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Notations and Terminologies Research Problem Previous Results Main Results References
ON THE METRIC DIMENSION OF T-FOLD WHEEL GRAPH
Deddy Rahmadi1
, Tri Atmojo Kusmayadi2
, Sri Kuntari3
deddyrahmadi07@gmail.com1 tri.atmojo.kusmayadi@gmail.com2
kuntari.uns@gmail.com3
Combinatorial Mathematics Research Group
Department of Mathematics, Faculty of Mathematics and Natural Sciences
Sebelas Maret University
Surakarta, Central Java, Indonesia
Yogyakarta, August, 21 2015
The 7th
SEAMS-UGM 2015

Notations and Terminologies

Graph

Graph
Let G be a connected and simple graph of order n.

Graph
The distance d(u, v) between two vertices u and v of G is the length of a
shortest u − v path in G.

Graph
Metric Dimension

Graph
Metric Dimension
For an ordered set W = {w1, w2, . . . , wk } ⊆ V(G) and a vertex v of G we
refer to the k-vector
r(v | W) = (d(v, w1), d(v, w2), ..., d(v, wk ))

Graph
Metric Dimension
r(v | W) = (d(v, w1), d(v, w2), ..., d(v, wk ))
as the metric representation of v with respect to W. The set W is called
resolving set for G if every pair of vertices u and v of G, r(u|W) = r(v|W) [3].

Graph
Metric Dimension
r(v | W) = (d(v, w1), d(v, w2), ..., d(v, wk ))
as the metric representation of v with respect to W. The set W is called
resolving set for G if every pair of vertices u and v of G, r(u|W) = r(v|W) [3].
A resolving set with minimum cardinality called basis and the number of
vertices in basis is called metric dimension denoted by dim(G) [1].

Research Problem
Given the Graph G.

Research Problem
Given the Graph G.
To determine the metric dimension of graph G.

Research Problem
Given the Graph G.
To determine the metric dimension of graph G.
In this case G is t-fold wheel graph.

Previous Results
Chartrand et al., 2000
dim(Kn) = n − 1

Previous Results
dim(Kn) = n − 1
dim(Pn) = 1

Previous Results
dim(Kn) = n − 1
dim(Pn) = 1
Buczkowski et al., 2003
dim(Wn) = 2n+2
5

Previous Results
dim(Kn) = n − 1
dim(Pn) = 1
Buczkowski et al., 2003
dim(Wn) = 2n+2
5
Caceres et al., 2005
dim(fn) = 2n+2
5

Main Results
T-Fold Wheel Graph

Main Results
T-Fold Wheel Graph
Wallis [4] deﬁned t-fold wheel graph as follows.

Main Results
T-Fold Wheel Graph
Wallis [4] deﬁned t-fold wheel graph as follows. Suppose G is the graph
derived from a wheel by duplicating the hub vertex one or more times. We
call G a t-fold wheel graph Wt,n if there are t hub vertices, each adjacent to all
rim vertices, and not adjacent to each other.

Main Results
T-Fold Wheel Graph
Wallis [4] deﬁned t-fold wheel graph as follows. Suppose G is the graph
derived from a wheel by duplicating the hub vertex one or more times. We
call G a t-fold wheel graph Wt,n if there are t hub vertices, each adjacent to all
rim vertices, and not adjacent to each other.
Figure: 2-fold wheel graph W2,4

Main Results
Theorem 1
For any integer t ≥ 2 and n ≥ 3, then
dim(Wt,n) =
t + 1, n = 3, 4, 5;
n+t−2
2
, n ≥ 6.

Main Results
Proof
We consider two cases based on the values of n.

Main Results
Proof
Case 1. n = 3, 4, 5.

Main Results
Proof
Case 1. n = 3, 4, 5.
1 We show that dim(Wt,n) ≥ t + 1.

Main Results
Proof
Case 1. n = 3, 4, 5.
Suppose on contrary that dim(Wt,n) = t. Let
W = {u1, u2, . . . , ut−2, v1, v2} for n = 3, 4 and
W = {u1, u2, . . . , ut−2, v1, v3} for n = 5 , there are two vertices
x, y ∈ V(Wt,n) such that r(x|W) = r(y|W) = (2, 2, . . . , 2, 1, 1), a
contradiction

Main Results
Proof
Case 1. n = 3, 4, 5.
W = {u1, u2, . . . , ut−2, v1, v2} for n = 3, 4 and
contradiction
2 We show that dim(Wt,n) ≤ t + 1.

Main Results
Proof
Case 1. n = 3, 4, 5.
W = {u1, u2, . . . , ut−2, v1, v2} for n = 3, 4 and
contradiction
For every vertices ui , vj ∈ V(Wt,n) where i = 1, 2, . . . , t − 1 and
j = 1, 2, 3. Let W = {u1, u2, . . . , ut−1, v1, v2} for n = 3, 4 and
W = {u1, u2, . . . , ut−2, v1, v3} for n = 5. We note that there are no two
vertices having the same representation implying dim(Wt,n) ≤ t + 1.

Main Results
Proof
Case 1. n = 3, 4, 5.
W = {u1, u2, . . . , ut−2, v1, v2} for n = 3, 4 and
contradiction
j = 1, 2, 3. Let W = {u1, u2, . . . , ut−1, v1, v2} for n = 3, 4 and
W = {u1, u2, . . . , ut−2, v1, v3} for n = 5. We note that there are no two
vertices having the same representation implying dim(Wt,n) ≤ t + 1.
This implies that dim(Wt,n) = t + 1.

