The question is: Let G be a connected graph with an even number of edges, a) show that G can be oriented so that the outdegree of each vertex is even. b) deduce that G admits a decomposition into paths of length two.
About part a), I know that the sum of outdegree is the same with the number of edges, which is even, but how can I get to that every vertex has even outdegree? Should I assume it's not the case then find a contradiction? I'm also not sure about how to use the fact that G is a connected graph.
No clue on part b).
Connected graph with an even number of edges
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graph-theory
3 Answers
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(a) Suppose $G$ is connected and has an even number of edges. Take an arbitrary orientation of $G.$ If any vertex has odd outdegree, then there are at least two vertices with odd outdegree, since the sum of the outdegrees is equal to the number of edges, which is even. Choose two vertices $u,v$ whose outdegrees are odd. Since $G$ is connected, there is a path $P$ from $u$ to $v.$ Reverse the orientation of each edge of $P.$ Observe that the outdegrees of $u$ and $v$ change parities from odd to even, while the parities of all other outdegrees are unchanged. In this way, the number of odd outdegrees is diminished by two. Repeat this process until all outdegrees are even.
(b) Suppose $G$ is oriented so that each vertex has even outdegree. For each vertex $v,$ partition the set of all edges with initial vertex $v$ into disjoint pairs, and make each of those pairs into a path of length two.
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For part b) we can induct as follows: assume a graph exists where a decomposition is NOT possible. Then there exists a graph with minimal edges for which it is not possible. Then this becomes quite simple as all we need to do is show that it is possible to 'remove', so to speak, a path of length two which leaves the graph connected (or breaks it into unconnected subgraphs which all satisfy the conditions, e.g even number of edges)
Still thinking about part a)
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The proof is by induction on the number of edges $e$.
If $e=0$, then there is at most one node, whose outdegree is $0$, which is even.
Suppose now we have a connected graph $G$ with even number of edges $e \geq 2$, and that for every connected graph with an even number of edges that is less than $e$ the claim holds; that is, that graph can be oriented in such a way that every node has even outdegree.
Since $G$ is connected, it has at least one node with two or more incident edges. Let $p$ be any such node and let $(p,q)$ and $(p,r)$ be any two of its incident edges.
Removing the two edges from $G$ may result in a graph that has one, two, or three components. We want to apply the inductive hypothesis to the components, but we need to overcome one problem: while all components may have even numbers of edges, it is also possible for two of them to have odd numbers of edges.
As an example, consider five nodes arranged in a straight line, and suppose we remove the two edges incident on the middle node. We are left with one component with no edges and two components with one edge each.
If this happens, then we add back one of the two edges to each of the components with odd number of edges. (Note that this is always possible.) This yields two or three graphs to which the inductive hypothesis applies. (Their numbers of edges are even and less than $e$.)
To finish the proof, we note that in case removing $(p,q)$ and $(p,r)$ from $G$ produces no components with odd numbers of edges, then the removed edges are oriented away from $p$. Otherwise, the orientation of $(p,q)$ and $(p,r)$ is decided by the recursive calls.
Part (b) is now quite straightforward, because each pair of edges oriented by the procedure is a path of length two. Together, they make up the entire edge set of $G$.