The Interval function, Ptolemaic, distance hereditary, bridged graphs and axiomatic characterizations
Abstract
In this paper we consider certain types of betweenness axioms on the interval function of a connected graph . We characterize the class of graphs for which satisfy these axioms. The class of graphs that we characterize include the important class of Ptolemaic graphs and some proper superclasses of Ptolemaic graphs: the distance hereditary graphs and the bridged graphs. We also provide axiomatic characterizations of the interval function of these classes of graphs using an arbitrary function known as transit function.
keywords:
transit function, interval function, betweenness axioms, Ptolemaic graphs, distance hereditary graphs, bridged graphsMSC:
[2020] 05C121 Introduction
Transit functions on discrete structures were introduced by Mulder [15] to generalize some basic notions in discrete geometry, amongst which convexity, interval and betweenness. A transit function on a non-empty set is a function to on satisfying the following three axioms:
-
, for all ,
-
, for all ,
-
, for all .
If is the vertex set of a graph , then we say that is a transit function on . Throughout this paper, we consider only finite, simple and connected graphs. The underlying graph of a transit function on is the graph with vertex set , where two distinct vertices and are joined by an edge if and only if .
A - shortest path in a connected graph is a -path in containing the minimum number of edges. The length of a shortest -path (that is, the number of edges in ) is the standard distance in . The interval function of a connected graph is the function defined with respect to the standard distance in as
: lies on some shortest - path in
The interval function is a classical example of a transit function on a graph ( we some times denote by , if there is no confusion for the graph ). It is easy to observe that the underlying graph of is isomorphic to . The term interval function was coined by Mulder in [14], where it is extensively studied using an axiomatic approach.
Nebeský initiated a very interesting problem on the interval function of a connected graph during the 1990s. The problem is the following: “ Is it possible to give a characterization of the interval function of a connected graph using a set of simple axioms (first - order axioms) defined on a transit function on ?”
Nebeský [17, 18] proved that there exists such a characterization for the interval function by using first - order axioms on an arbitrary transit function . In further papers that followed [19, 20, 21, 22], Nebeský improved the formulation and the proof of this characterization. Also, refer Mulder and Nebeský [16]. In [8], the axiomatic characterization of is extended to that of a disconnected graph. In all these characterizations, five essential axioms known as classical axioms are always required. These five classical axioms are and and three additional , , and defined as follows:
if and , then , | ||
if and then , | ||
if then |
The notation can be interpreted as is in between and . For example, the axiom can be interpreted as: if is between and , and is between and , then is between and . Similarly we can describe all other axioms. Hence we use the terminology of betweeness for an axiom on a transit function . The above interpretation was the motivation for the concept of betweenness in graphs using transit functions. It was formally introduced by Mulder in
[15] as those transit function that satisfy axioms and . Here the axiom is defined for every and a transit function
as follows:
The following implications can be easily verified for a transit function among axioms and .
-
1.
Axioms and implies axiom .
-
2.
Axioms and implies axiom which implies axiom ( that is , for a transit function implies implies )
The converse of the above implications need not hold. A transit function satisfying axioms and is known as a geometric transit function.
The problem of characterizing the interval function of an arbitrary graph can be adopted for different graph classes; viz., characterizing the interval function of special graph classes using a set of first - order axioms on an arbitrary transit function. Such a problem was first attempted by Sholander in [24] with a partial proof for characterizing the interval function of trees. Chvátal et al. [11] obtained the completion of this proof. Further new characterizations of the interval function of trees and block graphs are discussed in [1]. Axiomatic characterization of the interval function of median graphs, modular graphs, geodetic graphs, (claw, paw)-free graphs and bipartite graphs are respectively described in [9, 10, 14, 16, 19].
