Longest Paths in Partial 2-Trees - Semantic Scholar

Nonempty Intersection of Longest Paths in Series-Parallel Graphs Guantao Chen1,∗ , Julia Ehrenmüller2,† , Cristina G. Fernandes3,§ , Carl Georg Heise2,†,‡ , Songling Shan1,¶ , Ping Yang1 , and Amy Yates1,∗ 1

Department of Mathematics and Statistics, Georgia State University, United States, {gchen,sshan2,pyang5,ayates2}@gsu.edu 2 Institut für Mathematik, Technische Universität Hamburg-Harburg, Germany, {julia.ehrenmueller,carl.georg.heise}@tuhh.de 3 Institute of Mathematics and Statistics, University of São Paulo, Brazil, [email protected]

October 15, 2014

In 1966 Gallai asked whether all longest paths in a connected graph have nonempty intersection. This is not true in general and various counterexamples have been found. However, the answer to Gallai’s question is positive for several well-known classes of graphs, as for instance connected outerplanar graphs, connected split graphs, and 2-trees. A graph is series-parallel if it does not contain K4 as a minor. Series-parallel graphs are also known as partial 2-trees, which are arbitrary subgraphs of 2-trees. We present two independent proofs that every connected series-parallel graph has a vertex that is common to all of its longest paths. Since 2-trees are maximal series-parallel graphs, and outerplanar graphs are also seriesparallel, our result captures these two classes in one proof and strengthens them to a larger class of graphs. We also describe how this vertex can be found in linear time.

1 Introduction A path in a graph is a longest path if there exists no other path in the same graph that is strictly longer. The study of intersections of longest paths has a long history and, in particular, the ∗

This research was partially supported by NSA grant H98230-12-1-0239. The authors gratefully acknowledge the support of the Technische Universität München Graduate School’s Thematic Graduate Center TopMath. ‡ Partially supported by DFG grant GR 993/10-1. § Partially supported by CNPq Proc. 308523/2012-1 and 477203/2012-4, FAPESP 2013/03447-6, and a joint CAPES-DAAD project (415/PPP-Probral/po/D08/11629, Proj. no. 333/09). ¶ This research was partially supported by the B&B Fellowship of GSU. †

question of whether every connected graph has a vertex that is common to all of its longest paths was raised by Gallai in 1966 [11]. For some years it was not clear whether the answer is positive or negative until, finally, Walther [27] found a graph on 25 vertices that answers Gallai’s question negatively. Today, the smallest known graph answering Gallai’s question negatively is a graph on 12 vertices, found by Walther and Voss [28], and independently by Zamfirescu [30] (see Figure 1). To see that the depicted graph does not have a vertex common to all longest paths, one can identify the three leaves to obtain the Petersen graph. Since the Petersen graph is hypohamiltonian, meaning that it does not have a Hamiltonian cycle but every vertex-deleted subgraph is Hamiltonian, it follows first that the length of a longest path is at most 9 (and it is exactly 9) and second that the intersection of all longest paths is empty. Note that the length of a longest path in the depicted graph can be at most 10 since at most two of its three leaves can be contained in a longest path. But any path of length 10 in the depicted graph would correspond to a Hamiltonian cycle in the Petersen graph.

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8 7

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1

3 Figure6 1: The counterexample of Walther, Voss, and Zamfirescu. 4 1 9 10 These are by far not the only counterexamples. In fact, there are infinitely many (even planar) ones since every hypotraceable graph, meaning a graph having no Hamiltonian path whose all vertex-deleted subgraphs have a Hamiltonian path, is obviously a counterexample. Thomassen proved in [24] that there are infinitely many such graphs. 5

Since the answer to Gallai’s question is negative in general, it seems natural to restrict the problem to subsets of a fixed size of all longest paths. It is well-known [20] that any two longest paths of a connected graph share a common vertex. However, considering the intersection of more than two longest paths gets more intriguing. It is still unknown whether any three longest paths of every connected graph share a common vertex. Zamfirescu asked this question several times [26, 31] and it was mentioned at the 15th British Combinatorial Conference [8]. It is presented as a conjecture in [13] and as an open problem in the list collected by West [29]. Progress in this direction was made by de Rezende, Martin, Wakabayashi, and the second author [10], who proved that, if all nontrivial blocks of a connected graph are Hamiltonian, then any three longest paths of the graph share a vertex. Skupie´ n [23] showed that, for every p ≥ 7, there exists a connected graph such that p longest paths have no common vertex and every p−1 longest paths have a common vertex. Even though it seems as if the property of having a vertex common to all longest paths is too strong, there are some classes of connected graphs for which this property holds. A simple example is the class of trees since in a tree all longest paths contain its center(s). Moreover, Klavžar and Petkovšek [17] proved that the intersection of all longest paths of a (connected)

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split graph is nonempty. Furthermore, they showed that, if every block of a connected graph G is Hamilton-connected, almost Hamilton-connected, or a cycle, then there exists a vertex common to all its longest paths. The latter result implies immediately that the answer to Gallai’s question is positive for the class of (connected) cacti, where a graph is a cactus if and only if every block is either a simple cycle or a vertex or a single edge. In 2004, Balister, Gy˝ ori, Lehel, and Schelp showed a similar result in [3] for the class of circulararc graphs. There was a gap in their proof though, closed recently by Joos [15]. In 2013, de Rezende et al. proved the following two theorems. Theorem 1 ([10]) For every connected outerplanar graph, there exists a vertex common to all its longest paths. Theorem 2 ([10]) For every 2-tree, there exists a vertex common to all its longest paths. Theorem 1 is a strengthening of a theorem by Axenovich [2], which states that any three longest paths in a connected outerplanar graph share a vertex. A more comprehensive survey of the problem of intersections of longest paths and cycles in general can be found in [22]. In this paper we treat the general case of nonempty intersection of all longest paths and prove that the answer to Gallai’s question is positive for the class of connected series-parallel graphs, settling a question raised in [10]. Since not only all trees and cacti but also outerplanar graphs and 2-trees are series-parallel, our result gives a unified proof for Theorems 1 and 2 and generalizes them to a larger class of graphs. We present two proofs of our result. The first one, given in Section 3, heavily relies on the structure of a so called underlying 2-tree (a 2-tree that contains the series-parallel graph as a spanning subgraph), whereas the second one, presented in Section 5, is based on a graph decomposition method introduced by Tutte [25]. We present both proofs because they both seem intriguing for the study of series-parallel graphs. While the first one shows that one can prove surprisingly strong results for a series-parallel graph only using the structure of an underlying 2-tree, the second one provides insight into the structure of Tutte’s decomposition of a series-parallel graph. We strongly believe that both methods can be useful to tackle open problems on series-parallel graphs in the future. The rest of the paper is organized as follows. In Section 2, we give essential definitions and prove some statements that will be useful in what follows. In Section 3, we present the first proof of the main theorem. In Section 4, we present a specific way to decompose series-parallel graphs using a decomposition introduced by Tutte. We use this decomposition in Section 5 in our second proof of the main theorem. In Section 6, we describe a linear-time algorithm that finds a vertex contained in all longest paths in connected series-parallel graphs. Finally, in Section 7, we state several open problems concerning the intersection of longest paths in specific classes of graphs.

2 Preliminaries and definitions We start with a few basic definitions which we use in the subsequent part of our paper. All graphs in this paper are either simple graphs or have at least three edges sharing two end

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vertices. We write H ⊆ G if the graph H is a subgraph of the graph G. Also, we denote by V (G) the set of vertices of G, and by E(G) the set of edges of G. For a pair of vertices u and v of G, we denote by EG (u, v) the set of edges in G between u and v. Let G be a graph and s and t be two of its vertices. We say G is series-parallel with terminals s and t if it can be turned into K2 by a sequence of the following operations: replacement of a pair of parallel edges with a single edge that connects their common endpoints or replacement of a pair of edges incident to a vertex of degree 2 other than s or t with a single edge. A graph G is 2-terminal series-parallel if there are vertices s and t in G such that G is series-parallel with terminals s and t. A graph G is series-parallel if each of its 2-connected components is a 2terminal series-parallel graph. (See [7, Sec. 11.2].) A 2-tree can be defined in the following way. A single edge is a 2-tree. If T is not a single edge, then T is a 2-tree if and only if there exists a vertex v of degree 2 such that its neighbors are adjacent and T − v is also a 2-tree. A graph is a partial 2-tree if it is a subgraph of a 2-tree. (See [7, Sec. 11.1].) We say a partial 2-tree is trivial if it consists of a single vertex or a single edge. Note that every edge in a nontrivial 2-tree is contained in a triangle. It is well-known that a graph is a partial 2-tree if and only if it is K4 -minor-free. Partial 2-trees are exactly the series-parallel graphs, and are also known for being the graphs with tree width at most 2. (See [7, Sec. 11.1].) Next we present some notation we use in our proofs. The length of a path P, denoted by |P|, is the number of edges in P. Let L(G) denote the length of a longest path in the graph G. Let L (G) denote the set of all longest paths in G, that is, L (G) = {P | P is a path in G and |P| = L(G)}. If the graph G is clear from the context, we simply write L for L(G) and L for L (G). By the intersection P ∩ P 0 of two paths P and P 0 , we mean the intersection of their vertex sets. If v is a vertex of the path P, we write v ∈ P. If P and P 0 have a common endpoint x but no other common vertex, then the union P ∪ P 0 is simply defined as the path obtained by concatenating the path P and the path P 0 at the vertex x. A subpath of a path P is called a tail of P if it contains an endpoint of P. Given a vertex x in P, the path P can be split into two subpaths P 0 and P 00 such that P 0 ∩ P 00 = {x}; we call them tails of P starting at x. If |P 0 | ≥ |P 00 |, then P 0 is called a longer tail of P starting at x. Let H be a subgraph of G. If P ∩ H = {x}, then P is called a pending path of H with origin x. If P 0 is a tail of P starting at x and P 0 is also a pending path of H with origin x, we call it a pending tail of P on H with origin x. Given a second path Q such that Q ∩ P 6= ∅ and such that at least one endpoint x of P is not contained in Q, we define the bridge path P % x Q as the path starting at the endpoint x going along P until the first intersection with Q. For some subgraph H, we define P[H] to be the induced subgraph of P in H, that is, the collection of (maximal) subpaths of P that lie in H. Note that this might be more than one path. Next, we borrow some definitions and results presented by Tutte [25]. Let G be a graph and H be a subgraph of G. A vertex of attachment of H in G is a vertex of H that is incident to some

