Steiner intervals, geodesic intervals, and betweenness.

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Discrete Mathematics 309 (2009) 6114–6125
Contents lists available at ScienceDirect
Discrete Mathematics
journal homepage: www.elsevier.com/locate/disc
Steiner intervals, geodesic intervals, and betweenness$
Boštjan Brešara,1, Manoj Changatb, Joseph Mathewsc, Iztok Peterind,1,
Prasanth G. Narasimha-Shenoie, Aleksandra Tepeh Horvatd,∗,1
aFaculty of Natural Sciences and Mathematics, University of Maribor, Slovenia
bDepartment of Futures Studies, University of Kerala, Trivandrum-695034, India
cChingamparampil, Vazhapally, Changanassery-686 103, Kerala, India
dFaculty of Electrical Engineering and Computer Science, University of Maribor, Slovenia
eDepartment of Mathematics, Government College, Chittur, Palakkad-678 104, India
a r t i c l ei n f o
Article history:
Received 31 October 2008
Received in revised form 7 May 2009
Accepted 15 May 2009
Available online 11 June 2009
Keywords:
Steiner interval
Geodesic interval
Distance
Betweenness
Monotonicity
Block graph
a b s t r a c t
The concept of the k-Steiner interval is a natural generalization of the geodesic (binary)
interval. It is defined as a mapping S : V ×···×V −→ 2Vsuch that S(u1,...,uk) consists
of all vertices in G that lie on some Steiner tree with respect to a multiset W = {u1,...,uk}
of vertices from G. In this paper we obtain, for each k, a characterization of the class of
graphs in which every k-Steiner interval S has the so-called union property, which says
that S(u1,...,uk) coincides with the union of geodesic intervals I(ui,uj) between all pairs
from W. It turns out that, as soon as k > 3, this class coincides with the class of graphs in
which the k-Steiner interval enjoys the monotone axiom (m), respectively (b2) axiom, the
conditions from betweenness theory. Notably, S satisfies (m), if x1,...,xk∈ S(u1,...,uk)
implies S(x1,...,xk) ⊆ S(u1,...,uk), and S satisfies (b2) if x ∈ S(u1,u2,...,uk) implies
S(x,u2,...,uk) ⊆ S(u1,...,uk). In the case k = 3, these three classes are different, and
we give structural characterizations of graphs for which their Steiner interval S satisfies
the union property as well as the monotone axiom (m). We also prove several partial
observations on the class of graphs in which the 3-Steiner interval satisfies (b2), which
lead to the conjecture that these are precisely the graphs in which every block is a geodetic
graph with diameter at most two.
© 2009 Elsevier B.V. All rights reserved.
1. Introduction and preliminaries
For vertices u,v of a graph G, the interval I(u,v) in graphs is usually defined as the set of vertices lying on a geodesic
(shortestpath)betweenuandv.Butbesidesgeodesicstherearesomeothernotionsthatcanbeusedfordefininganinterval,
such as induced and detour paths. In this paper we will focus on yet another concept, the Steiner interval, and consider its
relation with geodesic intervals.
The Steiner tree problem is a well-known problem with several variations and applications. It can concern points in
Euclidean (or other metric) spaces, and vertices of weighted or non-weighted graphs [11], and has drawn much attention
$Work supported by the Ministry of Science of Slovenia and by the Ministry of Science and Technology of India under the bilateral India–Slovenia grants
BI-IN/06-07-002 and DST/INT/SLOV-P-03/05, respectively.
∗Corresponding address: Faculty of Electrical Engineering and Computer Science, University of Maribor, Smetanova ulica 17, 2000 Maribor, Slovenia.
Tel.: +386 2 220 7382, +386 41 912 698; fax: +386 2 220 7272.
E-mail addresses: bostjan.bresar@uni-mb.si (B. Brešar), mchangat@gmail.com (M. Changat), jose_chingam@yahoo.co.in (J. Mathews),
iztok.peterin@uni-mb.si (I. Peterin), gnprasanth@gmail.com (P.G. Narasimha-Shenoi), aleksandra.tepeh@uni-mb.si (A.T. Horvat).
1Authors are also with the Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia.
0012-365X/$ – see front matter © 2009 Elsevier B.V. All rights reserved.
doi:10.1016/j.disc.2009.05.022

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due to the development of approximation algorithms, see [1,16] and the references therein. In a non-weighted connected
graph G, a Steiner tree of a (multi)set W ⊆ V(G), is a minimum order tree in G that contains all vertices of W. The number
of edges in a Steiner tree T of W is called the Steiner distance of W, denoted d(W), while the size of T describes the number
of vertices in T (i.e. d(W) + 1). The k-Steiner interval is a mapping S : V × ··· × V −→ V such that S(u1,u2,...,uk)
consists of all vertices in G that lie on some Steiner tree with respect to {u1,...,uk}, where u1,...,ukare, not necessarily
distinct, vertices of G (in this way S is an extension of I, as S(u,v,...,v) = I(u,v)). (Note that, as above, we will simplify
the notation for S({u1,...,uk}) to S(u1,...,uk), where S denotes the k-Steiner interval, and u1,...,ukare not necessarily
distinct vertices of a graph.) Steiner intervals on ordinary vertex subsets have been studied in several papers [6,8–11,13,16,
20–22].
One of the main issues regarding Steiner intervals is related to connections between different variations of the geodetic
number, see the survey paper [5]. Chartrand and Zhang proposed a natural concept of the Steiner number of a graph [6],
and among several nice results ‘‘proved’’ an erroneous statement [6] regarding the connection between Steiner intervals
of a set of vertices, and the union of geodesic intervals between pairs of the vertices from the set. This error was observed
and corrected by Pelayo [22], and the development intrigued Hernando et al. [10] to raise the following problem: for which
graphs the Steiner interval of any set of vertices, whose union of geodesic intervals between pairs of vertices in the set is
the whole vertex set, also yields all vertices? Certainly this property holds for graphs in which S(W) ⊆ ∪u,v∈WI(u,v) for
all W ⊆ V(G), which was shown to be true in distance hereditary graphs [10,21] and in a more general family of 3-Steiner
distance hereditary graphs [8]. A characterization of these graphs (in which S(W) ⊆ ∪u,v∈WI(u,v) for all W ⊆ V(G))
remains open, and seems to be quite difficult. In this paper we consider the following stronger condition: given a fixed k, for
any multiset W of vertices with |W| = k,
S(W) =
u,v∈W
We call this the union property of the k-Steiner interval. When k = 2 the union property trivially holds in all graphs. We
prove in Section 2 that for any k greater than 3, the union property holds precisely in block graphs. The case k = 3 turns out
to be the most difficult and interesting, see Section 3. (See also [9,13,20] for other studies on Steiner intervals.)
