International Journal of Civil Engineering and Technology (IJCIET)
Volume 10, Issue 04, April 2019, pp. 1228-1238, Article ID: IJCIET_10_04_129
Available online at http://www.iaeme.com/ijciet/issues.asp?JType=IJCIET&VType=10&IType=04
ISSN Print: 0976-6308 and ISSN Online: 0976-6316
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THE COMPLEMENT METRIC DIMENSION OF
GRAPHS AND ITS OPERATIONS
L. Susilowati
Department of Mathematics, Faculty of Sciences and Technology, Universitas Airlangga,
Surabaya, Indonesia, Jl. Mulyorejo Surabaya, 60115
Slamin
Study Program of Information System, Universitas Jember,Jl. Kalimantan 37 Jember, 68121
A. Rosfiana
Department of Mathematics, Faculty of Sciences and Technology, Universitas Airlangga,
Surabaya, Indonesia, Jl. Mulyorejo Surabaya, 60115
ABSTRACT
Let G be a connected graph with vertex set V(G) and edge set E(G). The distance
between vertices u and v in G is denoted by d(u, v), which serves as the shortest path
length from u to v. Let 𝑊 = {𝑤1 , 𝑤2 , … , 𝑤𝑘 } ⊆ 𝑉(𝐺) be an ordered set, and v is a vertex
in G. The representation of v with respect to W is an ordered set 𝑘 − 𝑡𝑢𝑝𝑙𝑒, 𝑟(𝑣|𝑊) =
(𝑑(𝑣, 𝑤1 ), 𝑑(𝑣, 𝑤2 ), … , 𝑑(𝑣, 𝑤𝑘 )). The set W is called a resolving set for G if each
vertex in G has a different representation with respect to W. A resolving set containing
minimum cardinality is called a basis for G. The number of vertices in a basis of G is
called metric dimension of G, which is denoted by𝑑𝑖𝑚(𝐺). The 𝑆 ⊆ 𝑉(𝐺) is a
complement resolving set of G if there are two vertices𝑢, 𝑣 ∈ 𝑉(𝐺) ∖ 𝑆, such
that𝑟(𝑢|𝑆) = 𝑟(𝑣|𝑆). A complement basis of G is the complement resolving set
containing maximum cardinality. The number of vertices in a complement basis of G is
̅̅̅̅̅(𝐺). In this paper,
called complement metric dimension of G, which is denoted by 𝑑𝑖𝑚
we examined complement metric dimension of particular graphs and their
characteristics. Furthermore, we determined complement metric dimension of corona
and comb products graphs.
Key words: Metric dimension, basis, complement basis, complement metric
dimension.
Cite this Article: L. Susilowati, Slamin and A. Rosfiana, the Complement Metric
Dimension of Graphs and its Operations. International Journal of Civil Engineering
and Technology, 10(04), 2019, pp. 1228-1238
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1. INTRODUCTION
Metric dimension is one of the topics discussed in the ever-growing graph theory. According
to Saputro et al. [15], F. Harary built a concept of basis and metric dimension of graph in 1976.
The concept can be utilized to distinguish each vertex in a connected graph G by determining
its representation with respect to subset of vertex set of G. Harary defined the vertex
representation using metric concept of graph.
Chartrand et al. [3] defined metric dimension as follows. Let G be a connected graph with
vertex set V(G) and edge set E(G). The distance between vertices u and v in G is denoted by
d(u, v), which serves as the shortest path length from u to v. Let 𝑊 = {𝑤1 , 𝑤2 , … , 𝑤𝑘 } ⊆ 𝑉(𝐺)
be an ordered set, and v is a vertex in G. The representation of v with respect to W is an ordered
set 𝑘 − 𝑡𝑢𝑝𝑙𝑒, 𝑟(𝑣|𝑊) = (𝑑(𝑣, 𝑤1 ), 𝑑(𝑣, 𝑤2 ), … , 𝑑(𝑣, 𝑤𝑘 )). The set W is called a resolving
set for G if each vertex in G has a different representation with respect to W. A resolving set
containing minimum cardinality is called a basis for G. The number of vertices in a basis of G
is called metric dimension of G, which is denoted by dim(𝐺).
Chartrand, G. et. al. [3] found the characterization of graph with certain metric dimension.
Sebo and Tannier [14] presented the definition of strong metric dimension on graph and
Okamoto, et al. [10] developed the metric dimension concept into local metric dimension.
Okamoto, et al. [10] succeed in finding the characterization of graph with certain local metric
dimension. Baca, M., et al. [2] succeed in finding the metric dimension of regular bipartite
graph, Ali, M., et al. [1] found the metric dimension of some graphs containing the cycle.
Rodriguez, J.A., et al. [12] developed the strong metric dimension concept and found the strong
metric dimension of strong product graphs. Oellermann, et al. [9] found the strong metric
dimension graphs and digraphs. Rodriguez, et al. [11] found the local metric dimension of
corona product graphs. Iswadi, et al. [7] found the metric dimension of corona product graphs.
Rodriquez, et al. [13] continued the research for rooted product graphs. Dorota Kuziak et. al.
[8] found strong metric dimension on graph of corona product. Meanwhile, Susilowati, et al.
[16] found the similarity of metric dimension and local metric dimension of rooted product
graph.
If the concept of metric dimension applies a set such that each vertex in the graph can be
distinguished by the set, then the opposite concept is that a set ensuring at least two vertices
that are regarded the same by the set. Such set is called complement resolving set, and the
maximum cardinality of the complement resolving set is called complement metric dimension.
The concept of complement metric dimension can be utilized to formulate problems within
a project. Let there is a project that requires a team of workers along with the project leader. In
the project, there are at least two persons who can be selected as project leaders, with the criteria
of the two having similar views on the worker team. This is done so that if one of the project
leaders is absent, then there is a replacement of project leader who has similar perspective on
the worker team so that the project can still run well. The more worker teams that the project
leaders can recognize equally, the better, as more people can be prepared as a team. When
depicted in the graph, people are depicted as vertex, and two connected vertices if two persons
know each other. Let set S be the maximum set of workers who can work in the project, and
vertices u and v as the project execution leader. In this case, there are at least two leaders:
vertices u and v that can similarly recognize all workers in S, so that if one of the leaders cannot
perform their task, then there is other leader who can replace it. Since S is the maximum set of
workers who can work in the project, then if the project requires workers less than |𝑆| then the
project can still run well, because the availability of at least 2 project leaders and worker teams
from the subset of S. Therefore, this concept can measure the maximum ability to run a planned
project.
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the Complement Metric Dimension of Graphs and its Operations
Based on the concept of complement metric dimension, we built a definition of complement
metric dimension. Let G be a connected graph containing order more than 2 with vertex set
𝑉(𝐺) and edge set 𝐸(𝐺). The 𝑆 ⊆ 𝑉(𝐺) is complement resolving set of G if there are two
vertices𝑢, 𝑣 ∈ 𝑉(𝐺) ∖ 𝑆, such that 𝑟(𝑢|𝑆) = 𝑟(𝑣|𝑆). A complement basis of G is complement
resolving set containing maximum cardinality. The number of vertices in a complement basis
̅̅̅̅̅(𝐺).
of G is called complement metric dimension of G, which is denoted by 𝑑𝑖𝑚
In this paper, we examined complement metric dimension in particular graphs and their
characteristics. In addition, we also determined complement metric dimension of corona and
comb products graphs. We provided some definitions and vertex’s characteristics of graph to
support our discussion.
Definition 1 [10]. Two distinct vertices u and v of graph G is called twin if u and v have
the same neighbourhood in V(G) – {u,v}, and they are called true twin or false twin if u and v
are adjacent and twin or uand v are not adjacent and twin, respectively.
The following lemma describe the properties of twin that are discovered by Hernando et al
[6].
