Broadcasts in Graphs: Diametrical Trees
∗
L. Gemmrich
†‡and C.M. Mynhardt
‡Department of Mathematics and Statistics
University of Victoria, Victoria, BC, Canada
lgemmrich@gmail.com
;
kieka@uvic.caAbstract
A dominating broadcast on a graph = ( ) is a function : → {0 1 diam()} such that () ≤ () (the eccentricity of ) for all ∈ , and each ∈ is at distance at most () from a vertex with () ≥ 1. The upper broadcast domination number of is Γ() = max{P∈ () : is a minimal dominating broadcast on }. As shown by
Erwin in [D. Erwin, Cost domination in graphs, Doctoral dissertation, Western Michigan University, 2001], Γ() ≥ diam() for any graph .
We investigate trees whose upper broadcast domination number equal their diameter and, among more general results, characterize caterpillars with this property.
Keywords: broadcast on a graph, dominating broadcast, minimal dominating broadcast, up-per broadcast domination number
AMS 2010 Subject Classification Number: 05C69, 05C05, 05C12
1
Introduction
Suppose a telecommunications company has to provide radio coverage to a collection of geo-graphic regions. A single tower transmitting with a strength (or cost) of one unit can provide coverage to the region it is located in and all regions immediately adjacent to it. The company aims to minimize its expenses by erecting as few towers as possible. If we consider each region as a vertex of a graph , where two vertices are adjacent if their corresponding geographic regions are adjacent, then any dominating set (i.e. each vertex of belongs to or is adja-cent to a vertex in ) represents a suitable arrangement of radio towers, and a dominating set of minimum cardinality represents a minimum cost arrangement. However, if the company is able to build its towers with varying signal strength so that a tower may transmit its signal a greater distance, but at a proportionally greater cost, the total cost could be significantly less than for the former arrangement. This situation can be modelled with a broadcast on , as defined below.
∗To appear in the Australasian Journal of Combinatorics †Undergraduate research student
Unless stated otherwise, all graphs considered here are assumed to be simple, nontrivial and connected. For undefined graph theoretic concepts and terminology we refer the reader to [7] and [13].
A caterpillar is a tree of order at least three, the removal of whose leaves produces a path. We use standard notation for functions and write : → to denote the fact that is a function from to ; we also write = {( ()) : ∈ }. If and are functions with the same domain such that () ≤ () for each ∈ , we write ≤ . If in addition () () for at least one ∈ , we write .
As usual we denote the domination and upper domination numbers of a graph by () and Γ(), respectively. A broadcast on a graph = ( ) is a function : → {0 1 diam()} such that () ≤ () (the eccentricity of ) for all ∈ . A broadcast on is dominating if each ∈ is at distance at most () from a vertex with () ≥ 1, and minimal dominating if no broadcast 0 on with 0 is dominating. The cost of a broadcast is ( ) = P
∈ ().
The broadcast domination number of is
() = min{() : is a dominating broadcast on }
and the upper broadcast domination number of is
Γ() = max{() : is a minimal dominating broadcast on }
Broadcast domination was introduced by Erwin [11, 12], who proved the bounds
()≤ min{() rad()} ≤ max{Γ() diam()} ≤ Γ() (1)
for any graph . Graphs for which () = rad() are called radial graphs. Radial trees are
characterized in [16, 17]. The upper broadcast domination number Γ() is also studied in
[1, 2, 10, 21]. Other studies of broadcast domination can be found in [3, 4, 5, 6, 8, 9, 14, 15, 18, 19, 20, 22, 23, 24, 25].
Our purpose is to investigate trees whose upper broadcast domination number equals their diameter. Following the terminology for broadcast domination numbers, we call such trees diametrical trees. The characterization of diametrical trees is listed as an open problem in [21]. After presenting further definitions and known results in Section 2, we state a number of lemmas concerning properties of non-diametrical trees in Section 3. To avoid interrupting the flow of the proof of our main theorem, we defer the proofs of all lemmas to Section 5. A con-sequence of these lemmas is that a tree containing a path of length at least three, internally disjoint from a diametrical path, is non-diametrical. This result hints that the caterpillars may contain classes of diametrical trees, which is indeed the case. Our goal is to prove the charac-terization of diametrical caterpillars stated in Theorem 1.1 below, which we do in Section 4. We conclude with open problems in Section 6.
