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The Interval function, Ptolemaic, distance hereditary, bridged graphs and axiomatic characterizations

Manoj Changat111Supported by Department of Science and Technology, SERB (File No.MTR/2017/000238 “Axiomatics of Betweenness in Discrete Structures”). Department of Futures Studies,
University of Kerala,
Thiruvananthapuram - 695 581, India
email: [email protected]
Lekshmi Kamal K. Sheela222Supported by CSIR, Government India as a CSIR-JRF. Department of Futures Studies,
University of Kerala,
Thiruvananthapuram - 695 581, India
email: [email protected]
Prasanth G. Narasimha-Shenoi333Supported by Department of Science and Technology, SERB under their MATRICS Scheme (File No.MTR/2018/000012). Department of Mathematics,
Government College Chittur,
Palakkad - 678 104, India
email: [email protected]
Abstract

In this paper we consider certain types of betweenness axioms on the interval function IGI_{G} of a connected graph GG. We characterize the class of graphs for which IGI_{G} 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 graphs
MSC:
[2020] 05C12

1 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 VV is a function R:V×VR:V\times V to 2V2^{V} on VV satisfying the following three axioms:

(t1)(t1)

uR(u,v)u\in R(u,v), for all u,vVu,v\in V,

(t2)(t2)

R(u,v)=R(v,u)R(u,v)=R(v,u), for all u,vVu,v\in V,

(t3)(t3)

R(u,u)={u}R(u,u)=\{u\}, for all uVu\in V.

If VV is the vertex set of a graph GG, then we say that RR is a transit function on GG. Throughout this paper, we consider only finite, simple and connected graphs. The underlying graph GRG_{R} of a transit function RR on VV is the graph with vertex set VV, where two distinct vertices uu and vv are joined by an edge if and only if R(u,v)={u,v}R(u,v)=\{u,v\}.

A u,vu,v - shortest path in a connected graph G=(V,E)G=(V,E) is a u,vu,v-path in GG containing the minimum number of edges. The length of a shortest u,vu,v-path PP (that is, the number of edges in PP) is the standard distance in GG. The interval function IGI_{G} of a connected graph GG is the function IGI_{G} defined with respect to the standard distance dd in GG as

I:V×V:2VI:V\times V:\longrightarrow 2^{V}

IG(u,v)={wVI_{G}(u,v)=\{w\in V: ww lies on some shortest u,vu,v - path in G}={wV:d(u,w)+d(w,v)=d(u,v)}G\}=\{w\in V:d(u,w)+d(w,v)=d(u,v)\}

The interval function IGI_{G} is a classical example of a transit function on a graph ( we some times denote IGI_{G} by II, if there is no confusion for the graph GG). It is easy to observe that the underlying graph GIGG_{I_{G}} of IGI_{G} is isomorphic to GG. 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 II of a connected graph G=(V,E)G=(V,E) during the 1990s. The problem is the following: “ Is it possible to give a characterization of the interval function IGI_{G} of a connected graph GG using a set of simple axioms (first - order axioms) defined on a transit function RR on VV?” Nebeský [17, 18] proved that there exists such a characterization for the interval function I(u,v)I(u,v) by using first - order axioms on an arbitrary transit function RR. 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 IGI_{G} 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 (t1)(t1) and (t2)(t2) and three additional (b2)(b2), (b3)(b3), and (b4)(b4) defined as follows:

(b2)(b2) if xR(u,v)x\in R(u,v) and yR(u,x)y\in R(u,x), then yR(u,v)y\in R(u,v),
(b3)(b3) if xR(u,v)x\in R(u,v) and yR(u,x)y\in R(u,x) then xR(y,v)x\in R(y,v),
(b4)(b4) if xR(u,v)x\in R(u,v) then R(u,x)R(x,v)={x}R(u,x)\cap R(x,v)=\{x\}

The notation xR(u,v)x\in R(u,v) can be interpreted as xx is in between uu and vv. For example, the axiom (b2)(b2) can be interpreted as: if xx is between uu and vv, and yy is between uu and xx, then yy is between uu and vv. Similarly we can describe all other axioms. Hence we use the terminology of betweeness for an axiom on a transit function RR. 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 (b1)(b1) and (b2)(b2). Here the axiom (b1)(b1) is defined for every u,vVu,v\in V and a transit function RR as follows:
(b1)(b1) xR(u,v),xvvR(u,x)x\in R(u,v),x\neq v\Rightarrow v\not\in R(u,x)

The following implications can be easily verified for a transit function RR among axioms (t1),(t2),(t3),(b1),(b3)(t1),(t2),(t3),(b1),(b3) and (b4)(b4).

  • 1.

    Axioms (t1)(t1) and (b4)(b4) implies axiom (t3)(t3).

  • 2.

    Axioms (t1),(t2),(t3)(t1),(t2),(t3) and (b3)(b3) implies axiom (b4)(b4) which implies axiom (b1)(b1) ( that is , for a transit function R,(b3)R,(b3) implies (b4)(b4) implies (b1)(b1))

The converse of the above implications need not hold. A transit function RR satisfying axioms (b2)(b2) and (b3)(b3) 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 GG be a graph and HH a subgraph of GG. HH is called an isometric subgraph of GG if the distance dH(u,v)d_{H}(u,v) between any pair of vertices, u,vu,v in HH coincides with that of the distance dG(u,v)d_{G}(u,v). HH is called an induced subgraph if u,vu,v are vertices in HH such that uvuv is an edge in GG, then uvuv must be an edge in HH also. A graph GG is said to be HH-free, if GG has no induced subgraph isomorphic to HH. Let G1,G2,,GkG_{1},G_{2},\ldots,G_{k} be graphs. For a graph GG, we say that GG is G1,G2,,GkG_{1},G_{2},\ldots,G_{k}-free if GG has no induced subgraph isomorphic to GiG_{i}, i=1,,ki=1,\ldots,k. Chordal graph is an example of a graph GG which is defined with respect to an infinite number of forbidden induced subgraphs (GG is chordal if GG have no induced cycles CnC_{n} with length nn 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 G=(V,E)G=(V,E) is a bridged graph if GG has no isometric cycles of length greater than 33. Clearly the family of bridged graphs contain the family of chordal graphs. The graph GG is distance hereditary if the distances in any connected induced subgraph HH of GG are the same as in GG. Thus, any induced subgraph HH inherits the distances of GG. The graph GG is a Ptolemaic graph if GG 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 Cn,n>3C_{n},n>3.

In addition to the geometric axioms (b3)(b3) and (b2)(b2), we consider the following betweenness axioms (J0),(J2),(J2)(J0),(J2),(J2^{\prime}) and (J3)(J3^{\prime}) for a transit function RR on VV for proving the characterizations of these classes of graphs.

  • (J0)(J{0})

    : For any pair of distinct vertices u,v,x,yVu,v,x,y\in V we have xR(u,y),yR(x,v)xR(u,v)x\in R(u,y),y\in R(x,v)\Rightarrow x\in R(u,v).

  • (J0)(J{0^{\prime}})

    : xR(u,y),yR(x,v),R(u,y)R(x,v){u,x,y,v}xR(u,v)x\in R(u,y),y\in R(x,v),R(u,y)\cap R(x,v)\subseteq\{u,x,y,v\}\Rightarrow x\in R(u,v).

  • (J2)(J{2})

    : R(u,x)={u,x},R(x,v)={x,v},R(u,v){u,v}xR(u,v)R(u,x)=\{u,x\},R(x,v)=\{x,v\},R(u,v)\neq\{u,v\}\implies x\in R(u,v).

  • (J2)(J{2^{\prime}})

    : xR(u,y),yR(x,v),R(u,x)={u,x},R(x,y)={x,y},R(y,v)={y,v},R(u,v){u,v}xR(u,v)x\in R(u,y),y\in R(x,v),R(u,x)=\{u,x\},R(x,y)=\{x,y\},R(y,v)=\{y,v\},R(u,v)\neq\{u,v\}\Rightarrow x\in R(u,v)

  • (J3)(J{3^{\prime}})

    : xR(u,y),yR(x,v),R(x,y){x,y},R(u,v){u,v}xR(u,v)x\in R(u,y),y\in R(x,v),R(x,y)\neq\{x,y\},R(u,v)\neq\{u,v\}\Rightarrow x\in R(u,v).

