I Introduction
It is a wellknown fact that, in a connected graph, any two longest paths have a common vertex. In 1966, Gallai raised the following question: Does every connected graph contain a vertex that belongs to all of its longest paths? The answer to Gallai’s question is already known to be negative. Figure 1 shows the smallest known negative example, on 12 vertices, which was independently found by Walther and Voss [19] and Zamfirescu [20]. However, when we restrict ourselves to some specific classes of graphs, the answer to Gallai’s question turns out to be positive. For example, it is well known that any set of subtrees of a tree satisfies the Helly property. If we consider the set of subtrees consisting of the longest paths of the tree, since they are pairwise intersecting, we conclude that there is a vertex that belongs to all of them.
There are other graph classes which are known to have a positive answer to Gallai’s question. Klavžar and Petkovšek [16] proved that this is the case for split graphs, cacti, and graphs whose blocks are Hamiltonconnected, almost Hamiltonconnected or cycles. Balister et al. [2] and Joos [15] proved the same for the class of circular arc graphs. De Rezende et al. [7] proved that the answer to Gallai’s question is positive for 2trees and Chen et al. [6] extended this result for seriesparallel graphs, also known as partial 2trees. Chen [5] proved the same for graphs with matching number smaller than three, while Cerioli and Lima [4, 17] proved it for sparse graphs, free graphs, graphs that are the join of two other graphs and starlike graphs, a superclass of split graphs. Finally, Jobson et al. [14] proved it for dually chordal graphs and Golan and Shan [11] for free graphs.
A more general approach to Gallai’s question is to ask for the size of the smallest transversal of longest paths of a graph, that is, the smallest set of vertices that intersects every longest path. Given a graph , we denote the size of such a set by . In this direction, Rautenbach and Sereni [18] proved that for every connected graph on vertices, that for every connected planar graph on vertices, and that for every connected graph of treewidth at most .
In this work, we provide exact results and upper bounds on the value of when belongs to some specific classes of graphs. More specifically, we prove that:

for every connected chordal graph , where is the size of a maximum clique of .

for every connected bipartite permutation graph .

for every connected graph of treewidth at most .

for every connected full substar graph .
This paper is organized as follows. In the next section, we state the definitions and basic results that are going to be used throughout the text. In Sections III, IV, V, and VI, we consider, respectively, the class of chordal graphs, the class bipartite permutation graphs, the class of graphs of treewidth at most and the class of full substar graphs. Finally, in Section VII, we state the open problems to be considered in future work.
Ii Definitions and notation
All graphs considered are simple. Let be a vertex in a graph , we denote by the set of neighbors of in , and by the cardinality of . If the context is clear, we write simply and respectively. Let be a path in a graph . We denote by the length of , that is, the number of edges in . Given a path such that the only vertex it shares with is an extreme of both of them, we denote by the concatenation of and . For a vertex in , let and be the paths such that with . We refer to these two paths as the tails of . Given a path that contains vertices and , we denote by the tail of that does not contain and by the tail of that does not contain . Also, if the context is clear, we denote by the subpath of that has and as its extremes. Thus .
Let be a set of vertices of . Let be a path in that does not contain all vertices of and contains a vertex not in . We say that fences if all the vertices of are in a single component of , otherwise we say that crosses . Given a path that crosses and has both extremes not in , we say that is extremeseparated by when the extremes of are in different components of , and that is extremejoined by if its extremes are in the same component of .
For an integer , we say that touches if intersects at exactly vertices. A path is an corner path if 1touches . Let be an corner path. If is fenced by , we say that is an cornerfenced path. If crosses , we say that is an cornercrossing path. If two paths and touch at the same set of vertices, we say they are equivalent, otherwise they are nonequivalent.
If is fenced by , we denote by the set of vertices of the component of where lies. For a set of vertices not contained in , we denote by the set of vertices of the components of where lies. Two fenced paths and are componentdisjoint if . If is clear from the context, we just say they are componentdisjoint.
From now on, we use for the length of a longest path in . Also, remember that is the size of a maximum clique of .
A graph is called a minor of the graph if can be formed from by deleting edges and vertices and by contracting edges.
A tree decomposition [8, p. 337] of a graph is a pair , conformed by a tree and a collection of bags , that satisfies the following three conditions:


