How to get all combinations of n binary values where number of 1's are equal to or more than the number of 0's?

Question

I want to find a list of all possible combinations of 0’s and 1’s. The only condition is that the number of 1’s must be more than or equal to the number of 0’s. For example for n = 4 the output should be something like this:

[(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0), (1, 1, 1, 1)]

Is there an elegant way to do this?

Asked By: Mert Onur

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Source

Answer 1

You can use distinct_permutations:

from more_itertools import distinct_permutations

def get_combos(n):
    for i in range((n+1)//2, n + 1):
        for permutation in distinct_permutations([1] * i + [0] * (n - i), n):
            yield permutation
print(list(get_combos(4)))
# [(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0), (0, 1, 1, 1), (1, 0, 1, 1), (1, 1, 0, 1), (1, 1, 1, 0)]

Here, we simply consider the permutations of each sublist:

[0, 0, 1, 1]
[0, 1, 1, 1]
[1, 1, 1, 1]

Notice that for large n, the yield statement is very useful because you do not generate all permutations at once.

We need to use distinct_permutations because you use just 1’s and 0’s , so regular permutations will give you repeated elements.

If you don’t want to install another library, you can use:

from itertools import permutations

def get_combos(n):
    for i in range(n // 2 if n%2 == 0 else n//2 + 1, n):
        for permutation in permutations([1] * i + [0] * (n - i), n):
            yield permutation
print(set(get_combos(4)))
# {(0, 1, 0, 1), (0, 1, 1, 1), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 1, 0), (0, 1, 1, 0), (1, 0, 1, 0), (1, 0, 0, 1), (1, 1, 0, 1), (0, 0, 1, 1)}

as set will eliminate the repeated elements, at the cost of needing to process the entire set of permutations at once (ie., by calling set, you will consume the entire generator immediately, rather than drawing elements from it as you need them).

More details on distinct_permutations

It might not be clear why these are needed. Consider this list:

[1, 2]

permutations, by default, will tell you that all the permutations of this list are

(1, 2)

and

(2, 1)

However, permutations doesn’t bother checking what the elements are or if they are repeated, so it simply performs the swap as above and if the list is

[1, 1]

you’ll get back

[(1, 1), (1, 1)]

Answered By: Kraigolas

Answer 2

Building up on @Kraigolas excellent answer, we can even simplify the entire process by building one single initial list containing n // 2 zeros (0) and n ones (1). We then get the distinct permutations of length n only from that initial list

So for n = 4 we return distinct permutation of size 4 from the following list: [0, 0, 1, 1, 1, 1]

from more_itertools import distinct_permutations


def get_combos(n):
    return distinct_permutations([0] * (n // 2) + [1] * n, n)
    
    
print(list(get_combos(4)))
# [(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0), (1, 1, 1, 1)]

Answered By: Zeitounator

Answer 3

You could use itertools.product and filter (you keep at least half of them, so it’s not that wasteful):

from itertools import product

def get_combos(n):
    minimum = -(-n // 2)
    return [p
            for p in product([0, 1], repeat=n)
            if sum(p) >= minimum]

print(get_combos(4))

(Might be faster to use p.count(1).)

Or itertools.combinations for choosing k of the n indexes to set to 0:

from itertools import combinations

def get_combos(n):
    for k in range(n//2 + 1):
        for indexes in combinations(range(n), k):
            combo = [1] * n
            for i in indexes:
                combo[i] = 0
            yield tuple(combo)

print(list(get_combos(4)))

Answered By: Kelly Bundy

How to get all combinations of n binary values where number of 1's are equal to or more than the number of 0's?

Question:

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More details on distinct_permutations