My question was to prove using Tutte theorem that $3$-regular graphs and $2$-edge-connected graphs have perfect matching.

For the $3$-regular-matching, i found the solution by myself, using Tutte's thorem $q(G-S)$ $\leq$ $|$S$|$, for any $S$ in $V$. ( I took a odd component, and made the sum of vertex)

But for the $2$-edge-connected graphs i can't find the connection between the Tutte relation?

Any hints/ideas for the $2$-edge-connected part?

  • $\begingroup$ But it's false that any $2$-edge-connected graph has a perfect matching, e.g., consider any odd cycle. $\endgroup$ – Casteels Dec 15 '16 at 21:45
  • $\begingroup$ I put the question ambigue a bit, maybe. I don't need to prove for any, just that "$a$" 2-edge-connected graph has a perfect matching. $\endgroup$ – JohnK Dec 15 '16 at 22:27
  • $\begingroup$ Ok so take any even cycle. $\endgroup$ – Casteels Dec 15 '16 at 22:36
  • $\begingroup$ I think maybe what you are being asked is to prove that any graph that is both $3$-regular and $2$-edge-connected has a perfect matching. (Actually it's not true that any $3$-regular graph has a perfect matching). $\endgroup$ – Casteels Dec 15 '16 at 22:38
  • $\begingroup$ Yes, i think that it its. Now i realise, the connection between 3-regular and 2-edge-connected. Any example of graph, of how to use tutte theorem? $\endgroup$ – JohnK Dec 15 '16 at 22:56

Let $G$ be a 3-regular 2-connected graph. Take any $S\subseteq V(G)$. We aim to show that $q(G-S)\leq |S|$. Here, $q(G-S)$ is the number of odd components of $G-S$. Tutte then gives the required 1-factor. Let $C$ be an odd component of $G-S$. Consider $$\sum_{v\in V(C)}deg(v)=3|C|.$$ This sum is clearly odd, since $|C|$ is odd. Any edge with both endpoints in $C$ contributes 2 to the sum, whereas any edge with only one endpoint in $C$ contributes 1. As such, edges totally contained in $C$ must contribute only an even part of the sum. Furthermore, any edge with only one endpoint in $C$ must have its other endpoint in $S$, since $C$ is disconnected from the rest of $G$ when $S$ is removed. This fact combined with the fact that the edges contained totally in $C$ contribute only an even portion of the sum gives us that there must be an odd number of edges from $C$ to $S$.

Since $G$ is 2-connected, there cannot be only one edge, so each $C$ must have at least 3 edges to $S$. The total number of edges with an endpoint in $S$ can be at most $3|S|$ since $G$ is 3-regular. As such, $$3q(G-S)\leq 3|S|$$ which gives $$q(G-S)\leq |S|$$ Tutte's theorem gives that $G$ must have a 1 factor.


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