The second focus of this paper is on the concept of betweenness in graphs as introduced by Mulder in [18] (he implicitly
consideredthisnotionalreadyinhisbook[17]).Startingpointsofhisstudyweretwoverystrongpropertiesthatthegeodesic
interval I enjoys. Namely, (i) if x is between u and v (i.e. x ∈ I(u,v)) and x ?= u, then u is not between x and v, and, (ii) if x
is between u and v, and y is between u and x, then y is between u and v. Properties (i) and (ii) are usually denoted by (b1)
and (b2), respectively, and together they form betweenness axioms as defined in [18]. The interpretation of the betweenness
properties in this sense was first studied by Mulder and Morgana [15], where it was also proved that the induced path
interval J is a betweenness if and only if G is a house, hole, domino-free graph. A related property (not always satisfied by I)
is that if x and y are between u and v, and z is between x and y, then z is between u and v. This property is known as the
monotone axiom, which is again introduced formally in [18]. (In [26], van de Vel uses the term monotone law for what we
call the (b2) axiom.) The graphs in which the monotone axiom is always satisfied are known as interval monotone graphs
which were also introduced in [17], see also [2,14]. Clearly the monotone axiom always implies (b2), but the converse need
not hold and the characterization of interval monotone graphs is still an open problem. Betweenness in discrete structures
other than graphs has been studied much earlier, for example see [24].
As 2-Steiner intervals are precisely the geodesic intervals I, k-Steiner intervals form a generalization of the geodesic
interval, hence it is natural to look at the analogous concept of betweenness for k-Steiner intervals. The betweenness axioms
and the monotone axiom (m) can be generalized in a natural way from binary to k-ary functions (in particular from geodesic
intervals to k-Steiner intervals) as follows: for any u1,u2,...,uk,x,x1,x2,...,xk∈ V(G) which are not necessarily distinct,
(b1) x ∈ S(u1,u2,...,uk),x ?= u1⇒ u1?∈ S(x,u2,...,uk),
(b2) x ∈ S(u1,u2,...,uk) ⇒ S(x,u2,...,uk) ⊆ S(u1,u2,...,uk),
(m) x1,x2,...,xk∈ S(u1,u2,...,uk) ⇒ S(x1,x2,...,xk) ⊆ S(u1,u2,...,uk).
Somewhat surprisingly for the k-Steiner interval, where k > 2, the betweenness axioms are not satisfied in all graphs.
As we show in Section 3, in the case k = 3 the class of graphs in which the 3-Steiner interval has the union property (which
are the graphs in which each block is a clique or a 5-cycle) is properly contained in the class of graphs in which the 3-Steiner
interval satisfies the monotone axiom (m), which is in turn properly contained in the class of graphs in which the 3-Steiner
interval satisfies (b2). Example of graphs, for which the monotone axiom is satisfied for the 3-Steiner interval S but not
the union property are the graphs Mk, k ≥ 3, see Fig. 8. An example of a graph for which the 3-Steiner interval S satisfies
(b2), but not (m) is the famous Petersen graph, see Fig. 10. One can easily verify that (m) is not satisfied by the Petersen
graph. By using the labeling of vertices from Fig. 10, note that S(b,d,f) consists of all vertices in the graph except for w, yet
w ∈ S(x,y,z). Hence S(x,y,z) ?⊆ S(b,d,f), and (m) is not satisfied for S. On the other hand, for any k greater than 3 the
classes of graphs in which the k-Steiner interval satisfies the union property, the monotone axiom, and the (b2) axiom are
all the same, which is the main theorem in Section 2.
We conclude this section with the following lemma that considerably reduces the class of graphs in which the 3-Steiner
interval satisfies the union property. This property is even more restrictive for the k-Steiner interval where k > 3. Recall
that a subgraph H of a graph G is an isometric subgraph of G if for any pair of vertices u,v ∈ V(H), there exists a geodesic
?
I(u,v).

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Fig. 1. Forbidden isometric subgraphs from the lemma.
(i.e. a shortest path) in G between u and v that lies entirely in H. In other words, dH(u,v) = dG(u,v) for any u,v ∈ V(H),
where dGas usual denotes the shortest path distance in G (and similarly for dH). It is obvious that an isometric subgraph is
also an induced subgraph.
Lemma 1. Let G be a graph such that S(u,v,w) = I(u,v)∪ I(v,w)∪ I(u,w) for all (not necessarily distinct) u,v,w ∈ V(G).
Then G does not contain the diamond, C4, and Ct, for t ≥ 6 as an isometric subgraph. If, in addition, there exists an integer
k ≥ 4 such that S(u1,...,uk) = ∪i?=jI(ui,uj) for all (not necessarily distinct) u1,...,uk∈ V(G) then G does not contain an
induced C5.
Proof. For every graph from the list of forbidden subgraphs, one can find a triple u,v,w, such that I(u,w) is not a subset
of S(u,v,w) and so the condition from the lemma is violated. In Fig. 1 the appropriate u,v,w are depicted. Then S(u,v,w)
in each of the subgraphs is a path, while I(u,v) ∪ I(v,w) ∪ I(u,w) contains all vertices of the depicted subgraph. Vertex
x in each of these subgraphs belongs to I(u,w), and is not in S(u,v,w). This is clear in the case of K4− e and C4. Since
dC6({u,v,w}) = 3 and as x is not adjacent with either v or w, it is not on a Steiner tree for u,v, and w. Consider now
the vertices u,v, and w in the C7of Fig. 1. We show that dG({u,v,w}) = 4. Since these vertices form an independent
set, dG({u,v,w}) ≥ 3. However, no vertex is adjacent with all three of these vertices; otherwise, the C7in Fig. 1 is not an
isometric subgraph. So dG({u,v,w}) ≥ 4. Since the u − v path of length 2 of the C7in Fig. 1 followed by the v − w path of
length 2 is a tree with 4 edges that contains the vertices u,v, and w, it follows that dG({u,v,w}) ≤ 4. One can now argue
as for the C6that there is no vertex that is adjacent with both v and w and at least one of u or x, otherwise, the C7of Fig. 1 is
not an isometric subgraph.
