# Vertices of degree in edge-minimal, -edge-connected graphs

Halin [1] showed that every edge-minimal, -vertex-connected graph has a vertex of degree . In this note, we prove the analogue to Halin’s theorem for edge-minimal, -edge-connected graphs:

###### Theorem 1.

Let be an edge-minimal, -edge-connected graph. Then there are two nodes of degree in .

To prove Theorem 1, we first establish a link between edge-minimal, -edge-connected graphs and exactly -edge-connected graph [2] (Definition 7 and Proposition 9). The theorem is proved in the case of being an exactly -edge-connected graph (Proposition 16) and then transfered to edge-minimal graphs. Throughout, all graphs considered are multigraphs and all sets are multisets.

## The edge-connectivity equivalence relation.

###### Proposition 2.

The edge version of Menger’s theorem holds for multigraphs.

###### Proof.

Any multi-edge in the graph can be replaced by a length path to get a graph . There is then an obvious bijection between the paths in the original multigraph and the resulting graph . The result of the Menger theorem on can be immediatly applied to .∎

###### Definition 3.

Let be the relation between nodes and that holds if or there are edge-disjoint paths between and .

###### Proposition 4.

is an equivalence relation.

###### Proof.

is by definition reflexive and symmetric. Let be a graph, let satisfy and . If , or , then transitivity is obvious. Suppose that , and are distinct vertices and let be an edge set of cardinality . There are edge-disjoint paths from to in so, and are still connected in and so are and . So we have a path in . By Menger’s theorem, the set of paths between and in has cardinality at least and , and is transitive.∎

###### Proposition 5.

There are at most edge-disjoint paths between two equivalency classes of .

###### Proof.

Suppose we have edge-disjoint paths, , , between and , two distinct equivalency classes of in graph . Let , and , be the endpoints in and respectevily of edge-disjoint paths. (Note that the are not all necessary distinct. This is true for the as well.) Let be a set of edges. Then, in , at least one path, say , was not disconnected. Because , and are not disconnected. Similarly, and are not disconnected. So we have a path . By Menger’s theorem, , which is a contradiction. ∎

## Relationship between exact connectivity and minimality.

###### Proposition 6.

Let be a -edge-connected graph. is edge minimal if and only if for any adjacent vertices , there are at most edge disjoint paths.

###### Proof.

Let such that there are edge-disjoint paths. In , there are at least edge-disjoint paths. Let and be any two vertices, not necessarily distinct from and . There are edge-disjoint paths in . So, depending on whether or not edge is on one of these paths, in there are either edge-disjoint paths, or there are edge-disjoint paths and a path edge disjoint from the paths. There is a similar situation in between , and . Let be a set of distinct edges of . In , there is a or path, a or path and a path. Hence there is a path and is not a separating set. By Menger’s theorem, is -edge-connected, and is not edge minimal. Conversely, suppose that for any edge there are at most edge-disjoint paths. Then in there are at most edge disjoint paths. So is edge minimal.∎

###### Definition 7.

A graph is called exactly -edge-connected if there are exactly edge disjoint paths between any two nodes .

###### Definition 8.

Let be a -edge-connected graph. Define to be a graph where the vertices are the equivalency classes of on and there is an edge for every edge with and .

###### Proposition 9.

If is an edge-minimal, -edge-connected graph then is not trivial and is exactly -edge-connected.

###### Proof.

If is edge minimal, it is not -edge-connected, so has more than one equivalency class, and is not trivial. is -edge-connected like . By Proposition 5, it is exactly -edge-connected.∎

###### Proposition 10.

Let be an edge-minimal, -edge-connected graph, and let be an equivalence class of . Then, for any , .

###### Proof.

Let . By Proposition 6, if , then is not an edge in . So every neighbor of is not in and by construction . ∎

## Proof of Theorem 1.

###### Definition 11.

An edge cut of a graph is called trivial if one of the components of is the trivial graph.

###### Definition 12.

A -regular graph is a graph where all vertices have the same degree . A quasi -regular graph is a graph where at most one vertex has a degree different than .

###### Lemma 13.

An exactly -edge connected graph which has only trivial cuts is quasi -regular.

###### Proof.

Suppose there exists two vertices and of degree greater than . There exists a minmum cut separating and and this cut cannot be trivial.∎

###### Definition 14 (Vertex splitting).

Let be an exactly -edge connected graph and be a non-trivial minimum cut. Construct and by adding two new vertices and attached respectively to and by new edges to the vertices adjacent to . Formally, let and

The pair is called a vertex splitting of with respect to .

###### Proposition 15.

Let be an exactly -edge-connected graph, and let be a non-trivial minimum cut. and obtained from vertex splitting with respect to are exactly -edge-connected.

###### Proposition 16.

Let be an exactly -edge-connected graph. Then there are two vertices of degree in .

###### Proof.

We proceed by induction on the number of non-trivial minimum cuts in . If has no non-trivial minimum cuts, then it is quasi -regular and has at least 2 nodes of degree . Let be a non-trivial minimum cut and let and be the vertex splitting graphs induced by . Call and the new vertices (). Suppose were a non-trivial minimum cut in . Construct a corresponding non-trivial minimum cut in by changing any edge used by that is adjacent to to the corresponding edge in . So to any non-trivial minimum cut of or corresponds a distinct non-trivial minimum cut in . But no non-trivial cut in or corresponds to the cut (it would be a trivial cut in and ). So both and have fewer non-trivial minimum cuts than . By induction and have 2 nodes of degree , including and . So, has the same vertices as and , except for and , with the same degree. Hence it has 2 nodes of degree . ∎

###### Proof of Theorem 1.

By Proposition 16, has two vertices of degree . Let be an equivalence class of . If contains two distinct vertices and of , then there are at least edge disjoint paths in and both and have a degree . By Proposition 10, also has degree in . Hence, the vertices of degree in correspond to equivalence classes that must each contain at most one vertex of of degree . Therefore, has two vertices of degree . ∎

## References

- [1] R. Halin. A theorem on -connected graphs. Journal of Combinatorial Theory, 7:150–154, 1969.
- [2] Carl Kingsford and Guillaume Marçais. A synthesis for exactly 3-edge-connected graphs. Submitted to FOCS 2009, May 2009, arXiv:0905.1053 [math.CO].