a problem wilth C program

[the_shark]

Hi friends...

I have a problem with some c codes and I think that you will be Capable to point it up for me.
However, I do not want anybody to post the correct and complete code to me .

I want to enter a string of numbers and the program will check whether the entered string forms a multiplicative identity n*n matrix which means all elements are zero except for the diagonal elements
For example, 3*3 multiplicative identity
1 0 0
0 1 0
0 0 1

I think but I am not pretty sure that the problem is from the outer for-loop. But still I cannot figure the error out and here is my code fragment written in C:
======================================================

#include <stdio.h>
#include <string.h>
int main()
{
    int size,track=0,i,j ,sum=0;
    char string[400];
    printf("Enter the size of the matrix: ");
    scanf("%d",&size);
    gets(string);
    printf("Enter the %d by %d matrix (in row-order) ",size,size);
    gets(string);
    printf("the matrix elements are = %s \n", string);   
//notice that the entered string should be Separated by single space only


	for(j=0;j<(2*size*size-1);j=(j+2*size-1)) //go row by row 
	{
		for (i=0;i<2*size;i=i+2)    //go value by value in a single row
		{		
		if (string[j]!=1 && j==track || string[j]!=0 && j !=track)
                	sum= sum;
			

		else if ((string[j]==1 && j==track) || (string[j]==0 && j !=track))
                	sum++;
                }
              track++;
        	
	}
	if (sum==(size*size) )
		printf("This matrix is   a multiplicative identity\n");
	else
		printf("This matrix is  not a multiplicative identity\n");
    return 0;
}

=========================================
thanks.
and Remember I do not want to see any complete structure of the solution

the_shark

<< moderator edit: added code tags: [code][/code] >>

[Narue]

The most obvious problem is that you're testing for 0 or 1 instead of '0' or '1'. Remember the the contents of your string are characters, not integers. You're also going about the problem in a complicated way, but because you don't want help with that part, I won't show you a simpler solution.

[the_shark]

Hi Narue,

First of all, I would like to thank you for editing my post and making the code more readable.

Second of all, I did follow your suggestions as well as other modifications and now my program is working properly. :cool:

Third of all, I though my way to develop the problem idea was the only way to do so. However, since you are more expert than me, I understood that the problem can be solved in simpler way.

Can you write down the id :idea: ea of the other way?

Quick note:
I am a beginner in C.
So my knowledge includes pointers, functions, string, arrays, structures, loops and condition statements. :o

I hope that your solution will be within my knowledge-base. :cheesy:

Again, thanks for the cooperation.

[varunrathi]

use double dimensional array.

[Narue]

>Can you write down the idea of the other way?
Look carefully at the data. You have a 1 followed by 0's, then another 1 and the same amount of 0's. If you count the 0's then you'll see that it's N + 1 where N is the size of the matrix (ie. NxN, or 3x3).

To verify that the matrix is valid, you need all 0's except for the first, and every (N + 1)st beyond that. Those values must be 1. This gives you an immediate solution for counting the 1's:

size_t len = strlen ( s );
for ( i = 0; i < len; i += N + 1 ) {
  if ( s[i] != '1' )
    invalid();
}

Now the only problem is making sure that the other values are all 0. The simplest way to do this is a trivial loop:

for ( i = 0; i < len; i++ ) {
  if ( s[i] != '0' )
    invalid();
}

But because there are 1's, that won't work unless you make sure that the 1's won't be counted, or won't make a difference if they are. You're probably ahead of me by now, but consider this:

size_t len = strlen ( s );
for ( i = 0; i < len; i += N + 1 ) {
  if ( s[i] != '1' )
    invalid();
  s[i] = '0'; // Turn a 1 into a 0
}

for ( i = 0; i < len; i++ ) {
  if ( s[i] != '0' )
    invalid();
}

This solution is easy to understand, easy to verify, and has linear complexity.

[the_shark]

ahhhhhhha

The solution is much easier than I thought; I'll try to practice such ideas in the future.

Thanks Narue and varunrathi. :)