Theorem 1.1 A caterpillar with diametrical path : 0 1 is diametrical if and only
if
() each ∈ {1 − 1}, is adjacent to at most two leaves,
() for any ∈ {1 − 2}, min{deg() deg(+1)} = 2,
() whenever and , are adjacent to at least two leaves each, there exists an index
2
Definitions and Known Results
For a broadcast on a graph = ( ), define + = { ∈ : () 0}. The vertices in + are called broadcast vertices. A vertex hears the broadcast from some vertex ∈ +, and -dominates , if the distance ( ) ≤ (). An edge hears if both and hear from the same vertex ∈ +. A vertex ∈ + overdominates a vertex if ( ) (). For ∈ +, define the
• -neighbourhood of as [] ={ ∈ () : ( ) ≤ ()}
• -boundary of as () ={ ∈ () : ( ) = ()}
• -private neighbourhood of as PN() ={ ∈ [] : ∈ [] for all ∈ +− {}},
• -private boundary of as PB() ={ ∈ [] : is not dominated by ( −{( ())})∪
{( () − 1)}}.
Note that if () = 1, then PB() = PN(), and if () ≥ 2, then PB() = ()∩
PN(). For example, consider the tree in Figure 1. The broadcast defined by () =
4 () = 2 () = 3 () = 1 and () = 0 otherwise is a dominating broadcast such that PB() ={0} for each ∈ { }, and PB() = ∅.
The property that makes a dominating broadcast minimal dominating, determined in [11] and stated in [21] in terms of private boundaries, is essential in the study of upper broadcast numbers. We state it again here.
Proposition 2.1 [11] A dominating broadcast is a minimal dominating broadcast if and only if PB()6= ∅ for each ∈ +.
By Proposition 2.1 the broadcast in Figure 1, although dominating, is not minimal dom-inating. The broadcast 0 = ( − {( 2)}) ∪ {( 0)} is a minimal dominating broadcast on .
In general it is not true that if is a dominating broadcast on a graph , then some broadcast 0 with 0 ≤ is a minimal dominating broadcast on , nor is it necessarily true that if is a
broadcast on such that PB()6= ∅ for each ∈ +, then some broadcast 0 with ≤ 0 is
a minimal dominating broadcast on . Consider the tree and broadcast shown in Figure 2. Here, PB() = {0} for each ∈ { } and is not -dominated. Moreover, cannot
4 w' u' v w 1 u z' 2 z = 3 T
Figure 1: A tree with a dominating broadcast such that PB() = {0} for each ∈
4 v' v w 1 u 3 z 2 T y w'
Figure 2: A tree with a broadcast such that PB()6= ∅ for each ∈ +, but cannot
be extended to a minimal dominating broadcast.
be extended to a broadcast that dominates without leaving or with an empty private boundary.
It is well known that any independent set of vertices in a graph can be extended to a maximal (but not necessarily maximum) independent set of , and that a maximal independent set is also a minimal dominating set (cf. [13, pp. 70 — 71]). Denoting the cardinality of a maximum independent set of by (), it follows that () ≤ Γ() for all graphs .
Remark 2.2 [11] The characteristic function of a minimal dominating set in a graph is a minimal dominating broadcast on . Hence Γ()≥ Γ() ≥ () for any graph .
Proposition 2.3 [11] If is a broadcast on a graph and for each ∈ {1 2} we have ∈ +,
0 ∈ PB(), where 1 6= 2, and is a — 0 geodesic, then 1 and 2 are disjoint.
Using Proposition 2.3, Erwin [11] shows that Γ() ≤ |()| for any graph , and together
with the lower bound (1) this implies that Γ() = − 1 for each ≥ 2. Proposition 2.3 is
used frequently in the proofs in Section 5.
3
Non-Diametrical Trees
In this section we state a number of sufficient conditions for a tree to be non-diametrical. The proofs are given in Section 5. We assume throughout that has diameter and a diametrical path : 0 1 . For each ∈ {0 }, let be the subtree of induced by all vertices
that are connected to by paths that are internally disjoint from . Note that = 1 if and
only if ∈ {0 }, or ∈ {1 − 1} and deg() = 2. For example, in the tree in Figure
3, 2 ∼= 13 4 ∼= 2 6 ∼= 4 and = 1 for ∈ {0 1 3 5 7 8}.