Similarly, when G contains an isometric subgraph C2k : u1u2...u2ku1or C2k+1 : u1u2...u2k+1u1, for k ≥ 4, we take
u = u1,v = uk,w = uk+1, resp. w = uk+2in the odd case, and verify for x = u2k, resp. x = u2k+1, that x is not in S(u,v,w),
while clearly it is in I(u,w) = S(u,w,w). Since cycles are isometric one can infer that any Steiner tree that contains u,v,w
and x is of size at least k+ 2 (for odd cycles k+ 3), while there exists a Steiner tree for {u,v,w} of size k+ 1 (for odd cycles
k + 2). The details of the verification are left to the reader.
For the second part, consider the case when the k-Steiner interval satisfies the union property where k > 3. Note that
the diamond, and the cycles that cannot be isometric in the case k = 3 also cannot be isometric for k > 3. Let u1,...,u5
be the vertices of an induced 5-cycle in G, ordered in the natural way. Note that an induced C5is also isometric. Then
S(u1,...,u1,u2,u3,u4) = {u1,u2,u3,u4} while u5 ∈ I(u1,u4), which is a contradiction. Hence an isometric C5is not
possible in this case.
?
2. Case k > 3
A block of a graph is a maximal connected subgraph without a cut vertex. A graph G is called a block graph if and only if
every block of G is a clique. Note that block graphs are precisely chordal diamond-free graphs, hence they are also the graphs
in which there are no induced diamonds and no isometric cycles of length more than 3, cf. [4].
Lemma 2. Let G be a block graph, U = {u1,u2,...,uk} a multiset of vertices of V(G) and C = {c | c ?∈ U is a cut vertex of G
which lies on some shortest ui,uj-path}. Then S(u1,u2,...,uk) = U ∪ C.
Proof. Let G,U and C be as in the statement of the lemma. Let x ∈ U ∪ C. If x = uifor some i then obviously
x ∈ S(u1,u2,...,uk). Now, suppose x is a cut vertex of G which lies on some shortest ui,uj-path. Since uiand ujbelong
to different connected components of the graph G − x, x lies on every Steiner tree for U and is thus in S(U). Observe that
we can find a Steiner tree containing exactly the vertices of U and C, thus we derive that d(U) = k + |C| − 1. From this
we deduce that S(U) cannot include any additional vertex besides vertices of U and C, which completes the proof of the
lemma.
?
Theorem 3. Let G be a connected graph and k > 3. The following statements are equivalent:
(i) G is a block graph,
(ii) the k-Steiner interval on G satisfies (m),

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(iii) the k-Steiner interval on G satisfies (b2),
(iv) the k-Steiner interval on G satisfies the union property.
Proof. (i) ⇒ (ii). Let G be a block graph. We prove that S(x1,...,xk) ⊆ S(u1,...,uk) for every x1,x2,...,xk ∈ S(u1,
u2,...,uk). Let a ∈ S(x1,x2,...,xk). By Lemma 2, a is either a vertex from {x1,x2,...,xk}, or it is a cut vertex of G that lies
on a shortest path between two vertices from {x1,x2,...,xk}.
If a = xi, for some i, then a ∈ S(u1,...,uk) as desired, since xi∈ S(u1,u2,...,uk). So assume that a is a cut vertex lying
on a shortest path between xiand xjfor some i,j. We have the following three cases concerning xiand xjwith respect to
U = {u1,u2,...,uk}.
Case 1. xi= uk, and xj= u?, for some uk,u?∈ U.
In this case a ∈ S(u1,u2,...,uk) by Lemma 2.
Case 2. One of xiand xjis in U and the other is not.
We can choose the notation such that xi= umfor some um∈ U and xj?= u?, for any u?∈ U. Hence a lies on a shortest
path between xjand um. Since xj?= u?for any ?, xjis a cut vertex of G which lies on some shortest ui,uj-path. Hence G − xj
is a graph with at least two connected components, where uiis in one and ujin another. Assume without loss of generality
that umlies in some other component as ui. Then we have d(ui,um) = d(ui,xj)+d(xj,um) = d(ui,xj)+d(xj,a)+d(a,um) =
d(ui,a) + d(a,um), since a lies between xjand umand is thus in the same component as um. We derive that a is a cut vertex
of G that lies on a shortest ui,um-path and hence a ∈ S(u1,u2,...,uk).
Case 3. Both xiand xjare not from U.
Since xi,xj?∈ U, both xiand xjare cut vertices lying on some up,urand uq,usshortest path, respectively, which may be
distinct or the same. If both xiand xjare vertices lying on the same shortest path, say on the shortest path between upand
ur, then a is also a cut vertex lying on the same shortest path and the theorem is proved. Otherwise note that xiand xjare
not in the same block (since a is on the shortest xi,xj-path). Let upbe in a different connected component of G−xias xj, and
let uqbe in a different connected component of G−xjas xi. Then the vertices a,xi, and xjlie on a shortest up,uq-path, which
proves a ∈ S(u1,u2,...,uk).
(ii) ⇒ (iii) As noted in Section 1, this implication always holds.
(iii) ⇒ (i). Suppose the k-Steiner interval satisfies the (b2) axiom in G and suppose G is not a block graph. Then clearly
G contains an induced diamond or it contains an isometric Ct, t ≥ 4. First suppose G has an induced K4− e with vertices
u,w having degree 2 with respect to the diamond, and v is another vertex of the diamond. Then S(u,...,u,w) is not a
subset of S(u,...,u,v,w). Hence S does not satisfy (b2), a contradiction. Now suppose G has an isometric Ctwith vertices
u1,...,ut, t ≥ 4. If t = 2m then S(u1,u1,...,u1,um+1) is not a subset of S(u2,u1,...,u1,um+1), and if t = 2m + 1, then
S(u1,...,u1,um+1,um+2) is not a subset of S(u2,u1,...,u1,um+1,um+2); both are contradictions.
(i) ⇒ (iv) By Lemma 2, S(U) = U ∪ C. Note that U ∪ C equals?
i?=jI(ui,uj) for every multiset U = {u1,u2,...,uk} of vertices in G (and thus also for the
3-Steiner interval in which there are at most 3 distinct vertices in U). Hence by Lemma 1 we derive that every block in G
contains no isometric cycles of length greater than 3. Hence a shortest cycle in every block is a triangle, and since there are
noinduceddiamonds,andnogreaterisometriccycles(bythesamelemmaagain)oneeasilyinfersthateachblockisaclique.
