how to generate sparse matrix in c
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Answer / aman celly
//PROGRAM TO CREATE A REPRESENTATION OF SPARSE MATRIX AND
PERFORM OPERATIONS
#include<stdio.h>
#include<conio.h>
void implement(int[][10],int,int,int[][10]);
void add(int[][10],int,int,int[][10],int[][10]);
void sub(int[][10],int,int,int[][10],int[][10]);
void transpose(int[][10],int,int[][10]);
void main()
{
int a[10][10];
int b[10][10];
int c[10][10];
int d[10][10];
int e[50][10];
int row,col,k,l,ch;
char che;
clrscr();
printf("enter the no. of rows\t:");
scanf("%d",&row);
printf("\n enter the no. of coloms\t:");
scanf("%d",&col);
printf("\n enter the elements of sparse matrix \t:");
for(k=0;k<row;k++)
{
for(l=0;l<col;l++)
{
printf("\n element at [%d][%d]\t",k,l);
scanf("%d",&a[k][l]);
}
}
printf("sparse matrix is \n");
for(k=0;k<row;k++)
{
for(l=0;l<col;l++)
{
printf("%d\t",a[k][l]);
}
printf("\n");
}
do
{
printf("\n CYNET MENU\n");
printf("1:REPRESENTATION\n");
printf("2:ADDITION\n");
printf("3:SUBTRACTION\n");
printf("4:TRANSPOSE\n");
printf("5:EXIT\n");
printf("enter your choice:");
scanf("%d",&ch);
switch(ch)
{
case 1:
implement(a,row,col,b);
break;
case 2:
printf("\n enter the elements of second sparse matrix \t:");
for(k=0;k<row;k++)
{
for(l=0;l<col;l++)
{
printf("\n element at [%d][%d]\t",k,l);
scanf("%d",&c[k][l]);
}
}
printf("sparse matrix is \n");
for(k=0;k<row;k++)
{
for(l=0;l<col;l++)
{
printf("%d\t",c[k][l]);
}
printf("\n");
}
implement(a,row,col,b);
implement(c,row,col,d);
add(b,row,col,d,e);
break;
case 3:
printf("\n enter the elements of second sparse matrix \t:");
for(k=0;k<row;k++)
{
for(l=0;l<col;l++)
{
printf("\n element at [%d][%d]\t",k,l);
scanf("%d",&c[k][l]);
}
}
printf("sparse matrix is \n");
for(k=0;k<row;k++)
{
for(l=0;l<col;l++)
{
printf("%d\t",c[k][l]);
}
printf("\n");
}
implement(a,row,col,b);
implement(c,row,col,d);
sub(b,row,col,d,e);
break;
case 4:
implement(a,row,col,b);
transpose(b,col,c);
break;
case 5:
exit();
default:
printf("you entered wrong choice\n");
}
printf("do you want to continue(y\Y):");
che=getche();
//clrscr();
}while(che=='y'||che=='Y');
getch();
}
void implement(int a[][10],int row,int col,int b[][10])
{
int g=1,nz=0,k,l;
for(k=0;k<row;k++)
{
for(l=0;l<col;l++)
{
if(a[k][l]!=0)
{
nz=nz+1;
}
}
}
b[0][0]=row;
b[0][1]=col;
b[0][2]=nz;
for(k=0;k<row;k++)
{
for(l=0;l<col;l++)
{
if(a[k][l]!=0)
{
b[g][0]=k;
b[g][1]=l;
b[g][2]=a[k][l];
g++;
}
}
}
printf("implementation of sparse matrix is\n");
for(k=0;k<g;k++)
{
for(l=0;l<3;l++)
{
printf("%d\t",b[k][l]);
}
printf("\n");
}
}
void add(int b[][10],int row,int col,int d[][10],int e[]
[10])
{
int p1=1,p2=1,i=1;
int k,l;
if(b[0][0]!=d[0][0])
{
printf("addition is not possible\n");
}
else
{
while(p1<=b[0][2]&&p2<=d[0][2])
{
if(b[p1][0]==d[p2][0])
{
if(b[p1][1]<d[p2][1])
{
e[i][0]=b[p1][0];
e[i][1]=b[p1][1];
e[i][2]=b[p1][2];
i++;
p1++;
}
else if(b[p1][1]>d[p2][1])
{
e[i][0]=d[p2][0];
e[i][1]=d[p2][1];
e[i][2]=d[p2][2];
i++;
p2++;
}
else if(b[p1][1]==d[p2][1])
{
e[i][0]=d[p1][0];
e[i][1]=d[p1][1];
e[i][2]=b[p1][2]+d[p2][2];
if(e[i][2]!=0)
{
i++;
}
p1++;
p2++;
}
}
else if(b[p1][0]<d[p2][0])
{
e[i][0]=b[p1][0];
e[i][1]=b[p1][1];
e[i][2]=b[p1][2];
i++;
p1++;
}
else if(b[p1][0]>d[p2][0])
{
e[i][0]=d[p2][0];
e[i][1]=d[p2][1];
e[i][2]=d[p2][2];
i++;
p2++;
}
}
if(p1!=b[0][2])
{
while(p1<=b[0][2])
{
e[i][0]=b[p1][0];
e[i][1]=b[p1][1];