A stem of a tree À 2 is a vertex adjacent to a leaf and a strong stem is a stem that is
adjacent to at least two leaves; in Figure 3, 1 4 and 6 are (not the only) examples of stems.
The complete bipartite graph 1 ≥ 1, is also called a star. Thus a tree with diametrical
path as above is a caterpillar if each is either a star or 1.
Lemma 3.1 Let be a tree with diameter ≥ 3 and diametrical path : 0 1 . If
there exists an ∈ {1 − 2} such that each of and +1 is adjacent to a leaf other than
T
v
v0 2 v4 v6 v8
Figure 3: A tree and diametrical path : 0 1 8 such that 2 ∼= 13 4 ∼= 2 6 ∼= 4
and = 1 otherwise.
Lemma 3.2 If there exists a subscript ∈ {2 − 2} such that has an independent set
of cardinality 3 that dominates but does not contain , or if max{deg(1) deg(−1)} ≥ 4, then
Γ( ) diam( ).
Lemma 3.3 If there exists a subscript ∈ {2 − 2} such that has an independent set
of cardinality 2 that does not dominate , then Γ( ) diam( ).
Lemma 3.4 If diam()≥ 4 for some , or if diam() = 3 and is a peripheral vertex of ,
then Γ( ) diam( ).
By Lemmas 3.2 — 3.4, if is a diametrical tree, then each is isomorphic to either 1, 2,
3 with either a leaf or the stem of 3, or 4 with being a stem of 4. Thus, diametrical
trees are “nearly” caterpillars. We henceforth restrict our investigation to caterpillars. By Lemma 3.1, if ∼= 2, we may assume that neither −1 nor +1 is isomorphic to 2. If
∼= 3 with being a leaf of 3, or if ∼= 4, then is not a caterpillar and we ignore these
cases. We give one more sufficient condition for a caterpillar to be non-diametrical.
Lemma 3.5 Let be a caterpillar with diametrical path : 0 1 . If two vertices
+2 are strong stems, for some ≥ 1 and some integer such that + 2 ≤ − 1, and +2
is a stem for each ∈ {1 − 1}, then Γ( ) .
4
Diametrical Caterpillars
If is a diametrical caterpillar, then does not satisfy the hypothesis of any of Lemmas 3.1 — 3.5. In this section we show that the converse is also true: If the caterpillar does not satisfy the hypothesis of any of Lemmas 3.1 — 3.5, then is diametrical. The negation of these hypotheses, applied to caterpillars, gives the characterization of diametrical caterpillars stated in Theorem 1.1, which we restate here for convenience.
Theorem 1.1 A caterpillar with diametrical path : 0 1 is diametrical if and
only if
() each ∈ {1 − 1}, is adjacent to at most two leaves,
() whenever and , are strong stems, there exists an index , such
that deg() = deg(+1) = 2.
Proof. Suppose is a diametrical caterpillar. By Lemma 3.2, each ∈ {2 − 2}, is
adjacent to at most two leaves, while 1 and −1 are adjacent to at most one leaf other than
0 and , respectively, hence () holds. Similarly, condition () follows directly from Lemma
3.1. For (), condition () already implies that of any two consecutive internal vertices of , at least one has degree 2. Lemma 3.5 now implies that if and are both strong stems, then
some pair of consecutive strong stems between and (inclusive) are separated by at least
two vertices of degree 2. Hence () holds.
For the converse, note that the only caterpillars of diameter three or less that satisfy con-ditions () — () are 3 4 and the tree obtained by joining a new leaf to a stem of 4. It
is easy to verify that they are diametric. Assume that Theorem 1.1 is false and let be a smallest non-diametrical caterpillar that satisfies () — (). Then has diameter at least four. We state two more lemmas, the proofs of which are also given in Section 5.
Lemma 4.1 No vertex of is a strong stem.
Lemma 4.2 No vertex ∈ {2 − 2}, is adjacent to a leaf.
By Lemmas 4.1 and 4.2, = +1, which is impossible because Erwin [11] showed that
Γ() = − 1 = diam() for all ≥ 2. ¥
5
Proofs of Lemmas
This section contains the proofs of Lemmas 3.1 — 4.2, restated here for convenience.
Lemma 3.1 Let be a tree with diameter ≥ 3 and diametrical path : 0 1 .
If some and +1, ∈ {1 − 2}, are adjacent to leaves other than 0 or , then
Γ( ) diam( ).