In this paper, we continue the approach of characterizing the interval function of some related classes of graphs, namely, distance hereditary graphs, Ptolemaic graphs and bridged graphs. We fix the graph theoretical notations and terminology used in this paper. Let be a graph and a subgraph of . is called an isometric subgraph of if the distance between any pair of vertices, in coincides with that of the distance . is called an induced subgraph if are vertices in such that is an edge in , then must be an edge in also. A graph is said to be -free, if has no induced subgraph isomorphic to . Let be graphs. For a graph , we say that is -free if has no induced subgraph isomorphic to , . Chordal graph is an example of a graph which is defined with respect to an infinite number of forbidden induced subgraphs ( is chordal if have no induced cycles with length more than three). There are several graphs that can be defined or characterized by a list of forbidden induced subgraphs or isometric subgraphs. See the survey by Brandstädt et al. [3] and the information system [12], for such graph families. A graph is a bridged graph if has no isometric cycles of length greater than . Clearly the family of bridged graphs contain the family of chordal graphs. The graph is distance hereditary if the distances in any connected induced subgraph of are the same as in . Thus, any induced subgraph inherits the distances of . The graph is a Ptolemaic graph if is both chordal and distance hereditary. Both Ptolemaic graphs and distance hereditary graphs possess a characterization in terms of a list of forbidden induced subgraphs, while bridged graphs by definition itself possess an infinite list of forbidden isometric subgraphs. In this paper, our idea is to find suitable axioms that fail on every forbidden induced subgraph for the Ptolemaic and distance hereditary graphs, while that for the bridged graph is to find an axiom that fails on all of its forbidden isometric subgraphs, namely on all isometric cycles .
In addition to the geometric axioms and , we consider the following betweenness axioms and for a transit function on for proving the characterizations of these classes of graphs.
-
: For any pair of distinct vertices we have .
-
: .
-
: .
-
:
-
: .
From the definition of the axioms, we observe the following. The axiom is a simple betweenness axiom which is always satisfied by the interval function . The axiom is a natural extension of . We provide examples in the respective sections for the independence of the axioms and and . The axioms and were first considered in [6] and later in [4]. The axiom first appeared in [24] for characterizing the interval function of trees. The axiom is discussed in [4] for characterizing the interval function of a Ptolemaic graph . We may observe that both the family of bridged graphs and distance hereditary graphs is a strict super class of the family of Ptolemaic graphs,, that is, and and and coincide only in . But, and , This relation is also reflected in the implications between the axioms and . From the definitions, we have the axiom implies axioms and , and also implies , while the reverse implications are not true. In other words, axioms and are weaker axioms than and hence graphs whose interval function satisfy axioms and will be a super class of graphs whose interval function satisfies . Similarly graphs whose interval function satisfies axiom will be a super class of graphs whose interval function satisfies . See Figure 1 for the relationships between the family , and .

We organize the results as follows. In Section 2, we characterize the interval function of a distance hereditary graph, Ptolemaic graph in Section 3, bridged graph in Section 4 and in Section 5, a discussion of the so called induced path transit function for the distance hereditary graphs respectively.
2 Interval function of Distance hereditary graphs
For the axiomatic characterization of the interval function of distance hereditary graphs, we require the axioms and . First we show that these axioms are independent with the Examples 1 and 2 below. Also it is clear from the Figures 2 and 3 that the axioms and are independent. The interval function doesn’t satisfy axiom , while satisfy axiom for the graphs in Figure 2. Also doesn’t satisfy axiom , while satisfy axiom for the graphs in Figure 3. By an - fan - free graph , we mean that is free from the House graph, the Hole graph (cycles , ), the Domino graph and the -fan (See the Figures 2 and 3 for these graphs).
Example 1 ( but not ).
Let . Define a transit function on as follows. and . We can easily see that satisfies . But we can see that and but . So does not satisfy .
Example 2 ( but not ).
Let . Define a transit function on as follows: and . We can see that satisfies . But , but . Hence does not satisfy .
Proposition 1.
[4] For every graph , satisfies the axiom if and only if is house, , -fan free.
Lemma 1.
[6] Let be a transit function on a non-empty finite set satisfying the axioms and with underlying graph . Then is -free.
Bandelt and Mulder obtained a forbidden induced subgraph characterization of distance hereditary graphs in [2]. We quote the theorem as
Theorem 1.
[2] A graph is distance hereditary if and only if is - fan-free.
We state a related result from [4] using the axiom defined for a transit function as
Note that a graph is the graph obtained by adding a pendent edge on an induced 4-cycle, . It follows from the definition that axiom implies both the axioms and , but the reverse implications are not true. We quote the result.
Theorem 2.
[4] For every graph , satisfies the axiom if and only if is - fan - free graph.