From the definition of the axioms, we observe the following. The axiom (J2)(J2) is a simple betweenness axiom which is always satisfied by the interval function II. The axiom (J2)(J2^{\prime}) is a natural extension of (J2)(J2). We provide examples in the respective sections for the independence of the axioms (J2)(J2^{\prime}) and (J3)(J3^{\prime}) and (J0)(J0^{\prime}). The axioms (J2)(J2^{\prime}) and (J3)(J3^{\prime}) were first considered in [6] and later in [4]. The axiom (J0)(J0) 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 GG. We may observe that both the family of bridged graphs 𝒢\mathcal{BG} and distance hereditary graphs 𝒟\mathcal{DH} is a strict super class of the family of Ptolemaic graphs,𝒫𝒢\mathcal{PG}, that is, 𝒫𝒢𝒟\mathcal{PG}\subsetneq\mathcal{DH} and 𝒫𝒢𝒢\mathcal{PG}\subsetneq\mathcal{BG} and 𝒢\mathcal{BG} and 𝒟\mathcal{DH} coincide only in 𝒫𝒢\mathcal{PG}. But, 𝒢𝒟\mathcal{BG}\nsubseteq\mathcal{DH} and 𝒟𝒢\mathcal{DH}\nsubseteq\mathcal{BG}, This relation is also reflected in the implications between the axioms (J0),(J2),(J3)(J0),(J2^{\prime}),(J3^{\prime}) and (J0)(J0^{\prime}). From the definitions, we have the axiom (J0)(J0) implies axioms (J2)(J2^{\prime}) and (J3)(J3^{\prime}), and also (J0)(J0) implies (J0)(J0^{\prime}), while the reverse implications are not true. In other words, axioms (J2),(J3)(J2^{\prime}),(J3^{\prime}) and (J0)(J0^{\prime}) are weaker axioms than J(0)J(0) and hence graphs whose interval function satisfy axioms (J2)(J2^{\prime}) and (J3)(J3^{\prime}) will be a super class of graphs whose interval function satisfies (J0)(J0). Similarly graphs whose interval function satisfies axiom (J0)(J0^{\prime}) will be a super class of graphs whose interval function satisfies (J0)(J0). See Figure 1 for the relationships between the family 𝒫𝒢\mathcal{PG}, 𝒟\mathcal{DH} and 𝒢\mathcal{BG}.

Refer to caption
Figure 1: Relation between 𝒫𝒢\mathcal{PG}, 𝒟\mathcal{DH} and 𝒢\mathcal{BG}

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 (J2)(J2^{\prime}) and (J3)(J3^{\prime}). 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 (J2)(J2^{\prime}) and (J3)(J3^{\prime}) are independent. The interval function IGI_{G} doesn’t satisfy axiom (J2)(J2^{\prime}), while IGI_{G} satisfy axiom (J3)(J3^{\prime}) for the graphs in Figure 2. Also IGI_{G} doesn’t satisfy axiom (J3)(J3^{\prime}), while IGI_{G} satisfy axiom (J2)(J2^{\prime}) for the graphs in Figure 3. By an HHD3HHD3 - fan - free graph GG, we mean that GG is free from the House graph, the Hole graph (cycles CnC_{n}, n5n\geq 5), the Domino graph and the 33-fan (See the Figures 2 and 3 for these graphs).

Example 1 ((J2)(J2^{\prime}) but not (J3)(J3^{\prime}) ).

  
Let V={u,v,w,x,y,z}V=\{u,v,w,x,y,z\}. Define a transit function RR on VV as follows. R(u,x)={u,x},R(u,w)={u,x,z,w}=R(x,z),R(u,z)={u,z},R(u,v)={u,z,v},R(x,y)={x,w,y},R(u,y)=V=R(x,v),R(x,w)={x,w},R(z,w)={z,w},R(z,y)={z,w,y,v}=R(w,v),R(z,v)={z,v},R(w,y)={w,y},R(y,v)={y,v}R(u,x)=\{u,x\},R(u,w)=\{u,x,z,w\}=R(x,z),R(u,z)=\{u,z\},R(u,v)=\{u,z,v\},R(x,y)=\{x,w,y\},R(u,y)=V=R(x,v),R(x,w)=\{x,w\},R(z,w)=\{z,w\},R(z,y)=\{z,w,y,v\}=R(w,v),R(z,v)=\{z,v\},R(w,y)=\{w,y\},R(y,v)=\{y,v\} and R(x,x)={x}R(x,x)=\{x\}. We can easily see that RR satisfies (J2)(J2^{\prime}). But we can see that xR(u,y),yR(x,v),R(x,y){x,y}x\in R(u,y),y\in R(x,v),R(x,y)\neq\{x,y\} and R(u,v){u,v}R(u,v)\neq\{u,v\} but xR(u,v)x\notin R(u,v). So RR does not satisfy (J3)(J3^{\prime}).

Example 2 ((J3)(J3^{\prime}) but not (J2)(J2^{\prime}) ).

  
Let V={x,y,u,v,w}V=\{x,y,u,v,w\}. Define a transit function RR on VV as follows:R(u,x)={u,x},R(u,y)={u,x,y},R(u,v)={u,w,v},R(u,w)={u,w},R(y,w)={y,x,v,w}=R(x,v),R(y,v)={v,y},R(x,y)={x,y},R(x,w)={x,w},R(v,w)={v,w}R(u,x)=\{u,x\},R(u,y)=\{u,x,y\},R(u,v)=\{u,w,v\},R(u,w)=\{u,w\},R(y,w)=\{y,x,v,w\}=R(x,v),R(y,v)=\{v,y\},R(x,y)=\{x,y\},R(x,w)=\{x,w\},R(v,w)=\{v,w\} and R(x,x)={x}R(x,x)=\{x\}. We can see that RR satisfies (J3)(J3^{\prime}). But xR(u,y),yR(x,v),R(u,x)={u,x},R(x,y)={x,y},R(y,v)={y,v}x\in R(u,y),y\in R(x,v),R(u,x)=\{u,x\},R(x,y)=\{x,y\},R(y,v)=\{y,v\}, but xR(u,v)x\notin R(u,v). Hence RR does not satisfy (J2)(J2^{\prime}).

The following results are proved in [4] and [6].

Proposition 1.

[4] For every graph GG, IGI_{G} satisfies the (J2)(J2^{\prime}) axiom if and only if GG is house, C5C_{5}, 33-fan free.

Lemma 1.

[6] Let RR be a transit function on a non-empty finite set VV satisfying the axioms (b1),(J2),(J2)(b1),(J2),(J2^{\prime}) and (J3)(J3^{\prime}) with underlying graph GRG_{R}. Then GRG_{R} is HHDHHD -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 GG is distance hereditary if and only if GG is HHD3HHD3 - fan-free.

We state a related result from [4] using the axiom (J3)(J3) defined for a transit function RR as

x′′R(u,y),yR(x,v),xy,R(u,v){u,v}xR(u,v)′′{}^{\prime\prime}x\in R(u,y),y\in R(x,v),x\neq y,R(u,v)\neq\{u,v\}\Rightarrow x\in R(u,v)^{\prime\prime}

Note that a PP graph is the graph obtained by adding a pendent edge on an induced 4-cycle, C4C_{4}. It follows from the definition that axiom (J3)(J3) implies both the axioms (J2)(J2^{\prime}) and (J3)(J3^{\prime}), but the reverse implications are not true. We quote the result.

Theorem 2.

[4] For every graph GG, IGI_{G} satisfies the (J3)(J3) axiom if and only if GG is HHP3HHP3 - fan - free graph.

It may be observed that a PP graph is a distance hereditary graph and hence the class of HHP3HHP3 - fan - free graphs is a proper subclass of the class of HHD3HHD3 - 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 (J2)(J2^{\prime}) and (J3)(J3^{\prime}) are weaker axioms than the axiom (J3)(J3).

Theorem 3.

Let GG be a connected graph. The interval function IGI_{G} satisfy the axioms (J2)(J2^{\prime}) and (J3)(J3^{\prime}) if and only if GG is a distance hereditary graph.

Proof.
\psscalebox

1 yyxxuuvvyyxxuuvvvvyyxxuu

Figure 2: House, C5C_{5}, 33 - fan
\psscalebox

1 xxyyuuvvxxuuyyvvxxuuyyvv

Figure 3: Cn,n6C_{n},n\geq 6, Domino, C6C_{6} cycle with a path joining two diametrical vertices

