On the number of resolving pairs in graphs
Peter Dankelmann
Department of Pure and Applied Mathematics
University of Johannesburg
PO Box 524
Auckland Park 2006
South Africa
email: [email protected]
phone: +27-11-5594320
fax: +27-11-5592874
May 14, 2014
1
On the number of resolving pairs in graphs
Peter Dankelmann, University of Johannesburg
Abstract
Let G be a connected graph. An unordered pair {x, y} of vertices of
G is a resolving pair if no vertex of G has the same distance to x and y.
It was conjectured in [8] that the number of resolving pairs of a connected
2
graph of order n is bounded above by b n4 c. In this note we show that there
exists a constant c > 0such that for sufficiently large n every graph of order
n contains at most n2 − cn3/2 resolving pairs. We further show that the
exponent 3/2 is best possible by exhibiting an infinite sequence of graphs
with at least n2 − c1 n3/2 log n resolving pairs, where c1 is a positive constant,
thus disproving the above conjecture.
Keywords: distance; resolving pair; resolving set; metric dimension
1
Introduction and main result
A vertex v of a connected graph G is said to resolve a pair x, y of vertices if d(v, x) 6=
d(v, y). A set W of vertices of G is a resolving set of G if every pair of vertices of
G is resolved by some vertex of W . The metric dimension of a connected graph
G is defined as the minimum cardinality of a resolving set. So if W is a resolving
set of G, then every vertex of G can be uniquely identified from its distances to all
vertices of W . Resolving sets and the metric dimension have applications in sensor
placement in networks, where a contamination can be located if the sensors are able
to detect the distance to the contamination, and the locations of the sensors form a
resolving set.
Resolving sets and the metric dimension of connected graphs were introduced by
Slater [7], Harary and Melter [4], and Chartrand, Eroh, Johnson and Oellermann
[3]. An excellent survey of results on the metric dimension can be found in [1]. For
more results see, for example, [2, 5, 6].
In the context of metric dimension the notion of a resolving pair was introduced
by Tomescu and Imran [8]. An unordered pair of, not necessarily adjacent, vertices
{x, y} of a graph G is a resolving pair if every vertex of G resolves the pair {x, y},
i.e., if d(v, x) 6= d(v, y) for every vertex v of G. In the context of sensor placement in
networks, a sensor that can detect the distance to a contamination, will distinguish
between a contamination in x and a contamination in y, no matter in which vertex
it is placed.
It is clear that in general not every pair of vertices of a graph is a resolving pair,
for example if d(x, y) is even, then {x, y} is not a resolving pair of G.
In connected bipartite graphs the resolving pairs are the pairs of vertices from
2
different partite sets (see [9]), hence a bipartite graph of order n has at most b n4 c
resolving pairs. In [8], Tomescu and Imran showed that this bound on the number
2
of resolving pairs also holds for graphs of diameter 2, and conjectured that it holds
for all connected graphs.
Conjecture 1. (Tomescu, Imran [8])
If G is a connected graph of order n, then the number of resolving pairs of G is at
2
most b n4 c. Equality holds if and only if G = Kbn/2c,dn/2e .
In this note we show that only a much weaker upper bound on the number
of resolving pairs in a graph of given order holds. We show that there exists a
constant c > 0 suchthat for sufficiently large n every connected graph of order n
contains at most n2 − cn3/2 resolving pairs. The exponent 3/2 is best possible;
this
by by exhibiting an infinite sequence of graphs with at least
is demonstrated
n
3/2
−
c
n
log
n
resolving
pairs, where c1 is a positive constant. The latter result
1
2
disproves the above conjecture, and shows that there
exists no > 0 such that every
n
connected graph of order n has at most (1 − ) 2 resolving pairs.
Our notation is as follows. If G is a finite graph then n(G) denotes the order
of G, i.e., the number of vertices in G. We consider only connected graphs. The
distance dG (u, v) between two vertices u and v of G is the minimum number of edges
on a u − v path. We omit the argument or subscript G where the graph is clear
from the context. The eccentricity of a vertex v is the distance to a vertex farthest
from v, and the diameter of G is the largest of all distances between vertices of G.
