识别不相交（超）集

Question

I am looking for an algorithm to identify non-intersecting (super-)sets in a set of sets.

Lets, assume I have a set of sets containing the sets A, B, C and D, ie {A, B, C, D}. Each set may or may not intersect some or all of the other sets.

I would like to identify non-intersecting (super-)sets.

Examples:

• If A & B intersect and C & D intersect but (A union B) does not intersect (C union D), I would like the output of {(A union B), (C union D)}
• If only C & D intersect, I would like the output {A, B, (C union D)}

I am sure this problem has long been solved. Can somebody point me in the right direction?

Even better would be of course if somebody had already done the work and had an implementation in python they were willing to share. :-)

Answer 1

I would turn this from a set problem into a graph problem by constructing a graph whose nodes are the graphs with edges connecting sets with an intersection.

Here is some code that does it. It takes a dictionary mapping the name of the set to the set. It returns an array of sets of set names that connect.

def set_supersets (sets_by_label):
    element_mappings = {}
    for label, this_set in sets_by_label.items():
        for elt in this_set:
            if elt not in element_mappings:
                element_mappings[elt] = set()
            element_mappings[elt].add(label)
    graph_conn = {}
    for elt, sets in element_mappings.items():
        for s in sets:
            if s not in graph_conn:
                graph_conn[s] = set()
            for t in sets:
                if t != s:
                    graph_conn[s].add(t)

    seen = set()
    answer = []
    for s, sets in graph_conn.items():
        if s not in seen:
            todo = [s]
            this_group = set()
            while 0 < len(todo):
                t = todo.pop()
                if t not in seen:
                    this_group.add(t)
                    seen.add(t)
                    for u in graph_conn[t]:
                        todo.append(u)
            answer.append(this_group)
    return answer

print(set_supersets({
    "A": set([1, 2]),
    "B": set([1, 3]),
    "C": set([4, 5]),
    "D": set([3, 6])
}))