One-Way Trail Orientations Anders Aamand1 University of Copenhagen, Copenhagen, Denmark [email protected] https://orcid.org/0000-0002-0402-0514 Niklas Hjuler2 University of Copenhagen, Copenhagen, Denmark [email protected] https://orcid.org/0000-0002-0815-670X Jacob Holm3 University of Copenhagen, Copenhagen, Denmark [email protected] https://orcid.org/0000-0001-6997-9251 Eva Rotenberg4 Technical University of Denmark, Lyngby, Denmark [email protected] https://orcid.org/0000-0001-5853-7909 Abstract Given a graph, does there exist an orientation of the edges such that the resulting directed graph is strongly connected? Robbins’ theorem [Robbins, Am. Math. Monthly, 1939] asserts that such an orientation exists if and only if the graph is 2-edge connected. A natural extension of this problem is the following: Suppose that the edges of the graph are partitioned into trails. Can the trails be oriented consistently such that the resulting directed graph is strongly connected? We show that 2-edge connectivity is again a sufficient condition and we provide a linear time algorithm for finding such an orientation. The generalised Robbins’ theorem [Boesch, Am. Math. Monthly, 1980] for mixed multigraphs asserts that the undirected edges of a mixed multigraph can be oriented to make the resulting directed graph strongly connected exactly when the mixed graph is strongly connected and the underlying graph is bridgeless. We consider the natural extension where the undirected edges of a mixed multigraph are partitioned into trails. It turns out that in this case the condition of the generalised Robbin’s Theorem is not sufficient. However, we show that as long as each cut either contains at least 2 undirected edges or directed edges in both directions, there exists an orientation of the trails such that the resulting directed graph is strongly connected. Moreover, if the condition is satisfied, we may start by orienting an arbitrary trail in an arbitrary direction. Using this result one obtains a very simple polynomial time algorithm for finding a strong trail orientation if it exists, both in the undirected and the mixed setting. 1 Research supported by Thorup’s Advanced Grant DFF-0602-02499B from the Danish Council for Independent Research and by his Investigator Grant 16582, Basic Algorithms Research Copenhagen (BARC), from the VILLUM Foundation. 2 This work is supported by the Innovation Fund Denmark through the DABAI project. 3 Research supported by Thorup’s Advanced Grant DFF-0602-02499B from the Danish Council for Independent Research and by his Investigator Grant 16582, Basic Algorithms Research Copenhagen (BARC), from the VILLUM Foundation. 4 This research was conducted during the fourth author’s time as a PhD student at University of Copenhagen. © Anders Aamand, Niklas Hjuler, Jacob Holm, and Eva Rotenberg; E A licensed under Creative Commons License CC-BY T C 45th International Colloquium on Automata, Languages, and Programming (ICALP 2018). S Editors: Ioannis Chatzigiannakis, Christos Kaklamanis, Dániel Marx, and Donald Sannella; Article No. 6; pp. 6:1–6:13 Leibniz International Proceedings in Informatics Schloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany 6:2 One-Way Trail Orientations 2012 ACM Subject Classification Mathematics of computing → Graph algorithms, Mathemat- ics of computing → Paths and connectivity problems Keywords and phrases Graph algorithms, Robbins’ theorem, Graph orientation Digital Object Identifier 10.4230/LIPIcs.ICALP.2018.6 1 Introduction and motivation Suppose that the mayor of a small town decides to make all the streets one-way such that it is possible to get from any place to any other place without violating the orientations of the streets5. If all the streets are initially two-way then Robbins’ theorem [10] asserts that this can be done exactly when the corresponding graph is 2-edge connected. If, on the other hand some of the streets were already one-way in the beginning then the generalised Robbins’ theorem by Boesch [1] states that it can be done exactly when the corresponding “mixed” graph is strongly connected and the underlying graph is bridgeless. However, the proofs of both of these results assume that every street of the city corresponds to exactly one edge in the graph. This assumption hardly holds in any city in the world and therefore a more natural assumption is that every street corresponds to a trail (informally, a potentially self-crossing path) in the graph and that the edges of each trail must be oriented consistently6. In this paper we consider such graphs having their edges partitioned into trails. We prove that the trails can be oriented to make the resulting directed graph strongly connected exactly if the initial graph is 2-edge connected (note that this is precisely