If f (x) and g(x) are two real functions, then we write f = O(g) if there exists a
constant c ∈ R such that f (x) ≤ cg(x) for all sufficiently large x.
We begin with an upper bound on the number of resolving pairs of a graph in
terms of its order.
Theorem 1. For
every > 0, every connected graph of sufficiently large order n
has at most n2 − ( 41 − )n3/2 resolving pairs.
Proof: Let G be a connected graph of order n and diameter d and let > 0 be
arbitrarily small.
√
Case 1: d ≥ 2 n + 2.
√
Then each vertex v has√eccentricity at least 12 d ≥ n + 1 and is hence at even
distance from at least 21 n other vertices. Summing over all v ∈ V (G), and noting
that each pair
of vertices at even distance is counted twice, we conclude that G has
1 √
at least 4 n n pairs of vertices at even distance. Since a pair of vertices at even
distance
is not a resolving pair, the number of resolving pairs of G is not more than
n 3/2
n
− 4 n , as desired.
2
√
Case 2: d < 2 n + 2.
Let v be a vertex of eccentricity d and let Vi be the set of vertices at distance exactly
i from v for i = 0, 1, . . . , d. Then no pair of vertices u, w ∈ Vi form a resolving pair
since d(u, v0 ) = d(w, v0 ) = i. Hence
X
d n
|Vi |
|R| ≤
−
.
2
2
i=1
3
P
Now the real function f (x) = x2 is convex up, and the |Vi | satisfy di=1 |Vi | = n − 1
P
and |Vi | ≥ 1 for i = 1, 2, . . . , d. Hence, by Jensen’s inequality the sum di=1 |V2i |
is minimised subject to the above constraints if all |Vi | equal n−1
. Therefore, the
d
number of pairs which are not resolving in G is at least
n − 1 n − 1
(n − 1)/d
n − 1n − 1
1
√
d
=
−1 ≥
− 1 ≥ ( − )n3/2 ,
2
2
d
2
4
2( n + 1)
2
provided n is sufficiently large. The theorem follows.
In the rest of this note we show that the above bound is close to being best
possible, in the sense that the exponent 3/2 in the above bound cannot be improved.
Given a positive integer r ≥ 2, we recursively define an infinite sequence H1 , H2 , . . .
i
of graphs as follows. For i = 1, 2, . . . , r let Pi : v1i , v2i , . . . , v2i
be disjoint paths of
order 2i. Let H1 be the graph obtained from the union of the Pi by identifying their
initial vertices v11 , v12 , . . . , v1r to a new vertex u, and adding edges between all of its
neighbours v21 , v22 , . . . , v2r so that the neighbours of u form a complete graph. We root
r
H1 at u and declare the terminal vertices of the Pi , i.e., the vertices v21 , v42 , . . . , v2r
2
to be special vertices. Then H1 is a graph of order r + 1 with r special vertices and
rooted at u.
r−1
0
1
For k ≥ 2 we construct Hk as follows. Let Hk−1
, Hk−1
, . . . , Hk−1
be r disjoint
copies of Hk−1 , and for i = 1, 2, . . . , r − 1 let Pi : v1i , v2i , . . . , v2irk−1 be a path of order
i
2irk−1 . For i = 1, 2, . . . , r − 1 we join the terminal vertex v2ir
k−1 of Pi to the root of
r−1
1 2
i
Hk−1 , identify the initial vertices v1 , v1 , . . . , v1 of the paths Pi , i = 1, 2, . . . , r − 1,
0
with the root of Hk−1
to obtain a vertex u at which we root the new graph Hk .
Furthermore we add edges between all neighbours of u to make the neighbourhood
of u complete. We define each vertex that is special in one of the copies of Hk−1 to
be special in Hk . Figure 1 shows how for r = 2 the graph H3 is obtained from two
copies of H2 and a path of order 8. The root is solid black, and the special vertices
are solid grey.