We use the fact from Theorem 1 that distance hereditary graphs are precisely HHD3HHD3 - fan - free graphs for the proof. Suppose that the interval function IGI_{G} of GG satisfy the axioms (J2)(J2^{\prime}) and (J3)(J3^{\prime}). To prove that GG is HHD3HHD3 - fan - free, assume the contrary that GG contains a house, a hole, or a domino or a 33-fan as an induced subgraph. A graph with an induced house or 33 - fan or a C5C_{5} doesn’t satisfy (J2)(J2^{\prime}) ( The vertices u,x,y,vu,x,y,v in Figure 2 doesn’t satisfy the axiom (J2)(J2^{\prime})). For an isometric hole CnC_{n}, n6n\geq 6, we choose vertices u,x,y,vu,x,y,v as shown in Figure 3, to prove that (J3)(J3^{\prime}) is violated. If GG has a domino DD, which is an isometric subgraph of GG with vertices as shown in Figure 3, then xIG(u,v)x\notin I_{G}(u,v). If DD is not isometric, then there is a vertex zz adjacent to both uu and yy or vv and xx. In this case, the graph induced by u,a,x,y,zu,a,x,y,z is either a C5C_{5} or a house or a 33-fan.
Let Cn,n6C_{n},n\geq 6 be a hole in GG that is not isometric and assume that nn is minimum. Clearly n6n\geq 6 and there exist u,vV(Cn)u,v\in V(C_{n}) such that dG(u,v)<dCn(u,v)d_{G}(u,v)<d_{C_{n}}(u,v). Let P=p0,,pkP=p_{0},\ldots,p_{k} be a u,vu,v-geodesic; we may choose PP such that kk is minimum. Let QQ and RR be the u,vu,v-paths on CnC_{n} where Q=q0,,qQ=q_{0},\ldots,q_{\ell}, R=r0,,rmR=r_{0},\ldots,r_{m} and u=p0=q0=r0u=p_{0}=q_{0}=r_{0} and v=pk=q=rmv=p_{k}=q_{\ell}=r_{m}. Clearly 2k<2\leq k<\ell and k<mk<m and we may assume without loss of generality that m\ell\leq m. Moreover, we can choose PP such that the cycle CC induced by V(P)V(Q)V(P)\cup V(Q) has minimum length. In particular, by the choice of CC, pk1p_{k-1} is not adjacent to any of {qk+1,,q1}\{q_{k+1},\ldots,q_{\ell-1}\} whenever >k+1\ell>k+1. By minimality of kk, pip_{i}, i{1,,k2}i\in\{1,\ldots,k-2\} can be adjacent to qjq_{j} (or rjr_{j}) only if i=ji=j or j=i+1j=i+1. If k+3\ell\geq k+3, then {pk1,pk,qk,qk+1,,q1}\{p_{k-1},p_{k},q_{k},q_{k+1},\ldots,q_{\ell-1}\} form (together with possibly some additional vertices of PP or QQ) an induced hole shorter than nn, which is not possible. If =k+2\ell=k+2, then pk1p_{k-1} and qkq_{k} must be adjacent; otherwise we have a shorter hole than nn on vertices {pk1,pk,qk,qk+1}\{p_{k-1},p_{k},q_{k},q_{k+1}\} together with pk2p_{k-2} or qk1q_{k-1} (and possibly some other vertices of PP or QQ). But then A={pk1,pk,qk,qk+1}A=\{p_{k-1},p_{k},q_{k},q_{k+1}\} form a 44-cycle and we have an induced domino on Aqk1,pk2A\cup q_{k-1},p_{k-2} when pk1qk1,pk2qk2E(G)p_{k-1}q_{k-1},p_{k-2}q_{k-2}\notin E(G) and an induced house on the same vertices when pk1qk1E(G)p_{k-1}q_{k-1}\in E(G). Let now =k+1\ell=k+1. First note that p1p_{1} must be adjacent to at least one of {q1,q2}\{q_{1},q_{2}\} and of {r1,r2}\{r_{1},r_{2}\} by the minimality of nn, since there are no isometric holes. Let first k=2k=2. If either q1q_{1} or q2q_{2} is not adjacent to p1p_{1}, we have an induced house as a subgraph. Otherwise we have an induced 33 - fan on {p1,q0,q1,q2,q3}\{p_{1},q_{0},q_{1},q_{2},q_{3}\}. Let now k>2k>2. By the above, p2p_{2} is not adjacent to q1q_{1} or to r1r_{1} and p1p_{1} is not adjacent to q3q_{3} or r3r_{3}. Suppose that at least one of q2q_{2} or r2r_{2}, say q2q_{2}, is adjacent to p1p_{1}. If both q1q_{1} and r1r_{1} are adjacent to p1p_{1}, we have an induced 33 - fan on {p1,r1,q0,q1,q2}\{p_{1},r_{1},q_{0},q_{1},q_{2}\}. If q1q_{1} is adjacent to p1p_{1} but r1r_{1} is not, we have an induced house on {p1,q0,q1,r1,r2}\{p_{1},q_{0},q_{1},r_{1},r_{2}\}. If both q1q_{1} and r1r_{1} are not adjacent to p1p_{1} , r2r_{2} adjacent to p1p_{1} and q2q_{2} adjacent to p1p_{1}, we have an induced domino on {p1,q0,q1,q2,r1,r2}\{p_{1},q_{0},q_{1},q_{2},r_{1},r_{2}\}. Hence q2p1E(G)q_{2}p_{1}\notin E(G) but q1p1E(G)q_{1}p_{1}\in E(G) and by a similar argument r2p1E(G)r_{2}p_{1}\notin E(G) but r1p1E(G)r_{1}p_{1}\in E(G). Since q2p1,q1p2E(G)q_{2}p_{1},q_{1}p_{2}\notin E(G), p2p_{2} and q2q_{2} must be adjacent since there are no isometric holes or by the minimality of nn. This yields an induced house on {p2,p1,q0,q1,q2}\{p_{2},p_{1},q_{0},q_{1},q_{2}\}. Finally, If the induced CnC_{n} is not isometric with n=6n=6, the only case left is the one in Figure 3, and by choosing the vertices u,x,y,vu,x,y,v as in the figure, it follows that the axiom (J3)(J3^{\prime}) is violated. Therefore, when GG has an induced house, hole, domino or 33 - fan, either the axiom (J2)(J2^{\prime}) or (J3)(J3^{\prime}) is violated.

Conversely assume that axiom (J2)(J2^{\prime}) or (J3)(J3^{\prime}) is not satisfied by the interval function IGI_{G} of GG. It is already known from Proposition 1 that if axiom (J2)(J2^{\prime}) is not satisfied, then GG has an induced C5C_{5}, House or 33 - fan. Now suppose axiom (J3)(J3^{\prime}) is not satisfied. Then there exists distinct vertices u,x,y,vu,x,y,v in VV such that xIG(u,y),yIG(x,v)x\in I_{G}(u,y),y\in I_{G}(x,v) , IG(x,y){x,y}I_{G}(x,y)\neq\{x,y\}, IG(u,v){u,v}I_{G}(u,v)\neq\{u,v\} and xIG(u,v)x\notin I_{G}(u,v). Let PP be a u,yu,y-geodesic containing xx and QQ be a x,vx,v - geodesic containing yy. We claim that

uPxPyQvu\rightarrow P\rightarrow x\rightarrow P\rightarrow y\rightarrow Q\rightarrow v

is a u,vu,v-path.

Since dP(x,y)=dQ(x,y)d_{P}(x,y)=d_{Q}(x,y) we see that xPyQvx\rightarrow P\rightarrow y\rightarrow Q\rightarrow v is a x,vx,v geodesic. Therefore (xPy)(yQv)={y}(x\rightarrow P\rightarrow y)\cap(y\rightarrow Q\rightarrow v)=\{y\}, for otherwise we may find a shorter path from xx to vv.

Now we claim that (uPx)(yQv)=(u\rightarrow P\rightarrow x)\cap(y\rightarrow Q\rightarrow v)=\emptyset. Assume on the contrary, that w(uPx)(yQv)w\in(u\rightarrow P\rightarrow x)\cap(y\rightarrow Q\rightarrow v), and let ww be the last vertex of the intersection (when going from uu to xx along uPxu\rightarrow P\rightarrow x path). Then dP(w,y)=dQ(y,w)d_{P}(w,y)=d_{Q}(y,w), and since wyw\neq y we find that dP(w,x)<dQ(w,y)d_{P}(w,x)<d_{Q}(w,y). Hence we have that the path xPwQvx\rightarrow P\rightarrow w\rightarrow Q\rightarrow v is shorter than xQywQvx\rightarrow Q\rightarrow y\rightarrow w\rightarrow Q\rightarrow v, a contradiction.

Hence uPxPyQvu\rightarrow P\rightarrow x\rightarrow P\rightarrow y\rightarrow Q\rightarrow v is a u,vu,v-path. Since xIG(u,v)x\notin I_{G}(u,v), uPxPyQvu\rightarrow P\rightarrow x\rightarrow P\rightarrow y\rightarrow Q\rightarrow v is not an u,vu,v geodesic. If RR is any u,vu,v-geodesic, then xV(R)x\notin V(R). Fix an u,vu,v-geodesic RR. Let aa be the last vertex on PP before xx that is on RR and bb be the first vertex on QQ after yy that is on RR. Note that such vertices always exists, since uPRu\in P\cap R and vQRv\in Q\cap R. On the other hand, note that bb can be equal to yy, but axa\neq x. Label vertices of the path aPyQba\rightarrow P\rightarrow y\rightarrow Q\rightarrow b by a=b0,b1,b=ba=b_{0},b_{1},\ldots b_{\ell}=b. Label vertices of the a,ba,b-subpath of RR as a=a0,a1,,ak=ba=a_{0},a_{1},\ldots,a_{k}=b. Clearly 4\ell\geq 4 and k2k\geq 2 and k<k<\ell. Path b0b1bb_{0}b_{1}\ldots b_{\ell} is not necessarily an induced path. If not, we choose among all chords bibjb_{i}b_{j} the one with maximal jij-i and replace the part bibjb_{i}\ldots b_{j} by this chord. Vertices of this new path are denoted by a=c0,c1,,ct=ak=ba=c_{0},c_{1},\ldots,c_{t}=a_{k}=b, where still t>k2t>k\geq 2 by the choice of aa and bb. But a0a1aka_{0}a_{1}\ldots a_{k} is an induced path, since it is a shortest path. Note that c1c_{1} is not adjacent to a2,,aka_{2},\ldots,a_{k} by the choice of aa. Hence c1c_{1} can be adjacent only to a1a_{1}. Similarly chc_{h}, for h2h\geq 2, cannot be adjacent to ah+1,,aka_{h+1},\ldots,a_{k}. We consider the following two cases.

Case 1: a1c1E(G)a_{1}c_{1}\notin E(G).
If also a1c2E(G)a_{1}c_{2}\notin E(G), we have an induced cycle of length 6\geq 6. So let a1c2E(G)a_{1}c_{2}\in E(G). Also a2c2E(G)a_{2}c_{2}\notin E(G) and a1c3E(G)a_{1}c_{3}\notin E(G), otherwise we have a house on vertices c0,c1,c2,a1c_{0},c_{1},c_{2},a_{1}, and a2a_{2} or c3c_{3} respectively. This implies that c3a2c_{3}a_{2} is an edge and we have an induced domino (c1,c2,c3,a2,a1,a0c_{1},c_{2},c_{3},a_{2},a_{1},a_{0}), otherwise we have an induced cycle of length 6\geq 6.