the condition of Robbins’ theorem). Not only do we show that the strong trail orientation problem in undirected 2-edge connected graphs always has a solution, we also provide a linear time algorithm for finding such an orientation. In doing so, we use an interesting combination of techniques that allow us to reduce to a graph with a number of 3-edge connected components that is linear in the number of edges. Using that the average size of these components is constant and that we can piece together solutions for the individual components we obtain an efficient algorithm. Finally, we will consider the generalised Robbins’ theorem in this new setting by allowing some edges to be oriented initially and supposing that the remaining edges are partitioned into trails. We will show that if each cut (V1,V2) in the graph has either at least 2 undirected edges going between V1 and V2 or at least 1 directed edge in each direction then it is possible to orient the trails making the resulting graph strongly connected. In fact, we show that if this condition is satisfied we may start by orienting an arbitrary trail in an arbitrary direction. Although this condition is not necessary it does give a simple algorithm for finding a trail orientation if it exists. Indeed, initially the graph may contain undirected edges that are forced in one direction by some cut. For finding a trail orientation if it exists we can thus orient forced trails in the forced direction. If there are no forced trails we orient any trail arbitrarily. 5 The motivation for doing so is that the streets of the town are very narrow and thus it is a great hassle when two cars unexpectedly meet. 6 This version of the problem was given to us through personal communication with Professor Robert E. Tarjan. A. Aamand, N. Hjuler, J. Holm, and E. Rotenberg 6:3 Figure 1 The graph is strongly connected and the underlying graph is 2-edge connected, but irrespective of the choice of orientation of the red trail, the graph will no longer be strongly connected. Note that in the mixed setting the feasibility depends on the trail decomposition which is not the case for the other results. That the condition from the generalised Robbins’ theorem is not sufficient can be seen from Figure 1. Earlier methods Several methods have already been applied for solving orientation problems in graphs where the goal is to make the resulting graph strongly connected. One approach used by Robbins [10] is to use that a 2-edge connected graph has an ear-decomposition. An ear decomposition of a graph is a partition of the set of edges into a cycle C and paths P1,...,Pt such that Pi has its two endpoints but none of its internal Si−1 vertices on C ∪ j=1 Pj . Given the existence of an ear decomposition of a 2-edge connected graph it is easy to prove Robbins’ theorem. Indeed, any choice of consistent orientations of the paths and the cycle gives a strongly connected graph. A second approach introduced by Tarjan [4] gives another simple proof of Robbins’ theorem. One can create a DFS tree in the graph rooted at a vertex v and orient all edges in the DFS tree away from v. The remaining edges are all back edges (see [4]) and are oriented towards v. It is easily verified that this gives a strong orientation if the graph is 2-edge connected. A similar approach was used by Chung et al. [2] in the context of the generalized Robbins’ theorem for mixed multigraphs. The above methods not only prove Robbins’ theorem, they also provide linear time algorithms for finding strong orientations of undirected or mixed multigraphs. However, none of the above methods have proven fruitful in our case. In case of the ear decomposition we would need one that is somehow compatible with the partitioning into trails, and this seems hard to guarantee. Similar problems appear when trying a DFS-approach. Neither does the proof by Boesch [1] of Robbins’ theorem for mixed multigraphs generalise to prove our result. Most importantly, the corresponding theorem would no longer be true for trail orientations as is shown by the example in Figure 1. Since the classical linear time algorithms rely on ear-decompositions and DFS searches, and since these approaches do not immediately work for trail partitions, our linear time algorithm will be a completely new approach to solving orientation problems. Structure of the paper The structure of this paper is as follows. In section 3 we prove our generalisation of Robbins’ theorem for undirected graphs partitioned into trails. In section 4 we study the case of mixed graphs. Finally in section 5 we provide our linear time algorithm for trail orientation in an undirected graph. I C A L P 2 0 1 8 6:4 One-Way Trail Orientations 2 Preliminaries Let us briefly review the concepts from graph theory that we will need.