This method randomly increments or decrements (with equal probability) until either -3 or 3 are reached. I'd like to maintain printing the value of pos on each iteration, and printing the largest positive number max reached at the end of the loop. How can I improve this method using the Random() function? I know this sounds pretty ambiguous, but I'd like to know if I can make rand.nextInt(2) == 0 an easier condition to understand? Any suggestions/improvements are welcome!

public static void randomWalk() {
    Random rand = new Random();
    int pos = 0;
    int max = 0;
    System.out.println("position = " + pos);
    while (pos > -3 && pos < 3) {
        if (rand.nextInt(2) == 0) {
            if (pos > 0) {
                max = pos;
        } else {
        System.out.println("position = " + (int) pos);
    System.out.println("max position = " + max);

2 Answers 2


There's a minor bug there:

        if (pos > 0) {
            max = pos;

should test pos > max instead.


max = Math.max(max, pos)

to make it shorter.

I'd like to know if I can make rand.nextInt(2) == 0 an easier condition to understand?

Use rand.nextBoolean(). For a probability of exactly 50%, it's perfect.

For other probabilities, you can use

 rand.nextDouble() < probability

Another possibility is simply

pos += 2 * rand.nextInt(2) - 1;
max = Math.max(max, pos);


pos += rand.nextBoolean() ? 1 : -1;
max = Math.max(max, pos);


private static final int[] DELTAS = {+1, -1};

pos += DELTAS[rand.nextInt(DELTAS.length)];
max = Math.max(max, pos);

without any conditionals. You may consider it tricky, but it isn't. Obviously, my last solution shouldn't be used for such a simple case.

Concerning the computation of max it may be slightly less efficient since it gets executed even if the value decreases. But this doesn't matter unless you need to optimize heavily (your print is many orders of magnitude slower than this).

  • \$\begingroup\$ Thanks! I'm glad there is an easier way to determine probability, and that gives me a different perspective on what a boolean can be, rather than just true or false. \$\endgroup\$
    – cody.codes
    Commented Jun 30, 2015 at 1:05


pos > -3 && pos < 3

Can be

Math.abs(pos) < 3

Random choice

Java Random had method for sampling boolean values nextBoolean


This line

if (pos > 0) {

do not give you

the largest positive number max reached at the end of the loop


You don't need to make loop at all. You need random generator that will provide 2 values at each iteration:

  • max positive number
  • 3 or -3

This can be done using random in following way

long sample = Math.abs(rand.nextLong());
int pos, max;
if (sample % 2 == 0) {
    pos = 3;
    max = 3;
} else {
    pos = -3;
    max = int(sample % 3);
  • \$\begingroup\$ Concerning pos, you're right. Concerning max, you're most probably wrong as not all values are equiprobable. Even if they were, you're missing the case max==3. Concerning the pair (pos, max), as the probabilities are'nt independent, for example pos==-3 and max==2 is rather improbable. And pos==3 and max<3 is plain impossible. \$\endgroup\$
    – maaartinus
    Commented Jun 27, 2015 at 19:45
  • \$\begingroup\$ @maaartinus max never be 3, read TC's code. I'm so sorry I can't downvote this comment. pastebin.com/nnyTp0CK \$\endgroup\$
    – outoftime
    Commented Jun 27, 2015 at 20:54
  • 3
    \$\begingroup\$ Why shouldn't it be 3. First gets pos incremented to 3, then max computed, and then the loop exits. However, it's all irrelevant, as you can't shortcut the max computation to something that trivial. -1 This simply can't work. \$\endgroup\$
    – maaartinus
    Commented Jun 27, 2015 at 21:10
  • \$\begingroup\$ @maaartinus Now I see problem with max value. Fixed. \$\endgroup\$
    – outoftime
    Commented Jun 27, 2015 at 21:17
  • \$\begingroup\$ @maaartinus but I'm not agree that part of uniform distribution can be not uniform. This relate to max and pos values, as they are both parts of uniformly distributed values of sample \$\endgroup\$
    – outoftime
    Commented Jun 27, 2015 at 21:21

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