e[i][2]=b[p1][2];
i++;
p1++;
}
}
else if(p2!=d[0][2])
{
while(p2<=d[0][2])
{
e[i][0]=d[p2][0];
e[i][1]=d[p2][1];
e[i][2]=d[p2][2];
i++;
p2++;
}
}
e[0][0]=row;
e[0][1]=col;
e[0][2]=i-1;
printf("matrix after addition\n");
for(k=0;k<i;k++)
{
for(l=0;l<3;l++)
{
printf("%d\t",e[k][l]);
}
printf("\n");
} }
}
void sub(int b[][10],int row,int col,int d[][10],int e[]
[10])
{
int p1=1,p2=1,i=1;
int k,l;
if(b[0][0]!=d[0][0])
{
printf("subtraction is not possible\n");
}
else
{
while(p1<=b[0][2]&&p2<=d[0][2])
{
if(b[p1][0]==d[p2][0])
{
if(b[p1][1]<d[p2][1])
{
e[i][0]=b[p1][0];
e[i][1]=b[p1][1];
e[i][2]=b[p1][2];
i++;
p1++;
}
else if(b[p1][1]>d[p2][1])
{
e[i][0]=d[p2][0];
e[i][1]=d[p2][1];
e[i][2]=d[p2][2];
i++;
p2++;
}
else if(b[p1][1]==d[p2][1])
{
e[i][0]=d[p1][0];
e[i][1]=d[p1][1];
e[i][2]=b[p1][2]-d[p2][2];
if(e[i][2]!=0)
{
i++;
}
p1++;
p2++;
}
}
else if(b[p1][0]<d[p2][0])
{
e[i][0]=b[p1][0];
e[i][1]=b[p1][1];
e[i][2]=b[p1][2];
i++;
p1++;
}
else if(b[p1][0]>d[p2][0])
{
e[i][0]=d[p2][0];
e[i][1]=d[p2][1];
e[i][2]=d[p2][2];
i++;
p2++;
}
}
if(p1!=b[0][2])
{
while(p1<=b[0][2])
{
e[i][0]=b[p1][0];
e[i][1]=b[p1][1];
e[i][2]=b[p1][2];
i++;
p1++;
}
}
else if(p2!=d[0][2])
{
while(p2<=d[0][2])
{
e[i][0]=d[p2][0];
e[i][1]=d[p2][1];
e[i][2]=d[p2][2];
i++;
p2++;
}
}
e[0][0]=row;
e[0][1]=col;
e[0][2]=i-1;
printf("matrix after subtraction\n");
for(k=0;k<=i;k++)
{
for(l=0;l<3;l++)
{
printf("%d\t",e[k][l]);
}
printf("\n");
} }
}
void transpose(int b[][10],int col,int c[][10])
{
int i,j,k,temp;
for(i=0;i<=b[0][2];i++)
{
c[i][0]=b[i][1];
c[i][1]=b[i][0];
c[i][2]=b[i][2];
}
for(i=1;i<=b[0][2];i++)
{
for(j=i+1;j<=b[0][2];j++)
{
if(c[i][0]>c[j][0])
{
for(k=0;k<3;k++)
{
temp=c[i][k];
c[i][k]=c[j][k];
c[j][k]=temp;
}
}
}
}
printf("transpose of matrix is\n");
for(i=0;i<=c[0][2];i++)
{
for(j=0;j<3;j++)
{
printf("%d\t",c[i][j]);
}
printf("\n");
}
}
| Is This Answer Correct ? | 107 Yes | 41 No |
Answer / d. prashant
/* Program of sparse matrix for 3-tuple method using array*/
#include
#define srow 50
#define mrow 20
#define mcolumn 20
main()
{
int mat[mrow][mcolumn],sparse[srow][3];
int i,j,nzero=0,mr,mc,sr,s;
printf("Enter number of rows : ");
scanf("%d",&mr);
printf("Enter number of columns : ");
scanf("%d",&mc);
for(i=0;i for(j=0;j {
printf("Enter element for row %d,column %d : ",i+1,j+1);
scanf("%d",&mat[i][j]);
}
printf("Entered matrix is : \n");
for(i=0;i {
for(j=0;j {
printf("%6d",mat[i][j]);
if(mat[i][j]!=0)
nzero++;
}
printf("\n");
}
sr=nzero+1;
sparse[0][0]=mr;
sparse[0][1]=mc;
sparse[0][2]=nzero;
s=1;
for(i=0;i for(j=0;j {
if(mat[i][j]!=0)
{
sparse[s][0]=i+1;
sparse[s][1]=j+1;
sparse[s][2]=mat [i][j];
s++;
}
}
printf("Sparse matrix is :\n");
for(i=0;i {
for(j=0;j<3;j++)
printf("%5d",sparse[i][j]);
printf("\n");
}
}/*End of main()*/
| Is This Answer Correct ? | 57 Yes | 50 No |
Answer / it career point
//program to 3-tuple representation
#include<stdio.>
#include<conio.h>
void main()
{
int a[5][5],row,columns,i,j;
printf(Ënter the order of the matrix.(5*5)");
scanf("%d%d",&row,&columns);
printf("Enter the element of the matrix\n");
for(i=0;i<row;i++)
for(j=0;j<columns;j++)
{
scanf("%d", &a[i][j]);
}
printf("3-tuple representation");
for(i=0;i<row;i++)
for(j=0;j<column;j++)
{
if(a[i][j]!=0)
{
printf("%d %d %d", (i+1),(j+1),a[i][j]);
}
}
getch();
}
| Is This Answer Correct ? | 41 Yes | 42 No |
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C 