Main Results
Proof
Case 2. n ≥ 6.
We show that dim(Wt,n) ≥ n+t−2
2
. We consider the two cases.

Main Results
Proof
Case 2. n ≥ 6.
2
1 When n is even.

Main Results
Proof
Case 2. n ≥ 6.
2
1 When n is even.
Suppose on contrary that dim(Wt,n) < n+t−2
2
. Let
W = {u1, u2, . . . , ut−2, v1, v3, . . . , vn−3}, there are two vertices
x, y ∈ V(Wt,n) such that r(x|W) = r(y|W) = (2, 2, . . . , 2, 1, 1, . . . , 1), a
contradiction.

Main Results
Proof
Case 2. n ≥ 6.
2
1 When n is even.
2
. Let
contradiction.
2 When n is odd.

Main Results
Proof
Case 2. n ≥ 6.
2
1 When n is even.
2
. Let
contradiction.
2 When n is odd.
2
. Let
contradiction.

Main Results
Proof
We show that dim(Wt,n) ≤ n+t−2
2

Main Results
Proof
2
1 When n is even.

Main Results
Proof
2
1 When n is even.
j = 1, 2, 3. Let W = {u1, u2, . . . , ut−1, v1, v3, . . . , vn−3}. We note that
there are no two vertices having the same representation implying
dim(Wt,n) ≤ n+t−2
2
.

Main Results
Proof
2
1 When n is even.
2
.
2 When n is odd.

Main Results
Proof
2
1 When n is even.
2
.
2 When n is odd.
2
.

Main Results
Proof
2
1 When n is even.
2
.
2 When n is odd.
2
.
This implies that dim(Wt,n) = n+t−2
2
.

Example

Example
u1 u2 v1 v2 v3 v4
u1 0 2 1 1 1 1
u2 2 0 1 1 1 1
v1 1 1 0 1 2 1
v2 1 1 1 0 1 2
v3 1 1 2 1 0 1
v4 1 1 1 2 1 0

Example
W1 = {v1, v2}

Example
W1 = {v1, v2}
r(u1|W1) = (1, 1)
r(u2|W1) = (1, 1)
r(v1|W1) = (0, 1)
r(v2|W1) = (1, 0)
r(v3|W1) = (2, 1)
r(v4|W1) = (1, 2)

Example
W1 = {v1, v2}
r(u1|W1) = (1, 1)
r(u2|W1) = (1, 1)
r(v1|W1) = (0, 1)
r(v2|W1) = (1, 0)
r(v3|W1) = (2, 1)
r(v4|W1) = (1, 2)
r(u1|W1) = r(u2|W1)

Example
W1 = {v1, v2}
r(u1|W1) = (1, 1)
r(u2|W1) = (1, 1)
r(v1|W1) = (0, 1)
r(v2|W1) = (1, 0)
r(v3|W1) = (2, 1)
r(v4|W1) = (1, 2)
r(u1|W1) = r(u2|W1) → W1 is not resolving set.

Example
W2 = {u1, v1, v2}

Example
W2 = {u1, v1, v2}
r(u1|W2) = (0, 1, 1)
r(u2|W2) = (2, 1, 1)
r(v1|W2) = (1, 0, 1)
r(v2|W2) = (1, 1, 0)
r(v3|W2) = (1, 2, 1)
r(v4|W2) = (1, 1, 2)

Example
W2 = {u1, v1, v2}
r(u1|W2) = (0, 1, 1)
r(u2|W2) = (2, 1, 1)
r(v1|W2) = (1, 0, 1)
r(v2|W2) = (1, 1, 0)
r(v3|W2) = (1, 2, 1)
r(v4|W2) = (1, 1, 2)
We get the distinct representation of every vertices respect to W2

Example
W2 = {u1, v1, v2}
r(u1|W2) = (0, 1, 1)
r(u2|W2) = (2, 1, 1)
r(v1|W2) = (1, 0, 1)
r(v2|W2) = (1, 1, 0)
r(v3|W2) = (1, 2, 1)
r(v4|W2) = (1, 1, 2)
We get the distinct representation of every vertices respect to W2 → W2 is
resolving set

Example
W2 = {u1, v1, v2}
r(u1|W2) = (0, 1, 1)
r(u2|W2) = (2, 1, 1)
r(v1|W2) = (1, 0, 1)
r(v2|W2) = (1, 1, 0)
r(v3|W2) = (1, 2, 1)
r(v4|W2) = (1, 1, 2)
We get the distinct representation of every vertices respect to W2 → W2 is
resolving set → dim(W2,4) = 3.

Buczkowski, P. S., Chartrand, G., Poisson, C., and Zhang, P., On
k-dimensional Graphs and their bases, Periodica Math. Hung. 46(1),
9-15, 2003
Caceres, J., Hernando, C., Mora, M., Pelayo, I. M., Puertas, M. L.,
Seara, C., and Wood, D. R., On the Metric Dimension of Some Families
of Graphs, Electronic Notes Discrete Math. 22, 129-133, 2005
Chartrand, G., Eroh L., Johnson, M. A., and Oellermann, O. R.,
Resolvability in Graphs and Metric Dimension of a Graph, Discrete Appl.
Math. 105, 99-113, 2000
Wallis, W. D., Magic Graph, Birkhauser, Boston, Basel, Berlin, 2001

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