Does “n * (rand() / RAND_MAX)” make a skewed random number distribution?

问题

I\'d like to find an unskewed way of getting random numbers in C (although at most I\'m going to be using it for values of 0-20, and more likely only 0-8). I\'ve seen this formula but after running some tests I\'m not sure if it\'s skewed or not. Any help?

Here is the full function used:

int randNum() 
{ 
    return 1 + (int) (10.0 * (rand() / (RAND_MAX + 1.0)));
}

I seeded it using:

unsigned int iseed = (unsigned int)time(NULL);
srand (iseed);

The one suggested below refuses to work for me I tried

int greek; 
for (j=0; j<50000; j++) 
{ 
greek =rand_lim(5); 
printf(\"%d, \" greek); 
greek =(int) (NUM * (rand() / (RAND_MAX + 1.0))); 
int togo=number[greek]; 
number[greek]=togo+1; 
}

and it stops working and gives me the same number 50000 times when I comment out printf.

回答1:

Yes, it's skewed, unless your RAND_MAX happens to be a multiple of 10.

If you take the numbers from 0 to RAND_MAX, and try to divide them into 10 piles, you really have only three possibilities:

RAND_MAX is a multiple of 10, and the piles come out even.
RAND_MAX is not a multiple of 10, and the piles come out uneven.
You split it into uneven groups to start with, but throw away all the "extras" that would make it uneven.

You rarely have control over RAND_MAX, and it's often a prime number anyway. That really only leaves 2 and 3 as possibilities.

The third option looks roughly like this: [Edit: After some thought, I've revised this to produce numbers in the range 0...(limit-1), to fit with the way most things in C and C++ work. This also simplifies the code (a tiny bit).

int rand_lim(int limit) {
/* return a random number in the range [0..limit)
 */

    int divisor = RAND_MAX/limit;
    int retval;

    do { 
        retval = rand() / divisor;
    } while (retval == limit);

    return retval;
}

For anybody who questions whether this method might leave some skew, I also wrote a rather different version, purely for testing. This one uses a decidedly non-random generator with a very limited range, so we can simply iterate through every number in the range. It looks like this:

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

#define MAX 1009

int next_val() {
    // just return consecutive numbers
    static int v=0;

    return v++;
}

int lim(int limit) {
    int divisor = MAX/limit;
    int retval;

    do {
        retval = next_val() / divisor;
    } while (retval == limit);

    return retval;
}

#define LIMIT 10

int main() {

    // we'll allocate extra space at the end of the array:
    int buckets[LIMIT+2] = {0};
    int i;

    for (i=0; i<MAX; i++)
        ++buckets[lim(LIMIT)];

    // and print one beyond what *should* be generated
    for (i=0; i<LIMIT+1; i++)
        printf("%2d: %d\n", i, buckets[i]);
}

So, we're starting with numbers from 0 to 1009 (1009 is prime, so it won't be an exact multiple of any range we choose). So, we're starting with 1009 numbers, and splitting it into 10 buckets. That should give 100 in each bucket, and the 9 leftovers (so to speak) get "eaten" by the do/while loop. As it's written right now, it allocates and prints out an extra bucket. When I run it, I get exactly 100 in each of buckets 0..9, and 0 in bucket 10. If I comment out the do/while loop, I see 100 in each of 0..9, and 9 in bucket 10.

Just to be sure, I've re-run the test with various other numbers for both the range produced (mostly used prime numbers), and the number of buckets. So far, I haven't been able to get it to produce skewed results for any range (as long as the do/while loop is enabled, of course).

One other detail: there is a reason I used division instead of remainder in this algorithm. With a good (or even decent) implementation of rand() it's irrelevant, but when you clamp numbers to a range using division, you keep the upper bits of the input. When you do it with remainder, you keep the lower bits of the input. As it happens, with a typical linear congruential pseudo-random number generator, the lower bits tend to be less random than the upper bits. A reasonable implementation will throw out a number of the least significant bits already, rendering this irrelevant. On the other hand, there are some pretty poor implementations of rand around, and with most of them, you end up with better quality of output by using division rather than remainder.

I should also point out that there are generators that do roughly the opposite -- the lower bits are more random than the upper bits. At least in my experience, these are quite uncommon. That with which the upper bits are more random are considerably more common.

来源：https://stackoverflow.com/questions/10219355/does-n-rand-rand-max-make-a-skewed-random-number-distribution

标签

random

numbers

skew