Before I start doing my own problems where G&J *don’t* give proofs, here’s a really simple extension of Hamiltonian Circuit that will be useful.

**The** **problem: **Hamiltonian Path

**The description:** Given a graph G=(V,E), where |V| = N. Can I create an ordering <v_{1}, v_{2}…v_{n}>of the vertices where an edge exists between each (v_{i}, v_{i+1}) in the sequence?

(This is the Hamiltonian Circuit problem without the restriction that I return to the start)

EDIT 4/24/15. My original reduction doesn’t work. I’ve left it below. Thanks to Fernando Berzal for pointing it out to me. I’ve added his answer, which looks correct.

In all honesty, the best part about being wrong is that it means people actually read these things!

**The new Reduction: **From HC. Given a graph G =(V,E), an instance of HC, pick any vertex v.

We’re going to build a graph G’ =(V’, E’)

V’ = All of the vertices in V, plus a new vertex v’ (which is a “copy” of v)

E’ = All of the edges in E, plus an edge (v’, w) if there was an edge (v,w) in E.

So, G’ is G, with one vertex (v) “duplicated” as v’. v’ has edges to all of the places v has.

We want to see if there’s an HP from v to v’.

If G’ has an HP from v to v’, then the last edge in the path goes from some vertex u to v’. If we replace that edge with the edge (u,v) we get a HC in G.

If G has an HC, then we can follow that cycle starting from v. Instead of the last edge coming back to v, follow it to v’. This will give us a HP in G’

**My reduction that doesn’t work**: From HC. Given a graph G_{HC}, an instance of HC, we build G_{HP} to be the exact same graph with 2 vertices added- v_{start} and v_{end}. We add two edges, one from v_{start} to some vertex in G_{HC}, and one from v_{end} to the same vertex in G_{HC}. The idea is that a Hamiltonian path in this graph would start at v_{start}, travel through the HC in the given graph (if it exists), and end at v_{end}.

This doesn’t work because if the two new vertices both connect to the same vertex then we hit it twice in our HP, which breaks the rules.

**Difficulty**: 2. This is a good “simple reduction” for students to work on. I don’t use it because it’s just a variant on the same problem, and I like to drive home how reductions tie very different-looking problems together. But I do like how it’s simple, but does make you make some changes to the original problem.

(I think that I’m going to upgrade this difficulty to 4 now, since I can see a lot of students making the same mistake I did)

Dave Burchillin the new reduction how do you know that the HP found will be from v to v’, what if there are other HPs in G’. You could add a vertex v” and an edge (v”, v’), that way v” must be the start of any HP as it only has outgoing edges. I’m not sure what the edge (v’, w) is for maybe it addresses this in someway I don’t see right now.

Sean T. McCullochPost authorThere are _lots_ of edges from v’ to w, one for each edge incident on v.

I guess you can think of this as a reduction to “HP with designated endpoints”, but it’s pretty easy to go from there to regular HP.