Thus G is a block graph.
?
i?=jI(ui,uj), since every vertex in I(ui,uj) \ {ui,uj} for
i ?= j is a cut vertex and thus either in U or in C.
(iv) ⇒ (i) Suppose S(U) =?
3. Case k = 3
3.1. Graphs in which the 3-Steiner interval satisfies the union property
We begin this section with the structural result about 3-Steiner intervals in the class of geodetic graphs. A graph is called
geodetic if there is a unique shortest path (alias geodesic) between every pair of vertices. These graphs were considered by
several authors, see for instance [3,12,19,23,25].
Obvious examples of graphs, in which there is a unique geodesic between every pair of vertices, are odd cycles, trees and
complete graphs, but we shall come across several others during our study. Note that a graph is geodetic if and only if any
of its blocks is geodetic [25]. Now, take a triple of distinct vertices u,v,w, and consider the three geodesics between pairs
of the triple. By the structure of these graphs, the geodesics themselves form the corresponding intervals between the pairs.
Let u?be the last vertex that is common to the u − v and u − w geodesics (possibly u?= u), and similarly we define v?and
w?. Then there is a block B containing u?,v?and w?. Now, we are ready to state the following (straightforward) result which
has a similar role as Lemma 2, where Steiner intervals were studied in block graphs.
Lemma 4. Let G be a geodetic graph (i.e. a graph in which there is a unique geodesic between every pair of vertices). Let u,v, and
w be arbitrary distinct vertices of G, let u?, v?and w?be defined as above. Then S(u,v,w) = I(u,u?) ∪ I(v,v?) ∪ I(w,w?) ∪
S(u?,v?,w?).

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Fig. 2. An induced subgraph in Case 2.
It is easy to prove the lemma, so we will skip the proof. Note that in the above formula I(u,u?) is simply the geodesic
between u and u?. Although the above result is not interesting if there is only one block, it trivially holds for 2-connected
graphs. It also holds if two of the vertices from the triple u,v,w coincide.
For n,m ≥ 3 let Cm,ndenote the graph obtained from the cycles Cmand Cn, by amalgamating them along an edge of each
cycle. For example, C3,3is the diamond and C3,4is the house.
Theorem 5. Let G be a connected graph. Then S(u,v,w) = I(u,v) ∪ I(v,w) ∪ I(u,w) for all u,v,w ∈ V(G) if and only if
every block in G is either a complete graph or C5.
Proof. First, let G be a connected graph such that S(u,v,w) = I(u,v) ∪ I(v,w) ∪ I(u,w) for all u,v,w ∈ V(G). Let B be a
block in G different from K2. Therefore B contains a cycle. Let C be a shortest cycle contained in B. Note that C is then also
isometric in G. Hence by Lemma 1 it is only possible that C is isomorphic to C3or C5, hence we distinguish these two cases.
Case 1. C∼= C3. Let C?be a clique (maximal complete subgraph) that contains C. If C?equals B then the theorem follows.
Otherwise, let D be a shortest cycle in B which contains only two vertices of C?, say a and b. This cycle is again isometric.
Hence, by Lemma 1, it is isomorphic to C3or C5. Suppose D is isomorphic to C3. Then let a,b,d be its vertices, and let c be
a vertex of C?that is not adjacent to d. Such a vertex exists, since C?is a maximal complete subgraph. Then a,b,c,d induce
the diamond which is a contradiction with Lemma 1. Hence we may assume that D is isomorphic to C5.
Let c be a vertex of C?, distinct from a and b. We show that the union of D and c induces a subgraph, isomorphic to C3,5.
First it is clear that the neighbors a?and b?on D of a and b, respectively, are not adjacent to c, otherwise as above we obtain
a diamond. Also, the remaining vertex x of D cannot be adjacent to c, because that would imply that C5is not the shortest
cycle, containing two vertices from C?(namely x,c,b and b?would form a C4). Thus V(D)∪{c} induces a C3,5. Also V(D)∪{c}
induces an isometric subgraph, except in the case when there is a path of length 2 between x and c. Hence there are two
possibilities for induced subgraphs that can appear, see Fig. 5 where both graphs are depicted. First, in the graph on the left
side of Fig. 5, the Steiner interval S(a,c,x) is of size four, and so one easily finds that b and b?cannot be in S(a,c,x). On
the other hand b,b?∈ I(x,c), which yield a contradiction with the hypothesis. For the graph on the right side of Fig. 5, the
Steiner interval S(a?,b?,c) contains all vertices of the subgraph, while y ?∈ I(a?,b?) ∪ I(a?,c) ∪ I(b?,c).
We conclude that in the case C∼= C3, the block B is isomorphic to the complete graph C?.
Case 2. C∼= C5. If C is equal to B, the theorem follows. Otherwise, let D be a shortest cycle in B which contains at least two
vertices of C, and is not equal to C. Note that it could contain also three vertices of C. This cycle is also isometric, hence the
subgraph induced by its vertices can only be isomorphic to C5by Lemma 1 and minimality of C.
First let us consider the case when D shares three vertices with C. Then the graph in Fig. 2 is a subgraph of B. Note that
this is an induced subgraph in G. Indeed, there can be no edges between two vertices of C (respectively D), since they are
isometric cycles, and also there can be no edges between b (or c) and e (or f ) because we would get a shorter cycle than C5.
Note, as above, that the size of the Steiner tree for a, e and c is 5, and so all vertices are contained in S(a,e,c). On the other
hand, x is not in any of the three intervals, since we have d(a,e) = d(a,c) = d(c,e) = 2, and the subgraph is induced.
In the rest of the proof of this direction, we consider the case when D shares exactly two (adjacent) vertices with C. Then
?C ∪ D? contains a C8where a pair of antipodal vertices of the C8is adjacent. One can quickly check that only one other edge
in this subgraph is possible (using the fact that C and D are isometric and that C5is a shortest cycle), and this is the edge
between another pair of antipodal vertices that are not adjacent to the vertices of C ∩ D.
