Orientation of a graph

A choice of direction for each edge of an undirected graph, producing a directed graph.

An orientation of a graph keeps the same vertex set and the same individual edges. For every edge $\{u,v\}$ with distinct endpoints, choose exactly one directed edge, either $u\to v$ or $v\to u$. Replacing it by both arrows is a different construction: it doubles the edges and permits travel both ways along that connection. In an orientation of a graph, the return journey must use another route.^[ref]

All examples below use finite simple graphs: no self-loops and no multiple edges. For a multigraph, each individual nonloop edge receives its own direction, so two multiple edges can point in opposite directions. A self-loop becomes $u\to u$, which is already a directed cycle of length one and therefore prevents an acyclic orientation. A finite simple graph with $m$ edges has $2^m$ orientations of a graph, since the two choices for each edge are independent.

Think of vertices as junctions and edges as streets. An orientation of a graph makes each street one-way; reachability asks whether a legal journey exists from one specified junction to another. Two useful aims pull in different directions: an acyclic orientation rules out circular dependencies, while a strong orientation permits journeys between every ordered pair of junctions.

An acyclic orientation is an orientation of a graph whose resulting directed graph has no directed cycle, hence is a Directed acyclic graph. A directed cycle must follow every arrow around its closed route. A cycle in a graph in the skeleton of a directed graph only requires the corresponding undirected edges; their arrows need not agree around the route.

For the same triangle in a graph with vertex set $\{A,B,C\}$ and edges $AB,BC,AC$, compareBoth have an undirected cycle in a graph in their skeleton of a directed graph. Only $D_2$ has a directed cycle. In $D_1$, $C$ has no outgoing directed edge, so a route cannot continue from $C$ back to $A$.

Thus an acyclic orientation does not require its underlying undirected graph to be a forest. Every orientation of a graph of a forest is acyclic, because a directed cycle would give an undirected cycle in a graph. Conversely, every finite simple graph, including one with many cycles in a graph, admits at least one acyclic orientation by the following ordering construction. On this triangle in a graph, six of the eight orientations of a graph are acyclic; the other two follow the directed cycle in opposite senses.

Choose a strict total order $v_1<\cdots <v_n$ of the vertices of a finite simple graph. Direct every edge $v_i{v}_j$ from $v_i$ to $v_j$ when $i<j$. Along every directed path, the index strictly increases. Returning to the starting vertex would require an impossible strict increase back to the initial index, so this is an acyclic orientation. The displayed $D_1$ comes from $A<B<C$.

The chosen strict total order is a topological order of the resulting Directed acyclic graph. Conversely, every acyclic orientation of a finite simple graph has a topological order, and orienting its edges from earlier to later in that order recovers it. Different topological orders can induce the same acyclic orientation, since vertices with no edge between them need not impose a choice of arrow.^[ref]

An order respects reachability but does not create it: $u<v$ in a chosen topological order does not imply a directed path from $u$ to $v$. For example, orient the two edges of the graph path $A,B,C$ using $A<C<B$. The arrows are $A\to B$ and $C\to B$; there is no directed path from $A$ to $C$, despite $A<C$. In $D_1$, however, $A$ reaches $C$ both directly and via $B$. Following outgoing directed edges with graph search decides this directional reachability.^[ref]

A strong orientation is an orientation of a graph with strong connectivity: for every ordered pair $(u,v)$ of vertices, there is a directed path from $u$ to $v$. A length-zero directed path permits $u=v$. Ordinary connectivity of the underlying undirected graph is necessary, but does not ensure this two-way reachability after directions are chosen. The acyclic $D_1$ is not a strong orientation; the cyclic $D_2$ is, since a journey can continue around its directed cycle.

Attach a new vertex $D$ to that triangle in a graph by the single edge $CD$. This edge is a bridge in a graph. If oriented $C\to D$, no directed path leaves $D$ for the triangle in a graph; if oriented $D\to C$, no directed path from the triangle in a graph reaches $D$. Neither choice gives a strong orientation, even when the triangle in a graph itself is cyclically oriented. By contrast, the four-vertex cycle graph has the strong orientationFollowing this directed cycle reaches any requested vertex in at most three steps. Removing any one of its undirected edges leaves a graph path through all four vertices, so it has no bridge in a graph.

For more than one vertex, a strong orientation cannot also be an acyclic orientation: directed paths from $u$ to $v$ and back give a nonempty closed directed walk, from which removing repeated portions leaves a directed cycle. A directed graph containing one directed cycle need not have strong connectivity, as the attached $D$ shows.

A finite nonempty connected graph whose edges are undirected admits a strong orientation if and only if it has no bridge in a graph. This statement holds for finite multigraphs as well as finite simple graphs; self-loops do not help connect distinct vertices. The single-vertex case is allowed, with length-zero directed paths.^[ref]

Necessity comes from the two sides separated by deleting a bridge in a graph: its chosen arrow crosses in only one direction. For sufficiency, the cited constructive proof uses depth-first search to find a cycle in a graph, orients it cyclically, contracts it, and repeats. Undoing the contractions replaces junctions by directed cycles, retaining two-way reachability. Contraction can produce multiple edges, so these must remain distinct. This is a construction outline; the input graph is finite and bridgeless, and remaining edges within contracted pieces can receive arbitrary directions.^[ref]

A finite undirected graph can be oriented with outdegree at least an integer $k\ge 0$ at every vertex if and only if every vertex subset $U$ is incident to at least $k|U|$ distinct edges. The max-flow min-cut theorem and integral max-flow theorem prove sufficiency by assigning distinct edges to vertices through an incidence flow network; each assigned edge is oriented away from that vertex. Necessity follows by counting outgoing edges from $U$.