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edge of G that is not an edge of H. Let T be a 2-tree and let x and y be two adjacent vertices in T . An {x, y}-bridge in T is either the edge connecting x and y or a minimal subgraph B of T containing a vertex other than x and y and whose vertices of attachment are contained in {x, y}. We say the edge connecting x and y is a trivial {x, y}-bridge. For each common neighbor z of x and y, let B{x, y},z (T ) be the {x, y}-bridge in T containing z. If z1 6= z2 are two common neighbors of x and y, then B{x, y},z1 (T ) 6= B{x, y},z2 (T ), as otherwise T would contain a K4 -minor. Then, by Theorem I.51 [25], we have B{x, y},z1 (T ) ∩ B{x, y},z2 (T ) ⊆ {x, y}. The interior ◦ of the non-trivial bridge B{x, y},z (T ) is the graph B{x, (T ) = B{x, y},z (T ) − {x, y}. y},z In what follows, for a series-parallel graph G = (V, E), we let T (G) = (V, F ) denote an arbitrary but fixed 2-tree that contains G as a spanning subgraph. We say an underlying edge/triangle of G is an edge/triangle in T (G) independent of its existence in G. We denote by C{x, y},z (G) the maximal subgraph of G contained in B{x, y},z (T (G)). Note that such a subgraph of G may be disconnected. We call this subgraph the component of G generated by ◦ the underlying edge {x, y} in direction z. Similarly, C{x, (G) = C{x, y},z (G) − x − y is called y},z the interior of the component C{x, y},z (G). Again, if the graph is clear from the context, we write C{x, y},z = C{x, y},z (G). Also, let C{x, y} (G) = {C{x, y},z (G) : z is adjacent to x and y in T (G)} be the set of all components generated by the underlying edge {x, y}. Further, define C{x, y}|z (G) = C{x, y} (G)\{C{x, y},z (G)} for an underlying triangle {x, y, z}. A set of vertices W is called a Gallai set (for G) if W ∩ P 6= ∅ for all P ∈ L (G). If W is a Gallai set and if its vertices are pairwise connected by underlying edges, then |W | ≤ 3. In this case, we call this set an underlying Gallai edge and an underlying Gallai triangle when W has size two or three, respectively. We say a vertex v ∈ V is a Gallai vertex if {v} is a Gallai set. Note that the intersection of all longest paths of a graph G is nonempty if and only if G has a Gallai vertex. For an underlying edge {u, v}, let Luv = {P ∈ L | u, v ∈ P} and Luv = {P ∈ L | u ∈ P, v ∈ / P}. Similarly, for an underlying triangle ∆ = {u, v, w}, we define Luvw = {P ∈ L | u, v, w ∈ P}, Luvw = {P ∈ L | u, v ∈ P, w 6∈ P}, and Luvw = {P ∈ L | u ∈ P, v, w 6∈ P}. Moreover, let L(uvw) = {P ∈ Luvw | v is between u and w in P}. A block of G is a maximal 2-connected graph of G. For the purpose of this paper, we assume that K2 is 2-connected and call it the trivial 2-connected graph. A graph B1 v1 B2 v2 · · · vk−1 Bk which consists of an alternating sequence of blocks and cut vertices is called a block chain if each pair of consecutive blocks Bi and Bi+1 share the distinct cut vertex {vi } = V (Bi ) ∩ V (Bi+1 ) for i = 1, 2, . . . , k − 1. If G is a 2-connected graph, a 2-cut of G is a 2-vertex set W ⊆ V (G) such that G − W is not connected. A 2-vertex set {u, v} is called a 2-separation if {u, v} has at least three bridges, which is equivalent to have one of the following three conditions held: (1) G − {u, v} has at least three components, (2) G − {u, v} has exactly two components and |EG (u, v)| = 1, or (3) |EG (u, v)| ≥ 2 and |E(G)| ≥ 3. The last condition indicates that a 2-separation {u, v} is not a 2-cut only if there are multiple edges between u and v and |E(G)| ≥ 3.

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Following Tutte [25], we call a 2-vertex graph with at least three edges between them a bond. A Θ-graph is a simple graph consisting of three internally disjoint paths sharing two end vertices. We end this section with the following auxiliary results that will be useful in Section 3. Note that the first four results hold for general graphs, not only for series-parallel graphs. Proposition 3 ([20]) Any two longest paths in a connected graph share a common vertex. Lemma 4 In a graph, let P1 and P2 be two paths with tails R1 and R2 , respectively (that is, subpaths containing an endpoint of P1 or P2 ) such that R1 ∩ P2 = ∅ and R2 ∩ P1 = ∅. If there exists a connecting path P such that ∅ 6= P ∩ P1 ⊆ R1 and ∅ 6= P ∩ P2 ⊆ R2 , then P1 and P2 cannot both be longest paths. Proof. Assume for a contradiction that both P1 and P2 are longest paths. For i ∈ {1, 2}, let R0i denote the other tail of Pi so that Pi = R i ∪ R0i and R i and R0i intersect at only one vertex. By assumption, both R1 and R2 intersect P. Hence, there exist vertices x and y such that x ∈ R1 ∩P, y ∈ R2 ∩ P, and the interior of the subpath of P starting at x and ending in y does not contain vertices in P1 or P2 . Let Q 1 denote the path obtained from going along R01 , along R1 until x, along P until y, and then along R2 until the endpoint that is not in R02 . Let Q 2 denote the path obtained from going along R02 , along R2 until y, along P until x, and then along R1 until the endpoint that is not in R01 . Now, as |Q 1 | + |Q 2 | > |P1 | + |P2 |, we have that |Q 1 | > |P1 | or |Q 2 | > |P2 |, a contradiction. ƒ x

x R1

R1 R2

P2

y

R2

P P2

z P1

P1

y

P

Figure 2: Situation for Lemmas 4 and 5. Lemma 5 In a graph, let P1 and P2 be two paths that share a common vertex z and let R1 and R2 be two subpaths of P1 and P2 , respectively, both having z as an endpoint, such that R1 ∩ P2 = {z} R1 x and R2 ∩ P1 = {z}. If there exists a connecting path P such that z ∈ / P, ∅ 6= P ∩ P1 ⊆ R1 , and ∅ 6= P ∩ P2 ⊆ R2 , then P1 and P2 cannot both be longest y paths. P2 P P1 that bothz P and P are longest Proof. Assume for a contradiction paths. By assumption, 1 2 R2 both R1 and R2 intersect P in a vertex other than z. Hence, there exist vertices x and y distinct from z such that x ∈ R1 ∩ P, y ∈ R2 ∩ P, and the interior of the subpath of P starting at x ˜ 1 denote the path starting at z, and ending in y does not contain vertices in P1 or P2 . Let R ˜ 2 denote the path starting at z, going along R2 , and going along R1 , and ending in x, and let R ending in y. Let R01 and R02 denote the tails of P1 and P2 starting at z not containing R1 and R2 ,

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˜ 1 | ≥ |R ˜ 2 |, then by combining R0 , R ˜ 1 , the subpath of P starting at x and ending respectively. If |R 2 in y, and the tail of P2 starting at y and not containing z, we get a path strictly longer than P2 , ˜ 1 | < |R ˜ 2 |, then by combining R0 , R ˜ 2 , the subpath of P a contradiction. If, on the other hand, |R 1 starting at y and ending in x, xand the tail of P1 starting at x and not containing z, we get a x path strictly longer than P1 , a contradiction. ƒ R1 R2 R1 P y Observe that Lemma 4 is not R2 a consequence of Lemma 5. Indeed, in the situation of Lemma 4, P P2 vertex withzP might y starting2 from x and going along P1 , the first common not be the same as 2 P P1 the first common vertex of P2 with P1 , starting from y. Thus, the vertex z required in Lemma 5 P1 might not exist. At some points of the proofs in the next section, we are in a situation where one of the two lemmas above apply. The next corollary describes this situation.

