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We know that:

If we have N vertices To build a connected undirected graph, you'll need at least N-1 edges. Let M be the set of possible connected undirected graphs with N-1 edges.

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Can we prove or disprove that if there's an undirected connected graph with more than N-1 edges, it must contain one of the graphs in M? In other words, can we take one of the graphs in M and add edges to create this new graph?

(by "containing", I mean that it has all the edges of the other graph plus some more.)

NL3294
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  • "Can we prove or disprove that if there's an undirected connected graph with more than N-1 edges..." Are you talking about an undirected graph with N vertices and more than N-1 edges? – Sergey Kalinichenko Dec 09 '16 at 21:08
  • Yes, I'm only talking about undirected graphs, and the same N vertices in all cases. – NL3294 Dec 09 '16 at 22:34

2 Answers2

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Can we prove or disprove that if there's an undirected connected graph with more than N-1 edges, it must contain one of the graphs in M?

Assuming that undirected connected graph g with more than N-1 edges has N vertices, the answer is "yes".

You can prove it by constructing a Spanning Tree of g, which is a subgraph with N vertices and N-1 edges. The problem statea that M contains all such graphs, a spanning tree of g is a member of M. Since a spanning tree is constructed by removing edges from g, you can add these edges back, thus going from a member of M back to the original graph g.

Sergey Kalinichenko
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  • Ah, "spanning tree" was the word I forgot to Google. Been too long since doing this stuff. Thanks! – NL3294 Dec 09 '16 at 23:11
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No, this isn't necessarily the case. As an example, imagine a path graph with 2n nodes (and, therefore, 2n - 1 edges). Cut out the middle edge, splitting the graph into two connected components that are each path graphs. Both of these paths have n - 1 edges, but neither connected component is a connected graph with 2n - 2 edges.

templatetypedef
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