# Proof - a vertex in a path has an even number of edges [duplicate]

Given a simple undirected graph,

let's say we have a path of i edges that can repeat nodes but not edges

i.e. nodes may come up more than once in the path but not the edges.

NOTE : I never said this path has to contain all edges in the graph. So I'm not assuming a Eulerian circuit. Just an arbitrary path formed by a subset of the edge set of the graph that may have repeating vertices but not edges.

For a node in the path that is not an endpoint of the path (one that is in the "middle"), I think we can say that it is adjacent to an even number of edges in the path. Not adjacent to any edge in the graph, but those that belong to the path containing this node.

It seems very intuitive but I couldn't figure out a way to formally prove this. Any suggestions?

• I never stated the my case assumes a Eulerian circuit... is it still a duplicate the way I edited it? Commented Dec 20, 2017 at 15:48
• In this case, you can still use Euler Path Theorem because now the path we have has nothing to do with the other vertices and edges, in other words, we can isolate that path from the graph as a new graph. Let $G'(V',E')$ be the graph that is constructed by the vertices and edges of the path you have. Then you can use Euler Path Theorem on $G'$. Commented Dec 20, 2017 at 16:59
• That seems perfectly clear now. Thinking of the path as a subgraph seems like a great idea. Commented Dec 21, 2017 at 1:23
• I'm glad that it helped you :) Commented Dec 21, 2017 at 1:24

You can use the "Euler Path Theorem" directly. It is as following:

Let $G$ be a finite connected graph and $r$ be the number of vertices with odd degree (this means there are odd number of edges adjacent to that vertex). Then $G$ has an Euler path if and only if $r = 0$ or $r = 2$.

Now, in your question, we have a path already so we know that $r = 0$ or $r = 2$ by Euler Path Theorem. If $r = 0$, we are done because every vertex has the property you stated (and this is an Euler cycle). If $r = 2$, then these vertices of odd degrees are "starting vertex" and "terminal vertex" so vertices in middle have even degree. So the result follows.

from OP : look at the comment on the main post then look at this accepted answer

• I'm specifically more interested in the general case where there is no assumption about G having an Euler path, and we are just dealing with an arbitrary path in a simple undirected graph that may have vertices repeated. Commented Dec 18, 2017 at 12:33
• Well, if you pass every edge exactly once with repeated vertices, it is called Eulerian path as far as I know. Is what you are asking about passing every edge at most once? Commented Dec 18, 2017 at 12:37
• I don't think I ever said anywhere this path passes every edge exactly once... I just said that the edges in the path are not repeated, not that all edges of the graph are included in the path Commented Dec 18, 2017 at 13:13
• If that is the case, then your statement is not true I think. Suppose you have a graph $G(V,E)$ with $V(G) = \{1,2,3,4\}$ and $G(E) = \{\{1,2\},\{2,3\},\{2,4\}\}$, then according to your definition, we can have a path from $1$ to $3$ but although $2$ is in middle, degree of this vertex $d(2) = 3$ so there are odd number of edges adjacent to a mid vertex (Here you don't pass the edge $\{2,4\}$ and no edge is passed more than once). Commented Dec 18, 2017 at 13:18
• i must not have made myself clear so i edited the post again. Your example isn't exactly what I'm looking for because I said the node in the path is adjacent to edges "in the path". The path from 1 to 3 would not include the edge {2,4} so node 2 "in the path" is adjacent to {1,2} and {2,3}, which are "in the path" unlike {2,4} Commented Dec 20, 2017 at 15:47

A connected graph $G$ has an Eulerian circuit if and only if every vertex has even degree. Your observation from the title follows immediately from this standard characterization.