R1

x y

P2

P1

P

z R2

Figure 3: Situation for Corollary 6. Corollary 6 In a graph, let P1 and P2 be two paths that share a common vertex z and let R1 be a tail of P1 starting at z. Let R2 be a union of pairwise internally vertex disjoint subpaths of P2 (that is, they may have common endpoints) such that all paths in R2 have as one endpoint z or an endpoint of P2 . Suppose R1 ∩ P2 = {z} and R2 ∩ P1 ⊆ {z}. If there exists a connecting path P such that z∈ / P, ∅ 6= P ∩ P1 ⊆ R1 , and ∅ 6= P ∩ P2 ⊆ R2 , then P1 and P2 cannot both be longest paths. Proof. There exist vertices x ∈ P ∩ R1 and y ∈ P ∩ R2 such that the interior of the subpath P x, y of P starting at x and ending in y does not contain any other vertices in P1 or P2 . Let R02 be the path in R2 that contains y. If R02 contains z then the statement follows from Lemma 5 for longest paths P1 and P2 with their subpaths R1 and R02 , respectively, and connecting path P x, y . Otherwise, the statement follows from Lemma 4 again for longest paths P1 and P2 , tails R1 and R02 , and connecting path P x, y . ƒ The next two results are specific for series-parallel graphs. Lemma 7 Let ∆ = {v1 , v2 , v3 } be an underlying triangle in a connected series-parallel graph G. If R i is a path in G with vi as an endpoint and R i ∩ ∆ = {vi } for each i ∈ {1, 2, 3}, then only one of the sets R1 ∩R2 , R1 ∩R3 , and R2 ∩R3 can be nonempty. Furthermore, if R i ∩R j 6= ∅, then R i ∪R j ⊆ C for some component C ∈ C{vi ,v j }|vk . Proof. For the first statement, assume without loss of generality that R1 ∩ R2 6= ∅. Then we have a path S from v1 to v2 consisting of R1 % v1 R2 and the tail of R2 containing v2 . Note

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that this path does not use v3 or the underlying edge {v1 , v2 } because R i contains only vi in ∆ for i = 1, 2. If additionally R1 ∩ R3 6= ∅ or R2 ∩ R3 6= ∅, then v3 is connected to S by a path S 0 ∪ S 00 , where S 0 is the shortest tail of R3 from v3 to a vertex u in R1 ∪ R2 , and S 00 is a shortest path from u to S in the connected graph R1 ∪ R2 . Let x be the endpoint of S 00 in S. Observe that x is an internal vertex of S. So {x, v1 , v2 , v3 } determines a K4 -minor in T (G), a contradiction. For the second statement, suppose that R i and R j intersect. Obviously, H = (R i ∪ R j ) − {vi , v j } must lie in the interior of a {vi , v j }-bridge B of T (G). Also, the edge {vi , v j } is a cut set in T (G), separating vk from H. Otherwise we would have three paths as above, namely R i , R j , and the path from vk to H avoiding vi and v j , and at least two of the pairs within these three paths would intersect. Therefore B ∈ C{vi ,v j }|vk (T (G)) and thus R i ∪ R j ⊆ C = G[V (B)] ∈ C{vi ,v j }|vk (G). ƒ Lemma 8 Let ∆ = {v1 , v2 , v3 } be an underlying triangle in a connected series-parallel graph G, and R be a path in G with vi as endpoint and R ∩ ∆ = {vi }, for some i in {1, 2, 3}. Let j and k be such that {i, j, k} = {1, 2, 3}. If S1 is a path with endpoints vi and v j such that S1 ∩ ∆ = {vi , v j }, and S2 is a path with endpoints v j and vk such that S2 ∩ ∆ = {v j , vk }, then R ∩ S2 = ∅ and S1 ∩ S2 = {v j }. Proof. Assume for a contradiction that there is some vertex x ∈ R ∩ S2 . Split the path S2 at x j and look at the two tails S2 and S2k starting at x and ending at v j and vk , respectively. Now note j

that R, S2 , and S2k are three paths as in Lemma 7 and they all intersect at x, a contradiction. j

j

Similarly, if y ∈ (S1 ∩ S2 )\{v j }, split S2 analogously at y obtaining S2 and S2k . Then S1 − v j , S2 , and S2k are three paths as in Lemma 7 and they all intersect at y, again a contradiction. ƒ

3 First proof As we have already mentioned in Section 1, de Rezende at el. [10] proved that the intersection of all longest paths of a 2-tree is nonempty. In this section, we extend this result proving that all connected subgraphs of 2-trees, that is, all connected series-parallel graphs, have also this property. We proceed in four steps. First, we prove in Lemma 9 that there exists an underlying Gallai triangle. Then, we show in Lemma 10 that actually one underlying edge of this triangle is an underlying Gallai edge and there exists a component generated by this underlying edge that satisfies certain properties. In Lemma 12 we prove that either one of the endpoints of this underlying edge is a Gallai vertex or we can find an adjacent underlying Gallai edge and a strictly smaller component satisfying the same properties. By iterating, we end up with a Gallai vertex since we only consider finite graphs. Lemma 9 In every nontrivial connected series-parallel graph G, there exists an underlying Gallai triangle. Proof. Take any underlying triangle ∆0 of a nontrivial connected series-parallel graph G. Note that, in every connected series-parallel graph, every underlying edge is either a cut set of G or is contained in exactly one underlying triangle. Assume there exists a longest path P0 in G

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containing no vertex of ∆0 . Then there exists an underlying edge e0 ⊆ ∆0 , which is a cut set, and a vertex z0 ∈ / ∆0 such that z0 is adjacent to both endpoints of e0 in T (G) and P0 lies in the component generated by e0 in direction z0 , that is, P0 ⊆ Ce◦ ,z . By Proposition 3, all longest 0 0 paths must intersect P0 and so they have at least one vertex in Ce◦ ,z . Note that ∆1 = e0 ∪ {z0 } is 0 0 a triangle in T (G) and thus an underlying triangle in Ce0 ,z0 . Now either all longest paths contain a vertex of ∆1 and we are done, or there exist a longest path P1 , an underlying edge e1 ⊆ ∆1 , where e1 6= e0 and e1 is a cut set, and a vertex z1 ∈ / ∆1 such that z1 is adjacent to both endpoints of e1 in T (G) and P1 ⊆ Ce1 ,z1 . Note that Ce1 ,z1 ( Ce0 ,z0 , as P1 must intersect P0 in Ce◦ ,z again by 0 0 Proposition 3. Iteratively, obtain ∆2 and Ce2 ,z2 and eventually a strictly decreasing sequence of components Ce0 ,z0 ) Ce1 ,z1 ) Ce2 ,z2 ) · · · ) Cek ,zk . Since G is finite, this process ends with some triangle ∆ = ∆k such that all longest paths contain one vertex of ∆. ƒ Next we prove that one of the underlying edges of a Gallai triangle is an underlying Gallai edge and there exists a component generated by this underlying edge that satisfies certain properties. Lemma 10 For every connected series-parallel graph G = (V, E), there exists a Gallai vertex, or an underlying Gallai edge {u, v} and a component C ∈ C{u,v} such that, for every pair (P, P 0 ) ∈ (Luv × L vu ) ∪ (Luv × Luv ) ∪ (L vu × Luv ), there exists a vertex in C ◦ ∩ P ∩ P 0 . Before presenting the proof of Lemma 10, we prove an intermediate result stated in the next lemma. Throughout the next proofs, we keep Lemmas 7 and 8 in mind and use them implicitly whenever we claim that certain constructions are indeed paths and whenever we claim that a path lies in a certain component. For the proofs of Lemmas 11 and 12, we use the following notation. Every path P ∈ Luvw can be split at u and v, resulting in three subpaths. Let P (u) and P (v) be the tails of P starting at vertex u and vertex v, respectively. The remaining subpath, joining u and v, is denoted by P (u,v) . Analogously, a path P ∈ L(uvw) is split into P (u) , P (u,v) , P (v,w) , and P (w) (see Figure 4). w

w P(w) P(v,w)

P(u) u

v

P(u,v)

P(v)

P(u)

u

v P(u,v)

Figure 4: Splitting P ∈ Luvw at vertices u and v, and P ∈ L(uvw) at vertices u, v, and w. C

C

Lemma 11 w Let ∆ be an underlying Gallai trianglewin a nontrivial connected series-parallel graph C1 C1 G. If L x yz 6= ∅ for every x, y, z such that ∆ = {x, y, z}, then, for some u,𝑃 v, w such that {u, v, w} = ∆, v

u 𝑃v

𝑃u

9

u 𝑃v

v

𝑃u

there is a component C ∈ C{u,w}|v such that for every pair   (P, P 0 ) ∈ 

 [ {x, y,z}=∆

 L x yz × L xz y  ∪ (Luvw × Luvw ) ∪ (L vwu × Luvw )

there exists a vertex in C ◦ ∩ P ∩ P 0 . Proof (Lemma 11) Let P ∈ Luvw ∪Luwv ∪L vwu , where {u, v, w} = ∆, and x ∈ ∆ be such that x is in P and P (x) is as long as possible. Without loss of generality, we may assume P ∈ Luvw and x = u. In what follows, we use Pz to refer to an arbitrary path in L x yz where {x, y, z} = ∆. First note that P (u) intersects every Pu(w) , otherwise P (u) ∪ P (u,v) ∪ Pu(v,w) ∪ Pu(w) is a path of length strictly greater than L by the choice of P. Thus, P (u) must lie in a component C ∈ C{u,w}|v . We will prove that C has the property stated in the lemma. We start by proving that each path in Luvw intersects in C ◦ every path in L vwu , that is, we show that each Pw(u) intersects every Pu(w) . Observe that |Pw(u) | = |P (u) |, otherwise P (u) ∪ Pw(u,v) ∪ Pw(v) would be a path of length strictly greater than L. So the argument previously applied to P, now with Pw instead, implies that each Pw(u) intersects every Pu(w) . Now we prove that each path in Luvw ∪ L vwu intersects in C ◦ every path in Luwv ∪ Luvw . First note that each Pw(u) intersects Q(u,w) in C ◦ for every Q ∈ Luwv ∪ L(uwv) ∪ L(vuw) , otherwise Pw(u) ∪ Q(u,w) ∪ Pu(w,v) ∪ Pu(v) is a path of length strictly greater than L by the choice of P. As both Luwv and L vwu are nonempty, there exist at least one such Q and one such Pu . So Q and Pu(w) must intersect in C ◦ , otherwise we derive a contradiction from Lemma 5 for subpaths Pu(w) and Q(u,w) , z = w, and a connecting path contained in Pw(u) . Second, we prove that each Pw(u) and Pu(w) intersect every Q ∈ L(uvw) in C ◦ . Observe that |Q(u) | ≤ |P (u) |, otherwise either Q(u) ∪ P (u,v) ∪ P (v) or Q(u) ∪ Pv(u,w) ∪ Pv(w) is a path of length strictly greater than L by the choice of P. If |Q(u) | < |P (u) | = |Pw(u) |, then Pw(u) intersects Q(w) , otherwise Pw(u) ∪ Q(u,v) ∪ Q(v,w) ∪ Q(w) is a path of length strictly greater than L. So Pu(w) and Q must intersect in C ◦ , otherwise we derive a contradiction from Corollary 6 for longest paths Pu and Q with z = w, tail Pu(w) , subpaths Q[C], and connecting path Pw(u) . If |Q(u) | = |P (u) |, then Q(u) intersects Pu(w) , otherwise Q(u) ∪ Q(u,v) ∪ Pu(v,w) ∪ Pu(w) is a path of length strictly greater than L by the choice of P. So Pw(u) and Q must intersect in C ◦ , otherwise we derive a contradiction from Corollary 6 for longest paths Pw and Q with z = u, tail Pw(u) , subpaths Q[C], and connecting path Pu(w) . ƒ Proof (Lemma 10) If G is trivial, there exists a Gallai vertex. Otherwise, let ∆ = {u, v, w} be an underlying Gallai triangle, which exists by Lemma 9. First, we show that at least one of the edges of ∆ is an underlying Gallai edge. Assume for a contradiction that no edge of ∆ is an underlying Gallai edge, which means that there are three longest paths Pu ∈ Luvw , Pv ∈ L vuw , and Pw ∈ Lwuv . Thus, there exist three distinct components Cuv , Cuw , and C vw generated by the underlying edges of ∆ such that, for every x, y in ∆, all intersection points of Px and P y lie in the component C x y . Without loss of generality, let |Pu [Cuv ]| ≥ L/2. Then, by combining the paths Pu [Cuv ], Pu [Cuw ] %u Pw , and a longer tail of Pw , we obtain a path of length strictly greater than L, a contradiction. So, there exists an underlying Gallai edge in ∆.

