Is a list (potentially) divisible by another?


Say you have two lists A = [a_1, a_2, ..., a_n] and B = [b_1, b_2, ..., b_n] of integers. We say A is potentially-divisible by B if there is a permutation of B that makes a_i divisible by b_i for all i. The problem is then: is it possible to reorder (i.e. permute) B so that a_i is divisible by b_i for all i? For example, if you have

    A = [6, 12, 8]
    B = [3, 4, 6]

Then the answer would be True, as B can be reordered to be B = [3, 6, 4] and then we would have that a_1 / b_1 = 2, a_2 / b_2 = 2, and a_3 / b_3 = 2, all of which are integers, so A is potentially-divisible by B.

As an example which should output False, we could have:

    A = [10, 12, 6, 5, 21, 25]
    B = [2, 7, 5, 3, 12, 3]

The reason this is False is that we can't reorder B as 25 and 5 are in A, but the only divisor in B would be 5, so one would be left out.


Obviously the straightforward approach would be to get all the permutations of B and see if one would satisfy potential-divisibility , something along the lines of:

    import itertools
    def is_potentially_divisible(A, B):
      perms = itertools.permutations(B)
      divisible = lambda ls: all( x % y == 0 for x, y in zip(A, ls))
      return any(divisible(perm) for perm in perms)


What is the fastest way to know if a list is potentially-divisible by another list? Any thoughts? I was thinking if there's is a clever way to do this with primes , but I couldn't come up with a solution.

Much appreciated!

Edit : It's probably irrelevant to most of you, but for the sake of completeness, I'll explain my motivation. In Group Theory there is a conjecture on finite simple groups on whether or not there is a bijection from irreducible characters and conjugacy classes of the group such that every character degree divides the corresponding class size. For example, for U6(4) here are what A and B would look like. Pretty big lists, mind you!

Build bipartite graph structure - connect a[i] with all its divisors from b[]. enter image description here

Then find maximum matching) and check whether it is perfect matching (number of edges in matching is equal to the number of pairs (if graph is directed) or to doubled number).

Arbitrary chosen Kuhn algorithm implementation here.

@Eric Duminil made great concise Python implementation here

This approach has polynomial complexity from O(n^2) to O(n^3) depending on chosen matching algorithm and number of edges (division pairs) against factorial complexity for brute-force algorithm.


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