Intersection of Longest Paths in a Graph ?
Susanna F. de Rezende
1Cristina G. Fernandes
1Daniel M. Martin
2Yoshiko Wakabayashi
1Instituto de Matem´atica e Estat´ıstica, Universidade de S˜ao Paulo, Brazil 1 Universidade Federal do ABC, Brazil 2
Abstract
In 1966, Gallai asked whether every connected graph has a vertex that is common to all its longest paths. The answer to this question is negative. We prove that the answer is positive for outerplanar graphs. Another related question was raised in 1995 at the British Combinatorial Conference: Do any three longest paths in a connected graph have a vertex in common? We prove that, in a connected graph in which all non-trivial blocks are Hamiltonian, any three of its longest paths have a common vertex. Both of these results strengthen a recent result by Axenovich.
Keywords: longest path, intersection of longest paths, outerplanar graph.
1 Introduction
In 1966, during a coloquium on graph theory, Gallai [4] asked whether every connected graph has a vertex that appears in all its longest paths. A few years
? This research was partially supported by CNPq.
1 Email: {susanna,cris,yw}@ime.usp.br
2 Email: daniel.martin@ufabc.edu.br
later, Walther [7] answered this question negatively by exhibiting a counter- example on 25 vertices. Later, Walther [8] and Zamfirescu [9] independently found a smaller counterexample on 12 vertices (see Figure 1).
Fig. 1. The graph of Walther and Zamfirescu.
On the other hand, it is easy to prove that the answer is positive for trees.
In 1990, Klavˇzar and Petkovˇsek [5] proved that the answer is also positive for split graphs, cacti, and some other classes of graphs. More recently, in 2004, Balister, Gy˝ori, Lehel, and Schelp [3] established a similar result for the class of circular arc graphs. In this paper we prove the following result.
Theorem 1.1 If G is a connected outerplanar graph, then G has a vertex common to all its longest paths.
Since non-empty intersection of all longest paths seems to be a property exhibited by few classes of graphs, it is natural to consider the intersection of a smaller number of longest paths. In this case, while it is easy to prove that every two longest paths share a common vertex, for three longest paths, a satisfactory answer is not known. In fact, the following problem [1,10] remains open.
Problem 1.2 Do any three longest paths in a connected graph share a com- mon vertex?
The only known progress on Problem 1.2, excluding those on all longest paths mentioned above, was obtained by Axenovich [2], who proved the fol- lowing result.
Theorem 1.3 (Axenovich, 2009) If G is a connected outerplanar graph, then any three of its longest paths have a common vertex.
In Section3, we prove the following more general result.
Theorem 1.4 If G is a connected graph in which all non-trivial blocks are Hamiltonian, then any three of its longest paths have a common vertex.
Note that both Theorems 1.1 and 1.4 generalize Axenovich’s result, each in a different way.
In this paper, all graphs considered are undirected and simple. A graph is outerplanar if it has an embedding in the plane such that all vertices belong to the boundary of its outer face (the unbounded face).
If P is a path, then kPk denotes its length (number of edges), and P−1 denotes the reverse of P. If xand y are vertices of a cycleC embedded in the plane, we denote by Cx,y the section of C that goes clockwise from x toy.
IfB is a non-trivial block of a graphG, then we say that a non-trivial path inGis apending path(ofB) if it intersectsB in precisely one of its extremes, which we call its origin.
2 Longest Paths in Outerplanar Graphs
In this section we present the proof of Theorem 1.1. For that, we use the following result, obtained by Klavˇzar and Petkovˇsek [5].
Proposition 2.1 (Klavˇzar and Petkovˇsek, 1990) Let G be a connected graph. Then all longest paths in G have a common vertex if and only if for every block B of G all longest paths in G which use at least one edge of B have a common vertex.
Note that the above result implies that if G is a cactus, then all longest paths in G have a common vertex. A larger class of graphs containing the cacti is the class of outerplanar graphs. This class is captured by Theorem1.1.
Proof of Theorem1.1. LetGbe a graph as in the statement of Theorem1.1.
LetB be an arbitrary block of Gand PB be the set of the longest paths inG which use at least one edge of B. By Proposition 2.1, it suffices to prove that the paths in PB have a common vertex.
If B is a trivial block, then it is immediate that the paths in PB have a common vertex. So now suppose B is a non-trivial block. Let C be a Hamiltonian cycle in B, and letR? be a longest pending path (ofB). Denote by v the origin of R?. We shall prove that all paths in PB contain v.
Suppose there is a path P in PB that does not contain v. Consider an outerplanar embedding of G. Let x be a vertex in V(P)∩V(B) such that kCx,vk is minimum, and lety be a vertex in V(P)∩V(B) such that kCv,ykis minimum. Note that x6=y; otherwise, P would intersect B only at x.
Now letz be a vertex such thatxz ∈E(P)∩E(B) andkCy,zkis minimum.
Suppose y= z (see Figure 2(a)). In this case, consider the path P0 obtained from P by substituting the edge xy by the path Cx,y, i.e., P0 = (P −xy)∪ Cx,y. Observe that P0 is in fact a path, because Cx,y only intersectsP at the
vertices x and y. So kP0k=kP −xyk+kCx,yk=kPk −1 +kCx,vk+kCv,yk. But kCx,vk ≥ 1 and kCv,yk ≥ 1. Therefore, kP0k > kPk, which contradicts the fact that P is a longest path.
x
y=z P P0 v
(a)
x
P P0 v z
y
R∗
R
(b)
Fig. 2. Examples of the casesy =z andy 6=z respectively.
