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

This should do what you want.

rando will never generate the same X twice in a row, but I realized that it is possible (though seems unlikely, in that I never noticed it happen in the 10 or so times I ran without the extra check) that due to the potential discard of duplicate pairs it could happen upon a previous X. Oh! But I think I figured it out... will update my answer in a moment.

import random

X = [1,2,3,4,5]
Y = [1,2,3,4,5]


def rando(choice_one, choice_two):
    last_x = random.choice(choice_one)
    while True:
        yield last_x, random.choice(choice_two)
        possible_x = choice_one[:]
        possible_x.remove(last_x)
        last_x = random.choice(possible_x)


all_pairs = set(itertools.product(X, Y))
result = []
r = rando(X, Y)
while set(result) != all_pairs:
    pair = next(r)
    if pair not in result:
        if result and result[-1][0] == pair[0]:
            continue
        result.append(pair)

import pprint
pprint.pprint(result)

Answer 2

Here's a solution using NumPy

def generate_pairs(xs, ys):
    n = len(xs)
    m = len(ys)
    indices = np.arange(n)

    array = np.tile(ys, (n, 1))
    [np.random.shuffle(array[i]) for i in range(n)]

    counts = np.full_like(xs, m)
    i = -1

    for _ in range(n * m):
        weights = np.array(counts, dtype=float)
        if i != -1:
            weights[i] = 0
        weights /= np.sum(weights)

        i = np.random.choice(indices, p=weights)
        counts[i] -= 1
        pair = xs[i], array[i, counts[i]]
        yield pair

Here's a Jupyter notebook that explains how it works

Inside the loop, we have to copy the weights, add them up, and choose a random index using the weights. These are all linear in n. So the overall complexity to generate all pairs is O(n^2 m)

But the runtime is deterministic and overhead is low. And I'm fairly sure it generates all legal sequences with equal probability.

Answer 3

Here is my solution. First the tuples are chosen among the ones who have a different x value from the previous selected tuple. But I ve noticed that you have to prepare the final trick for the case you have only bad value tuples to place at end.

import random

num_x = 5
num_y = 5

all_ys = range(1,num_y+1)*num_x
all_xs = sorted(range(1,num_x+1)*num_y)

output = []

last_x = -1

for i in range(0,num_x*num_y):

    #get list of possible tuple to place    
    all_ind    = range(0,len(all_xs))
    all_ind_ok = [k for k in all_ind if all_xs[k]!=last_x]

    ind = random.choice(all_ind_ok)

    last_x = all_xs[ind]
    output.append([all_xs.pop(ind),all_ys.pop(ind)])


    if(all_xs.count(last_x)==len(all_xs)):#if only last_x tuples,
        break  

if len(all_xs)>0: # if there are still tuples they are randomly placed
    nb_to_place = len(all_xs)
    while(len(all_xs)>0):
        place = random.randint(0,len(output)-1)
        if output[place]==last_x:
            continue
        if place>0:
            if output[place-1]==last_x:
                continue
        output.insert(place,[all_xs.pop(),all_ys.pop()])

print output