Orientation of Edges on Graphs with Vertex Degree Constraints Suppose I have a graph $ G = (V,E) $ such that each vertex $ v \in V $ has degree 4. Can I always choose an orientation of edges (ie. arrows drawn on edges) such that each vertex has two incoming edges and two outgoing edges?
 A: Yes, for any connected component you take any Eulerian cycle (which exists because every vertex has even degree) and direct the edges according to the cycle.
Each time we visit the vertex we make one edge incoming (the one we just came by) and one edge outgoing (the one we will leave). Hence, a bit more general statement is true:

For any undirected graph $G = (V,E)$ with $\deg(v)$ being even for any $v \in V$, we can direct the edges so that $\deg_\text{in}(v) = \frac{\deg(v)}{2} = \deg_\text{out}(v)$.

I hope this helps $\ddot\smile$
A: Claim: Orientation can be chosen so that $indegree=outdegree=2$ for every node.
Since every node in $G$ has degree 4, there are no isolated nodes and we can say that G is comprised of one or more disjoint connected components. So it suffices to prove the claim for the general connected component, called $C$.

 [Edited to remove - based on false premise that there is a guaranteed maximal length cycle] Every node in $C$ has degree 4, so we can find a maximum length cycle in $C$, of length $m$. Then replace each one of the undirected edges with an edge directed in cyclic order. This contributes 1 to the indegree and outdegree of each node. Now disconnect all edges in this cycle to obtain a graph $G'$. 
Now the degree of each node in $G'$ is 2. So $G'$ consists of a disjoint set of connected components (there can be no isolated nodes or terminal nodes). Because any connected component $C' \in G'$ has nodes only of degree 2, $C'$ is a simple cycle. Hence, we can replace these undirected edges with edges directed in cyclic order. Again each node in every $C'$, and hence in their union $G'$, gains 1 in both its indegree and outdegree.

[Edited: replacement of original proof - for completeness only]
Since C is a connected component with nodes of degree four, it contains at least one cycle, $\Gamma_1$. Then replace the edges in $\Gamma_1$ by cyclically directing edges and form the next graph $C_1$ by removing these edges, i.e. $C_1 = C \backslash \Gamma_1$, and contructing $D_1$ from the nodes in $G$ and the directed edges removed. Similarly remove other cycles from $C_1$ until it is no longer a connected component, resulting in $C_2$, and augmenting $D_2$. Any resulting isolated nodes in $C_2$ correspond with nodes of indegree and outdegree 2 in $D_2$ (replacement of cycles always guarantees a balance). 
Repeat the procedure on the largest remaining connected component $C_3 \subset C_2$. All its nodes have even degree (2 or 4), so it contains a cycle, so $\dots$ 
Hence every node in the final iteration of $D$ has indegree and outdegree of 2. 