Lemma 2 [6]. If u and v are twin in graph G, then 𝑑(𝑢, 𝑥) = 𝑑(𝑣, 𝑥) for every vertex 𝑥
inV(G) – {u,v}.
2. COMPLEMENT METRIC DIMENSION OF GRAPHS
Based on the definition of complement metric dimension on a graph, the obtained results of
observation are as follows:
Observation 3. The minimum of complement metric dimension of a graph of order n is 1.
Observation 4. The maximum of complement metric dimension of a graph of order n is
𝑛– 2.
Lemma 5. Let 𝐺 be a connected graph. If for each 𝑊 ⊆ 𝑉(𝐺) where |𝑊| = 𝑘 is not a
complement resolving set of graph 𝐺, then for every 𝑆 ⊆ 𝑉(𝐺) where |𝑆| > 𝑘 is also not a
complement resolving set of graph 𝐺.
Proof. Let each 𝑊 ⊆ 𝑉(𝐺) where|𝑊| = 𝑘 is not a complement resolving set of graph 𝐺.
Suppose there is 𝑆 ⊆ 𝑉(𝐺) where |𝑆| > 𝑘 and 𝑆 is a complement resolving set of graph 𝐺.
Let |𝑆| = 𝑙 and 𝑆 = {𝑣1 , 𝑣2 , … 𝑣𝑙 }, then there are two vertices
𝑢, 𝑣 ∈ 𝑉(𝐺) such that
𝑟(𝑢|𝑆) = 𝑟(𝑣|𝑆). As a result, we obtain 𝑊 ′ = 𝑆 − {𝑣𝑖 |𝑖 = 1,2, … , 𝑙 − 𝑘} where |𝑊′| = 𝑘, so
that 𝑟(𝑢|𝑊 ′ ) = 𝑟(𝑣|𝑊 ′ ). It is mean that 𝑊 ′ is a complement resolving set of graph 𝐺.This is
contradicted by the statement that for each 𝑊 ⊆ 𝑉(𝐺) where |𝑊| = 𝑘is not a complement set
of graph 𝐺.∎
The following are the complement metric dimension on some special graphs, such as path
graph(𝑃𝑛 ), cycle graph (𝐶𝑛 ), star graph (𝑆𝑛 ), and complete graph (𝐾𝑛 ).
Lemma 6. Let 𝐺 be a connected graph of order 𝑛 > 2. If there is 𝑣 ∈ 𝑉(𝐺) where
̅̅̅̅̅(𝐺) ≥ 2.
𝑑𝑒𝑔(𝑣) ≥ 3, then 𝑑𝑖𝑚
Proof. Let 𝐺 be a connected graph of order 𝑛 > 2 and we have𝑣 ∈ 𝑉(𝐺) where deg(𝑣) ≥
3. Without loss of generality, letdeg(𝑣) = 3, then 𝑣 is adjacent to𝑣1 , 𝑣2 , 𝑣3 ∈ 𝑉(𝐺) or
𝑣𝑣1 , 𝑣𝑣2 , 𝑣𝑣3 ∈ 𝐸(𝐺).There are four possibilities that may occur at vertices 𝑣1 , 𝑣2 , 𝑣3 , namely:
(i) the three vertices are not adjacent to each other; (ii) there is a pair of vertices that are adjacent
to each other; (iii) there are two pairs of vertices that are adjacent to each other; (iv) the three
vertices are adjacent to each other.
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(i) Let 𝑣𝑣1 , 𝑣𝑣2 , 𝑣𝑣3 ∈ 𝐸(𝐺) and vertices 𝑣1 , 𝑣2 , 𝑣3 are not adjacent to each other. If 𝑆 =
{𝑣, 𝑣1 }, then 𝑟(𝑣2 |𝑆) = (1, 2) = 𝑟(𝑣3 |𝑆). Hence,𝑆 is a complement resolving set of graph 𝐺
̅̅̅̅̅(𝐺) ≥ 2.
and 𝑑𝑖𝑚
(ii) Let 𝑣1 , 𝑣𝑣2 , 𝑣𝑣3 ∈ 𝐸(𝐺) ; 𝑣1 𝑣2 ∈ 𝐸(𝐺); 𝑣1 𝑣3 , 𝑣2 𝑣3 ∉ 𝐸(𝐺). If 𝑆 = {𝑣, 𝑣3 }, then
̅̅̅̅̅(𝐺) ≥
𝑟(𝑣1 |𝑆) = (1, 2) = 𝑟(𝑣2 |𝑆). Hence, 𝑆 is a complement resolving set of graph 𝐺 and 𝑑𝑖𝑚
2.
(iii)Let 𝑣1 , 𝑣𝑣2 , 𝑣𝑣3 ∈ 𝐸(𝐺) ; 𝑣1 𝑣2 , 𝑣2 𝑣3 ∈ 𝐸(𝐺); 𝑣1 𝑣3 ∉ 𝐸(𝐺). If 𝑆 = {𝑣1 , 𝑣3 }, then
̅̅̅̅̅(𝐺) ≥
𝑟(𝑣|𝑆) = (1, 1) = 𝑟(𝑣2 |𝑆). Hence, 𝑆 is a complement resolving set of graph 𝐺 and 𝑑𝑖𝑚
2.
(iv) Let 𝑣1 , 𝑣𝑣2 , 𝑣𝑣3 ∈ 𝐸(𝐺) ; 𝑣1 𝑣2 , 𝑣1 𝑣3 , 𝑣2 𝑣3 ∈ 𝐸(𝐺). If 𝑆 = {𝑣1 , 𝑣3 }, then 𝑟(𝑣|𝑆) =
̅̅̅̅̅(𝐺) ≥ 2.
(1, 1) = 𝑟(𝑣2 |𝑆). Hence, 𝑆 is a complement resolving set of graph 𝐺 and 𝑑𝑖𝑚
̅̅̅̅̅(𝐺) ≥
Based on the description above, if we have𝑣 ∈ 𝑉(𝐺) where deg(𝑣) ≥ 3, then 𝑑𝑖𝑚
2. ∎
The complement metric dimension of path graph is presented as below.
̅̅̅̅̅(𝑃𝑛 ) = 1, for 𝑛 > 2.
Theorem 7.
𝑑𝑖𝑚
Proof. Let (𝑃𝑛 ) = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛}, 𝐸(𝑃𝑛 ) = {𝑣𝑖 𝑣𝑖+1 |𝑖 = 1,2, … , 𝑛 − 1},the path of
graph 𝑃𝑛 is𝑣1 , 𝑣2 , … , 𝑣𝑛−1 , 𝑣𝑛 , and if 𝑆 = {𝑣2 }, then we have𝑣1 , 𝑣3 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆. Hence,
𝑟(𝑣1 |𝑆) = (1) = 𝑟(𝑣3 |𝑆).
Furthermore, we will show that 𝑆 is a complement resolving set with maximum cardinality.
Choose any 𝑆 ′ ⊆ 𝑉(𝑃𝑛 ) where |𝑆 ′ | > 𝑆. Without loss of generality, let |𝑆 ′ | = 2. Let 𝑆 ′ =
{𝑥, 𝑦}, then there are two possibilities, namely: (i) one of the vertices is the endpoint, so that
𝑥 = 𝑣1 and 𝑦 = 𝑣𝑚 for 𝑚 = 2, 3, … , 𝑛 − 1 ; (ii) both vertices are not the endpoint, so that 𝑥 =
𝑣𝑖 and 𝑦 = 𝑣𝑗 , 𝑖, 𝑗 = 2,3, … , 𝑛 − 1
(i) Let 𝑆 ′ = {𝑣1 , 𝑣𝑚 } where 𝑚 = 2, 3, … , 𝑛 − 1. Choose any 𝑣𝑘 , 𝑣𝑙 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆′ where
𝑘 < 𝑙, then 𝑑(𝑣𝑘 , 𝑣1 ) < 𝑑(𝑣𝑙 , 𝑣1 ). Hence, 𝑟(𝑣𝑘 |𝑆′) ≠ 𝑟(𝑣𝑙 |𝑆′). Thus, 𝑆 ′ = {𝑣1 , 𝑣𝑚 } where
𝑚 = 2, 3, … , 𝑛 − 1 is not a complement resolving set of graph 𝑃𝑛 .
