Consider a simple connected graph $G$ with n vertices and n edges $(n>2)$. Then, which of the following statements are true?

  1. $G$ has no cycles
  2. The graph obtained by removing any edge from $G$ is not connected
  3. $G$ has at least one cycle
  4. The graph obtained by removing any two edges from $G$ is not connected

My attempt :

  1. always false
  2. not always true
  3. true (since exactly one is subset of at least one !).
  4. always true

Can you explain in formal way, please?

  • 1
    $\begingroup$ I think you are right: Option 3 is always correct. And no, it may not be an Euler graph. Consider a tree where every node is connected to a unique root, and add one more edge. $\endgroup$ Sep 10 '15 at 11:02
  • 1
    $\begingroup$ The third option is correct, otherwise you would deal with a forset ( a disjoint union of trees). But a standard exercize shows that in a tree $|V|-|E|=1$, so you cannot have the same number of edges and vertices. $\endgroup$
    – Crostul
    Sep 10 '15 at 11:03
  • $\begingroup$ Yes , here given graph is connected . I'm able to find the exactly one cycle such given graph . So , on basis only "exactly one is subset of at least one" , option 3 is true , but there is no graph such that contain cycle more than exactly one ! Am I right ? or that graphs are exists ? $\endgroup$
    – ً ً
    Sep 10 '15 at 12:03
  • $\begingroup$ It is exactly one, as you say. One way to see it is to suppose that by adding an $n$-th edge to a graph with $n - 1$ edges, you get two cycles (which couldn't have been there before). Then the two cycles share the inserted edge, and you should then be able to see a cycle that doesn't contain this edge. $\endgroup$ Sep 10 '15 at 21:16
  • $\begingroup$ @ManuelLafond , as you "by adding an n-th edge to a graph with n−1 edges, you get two cycles" , how ? Is order of cycle matter here ? $\endgroup$
    – ً ً
    Sep 11 '15 at 5:49

1) Always false. A graph with no cycles (also known as a tree) has $n$ nodes and $n-1$ edges. Add one more edge and you're going to make a cycle somewhere. You can prove this by induction.

2) Not necessarily true. Easy counterexample: a graph with $n$ nodes and $n$ edges that forms a circle (i.e. a single cycle). Take out an edge anywhere and the graph is still connected.

3) Always true, see (1). If "G has no cycle" is always false, then it always has at least one cycle.

4) Always true. Think about it like this: if it's a connected graph with $n$ vertices and $n$ edges, and you remove one edge, then you have $n-1$ edges. If it's still connected, then it's a tree. (If it is not connected, then we're done.) Now you have a tree. Remove any edge from a tree, and your tree is split into two connected components.


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