Composición de matrices MaxMin

I'm working on fuzzy classifications. Because I'm not a real software developer... just realized that I'm not able to implement, in C language, the max-min composition on matrices.

I'll try to be much more clear.

Suppose you've got a square matrix just like this:

float matrix[2][2] = {
                        { 1.0, 0.4 },
                        { 0.4, 1.0 }
                     };

The max-min composition of "matrix" by itself, yelds

result_matrix[2][2] = {
                         { 1.0, 0.4 }
                         { 0.4, 1.0 }
                      };

p.ej

first perform

min(1.0, 0.4) = 0.4 /* first row */
min(1.0, 0.4) = 0.4 /* first col */

luego

max(0.4, 0.4) = 0.4 that's the element at (0, 0) of the new matrix.

p.ej

min(1.0, 0.4) = 0.4; /* first row */
min(0.4, 1.0) = 0.4; /* second col */

max(0.4, 0.4) = 0.4;

element at (0, 1)

What I need to implement, in C language, is a sort of "rows by cols" check on a matrix.

I really dunno how to do this.

Algunas sugerencias ?

Gracias de antemano.

preguntado el 05 de mayo de 13 a las 18:05

4 Respuestas

Just figured out that, to solve this, I should follow an approach similar to the one used for matrices multiplication.

En lugar de

resultMatrix[i][j] += firstMatrix[i][k] * firstMatrix[k][j];

necesito algo como esto

resultMatrix[i][j] = MAX(MIN(firstMatrix[i][k], firstMatrix[k][j]));

Dónde MIN(firstMatrix[i][k], firstMatrix[k][j]) is just another array.

Supongo.

contestado el 06 de mayo de 13 a las 16:05

¡Resuelto!

Here the source code for max-min composition.

#include <stdio.h>
#include <stdlib.h>

float get_max(float a[], int num_elements);

int main(int argc, char **argv)
{
   int i;

   int j;

   int k;

   float firstMatrix[4][4] = {
                                { 1, 0.6, 0.3, 0.8 },
                                { 0.6, 1, 0.1, 0.4 },
                                { 0.3, 0.1, 1, 0.5 },
                                { 0.8, 0.4, 0.5, 1 }
                             };


   float resultMatrix[4][4];

   float min_array[4];

      for (i = 0; i < 4; i++)
      {
            for (j = 0; j < 4; j++)
            {
                  for (k = 0; k < 4; k++)
                  {
                        if (firstMatrix[i][k] <= firstMatrix[k][j])
                        { min_array[k] = firstMatrix[i][k]; }

                        else
                        { min_array[k] = firstMatrix[k][j]; }

                     resultMatrix[i][j] = get_max(min_array, 4);                                          
                  }

               fprintf(stdout, "%.1f ", resultMatrix[i][j]);                 
            }

         fprintf(stdout, "\n");
      }

   return 0;
}

float get_max(float a[], int num_elements)
{
   int i;

   float max = 0.0;

      for (i = 0; i < num_elements; i++)
      {
        if (a[i] > max)
        { max = a[i]; }
      }

   return(max);
}

contestado el 06 de mayo de 13 a las 20:05

You really don't need three answers for this question; you should edit esta respuesta with the contents of this answer. - LittleBobbyTables - Au Revoir

Could you be a bit more specific as to what rows/columns you are comparing? If I knew exactly what you were comparing and in what order I could help you out more. This is what I can give you so far. Here is a function for finding the minimum value of an array

float min(int n, float *array)
{
    int i;
    float minval;

    minval = *array;

    for (i=1; i<n;i++)
    {
        if (array[i] < minval)
        minval = array[i];
    }

    return minval;
}

To find the minimum value of the first row you would use

x = min(2, *matrix);

For other rows

x = min(NUMCOLS, *matrix+row*NUMCOLS + column));

And to access the elements of the matrix by rows over columns use two nested for loops

for (i = 0; i < 2; i++) 
    for (j = 0; j < 2; j++) 
        printf("%f\n", matrix[i][j]);

Echa un vistazo a Row-major_order

contestado el 05 de mayo de 13 a las 23:05

I guess this will help.

The "algorithm" should calculates each new element of a new matrix.

Por cierto...

You have an input matrix such as firstMatrix.

Step one: get the first element from the first row of firstMatrix and the first element of the first col of firstMatrix: firstMatrix[0][0] y firstMatrix[0][0]

Step two: repeat "Step one" for each row and col of firstMatrix:

/* row */
temp_array_1[i] = firstMatrix[i][j]

y

/* col */
temp_array_2[i] = firstMatrix[j][i]

Paso tres:

for (i = 0; i < 4; i++)
{
   if (temp_array_1[i] <= temp_array_2[i])
   { min_array[i] = temp_array_1[i]; }

   else
   { min_array[i] = temp_array_2[i]; }
}

Step four: get the max value in min_array[i].

Here follows my source code...

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv)
{
   int i;

   int j;

   float firstMatrix[4][4] = {
                                { 1, 0.6, 0.3, 0.8 },
                                { 0.6, 1, 0.1, 0.4 },
                                { 0.3, 0.1, 1, 0.5 },
                                { 0.8, 0.4, 0.5, 1 }
                             };


   float max = 0.0;

   float temp_array_1[4];

   float temp_array_2[4];

   float min_array[4];


      for (i = 0; i < 4; i++)
      {
         /* row */
         temp_array_1[i] = firstMatrix[0][i];

         /* col */
         temp_array_2[i] = firstMatrix[i][0];
      }

      for (i = 0; i < 4; i++)
      {
            if (temp_array_1[i] <= temp_array_2[i])
            { min_array[i] = temp_array_1[i]; }

            else
            { min_array[i] = temp_array_2[i]; }

            for (i = 0; i < 4; i++)
            {         
                  if (min_array[i] > max)
                  { max = min_array[i]; }         
            }
      }

   fprintf(stdout, "\nMax element: %.1f\n", max);

   return 0;
}

The point is that I'm not able to "iterate" this for each element of firstMatrix.

The output matrix that comes out, from this pandemonium is this:

outputMatrix[4][4] = {
                     { 1.0, 0.6, 0.5, 0.8 },
                     { 0.6, 1.0, 0.4, 0.6 },
                     { 0.5, 0.4, 1.0, 0.5 },
                     { 0.8, 0.6, 0.5, 1.0 }
                  }

contestado el 06 de mayo de 13 a las 12:05

Is this an answer, or a response to Roberto Gomez? If the latter, you should edit your question with this information, not post it as a new answer. - LittleBobbyTables - Au Revoir

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