The Petersen graph has 10 vertices and 15 edges, and the complete graph on 10 vertices has 45 edges. However, Allen Schwenk and (independently) O. P. Lossers (Jack van Lint’s problem-solving seminar in Eindhoven) showed that you can’t partition the edges of the complete graph into three Petersen graphs; you can pack two Petersen graphs edge-disjointly, but not three.

I described this here, and went on to explain how Sebi Cioaba and I showed that for any *m* ≥ 2 it is possible to cover the edges of the complete graph *m* times with 3*m* Petersens. In particular, you can cover the edges twice with six Petersen graphs.

This means that five Petersen graphs suffice to cover the edges of the complete graph, since we may omit one of the six. We cannot, however, omit two, since any two have edges in common.

So the minimum number of Petersen graphs required to cover all the edges is either 4 or 5.

In fact it is 4. If you apply four random permutations to the vertices of the Petersen graph you will quickly come up with a way of covering all edges.

But is there a nice way to see this? In particular, is it possible to arrange four Petersen graphs so that they cover each edge once or twice (so that deleting a complete graph leaves a trivalent simple graph remaining)? If so, what is this graph? If not, why not?

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About Peter Cameron

I count all the things that need to be counted.

Here is (hopefully) an answer to your question of if you can cover each edge once or twice:

gap> perms := [ (), (1,3,5)(2,10,6,4,9), (1,7)(2,9,8,4,6,10)(3,5), (1,9,7,10)(2,6,8,4,5,3) ];;

gap> peterson := [ [1,2],[2,3],[3,4],[4,5],[1,5], [6,7],[7,8],[8,9],[9,10],[6,10], [1,6],[2,7],[3,8],[4,9],[5,10]];;

gap> images := List(perms, x -> OnSetsSets(peterson, x));;

gap> List(Combinations([1..10], 2), edge -> Length(Filtered(images, x -> edge in x)));

[ 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1 ]

Wrong graph! Let me try with the right one…

gap> peterson := [ [1,2],[2,3],[3,4],[4,5],[1,5], [6,8],[7,9],[8,10],[6,9],[7,10], [1,6],[2,7],[3,8],[4,9],[5,10]];;

gap> perms := [ (), (1,7,6,5,3,2,10)(4,9,8), (1,7,10,9,4,8,5)(2,6,3), (1,7,3,2,8,5,4,9,10,6)];;

gap> images := List(perms, x -> OnSetsSets(peterson, x));;

gap> List(Combinations([1..10], 2), edge -> Length(Filtered(images, x -> edge in x)));

[ 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2 ]

Thanks Chris. I probably won’t have time to look at it today…