for every , there exists a bag such that

if a vertex is in two different bags , then is also in any bag such that is on the (unique) path from to in .
The width of is the number
and the treewidth of is the minimum width of any tree decomposition of .
A graph is called chordal if every induced cycle has length three. Next we present some basic properties on tree decompositions for general and chordal graphs. We fix a graph and a tree decomposition of . Proposition 1 is due to Bodlaender [3]. Gross [12] presented a proof for it and refers to tree decompositions such as in Proposition 1 as full tree decompositions. The tree decomposition mentioned in Proposition 2 is also called clique tree and it was introduced by Gavril [10]. Proposition 3 is a direct consequence of Corollary 12.3.12 of the book of Diestel [8].
Proposition 1.
If is the treewidth of a graph , then has a tree decomposition of width such that for every , and for every .
Proposition 2.
Every chordal graph has a tree decomposition such that the bags of are the maximal cliques of .
Proposition 3.
For every chordal graph , .
Given two different nodes , of , we denote by the component of where lies. We say that such component is a branch of at and that the components of are the branches of at [13]. Similarly, for a vertex , we denote by the branch of at such that . We also say that is in . Moreover, we can extend the notation and say that, if is a path fenced by for some , then , where is a vertex of . We also say that is in . Next we show some basic propositions of branches. Propositions 4 to 6 are used to justify that the previous two definitions are coherent. The first two of them appear in the work of Heinz [13].
Proposition 4.
Let be a node of and be a vertex of such that . Let and be nodes of . If , then and are in the same branch of at .
Proposition 5.
Let and be two vertices of , and let be a node of . If , , and and are not separated by in , then .
Proposition 6.
Let be a node of and be a path fenced by . For every two vertices and in , .
Proof.
By definition of fenced paths, and lie in the same component of , so they are not separated by in and we can apply Proposition 5. ∎
Proposition 7.
If is a path fenced by for some , then there exists a neighbor of in such that .
Proof.
Let be a vertex of (that exists by the definition of fenced). As , there exists a bag that contains . Let be the neighbor of in such that is in the (unique) path from to in . Then . ∎
Proposition 8 appears in the book of Diestel [8] as Lemma 12.3.1. Proposition 9 is a corollary of Proposition 8.
Proposition 8.
Let . Let and be the components of , with and . Then separates from in .
Proposition 9.
Let . Let and be vertices of with . If and , then and are not adjacent.
Proof.
Observe that is in a bag of because is in and . As is in a bag of , by Proposition 8, separates from . Hence, and are not adjacent. ∎
Proposition 10.
Let . Let be a path fenced by that 1touches such that , where . Then .
Proof.
Suppose by contradiction that there exists a vertex . As is fenced by , there exists a vertex . Moreover, . Let be the subpath of with and as extremes. Since 1touches , the subpath also 1touches . This implies that is internally disjoint from . As , , the subpath is disjoint from . But then we contradict Proposition 8, which says that separates , which is in a bag of , from , which is in a bag of . ∎
Proposition 11.
Let and . Let be a path fenced by that 1touches such that . Let be a path fenced by that 1touches at a vertex in . Then .
Proof.
Suppose by contradiction that . As , we must have that . By Proposition 10, , a contradiction. ∎
Iii Chordal graphs
We start by proving a lemma that is valid for every graph.
