Do connected graphs with more than N-1 edges always contain connected graph with N-1 edges?

Question

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.)

Answer 1

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.

Answer 2

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.