(ii) Let 𝑆 ′ = {𝑣𝑖 , 𝑣𝑗 } where 𝑖, 𝑗 = 2,3, … , 𝑛 − 1 and 𝑖 < 𝑗. Choose any 𝑣𝑘 , 𝑣𝑙 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆 ′
where 𝑘 < 𝑙, then there are six possibilities, namely: (a) 𝑘, 𝑙 < 𝑖; (b) 𝑘 < 𝑖; 𝑖 < 𝑙 < 𝑗; (c) 𝑘 <
𝑖; 𝑙 > 𝑗; (d) 𝑖 < 𝑘, 𝑙 < 𝑗; (e) 𝑖 < 𝑘 < 𝑗, 𝑙 > 𝑗; and (f) 𝑘, 𝑙 > 𝑗.
a) Choose any 𝑣𝑘 , 𝑣𝑙 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆′ where 𝑘 < 𝑙 < 𝑖 < 𝑗, then 𝑑(𝑣𝑙 , 𝑣𝑖 ) < 𝑑(𝑣𝑘 , 𝑣𝑖 ).
Hence, 𝑟(𝑣𝑘 |𝑆′) ≠ 𝑟(𝑣𝑙 |𝑆′).
b) Choose any 𝑣𝑘 , 𝑣𝑙 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆′ where 𝑘 < 𝑙 < 𝑖 < 𝑗, then 𝑑(𝑣𝑙 , 𝑣𝑗 ) < 𝑑(𝑣𝑘 , 𝑣𝑗 ).
Hence, 𝑟(𝑣𝑘 |𝑆′) ≠ 𝑟(𝑣𝑙 |𝑆′).
c) Choose any 𝑣𝑘 , 𝑣𝑙 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆′ where 𝑘 < 𝑖 < 𝑗 < 𝑙. Suppose 𝑑(𝑣𝑘 , 𝑣𝑖 ) = 𝑑(𝑣𝑙 , 𝑣𝑖 ),
then 𝑑(𝑣𝑙 , 𝑣𝑗 ) < 𝑑(𝑣𝑘 , 𝑣𝑗 ). Hence, 𝑟(𝑣𝑘 |𝑆′) ≠ 𝑟(𝑣𝑙 |𝑆′).
d) Choose any 𝑣𝑘 , 𝑣𝑙 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆′ where 𝑖 < 𝑘 < 𝑙 < 𝑗, then (𝑣𝑙 , 𝑣𝑗 ) < 𝑑(𝑣𝑘 , 𝑣𝑗 ). Hence,
𝑟(𝑣𝑘 |𝑆′) ≠ 𝑟(𝑣𝑙 |𝑆′).
e) Choose any 𝑣𝑘 , 𝑣𝑙 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆′ where 𝑖 < 𝑘 < 𝑗 < 𝑙, then 𝑑(𝑣𝑘 , 𝑣𝑖 ) < 𝑑(𝑣𝑙 , 𝑣𝑖 ).
Hence, 𝑟(𝑣𝑘 |𝑆′) ≠ 𝑟(𝑣𝑙 |𝑆′).
f)
Choose any 𝑣𝑘 , 𝑣𝑙 ∈ 𝑉(𝑃𝑛 ) ∖ 𝑆′ where 𝑖 < 𝑗 < 𝑘 < 𝑙, then 𝑑(𝑣𝑘 , 𝑣𝑖 ) < 𝑑(𝑣𝑙 , 𝑣𝑖 ).
Hence, 𝑟(𝑣𝑘 |𝑆′) ≠ 𝑟(𝑣𝑙 |𝑆′).
Based on the six possibilities above, 𝑆 ′ = {𝑣𝑖 , 𝑣𝑗 } where 𝑖, 𝑗 = 2,3, … , 𝑛 − 1 is not a
complement resolving set of graph 𝑃𝑛 .
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the Complement Metric Dimension of Graphs and its Operations
Thus, for each 𝑆 ′ ⊆ 𝑉(𝑃𝑛 ) where |𝑆 ′ | = 2 is not a complement resolving set of graph 𝑃𝑛 .
Based on Lemma 5, for each 𝑆 ′ ⊆ 𝑉(𝑃𝑛 ) where |𝑆 ′ | > 𝑆 is also is also not a complement
̅̅̅̅̅(𝑃𝑛 ) = 1. ∎
resolving set of graph 𝑃𝑛 . Therefore, 𝑆 is a complement base of graph 𝑃𝑛 and 𝑑𝑖𝑚
The complement metric dimension of cycle graph is presented as follow
̅̅̅̅̅(𝐶𝑛 ) = { 1, 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑 where 𝑛 > 2.
Theorem 8. 𝑑𝑖𝑚
2, 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛
Proof. Let graph 𝐶𝑛 has the order of 𝑛, then there are two possibilities where 𝑛 is odd or 𝑛
is even.
Case 1.If 𝑛 is odd, for𝑛 = 2𝑘 + 1, 𝑘 ≥ 1
Let 𝑉(𝐶𝑛 ) = {𝑣𝑖 |𝑖 = 1,2, … , 𝑘, 𝑘 + 1, … , 𝑛},
𝐸(𝐶𝑛 ) = {𝑣𝑖 𝑣𝑖+1 , 𝑣1 𝑣𝑛 |𝑖 = 1,2, … , 𝑛 − 1}, and choose 𝑆 = {𝑣1 }, then we have 𝑣2 , 𝑣𝑛 ∈
𝑉(𝐶𝑛 ). Hence, 𝑟(𝑣2 |𝑆) = (1) = 𝑟(𝑣𝑛 |𝑆).
Furthermore, we will show that 𝑆 is a complement resolving set with maximum cardinality.
Choose any 𝑆 ′ ⊆ 𝑉(𝐶𝑛 ) where |𝑆 ′ | > 𝑆. Without loss of generality, let |𝑆 ′ | = 2 where 𝑆 ′ =
{𝑥, 𝑦}, 𝑥, 𝑦 ∈ 𝑉(𝐺). Choose any 𝑆 ′ = {𝑥, 𝑦}, 𝑥, 𝑦 ∈ 𝑉(𝐺). If 𝑑(𝑢, 𝑥) = 𝑑(𝑣, 𝑥), then 𝑑(𝑢, 𝑦) ≠
𝑑(𝑣, 𝑦) or if 𝑑(𝑢, 𝑦) = 𝑑(𝑣, 𝑦), then
𝑑(𝑢, 𝑥) ≠ 𝑑(𝑣, 𝑥). Hence, 𝑟(𝑢|𝑆′) ≠ 𝑟(𝑣|𝑆′).
′
′
Thus, for each𝑆 ⊆ 𝑉(𝐶𝑛 ) where |𝑆 | = 2 is not a complement resolving set of graph 𝐶𝑛
where the 𝑛 is odd. Based on Lemma 5, for each 𝑆 ′ ⊆ 𝑉(𝐶𝑛 ) where|𝑆 ′ | > 𝑆 is also not a
complement resolving set of graph 𝐶𝑛 where the 𝑛 is odd. Therefore, 𝑆is a complement basis
̅̅̅̅̅(𝐶𝑛 ) = 1 for odd 𝑛.
of graph 𝐶𝑛 and 𝑑𝑖𝑚
Case 2. If 𝑛 is even, for𝑛 = 2𝑘, 𝑘 ≥ 2.