It may be observed that a graph is a distance hereditary graph and hence the class of - fan - free graphs is a proper subclass of the class of - fan - free graphs (distance hereditary graphs). The proof of the next theorem characterizing the class of distance hereditary graphs follows the same lines of ideas as in the proof of Theorem 2 with modifications since the axioms and are weaker axioms than the axiom .
Theorem 3.
Let be a connected graph. The interval function satisfy the axioms and if and only if is a distance hereditary graph.
Proof.
1
1
We use the fact from Theorem 1 that distance hereditary graphs are precisely - fan - free graphs for the proof.
Suppose that the interval function of satisfy the axioms and . To prove that is - fan - free, assume the contrary that contains a house, a hole, or a domino or a -fan as an induced subgraph. A graph with an induced house or - fan or a
doesn’t satisfy ( The vertices in Figure 2 doesn’t satisfy the axiom ).
For an isometric hole , , we choose vertices as shown in Figure
3, to prove that is violated. If has a domino ,
which is an isometric subgraph of with vertices as shown in Figure 3, then . If is not isometric, then there is
a vertex adjacent to both and or and . In this case, the graph
induced by is either a or a house or a -fan.
Let be a hole in that is not isometric and assume that is minimum. Clearly and there exist
such that . Let be a -geodesic; we may choose such that is minimum. Let and
be the -paths on where , and and . Clearly
and and we may assume without loss of generality that . Moreover, we can choose such that
the cycle induced by has minimum length. In particular, by the choice of , is not adjacent to any of
whenever .
By minimality of , , can be
adjacent to (or ) only if or .
If , then form (together with possibly some additional vertices of or ) an
induced hole shorter than , which is not possible. If , then and must be adjacent; otherwise we have a
shorter hole than on vertices together with or (and possibly some other vertices of or ). But then
form a -cycle and we have an induced domino on when and an induced house on the same vertices when .
Let now . First note that must be adjacent to at least one of and of by the minimality of , since there are no isometric holes. Let first . If either or is not adjacent to ,
we have an induced house as a subgraph. Otherwise we have an induced - fan on .
Let now . By the above, is not adjacent to or to and is not adjacent to or . Suppose that at least one of
or , say , is adjacent to . If both and are adjacent to , we have an induced - fan on . If is
adjacent to but is not, we have an induced house on . If both and are not adjacent to , adjacent to and adjacent to , we have an induced domino on . Hence but and by a similar argument but . Since , and must be adjacent since there are no isometric holes or by the minimality of . This yields an induced house on . Finally, If the induced is not isometric with , the only case left is the one in Figure 3, and by choosing the vertices as in the figure, it follows that the axiom is violated. Therefore, when has an induced house, hole, domino or - fan, either the axiom or is violated.
Conversely assume that axiom or is not satisfied by the interval function of . It is already known from Proposition 1 that if axiom is not satisfied, then has an induced , House or - fan. Now suppose axiom is not satisfied. Then there exists distinct vertices in such that , , and . Let be a -geodesic containing and be a - geodesic containing . We claim that
is a -path.
Since we see that is a geodesic. Therefore , for otherwise we may find a shorter path from to .
Now we claim that . Assume on the contrary, that , and let be the last vertex of the intersection (when going from to along path). Then , and since we find that . Hence we have that the path is shorter than , a contradiction.
Hence is a -path. Since , is not an geodesic. If is any
-geodesic, then . Fix an -geodesic . Let
be the last vertex on before that is on and be
the first vertex on after that is on . Note that such
vertices always exists, since and . On
the other hand, note that can be equal to , but .
Label vertices of the path by . Label vertices of
the -subpath of as . Clearly
and and . Path is not necessarily an induced path. If not, we choose
among all chords the one with maximal and replace the
part by this chord. Vertices of this new path are
denoted by , where still by
the choice of and . But is an induced
path, since it is a shortest path. Note that is not adjacent
to by the choice of . Hence can be
adjacent only to . Similarly , for , cannot be
adjacent to . We consider the following two
cases.
Case 1: .
If also , we have an induced cycle of length . So let . Also and
, otherwise we have a house on vertices
, and or respectively. This implies
that is an edge and we have an induced domino
(), otherwise we have an induced cycle of length
.