Now suppose y6=z (see Figure 2(b)). Let P1 and P2 be the sections of P such that P = P1 ·P2, V(P1)∩V(P2) = {z}, x ∈ V(P1) and y ∈ V(P2).
Since G is outerplanar, P2 uses only vertices of Cy,z and possibly a pending path, say R. Now, consider the path P0 = P1 · Cv,z−1 · R?. Then, kP0k = kP1k+kCv,yk+kCy,zk+kR?k. SincekCy,zk ≥ kP2k − kRk,kR?k ≥ kRk, and kCv,yk>0, then kP0k>kP1k+kP2k=kPk, a contradiction to the fact that P is a longest path. Thus P must containv, and this completes the proof. 2
3 Intersection of Three Longest Paths
In this section we focus on the intersection of three longest paths. Theorem1.4 shows a class of graphs for which any three longest paths have a common vertex. In particular, this class includes the outerplanar graphs.
Sketch of the proof of Theorem 1.4. LetP be a set of three longest paths in G. It can be shown that there is a block B of Gthat contains at least one vertex of each path in P. If one of the paths in P contains all vertices of B, then the result follows easily. Thus we can assume B is non-trivial and all paths in P use two pending paths of B.
Henceforth, we shall consider only pending paths of B that are contained in the paths of P. If all these pending paths together intersect B in only two vertices, then the result is obvious. So, let v1, v2 and v3 be origins of three disjoint pending paths that are as long as possible.
A longest pending path of B that has vi as its origin will be called a vi- pending path. Suppose a vi-pending path is at least as long as a vj-pending
path if i < j. Let C be a Hamiltonian cycle in B embedded in the plane in such a way that v1, v2 and v3 appear clockwise in C.
Case 1: There is at most one path in P containing both a v1-pending path and a v2-pending path.
Let Pa be a path in P that contains a v1-pending path, and its other pending path is as long as possible. Let Pc be a path in P that does not contain a v1-pending path, and let Pb be the remaining path inP.
Fori∈ {1,2,3}, let Ri be avi-pending path, and for x∈ {a, b, c}, denote by Tx and Tx0 the two pending paths contained in Px. Thus, we have
kR1k+kR2k ≥ kTak+kTa0k, kR1k+kR3k ≥ kTbk+kTb0k, (1)
kR2k+kR3k ≥ kTck+kTc0k. Now consider the paths
Qa=R−12 ·Cv2,v3 ·Cv3,v1 ·R1, Qb =R−11 ·Cv1,v2 ·Cv2,v3 ·R3, Qc=R−13 ·Cv3,v1 ·Cv1,v2 ·R2. SincePa, Pb,Pc are longest paths, it follows that (2) kPxk ≥ kQxk, for x∈ {a, b, c}.
For a pathPx, letC(Px) denote the subgraph obtained fromPx by remov- ing all vertices not in C. Observe thatC(Px) must be a path. Combining the fact that kPxk=kTxk+kC(Px)k+kTx0k with (1) and (2), we obtain
kC(Pa)k ≥ kCv2,v3k+kCv3,v1k, kC(Pb)k ≥ kCv1,v2k+kCv2,v3k, kC(Pc)k ≥ kCv3,v1k+kCv1,v2k. Adding both sides of the last three inequalities yields
kC(Pa)k+kC(Pb)k+kC(Pc)k ≥2kCk. Thus, there must exist a vertex common to Pa, Pb and Pc.
Case 2: There are (exactly) two paths in P containing both a v1-pending path and a v2-pending path.
Due to space constraints we omit the proof of this case. 2
4 Concluding Remarks
A natural question concerning Theorem 1.4 is whether one can replace “three longest paths” for “all longest paths” and guarantee that the correspond- ing statement holds. The answer is no, as shown by the graph in Figure 1.
Skupie´n [6], in 1996, proved a result related to this: for every k≥7, there is a connected graph in which some k longest paths have no common vertex, but every k−1 longest paths have a common vertex.
It would be interesting to find other classes of graphs for which all (or any three) longest paths have a common vertex. Also, in view of the result of Skupie´n [6], it would be interesting to either prove or disprove that every six (or less) longest paths in a connected graph have a common vertex. Other related open problems were proposed by Zamfirescu [10].
References
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[3] Balister, P., E. Gy¨ori, J. Lehel and R. Schelp, Longest paths in circular arc graphs, Combin. Probab. Comput.13(2004), pp. 311–317.
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[5] Klavˇzar, S. and M. Petkovˇsek, Graphs with non empty intersection of longest paths, Ars Combin.29(1990), pp. 13–52.
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[7] Walther, H., Uber die Nichtexistenz eines Knotenpunktes, durch den alle¨ l¨angsten Wege eines Graphen gehen, J. Combinatorial Theory6(1969), pp. 1–6.
[8] Walther, H. and H.-J. Voss, Uber Kreise in Graphen, VEB Deutscher Verlag¨ der Wissenschaften (1974).
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