Dynamic Programming. Nothing to do with dynamic and nothing to do with programming.

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1 Dynamic Programming

2 Deliverables Dynamic Programming basics Binomial Coefficients Weighted Interval Scheduling Matrix Multiplication /1 Knapsack Longest Common Subsequence 6/12/212 6:56 PM gdeepak.com 2

3 Dynamic Programming Nothing to do with dynamic and nothing to do with programming. Actually the name should be bottom up approach without repeated calculations 6/12/212 6:56 PM gdeepak.com 3

4 Dynamic Programming paradigm Difference between Dynamic Programming and Divide and Conquer is that the sub problems in D&C are disjoint and distinct but in DP they are overlapping. Dynamic programming always gives a optimal solution as compared to the greedy paradigm which fails in some cases. Recursively solve sub problems, starting from the trivial case, and save their solutions in memory. In the end we get solution of the whole problem Optimal Substructure of the problems: The solution to the problem must be a composition of sub problem solutions 6/12/212 6:56 PM gdeepak.com 4

5 Dynamic Programming paradigm Divide and Conquer is top-down while DP is bottom up. It works when the number of substructures are polynomial in size of the input. Recursion uses a stack while DP uses a queue whenever required. In many cases we start with the base case and upper level sub problems are generated after doing some work on lower levels. 6/12/212 6:56 PM gdeepak.com 5

6 Pascal triangle or Binomial Coefficients 6/12/212 6:56 PM gdeepak.com 6

7 Pascal triangle or Binomial Coefficients 6/12/212 6:56 PM gdeepak.com 7

8 Pascal triangle or Binomial Coefficients 6/12/212 6:56 PM gdeepak.com 8

9 Binomial Coefficients n where n is the row of Pascal triangle m and m is the column of Pascal triangle. = n 1 m + n 1 m 1 Recursive Algorithm B(n, m) { If m= or m=n then set 1 else ret B(n-1,m) + B(n-1,m-1) } 6/12/212 6:56 PM gdeepak.com 9

10 Analysis Recurrence is T(size) = T(size-1)+T(size-2) This looks like Fibonacci equation. Nth Fibonacci number is an exponential, so we will end up in getting exponential number of calculations in doing recursive algorithm. If you want to calculate B(6,3) it will calculate B(3,2) 3 times, B(3,1) 3 times, B(2,1) 5 times, B(1,1) 6 times and B(1,) 6 times It can be done with an iterative function of factorials in linear time. 6/12/212 6:56 PM gdeepak.com 1

11 Weighted Interval Scheduling Greedy algorithm works if all weights are 1 and consider jobs in ascending order of finish time and add job to subset if it is compatible with previously chosen jobs. Greedy algorithm fails if arbitrary weights are allowed weight = 999 b weight = 1 a 6/12/212 6:56 PM gdeepak.com 11

12 Weighted Interval Scheduling Label jobs by finishing time: f 1 f 2... f n.p(j) = largest index i < j such that job i is compatible with j. p(8) = 5, p(7) = 3, p(2) = Time 6/12/212 6:56 PM gdeepak.com 12

13 Dynamic Programming: Key Technique OPT(j) = value of optimal solution to the problem consisting of job requests 1, 2,..., j. 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) Case 2: OPT does not select job j. must include optimal solution to problem consisting of remaining compatible jobs 1, 2,..., j-1 if j OPT( j) max v j OPT( p( j)), OPT( j 1) otherwise 6/12/212 6:56 PM gdeepak.com 13

14 WIS Brute Force 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= then return else return max(v j +Compute-opt(p(j)),Compute-opt(j-1)) } 6/12/212 6:56 PM gdeepak.com 14

15 Example of Weighted Interval Scheduling 6/12/212 6:56 PM gdeepak.com 15

16 Exponential Growth 6/12/212 6:56 PM gdeepak.com 16

17 Worst Case Instance 6/12/212 6:56 PM gdeepak.com 17

18 WIS 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) Memo-WIS(j) { if j= then return else if M[j] is not empty then return M[j] else M[j]= max(v j + Memo-WIS(p(j)), Memo-WIS(j-1)) return M[j] } 6/12/212 6:56 PM gdeepak.com 18

