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The final paragraph states Tutte's theorem on perfect matchings in terms of the number of components of odd integer order after deleting a vertex subset. The worked perfect matching theorem for a cubic graph with at most two bridges uses this criterion directly.

Use Tutte's theorem on perfect matchings: a finite graph has a perfect matching if and only if, for every $S\subseteq V$, the number $q(S)$ of connected components of a graph of odd integer order in $G-S$ is at most $|S|$.

First, the handshaking lemma gives $3|V|=2|E|$, so $|V|$ is an even integer. Fix $S\ne \varnothing$. For each connected component of a graph $C$ of odd integer order in $G-S$, let $d(C)$ count its edges to $S$. Counting vertex degrees inside $C$ givesThus $d(C)$ is an odd integer; it is positive because $G$ is a connected graph. If $d(C)=1$, its unique outgoing edge is a bridge. Distinct such components yield distinct bridges, so there are at most two of them. Every other such component has $d(C)\ge 3$. ThereforeThe last inequality holds because each vertex of $S$ has vertex degree three. Furthermore,since components of even integer order do not affect parity, and $|V|$ is an even integer. If $q(S)>|S|$, this parity forces $q(S)\ge |S|+2$, contradicting the bound $3q(S)\le 3|S|+4$.

For $S=\varnothing$, the only component is the whole connected graph, of even integer order, so $q(\varnothing )=0$. Thus every $S$ satisfies the criterion, and $G$ has a perfect matching. This proves the perfect matching theorem for a cubic graph with at most two bridges directly from Tutte's theorem on perfect matchings.

Every finite connected cubic graph with at most two bridges has a perfect matching.

Proof. The handshaking lemma implies that the number $n$ of vertices is an even integer. Fix a nonempty subset $S$ of the vertex set, and let $q$ count the components of $G-S$ of odd integer order. For such a component $C$, counting vertex degrees gives $3|C|=2|E(C)|+d(C)$, where $d(C)$ counts its edges to $S$. Thus $d(C)$ is a positive odd integer. If it is one, the unique outgoing edge is a bridge; this happens for at most two components. For all others $d(C)\ge 3$. Therefore $3q-4\le 3|S|$. Meanwhile $q\equiv n-|S|\equiv |S|(\mathrm{mod}2)$. If $q>|S|$, the same parity would force $q\ge |S|+2$, contradicting the preceding bound. For $S=\varnothing$ there is one component of even integer order and hence $q=0$. Apply Tutte's theorem on perfect matchings.

A mathematician whose work includes graph theory and matroids; Tutte's theorem on perfect matchings is one of his fundamental results.

University of St Andrews biography and portrait collection.