Proof. Suppose the hypothesis of the lemma is satisfied. Say is adjacent to the leaf and
+1 is adjacent to the leaf 0. Define the broadcast by (0) = + 1, () = − and
() = 0 otherwise. Then ∈ PB(0) and 0 ∈ PB(), hence PB()6= ∅ for all ∈ +. If
is also dominating, let = ; otherwise, let 0 be the subgraph of induced by all vertices
that are not -dominated, let be a maximal independent set of 0 and define the broadcast by () = () if ∈ ( ) − (0), () = 1 if ∈ and () = 0 if ∈ (0)− . By
definition, is a dominating broadcast on . Since and 0are leaves, no vertex in is adjacent
to or 0, hence ∈ PB
(0) and 0 ∈ PB(). Since no vertex in hears the broadcast ,
∈ PB() for each ∈ . Hence, by Proposition 2.1, is a minimal dominating broadcast.
Moreover, ( ) ≥ + 1 + − = + 1 and the result follows. ¥ The proof of the next lemma is illustrated in Figure 4.
Lemma 3.2 If there exists a subscript ∈ {2 − 2} such that has an independent set
of cardinality 3 that dominates but does not contain , or if max{(1) (−1)} ≥ 2, then
v v0 v1 2 v6 1 4 (b) v v0 2 v3 v6 1 2 v4 (a) 2 1 1 1 1 1 1 (c) v v0 2 v3 v6 1 1 v4 2 1 1 1
Figure 4: An illustration of the proof of Lemma 3.2.
Proof. We may assume that does not satisfy the hypothesis of Lemma 3.1, otherwise we are done. Suppose (1) = ≥ 2. See Figure 4(a). Since 0 is a peripheral vertex of , no
vertex of 1 is at distance greater than one from 1. Hence 1 = 1and, by Lemma 3.1, 2 is
not adjacent to a leaf. Let be the set consisting of 0 and the leaves of 1, and define the
broadcast by () = − 2, () = 1 if ∈ and () = 0 otherwise. Then 2 ∈ PB()and
∈ PB() for each ∈ , hence PB()6= ∅ for all ∈ +. If is dominating, let = ,
otherwise let 0 be the subgraph of induced by all vertices that are not dominated by .
Since 2 is not adjacent to a leaf, there exists a maximal independent set of 0 that does not
contain a vertex adjacent to 2. Define the broadcast by () = () if ∈ ( ) − (0),
() = 1if ∈ and () = 0 if ∈ (0)− . Then
2 ∈ PB()and ∈ PB()for each
∈ ∪ , so is a minimal dominating broadcast on with () ≥ + 1 + − 2 . Hence Γ( ) .
If (−1)≥ 2 the result follows similarly. Hence assume some ∈ {2 − 2}, has
an independent set of cardinality 3 that dominates but does not contain . Then has a
maximal independent set of cardinality ≥ 3 such that ∈ . Define the broadcast by
(0) = − 1, () = − − 1, () = 1 if ∈ and () = 0 otherwise. Since ∈ ,
−1∈ PB(0) and +1 ∈ PB(). In addition, ∈ PB() for each ∈ . If ≥ 3 and −1
is adjacent to a leaf, then we may assume, by Lemma 3.1, that −2 is not adjacent to a leaf (other than 0 if = 3). Similarly, if ≤ − 3 and +1 is adjacent to a leaf, we may assume
that +2 is not adjacent to a leaf (other than if = − 3). Let 0 be the subgraph of
induced by the vertices that are not dominated by and choose a maximal independent set of 0 as follows.
• If 0 has a maximal independent set that does not contain a vertex adjacent to
−1or to
+1, let be such a set. See Figure 4(b).
• If each maximal independent set of 0 contains a vertex adjacent to
then −1 (or +1) is adjacent to a leaf. Then −2(or +2) is not adjacent to a leaf, and
there exists a maximal independent set of 0 that contains no vertex adjacent to −2 (or +2); let be such a set. See Figure 4(c).
Define the broadcast on as follows. If neither −1 nor +1 is adjacent to a leaf, let
() = ⎧ ⎨ ⎩ () if ∈ ( ) − (0) 1 if ∈ 0 otherwise.
Then is a dominating broadcast such that ( ) ≥ − 1 + − − 1 + , −1∈ PB(0),
+1∈ PB() and ∈ PB() for each ∈ ∪ .