Lemma 12.
Let be a graph with a clique . Let be the set of all longest paths in that cross , 2touch , and are extremejoined by . There are at most two nonequivalent paths in .
Proof.
Suppose by contradiction that there are (at least) three nonequivalent longest paths , , and in . Say , , and , where , , and are pairwise distinct but not necessarily pairwise disjoint. We may assume that either or . If is componentdisjoint from (and from ), and is componentdisjoint from (and from ), then and are paths whose lengths sum more than , a contradiction, as at least one of them would have length greater than . So,
(1) 
Applying the same reasoning to paths and , and to paths and , we conclude that
(2) 
and that
(3) 
Also, as , , and cross , from (1), (2), and (3), we have that
(4) 
Without loss of generality, we may assume that . (Otherwise, interchange with , and with .) See Figure 2(a). Now, if , then, by (4), , and thus, by (3), . But then , and we contradict (4). Hence, , and, by (2), . Similarly, one can deduce that . Thus, by (3), , and, again, we can deduce that . As , , and are extremejoined, we conclude that
(5) 
(6) 
(7) 
See Figure 2(b).
The previous lemma examines how longest paths that are extremejoined by a clique behave. The following lemma examines the case in which the longest paths are extremeseparated. Observe that, in both cases, we are only considering longest paths that cross the clique and touch it at most twice.
Lemma 13.
Let be a graph with a clique and let be the set of all longest paths in that are extremeseparated and touch at most twice. Every two elements of have a common vertex of .
Proof.
Let and be two arbitrary paths in . Suppose by contradiction that . As is extremeseparated by , path crosses and therefore either 1touches or 2touches . To address these two possibilities at once, let and be such that touches at and , with if 2touches . Also, if , then let and be different tails of and let be the path consisting of only the vertex . Let and , and possibly , , and , be defined similarly for .
As both and are extremeseparated by , the tail is componentdisjoint from at least one in . Analogously, is componentdisjoint from at least one in , is componentdisjoint from at least one in and is componentdisjoint from at least one in . We may assume without loss of generality that and are componentdisjoint and that and are componentdisjoint. (Otherwise interchange and .) Observe also that is componentdisjoint from at least one in . Without loss of generality, assume that is componentdisjoint from . (Otherwise interchange and , and and simultaneously.)
Note that is componentdisjoint from at least one in . First suppose that is componentdisjoint from . (See a representation of the interactions between the parts of and in Figure 3(a).) Then, one of the paths or is longer than , a contradiction. Now suppose that is not componentdisjoint from , that is, , and thus is componentdisjoint from . If and are componentdisjoint (see Figure 3(b)), then one of or is longer than , a contradiction. If and are not componentdisjoint, that is, , then, as and and are componentdisjoint, we have that is componentdisjoint from (see Figure 3(c)). Thus, one of the paths or is longer than , a contradiction. ∎
The following lemma synthesizes the two previous lemmas. It says that, for every clique, when the transversal is not in it, we would have a longest path that is fenced by the clique. Observe that the lemma is valid only for chordal graphs. Remember that is the size of a maximum clique in . A clique is a subset of vertices in that are pairwise adjacent.
Lemma 14.
Let be a connected chordal graph with a clique . One of the following is true:

