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I recently needed to generate a series of unique random numbers (non-repeated) within a bounded range inside of the Linux kernel. The code I came up with is below. I'd appreciate any feedback -- again, please keep in mind that it's within the Linux kernel, so the standard C library is out of the question.

#include <linux/kernel.h>
#include <linux/random.h>
#include <linux/slab.h>

static int mod(int, int);
static int _is_repeated_int(int *, int, int);

/* proj2_create_randints - creates a buffer of random integers
 * @dstruct_size: number of random integers
 * @max_range: maximum range of values to generate
 * 
 * Allocates a buffer that can hold @dstruct_size random integers. If 
 * @max_range is greater than 0, mods all values by @max_value to ensure
 * they fall within the range. If @dstruct_size is greater than @max_range,
 * returns NULL. The buffer returned is guaranteed to have no duplicate values.
 *
 * On successs, it fills the buffer and returns a pointer to the buffer;
 * otherwise returns NULL.
 */
int *proj2_create_randints(int dstruct_size, int max_range)
{
    int *randint;
    int tmp, idx;

    if ((max_range != 0) && (dstruct_size > max_range))
        return NULL;

    randint = kmalloc(dstruct_size * sizeof(*randint), GFP_KERNEL);
    if (randint != NULL) {
        idx = 0;
        while (idx < dstruct_size) {
            tmp = get_random_int();
            if (max_range > 0)
                tmp = mod(tmp, max_range);
            /* Check if 'tmp' is already in array */
            if (!_is_repeated_int(randint, idx, tmp)) {
                randint[idx] = tmp;
                idx++;
            }
        }

    }
    return randint;
}

/* mod - returns x modulo m
 * @x: value to reduce
 * @m: modulo value
 *
 * Guaranteed to be positive.
 */
static int mod(int x, int m)
{
    return (x%m + m)%m;
}

/* _is_repeated_int - checks array against repeated values
 * @array: array of integers to check
 * @size: size of @array
 * @new_value: value to check array for
 */
static int _is_repeated_int(int *array, int size, int new_value)
{
    int i;

    if (size == 0)
        return 0;

    for (i = 0; i < size; i++) {
        if (array[i] == new_value) {
            return 1;
        }
    }

    return 0;
}
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2 Answers 2

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What are the valid input ranges? The documentation flags two special cases and the code handles them as special cases, but

  1. I don't see any checks for dstruct_size < 0
  2. If max_range is very large, the implementation of mod is incorrect. Consider e.g. mod(1 << 30, (1 << 30) + 1): the addition overflows into a negative value.

On the assumption that the intention is to guarantee that max_range will be zero or positive, why int instead of uint? Using uint would encode the guarantee of non-negative return values, and would remove the need for the mod helper.


            /* Check if 'tmp' is already in array */
            if (!_is_repeated_int(randint, idx, tmp)) {
                randint[idx] = tmp;
                idx++;
            }

is very inefficient. Consider the worst case: dstruct_size == max_range. The expected number of calls to get_random_int will be quadratic in max_range, and the overall running time of the method will be cubic.

A simple improvement which requires only dstruct_size calls to get_random_int and overall quadratic time is the following (pseudocode):

for (idx = 0; idx < dstruct_size; idx++) {
    tmp = (uint)get_random_int() % (max_range - idx);
    for (cmpIdx = 0; cmpIdx < idx; cmpIdx++) {
        // Discussed below
        if (tmp >= sorted(randint)[cmpIdx]) tmp++;
    }
    randint[idx] = tmp;
}

Obviously it needs minor amendment to handle the case max_range = 0, but that's not too difficult, especially if working with uint.

The big issue is that, as JS1 pointed out in comments, the loop which increments to avoid collisions needs to process the previous numbers in ascending order. There are two ways to do this: either keep a second, temporary, buffer with the numbers sorted; or sort randint as we go and then Fisher-Yates shuffle at the end. One uses twice as much memory; the other uses twice as many calls to get_random_int(). Both use insertion sort, and leave the overall running time as quadratic.

The Fisher-Yates option, however, does allow a fairly tricky elegant insertion.

for (i = 0; i < dstruct_size; i++) {
    tmp = (uint)get_random_int() % (max_range - i);
    for (j = 0; j < i; j++) {
        if (tmp >= randint[j]) tmp++;
        else {
            tmp2 = tmp;
            tmp = randint[j];
            randint[j] = tmp2;
        }
    }
    randint[i] = tmp;
}
// TODO Now Fisher-Yates shuffle

Observe that because randint contains sorted distinct values, once we find the insertion point for tmp we will never again hit the first case. In fact, that means we could optimise the inner loop and assignment to randint[i] as

    for (j = 0; j < i && tmp >= randint[j]; j++, tmp++) {
        // Empty loop body
    }
    for (; j <= i; j++) {
        tmp2 = tmp;
        tmp = randint[j];
        randint[j] = tmp2;
    }

I've followed your code in taking get_random_int() modulo a value, and that's ok if you don't need uniformity. However, in some cases the fact that get_random_int() % m is more likely to be 0 than m-1 (unless m is a power of two) is a problem. In that case you'd want an auxiliary get_uniform_random_int(m).

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  • \$\begingroup\$ Your proposed quadratic solution won't work. That inner loop will only work if the array is in sorted order. Imagine generating these numbers: 2, 1, 1. For that last 1, your loop will convert it to a 2, which is a repeat. \$\endgroup\$
    – JS1
    Feb 10, 2017 at 10:25
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Addendum to Peter Taylor's answer

This is more of a comment on Peter Taylor's answer which won't fit in a comment. You can rewrite his final answer in a single inner loop instead of two, if you do this:

// n = number of random numbers to generate
// m = random number range
for (int i = 0; i < n; i++) {
    int j;
    int r = rand() % (m - i);
    for (j=i-1; j >= 0 && r + j < arr[j]; j--) {
        arr[j+1] = arr[j];
    }
    arr[j+1] = r + (j+1);
}
// Now do a Fisher Yates shuffle.

The original version had two phases: one to find the proper insertion spot and one to move all the rest of the values backward. This version works from the back to the front, so it does everything in one phase. It moves all the values backward until it reaches the insertion spot, and then inserts the new value in that spot. It uses the observation that at index j, the tentative random number r will have the value r + j because it will have been increased by skipping all the j numbers before it in the array.

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