Making a Transitive Pair Set using STL C++

Question

I'd like to create a set which has transitive pairs. My input will be of the form pair<int, int> and I need a set which has all transitive pairs for the given inputs. For example, if I have pairs {1, 2} {2, 1} {2, 3} {3,4} as inputs, then I need to have a SET which has pairs { {1,2}, {2,1}, {1, 3}, {3, 4}, {1, 4}, {2, 4} }. I also need to find whether a given pair is a member of this transitive set. Is there any built-in data structure/STL library that will allow me to accomplish this in C++?

Answer 1

You have a graph and you want to turn it into a free category, AKA the transitive closure of that graph.

Your int pair elements are vertices, the pairs themselves are edges, that's your graph. Now a free category on a graph is that graph with an edge composition law, and all additional edges that can be produced by the law. The law says that

• if there is an edge (or "arrow" as the category theory names it) f that runs from a to b
• and an edge g that runs from b to c
• then there's another edge denoted f∘g that runs from a to c
• the composition is associative ((f∘g)∘h = f∘(g∘h))

(Every category also have unit edges, denoted 1a that run from a to a for every vertex a, and a law which says that for every arrow f that runs from a to b, 1a∘f = f∘1b= f, but we are not talking about those).

_{The general education part of this answer is hereby completed.}

There are no relevant algorithms in the C++ standard library, but you may want to check boost::graph or any other graph-oriented library. boost::graph has the transitive_closure method which is just what you want.

Answer 2

You can represent your set with:

set<pair<int, int>> myset;

The transitive closure could then be computed with:

void transitive_closure(set<pair<int, int>>& s)
{
    set<pair<int, int>> a;   // missing nodes to add
    for (auto i = s.cbegin(); i != s.cend(); i++)
    {
        for (auto j = i; ++j != s.cend(); j)
        {
            if (i->second == j->first 
                  && i->first != j->second 
                  && s.count(make_pair(i->first, j->second))==0 )
                a.insert(make_pair(i->first, j->second));
        }
    }
    if (!a.empty()) {
        for (auto p : a)
            s.insert(p);
        transitive_closure(s);
    }
}

It's not the most optimized algorithm, because the recursion will repeat some comparisons. But it works. By he way, I think you've forgotten {2, 3} in your result set.

To check if a pair belongs to the set, just check if s.count(make_pair(i->first, j->second)!=0