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Strongly Pfaffian Graphs

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Strongly Pfaffian Graphs EUROCOMB 2021 Strongly Pfaffian Graphs Maximilian Gorsky, Raphael Steiner, and Sebastian Wiederrecht Technische Universit¨atBerlin [email protected] [email protected] [email protected] Abstract: An orientation of a graph G is Pfaffian if every even cycle C such that G−V (C) has a perfect matching has an odd number of edges oriented in either direction of traversal. Graphs that admit a Pfaffian orientation permit counting the number of their perfect matchings in polynomial time. We consider a strengthening of Pfaffian orientations. An orientation of G is strongly Pfaffian if every even cycle has an odd number of edges directed in either direction of the cycle. We show that there exist two graphs S1 and S2 such that a graph G admits a strongly Pfaffian orientation if and only if it does not contain a graph H as a subgraph which can be obtained from S1 or S2 by subdividing every edge an even number of times. Combining our main results with a result of Kawarabayashi et al. we show that given any graph the tasks of recognising whether the graph admits a strongly Pfaffian orientation and constructing such an orientation provided it exists can be solved in polynomial time. 1 Introduction All graphs considered in this article are finite and do not contain loops. We also exclude parallel edges, unless we explicitly address the graphs as multi-graphs. Let G be a graph and F ⊆ E(G) be a set of edges. F is called a matching if no two edges in F share an endpoint, a matching is perfect if every vertex of G is contained in some edge of F . A subgraph H of G is conformal if G − V (H) has a perfect matching, and finally an orientation of G is a digraph G~ such that (u; v) or (v; u) is an edge of G~ if and only if uv 2 E(G), and G~ does not contain (u; v) and (v; u) at the same time. An even cycle C of G is oddly oriented by an orientation G~ if it has an odd number of edges directed in either direction around C. An orientation G~ of G is Pfaffian if G has a perfect matching and every even conformal cycle C is oddly oriented by G~ . A graph G is called Pfaffian if it admits a Pfaffian orientation. Pfaffian orientations are significant as, given that a graph G admits a Pfaffian orien- tation, the number of perfect matchings of G can be computed in polynomial time. In general, the problem of counting the number of perfect matchings in a graph is polynomial-time equivalent to computing the permanent of a matrix, which is known to be ]P-hard [11]. By utilising a deep theorem of Lov´asz[5], Vazirani and Yannakakis [12] proved that, in terms of complexity, recognising a Pfaffian graph and finding a Pfaffian orientation can be seen as the same problem. Theorem 1 ([12]) The decision problems `Is a given orientation of a graph Pfaffian?’ and `Is a given graph Pfaffian?’ are polynomial-time equivalent. 1 Research Perspectives CRM Barcelona, Summer 20XX, vol. XX, in Trends in Mathematics Springer-Birkh¨auser,Basel 2 Strongly Pfaffian Graphs Let H be a graph and e 2 E(H) be some edge of H. We say that a graph H0 is obtained from H by subdividing, or respectively bisubdividing, the edge e, if H0 can be obtained from H be replacing e with a path of positive length, or respectively a path of odd length (possibly length one), whose endpoints coincide with the endpoints of e and whose internal vertices do not belong to H. A graph H00 is a emphsubdivision, or respectively a bisubdivision, of H if it can be obtained by subdividing, or respectively bisubdividing, all edges of H. Notably, bisubdivision preserves path- and cycle-parities. There exists a precise characterisation of Pfaffian bipartite graphs in terms of forbid- den bisubdivisions. Theorem 2 ([4]) A bipartite graph G is Pfaffian if and only if it does not contain a conformal bisubdivision of K3;3. While Theorem 2 characterises all Pfaffian bipartite graphs it does not immediately yield a polynomial time recognition algorithm. Such an algorithm was later found by McCuaig [6] and, independently, by Robertson et al. [8]. The case of non-bipartite Pfaffian graphs appears to be more illusive, and it remains an important open problem in graph theory whether Pfaffian graphs can be recognised in polynomial time. For several graph classes including non-bipartite graphs characteri- sations of Pfaffian graphs in these classes are known (see [2, 1]), but whether they can be recognised in polynomial time appears to be open. Moreover, there is no hope that all non-bipartite Pfaffian graphs can be described by excluding a finite number of minimal obstructions in a fashion similar to Theorem 2 [7]. For more Information on Pfaffian orientations and related problems we refer the reader to [6, 9]. 