problem in generating non repeated random numbers

[xshashiy]

i want to generate non repeated random numbers from 0 to 156. here is the code for the purpose. but when i try to generate the random numbers more then 1 time by using a for loop, the same pattern on numbers repeats. how to solve this problem ?

#include "sys/types.h"
#include "stdio.h"
#include "time.h"
#define MAX 2000
#define N 156
main( )
{
    int array[BIG_SIZE],r;
    int n = 0; int count_check,count_gen ,i;
    time_t t1;
    (void) time(&t1);
    srand48((long) t1);
      for (count_gen=0;count_gen<=MAX;count_gen++)
      {
        r = lrand48()%N;
        for ( count_check = 0; count_check < n; count_check++ )
        {
          if ( r == array[count_check] )break;
        }
      if ( count_check == n ) array[n++] = r;
      }
    for(i=0;i<N;i++)
    printf("%d\n",array[i]);
}

regards
xshashiy

[winbatch]

You are using time for purposes of generating the random number, but only once before the loop. You should probably reset the time each part of the loop.

(void) time(&t1);

(Though you may continue to have the same problem as I think time only holds seconds... )

Also, what you posted won't compile - BIG_SIZE is not defined, etc.

[Narue]

Since you already know the range of the sequence, and it's small, and you want non-repeating values, a random shuffle would be a better option than trying to check whether each new number was already generated or not:

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

#define N 157

int main ( void )
{
  int seq[N];
  int i;

  srand ( (unsigned)time ( NULL ) );

  /* Initialize seq to an ordered range */
  for ( i = 0; i < N; i++ )
    seq[i] = i;

  /* Random shuffle */
  for ( i = 0; i < N - 1; i++ ) {
    int r = ( rand() % ( N - i ) + 1 );
    int save = seq[i];
    seq[i] = seq[i + r];
    seq[i + r] = save;
  }

  /* Test the sequence */
  for ( i = 0; i < N; i++ )
    printf ( "%d ", seq[i] );

  return 0;
}

Not only is it easier to verify, it's also considerably faster.

[Rashakil Fol]

Sometimes the standard library makes life easier:

#include <algorithm>

/* Random shuffle */
std::random_shuffle(seq, seq + N);

Obviously not useful since we already have an implementation, but useful if you'll need it elsewhere.

[Narue]

>Sometimes the standard library makes life easier
Indeed. But since the original problem was clearly written in C, the C++ standard library is hardly useful. You might argue that C compiles as C++, so it's not a problem, but since you would be wrong, I recommend that you don't bother, and save yourself the intellectual manbeating I'll give you if you try. :D

[Rashakil Fol]

Oh man, what was I thinking?

Never mind then. :o

(It's not a question of whether C can compile to C++; C++ simply can't be compiled as C! ;-) )

[DKdontKNO]

Oh man you rock. I'm a noob, and getting lost in following how the random shuffle for loop works...was wondering if you could explain line by line?

Since you already know the range of the sequence, and it's small, and you want non-repeating values, a random shuffle would be a better option than trying to check whether each new number was already generated or not:

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

#define N 157

int main ( void )
{
  int seq[N];
  int i;

  srand ( (unsigned)time ( NULL ) );

  /* Initialize seq to an ordered range */
  for ( i = 0; i < N; i++ )
    seq[i] = i;

  /* Random shuffle */
  for ( i = 0; i < N - 1; i++ ) {
    int r = ( rand() % ( N - i ) + 1 );
    int save = seq[i];
    seq[i] = seq[i + r];
    seq[i + r] = save;
  }

  /* Test the sequence */
  for ( i = 0; i < N; i++ )
    printf ( "%d ", seq[i] );

  return 0;
}

Not only is it easier to verify, it's also considerably faster.