10

If all edges of ∆ are underlying Gallai edges, then Luvw = L vuw = Lwuv = ∅. If moreover at least one among Luvw , Luwv , L vwu is empty, then one of the vertices in ∆ is a Gallai vertex. Otherwise, we are in the situation of Lemma 11 and the statement of the lemma follows immediately. Without loss of generality, we may assume {u, v} is an underlying Gallai edge. Hence, Lwuv = ∅. Let Pu ∈ Luvw ∪ L vuw and x ∈ {u, v} be such that Pu has a tail starting at x that is as long as possible. Without loss of generality, we may assume x = u and thus Pu ∈ Luvw . Let Pu0 be such a longer tail starting at u. (If both tails of Pu starting at u have same length, choose Pu0 to be any one of them.) Let Pu00 be the other tail of Pu . As Luvw is nonempty, {v, w} is not an underlying Gallai edge. If all longest paths contain u, then we are done since u is a Gallai vertex. Otherwise, L vu is nonempty. Each Pv ∈ L vu intersects Pu0 because otherwise, by combining Pu0 , Pu00 %u Pv , and a longer tail of Pv , we get a path of length strictly greater than L by the choice of Pu . If {u, v} is the only underlying Gallai edge in ∆, then L vuw 6= ∅ and Pu0 has to lie in a component C ∈ C{u,v}|w , so that it intersects every Pv ∈ L vuw . Otherwise, {u, w} is also an underlying Gallai edge and Pu0 lies in a component C either in C{u,v}|w or in C{u,w}|v . (Note that in this case L vwu 6= ∅.) Without loss of generality, we assume C ∈ C{u,v}|w . We claim that C has the desired properties. First, we prove that each path in L vu intersects in C ◦ every path in Luv ∪ Luv . Let Pv ∈ L vu . Suppose that there exists a path Q ∈ Luv ∪ Luv such that Pv does not intersect Q in C ◦ . Either Q intersects Pu0 in C ◦ , or Pu0 and both tails of Q starting at u have length L/2 (see Figure 5). In the former case, since Pu0 intersects both Pv and Q in C ◦ , we can apply Lemma 4 if Q ∈ Luv (with paths Pv and Q and connecting path Pu0 ) or Corollary 6 if Q ∈ Luv (with paths Pv and Q, z = v, a suitable tail of Pv starting at v, subpaths Q[C], and a connecting path contained in Pu0 ) deriving a contradiction. So, suppose now that Pu0 and both tails of Q starting at u have equal length. The path Pv intersects both tails of Q starting at u because otherwise such a tail of Q, Pu0 %u Pv , and a longer tail of Pv would be a path of length strictly greater than L. Therefore, and since the combination of Pu0 with one tail of Q starting at u and the combination of Pu0 with the other tail of Q starting at u are both longest paths, we can apply Lemma 5 with a connecting path contained in Pv and z = u, deriving again a contradiction. Hence, each path in L vu intersects in C ◦ every path in Luv ∪ Luv . Next, we prove that each path in Luv intersects in C ◦ every path in Luv . If Luv 6= ∅, let P be a path in Luv and Q u in Luv . Assume that P does not intersect Q u in C ◦ . Let Pv ∈ L vu and note that such a longest path must exist. Since Pv intersects Q u in C ◦ and P in C ◦ , we derive a contradiction longest paths Q u and P with z = u, a suitable tail of Q u starting at u, subpaths P[C], and a connecting path contained in Pv [C]. ƒ Lemma 12 In a nontrivial connected series-parallel graph G, let e = {u, v} be an underlying Gallai edge and C ∈ Ce be a component such that all pairs of longest paths mutually intersect in at least one vertex of C and all pairs of longest paths in Luv × L vu , Luv × Luv , and L vu × Luv mutually intersect in C ◦ (as in Lemma 10). Let w be the unique vertex in C adjacent in T (G) to both u and v. Then u, v, or w is a Gallai vertex, or there is an underlying edge f incident to u or v and a component C1 ∈ C f , C1 ( C, with the properties of Lemma 10.

11

‚

Pu

C

‚

C

Pu 𝐿

Pv

v

u

2

Pv

v

u 𝐿

Q 𝐿

Q

2

2

Figure 5: Left: Pv and Q do not intersect in C (apply either Lemma 4 or Corollary 6); Right: Pv intersects both tails of Q outside of C (apply Lemma 5). Proof. Let ∆ = {u, v, w}. Assume neither u nor v are Gallai vertices (otherwise there is nothing more to prove). Thus, both Luv and L vu are nonempty. All pairs of paths in Luv × L vu intersect in C ◦ by the assumptions of the lemma. By Lemma 7, all paths in Luv or all paths in L vu must w w contain w since C ∈ / C{u,v}|w . Without loss of generality we may assume that all paths in Luv contain w. Therefore, Luvw = ∅ and {v, w} isP(w) an underlying Gallai edge. P(v,w)there exists a path in L that We distinguish two cases. First, we consider the case in which vu

(u) contain w (that is, a path doesPnot L vuw ) and then the case in which all paths in L vu 3 P3(v) in (u) P v v contain w (that is, L = ∅). In both cases there exists a Gallai vertex, or vuw u u we show that either P(u,v) (u,v) P an underlying Gallai edge and a component strictly smaller than C that fulfill the requirements of Lemma 12.

C

C

w

w C1

C1 𝑃

v

u 𝑃v

u 𝑃v

𝑃u

v

𝑃u

Figure 6: The scenario of the proof of Lemma 12: Case 1 (left) and Case 2 (right). Case 1. The set L vuw is nonempty. For every path Pu ∈ Luv = Luwv , the tail Pu(w) must intersect every path in L vuw by assumption. Let C1 ( C, C1 ∈ C{v,w}|u be the unique component where they mutually intersect.

12

We claim that the underlying edge {v, w} together with the component C1 fulfills the requirements of Lemma 12. If Luvw 6= ∅, let P ∈ Luvw and Pv ∈ L vuw . Assume for a contradiction that there exists a path Pu ∈ Luv such that P does not intersect Pu in C1◦ . Note that Pv and P must intersect in C ◦ by assumption and hence Pu and P (v) are disjoint. Since Pv intersects Pu in C1◦ and P in Cu (at least in vertex v), we can apply Lemma 4 (with paths Pu and P, tails Pu(w) and P (v) , and a connecting path contained in Pv [C1 ]) to derive a contradiction. Next, we prove that every path in L vwu ∪Luvw intersects every path in L vuw ∪Luv ∪Luvw in C1◦ . Let P ∈ L vwu ∪ Luvw and Pv ∈ L vuw . Note that Pv intersects P in C ◦ by assumption if P ∈ Luvw . Otherwise, P ∈ L vwu , and they also intersect in C ◦ , or not both Pv and P could be longest paths by Lemma 5 for z = v, a suitable tail of Pv [C1 ], tail P (v,w) ∪ P (w) , and connecting path Pu(w) for some Pu ∈ Luv . Furthermore, P must intersect Pv in C1◦ . Otherwise, Pv would have a tail starting at v completely in a component C 0 ∈ C{v,w}|u , C 0 6= C1 . Then |Pv [C 0 ]| < L/2 since Pv [C 0 ] is disjoint from Pu and so Pv [C 0 ] ∪ Pv [C1 ] % v Pu , and a longer tail of Pu would be a path of length strictly greater than L. But now, by combining Pv [C1 ] and a longer tail of P starting at v, we get a path of length strictly greater than L, a contradiction. For every Pu ∈ Luv , by applying Corollary 6 (with paths Pu and P, z = w, tail Pu(w) , subpaths P[C1 ], and connecting path Pv [C1 ]), we can deduce that P intersects Pu in C1◦ . Every Pw ∈ Luvw must intersect P in C1◦ , otherwise we get a contradiction by applying Corollary 6 (with paths P and Pw , z = v, tail Pw [C1 ], subpaths P[C1 ], and connecting path Pu(w) ). Case 2. The set L vuw is empty. If the set Luvw is empty, then all longest paths contain w, therefore w is a Gallai vertex and the requirements of Lemma 12 are fulfilled. So, from now on, we assume that the set Luvw is nonempty. First, we prove that, for each P ∈ Luvw , either P (v) intersects Pu(w) for every Pu ∈ Luv = Luwv , or P (u) intersects Pv(w) for every Pv ∈ L vu = L vwu . Assume for a contradiction that there exist P ∈ Luvw , Pu ∈ Luv , and Pv ∈ L vu such that P (u) does not intersect Pv(w) and P (v) does not intersect Pu(w) . By the assumptions of the lemma, P has to intersect both Pu and Pv in C ◦ . Therefore, P intersects Pu in the interior of some component of C{u,w}|v and Pv in the interior of some component of C{v,w}|u . (See Figure 7.) First, suppose that Pu(u) and Pv(v) do not intersect. By combining P (v) , P (v,u) % v Pu , and a longer tail of Pu , we get a path that cannot be of length strictly greater than L, hence |P (v) | < L/2. Combining P (u) ∪ P (u,v) ∪ Pv(v,w) ∪ Pv(w) would be a path of length strictly greater than L if |Pv(v) | ≤ L/2. However, by combining Pv(v) , Pv(v,w) % v Pu , and a longer tail of Pu , we get a path of length strictly greater than L, a contradiction. If, on the other hand, Pu(u) and Pv(v) intersect, then P (u) and Pu(u) are disjoint except for the vertex u. Note that |P (u) | = |Pu(u,w) ∪ Pu(w) | since otherwise P and Pu would not be longest paths. The combination of Pu(u) and P (u) is therefore a longest path in Luvw , which is a contradiction. Therefore, P (u) intersects Pv(w) , or P (v) intersects Pu(w) . We claim that if R(u) , for some longest path R in Luvw , does not intersect Pv(w) for some path Pv ∈ L vu , then for every longest path P ∈ Luvw , the tail P (v) intersects Pu(w) for every path Pu ∈ Luv . Indeed, let P be a path in Luvw . Assume for a contradiction that there

