Problem in storing coordinates of a pixel

Posted on 2003-11-19
Last Modified: 2010-04-01
hi! all
      i'm writng a code for image compression in which  i have to read the data from file and store it in a matrix form. Second step involves sorting of can be sort but i've to store their positions not the values in an array (acc. to the sorted values).

Suppose 64(3rd row & 5th col) is the max value in the matrix...then i've to store 3,5 position of 64 . Now problem is how do i store that. i think it's pretty simple but i'm not bale to get it.
This is done so that value's original location is not lost...since it's an image & after whatever operation we need to send that vlue back to its original position so as to get the original image back.

What kind of structure do i need to use.
Please if someone cud help me solving this prob.
thanks in advance
Question by:s_t_1979
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LVL 13

Expert Comment

ID: 9777947
Is this homework?
LVL 39

Expert Comment

ID: 9778110
Try this:

int* sortArray(int* array, int size)
   // check for minimal two items
   if (size <= 1)
       return NULL;

   // create integer array for the sorted indices
   int*     idxArr  = new int[size];  
   int      nn      = 0;       // nn is the number of already sorted items
   int      n1      = 0;       // n1 is the lower range index when searching binary
   int      n2      = 0;       // n2 is the upper range index when searching binary
   int      n       = 0;       // n is the middle of the binary range
   int      i;                 // loop counter

   // loop all items and build a sorted index array  
   for (i = 0; i < size; i++)
       // i1 is the new item which index must be inserted to the index array
       int i1 = array[i];
       // init the binary range to the current size of the already sorted index array
       n1    = 0;
       n2    = nn-1;

       // binary search
       while (n1 <= n2)
           // compute the middle of the current range for dividing the current range into two parts
           n     = (n1 + n2 + 1)/2;
           // get the item to compare
           int i2 = array[idxArr[n]];
           // compare both items
           // if the left item is less we make the lower part current
           if (i1 < i2)
               n2   = n - 1;
           // if the left item is greater or equal we make the upper part current
               n1 = n + 1;

       // at end of the previous loop n1 is the new position where the current item index (i)
       // has to be inserted to get the index array sorted

       // if this position is not outside the current size (that means we have not to change the order)
       // we move all indices on the right side for one position using memmove
       if (n1 < nn)
           memmove(&idxArr[n1+1], &idxArr[n1], (nn-n1)*sizeof(int));    
       idxArr[n1] = i;
   return idxArr;

#define MAXX 100
#define MAXY 100

int main(int argc, char* argv[])
    int matrix[MAXY][MAXX] = { 0 };

    // read matrix from somewhere

    if (readMatrix(matrix, ...))
        int* array  = (int*)matrix;
        int  size   = MAXX*MAXY;

        int* idxArr = sortArray(array, size);

        // now you may store the index array to a file
        // it is as big as the original image file (if the values to sort are integers)

        writeSortedIndex(idxArr, ...);

        // if you need the coordinates e. g. of the biggest value you may calculate
        int idx = idxArr[size-1];

        int x      = idx % MAXX;
        int y      = idx / MAXX;

        // then, you got the max value by
        int maxInt = matrix[x][y];

I assumed that your values are integers.

Hope, that helps


Accepted Solution

dhyanesh earned 100 total points
ID: 9779128

You can use an array of structures. Suppose you have M rows and N columns, then

     int x;
     int y;

Thus this array then can be accessed easily as array[0].x or array[0].y which will give x and y of first element.

Thus you can store the x and y values easily and access them.


Author Comment

ID: 9785128
thanks a lot Dhyanesh...i knew ans will be simple...but was not able to do it. sometimes u tend to forget basic things. Once again thanks a lot.

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