Length EdgeDisjoint Paths
and Partial Orientation
Abstract
In 2003, it was claimed that the following problem was solvable in polynomial time: do there exist edgedisjoint paths of length exactly between vertices and in a given graph? The proof was flawed, and we show that this problem is NPhard even if we disallow multiple edges. We use a reduction from Partial Orientation, a problem recently shown by Pálvölgyi to be NPhard.
Mathematics Subject Classification: 05C38, 05C40, 68Q25
Keywords: edgedisjoint paths, NPhardness, NPcompleteness, network flow, directed graph, oriented graph, degree sequence
In [2], Bley discussed the problem Max EdgeDisjoint ExactLength Paths, abbreviated : given an undirected multigraph, and two vertices and , do there exist edgedisjoint paths between and of length exactly ? He used a reduction to network flow to claim that was solvable in polynomial time, but the reduction was flawed, as we describe in [1], where we also showed that is NPhard. For completeness we duplicate the proof below, as Theorem 1.
In this note, we show that the problem remains NPhard even if we require the input graph to be simple. We call this restricted problem Simple Max EdgeDisjoint Exact3Length Paths, or . To show that and are NPhard, we use a polynomial reduction from the problem Partial Orientation, shown to be NPhard by Pálvölgyi in [3]: given a graph, can we replace some of the edges by directed edges, such that each vertex has prescribed in, out, and undirecteddegree?
Theorem 1.
The problem is NPhard.
Proof.
Let be the given graph in an instance of Partial Orientation. We construct a graph , the input to , as follows. To we add two new vertices and . Let be adjacent to each vertex of with multiplicity equal to the prescribed outdegree of , and let be adjacent to with multiplicity equal to the prescribed indegree of . Then the sum of the prescribed outdegrees is the degree of , and the sum of the prescribed indegrees is the degree of . If these sums are not the same, then trivially the instance of Partial Orientation has no solution, so we assume the degrees of and are equal. As input to , we set equal to the degree of and .
Now, any solution of Partial Orientation on corresponds to a solution of on , and vice versa. We simply correspond each directed edge in Partial Orientation with a path in . This is a polynomial reduction from Partial Orientation to , and thus shows that is NPhard. ∎
Note that this proof requires allowing duplicate edges incident to and to . Below we show that the problem , which disallows duplicate edges, is still NPhard.
Theorem 2.
The problem Partial Orientation is still NPhard even if the prescribed indegree and prescribed outdegree of each vertex must be at most .
The reduction presented in the proof of Theorem 1 immediately implies the following corollary:
Corollary 3.
The problem is NPhard.
To prove Theorem 2, we modify Pálvölgyi’s reduction from 3SAT to avoid using vertices of prescribed in or outdegree greater than . For each vertex in the graph , he denotes the prescribed indegree by , the prescribed outdegree by , and the remaining prescribed undirecteddegree by . He labels each vertex with a string of letters , , and , each with multiplicity equal to the value of that function at that vertex.
We will construct in parts, and within each part, we will assign the prescribed degrees to each vertex before having constructed all the edges incident to that vertex. Thus for convenience, by a “subgraph” of we mean the induced subgraph on a collection of vertices, together with the prescribed in, out, and undirecteddegrees in of those vertices, and any edges incident to those vertices with unspecified other endpoints in . We temporarily imagine each of these edges to have only one endpoint, and call them “unfinished edges” with respect to the subgraph. An “orientation” of a graph or subgraph chooses each edge (including any unfinished edges) to be oriented in one direction or the other, or to be undirected, in a way that satisfies the prescribed in, out, and undirecteddegrees of the vertices.
Lemma 4 (Pálvölgyi [3]).
For any , there exists a tree subgraph with the following properties:

Every vertex has degree type , , , or .

The subgraph has only two possible orientations, called “true” and “false”.

Among the unfinished edges, there are edges that point (down) away from the tree in the “true” orientation and (up) toward the tree in the “false” orientation, and unfinished edges that point (up) toward the tree in the “true” orientation and (down) away from the tree in the “false” orientation.
The size of this subgraph is linear in .
Pálvölgyi uses one tree subgraph for each variable of the given instance of 3SAT. He chooses larger than both the number of times appears in any clause and the number of times appears in any clause. Thus he can reserve one unfinished edge for each time appears in a clause and one unfinished edge for each time appears in a clause. Each clause is represented by a subgraph containing the other end of each of three edges or , corresponding to the three variables or that appear in that clause. Pálvölgyi’s subgraph for the clause is the only place in the construction where the prescribed in and outdegrees are ever greater than , so we propose a substitute subgraph, as follows.
Lemma 5.
There exists a subgraph including three unfinished edges , , and such that the assignment of at least one of these three to point (down) toward the subgraph and the others (up) away from the subgraph can be extended to an orientation of the whole subgraph, but the assignment of all three to point (up) away from the subgraph cannot. Furthermore, the subgraph can be constructed so that and for every vertex of the subgraph.
Proof.
We exhibit this subgraph in Figure 1, with , , and extending up above the subgraph.
Figure 2 shows the orientations in which at least one of , , and points down and the others point up.
∎
Note that , , and must point either up or down because of their status as or edges in the tree subgraphs. Thus we may ignore any orientations of the clause subgraph in which , , or is undirected, because these orientations are not possible when the subgraph is connected to the whole graph.
At this point in the construction, we have a tree subgraph for each variable, and a clause subgraph for each clause, with the subgraphs connected according to which variables appear in each clause. In order to resolve any remaining unfinished edges in the graph, we complete the construction as Pálvölgyi does: he takes the graph constructed so far and adds the mirrored reflection of it. That is, start by duplicating the construction. Then, for each new vertex corresponding to existing vertex , put , , and . Finally, add an edge between and if and only if was incident to an unfinished edge. This completes the construction. As in Pálvölgyi’s proof, the constructed instance of Partial Orientation has a solution if and only if the given instance of 3SAT has a solution, and our modified construction does not use any vertices of prescribed indegree or outdegree greater than .
Acknowledgments
This research was supervised by Garth Isaak at the Lafayette College REU, which was supported by Lafayette College and the National Science Foundation (grant number DMS 0552825). Thanks also to Brad Alpert for helping to edit the writing.
References
 [1] Hannah Alpert and Jennifer Iglesias. Length edgedisjoint paths is NPhard. Comput. Complexity, 2012. In press.
 [2] Andreas Bley. On the complexity of vertexdisjoint lengthrestricted path problems. Comput. Complexity, 12(34):131–149, 2003.
 [3] Dömötör Pálvölgyi. Deciding soccer scores and partial orientations of graphs. Acta Univ. Sapientiae Math., 1(1):35–42, 2009.