Returning a random even number

Question

I have the following methods. The method rnd, returns a single random integer between two bounds:

  /* Create next batch of 55 random numbers */
  void advance_random (){
int j1;
double new_random;
for(j1=0; j1<24; j1++){
  new_random = oldrand[j1]-oldrand[j1+31];
  if(new_random<0.0){
    new_random = new_random+1.0;
  }
  oldrand[j1] = new_random;
}
for(j1=24; j1<55; j1++){
  new_random = oldrand[j1]-oldrand[j1-24];
  if(new_random<0.0){
    new_random = new_random+1.0;
  }
  oldrand[j1] = new_random;
}
 } //advance_ramdom

  /* Fetch a single random number between 0.0 and 1.0 */
  double randomperc(){
jrand++;
if(jrand>=55){
  jrand = 1;
  advance_random();
}
return((double)oldrand[jrand]);
  } //randomPerc

  /* Fetch a single random integer between low and high including the bounds */
  synchronized int rnd (int low, int high){
int res;
if (low >= high){
  res = low;
} else {
  res = low + (int)(randomperc()*(high-low+1));
  if (res > high){
    res = high;
  }
}
return (res);
  } // rnd

How do I modify this so that the number returned mod2 =0?

Thanks

http://eternallyconfuzzled.com/arts/jsw_art_rand.aspx – sehe Jul 05 '11 at 11:39 — sehe, Jul 05 '11 at 11:39

Armen Tsirunyan · Accepted Answer · 2011-07-05T11:32:02.833

8

if you can get a random number in range [a, b] then all you have to do is get a random number in the range [(a+1)/2, b/2] and multiply it by 2 to get a random even number in range [a, b]

edited Jul 05 '11 at 11:32

answered Jul 05 '11 at 10:45

Armen Tsirunyan

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You need to round up the lower bound, otherwise an even between 3 and 5 can return 2. `2 * rnd((low+1)/2, high/2)` – Peter Lawrey Jul 05 '11 at 11:15

score 5 · Answer 2 · answered Jul 05 '11 at 10:46

5

Use a bit-mask to force the least-significant bit to be zero:

x = x & ~1;

answered Jul 05 '11 at 10:46

Oliver Charlesworth

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Note that the range should be adapted before to ensure a random distribution. For example, if the range is [2, 4], you'll get twice more 2s than 4s. You should thus extend the range to [2, 5] to get a random distribution between 2 and 4. This becomes negligible for very large ranges, of course. And if the range is [3, n], you'll get some 2s, which are out of range. – JB Nizet Jul 05 '11 at 10:55

score 2 · Answer 3 · answered Jul 05 '11 at 10:46

2

Multiply the result you get from your code by two at the end - still random, and divisble by 2!

answered Jul 05 '11 at 10:46

Jon Egerton

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You would need to be careful about overflow; specifically, you should reduce the range of your raw random numbers. – Oliver Charlesworth Jul 05 '11 at 10:49

score 1 · Answer 4 · answered Jul 05 '11 at 10:47

1

how about using:

return res & ~1;

answered Jul 05 '11 at 10:47

qbert220

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score 0 · Answer 5 · edited May 23 '17 at 12:17

In Java 1.7 or later, I would use ThreadLocalRandom:

import java.util.concurrent.ThreadLocalRandom;

// Get even random number within range [min, max]
// Start with an even minimum and add random even number from the remaining range
public static int randEvenInt(int min, int max) {
    if (min % 2 != 0) ++min;
    return min + 2*ThreadLocalRandom.current().nextInt((max-min)/2+1);
}

// Get odd random number within range [min, max]
// Start with an odd minimum and add random even number from the remaining range
public static int randOddInt(int min, int max) {
    if (min % 2 == 0) ++min;
    return min + 2*ThreadLocalRandom.current().nextInt((max-min)/2+1);
}

The reason to use ThreadLocalRandom is explained here. Also note, that the reason we +1 to the input to ThreadLocalRandom.nextInt() is to make sure the max is included in the range.

Returning a random even number

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