A perfect matching (red edges), in the Petersen graph. Since the Petersen graph is cubic and bridgeless, it meets the conditions of Petersen's theorem.

A cubic (but not bridgeless) graph with no perfect matching, showing that the bridgeless condition in Petersen's theorem cannot be omitted

In the mathematical discipline of graph theory, Petersen's theorem, named after Julius Petersen, is one of the earliest results in graph theory and can be stated as follows:

Petersen's Theorem. Every cubic, bridgeless graph contains a perfect matching.^[1]

In other words, if a graph has exactly three edges at each vertex, and every edge belongs to a cycle, then it has a set of edges that touches every vertex exactly once.

Proof

We show that for every cubic, bridgeless graph G = (V, E) we have that for every set U ⊆ V the number of connected components in the graph induced by V − U with an odd number of vertices is at most the cardinality of U. Then by Tutte theorem G contains a perfect matching.

Let G_i be a component with an odd number of vertices in the graph induced by the vertex set V − U. Let V_i denote the vertices of G_i and let m_i denote the number of edges of G with one vertex in V_i and one vertex in U. By a simple double counting argument we have that

${\displaystyle \sum \nolimits _{v\in V_{i}}\deg _{G}(v)=2|E_{i}|+m_{i},}$

where E_i is the set of edges of G_i with both vertices in V_i. Since

${\displaystyle \sum \nolimits _{v\in V_{i}}\deg _{G}(v)=3|V_{i}|}$

is an odd number and 2|E_i| is an even number it follows that m_i has to be an odd number. Moreover, since G is bridgeless we have that m_i ≥ 3.

Let m be the number of edges in G with one vertex in U and one vertex in the graph induced by V − U. Every component with an odd number of vertices contributes at least 3 edges to m, and these are unique, therefore, the number of such components is at most m/3. In the worst case, every edge with one vertex in U contributes to m, and therefore m ≤ 3|U|. We get

${\displaystyle |U|\geq {\frac {1}{3}}m\geq |\{{\text{components with an odd number of vertices}}\}|,}$

which shows that the condition of Tutte theorem holds.

History

The theorem is due to Julius Petersen, a Danish mathematician. It can be considered as one of the first results in graph theory. The theorem appears first in the 1891 article "Die Theorie der regulären graphs".^[1] By today's standards Petersen's proof of the theorem is complicated. A series of simplifications of the proof culminated in the proofs by Frink (1926) and König (1936).

In modern textbooks Petersen's theorem is covered as an application of Tutte's theorem.

Applications

• In a cubic graph with a perfect matching, the edges that are not in the perfect matching form a 2-factor. By orienting the 2-factor, the edges of the perfect matching can be extended to paths of length three, say by taking the outward-oriented edges. This shows that every cubic, bridgeless graph decomposes into edge-disjoint paths of length three.^[2]
• Petersen's theorem can also be applied to show that every maximal planar graph can be decomposed into a set of edge-disjoint paths of length three. In this case, the dual graph is cubic and bridgeless, so by Petersen's theorem it has a matching, which corresponds in the original graph to a pairing of adjacent triangle faces. Each pair of triangles gives a path of length three that includes the edge connecting the triangles together with two of the four remaining triangle edges.^[3]
• By applying Petersen's theorem to the dual graph of a triangle mesh and connecting pairs of triangles that are not matched, one can decompose the mesh into cyclic strips of triangles. With some further transformations it can be turned into a single strip, and hence gives a method for transforming a triangle mesh such that its dual graph becomes hamiltonian.^[4]

Extensions

Notes

References