Why does rand() repeat numbers far more often on Linux than Mac?
Apr 24
I was implementing a hashmap in C as part of a project I'm working on and using random inserts to test it. I noticed that rand() on Linux seems to repeat numbers far more often than on Mac. RAND_MAX is 2147483647/0x7FFFFFFF on both platforms. I've reduced it to this test program that makes a byte array RAND_MAX+1-long, generates RAND_MAX random numbers, notes if each is a duplicate, and checks it off the list as seen.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
int main() {
size_t size = ((size_t)RAND_MAX) + 1;
char *randoms = calloc(size, sizeof(char));
int dups = 0;
srand(time(0));
for (int i = 0; i < RAND_MAX; i++) {
int r = rand();
if (randoms[r]) {
// printf("duplicate at %d\n", r);
dups++;
}
randoms[r] = 1;
}
printf("duplicates: %d\n", dups);
}
Linux consistently generates around 790 million duplicates. Mac consistently only generates one, so it loops through every random number that it can generate almost without repeating. Can anyone please explain to me how this works? I can't tell anything different from the man pages, can't tell which RNG each is using, and can't find anything online. Thanks!
linux1 answer
Accepted answer · original discussion
Apr 24
While at first it may sound like the macOS rand() is somehow better for not repeating any numbers, one should note that with this amount of numbers generated it is expected to see plenty of duplicates (in fact, around 790 million, or (231-1)/e). Likewise iterating through the numbers in sequence would also produce no duplicates, but wouldn't be considered very random. So the Linux rand() implementation is in this test indistinguishable from a true random source, whereas the macOS rand() is not.
Another thing that appears surprising at first glance is how the macOS rand() can manage to avoid duplicates so well. Looking at its source code, we find the implementation to be as follows:
/*
* Compute x = (7^5 * x) mod (2^31 - 1)
* without overflowing 31 bits:
* (2^31 - 1) = 127773 * (7^5) + 2836
* From "Random number generators: good ones are hard to find",
* Park and Miller, Communications of the ACM, vol. 31, no. 10,
* October 1988, p. 1195.
*/
long hi, lo, x;
/* Can't be initialized with 0, so use another value. */
if (*ctx == 0)
*ctx = 123459876;
hi = *ctx / 127773;
lo = *ctx % 127773;
x = 16807 * lo - 2836 * hi;
if (x < 0)
x += 0x7fffffff;
return ((*ctx = x) % ((unsigned long) RAND_MAX + 1));
This does indeed result in all numbers between 1 and RAND_MAX, inclusive, exactly once, before the sequence repeats again. Since the next state is based on multiplication, the state can never be zero (or all future states would also be zero). Thus the repeated number you see is the first one, and zero is the one that is never returned.
Apple has been promoting the use of better random number generators in their documentation and examples for at least as long as macOS (or OS X) has existed, so the quality of rand() is probably not deemed important, and they've just stuck with one of the simplest pseudorandom generators available. (As you noted, their rand() is even commented with a recommendation to use arc4random() instead.)
On a related note, the simplest pseudorandom number generator I could find that produces decent results in this (and many other) tests for randomness is xorshift*:
uint64_t x = *ctx;
x ^= x >> 12;
x ^= x << 25;
x ^= x >> 27;
*ctx = x;
return (x * 0x2545F4914F6CDD1DUL) >> 33;
This implementation results in almost exactly 790 million duplicates in your test.
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Apr 24