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rmtogether

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find max value after some calculation




#include <stdio.h>


float matrix [3][3]={{0.5,0.25,0.25},
                     {0.375,0.125,0.375},
                     {0.125,0.675,0.375}};

main (){
    int i,j,k;

     for (i=0; i<3; i++){
          printf("\n");
         for (j=0; j<3; j++) {
             printf("\n");
            for (k=0; k<3; k++){
              printf("Prob(%d%d%d)=%.5f\n",i,j,k,matrix[i][j]*matrix[j][k]);   ------------>? how to find max value
           
            }//end third for
                       
        }// end second for
    }// end first for


    system("pause");
}
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ozo
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#include <stdio.h>
#include <float.h>


float matrix [3][3]={{0.5,0.25,0.25},
                     {0.375,0.125,0.375},
                     {0.125,0.675,0.375}};

main (){
    int i,j,k;
    float max = -FLT_MAX;

     for (i=0; i<3; i++){
          printf("\n");
         for (j=0; j<3; j++) {
             printf("\n");
            for (k=0; k<3; k++){
   
               if( max < matrix[i][j]*matrix[j][k] ){ max = matrix[i][j]*matrix[j][k]; }
              printf("Prob(%d%d%d)=%.5f\n",i,j,k,matrix[i][j]*matrix[j][k]);   ------------>? how to find max valu
           
            }//end third for
                       
        }// end second for
    }// end first for


    system("pause");
}
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rmtogether

ASKER

thanks,

what is "-FLT_MAX" means?
It is the smallest floating point number
thanks,

Can I ask you the last question? Since the max value is happen in more than one pleace, can I print a list of them?

for example,
in this program, the max value  is 0.25313 and happen in Prob (121), prob (210), prob(221). I would like have an end summary to show something like

   printf("max prob is: %.5f\n",max);
   printf("they are.....")---------------------------------? list of item has max value





#include <stdio.h>
#include <float.h>


float matrix [3][3]={{0.5,0.25,0.25},
                     {0.375,0.125,0.375},
                     {0.125,0.675,0.375}};

main (){
    int i,j,k;
    float max = -FLT_MAX;

     for (i=0; i<3; i++){
          printf("\n");
         for (j=0; j<3; j++) {
             printf("\n");
            for (k=0; k<3; k++){
   
               if( max < matrix[i][j]*matrix[j][k] ) {max = matrix[i][j]*matrix[j][k];
               }
                 printf("Prob(%d%d%d)=%.5f\n",i,j,k,matrix[i][j]*matrix[j][k]);  
           
            }//end third for
                       
        }// end second for
    }// end first for
   
   printf("\n");
   printf("max prob is: %.5f\n",max);
   printf("they are.....")---------------------------------? list of item has max value

    system("pause");
}
It's generally not safe to rely on floating point to be the same when they happen in more than one place but in this case, all the values in your matrix are dyadic rationals, which can be exactly represented in floating point numbers when FLT_RADIX==2

      
   printf("max prob is: %.5f\n",max);
   printf("they are.....")
     for (i=0; i<3; i++){
         for (j=0; j<3; j++) {
            for (k=0; k<3; k++){
   
               if( max == matrix[i][j]*matrix[j][k] ){  printf("Prob(%d%d%d)\n",i,j,k);  }
           
            }//end third for
                       
        }// end second for
    }// end first for
   
hi, ozo

I put the code like below, but seems not work.. could you help me about it?

#include <stdio.h>
#include <float.h>


float matrix [3][3]={{0.5,0.25,0.25},
                     {0.375,0.125,0.375},
                     {0.125,0.675,0.375}};

main (){
    int i,j,k;
    float max = -FLT_MAX;

     for (i=0; i<3; i++){
          printf("\n");
         for (j=0; j<3; j++) {
             printf("\n");
            for (k=0; k<3; k++){
   
               if( max < matrix[i][j]*matrix[j][k] ) {max = matrix[i][j]*matrix[j][k];
               }
                 printf("Prob(%d%d%d)=%.5f\n",i,j,k,matrix[i][j]*matrix[j][k]);  
           
            }//end third for
                       
        }// end second for
    }// end first for
   
   
   printf("max prob is: %.5f\n",max);
   printf("they are.....");
     for (i=0; i<3; i++){
         for (j=0; j<3; j++) {
            for (k=0; k<3; k++){
   
               if( max == matrix[i][j]*matrix[j][k] ){  printf("Prob(%d%d%d)\n",i,j,k);  }
           
            }//end third for
                       
        }// end second for
    }// end first for

    system("pause");
}
What does not seem to work?
there is  nothing showing after "they are...."

I don't see anything printed from this line

   if( max == matrix[i][j]*matrix[j][k] ){  printf("Prob(%d%d%d)\n",i,j,k);  }
it works after I change  float max = -FLT_MAX; to

double max = -FLT_MAX;
This looks like an instance of the dangers of comparing floating point numbers for equality.
As I look more carefully at
float matrix [3][3]={{0.5,0.25,0.25},
                     {0.375,0.125,0.375},
                     {0.125,0.675,0.375}};
I see that you have 0.675 which is not represented exactly in floating point numbers with FLT_RADIX==2
instead of 0.625 which could have been represented exactly, so there could be round off errors that keep the comparison from being exact.

one way to handle it would be to use exact integer arithmetic with
long matrix [3][3]={{500,250,250},
                     {375,125,375},
                     {125,675,375}};
or to have some error tolerance
like
epsilon=0.0001;
 if( fabs(max - matrix[i][j]*matrix[j][k]) < epsilon ){  printf("Prob(%d%d%d)\n",i,j,k);  }


thanks,

(1) what does FLT_RADIX==2 means?
(2) if I change matrix to long and make
     printf to printf("Prob(%d%d%d)=%f\n",i,j,k,matrix[i][j]*matrix[j][k]*0.000001), do I still have round off problem?
A floating-point number (x) is defined by the following model:

x =s * b^e *
p
&#931; f[k]*b^&#8722;k , emin &#8804; e &#8804; emax
k=1

s sign (±1)
b base or radix of exponent representation (an integer > 1)
e exponent (an integer between a minimum emin and a maximum
p precision (the number of base-b digits in the significand)
f[k] nonnegative integers less than b (the significand digits)

FLT_RADIX represents b
ASKER CERTIFIED SOLUTION
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ozo
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