Dynamic Programming. Andreas Klappenecker. [partially based on slides by Prof. Welch] Monday, September 24, 2012

Transcription

1 Dynmic Progrmming Andres Klppenecker [prtilly bsed on slides by Prof. Welch] 1

2 Dynmic Progrmming Optiml substructure An optiml solution to the problem contins within it optiml solutions to subproblems. Overlpping subproblems The spce of subproblem is smll so tht the recursive lgorithm hs to solve the sme problems over nd over. 2

3 Giving Optiml Chnge 3

4 Motivtion We hve discussed greedy lgorithm for giving chnge. However, the greedy lgorithm is not optiml for ll denomintions. Cn we design n lgorithm tht will give the minimum number of coins s chnge for ny given mount? Answer: Yes, using dynmic progrmming. 4

5 Dynmic Progrmming Tsk For dynmic progrmming, we hve to find some subproblems tht might help in solving the coin-chnge problem. Ide: Vry mount Restrict the vilble coins 5

6 Initil Set Up Suppose we wnt to compute the minimum number of coins with vlues v[1]>v[2]> >v[n]=1 to give chnge for n mount C. Let us cll the (i,j)-problem the problem of computing minimum number of coins with vlues v[i]>v[i+1]> >v[n]=1 to give chnge for n mount 1<= j <= C. The originl problem is the (1,C)-problem. 6

7 Tbultion Let m[i][j] denote the solution to the (i,j)-problem. Thus, m[i][j] denotes the minimum number of coins to mke chnge for the mount j using coins with vlues v[i],,v[n]. 7

8 Tbultion Exmple Denomintion v[1]=1, v[2]=6, v[3]=1 Tble of m[i][j] vlues: 8

9 A Simple Observtion In clculting m[i][j], notice tht: )Suppose we do not use the coin with vlue v[i] in the solution of the (i,j)-problem, then m[i][j] = m[i+1][j] b)suppose we use the coin with vlue v[i] in the solution of the (i,j)-problem, then m[i][j] = 1 + m[i][ j-v[i] ] 9

10 Tbultion Exmple Denomintion v[1]=1, v[2]=6, v[3]=1 Tble of m[i][j] vlues: 1

11 Recurrence We either use coin with vlue v[i] in the solution or we don t. m[i][j] = m[i+1][j] if v[i]>j min{ m[i+1][j], 1+m[i][ j-v[i] ] } otherwise 11

12 DP Coin-Chnge Algorithm Dynmic_Coin_Chnge(C,v,n) llocte rry m[1..n][..c]; for(i = ; i<=c, i++) m[n][i] = i; // mke chnge for mount i using coins of vlue v[n]=1. for(i=n-1; i=>1; i--) { // successively llow lrger number coin vlues for(j=; j<=c; j++) { // clc vlues of the rry. if( v[i]>j m[i+1][j]<1+m[i][j v[i]] ) m[i][j] = m[i+1][j]; // lrge coin does not help else m[i][j] = 1+m[i][j v[i]]; // } } return &m; 12

13 Question The previous lgorithm llows us to find the minimum number of coins. How cn you modify the lgorithm to ctully compute the chnge (i.e., the multiplicities of the coins)? 13

14 Mtrix Chin Algorithm 14

15 Mtrices An n x m mtrix A over the rel numbers is rectngulr rry of nm rel numbers tht re rrnged in n rows nd m columns. For exmple, 3 x 2 mtrix A hs 6 entries A = where ech of the entries ij is rel 15

16 Definition of Mtrix Multipliction Let A be n n x m mtrix B n m x p mtrix The product of A nd B is n x p mtrix AB whose (i,j)-th entry is k=1 m ik b kj In other words, we multiply the entries of the i-th row of A with the entries of the j-th column of B nd dd them up. 16

17 Mtrix Multipliction [Imges courtesy of Wikipedi] 17

18 Complexity of Nïve Mtrix Multiplying non-squre mtrices: A is n x m, B is m x p AB is n x p mtrix [ whose (i,j) entry is ik b kj ] Computing the product AB tkes nmp sclr multiplictions n(m-1)p sclr dditions 18 if we tke bsic mtrix multipliction lgorithm.

19 Mtrix Chin Order Problem 19

20 Mtrix Chin Order Problem Mtrix multipliction is ssocitive, mening tht (AB)C = A(BC). 19

21 Mtrix Chin Order Problem Mtrix multipliction is ssocitive, mening tht (AB)C = A(BC). Therefore, we hve choice of forming the product of severl mtrices. 19

22 Mtrix Chin Order Problem Mtrix multipliction is ssocitive, mening tht (AB)C = A(BC). Therefore, we hve choice of forming the product of severl mtrices. Wht is the lest expensive wy to form the product of severl mtrices if the nïve mtrix multipliction lgorithm is used? 19

23 Mtrix Chin Order Problem Mtrix multipliction is ssocitive, mening tht (AB)C = A(BC). Therefore, we hve choice of forming the product of severl mtrices. Wht is the lest expensive wy to form the product of severl mtrices if the nïve mtrix multipliction lgorithm is used? [Use number of sclr multiplictions s cost.] 19