Case 2: a1c1E(G)a_{1}c_{1}\in E(G).
If c3=a2c_{3}=a_{2}, we have a house if a1c2E(G)a_{1}c_{2}\notin E(G) or a 33 - fan on vertices a0,a1,c1,c2a_{0},a_{1},c_{1},c_{2}, and c3c_{3}. Hence c3a2c_{3}\neq a_{2} and also c3a1E(G)c_{3}a_{1}\notin E(G). We get an induced cycle of length 5\geq 5, if c2c_{2} is not adjacent to at least one of a1a_{1} or a2a_{2}. If first c2a2E(G)c_{2}a_{2}\in E(G) and c2a1E(G)c_{2}a_{1}\notin E(G), we get a house on vertices a0,a1,c1,a2a_{0},a_{1},c_{1},a_{2}, and c2c_{2}. Let now c2a2E(G)c_{2}a_{2}\notin E(G) and c2a1E(G)c_{2}a_{1}\in E(G). Since c3a2c_{3}\neq a_{2}, there exists c4c_{4}. If c4a1E(G)c_{4}a_{1}\in E(G), we get a house on vertices a1,c1,c2,c3a_{1},c_{1},c_{2},c_{3}, and c4c_{4}. Also c3a2E(G)c_{3}a_{2}\notin E(G), since otherwise we get a house on vertices a1,c1,c2,c3a_{1},c_{1},c_{2},c_{3}, and a2a_{2}. But now we have an induced path c4c3c2a1a2c_{4}c_{3}c_{2}a_{1}a_{2} which lead to an induced hole. Finally, if c2a2E(G)c_{2}a_{2}\in E(G) and c2a1E(G)c_{2}a_{1}\in E(G), we get a 33 - fan on vertices a0,a1,c1,a2a_{0},a_{1},c_{1},a_{2}, and c2c_{2}. Thus in all cases, we get either an induced house, hole, domino, or 33 - fan, and thus the proof is completed. ∎

We need the following Lemma

Lemma 2.

Let RR be a transit function on a non-empty finite set VV satisfying the axioms (b3),(J2),(J2)(b3),(J2),(J2^{\prime}) and (J3)(J3^{\prime}) with underlying graph GRG_{R}. Then GRG_{R} is HHD3HHD3 - fan - free.

Proof.

Since for a transit function axiom (b3)(b3) implies axiom (b1)(b1), by Lemma 1, GRG_{R} is HHDHHD-free. We prove that GRG_{R} is also 33 - fan - free. Suppose on the contrary, GRG_{R} contains a 33 - fan with vertices u,v,x,y,zu,v,x,y,z as an induced subgraph. Let u,x,y,vu,x,y,v be the path of length three and zz be the vertex adjacent to all of u,v,x,yu,v,x,y. Since R(u,x)={u,x},R(x,y)={x,y},R(y,v)={y,v}R(u,x)=\{u,x\},R(x,y)=\{x,y\},R(y,v)=\{y,v\} and R(u,y){u,y},R(x,v){x,v}R(u,y)\neq\{u,y\},R(x,v)\neq\{x,v\}, we have by axiom (J2),xR(u,y)(J2),x\in R(u,y), and yR(x,v)y\in R(x,v). Again since R(u,v){u,v}R(u,v)\neq\{u,v\} by (J2),xR(u,v)(J2^{\prime}),x\in R(u,v). Again, R(x,z)={x,z}R(x,z)=\{x,z\} and R(z,v)={z,v}R(z,v)=\{z,v\}, R(x,v){x,v}R(x,v)\neq\{x,v\}, by axiom (J2)(J2), we have zR(x,v)z\in R(x,v). Now, we have xR(u,v)x\in R(u,v) and zR(x,v)z\in R(x,v). Hence by axiom (b3)(b3), we have xR(u,z)x\in R(u,z), 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 IGI_{G} of a Ptolemaic graph GG, the essential axiom is (J0)(J0). Ptolemaic graphs are chordal graphs that are 33 - fan - free. Changat et al. in [4] characterized the graphs for which the interval function satisfies the axiom (J0)(J0) as follows.

Theorem 4.

[4] Let GG be a graph. The interval function IGI_{G} satisfies the axiom (J0)(J0) if and only if GG is a Ptolemaic graph.

Theorem 5.

Let RR be any transit function defined on a non-empty set VV. If RR satisfies (J0)(J0) and (J2)(J2) then the underlying graph GRG_{R} of RR is CnC_{n}-free for n4n\geq 4.

Proof.

Let RR be a transit function satisfying (J0)(J0) and (J2)(J2). Let GRG_{R} contains an induced cycle say Cn=u1u2unu1C_{n}=u_{1}u_{2}\ldots u_{n}u_{1}. Without loss of generality assume CnC_{n} is the minimum such cycle (in the sense that length of the induced cycle is as small as possible). We prove for every k4k\geq 4.
Case: k=4k=4.
Now since R(u1,u2)={u1,u2}R(u_{1},u_{2})=\{u_{1},u_{2}\} and R(u2,u3)={u2,u3}R(u_{2},u_{3})=\{u_{2},u_{3}\}, By (J2)(J2) axiom we have u2R(u1,u3)u_{2}\in R(u_{1},u_{3}) in a similar fashion we can show that u3R(u2,u4)u_{3}\in R(u_{2},u_{4}). Since RR satisfies (J0)-axiom we have u2R(u1,u4)u_{2}\in R(u_{1},u_{4}), which is a contradiction as R(u1,u4)={u1,u4}R(u_{1},u_{4})=\{u_{1},u_{4}\}.
Case: k=5k=5.
As in the above case, we can see that u4R(u5,u3)u_{4}\in R(u_{5},u_{3}), this together with u3R(u4,u1)u_{3}\in R(u_{4},u_{1}) we have u4R(u5,u1)u_{4}\in R(u_{5},u_{1}), which is again a contradiction.
Case: k6k\geq 6.
By repeated application of (J2)(J2)-axiom, as in the above two cases, un1R(un,un2)u_{n-1}\in R(u_{n},u_{n-2}) with un2R(un1,u1)u_{n-2}\in R(u_{n-1},u_{1}), by applying (J2)(J2), we can see that un1R(un,u1)u_{n-1}\in R(u_{n},u_{1}). Here again a contradiction.

Hence, in all cases, we can see that GRG_{R} does not contain CnC_{n} as an induced subgraph This completes theorem. ∎

The following straightforward Lemma for the connectedness of the underlying graph GRG_{R} of a transit function RR is proved in [6].

Lemma 3.

[6] If the transit function RR on a non-empty set VV satisfies axioms (b1)(b1) and (b2)(b2), then the underlying graph GRG_{R} of RR is connected.

We have the following lemma.

Lemma 4.

If RR is a transit function on VV satisfying the axioms (J0)(J0) and (b3)(b3), then RR satisfies axiom (b2)(b2) and GRG_{R} is connected.

Proof.

Let RR satisfies axioms (J0)(J0) and (b3)(b3). To prove RR satisfies (b2)(b2). Since RR satisfies (J0)(J0), For u,v,x,yVu,v,x,y\in V, let xR(u,v)x\in R(u,v), and yR(u,x)y\in R(u,x). Since RR satisfies (b3)(b3), we have xR(u,v),yR(u,x)xR(y,v)x\in R(u,v),y\in R(u,x)\implies x\in R(y,v). Now yR(u,x),xR(y,v)y\in R(u,x),x\in R(y,v) and so by axiom (J0)(J0) , we have yR(u,v)y\in R(u,v), which implies that RR satisfies (b2)(b2). Connectedness of GRG_{R} follows from Lemma 3, since RR satisfies axioms (b1)(b1) and (b2)(b2) as axiom (b3)(b3) implies axiom (b1)(b1). ∎

Example 3 ((J0),(J2)(J0),(J2) and (b2)(b2) but not (b3)(b3)).

  
Let V={u,v,w,x,y}V=\{u,v,w,x,y\}. Let R:V×V2VR:V\times V\rightarrow 2^{V} be defined as follows. R(u,v)=V,R(u,x)={u,y,w,x},R(w,v)={x,w,y,v}R(u,v)=V,R(u,x)=\{u,y,w,x\},R(w,v)=\{x,w,y,v\} and in all other cases R(a,b)={a,b}R(a,b)=\{a,b\}. Since GRG_{R} is a 33 - fan, RR satisfies (J0)(J0) and (J2)(J2). Next to show RR satisfies (b2)(b2) axiom. Since R(u,v)=VR(u,v)=V, we can see that R(u,x)R(u,v)R(u,x)\subseteq R(u,v) for all xR(u,v)x\in R(u,v) so that for this pair RR satisfies (b2)(b2) axiom. Now consider R(u,x)R(u,x), we can see that a(u,x)R(u,x)a(\neq u,x)\in R(u,x) we have R(u,a)={u,a}R(u,a)=\{u,a\} and R(x,a)={x,a}R(x,a)=\{x,a\} which implies that RR satisfies (b2)(b2) axiom for this pair too. The case is similar for R(w,v)R(w,v). All other pairs corresponds to edges. Hence we can see that RR satisfies (b2)(b2)-axiom. Now xR(u,v),yR(u,x)x\in R(u,v),y\in R(u,x) but xR(y,v)={y,v}x\notin R(y,v)=\{y,v\}, and RR violates (b3)(b3) axiom.