Let 𝑉(𝐶𝑛 ) = {𝑣𝑖 |𝑖 = 1,2, … , 𝑘, 𝑘 + 1, … , 𝑛},
𝐸(𝐶𝑛 ) = {𝑣𝑖 𝑣𝑖+1 , 𝑣1 𝑣𝑛 |𝑖 = 1,2, … , 𝑛 − 1},
the
cycle
of
𝐶𝑛 is
𝑣1 , 𝑣2 , … , 𝑣𝑘−1 , 𝑣𝑘 , 𝑣𝑘+1 , … , 𝑣𝑛 , 𝑣1 . Choose 𝑆 = {𝑣1 , 𝑣𝑘+1 }, then we have 𝑣2 , 𝑣𝑛 ∈ 𝑉(𝐶𝑛 ).
Hence, 𝑟(𝑣2 |𝑆) = (1, 𝑘 − 1) = 𝑟(𝑣𝑛 |𝑆).
Furthermore, we will show that 𝑆 is a complement resolving set with maximum cardinality.
Choose any 𝑆 ′ ⊆ 𝑉(𝐶𝑛 ) where |𝑆 ′ | > 𝑆. Without loss of generality, let |𝑆 ′ | = 3 where 𝑆 ′ =
{𝑥, 𝑦, 𝑧}, 𝑥, 𝑦, 𝑧 ∈ 𝑉(𝐶𝑛 ) − 𝑆. Choose any 𝑢, 𝑣 ∈ 𝑉(𝐺) − 𝑆′. If 𝑑(𝑢, 𝑥) = 𝑑(𝑣, 𝑥) and
𝑑(𝑢, 𝑦) = 𝑑(𝑣, 𝑦), then 𝑑(𝑢, 𝑧) ≠ 𝑑(𝑣, 𝑧). Hence, 𝑟(𝑢|𝑆′) ≠ 𝑟(𝑣|𝑆′).
Thus, for every 𝑆 ′ ⊆ 𝑉(𝐶𝑛 ) where |𝑆 ′ | = 3 is not a complement resolving set of graph 𝐶𝑛
where the 𝑛 is even. Based on Lemma 5, for every 𝑆 ′ ⊆ 𝑉(𝐶𝑛 ) where |𝑆 ′ | > 𝑆 is also not a
complement resolving set of graph 𝐶𝑛 where the 𝑛 is even. Therefore, 𝑆 is a complement basis
̅̅̅̅̅(𝐶𝑛 ) = 2 for even 𝑛. ∎
of graph𝐶𝑛 and 𝑑𝑖𝑚
The complement metric dimension of star graph is presented as follow.
̅̅̅̅̅(𝑆𝑛 ) = 𝑛 − 2, for 𝑛 > 2.
Theorem 9.
𝑑𝑖𝑚
Proof.
Let 𝑉(𝑆𝑛 ) = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛}, deg 𝑣1 = 𝑛 − 1, 𝐸(𝑆𝑛 ) = {𝑣1 𝑣𝑖 |𝑖 = 2, … , 𝑛}.
Choose𝑆 = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛 − 2}. Since 𝑣1 𝑣𝑖 ∈ 𝐸(𝑆𝑛 ), then we have 𝑣𝑛−1 , 𝑣𝑛 ∈ 𝑉(𝑆𝑛 ).
Hence, 𝑟(𝑣𝑛−1 |𝑆) = (1, 2, 2, … ,2) = 𝑟(𝑣𝑛 |𝑆).
Based on Observation 4, 𝑆 is a complement resolving set with maximum cardinality.
̅̅̅̅̅(𝑆𝑛 ) = 𝑛 − 2. ∎
Therefore, 𝑆 is a complement basis of graph 𝑆𝑛 and 𝑑𝑖𝑚
The complement metric dimension of complete graph is presented as follow.
̅̅̅̅̅(𝐾𝑛 ) = 𝑛 − 2, for 𝑛 > 2.
Theorem 10.
𝑑𝑖𝑚
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Proof.
Let 𝑉(𝐾𝑛 ) = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛}, 𝐸(𝐾𝑛 ) = {𝑣𝑖 𝑣𝑗 | 𝑖, 𝑗 = 1,2, … , 𝑛, 𝑖 ≠ 𝑗 }.
Choose 𝑆 = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛 − 2}. Since every two vertices of graph 𝐾𝑛 is adjacent, then we
have 𝑣𝑛−1 , 𝑣𝑛 ∈ 𝑉(𝐾𝑛 ). Hence, 𝑟(𝑣𝑛−1 |𝑆) = (1, 1, … ,1) = 𝑟(𝑣𝑛 |𝑆).
Based on Observation 4, 𝑆 is a complement resolving set with maximum cardinality.
̅̅̅̅̅(𝐾𝑛 ) = 𝑛 − 2. ∎
Therefore, 𝑆 is a complement basis of graph 𝐾𝑛 and 𝑑𝑖𝑚
The following is the graph characteristic with complement metric dimension is one.
̅̅̅̅̅(𝐺) = 1 if and only if
Theorem 11.
Let 𝐺 be a connected graph of order 𝑛 > 2, 𝑑𝑖𝑚
𝐺 = 𝑃𝑛 or 𝐺 = 𝐶𝑚 where 𝑚 is odd.
Proof. Let 𝐺 is a connected graph of order 𝑛 > 2 and 𝑉(𝐺) = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛}.
̅̅̅̅̅(𝐺) = 1 and 𝐺 ≠ 𝑃𝑛 , then there are two cases:
(A). Let 𝑑𝑖𝑚
𝐺 is a graph where we have 𝑣 ∈ 𝑉(𝐺) where deg(𝑣) ≥ 3; or
𝐺 is a graph where in every 𝑣 ∈ 𝑉 (𝐺 ) where deg(𝑣) = 2.
̅̅̅̅̅(𝐺) ≥ 2. It is contradictory to 𝑑𝑖𝑚
̅̅̅̅̅(𝐺) =
Case (i) is not possible in regard to Lemma 6, 𝑑𝑖𝑚
1.
Case (ii), let in every 𝑣 ∈ 𝑉(𝐺) applies deg(𝑣) = 2, then 𝐺 is a cycle graph. Based on
̅̅̅̅̅(𝐺) = 1, then 𝐺 = 𝐶𝑚 where 𝑚 is odd.
Theorem 8 and since 𝑑𝑖𝑚
̅̅̅̅̅(𝐺) = 1 and 𝐺 ≠ 𝑃𝑛 , then
Based on the description of case (i) and (ii), we obtain that if 𝑑𝑖𝑚
𝐺 = 𝐶𝑚 where 𝑚 is odd.
̅̅̅̅̅(𝐺) = 1 and 𝐺 ≠ 𝐶𝑛 where 𝑛 is odd, then there are two cases:
(B). Let 𝑑𝑖𝑚
(i) 𝐺 is a cycle graph containing even order; or
(ii) 𝐺 is a graph with 𝑣 ∈ 𝑉(𝐺) where deg(𝑣) ≠ 2.
̅̅̅̅̅(𝐶𝑛 ) = 2 where 𝑛 is even. It is
Case (i) is not possible in regard to Theorem 8, 𝑑𝑖𝑚
̅̅̅̅̅(𝐺) = 1.
contradictory to 𝑑𝑖𝑚
Case (ii), let 𝐺 is a graph with 𝑣 ∈ 𝑉(𝐺) where deg(𝑣) ≠ 2, then there are two cases,
namely (a) deg(𝑣) > 2 or (b) deg(𝑣) < 2.
(a). We have 𝑣 ∈ 𝑉(𝐺) where deg(𝑣) > 2. It is not possible in regard to Lemma 6,
̅̅̅̅̅
̅̅̅̅̅(𝐺) = 1.
