Question

I’m trying to figure out a way to create random numbers that “feel” random over short sequences. This is for a quiz game, where there are four possible choices, and the software needs to pick one of the four spots in which to put the correct answer before filling in the other three with distractors.

Obviously, arc4random % 4 will create more than sufficiently random results over a long sequence, but in a short sequence its entirely possible (and a frequent occurrence!) to have five or six of the same number come back in a row. This is what I’m aiming to avoid.

I also don’t want to simply say “never pick the same square twice,” because that results in only three possible answers for every question but the first. Currently I’m doing something like this:

bool acceptable = NO;
do {
  currentAnswer = arc4random() % 4;
  if (currentAnswer == lastAnswer) {
    if (arc4random() % 4 == 0) {
      acceptable = YES;
    }
  } else {
    acceptable = YES;
  }
} while (!acceptable);

Is there a better solution to this that I’m overlooking?

Answer 1

If your question was how to compute currentAnswer using your example’s probabilities non-iteratively, Guffa has your answer.

If the question is how to avoid random-clustering without violating equiprobability and you know the upper bound of the length of the list, then consider the following algorithm which is kind of like un-sorting:

from random import randrange
# randrange(a, b) yields a <= N < b

def decluster():
    for i in range(seq_len):
        j = (i + 1) % seq_len
        if seq[i] == seq[j]:
            i_swap = randrange(i, seq_len) # is best lower bound 0, i, j?
            if seq[j] != seq[i_swap]:
                print 'swap', j, i_swap, (seq[j], seq[i_swap])
                seq[j], seq[i_swap] = seq[i_swap], seq[j]

seq_len = 20
seq = [randrange(1, 5) for _ in range(seq_len)]; print seq
decluster(); print seq
decluster(); print seq

where any relation to actual working Python code is purely coincidental. I’m pretty sure the prior-probabilities are maintained, and it does seem break clusters (and occasionally adds some). But I’m pretty sleepy so this is for amusement purposes only.