24 Why Order Mtters 2

25 Why Order Mtters Suppose we hve 4 mtrices: A: 3 x 1 B: 1 x 4 C: 4 x 1 D: 1 x 25 2

26 Why Order Mtters Suppose we hve 4 mtrices: A: 3 x 1 B: 1 x 4 C: 4 x 1 D: 1 x 25 ((AB)(CD)) : requires 41,2 mults. 2

27 Why Order Mtters Suppose we hve 4 mtrices: A: 3 x 1 B: 1 x 4 C: 4 x 1 D: 1 x 25 ((AB)(CD)) : requires 41,2 mults. (A((BC)D)) : requires 14 mults. 2

28 Mtrix Chin Order Problem Given mtrices A 1, A 2,, A n, where A i is d i-1 x d i mtrix. [1] Wht is minimum number of sclr multiplictions required to compute the product A 1 A 2 A n? [2] Wht order of mtrix multiplictions chieves this minimum? We focus on question [1]; We will briefly sketch n nswer to [2]. 21

29 A Possible Solution Try ll possibilities nd choose the best one. Drwbck: There re too mny of them (exponentil in the number of mtrices to be multiplied) Need to be more clever - try dynmic progrmming! 22

30 Step 1: Develop Recursive Define M(i,j) to be the minimum number of multiplictions needed to compute A i A i+1 A j Gol: Find M(1,n). Bsis: M(i,i) =. Recursion: How cn one define M(i,j) recursively? 23

31 Defining M(i,j) Recursively Consider ll possible wys to split A i through A j into two pieces. Compre the costs of ll these splits: best cse cost for computing the product of the two pieces plus the cost of multiplying the two products Tke the best one M(i,j) = min k (M(i,k) + M(k+1,j) + d i-1 d k d j ) 24

32 Defining M(i,j) Recursively (A i A k ) (A k+1 A j ) P 1 P 2 minimum cost to compute P 1 is M(i,k) minimum cost to compute P 2 is M(k+1,j) cost to compute P 1 P 2 is d i-1 d k d j 25

33 Step 2: Find Dependencies M:

34 Step 2: Find Dependencies M: GOAL! 26

35 Step 2: Find Dependencies M: GOAL! 26

36 Step 2: Find Dependencies M: GOAL! 26

37 Step 2: Find Dependencies M: GOAL! computing the pink squre requires the purple ones: to the left nd below. 26

38 Defining the Dependencies Computing M(i,j) uses everything in sme row to the left: M(i,i), M(i,i+1),, M(i,j-1) nd everything in sme column below: M(i,j), M(i+1,j),,M(j,j) 27

39 Step 3: Identify Order for Recll the dependencies between subproblems just found Solve the subproblems (i.e., fill in the tble entries) this wy: go long the digonl strt just bove the min digonl end in the upper right corner (gol) 28

40 Order for Solving Subproblems M:

41 Order for Solving Subproblems M:

42 Order for Solving Subproblems M:

43 Order for Solving Subproblems M:

44 Order for Solving Subproblems M:

45 Pseudocode for i := 1 to n do M[i,i] := for d := 1 to n-1 do // digonls for i := 1 to n-d to // rows w/ n entry on d-th digonl j := i + d // column corresp. to row i on d-th digonl M[i,j] := infinity for k := i to j-1 to M[i,j] := min(m[i,j], M[i,k]+M[k+1,j]+d i-1 d k d j ) endfor endfor endfor 3

46 Pseudocode for i := 1 to n do M[i,i] := for d := 1 to n-1 do // digonls for i := 1 to n-d to // rows w/ n entry on d-th digonl j := i + d // column corresp. to row i on d-th digonl M[i,j] := infinity for k := i to j-1 to M[i,j] := min(m[i,j], M[i,k]+M[k+1,j]+d i-1 d k d j ) endfor endfor running time O(n 3 ) endfor 3

47 Pseudocode for i := 1 to n do M[i,i] := for d := 1 to n-1 do // digonls for i := 1 to n-d to // rows w/ n entry on d-th digonl j := i + d // column corresp. to row i on d-th digonl py ttention here M[i,j] := infinity to remember ctul for k := i to j-1 to sequence of mults. M[i,j] := min(m[i,j], M[i,k]+M[k+1,j]+d i-1 d k d j ) endfor endfor endfor running time O(n 3 ) 3

48 Exmple M: , 1: A is 3x1 2: B is 1x4 3: C is 4x1 4: D is 1x

49 Keeping Trck of the Order It's fine to know the cost of the chepest order, but wht is tht chepest order? Keep nother rry S nd updte it when computing the minimum cost in the inner loop After M nd S hve been filled in, then cll recursive lgorithm on S to print out the ctul order 32

50 Modified Pseudocode for i := 1 to n do M[i,i] := for d := 1 to n-1 do // digonls for i := 1 to n-d to // rows w/ n entry on d-th digonl j := i + d M[i,j] := infinity for k := i to j-1 to // column corresponding to row i on d-th digonl M[i,j] := min(m[i,j], M[i,k]+M[k+1,j]+d i-1 d k d j ) if previous line chnged vlue of M[i,j] then S[i,j] := k endfor endfor endfor keep trck of chepest split point found so fr: between A k nd A k+1 33

51 Exmple M: S: , : A is 3x1 2: B is 1x4 3: C is 4x1 4: D is 1x25 34

52 Using S to Print Best Ordering Cll Print(S,1,n) to get the entire ordering. Print(S,i,j): if i = j then output "A" + i //+ is string conct else k := S[i,j] output "(" + Print(S,i,k) + Print(S,k+1,j) + ")" 35