𝑑𝑖𝑚(𝐺) ≥ 2. It is contradictory to 𝑑𝑖𝑚
(b). We have 𝑣 ∈ 𝑉(𝐺) where deg(𝑣) < 2, then deg(𝑣) = 1. There are two possibilities:
(i) Each vertex 𝑣 ∈ 𝑉(𝐺), deg(𝑣) = 1. It is not possible because if this happens, graph 𝐺
is not connected. This is contradictory to connected graph 𝐺.
(ii) We have vertex 𝑢 ≠ 𝑣, so deg(𝑢) = 2. Since graph 𝐺 is connected, and the degree of
̅̅̅̅̅(𝐺) = 1.
vertex in 𝐺 are 1 and 2, then 𝐺 is a track graph. Based on Theorem 7, 𝑑𝑖𝑚
̅̅̅̅̅(𝐺) = 1 and 𝐺 ≠ 𝐶𝑚 where 𝑚 is odd,
Based on the description of (b), we obtain that if 𝑑𝑖𝑚
then 𝐺 = 𝑃𝑛 .
From the case of (A) and (B), it can be concluded that if ̅̅̅̅̅̅
𝑑𝑖𝑚(𝐺) = 1, then 𝐺 = 𝑃𝑛 or 𝐺 =
𝐶𝑚 where 𝑚 is odd.
̅̅̅̅̅(𝐺) = 1 based on
Otherwise, let 𝐺 = 𝑃𝑛 or 𝐺 = 𝐶𝑚 where 𝑚 is odd, then we obtain 𝑑𝑖𝑚
Theorem 7 and Theorem 8.∎
The following is the graph characteristic with complement metric dimension is two least
than its order.
̅̅̅̅̅(𝐺) = 𝑛 − 2 if and only
Theorem 12.
Let 𝐺 be a connected graph of order 𝑛 > 2. 𝑑𝑖𝑚
if 𝐺 have two twin vertices.
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Proof. Let 𝐺 be a connected graph of order 𝑛 and 𝑉(𝐺) = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛}.
̅̅̅̅̅(𝐺) = 𝑛 − 2. Suppose there is no 𝑢, 𝑣 ∈ 𝑉(𝐺), such that 𝑢 and 𝑣 are twin. Choose
Let 𝑑𝑖𝑚
any 𝑢, 𝑣 ∈ 𝑉(𝐺) applies 𝑁𝐺 (𝑢) ≠ 𝑁𝐺 (𝑣).
̅̅̅̅̅(𝐺) = 𝑛 − 2, then we have 𝑆 ⊆ 𝑉(𝐺) where |𝑆| = 𝑛 − 2 which is the
Since 𝑑𝑖𝑚
complement basis of graph 𝐺. Without loss of generality, let 𝑆 = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛 − 2}, then
we have 𝑣𝑛−1 , 𝑣𝑛 ∈ 𝑉(𝐺) − 𝑆 where 𝑟(𝑣𝑛−1 |𝑆) = 𝑟(𝑣𝑛 |𝑆). Hence, (𝑣𝑛−1 , 𝑣𝑖 ) = 𝑑(𝑣𝑛 , 𝑣𝑖 ) for
𝑖 = 1,2, … , 𝑛 − 2. Since for every 𝑢, 𝑣 ∈ 𝑉(𝐺) applies 𝑁𝐺 (𝑢) ≠ 𝑁𝐺 (𝑣), then 𝑁𝐺 (𝑣𝑛−1 ) ≠
𝑁𝐺 (𝑣𝑛 ). Hence, we have 𝑗 ∈ {1,2, … , 𝑛 − 2} such that 𝑣𝑗 ∈ 𝑁𝐺 (𝑣𝑛−1 ), but 𝑣𝑗 ∉ 𝑁𝐺 (𝑣𝑛 ) or
viceversa.
Let 𝑣𝑗 ∈ 𝑁𝐺 (𝑣𝑛−1 ), 𝑣𝑗 ∉ 𝑁𝐺 (𝑣𝑛 ), 𝑗 ∈ {1,2, … , 𝑛 − 2}, then 𝑑(𝑣𝑛−1 , 𝑣𝑗 ) = 1 and
𝑑(𝑣𝑛 , 𝑣𝑗 ) > 1. Hence, 𝑑(𝑣𝑛−1 , 𝑣𝑗 ) ≠ 𝑑(𝑣𝑛 , 𝑣𝑗 ). It is contradictory to the statement where
̅̅̅̅̅(𝐺) = 𝑛 − 2, then 𝐺h ave two
𝑑(𝑣𝑛−1 , 𝑣𝑖 ) = 𝑑(𝑣𝑛 , 𝑣𝑖 ) for 𝑖 = 1,2, … , 𝑛 − 2. Therefore, if 𝑑𝑖𝑚
twin vertices.
Otherwise, let 𝐺 have two twin vertices. Based on Lemma 1, 𝑑(𝑢, 𝑥) = 𝑑(𝑣, 𝑥) for every
𝑑(𝑢, 𝑥) = 𝑑(𝑣, 𝑥). Let = 𝑉(𝐺) − {𝑢, 𝑣}, then |𝑆| = |𝑉(𝐺)| − 2 = 𝑛 − 2. Since 𝑑(𝑢, 𝑥) =
𝑑(𝑣, 𝑥) for every 𝑥 ∈ 𝑉(𝐺) − {𝑢, 𝑣}, then 𝑟(𝑢|𝑆) = 𝑟(𝑣|𝑆). Hence, 𝑆 is a complement basis
̅̅̅̅̅(𝐺) = 𝑛 − 2. ∎
of graph 𝐺and 𝑑𝑖𝑚
3. COMPLEMENT METRIC DIMENSION OF CORONA PRODUCT
GRAPH
Frucht [4] provided the definition of the corona product. Let 𝐺 be a connected graph of order
𝑛 and 𝐻 (not necessarily connected) be a graph of order at least two. The graph 𝐺 corona 𝐻,
denoted by 𝐺⨀𝐻, is a graph that is obtained by taking 𝑛 copies of graph 𝐻1 , 𝐻2 , … , 𝐻𝑛 of 𝐻
and connecting the 𝑖-th vertex of 𝐺 to all vertices of 𝐻𝑖 . The theorem of complement metric
dimension of corona product graph is presented below.
Theorem 13. Let 𝐺 be a connected graph of order 𝑛1 and 𝐻 be a connected graph of order
𝑛2 ≥ 2. Then
̅̅̅̅̅(𝐺⨀𝐻) = (𝑛1 − 1)(1 + 𝑛2 ) + 𝑑𝑖𝑚
̅̅̅̅̅(𝐾1 + 𝐻).
𝑑𝑖𝑚
Proof. Let 𝐺 and 𝐻 be connected graphs of order 𝑛1 and 𝑛2 , respectively. Let 𝑉(𝐺) =
{𝑢𝑖 |𝑖 = 1,2, … , 𝑛1 }, 𝑉(𝐻) = {𝑣𝑖 |𝑖 = 1,2, … , 𝑛2 } for 𝑛2 ≥ 2, 𝐻𝑖 be 𝑖-th copy of 𝐻 for 𝑖 =
1,2, … , 𝑛1 , and 𝑉(𝐻𝑖 ) = {𝑣𝑖𝑗 |𝑗 = 1,2, … , 𝑛2 } for 𝑖 = 1,2, … , 𝑛1 . Let
𝑉(𝐺⨀𝐻) =
𝑛1
𝑉(𝐺) ⋃𝑖=1 𝑉(𝐻𝑖 ) and
𝑛1
𝐸(𝐺⨀𝐻) = 𝐸(𝐺) ⋃𝑖=1
𝐸(𝐻𝑖 ) ⋃ {𝑢𝑖 𝑣𝑖𝑗 | 𝑢𝑖 ∈ 𝑉(𝐺), 𝑣𝑖𝑗 ∈ 𝑉(𝐻𝑖 )}. Let 𝑆𝑖 is a basis
𝑛1
complement of graph ⟨𝑢𝑖 ⟩ + 𝐻𝑖 , Choose 𝑆 = 𝑆1 ⋃𝑖=2
𝑉(𝐻𝑖 ) ∪ 𝑉(𝐺) − {𝑢1 }. Since for every
|𝑆| = |𝑆1 | +
𝑖 ∈ {1,2, … , 𝑛1 } applies ⟨𝑢𝑖 ⟩ + 𝐻𝑖 ≅ 𝐾1 + 𝐻 and 𝐻𝑖 ≅ 𝐻, then
(𝑛1 − 1)|𝑉(𝐻)| + |𝑉(𝐺)| − 1 or
̅̅̅̅̅(𝐾1 + 𝐻 ) + (𝑛1 − 1)𝑛2 + (𝑛1 − 1).
