Question

This algorithm solves Hamiltonian path problem. G is a unoriented graph, v starting vertex,
G.size() size of the graph, G.get(v).gV all the neighbor verices of the current vertex.

static private void dfs(HashMap<Integer, Virsune> G, int v) {
    path.push(v);
    // add v to the current path
    onPath[v] = true;

    if (path.size() == G.size()) {
        System.out.println(path);

        Integer[] tmp = new Integer[G.size()];
        System.arraycopy(path.toArray(), 0, tmp, 0, path.size());
        hamPaths.add(tmp);
    }

    for (int w : G.get(v).gV) {
        if (!onPath[w]) {
            dfs(G, w);
        }
    }

    path.pop();
    onPath[v] = false;

} 
   // main method
   dfs(G,0);

Can I just say that complexity of this algorithm is O(n!) ?

Answer 1

This is algorithm is enumerating all the paths of the graph.

If you’re enumerating all the paths in a graph, this should give you a hint at the runtime. In a complete graph, there are indeed n! paths, so this is a lower bound. I’ll leave it to you to say if it’s an upper bound as well.

FWIW – the problem is solvable in O(2^n)