2 Main Results Inspired by the inherent complexity of Pfaffian orientations, especially in non-bipartite graphs, we investigate a stronger, and therefore more restrictive version of Pfaffian ori- entations. Definition 3 Let G be a graph. An orientation G~ of G is strongly Pfaffian if every even cycle C of G is oddly oriented by G~ . A graph G that admits a strongly Pfaffian orientation is called strongly Pfaffian. S1 S2 Figure 1. The two obstructions for strongly Pfaffian graphs. Surprisingly, for the class of strongly Pfaffian graphs it suffices to exclude bisubdivi- sions of S1 and S2 (see Figure 1), as subgraphs to characterise the entire class. Theorem 4 A graph G is strongly Pfaffian if and only if it does not contain a bisubdi- vision of S1 or S2 as a subgraph. And furthermore, thanks to the results in [3], Theorem 4 allows us to deduce that the task of recognising strongly Pfaffian graphs can be performed in polynomial time, since we can simply check for the existence of the bisubdivided versions of S1 and S2. EUROCOMB 2021 3 Theorem 5 ([3]) Let H be a fixed graph, and for each edge e 2 E(H) let a value p(e) 2 f0; 1g be fixed. There exists a polynomial time algorithm for testing if a given graph G contains a subdivision of H in which for every e 2 E(H) the length of the subdivision-path representing e is congruent to p(e) modulo 2. Corollary 6 Strongly Pfaffian graphs can be recognised in polynomial time. Similar to Theorem 1, we can additionally show the following. Theorem 7 Given a strongly Pfaffian graph, a strongly Pfaffian orientation can be constructed in polynomial time. Consequently given a graph G, we can test whether it is strongly Pfaffian and con- struct a strongly Pfaffian orientation in polynomial time. 3 A Structural Characterisation of Strongly Pfaffian Graphs Let S be the class of graphs which do not contain bisubdivisions of S1 or S2 as subgraphs. In order to show Theorem 4, we need to show that (1) every strongly Pfaffian graph is contained in S and (2) every graph in S admits a strongly Pfaffian orientation. An important tool when working with strongly Pfaffian orientations is the switching operation. For this let G~ be an orientation of a graph G and let v 2 V (G). Then switching at v in G~ means reversing all arcs of G~ incident with v to obtain a modified orientation of G. We say that two orientations of a graph are switching-equivalent if one can be obtained from the other by a sequence of switchings. Observe that if G~ 1 and G~ 2 are switching-equivalent orientations of the same graph, then G~ 1 is strongly Pfaffian if and only the same holds for G~ 2. Using this fact we can give a short proof of (1). Lemma 8 If G is a strongly Pfaffian graph, then G 2 S. Proof. Since every subgraph of a strongly Pfaffian graph is strongly Pfaffian, it suffices to show that no bisubdivision of S1 or S2 is strongly Pfaffian. First let G be a bisubdivision of S1 and suppose towards a contradiction that a strongly Pfaffian orientation G~ of G exists. Let u; v be the unique vertices of degree three in G and let P1;P2;P3 be the three disjoint u-v-paths in G. Since G is a bisubdi- vision of S1, all three paths P1;P2;P3 have even length. Pause to note that by applying vertex-switchings, we may find a strongly Pfaffian orientation G~ 0 of G that is switching- 0 equivalent to G~ such that P1;P2 − v; P3 − v each form directed paths in G~ starting at 0 u. Since the cycles P1 [ P2 and P1 [ P3 must be oddly oriented by G~ , the last edges of both P2 and P3 must start in v. However, this means that the cycle P2 [ P3 has an even number of edges oriented in either direction in G~ 0, a contradiction showing that G is not strongly Pfaffian. Next suppose that G is a bisubdivision of S2. Let v1; v2; v3; v4 be the vertices of degree three in G and for i; j 2 f1; 2; 3; 4g, let Pi;j be the subdivision-path of G connecting vi and vj. Possibly after relabelling, we may assume that Pj;4 is even for j 2 f1; 2; 3g and P1;2;P2;3;P1;3 are odd. Suppose towards a contradiction that there exists a strongly Pfaffian orientation G~ of G.
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