Question

(This is derived from a recently completed programming contest)

You are given G, a connected graph with N nodes and N-1 edges.

(Notice that this implies G forms a tree.)

Each edge of G is directed. (not necessarily upward to any root)

For each vertex v of G it is possible to invert zero or more edges such that there is a directed path from every other vertex w to v. Let the minimum possible number of edge inversions to achieve this be f(v).

By what linear or loglinear algorithm can we determine the subset of vertexes that have the minimal overall f(v) (including the value of f(v) of those vertexes)?

For example consider the 4 vertex graph with these edges:

A<--B
C<--B
D<--B

The value of f(A) = 2, f(B) = 3, f(C) = 2 and f(D) = 2…

..so therefore the desired output is {A,C,D} and 2

(note we only need to calculate the f(v) of vertexes that have a minimal f(v) – not all of them)

Code:

For posterity here is the code of solution:

int main()
{
    struct Edge
    {
        bool fwd;
        int dest;
    };

    int n;
    cin >> n;

    vector<vector<Edge>> V(n+1);

    rep(i, n-1)
    {
        int src, dest;
        scanf("%d %d", &src, &dest);

        V[src].push_back(Edge{true, dest});
        V[dest].push_back(Edge{false, src});
    }

    vector<int> F(n+1, -1);

    vector<bool> done(n+1, false);

    vector<int> todo;
    todo.push_back(1);
    done[1] = true;
    F[1] = 0;

    while (!todo.empty())
    {
        int next = todo.back();
        todo.pop_back();

        for (Edge e : V[next])
        {
            if (done[e.dest])
                continue;

            if (!e.fwd)
                F[1]++;
            done[e.dest] = true;
            todo.push_back(e.dest);
        }
    }

    todo.push_back(1);

    while (!todo.empty())
    {
        int next = todo.back();
        todo.pop_back();

        for (Edge e : V[next])
        {
            if (F[e.dest] != -1)
                continue;

            if (e.fwd)
                F[e.dest] = F[next] + 1;
            else
                F[e.dest] = F[next] - 1;

            todo.push_back(e.dest);
        }
    }

    int minf = INT_MAX;

    rep(i,1,n)
        chmin(minf, F[i]);

    cout << minf << endl;

    rep(i,1,n)
        if (F[i] == minf)
            cout << i << " ";
    cout << endl;

}

Answer 1

I think that the following algorithm works correctly, and it certainly works in linear time.

The motivation for this algorithm is the following. Let’s suppose that you already know the value of f(v) for some single node v. Now, consider any node u adjacent to v. If we want to compute the value of f(u), we can reuse some of the information from f(v) in order to compute it. Note that in order to get from any node w in the graph to u, one of two cases must happen:

1. That path passes through the edge connecting u and v. In that case, the way that we get from w to u is to go from w to v, then to follow the edge from v to u.
2. That path does not pass through the edge connecting u and v. In that case, the way that we get from w to u is the exact same way that we got from w to v, except that we stop as soon as we get to u.

The reason that this observation is important is that it means that if we know the number of edges we’d flip to get from any node to v, we can easily modify it to get the set of edges that we’d flip to get from any node to u. Specifically, it’s going to be the same set of edges as before, except that we want to direct the edge connecting u and v so that it connects v to u rather than the other way around.

If the edge from u to v is initially directed (u, v), then we have to flip all the normal edges we flipped to get every node pointing at v, plus one more edge to get v pointed back at u. Thus f(u) = f(v) + 1. Otherwise, if the edge is originally directed (v, u), then the set of edges that we’d flip would be the same as before (pointing everything at v), except that we wouldn’t flip the edge (v, u). Thus f(u) = f(v) – 1.

Consequently, once we know the value of f for a single node v, we can compute it for each adjacent node u as follows:

f(u) = f(v) + 1    if (u, v) is an edge.
f(u) = f(v) - 1    otherwise

This means that we can compute f(v) for all nodes v as follows:

1. Compute f(v) for some initial node v, chosen arbitrarily.
2. Do a DFS starting from v. When reaching a node u, compute its f score using the above logic.

All that’s left to do is to compute f(v) for some initial node. To do this, we can run a DFS from v outward. Every time we see an edge pointed the wrong way, we have to flip it. Thus the initial value of f(v) is given by the number of wrong-pointing edges we find during the initial DFS.

We thus can compute the f score for each node in O(n) time by doing an initial DFS to compute f(v) for the initial node, then a secondary DFS to compute f(u) for each other node u. You can then for-loop over each of the n f-scores to find the minimum score, then do one more loop to find all values with that f-score. Each of these steps takes O(n) time, so the overall algorithm takes O(n) time as well.

Hope this helps! This was an awesome problem!