If −1 is adjacent to a leaf and +1 is not, let
() = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ − 2 if = 0 () if ∈ ( − 0)− (0) 1 if ∈ 0 otherwise.
Then || ≥ 1 and −1 hears from an adjacent leaf. Hence is a dominating broadcast such
that ( ) ≥ − 2 + − − 1 + + || −2∈ PB(0) +1 ∈ PB() and ∈ PB()
for each ∈ ∪ .
Similarly, if +1 is adjacent to a leaf and −1 is not, let
() = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ − − 2 if = () if ∈ ( − )− (0) 1 if ∈ 0 otherwise.
Finally, if both −1 and +1 are adjacent to leaves, define by
() = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ − 2 if = 0 − − 2 if = () if ∈ ( − {0 }) − (0) 1 if ∈ 0 otherwise.
Now || ≥ 2 and is a dominating broadcast such that () ≥ − 2 + − − 2 + + || ≥ − 4 + + || , −2∈ PB(0) +2∈ PB() and ∈ PB()if ∈ ∪ .
Hence in each case is a minimal dominating broadcast such that ( ) , which implies that Γ( ) diam( ) ¥
Lemma 3.3 If there exists a subscript ∈ {2 − 2} such that has an independent set
of cardinality 2 that does not dominate , then Γ( ) diam( ).
Proof. Suppose has an independent set of cardinality 2 that does not dominate . If
from Lemma 3.2. Hence assume this is not the case (in particular, is not a stem) and let
be a maximal independent set of cardinality ≥ 2 of − containing no vertex adjacent to
. Define the broadcast on by (0) = () = − − 1 () = 1 for each ∈ and
() = 0 otherwise. Note that ∈ PB(0), +1 ∈ PB() ∈ PB() for each ∈ and
() ≥ + − − 1 + . We can now proceed as in the proof of Lemma 3.2 to construct a minimal dominating broadcast on such that ( ) ≥ () to obtain that Γ( ) .
The details are omitted.¥
Lemma 3.4 If diam()≥ 4 for some , or if diam() = 3 and is a peripheral vertex of
, then Γ( ) diam( ).
Proof. If diam() ≥ 5, then contains a subgraph isomorphic to 6, which, regardless of
which vertex of 6 corresponds to , has an independent set of cardinality 3 that dominates but
does not contain , and the result follows from Lemma 3.2. If diam() = 4and corresponds
to a stem of a subgraph isomorphic to 5, the result follows similarly.
Suppose diam() = ∈ {3 4} and is a peripheral vertex of . Then is not a stem.
Let be a vertex of at distance from . Define the broadcast on by () = (0) =
− 1 () = − − 1 and () = 0 otherwise. Then ∈ PB() −1 ∈ PB(0) and
+1∈ PB(), while () = − 1 + − − 1 + . Possibly −1 or +1 is a stem, or both
are. We proceed as in the proof of Lemma 3.2 to show that Γ( ) .
Finally, suppose diam() = 4 and is the central vertex of a subgraph ∼= 5 of .
Let 1 and 2 be the leaves of and let be the stem of adjacent to 2. If is a
stem of the result again follows from Lemma 3.2, hence assume is not a stem. Define
the broadcast by (1) = 2 (2) = 1 (0) = − 1 () = − − 1 and () = 0
otherwise. Then ∈ PB(1) ∈ PB(2) −1 ∈ PB(0) and +1 ∈ PB(), while
() = − 1 + − − 1 + 3 . As before it (eventually) follows that Γ( ) .¥
Lemma 3.5 Let be a caterpillar with diametrical path : 0 1 . If two vertices
+2 are strong stems, for some ≥ 1 and some integer such that + 2 ≤ − 1, and
+2 is a stem for each ∈ {1 − 1}, then Γ( ) .
Proof. Let be the set of leaves adjacent to +2 ∈ {0 1 }, and = {+1 +3
+2−1}. Then ∪ is independent. By the hypothesis, || ≥ + 3 and so | ∪ | ≥ 2 + 3. By Lemma 3.1 we may assume that deg() = 2 for each ∈ , otherwise the result follows.
If = 1 and + 2 = − 1, then ∪ is a maximal independent set of of cardinality at least + 1. Let be the characteristic function of ∪ .
