filling an array with random number

Question

Hi there, I trying to fill an array of 20 ints with numbers from 1-20 in random sequence. here's my code:

 int lookup[20]={0}; 
 int array[20]={0};
 srand(time(NULL));
 for(int i=0;i<20;++i){ 
    bool done=false;
    while(!done){
      int n=rand()%20;
      if(lookup[n]==0){
          array[i]=n;
          lookup[n]=1;
          done=true;
      }
    }
 }

I've created a lookup array to check if the random number is not yet chosen and stored it in array. As you can see I've created 2 loops, one for traversing array and the while for choosing the random number. In every while loop iteration the number may reappear and causing another while loop. Is there faster way to do this?

Answer 1

You could fill the array in sequence and then shuffle it. That would prevent having to ever do more than 20 random number generations.

Fisher-Yates shuffle: can be done in O(n) time.

From wikipedia:

Properly implemented, the Fisher–Yates shuffle is unbiased, so that every permutation is equally likely. The modern version of the algorithm is also rather efficient, requiring only time proportional to the number of items being shuffled and no additional storage space.

Answer 2

I'd put the 20 random numbers in an array, then copy to another 20 element array, sort one array, and for each number in the sorted array, find the corresponding number in the unsorted array, and put the index of the sort there.

Answer 3

you could fill an array with numbers from 1 to 20 and use std::random_shuffle

note you don't need a vector for that matter a simple array will do.
example :

#include <iostream>
#include <algorithm>

using namespace std;

int main( void )
{
        int array[] = { 0, 1, 2, 3, 4 };

        srand( unsigned( time(NULL) ) );

        random_shuffle(array, array+5);

        for(int i=0; i<5; i++)
                cout << array[i] << endl;

        return 0;
}

Answer 4

There's the possibility in your code of a very long running loop. If you're inside the while(!done) loop there's no guarantee you'll ever finish. Obviously with an array of 20 elements this won't in practice be an issue but it could cause problems if you scale this up to many thousands of elements.

A more reliable solution would be to fill the array sequentially and then shuffle it afterwards.

Answer 5

Look at std::random_shuffle and std::vector.

Answer 6

If you can use containers, I'd just fill a std::set with the numbers from 1 to 20.

Then you can pull a random number from your set and insert it into the array.

Answer 7

Other possible way

int array[20]={0};
int values[20];
for(int i = 0; i < 20; i++)
    values[i] = i;
int left = 20;
for(int i = 0; i < 20; i++)
{
    int n = rand() % left;
    array[i] = values[n];
    left--;
    values[n] = values[left];
}

PS filled with nombers from 0 to 19 not from 1 to 20. Same as original

Answer 8

Code example for Fisher-Yates:

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

void shuffle(int array[], size_t size)
{
    if(size) while(--size)
    {
        size_t current = (size_t)(rand() / (1.0 + RAND_MAX) * (size + 1));
        int tmp = array[current];
        array[current] = array[size];
        array[size] = tmp;
    }
}

int main(void)
{
    srand((int)time(NULL));
    rand(); // throw away first value: necessary for some implementations

    int array[] = { 1, 2, 3, 4, 5 };
    shuffle(array, sizeof array / sizeof *array);

    size_t i = 0;
    for(; i < sizeof array / sizeof *array; ++i)
        printf("%i\n", array[i]);

    return 0;
}