Theorem 6.

Let RR be any transit function satisfying the axioms (b3),(J0)(b3),(J0) and (J2)(J2) then GRG_{R} is Ptolemaic and R(u,v)=I(u,v)R(u,v)=I(u,v).

Proof.

Since RR satisfies the axioms (b3),(J0)(b3),(J0) and (J2)(J2), we have that GRG_{R} is a chordal graph by Theorem 5. To prove that GRG_{R} is Ptolemaic, we have to show that GRG_{R} is 33 - fan - free. Suppose that GRG_{R} contains an induced 33 - fan with vertices u,x,y,v,zu,x,y,v,z with u,x,y,vu,x,y,v forming a path P4P_{4} and zz as the vertex adjacent to all the vertices u,x,y,vu,x,y,v. Since uxux and xyxy are adjacent and uyuy is not an edge, by (J2)(J2), xR(u,y)x\in R(u,y). Similarly yR(x,v)y\in R(x,v). Since RR is a transit function, by (t2)(t2), yR(v,x)y\in R(v,x) and xR(y,u)x\in R(y,u) and hence by (J0)(J0), yR(u,v)y\in R(u,v). Again, since uzuz and zyzy are edges and uyuy is not an edge, zR(u,y)z\in R(u,y). That is, yR(u,v)y\in R(u,v) and zR(u,y)z\in R(u,y), by (b3)(b3), we have yR(z,v)y\in R(z,v), which is not true as zvzv is an edge. That is, we have proved that GRG_{R} is a chordal graph which is 33 - fan - free and hence GRG_{R} is a Ptolemaic graph. By Lemma 4, RR satisfies axiom (b2)(b2) and (b1)(b1) and GRG_{R} is connected.

Now we prove that R(u,v)=I(u,v)R(u,v)=I(u,v) for all u,vVu,v\in V. We prove by induction on the distance between uu and vv.
Case when d(u,v)=2d(u,v)=2.
Let xI(u,v)x\in I(u,v) Hence we can see that ux,xvEux,xv\in E. That is, R(u,x)={u,x},R(x,v)={x,v}R(u,x)=\{u,x\},R(x,v)=\{x,v\} and R(u,v)={u,v}R(u,v)=\{u,v\}, since RR satisfies (J2),xR(u,v)(J2),x\in R(u,v). Therefore I(u,v)R(u,v)I(u,v)\subseteq R(u,v). Conversely suppose xR(u,v)x\in R(u,v). Suppose xI(u,v)x\notin I(u,v). Since d(u,v)=2d(u,v)=2 there exists at least one element yI(u,v)y\in I(u,v) such that uy,yvuy,yv are edges in GRG_{R}. By assumption, xx is not adjacent to both uu and vv. Assume that xuxu is not an edge. Since xR(u,v)x\in R(u,v) and RR satisfies (b2)(b2) and (b1),R(u,x)R(u,v)(b1),R(u,x)\subset R(u,v) with |R(u,x)|<|R(u,v)||R(u,x)|<|R(u,v)|. By applying axioms (b2)(b2) and (b1)(b1) continuously on R(u,x)R(u,x), we get vertices xi,xi+1,,xk,xk+1=xR(u,x)x_{i},x_{i+1},\ldots,x_{k},x_{k+1}=x\in R(u,x) such that R(xi,u)R(xi+1,u)R(x_{i},u)\subset R(x_{i+1},u) and |R(xi,u)|<|R(xi+1,u)||R(x_{i},u)|<|R(x_{i+1},u)|, for i=1,,ki=1,\ldots,k and since VV is finite, R(u,xi)={u,xi}R(u,x_{i})=\{u,x_{i}\}, for some ii, say i=1i=1. That is, we have vertices x1,x2,,xk,xk+1=xR(u,x)x_{1},x_{2},\ldots,x_{k},x_{k+1}=x\in R(u,x) with R(x1,u)={x1,u}R(x_{1},u)=\{x_{1},u\}. Let us assume that R(x1,y){x1,y}R(x_{1},y)\neq\{x_{1},y\}. That is x1yE(GR)x_{1}y\notin E(G_{R}). Consider vertices x1,u,y,vx_{1},u,y,v. By (J2),uR(x1,y)(J2),u\in R(x_{1},y) and since yR(u,v)y\in R(u,v), by (J0),uR(x1,v)(J0),u\in R(x_{1},v). That is, x1R(v,u),uR(v,x1)x_{1}\in R(v,u),u\in R(v,x_{1}) and hence by (b3),x1R(u,u)(b3),x_{1}\in R(u,u), a contradiction. Therefore R(x1,y)={x1,y}R(x_{1},y)=\{x_{1},y\}. This implies that yR(x1,v)y\in R(x_{1},v) by axiom (J2)(J2), provided R(x1,v){x1,v}R(x_{1},v)\neq\{x_{1},v\}, which implies that x1R(y,u)x_{1}\in R(y,u) by (b3)(b3), a contradiction since R(u,y)={u,y}R(u,y)=\{u,y\}. Therefore R(x1,v)={x1,v}R(x_{1},v)=\{x_{1},v\}. That is, we have xR(u,v),x1R(u,x)x\in R(u,v),x_{1}\in R(u,x) and hence by (b3),xR(x1,v)(b3),x\in R(x_{1},v), a final contradiction. Therefore R(u,x)={u,x}R(u,x)=\{u,x\}. Similarly, we can prove that R(v,x)={v,x}R(v,x)=\{v,x\}. xI(u,v)x\in I(u,v) and hence R(u,v)I(u,v)R(u,v)\subset I(u,v), which completes the proof when d(u,v)=2d(u,v)=2.

Let us assume that the result holds for all distances less than k>2k>2 and let uu,vv be two vertices such that d(u,v)=k>2d(u,v)=k>2. We first prove I(u,v)R(u,v)I(u,v)\subseteq R(u,v). Let xI(u,v)x\in I(u,v). Since d(u,v)>2d(u,v)>2, we can find another vertex yy in the shortest u,vu,v-path containing xx. Now since II satisfies (b2),I(u,x)I(u,v),I(x,v)I(u,v)(b2),I(u,x)\subseteq I(u,v),I(x,v)\subseteq I(u,v). So by induction we have I(u,x)=R(u,x)I(u,x)=R(u,x) and I(x,v)=R(x,v)I(x,v)=R(x,v). Also by (b3)(b3) axiom xI(u,y)=R(u,y)x\in I(u,y)=R(u,y), yI(x,v)=R(x,v)y\in I(x,v)=R(x,v). Then by (J0)(J0) axiom xR(u,v)x\in R(u,v). Hence I(u,v)R(u,v)I(u,v)\subseteq R(u,v). Let xR(u,v)x\in R(u,v). If possible let xI(u,v)x\notin I(u,v). Since xR(u,v)x\in R(u,v), by applying axioms (b1)(b1) and (b2)(b2) similarly as in the case of d(u,v)=2d(u,v)=2, we get vertices x1,x2,,xk,xk+1=xR(u,x)x_{1},x_{2},\ldots,x_{k},x_{k+1}=x\in R(u,x) with R(x1,u)={x1,u}R(x_{1},u)=\{x_{1},u\} such that R(xi,u)R(xi+1,u)R(x_{i},u)\subset R(x_{i+1},u) and |R(xi,u)|<|R(xi+1,u)||R(x_{i},u)|<|R(x_{i+1},u)|, for i=1,,ki=1,\ldots,k and R(x1,u)={x1,u}R(x_{1},u)=\{x_{1},u\}. Let yy be a vertex such that R(u,y)={u,y}R(u,y)=\{u,y\} and yIGR(u,v)y\in I_{G_{R}}(u,v). Similar to the case of d(u,v)=2d(u,v)=2, we can prove that R(x1,y)={x1,y}R(x_{1},y)=\{x_{1},y\}. That is u,x1,yu,x_{1},y form a c3c_{3} in GRG_{R}. Here there are two possibilities for d(x1,v)d(x_{1},v).
Case (i): d(x1,v)=kd(x_{1},v)=k. In this case, since d(u,v)=kd(u,v)=k and yy is on a shortest u,vu,v-path in GRG_{R} with d(y,v)=k1d(y,v)=k-1, we have that yy is on a shortest x1,vx_{1},v-path in GRG_{R}, that is, yIGR(x1,v)R(x1,v)y\in I_{G_{R}}(x_{1},v)\subseteq R(x_{1},v). Therefore, we have x1R(u,v),yR(x1,v)x_{1}\in R(u,v),y\in R(x_{1},v) and hence by (b3),x1R(y,u)(b3),x_{1}\in R(y,u), a contradiction as R(y,u)={y,u}R(y,u)=\{y,u\}.
Case(ii): d(x1,v)=k1d(x_{1},v)=k-1. In this case, x1IGR(u,v)x_{1}\in I_{G_{R}}(u,v). Since xR(u,v)x\in R(u,v) and so by (b2)(b2) axiom, R(x,v)R(u,v)R(x,v)\subseteq R(u,v). We have also xR(u,v),x1R(u,x)x\in R(u,v),x_{1}\in R(u,x) and hence by axiom (b3)(b3), we have xR(x1,v)=IGR(x1,v)x\in R(x_{1},v)=I_{G_{R}}(x_{1},v), by induction hypothesis. That is xIGR(x1,v)IGR(u,v)x\in I_{G_{R}}(x_{1},v)\subseteq I_{G_{R}}(u,v), since x1R(u,v)x_{1}\in R(u,v), 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 xR(u,v)IGR(u,v)x\in R(u,v)\subseteq I_{G_{R}}(u,v) and hence the theorem. ∎

The following examples show that the axioms (J0),(J2)(J0),(J2) and (b3)(b3) are independent.