Case 2: .
If , we have
a house if or a - fan on vertices
, and . Hence and also
. We get an induced cycle of length , if
is not adjacent to at least one of or . If first
and , we get a house on
vertices , and . Let now
and . Since , there exists . If
, we get a house on vertices , and
. Also , since otherwise we get a house on
vertices , and . But now we have an induced
path which lead to an induced hole. Finally, if
and , we get a - fan on vertices
, and .
Thus in all cases, we get either an induced house, hole, domino, or - fan, and thus the proof is completed.
∎
We need the following Lemma
Lemma 2.
Let be a transit function on a non-empty finite set satisfying the axioms and with underlying graph . Then is - fan - free.
Proof.
Since for a transit function axiom implies axiom , by Lemma 1, is -free. We prove that is also - fan - free. Suppose on the contrary, contains a - fan with vertices as an induced subgraph. Let be the path of length three and be the vertex adjacent to all of . Since and , we have by axiom , and . Again since by . Again, and , , by axiom , we have . Now, we have and . Hence by axiom , we have , which is a contradiction and hence the lemma is proved. ∎
3 Axiomatic characterization of the interval function of Ptolemaic graphs
For the axiomatic characterization of of a Ptolemaic graph , the essential axiom is . Ptolemaic graphs are chordal graphs that are - fan - free. Changat et al. in [4] characterized the graphs for which the interval function satisfies the axiom as follows.
Theorem 4.
[4] Let be a graph. The interval function satisfies the axiom if and only if is a Ptolemaic graph.
Theorem 5.
Let be any transit function defined on a non-empty set . If satisfies and then the underlying graph of is -free for .
Proof.
Let be a transit function satisfying and . Let contains an induced cycle say . Without loss of generality assume is the minimum such cycle (in the sense that length of the induced cycle is as small as possible). We prove for every .
Case: .
Now since and , By axiom we have in a similar fashion we can show that . Since satisfies (J0)-axiom we have , which is a contradiction as .
Case: .
As in the above case, we can see that , this together with we have , which is again a contradiction.
Case: .
By repeated application of -axiom, as in the above two cases, with , by applying , we can see that . Here again a contradiction.
Hence, in all cases, we can see that does not contain as an induced subgraph This completes theorem. ∎
The following straightforward Lemma for the connectedness of the underlying graph of a transit function is proved in [6].
Lemma 3.
[6] If the transit function on a non-empty set satisfies axioms and , then the underlying graph of is connected.
We have the following lemma.
Lemma 4.
If is a transit function on satisfying the axioms and , then satisfies axiom and is connected.
Proof.
Let satisfies axioms and . To prove satisfies . Since satisfies , For , let , and . Since satisfies , we have . Now and so by axiom , we have , which implies that satisfies . Connectedness of follows from Lemma 3, since satisfies axioms and as axiom implies axiom . ∎
Example 3 ( and but not ).
Let . Let be defined as follows. and in all other cases . Since is a - fan, satisfies and . Next to show satisfies axiom. Since , we can see that for all so that for this pair satisfies axiom. Now consider , we can see that we have and which implies that satisfies axiom for this pair too. The case is similar for . All other pairs corresponds to edges. Hence we can see that satisfies -axiom.
Now but , and violates axiom.
Theorem 6.
Let be any transit function satisfying the axioms and then is Ptolemaic and .
Proof.
Since satisfies the axioms and , we have that is a chordal graph by Theorem 5. To prove that is Ptolemaic, we have to show that is - fan - free. Suppose that contains an induced - fan with vertices with forming a path and as the vertex adjacent to all the vertices . Since and are adjacent and is not an edge, by , . Similarly . Since is a transit function, by , and and hence by , . Again, since and are edges and is not an edge, . That is, and , by , we have , which is not true as is an edge. That is, we have proved that is a chordal graph which is - fan - free and hence is a Ptolemaic graph. By Lemma 4, satisfies axiom and and is connected.
Now we prove that for all . We prove by induction on the distance between and .
Case when .
