt(p(j)), Compute-Opt(j-1)) 11

12 Weighted Interval Scheduling: Brute Force Observation. Recursive algorithm fails spectacularly because of redundant sub-problems exponential algorithms. Ex. Number of recursive calls for family of "layered" instances grows like Fibonacci sequence p(1) = 0, p(j) = j

13 Problem Recursive formula Recursive algorithm: Top-down approach Brute Force Memoizing Finding solution Running time Bottom-up approach. Define sub-problems Building up solutions Running time 13

14 Memoization Memoization. Store results of each sub-problem in a cache; lookup as needed. (Built-in in some Programming languages ) Input: n, s 1,,s n, f 1,,f n, v 1,,v n Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) for j = 1 to n M[j] = empty M[j] = 0 global array M-Compute-Opt(j) { if (M[j] is empty) M[j] = max(w j + M-Compute-Opt(p(j)), M-Compute-Opt(j-1)) return M[j] } 14

15 p(1) = 0, p(j) = j Input: n, s 1,,s n, f 1,,f n, v 1,,v n Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) for j = 1 to n M[j] = empty M[j] = 0 M-Compute-Opt(j) { if (M[j] is empty) M[j] = max(w j + M-Compute-Opt(p(j)), M-Compute-Opt(j-1)) return M[j] } 15

16 Running Time Claim. Memoized version of algorithm takes O(n log n) time. Sort by finish time: O(n log n). Computing p( ) : O(n) after sorting by start time. M-Compute-Opt(j): each invocation takes O(1) time and either (i) returns an existing value M[j] (ii) fills in one new entry M[j] and makes two recursive calls Progress measure Φ = # nonempty entries of M[]. initially Φ = 0, throughout Φ n. (ii) increases Φ by 1 at most 2n recursive calls. Overall running time of M-Compute-Opt(n) is O(n). Remark. O(n) if jobs are pre-sorted by start and finish times. 16

17 Finding a Solution Q. Dynamic programming algorithms computes optimal value. What if we want the solution itself? A. Do some post-processing. Run M-Compute-Opt(n) Run Find-Solution(n) Find-Solution(j) { if (j = 0) output nothing else if (v j + M[p(j)] > M[j-1]) print j Find-Solution(p(j)) else Find-Solution(j-1) } # of recursive calls n O(n). 17

18 Problem Recursive formula Recursive algorithm: Top-down approach Brute Force Memoizing Finding solution Running time Bottom-up approach. Define sub-problems Building up solutions Running time Standard, easier to explain! 18

19 Sub-problems: M[i] for every i (Optimal solution for problem containing intervals 1 to i) Building solutions: bottom-up! M[j] = max(v j + M[p(j)], M[j-1]) P(j)<j and j-1<j Running time: linear 19

20 Bottom-Up Bottom-up dynamic programming. Unwind recursion. Input: n, s 1,,s n, f 1,,f n, v 1,,v n Sort jobs by finish times so that f 1 f 2... f n. Compute p(1), p(2),, p(n) Iterative-Compute-Opt { M[0] = 0 for j = 1 to n M[j] = max(v j + M[p(j)], M[j-1]) } 20

21 Divide and conquer Top-down Bottom-up 21

22 Recursive algorithm: Top-down approach Brute Force Memoizing Finding solution Running time Bottom-up approach. Define sub-problems Using recursive formula to build up solutions 1. #sub-problems is polynomial 2. There is an order of the sub-problems 3. Compute big problem from smaller ones is easy. The main idea: Recursive formula is Big problem = f(sub-problems) Q:We need to find recursive formula first or define sub-problems? A: Chicken and egg question! Answered by practice. 22

23 Practice question Consider a similar problem: Weighted interval scheduling, with the condition that no point is covered by more than 2 intervals? Time

24 Dynamic Programming (cont.) Weighted Interval Scheduling 0 if j= 0 OPT( j)= max v j + OPT( p( j)), OPT( j 1) { } otherwise eneral Ingredient Recursive formula Same problems OPT(big problem )= f(opt(small prob_1),..., OPT( small prob_k)) Terminate: subproblems get smaller Running time is linear. Because: # sub-problems is linear, and it takes O(1) time to compute each step, Terminate: monoton decreasing in the structure of subproblems Running time: Need a bound on # sub-problems Computing at each recursion is easy. 24

