C Program for Finding Transpose of a Sparse Matrix Archives

Here you will get C program to discover transpose of a meagre matrix. Transpose of a matrix is gotten by exchanging lines and segments. In another manner, we can say that component in the I, j position gets put in the j, I position. Transpose of the matrix B1 is gotten as B2 by embeddings (i,j)th component of B1 as (j, i)th component in B2. Program #include #define MAX 20 void printsparse(int[][3]); void readsparse(int[][3]); void transpose(int[][3],int[][3]); int main() { int b1[MAX][3],b2[MAX][3],m,n; printf("Enter the size of matrix (rows,columns):"); scanf("%d%d",&m,&n); b1[0][0]=m; b1[0][1]=n; readsparse(b1); transpose(b1,b2); printsparse(b2); } void readsparse(int b[MAX][3]) { int i,t; printf("\nEnter no. of non-zero elements:"); scanf("%d",&t); b[0][2]=t; for(i=1;i<=t;i++) { printf("\nEnter the next triple(row,column,value):"); scanf("%d%d%d",&b[i][0],&b[i][1],&b[i][2]); } } void printsparse(int b[MAX][3]) { int i,n; n=b[0][2]; //no of 3-triples printf("\nAfter Transpose:\n"); printf("\nrow\t\tcolumn\t\tvalue\n"); for(i=0;i<=n;i++) printf("%d\t\t%d\t\t%d\n",b[i][0],b[i][1],b[i][2]); } void transpose(int b1[][3],int b2[][3]) { int i,j,k,n; b2[0][0]=b1[0][1]; b2[0][1]=b1[0][0]; b2[0][2]=b1[0][2]; k=1; n=b1[0][2]; for(i=0;i<b1[0][1];i++) for(j=1;j<=n;j++) //if a column number of current triple==i then insert the current triple in b2 if(i==b1[j][1]) { b2[k][0]=i; b2[k][1]=b1[j][0]; b2[k][2]=b1[j][2]; k++; } } Output Enter the size of matrix (rows,columns):3 4 Enter no. of non-zero elements:4 Enter the next triple(row,column,value):1 0 5 Enter the next triple(row,column,value):1 2 3 Enter the next triple(row,column,value):2 1 1 Enter the next triple(row,column,value):2 3 2 After Transpose: row column value 4 3 4 0 1 5 1 2 1 2 1 3 3 2 2