Theorem: Every regular graph of positive even degree has a spanning 2-regular subgraph.

This was taken from Corollary 5.10 of ETH Zurich's notes on graph theory. The proof constructs a Eulerian tour, splits the vertices into in and out vertices on the tour, then invokes Hall's theorem on the regular (and now bipartite graph) to get a perfect matching. This is joined together to form the spanning 2-regular subgraph.

While the proof seems relatively straightforward, I have two questions:

First, where is the 2-regular, spanning subgraph in this 4-regular graph? It seems to me that following the theorem it should have one, but I have been unable to identify a 2-regular, spanning subgraph in this relatively simple graph.

Second, doesn't this imply that every even degree, connected, regular graph has a Hamiltonian cycle?


1 Answer 1


A $2$-regular graph is a union of disjoint cycles.(So it doesn't have to be exactly one cycle)

You yourself have provided an example of a $4$-regular graph with a $2$-regular spanning subgraph but no hamiltonian cycle.

An example of a $2$-regular subgraph in your linked graph is the union of the following two cycles:

$(1,9,0,6,8,4,1)$ and $(2,10,3,5,7,2)$

enter image description here


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