Create Matrix_chain_multiplication.c

Author: madihamallick

Dynamic Programming/Matrix_chain_multiplication.c

@@ -0,0 +1,92 @@
+/* Find the minimum number of scaler multiplication for chain of matrix using
+dynamic programming approach and find the time complexity also. Consider the given
+set ofmatrices A1 , A2 , A3 ,A4 with p0 = 5 , p1 = 4 , p2 = 6, p3 = 2 , p4 = 7. */
+
+
+#include<stdio.h>
+#include<stdlib.h>
+int p[100], mt[100][100], st[100][100];
+void print_optimal(int i , int j){
+	if(i==j){
+		printf("A%d ", i);
+	}
+	else{
+		printf("(");
+		print_optimal(i, st[i][j]);
+		print_optimal(st[i][j] + 1, j);
+		printf(")");
+	}
+}
+int main(){
+	int i, j, k, m,n, l, q;
+	printf("Enter no. of elements in p: ", m);
+	scanf("%d",&m);
+	for(i=0;i<m;i++){
+		printf("Enter the value P %d ", i);
+		scanf("%d", &p[i]);
+		printf("\n");
+	}
+	n = m - 1;
+	for(i=1;i<=4;i++){
+		mt[i][i] = 0;
+		st[i][i] = 0;
+	}
+	for(l=2; l<=n;l++ ){
+		for(i=1;i<=(n-l+1);i++){
+			j = i+l-1;
+			mt[i][j] = 1000000;
+			for(k = i ; k<=(j-1); k++){
+				q = mt[i][k] + mt[k+1][j] + (p[i-1] * p[k] * p[j]);
+				if(q< mt[i][j]){
+					mt[i][j] = q;
+					st[i][j] = k;
+				}
+			}
+		}
+	}
+	printf("\n\n M Table \n\n");
+	for(i = 1; i<=4; i++){
+		for(j=1;j<=4;j++){
+			printf("%d 	", mt[i][j]);
+		}
+		printf("\n");
+	}
+	printf("\n\n S Table \n\n");
+	for(i = 1; i<=4; i++){
+		for(j=1;j<=4;j++){
+			printf("%d 	", st[i][j]);
+		}
+		printf("\n");
+	}
+	print_optimal(1 , n);
+	printf("\nHence minimum cost for multiplication will be = %d",
+	mt[1][n]);
+}
+
+
+/*OUTPUT
+Enter no. of elements in p: 5
+Enter the value P 0 - 5
+Enter the value P 1 - 4
+Enter the value P 2 - 6
+Enter the value P 3 - 2
+Enter the value P 4 - 7
+
+M Table 
+
+0  120  88  158
+0   0   48  104
+0   0    0   84
+0   0    0    0
+
+S Table
+
+0  1  1  3
+0  0  2  3
+0  0  0  3
+0  0  0  0
+
+((A1 (A2 A3)) A4)
+Hence minimum cost for multiplication will be = 158
+
+*/