Create a random order of (x, y) pairs, without repeating/subsequent x's

Question

Say I have a list of valid X = [1, 2, 3, 4, 5] and a list of valid Y = [1, 2, 3, 4, 5].

I need to generate all combinations of every element in

Answer 1

Distribute the x values (5 times each value) evenly across your output:

import random

def random_combo_without_x_repeats(xvals, yvals):
    # produce all valid combinations, but group by `x` and shuffle the `y`s
    grouped = [[x, random.sample(yvals, len(yvals))] for x in xvals]
    last_x = object()  # sentinel not equal to anything
    while grouped[0][1]:  # still `y`s left
        for _ in range(len(xvals)):
            # shuffle the `x`s, but skip any ordering that would
            # produce consecutive `x`s.
            random.shuffle(grouped)
            if grouped[0][0] != last_x:
                break
        else:
            # we tried to reshuffle N times, but ended up with the same `x` value
            # in the first position each time. This is pretty unlikely, but
            # if this happens we bail out and just reverse the order. That is
            # more than good enough.
            grouped = grouped[::-1]
        # yield a set of (x, y) pairs for each unique x
        # Pick one y (from the pre-shuffled groups per x
        for x, ys in grouped:
            yield x, ys.pop()
        last_x = x

This shuffles the y values per x first, then gives you a x, y combination for each x. The order in which the xs are yielded is shuffled each iteration, where you test for the restriction.

This is random, but you'll get all numbers between 1 and 5 in the x position before you'll see the same number again:

>>> list(random_combo_without_x_repeats(range(1, 6), range(1, 6)))
[(2, 1), (3, 2), (1, 5), (5, 1), (4, 1),
 (2, 4), (3, 1), (4, 3), (5, 5), (1, 4),
 (5, 2), (1, 1), (3, 3), (4, 4), (2, 5),
 (3, 5), (2, 3), (4, 2), (1, 2), (5, 4),
 (2, 2), (3, 4), (1, 3), (4, 5), (5, 3)]

(I manually grouped that into sets of 5). Overall, this makes for a pretty good random shuffling of a fixed input set with your restriction.

It is efficient too; because there is only a 1-in-N chance that you have to re-shuffle the x order, you should only see one reshuffle on average take place during a full run of the algorithm. The whole algorithm stays within O(N*M) boundaries therefor, pretty much ideal for something that produces N times M elements of output. Because we limit the reshuffling to N times at most before falling back to a simple reverse we avoid the (extremely unlikely) posibility of endlessly reshuffling.

The only drawback then is that it has to create N copies of the M y values up front.