.

There exists a longest path that does not touch .

There exists a vertex of such that there is a longest path that is fenced by and 1touches at . Moreover, no longest path that 1touches at crosses .

There exists an edge of such that there is a longest path that is fenced by and 2touches at the ends of . Moreover, no longest path that 2touches at the ends of crosses .
Proof.
We will prove that the negation of , , and implies . So, suppose that no clique of size is a longest path transversal in , that every longest path touches at least once, and that, if a vertex is such that some longest path touches at , then there exists a longest path that touches at and crosses . If , then either or holds. So we may assume that . If , then by Proposition 3, and holds by Chen et al. [6]. We conclude that and .
Case 1. There is a longest path that 1touches .
If then, as and do not hold, for every vertex in , there exists a longest path that 1touches at that vertex. Also, as is false, we may assume that each such path crosses , a contradiction to Lemma 13, because . So . As does not hold, and we are assuming that there is a longest path that 1touches , there exists a longest path that 1touches at a vertex and crosses . As does not apply, for every clique in containing , there exists a longest path that does not contain any vertex in that clique. If any of these longest paths 1touches at a vertex , then, as does not hold, there is a longest path that crosses at , contradicting Lemma 13. Hence, for every edge in not incident to , there exists a longest path that 2touches at the ends of that edge. Again, by Lemma 13, as crosses at , none of these paths is extremeseparated by . As , there are at least three such paths. By Lemma 12, one of these edges, call it , is such that no longest path crosses and 2touches at the ends of . Moreover, we know that there is a longest path that 2touches at the ends of and, by the previous discussion, that path is fenced by . So holds.
Case 2. Every longest path touches at least twice.
If , then any subset of vertices of of size is a longest path transversal, and would hold. Thus we may also assume that . As , for every edge of , there exists a longest path that 2touches at the ends of that edge. As , there are at least six nonequivalent longest paths that 2touch . Suppose by contradiction that does not hold. Hence, we may assume that these six paths cross . By Lemma 12, four of these longest paths are extremeseparated by . As at least two of the corresponding four edges of are disjoint, by Lemma 13, we have a contradiction. ∎
We can finally prove our main result.
Theorem 15.
For every connected chordal graph ,
Proof.
Suppose by contradiction that . Then, for every clique in , there exists a longest path fenced by as in , , or of Lemma 14. We create a directed graph , that admits antiparalell arcs, as follows. Let be a tree decomposition of . The nodes of are exactly the nodes of . Let be a node in and let be a longest path in fenced by that satisfies one of the conditions , , or of Lemma 14. By Proposition 7, there exists a neighbor of in such that . Hence is an arc in . Thus every node of is the tail of at least one arc in .
Let be the last arc of a maximal directed path in . As is a tree, is also an arc in , which implies that there exist two longest paths and in such that and , where is fenced by and is fenced by , and both satisfy one of the conditions , , or of Lemma 14.
From now on, we assume that is a tree decomposition of as in Proposition 2. Note that the bags containing vertices of are only in , and the bags containing vertices of are only in . As and are disjoint, . Let be a vertex such that
. Suppose for a moment that
contains and let be a neighbor of in . By Proposition 9, vertex cannot be in , so . This implies that is an edge in and, as is a clique, contains all vertices of , contradicting the fact that is fenced. So does not contain vertices in . By a similar reasoning, does not contain vertices in . Thus . As is connected,(8) 
This implies that and , therefore none of and satisfies condition of Lemma 14.
Suppose for a moment that . Then, as , there exists a longest path that does not contain any vertex of . As is connected, intersects . As does not contain vertices in and does not contain vertices in , we have that . As the bags containing vertices of are only in , has a vertex in a bag of . A similar reasoning, with instead of , shows that also has a vertex in a bag of . This is a contradiction to Proposition 8, as contains no vertex in . Hence . Moreover, as both and are maximal (and different), we conclude that and .
Remember that none of and satisfies condition of Lemma 14. So touches at least once and touches at least once. First suppose that 1touches at a vertex . That is, satisfies condition of Lemma 14. By (8),
So . That is, and only intersect each other at , which implies that divides both longest paths in half. Let and be the two tails of , and let and be the two tails of . Let and . As 1touches , we may assume without loss of generality that . Suppose that also 1touches . Then, we may assume without loss of generality that . But then is a longest path that 1touches at and crosses . As exists, condition of Lemma 14 is not satisfied, a contradiction. Now suppose that 2touches at . Note that or . If then is a longest path that 1touches at and crosses , again a contradiction. Hence, . But then is a longest path that 2touches and crosses . As exists, condition of Lemma 14 is not satisfied, again a contradiction. Therefore touches at least twice.
By a similar reasoning, we may conclude that touches at least twice. So both and must satisfy condition of Lemma 14. Suppose that 2touches at the ends of edge . First suppose that also 2touches at the same vertices. Then, , , and . If then is a longest path that 2touches and crosses . As exists, condition of Lemma 14 is not satisfied, a contradiction. Hence, and . Then is a longest path that 2touches and crosses , again a contradiction. Hence, we may assume that 2touches at the ends of an edge with . Then and are paths, yielding the final contradiction.
∎
The previous theorem implies the following results.
Corollary 16.
If is a tree or a 2tree, then
Corollary 17.
If is a 3tree or a connected chordal planar graph, then .
Iv Bipartite permutation graphs
Let and be two parallel lines in the plane. Consider two sets and of segments that joins a point in with a point in , such that the extremes of every two elements in are pairwise disjoint. Moreover, every two elements in do not intersect each other and every two elements in do not intersect each other.
Let be the function that maps the extreme in of a segment to the other extreme. That is, if is the extreme in of a segment in , then the other extreme is ; and if is the extreme in of a segment in , then the other extreme is . Consider an associated bipartite graph where if and only if the segments and intersect each other. We call the tuple ( a line representation of and a graph is called a bipartite permutation graph if it has a line representation. (See Figure 4.)
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