13

C 𝑃v

𝑃u

𝑃u

𝑃

v 𝐿

w

u

v

2

Figure 7: Left: Pu(u) and Pv(v) do not intersect; Right: Pu(u) and Pv(v) intersect. exists a path Pu ∈ Luv such that P (v) does not intersect Pu(w) . Note that by the latter paragraph Pu(w) must intersect R(v) and Pv(w) must intersect P (u) . If P (v) lies in some component of C{v,w}|u , then by Lemma 7 P (v) cannot intersect R(u) and R(u,v) − v, and R(v) cannot intersect P (u) and P (u,v) − v since R(v) also lies in a component of C{v,w}|u . Hence, we have |P (v) | = |R(v) | and Q = R(u) ∪ R(u,v) ∪ P (v) is a path in Luvw such that Q(u) does not intersect Pv(w) , and Q(v) does not intersect Pu(w) , a contradiction. Analogously, if R(u) lies in a component of C{u,w}|v , the path R(u) ∪ P (u,v) ∪ P (v) yields a contradiction. Thus, we may assume that both R(u) and P (v) lie in a component of C{u,v}|w and are therefore disjoint from P (u) except for u, and from R(v) except for v, respectively. Now, we get a contradiction from Lemma 4 for longest paths R and P, tails R(v) and P (u) , and a connecting path contained in (Pu(w) %w R) ∪ (Pv(w) %w P). Without loss of generality, we may assume that, for each P ∈ Luvw , the tail P (v) intersects Pu(w) for every Pu ∈ Luv . Let C1 ∈ C{v,w}|u be the unique component where they mutually intersect. Note that the underlying edge {v, w} is indeed an underlying Gallai edge since the set Luvw is empty. Let P ∈ Luvw and Pu ∈ Luv be arbitrary but fixed. Note that Pu intersects P in C1◦ . Assume for a contradiction that there exists a longest path Pv ∈ L vwu ∪ Luvw such that P and Pv do not intersect each other in C1◦ . Then not both P and Pv can be longest paths by Corollary 6 for z = v, tail P (v) , subpaths Pv [C1 ], and connecting path Pu(w) , a contradiction. Therefore, the underlying edge {v, w} together with the component C1 fulfills the requirements of Lemma 12.ƒ Theorem 13 For every connected series-parallel graph G, there exists a vertex v such that all longest paths in G contain v. Proof. This follows from Lemma 10 and by iteratively applying Lemma 12 since G is finite. ƒ

4 A decomposition of K4 -minor-free graphs We now proceed to prove Theorem 13 in a different manner, using a graph decomposition method introduced by Tutte [25]. Similar to decomposing a connected graph into blocks and cut vertices, Tutte introduced a method to decompose 2-connected graphs into 3-blocks: bonds, cycles, and 3-connected graphs. We modify this method particularly for series-parallel graphs.

14

For this, if there exists a path P = v0 v1 · · · vm , then, assuming that i ≤ j, let P [vi ,v j ] = vi vi+1 · · · v j , P (vi ,v j ) = P [vi ,v j ] − {vi , v j }, P [vi ,v j ) = P [vi ,v j ] − {v j }, and P (vi ,v j ] = P [vi ,v j ] − {vi }. Also, if there is a − → ← − cycle C = v0 v1 · · · vm−1 vm , where v0 = vm , let C [vi ,v j ] = vi vi+1 · · · v j and C [vi ,v j ] = v j v j+1 · · · vi , − → − → with the indices taken modulo m. Similarly, define C [vi ,v j ) and C (vi ,v j ) . The following observations imply that there are no 3-connected blocks (except K3 ) when applying the Tutte Decomposition Algorithm to a series-parallel graph. Lemma 14 Let G be a connected series-parallel graph. If a subgraph H of G is a Θ-graph consisting of [u,v]

[u,v]

[u,v]

(u,v)

(u,v)

(u,v)

three internally vertex-disjoint paths P1 , P2 , and P3 , then P1 , P2 , and P3 are in three different components of G − {u, v} provided they are not empty. Consequently, {u, v} is a 2-separation of G. (u,v)

Proof. Suppose to the contrary, there is a path Q[x, y] connecting x ∈ P1 [x, y]

(u,v) P3

that Q ∩ = ∅. Then by contracting we get a K4 , giving a contradiction.

(u,v) P1

to x,

(u,v) P2

to y,

[u,v) P3

(u,v)

and y ∈ P2

to u, and Q

, such

[x, y)

to x, ƒ

Lemma 15 Let G be a 2-connected series-parallel graph but not a bond, and let e = uv be an arbitrary edge of G. If {u, v} is not a 2-separation of G, then G − e is no longer 2-connected. In particular, if G has at least 4 vertices, G is not 3-connected. Proof. Suppose on the contrary that G − e is still 2-connected. By Menger’s Theorem, there exist two internally vertex-disjoint (u, v)-paths P and Q. Now, P ∪ Q ∪ {e} is a Θ-graph. By Lemma 14, the paths P (u,v) and Q(u,v) are in two different components of G − {u, v}, which in turn shows that {u, v} is a 2-separation of G, a contradiction. ƒ Lemma 16 Let G be a 2-connected graph and {u, v} be a 2-separation of G. There is no 2-separation {x, y} such that x and y are in different {u, v}-bridges of G. Proof. Since {u, v} is a 2-separation of G, there are three internally vertex-disjoint (u, v)-paths in G. So, for any two distinct vertices x and y in different {u, v}-bridges, G − {x, y} does not separate u and v. As G is 2-connected, each graph induced on the vertex set of a {u, v}-bridge together with the edge {u, v} is 2-connected, so {x, y} is not a 2-cut of G, which in turn shows that {x, y} is not a 2-separation of G. ƒ In the following, we apply the Tutte’s Decomposition Algorithm (TDA) (see Chapter 3 in [25]) specifically to series-parallel graphs. Given a 2-connected graph G, we apply TDA to produce a set D(G) of 3-blocks, a set ϕ(G) of virtual edges, and a rooted tree T3 (G) with vertex set D(G). This algorithm is applied on an ordered pair (G, e) where e is an arbitrary edge of G, but both D(G) and ϕ(G) are independent from the selection of edge e as Tutte [25] pointed out. Tutte’s Decomposition Algorithm (TDA) Let G be a nontrivial 2-connected series-parallel simple graph, and let e ∈ E(G) be an arbitrary edge of G with endpoints u and v. Perform the following operations.

