2

Let $k∈N$ and $G$ be a simple graph with $δ(G) = 2k$. Show that $G$ has a $(k + 1)$-edge-connected subgraph.

$δ(G)$ is defined as the minimum degree in the graph.

I thought about proceeding by induction on $k$ because I can see for a base case of $k=1$, the graph has every vertex of at least degree $2$, which means the graph is not a tree and thus contains a cycle. Cycles are always $2$-connected, so this cycle makes the desired $2$-edge-connected subgraph.

Then we assume the statement is true up to $k-1$ and show that it's true for $k$, but I don't see what to do next. And I'm not even sure if I did the right thing for the base case. How would I finish this proof? Or is there another/better way to do it without induction? Any help would be much appreciated.

Masie
  • 201
  • Your argument for k=1 is very neat but I do not think it is suitable for induction. I believe this problem is one where choosing the right result to aim for is crucial - see the answer below. –  Sep 13 '19 at 00:11

1 Answers1

3

We will prove the stronger result that if the sum of the degrees of the $n$ vertices of a simple graph is at least $2nk-2k+1$, then the graph has a $(k+1)$-edge-connected subgraph.

Consider such a graph and suppose it is not $(k+1)$-edge-connected. Then the vertices of the graph can be split into two subsets, $A$ and $B$, with at most $k$ edges connecting vertices of $A$ and vertices of $B$.

The sum of the degrees of vertices in at least one of $A$ and $B$, say $A$, must be at least $2mk-k+1$ where $m$ is the number of vertices in that subset. Deleting all the vertices of $B$ and the edges incident with these vertices then leaves a graph with $m$ vertices such that the sum of their degrees is at least $$(2mk-k+1)-k=2mk-2k+1.$$

The initial set up has now been reproduced in a subgraph with fewer vertices. This process cannot be continued indefinitely and so, at some point, we will obtain a subgraph which is $(k+1)$-edge-connected.

  • Very neat! How did you find the "right (induction) result to aim for"? Even if you gave me a hint to induct on the no. of nodes, I might still not find the $-2k +1$ correction term. Also BTW, minor typo in the standalone equation line: the RHS should say $2mk$. – antkam Sep 13 '19 at 02:54
  • 2
    Thanks for typo. I tried using the obvious '2nk' and found that it 'reduced'. I then investigated the reduction in the case k=2 and spotted what was happening. –  Sep 13 '19 at 06:54
  • Sorry I can't follow...Did you use induction on n or what? If so, then the induction should start with n$\geq$4k+1 to make graph have enough edges, but your latter reduction into smaller m makes no guarantee that m$\geq$4k+1? There should be something wrong with my comprehension, please help me with it – Silas Wang Feb 24 '22 at 10:12
  • In effect it is a proof by reductio ad absurdum. –  Feb 24 '22 at 11:02
  • I totally misunderstood...Now I see. Thank you. – Silas Wang Feb 24 '22 at 17:24
  • Sorry I'm coming late, but I don't understand how the sum of the degrees of the vertices in the subset A is at least $2mk−k+1$ – GodrickTheGrafted Apr 05 '22 at 14:30