Suppose that this edge exists, see Fig. 3, where the graph that appears as an induced subgraph is depicted. Let U =
{u,v,w}. Then the Steiner distance d(U) of U is 4 (it cannot be 3, since, say u andv cannot have another common neighbor),
hence S(u,v,w) contains all vertices of this subgraph. Now, x is not in I(u,v) ∪ I(v,w) ∪ I(u,w), otherwise the subgraph
of Fig. 3 is not induced.
Now, suppose that the union of C and D yields an induced subgraph (see the graph C5,5as depicted in Fig. 4(a)). Let u,v
and w be vertices of this subgraph as depicted in this figure. Let U = {u,v,w}. Note that d(U) ≤ 4. On the other hand,
since the subgraph is induced, d(U) ≥ 3. If d(U) = 3, then u and v have a common neighbor z, or u and w have a common
neighbor y (z ?= y since there are no triangles). In the first case we obtain as a subgraph in G the graph of Fig. 4(b). This is
also an induced subgraph in G which can be easily checked. Now, vertices a,b,c,d,u,z andv induce a subgraph isomorphic

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Fig. 3. An induced subgraph in Case 2.
abcd
Fig. 4. Induced subgraphs in Case 2.
to the graph of Fig. 2, which we have shown is not possible. Suppose u and w have a common neighbor y, as shown in
Fig. 4(c). Since we have shown that u and v have no common neighbor, dG(u,v) ≥ 3. So c ∈ I(u,v). But c ?∈ S(u,v,w), a
contradiction. Thus d(U) = 4.
Now,considerthedistanceinGbetweenuandw.Itcannotbeequalto1or2aswehavealreadyobserved.IfdG(u,w) = 3,
then, since the graph of Fig. 4(b) is not possible, G must contain the graph of Fig. 4(d) as a subgraph. One can check that this
subgraph is induced otherwise one of the earlier situations arises. In this case the vertices p and r belong to I(u,w) but they
cannot belong to the Steiner interval S(u,w,q). In the last case, when dG(u,w) = 4 (see the graph of Fig. 4(a)), b ∈ I(u,w)
but b ?∈ S(u,v,w), a final contradiction, and the proof of this direction is completed.
For theconverse, letGbe agraph inwhich everyblock is eithera completegraph orC5. Thenit is easyto seethat forevery
two vertices u,v ∈ V(G) there is a unique shortest path connecting u andv. So G is a geodetic graph. Consider three vertices
u,v,w ∈ V(G), and let u?,v?,w?be as defined prior to Lemma 4. Then by this lemma, S(u,v,w) is the union of the vertices
on the u,u?-geodesic, those on the v,v?-geodesic, those on the w,w?-geodesic and those in S(u?,v?,w?). The structure of
the Steiner interval S(u?,v?,w?) depends on whether B is a complete graph or C5and on whether some (or all) of u?,v?and
w?coincide, and can be easily analyzed. (For instance, if B is C5, then S(u?,v?,w?) is a path, if vertices of the triple lie on some
P3of the cycle, otherwise S(u?,v?,w?) is the entire cycle.) In all cases S(u?,v?,w?) = I(u?,v?) ∪ I(v?,w?) ∪ I(u?,w?) which
implies S(u,v,w) = I(u,v) ∪ I(v,w) ∪ I(u,w), as desired.
?
3.2. Graphs in which the 3-Steiner interval satisfies (m)
In this section we aim to characterize the graphs in which the 3-Steiner interval satisfies the (m) axiom (respectively,
the (b2) axiom). It turns out that the class of graphs with (b2) is strictly larger than the class of graphs with (m). We will
characterize the latter class of graphs, and at the same time look at the former (for which we will present some partial
observations that will lead to a conjecture about their structure).
Lemma 6. Let G be a graph in which the 3-Steiner interval satisfies(b2) axiom. Then G does not contain the diamond, C4, and Ct,
for t ≥ 6 as an isometric subgraph.
Proof. WerefertoFig.1andusesimilarargumentsasinLemma1.Thecondition(b2)isnotsatisfiedifGcontainsaninduced
subgraph isomorphic to a C4or a diamond as labeled in Fig. 1, since S(u,u,w) = I(u,w) is not a subset of S(v,u,w). For
cycles Ct, t ≥ 6 we can use the same notation as in Lemma 1. In the same way we derive that S(u,u,w) = I(u,w) is not a
subset of S(u,v,w) which concludes the proof.
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Fig. 5. Graphs C3,5and M3.
Fig. 6. A graph from the proof of Lemma 9.
Graphs C3,5and M3, depicted in Fig. 5, will play an important role in what follows. We will also use the notion of a
g-convex set in a graph G, which is a set of vertices W ⊆ V(G) such that for any u,v ∈ W, I(u,v) ⊆ W. Clearly, an
intersection of g-convex sets in a graph G is also a g-convex set, and the smallest g-convex set that contains a set U ⊂ V(G)
is called the convex hull of U.
Lemma 7. Let G be a graph in which the 3-Steiner interval satisfies condition (b2) and let H be a subgraph of G, isomorphic to
C3,5. Then the convex hull of H is either the complete graph, or H is an induced subgraph in G and its convex hull is isomorphic
to M3.
Proof. Let G be a graph in which the 3-Steiner interval satisfies the (b2) axiom. Suppose that there is a subgraph H in G
isomorphictoC3,5.SupposethattheconvexhullofH isnotaclique.ThenitisnothardtoseethatH isinducedinGotherwise
we obtain an isometric C4or diamond which is impossible by Lemma 6. Note that the distance between every two vertices
of H is at most two in H, except the distance between c and x which is 3 in H. We show that H cannot be isometric in G. Let
U = {a,c,x}. First observe that d(U) = 3 and that b ?∈ S(a,c,x). On the other hand b ∈ S(c,c,x), which is a contradiction
to the assumption that (b2) is satisfied in G. Hence dG(c,x) = 2, thus there exists a common neighbor y of c and x. The
resulting graph (isomorphic to M3) is g-convex in G, since if we connect any two vertices in M3that are at distance 2 by a
path of length 2 whose internal vertex is not in M3, we obtain a forbidden C4or diamond.
We follow with another property of graphs with (b2).
?
Lemma 8. Let G be a graph in which the 3-Steiner interval satisfies the condition(b2) and let a 6-cycle C be an induced subgraph
of G. Then every pair of antipodal vertices in C has a common neighbor (and all three neighbors are pairwise different).
Proof. Let C : abcdefa be a 6-cycle in G. Suppose (without loss of generality) that a and d do not have a common neighbor.