19 WIS Improved 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-wis(j) { M[]= for j=1,2,,n M[j]= max(v j + M(p(j)),M[j-1]) } 6/12/212 6:56 PM gdeepak.com 19

20 Example with Iterations 6/12/212 6:56 PM gdeepak.com 2

21 Complexity Analysis Running Time in O(nlogn), If we exclude sorting for finishing time then the remaining complexity is only O(n) 6/12/212 6:56 PM gdeepak.com 21

22 Multiplying two matrices For multiplication two matrices A and B number of columns of A must equal the number of rows of B. If A is a p q matrix and B is a q r matrix, the resulting matrix C is a p r matrix. The time to compute C is dominated by the number of scalar multiplications which is pqr Compute A=A *A 1 * *A n-1 A i is d i d i+1 Problem: How to parenthesize? Example B is 3 1 C is 1 5 D is 5 5 (B*C)*D takes = 1575 ops B*(C*D) takes = 4 ops 6/12/212 6:56 PM gdeepak.com 22

23 Matrix Multiplication Want to multiply matrices A x B x C x D x E We could parenthesize many ways (A x (B x (C x (D x E)))) ((((A x B) x C) x D) x E) Each different way has different number of multiplications A x B x C x D x E Calculate in advance the cost (multiplies) AB, BC, CD, DE Use those to find the cheapest way to form ABC, BCD, CDE and then ABCD BCDE From that derive best way to form ABCDE 6/12/212 6:56 PM gdeepak.com 23

24 Example Let matrices are 1x4,4x5,5x3,3x6,6x7,7x1 6/12/212 6:56 PM gdeepak.com 24

25 Example 6/12/212 6:56 PM gdeepak.com 25

26 Time complexity In the th Row or base row No of calucations - In the 1 st Row 1*n In the 2 nd Row- 2*n-1 In the 3 rd Row- 3* n-2 : In the nth Row n*1 Adding All the above n k=1 k(n k + 1) It comes out to be Ө(n 3 ) and the constant is 1/6 6/12/212 6:56 PM gdeepak.com 26

27 Keeping track of the solution Need to keep track of the Split point. For this we need to go backward from the final solution. Getting the actual sequence of solution is always recursive. For Example (M1M2) *(M3M4M5) ((M1)(M2)) * ((M3) (M4M5)) ((M1)(M2)) * ((M3) ((M4)(M5))) 6/12/212 6:56 PM gdeepak.com 27

28 /1 knapsack: Recursive Formulation S k : Set of items numbered 1 to k. Define B[k,w] = best selection from S k with weight exactly equal to w this does have sub problem optimality: B[ k, w] max{ B[ k B[ k 1, w] 1, w], B[ k 1, w w k ] b } k if w k else w 6/12/212 6:56 PM gdeepak.com 28

29 Running Time Running time: O(nW). Not a polynomial-time algorithm if W is large. This is a pseudo-polynomial time algorithm Algorithm 1Knapsack(S, W): Input: set S of items w/ benefit b i and weight w i ; max. weight W Output: benefit of best subset with weight at most W for w to W do B[w] for i = 1 to n B[i,] = for k 1 to n do for w W downto w k do if B[w-w k ]+b k > B[w] then B[w] B[w-w k ]+b k 6/12/212 6:56 PM gdeepak.com 29

30 Initial Setting of the /1 knapsack Item:W, P : 4,2 2: 2,3 3 : 2,6 4 : 6,25 5 :2,8 W=9 6/12/212 7:7 PM gdeepak.com 3

31 Considering Item Item:W, P : 4,2 2: 2,3 3 : 2,6 4 : 6,25 5 :2,8 W=9 6/12/212 7:8 PM gdeepak.com 31

32 Considering Item Item:W, P : 4,2 2: 2,3 3 : 2,6 4 : 6,25 5 :2,8 W=9 6/12/212 7:9 PM gdeepak.com 32

33 Considering Item Item:W, P : 4,2 2: 2,3 3 : 2,6 4 : 6,25 5 :2,8 W=9 6/12/212 7:9 PM gdeepak.com 33

34 Considering Item Item:W, P : 4,2 2: 2,3 3 : 2,6 4 : 6,25 5 :2,8 W=9 6/12/212 7:1 PM gdeepak.com 34

35 Considering Item Item:W, P : 4,2 2: 2,3 3 : 2,6 4 : 6,25 5 :2,8 W=9 6/12/212 7:1 PM gdeepak.com 35

36 Time Complexity Complexity is Ө(nM) where M is the size of the knapsack. It can be written as Ө(n2 s ) where s is the size of the input. It is not polynomial but exponential. 6/12/212 6:56 PM gdeepak.com 36

