The Partition Dimension of a Path Graph

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SAINSTEK: JURNAL SAINS DAN TEKNOLOGI

Publisher: AMSET IAIN Batusangkar and IAIN Batusangkar Press Website: http://ecampus.iainbatusangkar.ac.id/ojs/index.php/sainstek E-mail: [email protected]

December 2021 Vol 13 No 2

ISSN: 2085-8019 (p) ISSN: 2580-278X (e) pp : 66-72

The Partition Dimension of a Path Graph

Vivi Ramdhani

¹

*, Fathur Rahmi

¹

1Departement of Mathematics Education, FTIK IAIN Bukittinggi Jl. Gurun Aur, Kubang Putih, Kab. Agam, Sumatera Barat

*email: [email protected]

Article History

Received: 22 October 2021 Reviewed: 11 November 2021 Accepted: 3 December 2021 Published: 30 December 2021

Key Words

The Partition Dimension;

Path Graph; Resolving Partition.

Abstract

Resolving partition is part of graph theory. This article, explains about resolving partition of the path graph, 𝑃_𝑛with𝑛 ≥ 2. Given a connected graph 𝐺 and 𝑆 is a subset of 𝑉(𝐺), writen 𝑆 ⊆ 𝑉(𝐺).

Suppose there is 𝑣 ∈ 𝑉(𝐺), then the distance between 𝑣 and 𝑆 is denoted in the form 𝑑(𝑣, 𝑆) = 𝑚𝑖𝑛{𝑑(𝑣, 𝑥)|𝑥 ∈ 𝑆}. There is an ordered set of 𝑘-partitions of 𝑉(𝐺), writen 𝛱 = {𝑆₁, 𝑆₂, … , 𝑆_𝑘}, then the representation of 𝑣 with respect to 𝛱 is the 𝑟(𝑣|𝛱) = (𝑑(𝑣, 𝑆₁), 𝑑(𝑣, 𝑆₂), … , 𝑑(𝑣, 𝑆_𝑘)) can be obtained. The set of partitions of 𝛱 is called a resolving partition if the representation of each 𝑣 ∈ 𝑉(𝐺) to 𝛱 is different. The minimum cardinality of the solving𝑘-partition to 𝑉(𝐺) is called the partition dimension of G which is denoted by 𝑝𝑑(𝐺). Before getting the partition dimension of a path graph, the first step is to look for resolving partition of the graph. Some resolving partitions of path graph, 𝛱₁, 𝛱₂, 𝛱₃ with 𝛱₁= {𝑆₁, 𝑆₂}, 𝛱₂= {𝑆₁, 𝑆₂, 𝑆_3,} and 𝛱₃= {𝑆₁, 𝑆₂, 𝑆_3,𝑆₄, 𝑆₅} are obtained. Then, the partition dimension of the path graph which is the minimum cardinality of resolving partition, namely 𝑝𝑑(𝑃_𝑛) = 2.

INTRODUCTION

One part of mathematics is graph theory.

A graph is a set of objects that contains vertices and a line connecting these vertices which can be written as 𝐺 = (𝑉, 𝐸), where V is a nonempty set from vertices and 𝐸 is the set of edges connecting vertices on the graph (Goodaire &

Parmenter, 2002). According to (Chartrand &

Lesniak, 1979), each edge in a graph connects exactly two vertices and each vertices can have many edges connecting with other vertices.

Partition dimensions are one of the topics of graph theory. There are many graphs whose partition dimensions have not been found, so this is a challenge for researchers to examine. In addition, this topic is also related to applied mathematics that is applicable in everyday life, especially those related to the problem of finding the shortest path. There are many studies which

graph. Among them are studies that discuss about the partition dimension graph (Chartrand, Salehi, et al., 2000), studies related to the partition dimensions of infinite regular graphs (Javaid & Shokat, 2008) which also examine the partition dimension of several connected wheel graph. Research on the partition dimensions continues to grow along with the development of time that discusses various other types of graphs.

There is a study that examines the partition dimensions of a graph 𝐾₁+ 𝑚𝐶_𝑛, with 𝑚, 𝑛 ∈ 𝑁, 𝑛 ≥ 3 (Auliya, 2014). There is a study that examine the partition dimensions and the partition dimensions of the star graph resulting from the operation of two connected graphs (Alfarisi, 2017).

There is a path graph, 𝑃_𝑛. All vertices in the graph are partitioned become some partitions so that an ordered set of 𝑘-partitions is formed, which can be written in the form 𝛱 = {𝑆₁, 𝑆₂, … ,

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𝑆_𝑘}. Next, look for the minimum 𝑘 value from the set of 𝑘-partitions so that the representation of each vertices on each graph is 𝛱 different.