If = 1 and + 2 − 1, define the broadcast on by () = 1 if ∈ ∪ , () = − − 2 − 1 and () = 0 otherwise. Then ∈ PB() for each ∈ ∪ and
+2+1 ∈ PB(). Since +2 is a stem, we may assume that deg(+2+1) = 2, otherwise the
result holds by Lemma 3.1. Therefore is a dominating broadcast, thus a minimal dominating broadcast, and ( ) = | ∪ | + − − 2 − 1 ≥ + 1.
If 1 and + 2 = − 1, reverse the direction of and proceed as above. Hence assume 1 + 2 − 1. See Figure 5, where = 10 = 3 and = 2. As above we may assume that deg(−1) = deg(+2+1) = 2. Define the broadcast by (0) = − 1 () =
− − 2 − 1 () = 1 for each ∈ ∪ and () = 0 otherwise. Then is a dominating broadcast such that ( ) ≥ − 2 − 2 + 2 + 3 −1 ∈ PB(0) +2+1 ∈ PB() and
v v0 3 v5 v7 v10 2 2 1 1 1 1 1 1 1 v2 v8
Figure 5: An illustration of the proof of Lemma 3.5.
v v0 r f 4 b v0 vr u 5 b u g
Figure 6: A step in the proof of Lemma 5.1.
∈ PB() for each ∈ ∪ . Hence is a minimal dominating broadcast of such that
( ) . The result now follows. ¥
Before proving Lemmas 4.1 and 4.2 we state and prove two additional lemmas. If is a broadcast on and 0 is a subtree of , we define the restriction of to 0 to be the broadcast
0 = » 0 with +
0 = +∩ (0) and 0() = ()for all ∈ (0).
Lemma 5.1 Suppose is a smallest non-diametrical caterpillar that satisfies Theorem 11() — (). Let be a minimal dominating broadcast on such that ( ) diam( ). Then 0 ∈ +
or {0} = PB() for some ∈ +, and a similar result holds for .
Proof. Suppose the conclusion is false and say ∈ + broadcasts to 0, where 6= 0. Since
{0} 6= PB(), there exists ∈ PB()− {0}. Possibly is a leaf adjacent to 1, in which
case 0 ∈ PB() diam( − ) = diam( ) and is a minimal dominating broadcast on − .
But then − satisfies Theorem 1.1() — () and Γ( − ) diam( − ), contradicting the
choice of . Hence assume is not a leaf adjacent to 1.
Let ≥ 1 be the largest index such that lies on the − 0 path in . Possibly =
, otherwise is a leaf adjacent to . Since 0 is a peripheral vertex, broadcasts to all
vertices of for each = 0 , and each vertex in each such is overdominated by
. Therefore ∈ () for some . In addition, if lies on , then is not a stem,
otherwise the leaves adjacent to are not -dominated. Therefore also broadcasts to each vertex of each for ≤ ≤ . See Figure 6. But then the broadcast defined by (0) =
()− ( ) + (0 ) () = 0and () = () otherwise is also a dominating broadcast
such that ∈ PB(0) and PB() = PB() for all ∈ + − {0}, that is, is a minimal
dominating broadcast. Now () ≥ () − ( ) + (0 ) ≥ () − 1 + 1 = (). Hence
() = ( )if and only if = 1 and is a leaf adjacent to 1. In this case, − 0 also satisfies
() — () diam( − 0) = diam( ), and is also a minimal dominating broadcast on − 0,
contradicting the choice of . Hence () ( ) and we again have a contradiction, because ( ) = Γ( ) and no minimal dominating broadcast has cost greater than Γ( ). This proves
Lemma 5.2 Let be a smallest non-diametrical caterpillar that satisfies Theorem 11() — () and be a minimal dominating broadcast on such that ( ) diam( ). Then each leaf ∈ {0 } of is either a broadcast vertex or PB() ={} for some ∈ +.
Proof. Suppose the conclusion is false and ∈ {0 } is a leaf of that is neither a broadcast
vertex nor the only vertex in the private boundary of some ∈ +. Then − is a tree with
diameter that satisfies () — (), and is a minimal dominating broadcast on − as well, contrary to the choice of . ¥
We now return to Lemmas 4.1 and 4.2.
Lemma 4.1 If is a smallest non-diametrical caterpillar that satisfies Theorem 11() — (), then no vertex of is a strong stem.