Example 4 ((J0),(J2)(J0),(J2) but not (b3)(b3)).

  
Let V={a,b,c,d,e}V=\{a,b,c,d,e\} and define a transit function RR on VV as follows: R(a,b)={a,b},R(a,c)={a,c},R(a,d)={a,b,c,d},R(a,e)=V,R(b,c)={b,c},R(b,d)={b,d},R(b,e)={b,e},R(c,d)={c,d},R(c,e)={b,c,d,e},R(d,e)={d,e}R(a,b)=\{a,b\},R(a,c)=\{a,c\},R(a,d)=\{a,b,c,d\},R(a,e)=V,R(b,c)=\{b,c\},R(b,d)=\{b,d\},R(b,e)=\{b,e\},R(c,d)=\{c,d\},R(c,e)=\{b,c,d,e\},R(d,e)=\{d,e\}. We can see that RR satisfies (J0)(J0) and (J2)(J2). But dR(a,e),bR(a,d)d\in R(a,e),b\in R(a,d), but dR(b,e)d\notin R(b,e). Therefore RR doesnot satisfy the (b3)(b3) axiom.

Example 5 ((J2),(b3)(J2),(b3) but not (J0)(J0)).

  
Let V={a,b,c,d,e}V=\{a,b,c,d,e\} and define a transit function RR on VV as follows: R(a,b)={a,b},R(a,c)={a,c},R(a,d)={a,b,c,d},R(a,e)={a,b,e},R(b,c)={b,c},R(b,d)={b,d},R(b,e)={b,e},R(c,d)={c,d},R(c,e)={b,c,d,e},R(d,e)={d,e}R(a,b)=\{a,b\},R(a,c)=\{a,c\},R(a,d)=\{a,b,c,d\},R(a,e)=\{a,b,e\},R(b,c)=\{b,c\},R(b,d)=\{b,d\},R(b,e)=\{b,e\},R(c,d)=\{c,d\},R(c,e)=\{b,c,d,e\},R(d,e)=\{d,e\}. Here RR Satisfies (J2)(J2) and (b3)(b3). We can see that cR(a,d),dR(c,e)c\in R(a,d),d\in R(c,e) but cR(a,e)c\notin R(a,e). So RR doesnot satisfy (J0)(J0).

Example 6 ((J0),(b3)(J0),(b3) but not (J2)(J2)).

  
Let V={a,b,c,d,e}V=\{a,b,c,d,e\} and define a transit function RR on VV as follows: R(a,e)={a,e},R(b,e)={b,e},R(a,b)={a,b,c}R(a,e)=\{a,e\},R(b,e)=\{b,e\},R(a,b)=\{a,b,c\} and for all other pair R(x,y)={x,y}R(x,y)=\{x,y\} we can see that RR satisfies (J0),(b3)(J0),(b3) . But since eR(a,b)e\notin R(a,b) we can see that RR fails to satisfy (J2)(J2).

From Theorem 6 and Theorem 4, we have the following theorem characterizing the interval function of Ptolemaic graphs.

Theorem 7.

Let RR be a transit function on the vertex set VV of a connected graph GG. Then GG is a Ptolemaic graph and RR coincides the interval function IGI_{G} of GG if and only if RR satisfies the axioms (b3),(J0)(b3),(J0) and (J2)(J2) and the axiom R(u,v)={u,v}R(u,v)=\{u,v\} implies that uvE(G)uv\in E(G).

4 Interval function of bridged graphs

From the definitions of (J0)(J0) and (J0)(J0^{\prime}) it follows that (J0)(J0)(J0)\implies(J0^{\prime}). The example 7 shows that (J0)\centernot(J0)(J0^{\prime})\centernot\implies(J0).

Example 7 ((J0)\centernot(J0)(J0^{\prime})\centernot\implies(J0)).

  
Let V={a,b,c,d,e}V=\{a,b,c,d,e\} Let R:V×V2VR:V\times V\rightarrow 2^{V} defined as follows. R(a,e)={a,e},R(a,b)={a,b},R(b,e)={b,e},R(b,c)={b,c},R(c,e)={c,e},R(c,d)={c,d},R(d,e)={d,e},R(a,c)={a,b,c,e},R(a,d)={a,e,d},R(b,d)={b,c,d,e}R(a,e)=\{a,e\},R(a,b)=\{a,b\},R(b,e)=\{b,e\},R(b,c)=\{b,c\},R(c,e)=\{c,e\},R(c,d)=\{c,d\},R(d,e)=\{d,e\},R(a,c)=\{a,b,c,e\},R(a,d)=\{a,e,d\},R(b,d)=\{b,c,d,e\}. We can see that bR(a,c)b\in R(a,c) and cR(b,d)c\in R(b,d) but bR(a,d)b\notin R(a,d), so that RR doesnot satisfy (J0)(J0) axiom. We can see that there exists no u,v,x,yu,v,x,y and zz satisfying the assumptions of the axiom (J0)(J0^{\prime}) and hence the axiom (J0)(J0^{\prime}) follows trivially.

We now prove the theorem characterizing interval function of bridged graphs.

Theorem 8.

Let GG be a connected graph. The interval function IGI_{G} satisfies the axiom (J0)(J0^{\prime}) if and only if GG is a bridged graph.

Proof.

Let GG has an isometric cycle Ck=v1v2vkC_{k}=v_{1}v_{2}\ldots v_{k}, k>3k>3. If kk is odd, say k=2t+1,t2k=2t+1,t\geq 2, let u=v1,x=vt,y=vt+1,v=vt+2u=v_{1},x=v_{t},y=v_{t+1},v=v_{t+2}. Then it is easy to see that xIG(u,y)x\in I_{G}(u,y), yIG(x,v)y\in I_{G}(x,v) and IG(u,y)IG(x,v)={x,y}{u,x,y,v}I_{G}(u,y)\cap I_{G}(x,v)=\{x,y\}\subseteq\{u,x,y,v\}. If kk is even, say, k=2tk=2t, t2t\geq 2, let u=v1,x=vtu=v_{1},x=v_{t}, y=vt+1,v=vt+2y=v_{t+1},v=v_{t+2}. Then it is easy to see that xIG(u,y),yIG(x,v)x\in I_{G}(u,y),y\in I_{G}(x,v) and IG(u,y)IG(x,v)={x,y,v}{u,x,y,v}I_{G}(u,y)\cap I_{G}(x,v)=\{x,y,v\}\subseteq\{u,x,y,v\}. In both cases of kk being odd or even, xx is not on any shortest u,vu,v-path and hence xIG(u,v)x\notin I_{G}(u,v). This implies that If GG has an isometric cycle, then IGI_{G} doesn’t satisfy the axiom (J0)(J0^{\prime}), that is, we have proved that if IGI_{G} satisfies axiom (J0)(J0^{\prime}) implies that GG is bridged graph.

Conversely, if GG is a bridged graph, then we claim that IGI_{G} satisfy the axiom (J0)(J0^{\prime}). Suppose not. Then there exist vertices u,x,y,vu,x,y,v in GG satisfying the following. A u,yu,y-geodesic PP containing xx, an x,vx,v-geodesic QQ containing yy with IG(u,y)IG(y,v){u,x,y,v}I_{G}(u,y)\cap I_{G}(y,v)\subseteq\{u,x,y,v\} such that xx is not on any u,vu,v-geodesic in GG. Then xx and yy should be adjacent, since IG(u,y)IG(y,v){u,x,y,v}I_{G}(u,y)\cap I_{G}(y,v)\subseteq\{u,x,y,v\}

Using the same arguments as in the proof of Theorem 3, we derive the following.

  • i

    : uPxPyQvu\rightarrow P\rightarrow x\rightarrow P\rightarrow y\rightarrow Q\rightarrow v is a u,vu,v-path, say PP^{\prime}.

  • ii

    : The last vertex on PP before xx that is on a shortest u,vu,v-path RR containing xx is aa and the first vertex on QQ after yy that is on RR is bb.

  • iii

    : An a,ba,b-subpath of RR, Ra,b:a=z0z1zt1zt=bR_{a,b}:a=z_{0}z_{1}\ldots z_{t-1}z_{t}=b,(t1)(t\geq 1) and an a,ba,b induced path P′′:a=u0u1u=xu+1=yu+s=b,(+s2)P^{\prime\prime}:a=u_{0}u_{1}\ldots u_{\ell}=xu_{\ell+1}=y\ldots u_{\ell+s}=b,(\ell+s\geq 2) containing xx and yy, which is a subpath of of PP^{\prime}.