|𝑆| = 𝑑𝑖𝑚
Furthermore, we will show that 𝑆 is complement resolver of graph 𝐺⨀𝐻. Since 𝑆1 is a
basis complement of graph ⟨𝑢1 ⟩ + 𝐻1, then we have 𝑢, 𝑣 ∈ 𝑉(𝐻1 ) ∪ {𝑢1 } ∖ 𝑆1 . Hence,
𝑟(𝑢|𝑆1) = 𝑟(𝑣|𝑆1 ). Since 𝑢1 𝑢, 𝑢1 𝑣 ∈ 𝐸(𝐺⨀𝐻), then
𝑑(𝑢, 𝑢1 ) = 1 = 𝑑(𝑣, 𝑢1 ).
Hence, 𝑢1 ∈ 𝑆1 . Choose 𝑥 ∈ 𝑉(𝐻1 ),
𝑦 ∈ 𝑉(𝐺⨀𝐻) − 𝑉(𝐻1 ) − {𝑢1 }, then
𝑛1
𝑑(𝑥, 𝑦) = 𝑑(𝑥, 𝑢1 ) + 𝑑(𝑢1 , 𝑦). Hence,
𝑑(𝑢, 𝑦) = 𝑑(𝑣, 𝑦). Since 𝑆 = 𝑆1 ⋃𝑖=2
𝑉(𝐻𝑖 ) ∪
𝑉(𝐺) − {𝑢1 }, then 𝑟(𝑢|𝑆) = 𝑟(𝑣|𝑆). Hence, 𝑆 is a complement resolving set of graph 𝐺⨀𝐻.
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Afterward, we will show that 𝑆 is a complement resolving set of graph 𝐺⨀𝐻 with
maximum cardinality. Choose 𝑆′ ⊆ 𝑉(𝐺⨀𝐻) with |𝑆| < |𝑆′| < |𝑉(𝐺⨀𝐻)| − 2, then we have
̅̅̅̅̅(⟨𝑢𝑖 ⟩ + 𝐻𝑖 ). Choose
𝑊𝑖 = 𝑆′ ∩ 𝑉(⟨𝑢𝑖 ⟩ + 𝐻𝑖 ) for every 𝑖 ∈ {1,2, … , 𝑛1 }. Hence, |𝑊𝑖 | > 𝑑𝑖𝑚
𝑥, 𝑦 ∈ 𝑉(𝐺⨀𝐻) ∖ 𝑆′, then there are four possibilities namely: (i) 𝑥 ∈ 𝑉(𝐺), 𝑦 ∈ 𝑉(𝐻𝑖 ) (ii)
𝑥, 𝑦 ∈ 𝑉(𝐺) (iii) 𝑥, 𝑦 ∈ 𝑉(𝐻𝑖 ) or (iv) 𝑥 ∈ 𝑉(𝐻𝑖 ), 𝑦 ∈ 𝑉(𝐻𝑗 ).
I.
If 𝑥 ∈ 𝑉(𝐺) − 𝑆′ and 𝑦 ∈ 𝑉(𝐻𝑖 ) − 𝑆′, then there are 𝑙, 𝑖 ∈ {1,2, … , 𝑛1 }, 𝑗 ∈
{1,2, … , 𝑛2 }. Hence, 𝑥 = 𝑢𝑙 and 𝑦 = 𝑣𝑖𝑗 . There are two possibilities which can
occur, 𝑙 = 𝑖 or 𝑙 ≠ 𝑖.
a. Suppose 𝑢𝑙 ∈ 𝑉(𝐺), 𝑣𝑖𝑗 ∈ 𝑉(𝐻𝑖 ) with 𝑙 = 𝑖, then 𝑢𝑙 = 𝑢𝑖 and 𝑣𝑖𝑗 is adjacent
with 𝑢𝑖 . Choose 𝑤 ∈ 𝑊𝑚 with 𝑚 ≠ 𝑖, 𝑚 ∈ {1,2, … , 𝑛1 }, then 𝑑(𝑢𝑖 , 𝑤) <
𝑑(𝑣𝑖𝑗 , 𝑤). Hence, 𝑟(𝑢𝑖 |𝑊𝑚 ) ≠ 𝑟(𝑣𝑖𝑗 |𝑊𝑚 ). Since 𝑊𝑚 ⊆ 𝑆′, then 𝑟(𝑢𝑖 |𝑆′) ≠
𝑟(𝑣𝑖𝑗 |𝑆′).
b. Suppose 𝑢𝑙 ∈ 𝑉(𝐺), 𝑣𝑖𝑗 ∈ 𝑉(𝐻𝑖 ) with 𝑙 ≠ 𝑖 and choose 𝑤 ∈ 𝑊𝑙 . Since
𝑑(𝑢𝑙 , 𝑤) < 𝑑(𝑣𝑖𝑗 , 𝑤), then 𝑟(𝑢𝑙 |𝑊𝑙 ) ≠ 𝑟(𝑣𝑖𝑗 |𝑊𝑙 ). Since 𝑊𝑙 ⊆ 𝑆′, then
𝑟(𝑢𝑙 |𝑆′) ≠ 𝑟(𝑣𝑖𝑗 |𝑆′)
If 𝑥, 𝑦 ∈ 𝑉(𝐺) − 𝑆′, then there are 𝑖, 𝑗 ∈ {1,2, … , 𝑛1 } with 𝑖 ≠ 𝑗. Hence 𝑥 = 𝑢𝑖 and
𝑦 = 𝑢𝑗 . Choose 𝑤 ∈ 𝑊𝑖 , then 𝑑(𝑢𝑖 , 𝑤) < 𝑑(𝑢𝑗 , 𝑤). Hence 𝑟(𝑢𝑖 |𝑊𝑖 ) ≠ 𝑟(𝑢𝑗 |𝑊𝑖 ).
Since 𝑊𝑖 ⊆ 𝑆′, then 𝑟(𝑢𝑖 |𝑆′) ≠ 𝑟(𝑢𝑗 |𝑆′)
III.
Suppose 𝑥, 𝑦 ∈ 𝑉(𝐻𝑖 ) − 𝑆′ for every 𝑖 ∈ {1,2, … , 𝑛1 }, then 𝑥, 𝑦 ∈ 𝑉(⟨𝑢𝑖 ⟩ + 𝐻𝑖 ) −
̅̅̅̅̅(⟨𝑢𝑖 ⟩ + 𝐻𝑖 ), then
|𝑊𝑖 | > 𝑑𝑖𝑚
𝑆′. Since there is 𝑊𝑖 = 𝑆′ ∩ 𝑉(⟨𝑢𝑖 ⟩ + 𝐻𝑖 ) with
𝑟(𝑥|𝑊𝑖 ) ≠ 𝑟(𝑦|𝑊𝑖 ). Hence 𝑟(𝑥|𝑆′) ≠ 𝑟(𝑦|𝑆′).
IV.