Proof. Suppose, to the contrary, that some vertex of is a strong stem. Then = for
some , since is a caterpillar. Say is adjacent to the leaves and 0. Let be a minimal
dominating broadcast on such that ( ) diam( ). By Lemmas 5.1 and 5.2 we may assume that each leaf of is either a broadcast vertex, or the only vertex in the -private boundary of some vertex in +. Let be the vertex that broadcasts to .
Suppose 6= . Then PB() = {}. If 6= 0, then ( ) = ( 0) and we also have
0 ∈ PB
(), contrary to Lemma 5.2. Hence = 0 (0) = 2 and PB(0) = {}. Let
1 and 2 be the subtrees of − that contain 0 and , respectively. If ∈ {1 − 1},
assume without loss of generality that = − 1 and ignore 2. Since diam( ) ≥ 4, 1 is
nontrivial. By Theorem 1.1(), −1 is not a stem of , hence diam(1) = − 1. Since 0
broadcasts to −1 and PB(0) ={}, −1 also hears from some vertex ∈ +− {0}. Since
∈ PB(0), ∈ { } ∪ (2), hence ∈ (1). By Proposition 2.3 applied to 0 ∈ +,
PB()⊆ (1). Therefore » 1 is a minimal dominating broadcast on 1.
• If −2is not a stem of , then either −2 is adjacent to only one leaf in 1, namely −1,
in which case 1 satisfies Theorem 1.1() — (), or −2is adjacent to the two leaves −1
and 0 in 1, in which case 1 ∼= 3.
• On the other hand, if −2 is a stem of , then by Theorem 1.1() and the fact that is
adjacent to two leaves, −2 is adjacent to exactly one leaf in , so that it is adjacent to two leaves in 1. If −2 is the only strong stem of 1, then 1 satisfies Theorem 1.1() —
(). Hence suppose that for some 0 − 2, 0 is a strong stem (of 1 and of ). Since
() holds for , and deg(−2) deg() 2, there exists an index 0 − 2,
such that deg() = deg(+1) = 2. Therefore 1 satisfies Theorem 1.1() — () in
this case as well.
By the choice of , Γ(1) = diam(1) = − 1 in all cases. Since » 1 is a minimal
dominating broadcast on 1, ( » 1)≤ − 1. Similarly, if 1 , 2 satisfies Theorem
1.1() — () and » 2 is a minimal dominating function of 2 such that ( » 2) ≤
diam(2) = − − 1. But then () = ( » 1) + ( » 2) + 2 ≤ (or () = ( »
1) + 2≤ − 2 + 2 = , if = − 1), which is a contradiction because is non-diametrical.
Hence we may assume that = ; that is, is a broadcast vertex. If broadcasts to 0, we
0 without broadcasting to ). Then
∈ PB ()for each ∈ +. We may now define 1 and
2 as above and proceed as before to obtain a contradiction. ¥
Lemma 4.2 If is a smallest non-diametrical caterpillar that satisfies Theorem 11() — (), then no vertex ∈ {2 − 2}, is adjacent to a leaf.
Proof. Suppose, to the contrary, that some ∈ {2 −2}, is adjacent to a leaf and let
be the largest index in {2 −2} such that is a stem. By Lemma 4.1 we may assume that
has no strong stems. By Theorem 1.1(), deg(−1) = deg(+1) = 2. Let be the leaf
adjacent to and let be a minimal dominating broadcast on such that ( ) diam( ).
By Lemmas 5.1 and 5.2 we may assume that each of and is either a broadcast vertex or
the only vertex in the -private boundary of some vertex in +. We consider several cases. In each case we delete an edge to obtain subtrees of , each of which contains at most one strong stem. Since satisfies Theorem 1.1() — (), so do the subtrees. By the choice of , each subtree thus obtained is diametrical. We omit these details in the cases for the sake of brevity. Case 1 belongs to a private boundary and ∈ +. Then either ∈
+
and ∈ PB(), or
PB() ={} for a vertex 6= .
Case 1(a) {} = PB(). Then () = − + 1 and broadcasts to −1. Hence −1
does not belong to the private boundary of any vertex in +. Therefore −1 also hears from a vertex in +− {}. Also, { −1} ∩ + = ∅. Let 0 be the subtree of − −1 that
contains 0. For each vertex ∈ +∩ (0), Proposition 2.3 applied to and implies that
PB()⊆ (0). Therefore » 0 is a minimal dominating broadcast on 0. By the choice of
, ( » 0)≤ diam(0) = − 1. But now () = ( » 0) + (
)≤ − 1 + − + 1 = , a
contradiction.