37 Longest common subsequence (LCS) For a sequence X = x 1, x 2,, x n, a subsequence is a subset of the sequence defined by a set of increasing indices (i 1, i 2,, i k ) where 1 i 1 < i 2 < < i k n X = A B A C D A B A B ABA? 6/12/212 6:56 PM gdeepak.com 37

38 Longest common subsequence (LCS) For a sequence X = x 1, x 2,, x n, a subsequence is a subset of the sequence defined by a set of increasing indices (i 1, i 2,, i k ) where 1 i 1 < i 2 < < i k n X = A B A C D A B A B ABA 6/12/212 6:56 PM gdeepak.com 38

39 Longest common subsequence (LCS) For a sequence X = x 1, x 2,, x n, a subsequence is a subset of the sequence defined by a set of increasing indices (i 1, i 2,, i k ) where 1 i 1 < i 2 < < i k n X = A B A C D A B A B ACA? 6/12/212 6:56 PM gdeepak.com 39

40 Longest common subsequence (LCS) For a sequence X = x 1, x 2,, x n, a subsequence is a subset of the sequence defined by a set of increasing indices (i 1, i 2,, i k ) where 1 i 1 < i 2 < < i k n X = A B A C D A B A B ACA 6/12/212 6:56 PM gdeepak.com 4

41 Longest common subsequence (LCS) For a sequence X = x 1, x 2,, x n, a subsequence is a subset of the sequence defined by a set of increasing indices (i 1, i 2,, i k ) where 1 i 1 < i 2 < < i k n X = A B A C D A B A B DCA? 6/12/212 6:56 PM gdeepak.com 41

42 Longest common subsequence (LCS) For a sequence X = x 1, x 2,, x n, a subsequence is a subset of the sequence defined by a set of increasing indices (i 1, i 2,, i k ) where 1 i 1 < i 2 < < i k n X = A B A C D A B A B DCA 6/12/212 6:56 PM gdeepak.com 42

43 Longest common subsequence (LCS) For a sequence X = x 1, x 2,, x n, a subsequence is a subset of the sequence defined by a set of increasing indices (i 1, i 2,, i k ) where 1 i 1 < i 2 < < i k n X = A B A C D A B A B AADAA? 6/12/212 6:56 PM gdeepak.com 43

44 Longest common subsequence (LCS) For a sequence X = x 1, x 2,, x n, a subsequence is a subset of the sequence defined by a set of increasing indices (i 1, i 2,, i k ) where 1 i 1 < i 2 < < i k n X = A B A C D A B A B AADAA 6/12/212 6:56 PM gdeepak.com 44

45 LCS problem Given two sequences X and Y, a common subsequence is a subsequence that occurs in both X and Y Given two sequences X = x 1, x 2,, x n and Y = y 1, y 2,, y n, What is the longest common subsequence? X = A B C B D A B Y = B D C A B A 6/12/212 6:56 PM gdeepak.com 45

46 LCS problem Given two sequences X and Y, a common subsequence is a subsequence that occurs in both X and Y Given two sequences X = x 1, x 2,, x n and Y = y 1, y 2,, y n, What is the longest common subsequence? X = A B C B D A B Y = B D C A B A 6/12/212 6:56 PM gdeepak.com 46

47 Step 1: Define the problem with respect to subproblems X = A B C B D A B Y = B D C A B A 6/12/212 6:56 PM gdeepak.com 47

48 Step 1: Define the problem with respect to subproblems X = A B C B D A? Y = B D C A B? Is the last character part of the LCS? 6/12/212 6:56 PM gdeepak.com 48

49 Step 1: Define the problem with respect to subproblems X = A B C B D A? Y = B D C A B? Two cases: either the characters are the same or they re different 6/12/212 6:56 PM gdeepak.com 49

