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The vertices of an undirected graph are numbered 1,2,...4286. The edge (i,j) exists if |i-j| <= 3, where i!=j. Which statements are true?

  • the graph contains an Eulerian circle
  • the graph contains a perfect matching graph
  • the graph is Hamiltonian

I'm thinking that the last two are true, while the first one is false. Is this correct?

aneys
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  • Why do you think that? Please explain your rationale, so answerers can help you better by correcting your mistake (if there are such). – amit Jun 25 '21 at 15:03
  • All vertices need to have an even degree for it to be an Eulerian circle and vertex 1 has the degree 3, so it isn't an Eulerian circle. This is the only one I'm sure of. The other two are more based on intuition. For it to be a Hamiltonian graph I think you can go from even vertex to even vertex, then return from odd vertex to odd vertex @amit – aneys Jun 25 '21 at 15:21

1 Answers1

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The graph is not an Eulerian cycle since such a cycle requires that every node has an even degree, but in this graph, node 1 has an odd degree -- its neighbors are 2, 3 and 4.

The graph contains a perfect matching: one possible match would be the collection of edges that connects to +1, for each odd :

1─2, 3─4, 5─6, ... , 4285─4286

The graph is a Hamiltonian graph, as it contains the following Hamiltonian cycle:

1─2 ─ 4─5 ─ 7─8 ─ 10─11 ─ 13─14 ─ ...    ─ 4282─4283 ── 4285─4286   
 \                                                            /
  ─ 3 ─── 6 ─── 9 ──── 12 ──── ... ─── 4281 ─────── 4284 ─────

So at the bottom we have all multiples of 3, and at the top all the other values.

trincot
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  • Thanks for the response! And if I had an odd number instead of an even number, then it would only be a Hamiltonian graph, right? – aneys Jun 25 '21 at 20:30