Let Hence we can see that . That is, and , since satisfies . Therefore . Conversely suppose . Suppose . Since there exists at least one element such that are edges in . By assumption, is not adjacent to both and . Assume that is not an edge. Since and satisfies and with . By applying axioms and continuously on , we get vertices such that and , for and since is finite, , for some , say . That is, we have vertices with . Let us assume that . That is . Consider vertices . By and since , by . That is, and hence by , a contradiction. Therefore . This implies that by axiom , provided , which implies that by , a contradiction since . Therefore . That is, we have and hence by , a final contradiction. Therefore . Similarly, we can prove that . and hence , which completes the proof when .
Let us assume that the result holds for all distances less than and let , be two vertices such that . We first prove . Let . Since , we can find another vertex in the shortest -path containing . Now since satisfies . So by induction we have and . Also by axiom , . Then by axiom . Hence .
Let . If possible let . Since , by applying axioms and similarly as in the case of , we get vertices with such that and , for and .
Let be a vertex such that and . Similar to the case of , we can prove that . That is form a in . Here there are two possibilities for .
Case (i): .
In this case, since and is on a shortest -path in with , we have that is on a shortest -path in , that is, . Therefore, we have and hence by , a contradiction as .
Case(ii): .
In this case, . Since and so by axiom, . We have also and hence by axiom , we have , by induction hypothesis. That is , since , which is a contradiction to our assumption.
Therefore in all cases, we get contradictions to the assumption and hence our assumption is wrong, that is and hence the theorem.
∎
The following examples show that the axioms and are independent.
Example 4 ( but not ).
Let and define a transit function on as follows: . We can see that satisfies and . But , but . Therefore doesnot satisfy the axiom.
Example 5 ( but not ).
Let and define a transit function on as follows: . Here Satisfies and . We can see that but . So doesnot satisfy .
Example 6 ( but not ).
Let and define a transit function on as follows: and for all other pair we can see that satisfies . But since we can see that fails to satisfy .
From Theorem 6 and Theorem 4, we have the following theorem characterizing the interval function of Ptolemaic graphs.
Theorem 7.
Let be a transit function on the vertex set of a connected graph . Then is a Ptolemaic graph and coincides the interval function of if and only if satisfies the axioms and and the axiom implies that .
4 Interval function of bridged graphs
From the definitions of and it follows that . The example 7 shows that .
Example 7 ().
Let Let defined as follows. . We can see that and but , so that doesnot satisfy axiom. We can see that there exists no and satisfying the assumptions of the axiom and hence the axiom follows trivially.
We now prove the theorem characterizing interval function of bridged graphs.
Theorem 8.
Let be a connected graph. The interval function satisfies the axiom if and only if is a bridged graph.
Proof.
Let has an isometric cycle , . If is odd, say , let . Then it is easy to see that , and . If is even, say, , , let , . Then it is easy to see that and . In both cases of being odd or even, is not on any shortest -path and hence . This implies that If has an isometric cycle, then doesn’t satisfy the axiom , that is, we have proved that if satisfies axiom implies that is bridged graph.
Conversely, if is a bridged graph, then we claim that satisfy the axiom . Suppose not. Then there exist vertices in satisfying the following. A -geodesic containing , an -geodesic containing with such that is not on any -geodesic in . Then and should be adjacent, since
Using the same arguments as in the proof of Theorem 3, we derive the following.
-
i
: is a -path, say .
-
ii
: The last vertex on before that is on a shortest -path containing is and the first vertex on after that is on is .
-
iii
: An -subpath of , , and an induced path containing and , which is a subpath of of .
-
iv
: The cycle formed by the vertices of has length, , at least four.
Now, cannot be four, since is isometric, a contradiction to being a bridged graph, which implies that the length of the path , namely is strictly greater than one.
Since is a bridged graph, it follows that, there are chords from vertices , to vertices of so that the only isometric cycles are triangles.
Case 1: .
We claim that the index . Since is not a -geodesic containing ,
we have . If , since the only isometric cycles are triangles, we get is not a -geodesic containing or is not a -geodesic containing . There for . Also if ,, cannot be adjacent to for , since is a -geodesic containing , and if , cannot be adjacent to for , since is a -geodesic containing . Which implies that is adjacent to both and , otherwise there exist an induced - cycle on or . Then the path is also a shortest path and the path is also a -shortest path. This implies that the vertex , which is different from also belongs to , a contradiction to the hypothesis of the axiom .