50 Step 1: Define the problem with respect to subproblems X = A B C B D A A LCS Y = B D C A B A The characters are part of the LCS LCS If they re the same ( X, Y) LCS( X1... n 1, Y1... m 1) x n 6/12/212 6:56 PM gdeepak.com 5

51 Step 1: Define the problem with respect to subproblems X = A B C B D A B LCS Y = B D C A B A If they re different LCS( X, Y) LCS( X 1... n 1, Y) 6/12/212 6:56 PM gdeepak.com 51

52 Step 1: Define the problem with respect to subproblems X = A B C B D A B LCS Y = B D C A B A If they re different LCS( X, Y) LCS( X, Y1... m 1) 6/12/212 6:56 PM gdeepak.com 52

53 Step 1: Define the problem with respect to subproblems X = A B C B D A B Y = B D C A B A X = A B C B D A B? Y = B D C A B A If they re different 6/12/212 6:56 PM gdeepak.com 53

54 Step 1: Define the problem with respect to subproblems X = A B C B D A B Y = B D C A B A LCS( X, Y) 1 LCS( X1... n 1, Y1... m 1) max( LCS( X1... n 1, Y), LCS( X, Y 1... m 1 ) if x n y m otherwise 6/12/212 6:57 PM gdeepak.com 54

55 Step 2: Build the solution from the bottom up LCS( X, Y) 1 LCS( X1... n 1, Y1... m 1) max( LCS( X1... n 1, Y), LCS( X, Y 1... m 1 ) if x n y m otherwise What types of subproblem solutions do we need to store? LCS(X 1 j, Y 1 k ) two different indices 6/12/212 6:57 PM gdeepak.com 55

56 Step 2: Build the solution from the bottom up LCS( X, Y) 1 LCS( X1... n 1, Y1... m 1) max( LCS( X1... n 1, Y), LCS( X, Y 1... m 1 ) if x n y m otherwise What types of subproblem solutions do we need to store? LCS(X 1 j, Y 1 k ) LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j 6/12/212 6:57 PM gdeepak.com 56

57 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i j y j B D C A B A x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B 6/12/212 6:57 PM gdeepak.com 57

58 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B j y j B D C A B A 6/12/212 6:57 PM gdeepak.com 58

59 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B j y j B D C A B A? LCS(A, B) 6/12/212 6:57 PM gdeepak.com 59

60 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B j y j B D C A B A 6/12/212 6:57 PM gdeepak.com 6

61 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B j y j B D C A B A? LCS(A, BDCA) 6/12/212 6:57 PM gdeepak.com 61

62 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B j y j B D C A B A 1 LCS(A, BDCA) 6/12/212 6:57 PM gdeepak.com 62

63 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B j y j B D C A B A ? LCS(ABCB, BDCAB) 6/12/212 6:57 PM gdeepak.com 63

64 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B j y j B D C A B A LCS(ABCB, BDCAB) 6/12/212 6:57 PM gdeepak.com 64

65 LCS[ i, j] 1 LCS[ i max( LCS[ i 1, 1, j 1] j], LCS[ i, j 1] if x i y otherwise j i x i 1 A 2 B 3 C 4 B 5 D 6 A 7 B j y j B D C A B A Where s the final answer? 6/12/212 6:57 PM gdeepak.com 65

66 Questions, Comments and Suggestions 6/12/212 6:57 PM gdeepak.com 66

67 Question 1 Any Dynamic Programming algorithm with n sub problems will run in O(n) time 6/12/212 6:57 PM gdeepak.com 67

68 Question 2 Memoization is the basis for a top-down alternative to the usual bottom-up version of dynamic programming. 6/12/212 6:57 PM gdeepak.com 68

69 Question 3 The Floyd-Warshall algorithm solves the all-pairs shortestpaths problem using dynamic programming. 6/12/212 6:57 PM gdeepak.com 69

CMSC 451: Lecture 10 Dynamic Programming: Weighted Interval Scheduling Tuesday, Oct 3, 2017

CMSC 451: Lecture 10 Dynamic Programming: Weighted Interval Scheduling Tuesday, Oct 3, 2017 CMSC 45 CMSC 45: Lecture Dynamic Programming: Weighted Interval Scheduling Tuesday, Oct, Reading: Section. in KT. Dynamic Programming: In this lecture we begin our coverage of an important algorithm design