  • iv

    : The cycle CC formed by the vertices of Ra,bP′′R_{a,b}\cup P^{\prime\prime} has length, l(C)l(C), at least four.

Now, l(C)l(C) cannot be four, since CC is isometric, a contradiction to GG being a bridged graph, which implies that the length of the path Ra,bR_{a,b}, namely tt is strictly greater than one. Since GG is a bridged graph, it follows that, there are chords from vertices ziz_{i}, i=1,t1i=1,\ldots t-1 to vertices u1,,u=x,u+1=y,u+s1u_{1},\ldots,u_{\ell}=x,u_{\ell+1}=y,\ldots u_{\ell+s-1} of P′′P^{\prime\prime} so that the only isometric cycles are triangles.
Case 1: yby\neq b.
We claim that the index t=+s1t=\ell+s-1 . Since P′′P^{\prime\prime} is not a u,vu,v-geodesic containing xx, we have t+s1t\leq\ell+s-1. If t+s2t\leq\ell+s-2, since the only isometric cycles are triangles, we get PP is not a u,yu,y-geodesic containing xx or QQ is not a x,vx,v-geodesic containing yy. There for t=+s1t=\ell+s-1. Also if 1i11\leq i\leq\ell-1 ,1j1\leq j\leq\ell, ziz_{i} cannot be adjacent to uju_{j} for ji+2j\geq i+2 , since PP is a u,yu,y-geodesic containing xx, and if +1it1,j+s1\ell+1\leq i\leq t-1,\ell\leq j\leq\ell+s-1, ziz_{i} cannot be adjacent to uju_{j} for ji1j\leq i-1 , since QQ is a x,vx,v-geodesic containing yy. Which implies that zz_{\ell} is adjacent to both u=xu_{\ell}=x and u+1=yu_{\ell+1}=y, otherwise there exist an induced 44 - cycle on {z1,z,x,y}\{z_{\ell-1},z_{\ell},x,y\} or {z,x,y,z+1}\{z_{\ell},x,y,z_{\ell+1}\}. Then the path uaz1zyu\ldots az_{1}\ldots z_{\ell}y is also a u,yu,y shortest path and the path xzzt=bvxz_{\ell}\ldots z_{t}=b\ldots v is also a x,vx,v-shortest path. This implies that the vertex zz_{\ell}, which is different from u,x,y,vu,x,y,v also belongs to IG(u,y)IG(x,v)I_{G}(u,y)\cap I_{G}(x,v), a contradiction to the hypothesis of the axiom (J0)(J0^{\prime}).
Case 2: y=by=b.
Since PP is a u,yu,y-geodesic containing xx, t=+st=\ell+s. In this case IG(x,v)I_{G}(x,v) does not contain any of zi,i=1,2,,t1z_{i},i=1,2,\ldots,t-1. Which implies that IG(u,y)IG(x,v){u,x,y,v}I_{G}(u,y)\cap I_{G}(x,v)\subseteq\{u,x,y,v\}, a contradiction to the hypothesis of the axiom (J0)(J0^{\prime}). Therefore in both case IGI_{G} satisfies the axiom (J0)(J0^{\prime}), 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 GG, we get the induced path transit function JGJ_{G}. 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 JJ of a connected graph using a set of first - order axioms. Changat et al. in [6] characterized the induced path transit function axiomatically on HHDHHD-free and HHPHHP-free graphs.

Formally the induced path function JGJ_{G} of GG, is defined as

J(u,v)={ww lies on an induced u,v-path}.J(u,v)=\{w\mid w\mbox{ lies on an induced }u,v\mbox{-path}\}.

Next, we define an axiom (J1)(J1), which is used in the following discussions.

  1. (J1)(J1):

    wR(u,v),wu,vw\in R(u,v),w\neq u,v\Rightarrow there exists u1R(u,w)R(v,w),v1R(v,w)R(u,w)u_{1}\in R(u,w)\setminus R(v,w),v_{1}\in R(v,w)\setminus R(u,w), such that R(u1,w)={u1,w}R(u_{1},w)=\{u_{1},w\}, R(v1,w)={v1,w}R(v_{1},w)=\{v_{1},w\} and wR(u1,v1)w\in R(u_{1},v_{1}).

The following result is proved in [6].

Theorem 9.

[6] Let RR be a transit function on a non-empty finite set VV satisfying the axioms (b1),(b2),(J1),(J2),(J2)(b1),(b2),(J1),(J2),(J2^{\prime}) and (J3)(J3^{\prime}) with underlying graph GRG_{R}. Then GRG_{R} is HHDHHD-free and RR is precisely the induced path transit function of GRG_{R}.

We have the following proposition for a transit function RR on VV.

Proposition 2.

If a transit function RR on VV satisfies the axioms (b2)(b2) and (b3)(b3) then RR satisfies axiom (J1)(J1).

Proof.

Let VV be any non-empty set, and RR be a transit function defined on VV. We know that RR satisfies (b3)(b3) implies RR satisfies (b1)(b1). Let wR(u,v)w\in R(u,v). Since RR satisfies (b2)(b2), we can see that R(u,w)R(u,v)R(u,w)\subseteq R(u,v). Again since RR satisfies (b3)(b3), we can see that vR(u,w)v\notin R(u,w) (other wise RR will not satisfy the (b1)(b1) axiom). So we have R(u,w)R(u,v)R(u,w)\subsetneq R(u,v) and R(v,w)R(u,v)R(v,w)\subsetneq R(u,v).
Claim : R(u,w)R(v,w)R(u,w)\nsubseteq R(v,w) and R(v,w)R(u,w)R(v,w)\nsubseteq R(u,w).
If R(u,w)R(v,w)R(u,w)\subseteq R(v,w), then we can see that uR(v,w)u\in R(v,w) a contradiction to the fact that RR satisfies (b1)(b1) axiom. In a similar fashion if R(v,w)R(u,w)R(v,w)\subseteq R(u,w), we will get a contradiction.
So there exists a vertices x1R(u,w)R(v,w)x_{1}\in R(u,w)\setminus R(v,w) and y1R(v,w)R(u,w)y_{1}\in R(v,w)\setminus R(u,w). Consider R(x1,w)R(x_{1},w) and R(y1,w)R(y_{1},w). Since RR satisfies (b2),(b1)(b2),(b1), we have (as in the above lines) R(x1,w)R(w,u)R(u,v)R(x_{1},w)\subsetneq R(w,u)\subsetneq R(u,v) and R(y1,w)R(w,v)R(u,v)R(y_{1},w)\subsetneq R(w,v)\subsetneq R(u,v). Continuing like this we can get a sequence of vertices x1,x2,,xx_{1},x_{2},\ldots,x_{\ell} and y1,y2,,ymy_{1},y_{2},\ldots,y_{m} so that R(x,w)R(x1,w)R(x1,w)R(w,u)R(u,v)R(x_{\ell},w)\subsetneq R(x_{\ell-1},w)\subsetneq\ldots\subsetneq R(x_{1},w)\subsetneq R(w,u)\subsetneq R(u,v) and R(ym,w)R(ym1,w)R(y1,w)R(w,v)R(u,v)R(y_{m},w)\subsetneq R(y_{m-1},w)\subsetneq\ldots\subsetneq R(y_{1},w)\subsetneq R(w,v)\subsetneq R(u,v), with R(x,w)={x,w},R(ym,w)={ym,w}R(x_{\ell},w)=\{x_{\ell},w\},R(y_{m},w)=\{y_{m},w\}.
Without loss of generality we assume x=xx_{\ell}=x and ym=yy_{m}=y. Now since RR satisfies (b3)(b3) we have wR(u,v),xR(u,w)wR(x,v)w\in R(u,v),x\in R(u,w)\implies w\in R(x,v). Again using the (b3)(b3) axiom, wR(x,v),yR(w,v)wR(x,y)w\in R(x,v),y\in R(w,v)\implies w\in R(x,y). Hence RR satisfies the (J1)(J1) axiom. ∎

Proposition 3.

If a transit function RR on VV satisfies the axioms (J1),(b2)(J1),(b2) then RR satisfies (b1)(b1).

Proof.

Let VV be any non empty set. Let RR be any transit function defined on VV. Let RR satisfy the axioms (J1),(b2)(J1),(b2). If possible assume that RR doesnot satisfy (b1)(b1). Therefore there exists u,v,wu,v,w with wR(u,v)w\in R(u,v) and vR(u,w)v\in R(u,w). Since RR satisfies (b2)(b2), and vR(u,w)v\in R(u,w) we have R(u,v)R(u,w)R(u,v)\subseteq R(u,w). Again since wR(u,v)w\in R(u,v), we must have R(u,w)R(u,v)R(u,w)\subseteq R(u,v). So we have R(u,w)=R(u,v)R(u,w)=R(u,v). Now since RR satisfies (J1)(J1), there should exist an element yR(v,w)R(u,w)y\in R(v,w)\setminus R(u,w). Now we have R(v,w)R(u,v)R(v,w)\subseteq R(u,v) so we have R(v,w)R(u,w)=R(v,w)R(u,v)=R(v,w)\setminus R(u,w)=R(v,w)\setminus R(u,v)=\emptyset, a contradiction to the assumption of (J1)(J1). So our assumption is wrong and RR satisfies (b1)(b1). ∎

The example below shows that axioms (b2),(J1)(b2),(J1), doesn’t imply axiom (b3)(b3).