If 𝑥 ∈ 𝑉(𝐻𝑖 ) − 𝑆′ and 𝑦 ∈ 𝑉(𝐻𝑗 ) − 𝑆′ with 𝑖 ≠ 𝑗 and 𝑖, 𝑗 ∈ {1,2, … , 𝑛1 }, then we
have 𝑚, 𝑛 ∈ {2,3, … , 𝑛2 }. So 𝑥 = 𝑣𝑖𝑚 and 𝑦 = 𝑣𝑗𝑛 . Choose 𝑤 ∈ 𝑊𝑖 , then
𝑑(𝑣𝑖𝑚 , 𝑤) ≤ 2 and 𝑑(𝑣𝑗𝑛 , 𝑤) ≥ 3. Hence, 𝑑(𝑣𝑖𝑚 , 𝑤) < 𝑑(𝑣𝑗𝑛 , 𝑤). Thus,
𝑟(𝑣𝑖𝑚 |𝑊𝑖 ) ≠ 𝑟(𝑣𝑗𝑛 |𝑊𝑖 ) and 𝑟(𝑣𝑖𝑚 |𝑆′) ≠ 𝑟(𝑣𝑗𝑛 |𝑆′) because 𝑊𝑖 ⊆ 𝑆′.
According to the four possibilities mentioned above, 𝑆′ ⊆ 𝑉(𝐺⨀𝐻) with |𝑆| < |𝑆′| <
|𝑉(𝐺⨀𝐻)| − 2 is not a basis complement of graph 𝐺⨀𝐻. Hence, 𝑆 is a basis complement with
maximum cardinality. We can conclude that,
̅̅̅̅̅(𝐺⨀𝐻) = (𝑛1 − 1)(1 + 𝑛2 ) + 𝑑𝑖𝑚
̅̅̅̅̅(𝐾1 + 𝐻).
𝑑𝑖𝑚
II.
4. COMPLEMENT METRIC DIMENSION OF COMB PRODUCT
GRAPH
Comb product is a special case of the rooted product graph, which has been defined by Godsil
[5]. Let 𝐺 and 𝐻 be two connected graphs and 𝜊 be a root vertex of 𝐻. The comb product
between 𝐺 and 𝐻, denoted by 𝐺𝜊𝐻, is a graph obtained by taking one copy of 𝐺 and |𝑉(𝐺)|
copies of 𝐻 and grafting the 𝑖-th copy of 𝐻 at the vertex 𝜊 to the 𝑖-th vertex of 𝐺. The theorem
of complement metric dimension of operation graph of comb product is presented below.
Theorem 14. Let 𝐺 be a connected graph of order 𝑛, then
̅̅̅̅̅(𝐺𝜊𝐶𝑚 ) = 𝑑𝑖𝑚
̅̅̅̅̅(𝐶𝑚 ) + (𝑛 − 1)𝑚
𝑑𝑖𝑚
Proof. Let 𝐺 be a connected graph of order 𝑛, 𝑉(𝐺) = {𝑢𝑖 |𝑖 = 1,2, … , 𝑛},
𝑉(𝐶𝑚 ) =
{𝑣𝑖 |𝑖 = 1,2, … , 𝑚} with 𝑚 ≥ 3; 𝜊 is grafting vertex of graph 𝐶𝑚 .
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the Complement Metric Dimension of Graphs and its Operations
Let 𝐶𝑚 𝑖 be a copy of graph 𝐶𝑚 with 𝑖 = 1,2, … , 𝑛; 𝜊𝑖 is grafting vertex at 𝐶𝑚 𝑖 , 𝑉(𝐶𝑚 𝑖 ) =
{𝑣𝑖𝑗 |𝑗 = 1,2, … , 𝑚} with 𝑖 = 1,2, … , 𝑛; 𝑉(𝐺𝜊𝐶𝑚 ) = ⋃𝑛𝑖=1 𝑉(𝐶𝑚 𝑖 ), 𝑆𝑖 is a basis complement
of graph 𝐶𝑚 𝑖 which has grafting vertex 𝜊𝑖 .
Choose 𝑆 = 𝑆1 ⋃𝑛𝑖=2 𝑉(𝐶𝑚 𝑖 ). Since for every 𝑖 ∈ {1,2, … , 𝑛} applies 𝐶𝑚 𝑖 ≅ 𝐶𝑚 then |𝑆| =
̅̅̅̅̅(𝐶𝑚 ) + (𝑛 − 1)𝑚. Furthermore, we will show that 𝑆 is
|𝑆1 | + (𝑛 − 1)|𝑉(𝐶𝑚 )| or |𝑆| = 𝑑𝑖𝑚
a complement resolving set of graph 𝐺ο𝐶𝑚 . Since 𝑆1 is a basis complement of graph 𝐶𝑚 1, then
we have 𝑢, 𝑣 ∈ 𝑉(𝐶𝑚 1 ) ∖ 𝑆1. So 𝑟(𝑢|𝑆1 ) = 𝑟(𝑣|𝑆1). Choose 𝑥 ∈ 𝑉(𝐶𝑚 1 ), 𝑦 ∈ 𝑉(𝐺𝜊𝐶𝑚 ) −
𝑉(𝐶𝑚1 ), then 𝑑(𝑥, 𝑦) = 𝑑(𝑥, 𝑜1 ) + 𝑑(𝑜1 , 𝑦). Since 𝑟(𝑢|𝑆1 ) = 𝑟(𝑣|𝑆1) and 𝑜1 ∈ 𝑆1, then
𝑑(𝑢, 𝑜1 ) = 𝑑(𝑣, 𝑜1 ). So 𝑑(𝑢, 𝑦) = 𝑑(𝑣, 𝑦). Since 𝑆 = 𝑆1 ⋃𝑛𝑖=2 𝑉(𝐶𝑚 𝑖 ), then 𝑟(𝑢|𝑆) = 𝑟(𝑣|𝑆).
Thus, 𝑆 is a complement resolving set of graph 𝐺𝜊𝐶𝑚 .
Following this, we will show that 𝑆 is a complement resolving set of graph 𝐺𝜊𝐶𝑚 with
maximum cardinality. Choose 𝑆′ ⊆ 𝑉(𝐺𝜊𝐶𝑚 ) with |𝑆| < |𝑆′| < |𝑉(𝐺𝜊𝐶𝑚 )| − 2 then we have
𝑊𝑖 = 𝑆′ ∩ 𝑉(𝐶𝑚 𝑖 ) for every 𝑖 ∈ {1,2, … , 𝑛}. Hence, |𝑊𝑖 | > |𝑆𝑖 |. Choose 𝑥, 𝑦 ∈ 𝑉(𝐺𝜊𝐶𝑚 ) ∖ 𝑆′,
then there are four possibilities namely: (i) 𝑥 ∈ 𝑉(𝐺), 𝑦 ∈ 𝑉(𝐶𝑚 𝑖 ) (ii) 𝑥, 𝑦 ∈ 𝑉(𝐺) (iii) 𝑥, 𝑦 ∈
𝑉(𝐶𝑚 𝑖 ) or (iv) 𝑥 ∈ 𝑉(𝐶𝑚 𝑖 ), 𝑦 ∈ 𝑉(𝐶𝑚 𝑗 ).
I.
If 𝑥 ∈ 𝑉(𝐺) − 𝑆′ and 𝑦 ∈ 𝑉(𝐶𝑚 𝑖 ) − 𝑆′, then there are 𝑙, 𝑖 ∈ {1,2, … , 𝑛},
𝑗∈
{1,2, … , 𝑚}. So, 𝑥 = 𝑜𝑙 and 𝑦 = 𝑣𝑖𝑗 . There are two possibility, namely 𝑙 = 𝑖 or 𝑙 ≠
𝑖.