25 Some dynamic programming algorithms yesterday -Weighted interval scheduling -Segmented Least Square -Problems on tree structures -Knapsack -Weighted interval scheduling with two resources -RNA secondary structure -Sequence Alignment (Edit distance) tomorrow -Shortest path with negative cost with many applications

26 Problem on tree structures Party Problem: Problem : Invite some colleagues for a party Maximize: the number of interesting people Constraint: everyone should be having fun (do not invite a colleague and his direct boss at the same time) Input: tree with weight on vertices w(u) Output: maximum weighted independent set

27 Problem on tree structures T[v]= tree rooted at v. A[v]= max weight of an independent set in T[v] B[v]= max weight of an independent set in T[v] that does not include v. B[v]= sum of A[u] for all children u of v 5 A[v]= max {B[v], w(v)+ sum of B[u] for all children u of v}

28 Weighted interval scheduling with two resources Having two resources, try to maximize total weight of jobs can be served Find a set of intervals with max total weight, and at any time there are at most 2 intervals overlapping 1 DEF 1: OPT(j)= optimal solution of job1 to job j overlappings RECURSION: Can we express OPT(8) in terms of OPT(7),OPT(6)..? Time

29 1 2 3 overlappings t 8 1 {1,2,4,6 } is OPT of a problem with a more general type of constraints DEF 2: OPT(j, t) = max of problem with interval 1 to interval j with the constraint as above Number of the sub-problems is polynomial : O(n^2) 29

30 1 2 (j-1,t) (j,t) 3 j-1 (p(j,t),s) j Decreasing subproblems 2 s t 1 RECURSION: OPT(j,t) = max{ OPT(j-1,t), OPT(p(j,t), s ) } Where p(j,t) is the biggest index of that can be satisfied with j under the constraint s: is the beginning time of j. 30

31 6.5 RNA Secondary Structure

32 32 RNA Secondary Structure RNA. String B = b 1 b 2 b n over alphabet { A, C,, U }. Secondary structure. RNA is single-stranded so it tends to loop back and form base pairs with itself. This structure is essential for understanding behavior of molecule. U C A A A C A U A U U A A C A A C U A U C A U C C C Ex: UCAUUACAAUUAACAACUCUACCAA complementary base pairs: A-U, C-

33 RNA Secondary Structure Secondary structure. A set of pairs S = { (b i, b j ) } that satisfy: [Watson-Crick.] S is a matching and each pair in S is a Watson- Crick complement: A-U, U-A, C-, or -C. [No sharp turns.] The ends of each pair are separated by at least 4 intervening bases. If (b i, b j ) S, then i < j - 4. [Non-crossing.] If (b i, b j ) and (b k, b l ) are two pairs in S, then we cannot have i < k < j < l. Free energy. Usual hypothesis is that an RNA molecule will form the secondary structure with the optimum total free energy. approximate by number of base pairs oal. iven an RNA molecule B = b 1 b 2 b n, find a secondary structure S that maximizes the number of base pairs. 33

34 RNA Secondary Structure: Examples Examples. C U C U C C C U A U A U A U A U A U A base pair A U U C C A U A U C A U 4 A U U C C A U ok sharp turn crossing 34

35 RNA Secondary Structure: Subproblems First attempt. OPT(j) = maximum number of base pairs in a secondary structure of the substring b 1 b 2 b j. match b t and b n 1 t n Difficulty. Results in two sub-problems. Finding secondary structure in: b 1 b 2 b t-1. Finding secondary structure in: b t+1 b t+2 b n-1. OPT(t-1) need more sub-problems 35

36 Dynamic Programming Over Intervals Notation. OPT(i, j) = maximum number of base pairs in a secondary structure of the substring b i b i+1 b j. Case 1. If i j - 4. OPT(i, j) = 0 by no-sharp turns condition. Case 2. Base b j is not involved in a pair. OPT(i, j) = OPT(i, j-1) Case 3. Base b j pairs with b t for some i t < j - 4. non-crossing constraint decouples resulting sub-problems OPT(i, j) = 1 + max t { OPT(i, t-1) + OPT(t+1, j-1) } take max over t such that i t < j-4 and b t and b j are Watson-Crick complements Remark. Same core idea in CKY algorithm to parse context-free grammars. 36

37 Bottom Up Dynamic Programming Over Intervals Q. What order to solve the sub-problems? A. Do shortest intervals first. RNA(b 1,,b n ) { for k = 5, 6,, n-1 for i = 1, 2,, n-k j = i + k Compute M[i, j] i } return M[1, n] using recurrence j Running time. O(n 3 ). 37