Then dG(a,d) = 3, hence S(a,a,d) contains all vertices of C. But S(a,b,d) does not contain f . Hence the triple a,b,d does
not satisfy the condition (b2), a contradiction. Hence every pair of antipodal vertices in C has a common neighbor not on C.
Obviously all of them are pairwise different, otherwise we obtain an isometric C4or a diamond as induced subgraph which
is impossible by Lemma 6.
?
In the case of (m), Lemma 8 can be further strengthened.
Lemma 9. Let G be a graph in which the 3-Steiner interval satisfies condition (m). Then G does not contain an induced 6-cycle.
Proof. Suppose that C is an induced 6-cycle in G with vertices a,b,c,d,e,f . By Lemma 8 (we can use it since if a graph
satisfies condition (m) it satisfies also the condition (b2)) a and d have a common neighbor x, b and e have a common
neighbor y, and c and f have a common neighbor z. It is easy to check that the 6-cycle bczfeyb is induced in G, otherwise a
contradiction to Lemma 6 arises. Hence by Lemma 8, y and z have a common neighbor w, see Fig. 6.
Let U = {a,c,e}. We now show that d(U) = 4. It is easy to see that d(U) ?= 2 since C is induced, and d(U) ?= 3 otherwise
a,c, and e would have a common neighbor which would lead to an isometric C4or diamond, contrary to Lemma 6. From

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abc
Fig. 7. Cases from Lemma 10.
this it is easy to see that S(a,c,e) includes every vertex depicted in Fig. 6 except the vertex w (otherwise w would have
to be adjacent to at least one of the vertices a,c,e, but this again eventually leads to a contradiction with Lemma 6). Let
W = {y,z,e}. Observe that d(W) ≥ 3, since z cannot be adjacent to y or e. Also, since eywz is the path that contains all
vertices of W, we infer d(W) = 3. Hence S(y,z,e) is not contained in S(a,c,e) which contradicts the assumption that G
satisfies the condition (m).
?
Lemma 10. Let G be a graph in which the 3-Steiner interval satisfies condition(m). Then G does not contain the graph C5,5as an
induced subgraph (see Fig. 4 (a)).
Proof. Suppose that the graph isomorphic to the graph in Fig. 4(a) is an induced subgraph of G. Let u,v and w be vertices
of this graph as labeled in Fig. 4(a), and let U = {u,v,w}. Note that d(U) ≤ 4. Since the C5,5is induced, d(U) ≥ 3.
If d(U) = 3, then u,v,w have a common neighbor x or only u and w have a common neighbor y or only u and v have a
common neighbor z, see Fig. 7.
In the first case when the graph in Fig. 7(a) appears, we can (using Lemma 6) easily check that either this graph is induced
or the edge ax exists. If the graph in Fig. 7(a) is an induced subgraph of G, then also the 6-cycle xwebaux is induced, a
contradiction with Lemma 9. If the edge ax exists, then by Lemma 7 there exist vertices s and t, a common neighbor of u
and e, and d and w, respectively (note that s and t cannot coincide since otherwise we would obtain a forbidden isometric
4-cycle xutwx). But then the 6-cycle sudtwes is induced. Indeed, the following edges are not possible:
• sw (and ut by symmetry), otherwise swxus is an isometric 4-cycle, since the edge sx cannot exist,
• sd (and et by symmetry), otherwise the 6-cycle sdcvwes would be induced (s cannot be adjacent to any of c,v,w,
otherwise we obtain either an isometric 4-cycle or diamond),
• st, otherwise s,t,d,u induce a C4,
• uw, ue, dw and de, since, by the assumption, the graph in Fig. 4(a) is induced.
This implies a contradiction with Lemma 9.
Now suppose that the graph from Fig. 7(b) appears (where y and v are not adjacent, otherwise we obtain a graph
isomorphic to the graph 7(a)). Note that y and c cannot be adjacent either. If yd is not an edge, then vertices y,u,d,c,v,w
would induce a 6-cycle, a contradiction to Lemma 9. But if there is the edge yd, we get the graph from Fig. 7(a) for which we
have already proved that it is not possible.
The last case to consider is the case when the graph from Fig. 7(c) appears. In this case (since we assume that z andw are
not adjacent) two situations are possible: either the graph from Fig. 7(c) is an induced subgraph of G or there is an edge az.
In the former case, also uabcvzu is an induced 6-cycle, and if az exists then azvweba is an induced 6-cycle, a contradiction
to Lemma 9.
We derive that d(U) > 3, and so d(U) = 4. Hence dG(u,w) ≥ 3. If dG(u,w) = 3 this implies that the graph from Fig. 4(d)
is a subgraph of G (since the graph from Fig. 7(c) is forbidden), in which vertices p and r belong to I(u,w) = S(u,u,w)
but they cannot belong to S(u,w,q). In the case when dG(u,w) = 4, the graph in Fig. 4(a) is a subgraph of G, with
b ∈ I(u,w) = S(u,u,w) but b ?∈ S(u,v,w), the final contradiction.
Now we introduce graphs Mn, n > 1. They can be constructed from the complete graph Knwith vertices x1,...,xn
and a star K1,nwith x as its center and y1,...,ynas its leaves, by adding an edge between yiand xifor i = 1,...,n. It is
straightforward to verify that in these graphs, called Mn, the 3-Steiner interval satisfies (m) and hence (b2), see Fig. 8.
?
Theorem 11. Let G be a connected graph. Then the 3-Steiner interval satisfies condition (m) in G if and only if each block of G is
either a complete graph, or a graph isomorphic to Mnfor n ≥ 2.
Proof. LetGbeaconnectedgraphinwhichthe3-Steinerintervalsatisfiesthecondition(m),thatis,foranyu1,u2,u3∈ V(G)
and any x1,x2,x3∈ S(u1,u2,u3), we have S(x1,x2,x3) ⊆ S(u1,u2,u3).

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Fig. 8. Graphs with condition (m): M2, M3, M4, ....
Fig. 9. Graphs in the proof of Theorem 11.
By Lemma 6 G does not contain any C4, diamond or Ct, t ≥ 6, as an isometric subgraph. Let C be a smallest cycle C in a
block B of G. Then C must be a triangle or a 5-cycle.