Case 1(b) ∈ PB() 6= (possibly = ). Then broadcasts to , hence ∈ PB ().
By Proposition 2.3 and the choice of as the largest index such that 6= −1 is a stem, there
exists an index such that ∈ PB() (and thus () = − ). Evidently, then, the
edge −1 does not hear from any vertex in +. Let 0 be the subtree of − −1 that
contains 0. As in Case 1(a) we see that » 0 is a minimal dominating broadcast on 0.
If = + 1, then broadcasts to and but not to +1. (This is only possible if =
and () = 1.) In this case, diam(0) = + 1and (
) = − = − − 1.
If +1, i.e., −1 ≥ +1, then diam(0) = −1. In either case we obtain a contradiction
as before as in Case 1(a).
Case 2 ∈ PB() and ∈ PB(). By Lemma 5.1, 6= .
Case 2(a) ∈ PB(). Then () = 1. If PB() ={−1 }, delete the edge −2−1
and proceed as in Case 1(b) to get a contradiction. Thus, assume PB() ={}. Then −1
hears from some other vertex as well, hence » ( − )is a minimal dominating broadcast
on − . By the choice of , ( » ( − ))≤ − 1 and so () ≤ , a contradiction.
Case 2(b) {} = PB()for some 6= . Since does not broadcast to , Proposition 2.3
and the choice of imply that = for some ≥ + 1. Since ∈ PB(), () = − . Let
00 be the subtrees of −
−1 that contain 0 and , respectively. Then diam(00) = −
and +∩ (00) ={}. Since ≤ , () = − ≤ − .
If = , then is a path from to ∈ PB(). Otherwise, and, by
definition of , { } ⊆ PN(). In either case, Proposition 2.3 again implies that
PB()⊆ (0) for each ∈ +− {}. Hence » 0 is a minimal dominating broadcast on
0, so that by the choice of , ( » 0)≤ diam(0). If = + 1, then diam(0) = , while if + 1, then diam(0) = − 1. In either case () ≤ + (
)≤ , a contradiction.
Case 3 is a broadcast vertex and PB() ={} for some vertex 6= .
Case 3(a) PB() = {}. Then () = − + 1 ≥ 3. Let be the − path in and
let ∈ + − {}. Then ∼= ()+1. By Proposition 2.3, ∈ () for some ≤ − 1.
Also, PB()∩ ( ) = ∅. Thus, if 0 is the subtree of − −1 that contains 0, then
diam(0) = − 1 and » 0 is a minimal dominating broadcast on 0, which is a diametrical tree. Now ( ) = ( » 0) + ()≤ − 1 + − + 1 = , a contradiction.
Case 3(b) ∈ PB() and PB() = {}, where ∈ { } and 6= . By Proposition
2.3, = for some ≥ + 1. We now proceed as in Case 2(b) to obtain a contradiction.
Case 4 and are both broadcast vertices. If () = 1, then ∈ PB(). This is Case
2(a), hence assume () ≥ 2. Then PB() ={} for some such that + 1 ≤ ≤ − 2.
Evidently, then, the edge = −1 does not hear from any vertex. By deleting we proceed
as before to obtain a contradiction.
Since Cases 1 — 4 and their subcases cover all possibilities for and , the lemma follows.¥
This concludes the proofs of Lemmas 3.1 — 4.2, hence the proof of Theorem 1.1 is complete.
6
Open Problems
A characterization of diametrical caterpillars is presented in Theorem 1.1. In general, diametri-cal trees can have paths of length one or two, but not longer paths, that are internally disjoint from a diametrical path.
Problem 1 Characterize diametrical trees that contain at least one path of length two internally disjoint from a diametrical path.
Problem 2 Characterize trees with () Γ( ) = ( ) () Γ( ) = Γ( ).
Problem 3 Study other classes of graphs such that Γ() = diam().
Problem 4 [21] Determine the maximum ratio Γ()Γ() for () general graphs, () trees.
The stars 1 satisfy diam(1) = 2 and Γ(1) = , hence the ratio Γ() diam() is
unbounded.
The proof of Lemma 3.1 suggests the following problem.
Problem 5 If and are graphs and is an isometric subgraph of , is it true that Γ() ≤ Γ()?
Acknowledgements
The authors are indebted to the referees for several corrections and improvements to the paper. In particular, Problem 5 was suggested by one of them.
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