C programming Lecture 15. School of Mathematics Trinity College Dublin. Marina Krstic Marinkovic

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1 C programming 5613 Lecture 15 Marina Krstic Marinkovic School of Mathematics Trinity College Dublin Marina Krstic Marinkovic 1 / C programming

2 In a rucksack (knapsack) problem we are given a set of n items where each item i is specified by a weight wi and value(profit) pi We are also given a maximal capacity M: the size of the knapsack Item # Weight Value Determine which item (the nr. of each items) to place in a knapsack to maximise the profit P= p1+p2+ +pk, s.t. w1+w2+ +wk M value(profit) 0-1 knapsack problem (vs. fractional knapsack problem) Marina Krstic Marinkovic 2 / C programming

3 Step 4: Construction of optimal : Construct an optimal from Dynamic programming strategy: A problem solved by identifying a collection of subproblems and tackling them one by one: smallest first, using the answers to small problems to help figure out larger ones, until the whole lot of them is solved Marina Krstic Marinkovic 3 / C programming

4 Step 4: Construction of optimal : Construct an optimal from Construct a matrix V[0 n,0 M] For i=1 n and 0 w M the entry V[i,w] stores the maximum profit of any subset of items 1,2,,i of combined weight at most w if we can compute all the entries of this matrix, then the entry V[n,M] will contain the maximal profit of items that can fit into the sack > the to the problem Marina Krstic Marinkovic 4 / C programming

5 Step 4: Construction of optimal : Construct an optimal from Initial settings: set V[0,w]=0 for 0 w M -> no item set V[0,w]=- for w <0 -> impossible Recursive step: use V[i,w]=max(V[i-1,w],pi+V[i-1,w-wi]) for 1 i n, 0 w M Marina Krstic Marinkovic 5 / C programming

6 Step 4: Construction of optimal : Construct an optimal from Bottom: V[i,w] w= M i= n set V[0,w]=0 for 0 w M -> no item Fill in the table (matrix) bottom-up: use V[i,w]=max(V[i-1,w],pi+V[i-1,w-wi]) for 1 i n, 0 w M Marina Krstic Marinkovic 6 / C programming

7 Step 4: Construction of optimal : Construct an optimal from n=4, M=8 Item # Weight(wi) Value(pi) V[i,w] w= i= Bottom: set V[0,w]=0 for 0 w M -> no item Fill in the table (matrix) bottom-up: use V[i,w]=max(V[i-1,w],pi+V[i-1,w-wi]) for 1 i n, 0 w M Marina Krstic Marinkovic 7 / C programming

8 Step 4: Construction of optimal : Construct an optimal from n=4, M=8 Item # Weight(wi) Value(pi) V[i,w] w= i= The final output is V[4,8]=15 (compare with the from backtracking) The method does not (yet) tell which subset gives the optimal ((0,1,0,1) in this example) Marina Krstic Marinkovic 8 / C programming

9 Step 4: Construction of optimal : Construct an optimal from /*Pseudocode for dynamic programming of 0-1 knapsack problem*/ knapsack(w,p,n,m) for (w=0... M) V[0][w]=0; for (i=0... n) for (w=0... M) if (w[i] w) V[i][w]=maxV[i-1][w],pi+V[i-1][w-w[i]]; else V[i][w]=V[i-1][w]; return V[n,M]; Complexity: O(nW) From exponential to polynomial complexity: n2 n > nm Marina Krstic Marinkovic 9 / C programming

10 Step 4: Construction of optimal : Construct an optimal from This algorithm does not keep record of which subset of items gives optimal Auxiliary array: /*Pseudocode for dynamic programming of 0-1 knapsack problem*/ knapsack(w,p,n,m) for (w=0... M) V[0][w]=0; for (i=0... n) for (w=0... M) if (w[i] w) V[i][w]=maxV[i-1][w],pi+V[i-1][w-w[i]]; else V[i][w]=V[i-1][w]; return V[n,M]; keep[i, w] = ( 1 if we decide to pack i-th item, 0 otherwise Marina Krstic Marinkovic 10 / C programming

11 Step 4: Construction of optimal : Construct an optimal from /*Pseudocode for constructing optimal of 0-1 knapsack problem with dynamic programing approach*/ K=M; for (i=n... 0) if (keep[n][k]==1) print i; K=K-w[i]; If keep[n,m]=1 then n-th item is in the optimal >look up keep[n-1,m-w n ] If keep[n,m]=0 then n-th item is NOT in the optimal >look up keep[n-1,m] Marina Krstic Marinkovic 11 / C programming

12 /*Solving knapsack problem using dynamic programing*/ #include <stdio.h> #define max 100 int w[max],p[max]; int n,m; int V[max][max],keep[max][max]; void read_items() int i; printf("\n Enter total number of items: "); scanf("%d",&n); printf("\n Enter the Maximum capacity of the Rucksack: "); scanf("%d",&m); for(i=0;i<n;i++) printf("\n Enter the weight of the item %d : ",i+1); scanf("%d",&w[i]); printf("\n Enter the profit of the item %d : ", i+1); scanf("%d", &p[i]); void print_items() int i; float s=0.0; printf("\n\titem\tweight\tcost "); for(i=0;i<n;i++) printf("\n\t%d\t%d\t%d",i+1,w[i],p[i]); printf("\n\nthe Rucksack now holds following items:\n"); Marina Krstic Marinkovic 12 / C programming

13 int knapsack() /*filling in DP table: V[0...n][0...m]*/ int i,j; for (i=0;i<=n;i++) for (j=0;j<=m;j++) keep[i][j]=0; for (i=0;i<m;i++) V[0][i]=0; for (i=1;i<=n;i++) for (j=0;j<=m;j++) if (w[i-1]<=j) if (V[i-1][j] < (p[i-1]+v[i-1][j-w[i-1]])) V[i][j]=p[i-1]+V[i-1][j-w[i-1]]; keep[i][j]=1; else V[i][j]=V[i-1][j]; return V[n][m]; int reconstruct_optimal() int i,k; k=m; for (i=n;i>0;i--) if (keep[i][k]==1) printf("%d\n",i); k=k-w[i-1]; Marina Krstic Marinkovic 13 / C programming

14 void print_matrix() int i,j; printf("\n keep matrix:\n"); printf("i\\w "); for (j=0;j<=m;j++) printf("%d ",j); printf("\n"); for (i=0;i<=n;i++) printf("%d ",i); for (j=0;j<=m;j++) printf("%d ",keep[i][j]); printf("\n"); void main() read_items(); printf("\n The Rucksack is arranged in the order\n"); printf("\n Maximum Profit: %d \n\n",knapsack()); print_matrix(); print_items(); reconstruct_optimal(); Marina Krstic Marinkovic 14 / C programming