Case 1. C∼= C3. The proof of this case is the same as the proof of Case 1 of Theorem 5 up to the case where we encounter the
graph C3,5from Fig. 5. In this case, we use Lemma 7 and derive that M3as labeled in Fig. 5 appears as an induced subgraph
of G where C = abca. Let C?be a maximal clique that contains C. If there is a vertex d ∈ C?different from a,b,c, then again
by Lemma 7, there must be a vertex y?adjacent to both d and x. If y = y?, then the subgraph induced by b,c,d and y is the
diamond which is forbidden by Lemma 6. So y ?= y?. By repeating the same argument for every vertex c?
distinct vertex y?
size of the clique C?.
Now we prove that the resulting graph Mncoincides with the entire block B. Suppose to the contrary that there are some
other vertices in B. Hence, let Q be a shortest path between two vertices of Mnsuch that no vertex of Q (except its end
vertices) is a vertex of Mn. Because of the symmetry in Mnit is enough to look at the graph M3and distinguish between the
following possibilities:
iin C?, there exists a
i?= y in D which is a common neighbor of c?
iand x. Thus we obtain the graph Mn(see Fig. 8) where n is the
(i) Q is a path between vertices b and c. By Lemma 6, Q has length 4. Let its interior vertices be b??,x?and c?as labeled
in Fig. 9(i). We claim that the set of vertices S = {c,b,b?,x,y,b??,x?,c?} induces a graph isomorphic to C5,5. First note
that by Lemma 7 there exists y?that is a common neighbor of a and x?. Again by Lemma 7 we only have to check the
nonexistence of edges starting in c?,x?, or b??and ending in y,x, or b?. However all these edges cannot exist by the
minimality of Q. Now we arrive at a contradiction since by Lemma 10, C5,5cannot be induced in G.
(ii) Q is a path between vertices b and b?. By Lemma 6 and minimality, Q has length 2 or 4. Suppose first that its length
is 2, and let d be a common neighbor of b and b?. By Lemma 7 there exists a vertex z that is a common neighbor of d
and a?and again by the same lemma there exists v, a common neighbor of c and z, see the graph in Fig. 9(ii). Note that
ycvza?xy is a 6-cycle, and let H be the subgraph induced by its vertices. If any one of the edges yz,xv, and xz is in H,
it is a part of an induced C4or diamond which is forbidden. By Lemma 7 also all other edges of H are forbidden with
the possible exception of yv. Suppose yv ∈ E(H). But then yvzdb?xy is another 6-cycle which is induced contrary to
Lemma 9. Indeed, all other edges in H are forbidden either by Lemma 7 or they force an induced C4or a diamond (we
leave the details to the reader).
Suppose now that the length of Q is 4. Note that vertices of Q ∪ {a,a?,x} induce a subgraph that contains C5,5as
spanning subgraph. By Lemma 10, G cannot contain C5,5as an induced subgraph. Thus there must be an edge from a?
to a central vertex of Q. But this contradicts the minimality of Q.
(iii) Q is a path between vertices b?and x. Again the length of Q can only be 2 or 4 by the same reasons as above. Suppose
Q has length 2. Let d be the central vertex of Q. By Lemma 7 there exists a vertex z that is a common neighbor of a and
d. Then acyxda is a 6-cycle, and let H be the subgraph induced by its vertices. By Lemma 9, H is not induced, and using

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Lemma 7 we infer that only the edges yd,zc,cd, and zy could be added between vertices of H. However, the first two
edges force a diamond, while cd and zy force an induced C4or diamond, a contradiction in each case.
Suppose now that the length of Q is 4. Then vertices of V(Q)∪{y,c,b} induce a subgraph that is isomorphic to C5,5,
which cannot be induced by Lemma 10. Hence there must be an edge from c to the central vertex of Q, contrary to the
minimality of Q.
(iv) Q is a path between vertices b and x. By Lemma 6 and minimality of Q the length of Q is 3. Let x?and b??be the neighbors
of x and b, respectively, on Q. Then a?abb??x?xa is a 6-cycle in G which is either induced (which is a contradiction to
Lemma 9), or it is not induced (in which case we arrive at a contradiction to the minimality of Q). Hence such a path
cannot exist.
(v) Q is a path between vertices b?and c. In this case the length of Q must again be 3 (by Lemma 6 and minimality of Q).
Let c?and b??be the neighbors of c and b?, respectively, on Q. We derive that cc?b??b?xyc is a 6-cycle in G, and conclude
similarly as in the previous case that this is not possible.
(vi) Q is a path between vertices b?and a?. By the same reasoning as in case (iv) we find that the length of Q must be 3. Let
a??and b??be the neighbors of a?and b?, respectively, on Q. Now aa?a??b??b?ba is a 6-cycle in G and we conclude similarly
as in the previous two cases. Hence B is isomorphic to Mn.
Case 2. C∼= C5. If C is equal to B, the theorem is proved. Otherwise, let D be a shortest cycle in B that contains at least two
vertices of C, and is not equal to C (note that it might contain also three vertices of C). This cycle is also isometric, hence it
can only be isomorphic to C5by Lemma 6 and minimality of C.
First let us consider the case when D shares three vertices with C. Then the graph in Fig. 2 is a subgraph of G. In addition,
there can be no edges between two vertices of C (respectively D), since C and D are isometric cycles. Also there can be no
edges between b (or c) and e (or f ), as labeled in Fig. 2, because this would imply that there is a shorter cycle in B as C5. Hence
the graph in Fig. 2 is an isometric subgraph of G. But then abcdefa is an induced 6-cycle, a contradiction to Lemma 9.
The second and final case is that D shares exactly two (adjacent) vertices with C. Let H be the subgraph of G induced by
V(C) ∪ V(D). As C5,5cannot be an induced subgraph in G (by Lemma 10), we derive that the graph in Fig. 3 is a spanning
subgraph of H, and is in fact isomorphic to H (since no additional edges are possible between vertices in that figure). But
then H contains the graph from Fig. 2 as an induced subgraph, which in turn contains an induced 6-cycle, a contradiction
with Lemma 9.
