Random walk with equal probability

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  0) {
                max = pos;
            }
        } else {
            pos--;
        }
        System.out.println("position = " + (int) pos);
            
    }
    System.out.println("max position = " + max);
}

There's a minor bug there:

if (pos > 0) {
            max = pos;
        }

should test pos > max instead.

Use

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);

or

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

or

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).

StackExchange Code Review Q#94916, answer score: 12