H21
H22
Figure 1: The graph H3 for r = 2. The two copies of H2 are indicated.
4
Lemma 1. Let k ∈ N and Hk the graph defined above and let Sk be the set of special
vertices of Hk . Then
(i) n(Hk ) = krk+1 − (k − 1)rk + 1.
(ii) |Sk | = rk .
(iii) Every pair of special vertices of Hk is resolving.
Proof: (i) Denote the order of Hk by hk for k = 1, 2, . . .. Then it follows from
the construction of Hk that the hk satisfy the linear difference equation
hk = rhk−1 − r + 1 +
r−1
X
2irk−1 = rhk−1 − r + 1 + rk (r − 1)
i=1
for k = 2, 3, . . .. It is easy to verify that the above difference equation, in conjunction
with h1 = r2 + 1, has the unique solution
hk = krk+1 − (k − 1)rk + 1,
which is (i).
(ii) Let sk = |Sk | for k ∈ N. From the construction of Hk we get that sk = rsk−1 for
k ≥ 2. Since s1 = r, we conclude that sk = rk for k ∈ N, which is (ii).
(iii) Let v and w two distinct special vertices of Hk and let Pvw be the unique shortest
v − 2 path. Then Pvw has odd length and no vertex of Pvw has the same distance to
w
v
, whose elements
and Vvw
v and w. We partition the vertex set of Pvw into sets Vvw
are the vertices of Pvw closer to v than to w, and the vertices of Pvw closer to w than
to v, respectively. It is easy to verify that every component, C say, of Hk − V (Pvw )
v
w
has neighbours only in one of the two sets Vvw
or Vvw
. If some vertex of C has a
v
neighbour in Vvw , then all vertices of C are closer to v than to w in Hk , and if some
w
, then all vertices of C are closer to w than to v in
vertex of C has a neighbour in Vvw
Hk . Hence no vertex has the same distance to v and w, and so {v, w} is a resolving
pair of Hk .
2
Theorem 2. There exists a constant c ∈ R and an infinite
sequence
√ of connected
n
3/2
log n resolving
graphs such that every graph G of the sequence has at least 2 −cn
pairs.
Proof: We fix a constant r ∈ N with r ≥ 2 and let ck = krk for every k ∈ N.
Define Gk to be the graph obtained from Hk by attaching a set Rv of ck leaves to
each special vertex v of Hk , where the sets Rv are pairwise disjoint. If hk and sk
denote the number of vertices and the number of special vertices of Hk respectively,
and if we denote the order of Gk by nk , then we have for large k that
nk = hk + sk ck = (krk + kr − k + 1)rk + 1 = kr2k + O(krk ).
We now show that almost all pairs of vertices of Gk are resolving. Clearly,
2 1
nk
1
1
1
1 2k
= n2k − nk ≤ n2k =
kr + O(krk ) ≤ k 2 r4k + O(k 2 r3k ).
2
2
2
2
2
2
5
(1)
(2)
Since every pair {v, w} of distinct special vertices is a resolving pair of Hk , every
pair or the form {v 0 , w0 } with v 0 ∈ Rv and w ∈ Rw is a resolving pair of Gk . The
number of such pairs is
1
sk 2 1 2 2
(3)
ck = sk ck − O(sk c2k ) = k 2 r4k + O(k 2 r3k ).
2
2
2
Let Nk be the number of pairs of vertices of Gk that are not resolving. Subtracting
the last term of (3) from (2) we obtain that
Nk = O(k 2 r3k ).
√
3/2
Now (1) yields that k = O(log nk ), and so k 2 r3k = k 1/2 (kr2k )3/2 = O( log nk nk ).
Hence
p
3/2
Nk = O( log nk nk ),
2
which is the statement of the theorem.
As a consequence of Theorem
2 we obtain that for every real > 0 there exist
n
graphs with at least (1 − ) 2 resolving pairs, which disproves Conjecture 1.
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