15

O-0 If G is a cycle, let T3 (G) := K1 with vertex set D(G) := {G} and G being the root, and ϕ(G) := ∅. Otherwise, perform the following operations. O-A If {u, v} is a 2-separation of G, let G1 , G2 , . . . , Gk be all nontrivial {u, v}-bridges. For i = 1, 2, . . . , k, add a virtual edge ei between u and v, and let Gi := Gi + ei . Let Buv be a bond with vertex set {u, v} and edge set {e, e1 , e2 , . . . , ek }. Clearly each Gi is a 2-connected series-parallel graph with |E(Gi )| < |E(G)| and {u, v} is not a 2-separation of Gi . By applying the following operation O-B to the ordered pair (Gi , ei ) and to the resulting Sk smaller graphs, obtain T3 (Gi ), D(Gi ), and ϕ(Gi ). Let D(G) := ( i=1 D(Gi )) ∪ {Buv } and T3 (G) be a tree obtained from vertex set D(G) by adding an edge between Buv and the root of T3 (Gi ) for each i = 1, 2, . . . , k. Set Buv as the root of T3 (G). Let Sk ϕ(G) := ( i=1 ϕ(Gi )) ∪ {e1 , e2 , . . . , ek } be the set of virtual edges. O-B If {u, v} is not a 2-separation of G, then G − e is no longer 2-connected by Lemma 15. That means G − e is a nontrivial block chain, say G1 v1 G2 v2 · · · vk−1 Gk with u ∈ G1 , v ∈ Gk , u 6= v1 , and v 6= vk−1 . Rename u = v0 and v = vk . For i = 0, 1, . . . , k − 1, if Gi = vi vi+1 , let ei = vi vi+1 . Otherwise, let Fi,1 , Fi,2 , . . . , Fi,ki be all nontrivial {vi , vi+1 }-bridges in Gi , add ki +1 virtual edges ei and ei, j between vi and vi+1 for 1 ≤ j ≤ ki , let B vi vi+1 be a bond with vertex set {vi , vi+1 } and edge set E(vi , vi+1 ) ∪ {ei , ei,1 , . . . , ei,ki }, and let Gi, j = Fi, j ∪ {ei, j } for each j = 1, 2, . . . , ki . Let Cuv be the cycle induced by {e, e0 , e1 , . . . , ek−1 }. For each pair (i, j) with 0 ≤ i ≤ k − 1 and 1 ≤ j ≤ ki , it is readily seen that Gi, j is a 2-connected K4 -minor-free simple graph. Moreover, {vi , vi+1 } is not a 2-separation of Gi, j . By applying O-B to all pairs (Gi, j , ei, j ) for Gi 6= ei and the resulting smaller graphs, obtain T3 (Gi, j ), D(Gi, j ), and ϕ(Gi, j ) for Gi 6= ei . Let 

ki







 [ [   [  D(G) :=  D(Gi, j ) ∪ {Cuv } ∪  {B vi vi+1 } . i:Gi 6=ei j=1

i:Gi 6=ei

S S ki T3 (Gi, j ) (where the union on i are taken over Gi 6= ei ) Let T3 (G) be obtained from i j=1 by adding a vertex Cuv and B vi vi+1 and an edge between Cuv and each B vi vi+1 for each Gi 6= ei , and by adding an edge between B vi vi+1 to the root of each Gi, j provided Gi 6= ei . Set Cuv as the root of T3 (G). Let ki  [  [ ϕ(G) := {ei } ∪ ( ϕ(Gi, j ) ∪ {ei, j }) i:Gi 6=ei

j=1

be the set of virtual edges. Lemma 17 Let G be a 2-connected simple series-parallel graph and A = {x, y} be a set of two vertices of G. Then, A is the vertex set of a bond B ∈ D(G) if and only if A is a 2-separation of G. Proof. Lemma 17 is clearly true if G is a cycle. We assume that G is not a cycle and verify Lemma 17 according to the bonds produced in O-A or O-B.

16

In O-A, by Lemma 16, the bonds are Buv , and the ones resulted in Gi for i = 1, 2, . . . , k. Note that {u, v} is a 2-separation of G. Since u and v are adjacent in Gi and {u, v} is not a 2-separation of Gi , a pair of vertices in Gi is a 2-separation in Gi if and only if it is a 2-separation in G. Inductively, we can show that Lemma 17 holds. In O-B, by Lemma 16, the bonds are B vi vi+1 , and the ones resulted in Gi, j for each Gi 6= ei with 1 ≤ j ≤ ki . Since each Gi is 2-connected, there are two internally vertex-disjoint (vi , vi+1 )-paths P1 and P2 in Gi . Since these two paths and a (vi , vi+1 )-path along the other direction of Cuv form a Θ-graph, P1 (vi , vi+1 ) and P2 (vi , vi+1 ) are in different bridges of {vi , vi+1 } in Gi , which in turn shows that {vi , vi+1 } is a 2-separation of G. Since Gi, j is 2-connected and vi and vi+1 are adjacent in Gi, j for each 1 ≤ j ≤ ki , a pair of vertices in Gi, j form a 2-separation in Gi, j if and only if they form a 2-separation in G. Inductively, we can show that Lemma 17 holds in this case. ƒ Applying Lemma 17, we can show that D(G), ϕ(G), and T3 (G) are independent from the choice of the edge uv and are uniquely determined by G. In fact, Tutte [25] showed that this statement is true for every 2-connected simple graph without the condition of being K4 -minor-free. Lemma 18 For each vertex v ∈ V (G), let D v (G) denote the set of 3-blocks containing v and T3 [D v (G)] be the subgraph of T3 (G) induced by D v (G). Then T3 [D v (G)] is a subtree of T3 (G). Proof. Since T3 (G) is a tree, we only need to show that the subgraph induced by D v (G) is connected. Otherwise, assume that there exist two 3-blocks A, B ∈ D v (G) such that not all of the internal vertices of the (A, B)-path D1 (= A)D2 · · · Dm−1 Dm (= B) in T3 (G) are in D v (G). Let i > 1 be such that v ∈ V (Di−1 ) and v ∈ / V (Di ). Note that T3 (G) is a bipartite graph with one class containing all bonds and the other containing all cycles, so this path is an alternating path of bonds and cycles. Recall that V (B) ⊂ V (C) if bond B and cycle C are adjacent in T3 (G). So Di−1 is a cycle and Di is a bond. By Lemma 17, V (Di ) is a 2-separation of G. As v 6∈ V (Di ) and v ∈ V (Dm ), we have that i < m. Say V (Di ) = {u, w}. Since T3 (G) is a tree, D , D , . . . , Di−1 and Di+1 , Di+2 , . . . , Dm are in two different components of T3 (G) − Di . So Si−1 1 2 Sm j=1 V (D j ) and j=i+1 V (D j ) are in two different components of G − {u, w}, which contradicts that both contain the vertex v. ƒ We now expand our consideration from 2-connected graphs to connected graphs. Let G be a connected series-parallel graph. We obtain a Decomposition Tree, denoted as TG , by the following steps. Step 1. Decompose G into blocks and cut vertices (see [25]) and obtain a block-cut vertex tree, say T2 (G), by the following description. The tree T2 (G) is a bipartite graph with two partitions (U, V ) such that, for vertices in U, there is a 1-1 correspondence to the blocks of G and such that, for vertices in V , there is a 1-1 correspondence to the cut vertices of G. For any u ∈ U and v ∈ V , uv ∈ E(T2 (G)) if and only if the block corresponding to u contains the cut vertex corresponding to v in G. It is easy to see that T2 (G) is a tree. Note that each block

17

with at least 4 vertices is a 2-connected series-parallel graph but not 3-connected by Lemma 15. To simplify the notation, for each element X ∈ U, we use X either as a vertex of T2 (G) or as a 2-connected subgraph of G, with the meaning being clear from the context. Step 2. For each X ∈ U, we apply TDA to X and obtain T3 (X ), the rooted tree of X , with vertex set D(X ) and virtual edge set ϕ(X ). Notice that if X = K2 , then T3 (X ) = K1 , which is a single-vertex graph, and D(X ) consists of only cycles and bonds by the assumption that G is K4 -minor-free. For each X ∈ U and a cut vertex v ∈ X , recall D v (X ) = {B ∈ D(X ) | v ∈ B}, which we will call v-blocks in X . By Lemma 18, T3 [D v (X )] is a subtree of T3 (X ). Let D(G) =

[

D(X ),

X ∈U

which is a set consisting of K2 , cycles, and bonds obtained from Step 2. Recall that V is the set of cut vertices of G mentioned in Step 1. We modify T2 (G) into a graph TG with vertex set V ∪ D(G) through the step below: Step 3. For each X ∈ U ⊂ V (T2 (G)), replace X by the Tutte decomposition rooted tree T3 (X ) of it; then, for each v ∈ V , if vX ∈ E(T2 (G)), X is a block containing v. In this case, we let v be adjacent to a vertex in T3 [D v (X )]. (Note that to which specific vertex in T3 [D v (X )] we join v is not essential to our proof.) Denote the resulted tree by TG and call it a decomposition tree of G. Given a subgraph H of G, let VTG (H) = {X ∈ V (TG ) | X ∩ V (H) 6= ∅} and TH = TG [VTG (H)]. Particularly, when H = {v}, i.e. a single vertex, we simply denote T{v} by Tv . The following lemma implies that if H is connected, then TH is connected. Lemma 19 The following two statements hold. (1) For each vertex v ∈ V (G), Tv is a subtree of TG ; (2) For every edge e = uv ∈ E(G), Tu ∩ Tv 6= ∅. Proof. (1) If v is not a cut vertex of G, then v is contained in exactly one block of G. We thus know Tv is connected by Lemma 18. Assume v is a cut vertex. Hence, for each block X of G which contains v, TG [D v (X )] is a subtree. Then Tv is the graph obtained by taking the union of all TG [D v (X )] and adding v and edges joining v to a vertex of TG [D v (X )], for each of the subtrees TG [D v (X )]. It is easy to see that Tv is a connected graph. (2) For each edge e ∈ E(G), there exists exactly one block of G, say B, containing e. If B = e, then e ∈ Tu ∩ Tv . Otherwise, there are at least 3 vertices in B. If {u, v} is not a 2-separation of G, there is a unique cycle in D(B) containing e; otherwise, there is a bond in D(B) containing e. In either case, we have Tu ∩ Tv 6= ∅. ƒ