Case 2: .
Since is a -geodesic containing , . In this case does not contain any of . Which implies that , a contradiction to the hypothesis of the axiom .
Therefore in both case satisfies the axiom , which completes the proof of sufficiency part.
∎
5 Concluding Remarks
We conclude the paper by discussing another graph transit function, namely the induced path transit function for a distance hereditary graph. By replacing shortest paths by induced paths in a graph , we get the induced path transit function . This function is also well studied. For example, see the references; [5, 7, 13, 25]. Nebeský proved in [23] a very interesting result: there does not exist a characterization of the induced path function of a connected graph using a set of first - order axioms. Changat et al. in [6] characterized the induced path transit function axiomatically on -free and -free graphs.
Formally the induced path function of , is defined as
Next, we define an axiom , which is used in the following discussions.
-
:
there exists , such that , and .
The following result is proved in [6].
Theorem 9.
[6] Let be a transit function on a non-empty finite set satisfying the axioms and with underlying graph . Then is -free and is precisely the induced path transit function of .
We have the following proposition for a transit function on .
Proposition 2.
If a transit function on satisfies the axioms and then satisfies axiom .
Proof.
Let be any non-empty set, and be a transit function defined on . We know that satisfies implies satisfies . Let . Since satisfies , we can see that . Again since satisfies , we can see that (other wise will not satisfy the axiom). So we have and .
Claim : and .
If , then we can see that a contradiction to the fact that satisfies axiom. In a similar fashion if , we will get a contradiction.
So there exists a vertices and . Consider and . Since satisfies , we have (as in the above lines) and . Continuing like this we can get a sequence of vertices and so that and , with .
Without loss of generality we assume and .
Now since satisfies we have . Again using the axiom, . Hence satisfies the axiom.
∎
Proposition 3.
If a transit function on satisfies the axioms then satisfies .
Proof.
Let be any non empty set. Let be any transit function defined on . Let satisfy the axioms . If possible assume that doesnot satisfy . Therefore there exists with and . Since satisfies , and we have . Again since , we must have . So we have . Now since satisfies , there should exist an element . Now we have so we have , a contradiction to the assumption of . So our assumption is wrong and satisfies . ∎
The example below shows that axioms , doesn’t imply axiom .
Example 8 ( but not ).
Let . Define on as follows., . We can easily see that satisfies and . Now , but . So that fails to satisfy axiom.
The following examples establish that the axioms and are independent.
Example 9 ( but not ).
Let . Define a transit function on as follows: We can see that does not satisfy as but . But we can see that satisfies .
Example 10 ( but not ).
Let . Define a transit function on as follows: . We can see that satisfies . Now but we can see that , so doesnot satisfy the axiom.
Example 11 ( but not ).
Let . Define a transit function on as follows: . We can see that satisfies . We have but . Hence does not satisfy .
Example 12 ( but not ).
Let . Define a transit function on as follows: and for all other pair define . We can see that satisfies . But and . So does not satisfy .
Example 13 ( but not ).
Let . Define a transit function on as follows: . We can see that satisfies . But but . So does not satisfy .
We have already noted in the introductory section that axiom implies axiom , for any transit function .
Therefore, we replace by in Theorem 9 and using Lemma 2 and Proposition 2, we can reformulate Theorem 9 using a minimal set of independent axioms as
Theorem 10.
Let be a transit function on a non-empty finite set satisfying the axioms and with underlying graph . Then is -fan -free (distance hereditary graph) and is precisely the induced path transit function of .
A distance hereditary graph is precisely the graph in which every induced path is a shortest path and hence both the induced path transit function and the interval function coincide in . Therefore we have that Theorem 10 also holds for the interval function of . That is, we have
Theorem 11.
Let be any transit function satisfying the axioms and then is distance hereditary and coincides with the interval function of .
Also note that since axiom implies axiom and , we can use the same ideas in the proof of Theorem 6 to prove an independent proof for Theorem 11. Finally we have the following theorem characterizing the interval function of a distance hereditary graph from Theorem 3 and Theorem 11
Theorem 12.
Let be a transit function on the vertex set of a connected graph . Then is a distance hereditary graph and coincides the interval function of if and only if satisfies the axioms and the axiom implies that .
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