Let 𝑜𝑙 ∈ 𝑉(𝐺) − 𝑆′, 𝑣𝑖𝑗 ∈ 𝑉(𝐶𝑚 𝑖 ) − 𝑆′ with 𝑙 = 𝑖 then 𝑜𝑙 = 𝑜𝑖 . Choose 𝑤 ∈ 𝑊𝑘
with 𝑘 ≠ 𝑖, 𝑘 ∈ {1,2, … , 𝑛} then
𝑑(𝑜𝑖 , 𝑤) ≠ 𝑑(𝑣𝑖𝑗 , 𝑤) hence 𝑟(𝑜𝑖 |𝑊𝑘 ) ≠
𝑟(𝑣𝑖𝑗 |𝑊𝑘 ). Karena 𝑊𝑘 ⊆ 𝑆′ then 𝑟(𝑜𝑖 |𝑆 ′ ) ≠ 𝑟(𝑣𝑖𝑗 |𝑆 ′ ) or 𝑟(𝑥|𝑆′) ≠ 𝑟(𝑦|𝑆′).
Let 𝑜𝑙 ∈ 𝑉(𝐺) − 𝑆′, 𝑣𝑖𝑗 ∈ 𝑉(𝐶𝑚 𝑖 ) − 𝑆′ with 𝑙 ≠ 𝑖 and choose 𝑤 ∈ 𝑊𝑙 . Since
𝑑(𝑜𝑙 , 𝑤) < 𝑑(𝑣𝑖𝑗 , 𝑤), then 𝑟(𝑜𝑙 |𝑊𝑙 ) ≠ 𝑟(𝑣𝑖𝑗 |𝑊𝑙 ). Because 𝑊𝑙 ⊆ 𝑆′, then
𝑟(𝑜𝑙 |𝑆 ′ ) ≠ 𝑟(𝑣𝑖𝑗 |𝑆 ′ ) or 𝑟(𝑥|𝑆′) ≠ 𝑟(𝑦|𝑆′).
II.
Let 𝑥, 𝑦 ∈ 𝑉(𝐺) − 𝑆′, then we have 𝑖, 𝑗 ∈ {1,2, … , 𝑛} with 𝑖 ≠ 𝑗. Therefore,
𝑥=
𝑜𝑖 and 𝑦 = 𝑜𝑗 . Since 𝑑(𝑜𝑖 , 𝑜𝑗 ) ≠ 𝑑(𝑜𝑖 , 𝑜𝑖 ), then 𝑑(𝑜𝑖 , 𝑤) ≠ 𝑑(𝑜𝑗 , 𝑤) for every 𝑤 ∈
𝑊𝑖 . Hence, 𝑟(𝑜𝑖 |𝑊𝑖 ) ≠ 𝑟(𝑜𝑗 |𝑊𝑖 ). Since 𝑊𝑖 ⊆ 𝑆′, then
𝑟(𝑜𝑖 |𝑆′) ≠
𝑟(𝑜𝑗 |𝑆′) or 𝑟(𝑥|𝑆′) ≠ 𝑟(𝑦|𝑆′).
III.
Let 𝑥, 𝑦 ∈ 𝑉(𝐶𝑚 𝑖 ) − 𝑆′ for 𝑖 ∈ {1,2, … , 𝑛}. Because there is 𝑊𝑖 = 𝑆′ ∩ 𝑉(𝐶𝑚 𝑖 )
with |𝑊𝑖 | > |𝑆𝑖 |, then 𝑟(𝑥|𝑊𝑖 ) ≠ 𝑟(𝑦|𝑊𝑖 ). Since 𝑊𝑖 ⊆ 𝑆′, then
𝑟(𝑥|𝑆′) ≠
𝑟(𝑦|𝑆′).
IV.
Let 𝑥 ∈ 𝑉(𝐶𝑚 𝑖 ) − 𝑆′ and 𝑦 ∈ 𝑉 (𝐶𝑚 𝑗 ) − 𝑆′ with 𝑖 ≠ 𝑗; 𝑖, 𝑗 ∈ {1,2, … , 𝑛}, then we
have 𝑘, 𝑙 ∈ {1,2, … , 𝑚} so 𝑥 = 𝑣𝑖𝑘 and 𝑦 = 𝑣𝑗𝑙 . Since 𝑊𝑖 = 𝑆′ ∩ 𝑉(𝐶𝑚 𝑖 ) with
|𝑊𝑖 | > |𝑆𝑖 |, then 𝑟(𝑣𝑖𝑘 |𝑊𝑖 ) ≠ 𝑟(𝑜𝑖 |𝑊𝑖 ). Choose 𝑤 ∈ 𝑊𝑖 , then we make sure that
path 𝑣𝑗𝑙 − 𝑤 contains vertices 𝑜𝑖 and 𝑑(𝑣𝑗𝑙 , 𝑤) = 𝑑(𝑣𝑗𝑙 , 𝑜𝑖 ) + 𝑑(𝑜𝑖 , 𝑤). Since in
cycle graph 𝑑(𝑣𝑖𝑘 , 𝑤) − 𝑑(𝑜𝑖, 𝑤) = 𝑑(𝑣𝑗𝑙 , 𝑜𝑖 ) is not possible for every 𝑤 ∈ 𝑊𝑖 and
𝑟(𝑣𝑖𝑘 |𝑊𝑖 ) ≠ 𝑟(𝑜𝑖 |𝑊𝑖 ), then 𝑟(𝑣𝑖𝑘 |𝑊𝑖 ) ≠ 𝑟(𝑣𝑗𝑙 |𝑊𝑖 ). Consequently, 𝑟(𝑣𝑖𝑘 |𝑆′) ≠
𝑟(𝑣𝑗𝑙 |𝑆′) or 𝑟(𝑥|𝑆′) ≠ 𝑟(𝑦|𝑆′) because 𝑊𝑖 ⊆ 𝑆′.
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L. Susilowati, Slamin and A. Rosfiana
Based on the four possibilities mentioned above, then 𝑆′ ⊆ 𝑉(𝐺𝜊𝐶𝑚 ) with |𝑆| < |𝑆′| <
|𝑉(𝐺𝜊𝐶𝑚 )| − 2 is not a basis complement of graph 𝐺𝜊𝐶𝑚 . Hence, 𝑆 is a basis complement
̅̅̅̅̅(𝐺𝜊𝐶𝑚 ) = 𝑑𝑖𝑚
̅̅̅̅̅(𝐶𝑚 ) + (𝑛 − 1)𝑚.
with maximum cardinality. Thus, 𝑑𝑖𝑚
Conjecture 15. Let 𝐺 and 𝐻 be connected graphs of order 𝑛, 𝑚 ≥ 3, then
̅̅̅̅̅
𝑑𝑖𝑚(𝐻) + (𝑛 − 1)𝑚 ,
𝑖𝑓 𝑔𝑟𝑎𝑓𝑡𝑖𝑛𝑔 𝑣𝑒𝑟𝑡𝑒𝑥 𝑖𝑠 𝑎 𝑚𝑒𝑚𝑏𝑒𝑟
̅̅̅̅̅
𝑏𝑎𝑠𝑖𝑠 𝑐𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝐻
𝑑𝑖𝑚(𝐺𝜊𝐻) {
𝑚. ̅̅̅̅̅
𝑑𝑖𝑚(𝐺),
𝑓𝑜𝑟 𝑎𝑛𝑜𝑡ℎ𝑒𝑟
5. SUGGESTION
Further research could examine the complement metric dimension of the resulted graphs.
Inspired by the previous research conducted by Susilowati, et al. [17] and Susilowati, et al.
[18], this research can also be used as a source of information about commutative
characterization of comb and corona operation in complement metric dimension.
6. ACKNOWLEDGEMENTS
This research is supported by DRPM Indonesia through Penelitian Dasar Unggulan Perguruan
Tinggi Universitas Airlangga2018.
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