18

5 Second proof In this section we present a second proof of Theorem 13. It follows from a sequence of claims. Let G be a connected simple series-parallel graph and L be the set of all longest paths in G. Claim 20 We may assume that there exists a cycle C ∈ D(G) such that P ∩ C 6= ∅ for each P ∈ L . Proof. Let TG be a decomposition tree of G. For each longest path P, recall VTG (P) = {X ∈ V (TG ) | X ∩ V (P) 6= ∅} and TP = TG [VTG (P)]. By Lemma 19, TP is connected and thus a subtree of TG . For any two longest paths P and Q in L , we have TP ∩ TQ 6= ∅ since P ∩ Q 6= ∅. Let TL = {TP : P ∈ L }. It is well-known that a family of subtrees of a tree has the Helly T property (see problem 18 on p. 49 of [18]), so there is a vertex B ∈ V (TG ) such that B ∈ P∈L TP . By the construction of TG , there are four possibilities for B: a cut vertex of G, a block K2 of G, a bond, or a cycle from D(G). We may assume that B is a cut vertex, a cut edge, or a bond. T If B is a cut vertex of G, then B ⊂ P∈L P, so Theorem 13 holds. T If B = {x, y} is a cut edge of G, we may assume, without loss of generality, x ∈ P∈L P, so Theorem 13 holds. Indeed, we assume this, for otherwise there exist two longest paths P, Q ∈ L such that x ∈ P but y ∈ / P and x ∈ / Q with y ∈ Q. Since P ∩ Q 6= ∅, there is a vertex z ∈ P ∩ Q. Then {x, y, z} contains a triangle-minor, which contradicts {x, y} being a cut edge in G. Suppose B is a bond. Since G is a simple graph, let C be a cycle adjacent to B in TG . We have V (B) ⊆ V (C), which in turn gives Claim 20. ƒ In what follows, the notation C is reserved for a cycle C ∈ D(G) such that every longest path contains a vertex of C. We want to show that C contains a Gallai vertex. Let uv ∈ E(C). If {u, v} is not a 2-cut in G, let Guv be just the edge uv. Otherwise, if {u, v} is a 2-cut in G, then uv ∈ E(C) is a virtual edge while the possible real edge between u and v is in the corresponding bond. Following TDA and by Lemma 17, {u, v} is a 2-separation of G. In this case, let Guv be the subgraph of G obtained by deleting all components of G − {u, v} containing a vertex of C − {u, v}. Since {u, v} is a 2-separation, there are two (u, v)-paths R[u,v] and S [u,v] in Guv . By Lemma 14, paths R(u,v) and S (u,v) are in different components of G − {u, v}. We call R[u,v] and S [u,v] connectors of Guv . The following two claims follow directly from TDA. Claim 21 For any two distinct edges uv, pq ∈ E(C), we have V (Guv ) ∩ V (G pq ) ⊆ {u, v} ∩ {p, q} and E(Guv − {u, v}, G pq − {p, q}) = ∅ (see Figure 8). Claim 22 If P [u,v] is a path in G with P [u,v] ∩ C = {u, v}, then u and v are two consecutive vertices on C. Claim 23 If P ∈ L is a longest path in G such that P has at least one end, say u, on C, then both its predecessor u− and successor u+ along C are also on P.

19

z

y x u z

y

q

Guv v

p

Gpq

x u

Figure 8:qStructure of the cycle C.

Guv v

p Gpq Proof. Suppose, on the contrary, that u+ ∈ / P. If {u, u+ } ∈ E(G), then P ∪ {uu+ } is a strictly + + + + longer path, a contradiction. Otherwise, consider R[u,u ] and S [u,u ] . Since R(u,u ) and S (u,u ) + are in different components of G −{u, v}, one of them, say R(u,u ) is vertex-disjoint from P. Thus + P ∪ R[u,u ] is a strictly longer path than P, a contradiction. ƒ x y R [u,v]

u

v S R

S [u,v]

Figure 9: Two (u, v)-paths in Guv . − →[x, y] ← −[x, y] For any two vertices x, y ∈ C, we use C Π (resp. C Π ) to denote a path obtained from − →[x, y] ← − C (resp. C [x, y] ) by replacing each edge {u, v} ∈ E(C) by R[u,v] or S [u,v] whenever {u, v} is a 2-separation (see Figure 9.) Let L1 = {P|P ∈ L and P has at least one end vertex on C}. Claim 24 T For every P ∈ L1 , V (C) ⊂ V (P), so V (C) ⊂ P∈L1 V (P). Proof. Let P ∈ L1 with endpoint u on C, and set Q = P − {u}. By Claim 23, path Q contains both u+ and u− , but not u. By Claim 22, traveling along Q from u+ to u− one must go through all vertices in V (C) − u, so V (C) ⊂ V (P). ƒ Following Claim 24, we may assume L2 := L \ L T1 6= ∅. If |L2 |T≤ 1, then Theorem 13 holds. So, we may assume |L2 | ≥ 2. Moreover, we have P∈L P ∩ C = P∈L2 P ∩ C. Notice that each path in L2 has exactly two pending paths of C.

20

Claim 25 T If |P ∩ C| ≥ 2 for every P ∈ L2 , then P∈L P 6= ∅. Proof. Let L be a longest pending path of C, and let z be the origin (the common vertex of L T and C) of L. We claim z ∈ P∈L P. Suppose this is not true. By Claim 24, there exists P ∈ L2 such that z ∈ / P. Let P 0 and P 00 be the two pending tails of P on C with origins u1 and u2 , respectively. If P 0 ∩ L 6= ∅, then {u1 , z} ∈ E(C) and both P 0 and L are subgraphs of Gu1 z by Claim 21, which in turn shows that P 00 ∩ L = ∅. Similarly, if P 00 ∩ L 6= ∅, then P 0 ∩ L = ∅. We − → assume, without loss of generality, P 00 ∩ L = ∅. We also assume that the segment C [u1 ,u2 ] does ← −[u ,z) not contain the vertex z. Then, P [u1 ,u2 ] ∩ L = ∅. Note that the paths P 00 ∪ P [u2 ,u1 ] , C Π 1 , and L are internally vertex-disjoint. Concatenating these paths together, we obtain a strictly longer path, which gives a contradiction. ƒ Hence, we assume there exists P ∈ L2 sharing exactly one vertex with C. We let P ∩ C = {v}, and let P 0 and P 00 be the two tails of P split at v ending in v1 and v2 , respectively. Claim T 26 v ∈ Q∈L Q. Proof. Suppose to the contrary that there exists Q ∈ L such that v ∈ / Q. By Claim 24, neither of the two ends of Q is on C. Thus Q has two pending tails, denoted as Q0 = Q[w1 ,u1 ] and Q00 = Q[w2 ,u2 ] , of C with origins w1 and w2 , respectively. We assume, without loss of generality, − → the segment C [w1 ,w2 ] does not contain v. Note that w1 = w2 if and only if Q and C share exactly one vertex. We will distinguish a few cases according to which one of the four pending tails P 0 , P 00 , Q0 , and Q00 is the longest. By the symmetry of P 0 and P 00 and the symmetry of Q0 and Q00 , we only need to consider two cases according to whether P 0 or Q0 is the longest one among the four pending tails. In fact, since w1 = w2 may occur, the case that Q0 is the longest one is more general than the case in which P 0 is the longest one. So we assume that Q0 is the longest pending tail among P 0 , P 00 , Q0 , and Q00 . We claim that P 0 ∩ Q0 6= ∅ and P 00 ∩ Q0 6= ∅. Indeed, assume for a contradiction that Q0 and, without loss of generality, P 0 are disjoint. Also, assume that, without loss of generality, ← − C [v,w1 ] = x 0 (= v)x 1 · · · x m (= w1 ) contains at least three vertices. We consider the cases that both, only one, or none of the edges {x 0 , x 1 } and {x m−1 , x m } are 2-cuts. First, assume that both edges are 2-cuts. Since the two (x 0 , x 1 )-paths R(x 0 ,x 1 ) and S (x 0 ,x 1 ) are in two different {x 0 , x 1 }-bridges, we may assume R[x 0 ,x 1 ] and P 0 only share a common vertex x 0 . Similarly, we may assume that R[x m−1 ,x m ] and Q0 only share a common vertex x m . Then, P 0 , R(x 0 ,x 1 ) , ← −[x ,x ] C Π 1 m−1 , R(x m−1 ,x m ) , and Q0 are internally vertex-disjoint. By concatenating these paths, we get a path strictly longer than P, a contradiction. Without loss of generality assume that {x 0 , x 1 } is a 2-cut but {x m−1 , x m } is not. Then the two (x 0 , x 1 )-paths R(x 0 ,x 1 ) and S (x 0 ,x 1 ) are in two different {x 0 , x 1 }-bridges. This means that, without loss of generality, R[x 0 ,x 1 ] and P 0 ← −[x ,x ] only share a common vertex x 0 . Hence, P 0 ∪ R[x 0 ,x 1 ] ∪ C Π 1 m−1 ∪ {x m−1 , x m } ∪ Q0 is strictly longer than P, again a contradiction. If both {x 0 , x 1 } and {x m−1 , x m } are not 2-cuts, then ← −[x ,x ] P 0 {x 0 , x 1 } C Π 1 m−1 {x m−1 , x m }Q0 is strictly longer than P, a contradiction. Hence, P 0 ∩ Q0 6= ∅ and P 00 ∩ Q0 6= ∅. Along with TDA, this indicates that {w1 , v} ∈ E(C).

21

Let x (respectively y) be the first vertex along P 0[v,v1 ] (respectively P 00[v,v2 ] ) intersecting Q0 , that is, Q0 ∩ P 0[v,x] = {x} and Q0 ∩ P 00[v, y] = { y}. Moreover, we assume without loss of generality that x is between w1 and y on Q0 .

Q''

v

w2 w1

u2

Q' u1

x y v1

P'

P'' v2

Figure 10: Illustration of x and y along with pending tails P 0 ,P 00 , Q0 , and Q00 . Since Q0[w1 ,x] ∪ P 0[x,v] and Q0[w1 , y] ∪ P 00[ y,v] are two (w1 , v)-paths in Gw1 v , and the three pending tails Q0 , P 0 , and P 00 are in the same {v, w1 }-bridge in G, note that, if Q00 ∩ P 0 6= ∅, then w2 = w1 . − →[w ,v] Let Θ v x be the union of the paths P 0[v,x] , P 00[v, y] ∪Q0[ y,x] , and Q[x,w1 ] ∪Q[w1 ,w2 ] ∪ C Π 2 . Clearly, Θ v x is a Θ-graph. Applying Lemma 14 to Θ v x , we get P 00[ y,v2 ] ∩ Q0[w1 ,x] = ∅, Q00 ∩ P 00 = ∅, and Q00 ∩ P 0[v,x] = ∅.

