fast transpose of a sparse matrix

[thejokerguy]

hello Everyone!
I want to implement fast transpose of a sparse matrix. It's really hard for me to understand the algorithm explained in book.
Can anyone please explain in simple manner?
Thank you ! :)

[jnawrocki]

Simple. A matrix is made up of rows and columns. Given a matrix with 3 rows and 3 columns:
1 2 3
4 5 6
7 8 9

After transpose you have:
1 4 7
2 5 8
3 6 9

[jnawrocki]

Simple algorithm:

A ^= B
B ^= A
A ^= B

Swaps without temporary storage.

[thejokerguy]

Thanks for reply!
I know transpose of a matrix. I want to know about "fast transpose" method of a sparse matrix.

[jnawrocki]

Call a function transopose(*x, *y) and pass the address of the items to swap. You can use the xor method described in my previous post to do the swap. First call would swap item 2 in col 1 with item 2 in row 1. and so on.

[avnish.singh5]

void ftrans(int a[][3],int b[][3])
{
	int i,j,k=0,min,pos=0,c[20][3];
	for(i=0;i<=a[0][2];i++)
	{
		for(j=0;j<3;j++)
		c[i][j]=a[i][j];
	}
	b[k][0]=c[pos][1];
	b[k][1]=c[pos][0];
	b[k][2]=c[pos][2];
	for(j=c[0][2];j>0;j--)
	{
		min=c[0][1];
		for(i=c[0][2];i>0;i--)
		{
			  if(c[i][1]<=min&&c[i][2]!=0)
			  {
				min=c[i][1];
				pos=i;
			  }
		}
		k++;
		b[k][0]=c[pos][1];
		b[k][1]=c[pos][0];
		b[k][2]=c[pos][2];
		c[pos][2]=0;
	}
}

[avnish.singh5]

simple funda hai yr n it works only whn........your input function is as follows

void input(int a[][3],int r,int c)
{
	int n,i,j,k=0;
	a[0][0]=r;
	a[0][1]=c;
	a[0][2]=0;
	for(i=0;i<r;i++)
	{
		for(j=0;j<c;j++)
		{
			scanf("%d",&n);
			if(n!=0)
			{
				++k;
				a[k][0]=i;
				a[k][1]=j;
				a[k][2]=n;
				a[0][2]++;
			}
		}
	}
}