Common Membership Algorithm
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22-07-2019 - |
Question
People can belong to one or many groups. What is a good algorithm to output common memberships?
ie, Persons A and B are in Groups C, D, and E ... etc
My preferred language would be Ruby (or maybe Python), but any code or pseudocode would be greatly appreciated.
Solution
It's a very simple algorithm, actually (at least for reasonable numbers of users and groups).
Consider each user to be a set whose elements are the groups of which they are a member. To find the groups two users have in common, simply take the intersection of those two users' membership sets.
So, if Person A is in Group K, M, and N, and Person B is in K, N, and P, you would have the following sets:
A := {K, M, N}
B := {K, N, P}
intersect(A, B) = {K, N}
In Ruby, you can use the standard library class Set
to perform these calculations:
require 'set'
memberships_a = Set[:K, :M, :N]
memberships_b = Set[:K, :N, :P]
shared = memberships_a.intersection(memberships_b)
# you can also use the '&' operator as shorthand for 'intersection'
shared_2 = memberships_a & memberships_b
OTHER TIPS
Did you mean something like the below? (python):
>>> a_groups = set(["A", "B", "C"])
>>> b_groups = set(["B", "C", "D"])
>>> print a_groups & b_groups
set(['C', 'B'])
>>>
Are you trying to find anything in particular about the memberships? Or are you just trying to find all memberships... Ie:
A - No group
B - Groups 1, 2, 3
C - Groups 2, 5
D - Groups 2, 3, 4
If it's the latter, I don't think there's a special algorithm to do this; as long as verifying that a person is in a group takes O(1) your best bet is the O(M * N) brute force algorithm.
For each person O(N) {
Create a set for this person
For each group O(M) {
if the person is in the group, add this group to the set O(1) when using maps/hashed structures
}
output the set
}
If you're looking for the intersection of sets, there are many other algorithms out there... but this particular problem is nothing special.