For the converse, we will use Lemma 4. Notably, if every block of G is a complete subgraph or isomorphic to Mnthen
there is a unique geodesic between every pair of vertices in G (i.e. G is a geodetic graph). We also know that if G is Mn,
then the monotone condition is satisfied. Using Lemma 4 and the definition prior to this lemma, for any triple of vertices
u,v,w ∈ V(G) we have S(u,v,w) = I(u,u?) ∪ I(v,v?) ∪ I(w,w?) ∪ S(u?,v?,w?), where u?,v?w?all lie in the common
block B. Let x,y,z ∈ S(u,v,w). If x,y,z all belong to either I(u,u?), or I(v,v?) or I(w,w?), then S(x,y,z) will be a subset of
I(u,u?), or I(v,v?) or I(w,w?) as the case may be, and hence S(x,y,z) is a subset of S(u,v,w). If x,y,z ∈ S(u?,v?,w?), then
S(x,y,z) is contained in the block B and is also a subset of S(u,v,w) since every block of G satisfies (m). If two of x,y,z,
say x,y ∈ I(u,u?) (without loss of generality we can assume that y ∈ I(x,u?)) and z ∈ I(v,v?), then S(x,y,z) consists of
I(x,u?) ∪ I(z,v?) and a subset of S(u?,v?,w?) (which can be a proper subset or the whole of S(u?,v?,w?), depending upon
the structure of S(u?,v?,w?) and the nature of the vertices x,y,z). In this case also S(x,y,z) is contained in S(u,v,w). The
cases when x ∈ I(u,u?),y,z ∈ S(u?,v?,w?) and x ∈ I(u,u?),y ∈ I(v,v?),z ∈ S(u?,v?,w?) can be handled similarly as
above, and are left to the reader. The final case is when all of x,y,z lie in different intervals, say x ∈ I(u,u?), y ∈ I(v,v?) and
z ∈ I(w,w?). In this case, we have S(x,y,z) = I(x,u?)∪ I(y,v?)∪ I(z,w?)∪ S(u?,v?,w?). Here also, S(x,y,z) ⊆ S(u,v,w).
Thus the 3-Steiner interval satisfies condition (m) and the proof of the theorem is complete.
?
4. Concluding remarks
1. The most evident problem that arises from this paper is a structural characterization of graphs in which the 3-Steiner
interval satisfies (b2) axiom. Lemmas 6–8 already give a lot of information about the structure of these graphs. The main
distinction with axiom (m) comes in Lemma 9, where it is proved that C6cannot be an induced subgraph of graphs satisfying
(m). In the case of (b2) the 6-cycle can be induced, and the graph from Fig. 6 is a subgraph of the Petersen graph. It is
straightforward to check that in the Petersen graph the 3-Steiner interval satisfies the (b2) axiom. Moreover, we think
that this holds in all geodetic graphs with diameter 2, which is a much larger class of graphs (see Stemple [25] for a
characterization of these graphs and more examples). We think that the converse could also be true, and state this as the
following conjecture.
Conjecture 12. Let G be a connected graph. The 3-Steiner interval satisfies the condition (b2) in G if and only if each block of G
is a geodetic graph with diameter at most 2.
2. The Steiner set in a graph G is defined as a set W ⊆ V(G) whose Steiner interval S(W) equals V(G). As mentioned in
the introduction, an erroneous statement from [6] encouraged some further investigation in the area. The statement was
that every Steiner set is a geodetic set (i.e. a set W of vertices in a graph such that every vertex of a graph lies on a geodesic
interval between two vertices from W). However this is true in some graphs, but it is an open problem to determine a

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Fig. 10. The Petersen graph satisfies (b2), but not (m).
characterization for these graphs. A related (stronger) question could be more easy to solve (based on the results of this
paper): In which graphs G, for every set W ⊂ V(G), the Steiner interval S(W) is included in the union of geodesic intervals,
that is
S(W) ⊆
ui,uj∈W
We could also restrict the question to k-Steiner intervals for particular integers k, where k > 2. Also, the reversed inclusion
would be interesting: In which graphs G, for every set W ⊂ V(G),?
3.Wecaneasilyconsiderthe(b1)axiominthewaytheotherconditionswerehandledinthispaper.LetGbeagraphonat
least three vertices. For a complete graph G, we have S(u1,u2,...,uk) = {u1,u2,...,uk} and hence S satisfies (b1). On the
other hand, if G is a connected graph on at least 3 vertices that is not complete, then there exist vertices u,v,w with v ?= u
such that vu,uw ∈ E(G) but vw ?∈ E(G), then S(u,v,...,v,w) = {u,v,w} and thus v ∈ S(u,v,...,v,w) = {u,v,w},
but we have u ∈ S(v,v,...,v,w), hence (b1) is not satisfied.
Remark 13. Thek-Steinerinterval,wherek > 2,ofaconnectedgraphGsatisfiesthe(b1)axiomifandonlyifGisacomplete
graph.
?
I(ui,uj).
ui,uj∈WI(ui,uj) ⊆ S(W)? Observe that, for|W| = 3, the
Mn’s from Section 3 satisfy this property.
ThequestionremainsifonecancharacterizethegraphsinwhichtheSteinerintervalsatisfiesotherbetweennessaxioms,
as introduced in [15].
4. One could compare the Steiner intervals also with other natural binary intervals. In particular, the main question from
this paper could be posed for monophonic (induced path) interval J. It was shown by Hernando et al. [10] that every Steiner
set in a connected graph is a monophonic set — that is, vertices of a Steiner set W have the property that every vertex in G
lies on an induced path between two vertices from W. However the following natural question is open: in which graphs do
we have that for every set W
?
Furthermore, what is the relation of the resulting class with the house, hole, domino-free graphs? The latter class of graphs
was studied also with respect to another type of convexity (namely m3convexity), based on the so-called m3intervals [7].
5. When W is a set instead of a multiset, then the classes of graphs, in which the k-Steiner interval for W has the union
property, satisfies the monotone axiom (m), and satisfies betweenness axiom (b2), are in general not the same as what we
have obtained in this paper. For example, it can be easily verified that in any cycle Cnthe k-Steiner interval for k-sets satisfies
the monotone axiom (m) and hence the (b2) axiom. Also there are graphs other than complete graphs in which the k-Steiner
interval satisfies (b1) axiom; for instance the stars, the paws etc. Thus when no repetitions of the vertices are allowed, the
corresponding problem for k-Steiner intervals is another interesting problem. However the class of graphs in which the
3-Steiner interval satisfies the union property is the same class as what we obtained for the union property in this paper.
That is, Theorem 5 holds also for 3-Steiner intervals S(u,v,w) with distinct u,v,w.
S(W) =
ui,uj∈W
J(ui,uj)?
Acknowledgements
We are grateful to the anonymous referees for their numerous valuable comments.
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