Example 8 ((b2),(J1)(b2),(J1) but not (b3)(b3)).

  
Let V={u,v,x,y,z}V=\{u,v,x,y,z\}. Define RR on VV as follows.R(u,v)=V,R(u,y)={u,y},R(u,x)={u,y,z,x},R(u,z)={u,z},R(z,y)={z,y},R(z,x)={z,x}R(u,v)=V,R(u,y)=\{u,y\},R(u,x)=\{u,y,z,x\},R(u,z)=\{u,z\},R(z,y)=\{z,y\},R(z,x)=\{z,x\}, R(z,v)={z,y,x,v},R(x,v)={x,v},R(x,y)={x,y},R(y,v)={y,v}R(z,v)=\{z,y,x,v\},R(x,v)=\{x,v\},R(x,y)=\{x,y\},R(y,v)=\{y,v\}. We can easily see that RR satisfies (b2)(b2) and (J1)(J1). Now xR(u,v),yR(u,x)x\in R(u,v),y\in R(u,x), but xR(y,v)x\notin R(y,v). So that RR fails to satisfy (b3)(b3) axiom.

The following examples establish that the axioms (J2),(J2),(J3),(b3)(J2),(J2^{\prime}),(J3^{\prime}),(b3) and (b2)(b2) are independent.

Example 9 ((J2),(J2),(J3),(b3)(J2),(J2^{\prime}),(J3^{\prime}),(b3) but not (b2)(b2)).

  
Let V={a,b,c,d,e}V=\{a,b,c,d,e\}. Define a transit function RR on VV as follows: R(a,b)={a,b},R(a,c)={a,b,c},R(a,d)={a,b,c,d},R(a,e)={a,b,d,e},R(b,c)={b,c},R(b,d)={b,c,d},R(b,e)={b,c,d,e},R(c,d)={c,d},R(c,e)={c,d,e},R(d,e)={d,e}R(a,b)=\{a,b\},R(a,c)=\{a,b,c\},R(a,d)=\{a,b,c,d\},R(a,e)=\{a,b,d,e\},R(b,c)=\{b,c\},R(b,d)=\{b,c,d\},R(b,e)=\{b,c,d,e\},R(c,d)=\{c,d\},R(c,e)=\{c,d,e\},R(d,e)=\{d,e\} We can see that RR does not satisfy (b2)(b2) as dR(a,e)d\in R(a,e) but R(a,d)R(a,e)R(a,d)\nsubseteq R(a,e). But we can see that RR satisfies (J2),(J2),(J3),(b3)(J2),(J2^{\prime}),(J3^{\prime}),(b3).

Example 10 ((J2),(J2),(J3),(b2)(J2),(J2^{\prime}),(J3^{\prime}),(b2) but not (b3)(b3)).

  
Let V={a,b,c,d,e}V=\{a,b,c,d,e\}. Define a transit function RR on VV as follows: R(a,b)={a,b},R(a,c)={a,c},R(a,d)={a,b,c,d},R(a,e)=V,R(b,c)={b,c},R(b,d)={b,d},R(b,e)={b,e},R(c,d)={c,d},R(c,e)={c,b,d,e},R(d,e)={d,e}R(a,b)=\{a,b\},R(a,c)=\{a,c\},R(a,d)=\{a,b,c,d\},R(a,e)=V,R(b,c)=\{b,c\},R(b,d)=\{b,d\},R(b,e)=\{b,e\},R(c,d)=\{c,d\},R(c,e)=\{c,b,d,e\},R(d,e)=\{d,e\}. We can see that RR satisfies (J2),(J2),(J3),(b2)(J2),(J2^{\prime}),(J3^{\prime}),(b2). Now dR(a,e),bR(a,d)d\in R(a,e),b\in R(a,d) but we can see that dR(b,e)d\notin R(b,e), so RR doesnot satisfy the (b3)(b3) axiom.

Example 11 ((J2),(J3),(b2),(b3)(J2),(J3^{\prime}),(b2),(b3) but not (J2)(J2^{\prime})).

  
Let V={a,b,c,d,e}V=\{a,b,c,d,e\}. Define a transit function RR on VV as follows: R(a,b)={a,b},R(a,c)={a,c},R(a,d)={a,b,c,d},R(a,e)={a,b,e},R(b,c)={b,c},R(b,d)={b,d},R(b,e)={b,e},R(c,d)={c,d},R(c,e)={c,b,d,e},R(d,e)={d,e}R(a,b)=\{a,b\},R(a,c)=\{a,c\},R(a,d)=\{a,b,c,d\},R(a,e)=\{a,b,e\},R(b,c)=\{b,c\},R(b,d)=\{b,d\},R(b,e)=\{b,e\},R(c,d)=\{c,d\},R(c,e)=\{c,b,d,e\},R(d,e)=\{d,e\}. We can see that RR satisfies (J2),(J3),(b2),(b3)(J2),(J3^{\prime}),(b2),(b3). We have cR(a,d),dR(c,e),R(a,c)={a,c},R(a,d)={a,d},R(d,e)={d,e}c\in R(a,d),d\in R(c,e),R(a,c)=\{a,c\},R(a,d)=\{a,d\},R(d,e)=\{d,e\} but cR(a,e)={a,b,e}c\notin R(a,e)=\{a,b,e\}. Hence RR does not satisfy (J2)(J2^{\prime}).

Example 12 ((J2),(b2),(J3),(b3)(J2^{\prime}),(b2),(J3^{\prime}),(b3) but not (J2)(J2)).

  
Let V={a,b,c,d,e}V=\{a,b,c,d,e\}. Define a transit function RR on VV as follows: R(a,b)={a,b,c},R(a,e)={a,e},R(b,e)={b,e}R(a,b)=\{a,b,c\},R(a,e)=\{a,e\},R(b,e)=\{b,e\} and for all other pair define R(x,y)={x,y}R(x,y)=\{x,y\}. We can see that RR satisfies (J2),(J3),(b2),(b3)(J2^{\prime}),(J3^{\prime}),(b2),(b3). But R(a,e)={a,e},R(b,e)={b,e}R(a,e)=\{a,e\},R(b,e)=\{b,e\} and eR(a,b)={a,b,c}e\notin R(a,b)=\{a,b,c\}. So RR does not satisfy (J2)(J2).

Example 13 ((J2),(J2),(b2),(b3)(J2),(J2^{\prime}),(b2),(b3) but not (J3)(J3^{\prime})).

  
Let V={u,v,w,x,y,z}V=\{u,v,w,x,y,z\}. Define a transit function RR on VV as follows: R(u,x)={u,x},R(u,z)={u,x,z},R(u,y)=V=R(x,v)=R(z,w),R(u,v)={u,w,y},R(u,w)={u,w},R(x,z)={x,z},R(x,y)={x,z,y},R(x,w)={x,w},R(z,y)={z,y},R(z,v)={z,v},R(y,v)={y,v},R(y,w)={y,w},R(v,w)={v,w}R(u,x)=\{u,x\},R(u,z)=\{u,x,z\},R(u,y)=V=R(x,v)=R(z,w),R(u,v)=\{u,w,y\},R(u,w)=\{u,w\},R(x,z)=\{x,z\},R(x,y)=\{x,z,y\},R(x,w)=\{x,w\},R(z,y)=\{z,y\},R(z,v)=\{z,v\},R(y,v)=\{y,v\},R(y,w)=\{y,w\},R(v,w)=\{v,w\}. We can see that RR satisfies (J2),(J2),(b2),(b3)(J2),(J2^{\prime}),(b2),(b3). But xR(u,y),yR(x,v),R(x,y){x,y},R(u,v){u,v}x\in R(u,y),y\in R(x,v),R(x,y)\neq\{x,y\},R(u,v)\neq\{u,v\} but xR(u,v)x\notin R(u,v). So RR does not satisfy (J3)(J3^{\prime}).

We have already noted in the introductory section that axiom (b3)(b3) implies axiom (b1)(b1), for any transit function RR.

Therefore, we replace (b1)(b1) by (b3)(b3) 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 RR be a transit function on a non-empty finite set VV satisfying the axioms (b2),(b3),(J2),(J2)(b2),(b3),(J2),(J2^{\prime}) and (J3)(J3^{\prime}) with underlying graph GRG_{R}. Then GRG_{R} is HHD3HHD3-fan -free (distance hereditary graph) and RR is precisely the induced path transit function of GRG_{R}.

A distance hereditary graph GG 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 GG. Therefore we have that Theorem 10 also holds for the interval function of GRG_{R}. That is, we have

Theorem 11.

Let RR be any transit function satisfying the axioms (b2),(b3),(J2),(J2)(b2),(b3),(J2),(J2^{\prime}) and (J3)(J3^{\prime}) then GRG_{R} is distance hereditary and RR coincides with the interval function II of GRG_{R}.

Also note that since axiom (J0)(J0) implies axiom (J3)(J3^{\prime}) and (J2)(J2^{\prime}), 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 RR be a transit function on the vertex set VV of a connected graph GG. Then GG is a distance hereditary graph and RR coincides the interval function IGI_{G} of GG if and only if RR satisfies the axioms (b2),(b3),(J2),(J2),(J3)(b2),(b3),(J2),(J2^{\prime}),(J3^{\prime}) and the axiom R(u,v)={u,v}R(u,v)=\{u,v\} implies that uvE(G)uv\in E(G).

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