Θvx

v

w 1 = w2 Q'

y

x

u1 P''

Q''

v2

P' v1

u2

Figure 11: Illustration of the Θ-graph. We claim Q0(x,u1 ] ∩ P 0(x,v1 ] 6= ∅. Otherwise, let R1 = P 0[x,v1 ] ∪Q0[x,u1 ] and R2 = P 00[v2 ,v] ∪ P 0[v,x] ∪ Q0[x,w1 ] ∪ Q[w1 ,w2 ] ∪ Q00[w2 ,u2 ] be two walks. Since we assume Q0(x,u1 ] ∩ P 0(x,v1 ] = ∅, R1 is a path. Note that P 00[v2 ,v] ∪ P 0[v,x] = P [v2 ,x] and Q0[x,w1 ] ∪ Q[w1 ,w2 ] ∪ Q00[w2 ,u2 ] = Q[x,u2 ] . Applying Lemma 14 to Θ v x , we have that P [v2 ,x] and Q[x,u2 ] are internally vertex-disjoint, so R2 is also a path. Thus, we have |R1 | + |R2 | = |P| + |Q|.

22

By Claim 20 and the fact that R1 ∩ C = ∅, we know R1 is not a longest path, so |R2 | > |P|, which also gives a contradiction. Since Q0(x,u1 ] ∩ P 0(x,v1 ] 6= ∅, applying Lemma 14 to Θ v x again, we get P 0(x,v1 ] ∩ Q0[w1 ,x) = ∅. Since R(v,w1 ) and S (v,w1 ) belong to two different {v, w1 }-bridges, we may assume that R(v,w1 ) is − → − → not in the {v, w1 }-bridge containing x. Let C [w1 ,v] be the segment of C − vw1 . Clearly, C (w1 ,v) ← − contains at least one vertex, because C [v,w1 ] contains at least three vertices. − →[w ,v] Let R1 := P 0[v1 ,x] ∪ Q0[x,w1 ] ∪ C Π 1 ∪ P 00[v,v2 ] . Since (P 0[v1 ,x] ∪ P 00(v,v2 ] ) ∩ Q0(x,w1 ) = ∅ and all − →[w ,v] these three paths are internally vertex-disjoint from C Π 1 , R1 is indeed a path. We now define a walk R2 as follows: • If R(v,w1 ) ∩ Q00 = ∅, let R2 := Q0[u1 ,x] ∪ P 0[x,v] ∪ R[v,w1 ] ∪ Q[w1 ,w2 ] ∪ Q00[w2 ,u2 ] ; • If R(v,w) ∩ Q00 6= ∅ (in this case, w2 = w1 is adjacent to v in C), let R2 := Q0[u1 ,x] ∪ P 0[x,v] ∪ − →[w1 ,v] ∪ Q00[w2 ,u2 ] . CΠ Note that Q[w1 ,w2 ] ∪ Q00[w2 ,u2 ] = Q[w1 ,u2 ] , which are written separately in the definition of R2 for the purpose of emphasizing their locations. Since (Q0[u1 ,x) ∪ Q[w1 ,u2 ] ) ∩ R(v,w1 ) = ∅ in the first − →[w ,v] case and (Q0[u1 ,x) ∪ Q[w1 ,u2 ] ) ∩ C Π 1 = ∅ in the second case, and these corresponding paths are internally vertex-disjoint from P 0[w1 ,x] , R2 is also a path. By summing the lengths of R1 and R2 , we get |R1 | + |R2 | ≥ |P| + |Q| + |C| > |P| + |Q|, since |C| ≥ 3, which gives a contradiction to the assumption that both P and Q are longest paths. ƒ This completes our second proof of Theorem 13.

„

6 Algorithmic remarks For any hereditary class of graphs for which there is a polynomial-time algorithm that computes (the length of) a longest path, it is easy to derive a polynomial-time algorithm that finds all Gallai vertices. Indeed, one just has to compute the length L of a longest path in the given (connected) graph G, and then to check, for each vertex v, whether the length of a longest path in G − v remains the same. If not, v is a Gallai vertex. It is a well-known result that one can use dynamic programming to solve many combinatorial problems on graphs of bounded treewidth in polynomial or even linear time [1, 4]. In particular, Bodlaender [5, Thm. 2.2] claims a linear-time algorithm following these lines to find a longest path in a graph with bounded treewidth. (See also [6] on how to obtain in linear time a tree decomposition for graphs with bounded treewidth.) Therefore, using the idea described in the previous paragraph, one can find all Gallai vertices in time quadratic in the number of vertices of the given connected series-parallel graph.

23

In fact, one can do better by applying the same strategy used to compute the length of a longest path in a partial k-tree, but carrying more information during the process. Given a connected graph G of treewidth k, compute in linear time a “nice” tree decomposition for G (as done in [9] for instance). Then run a dynamic programming algorithm on top of this tree decomposition, to compute the length L of a longest path in G. Roughly speaking, this algorithm computes the length of longest parts of paths within the subgraph induced by the vertices in clusters already traversed of the tree decomposition, and puts together this information while going through the tree decomposition. Specifically, when visiting a node u of the tree decomposition, if Hu is the subgraph of G induced by the vertices in the cluster X u or clusters of nodes below u, for each different way that a path can behave in the cluster X u and in Hu , we have a configuration as the ones described in Figure 12 for the case in which X u has three vertices. 1 0 0

1 1

0

1 1

0

1 1

2 1

1

0

2 1

1

2 1

1

2 2

1

Figure 12: A sample of the configurations for a cluster with three vertices. The whole set of configurations has to consider the labels of the vertices in the cluster. The number of such configurations depends only on the treewidth. For each such configuration, the algorithm computes the length of a longest part of a path in Hu that “agrees” with that configuration. It does this using dynamic programming, that is, it computes such length for a node u and one of the configurations using the information that it already computed for the children of u in the tree. Some of the configurations of the children, together with new edges within X u , combine into each configuration for u. The combinations that give raise to the longest parts are the ones of interest, and give the length of a longest part for that configuration for u. In a first traversal of the tree decomposition, the value of L is computed. Now, as it is usual in dynamic programming, in a reverse traversal of the tree, retracing backwards what was done to find out L, one can mark, for each node and each configuration, if that configuration at that node gives raise to a path of length L in G. Once this is done for a node, the algorithm checks whether the configurations for that node that give raise to a longest path all contain one of the vertices in the cluster of that node. If so, this is a Gallai vertex. Otherwise the algorithm proceeds to the next node in the reverse traversal. This process finishes with a Gallai vertex as long as the graph has one such vertex. In particular, for partial 2-trees, this process will find a Gallai vertex in the reverse traversal as soon as it reaches the first cluster of the tree that contains a Gallai vertex. By proceeding with the reverse traversal in this way, one can find all Gallai vertices. For bounded k, the running time of this algorithm is linear in the number of vertices of the graph. (Note that the number of edges in a partial k-tree is at most kn, where n is the number of vertices in the partial k-tree.) Indeed, first computing a nice tree decomposition can be done in linear time. Second, the number of configurations depends only on k, and the processing of each node of the tree decomposition depends only on the number of configurations (and on the size of the cluster, which is bounded by k + 1 and thus also by the number of configurations). Therefore, for series-parallel graphs,

24

this algorithm finds a Gallai vertex (or even all Gallai vertices) in time that is linear in the number of vertices of the graph.

7 Related results and open questions There are several questions related to Gallai’s original question that remain open. For instance, it was asked [16, 31] whether there is a vertex common to all longest paths in all 4-connected graphs. This problem is open so far, and even the more general question for k-connected graphs with larger k has not been answered. There are 3-connected examples known for which Gallai’s question has a negative answer [12]. In [10], where a proof that all 2-trees have nonempty intersection of all longest paths was presented, it was asked whether the same holds for k-trees with larger values of k. As far as we know, this also has not yet been answered. In the present paper, we have proven that all connected subgraphs of 2-trees have nonempty intersection of all longest paths. We observe that the same does not hold for all subgraphs of 3-trees. Indeed, the counterexample by Walther, Voss, and Zamfirescu in [28, 30] is a connected spanning subgraph of a 3-tree (see Figure 13).

1

8 7

2

1

3

5 9

4

1

6 10

Figure 13: The counterexample of Walther, Voss, and Zamfirescu as a subgraph of a 3-tree. Underlying edges are dotted. The number next to each vertex indicates the sequence in which they are added to the 3-tree. In other words, Gallai’s question has a positive answer for connected graphs with treewidth at most 2 (series-parallel graphs), but a negative answer for connected graphs with treewidth at most 3. As series-parallel graphs are the class of K4 -minor-free graphs, one might also ask whether the answer is positive for all (connected) K5 -minor-free graphs, but there are planar counterexamples known [24]. As split graphs and 2-trees are chordal, a natural question raised by Balister et al. [3] is whether all longest paths share a vertex in all chordal graphs. Recently, Michel Habib (personal communication) suggested that the answer to Gallai’s question might be positive in co-comparability graphs. For this class of graphs, as well as for series-parallel graphs, there is a polynomial-time algorithm to compute a longest path [14]. (For chordal graphs, computing a longest path is NP-hard [19].) As already stated in Section 1, instead of looking at the intersection of all longest paths, Zamfirescu asked whether any p longest paths in an arbitrary connected graph contain a common

25

vertex. This is certainly true for p = 2, proven to be false [21, 23] for p ≥ 7, but still open for p in {3, 4, 5, 6}.

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, On linear time minor tests with depth-first search, Journal of Algorithms 14 (1993), no. 1, 1–23.

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