This article aims to determine the resolving partition and the partition dimensions of the path graph, 𝑃_𝑛.

Define resolving partition of a graph is related to the shortest path that can be formed on the graph. The shortest path is closely related to what is seen in our daily lives. For example, related to the delivery of packages by a package agent. Packages brought will be distributed to consumers who are in other places. If the package departs from city A to city B. There will be several routes that can be taken to get the package to city B. In order to save operational costs, the package agent who carries the goods will definitely choose which one is the shortest path to get to the destination quickly. That analogy can be used to estimate the usefulness of resolving partition of a graph which in the end can find the partition dimensions of the graph that is discussed in this article, namely the path graph, 𝑃_𝑛.

There is a graph, 𝐺 = (𝑉, 𝐸) is pairs of sets (𝑉, 𝐸) with 𝑉 is a nonempty set from vertices in graph 𝐺 and 𝐸 is the set of edges joining a pair of vertices on the graph (Parment

& Goodaire, 2002). Many vertices in 𝐺 called order from G which is denoted by |𝑉(𝐺)|, and many sides to 𝐺 called size from G which is denoted by |𝐸(𝐺)|. If side 𝑒 = 𝑢𝑣 with 𝑢, 𝑣 𝜖 𝑉(𝐺), then vertice 𝑢 is said to be adjacent to vertices 𝑣 and the other hand. In this case, the side 𝑒 is said to be related to the vertices 𝑢 and 𝑣.

A graph 𝐻 is a subgraph of 𝐺 if 𝑉(𝐻) ⊆ 𝑉(𝐺) dan 𝐸(𝐻) ⊆ 𝐸(𝐺). 𝐻 is said to be an induced subgraph of 𝐺 and denoted by 〈𝑈〉 if 𝑈 is a nonempty set of vertices in graph 𝐺 and has edges which adjacent that are the same as the edges in graph 𝐺. For more details, see Figure 1.

Walk from the vertices 𝑣₀ to 𝑣_𝑛in 𝐺 is infinite sequence 𝑣₀, 𝑒₁, 𝑣₁,𝑒₂, … , 𝑒_𝑛−1, 𝑣_𝑛 from the set of vertices and the set of edges in 𝐺 such that 𝑣_𝑖−1𝑣_𝑖𝜖 𝐸(𝐺) for 𝑖 = 1, 2, … , 𝑛. Walk from the vertices 𝑣₀ to 𝑣_𝑛 can be written to be from 𝑣₀𝑣_𝑛-walk (Chartrand & Oellermann, 1993). The length of a walk is the number of sides of the sequence (Chartrand & Oellermann, 1993). A walk is called a closed path if 𝑣₀ = 𝑣_𝑛 and called open walk if 𝑣₀≠ 𝑣_𝑛 (Chartrand

& Oellermann, 1993).

A walk is called a path if all the vertices and sides are different. Path from vertices 𝑣₀ to vertices 𝑣_𝑛 can be written to 𝑣₀𝑣_𝑛-path. The shortest path from 𝑣₀ to 𝑣_𝑛 is the minimum number of edges contained in 𝑣₀𝑣_𝑛-path (Chartrand & Oellermann, 1993). Graph 𝐺 is called a connected graph if every two vertices there is a path (Chartrand & Oellermann, 1993).

The distance between two vertices 𝑢 and 𝑣 in a connected graph 𝐺 is the many edges contained in the shortest path of the two vertices, denoted by 𝑑(𝑢, 𝑣) (Chartrand, Raines, et al., 2000). Two vertices are said to be adjacent if 𝑒 = 𝑢𝑣 and 𝑢, 𝑣 𝜖 𝑉(𝐺), vertices 𝑢 be adjacent with vertices 𝑣 and same the other hand.

Graph 𝐺 is called path graph, called 𝑃𝑛, if the graph just has one path with 𝒏 vertices and path of length 𝑛 − 1. Figure 2 is one of example from path graph.

Figure 1. Induced Subgraph

Figure 2. Path Graph, 𝑃₃

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Vivi Ramdhani,, et al.. The Partition Dimension of …

Here are some definitions that will be used to explain the results and discussion:

Definition 1. Let G be a connected graph and let S be a subset of 𝑉(𝐺). Next, suppose there is a vertices 𝑣 ∈ 𝑉(𝐺) then the distance between 𝑣 and 𝑆 is defined as 𝑑(𝑣, 𝑆) = 𝑚𝑖𝑛 {𝑑(𝑣, 𝑥)|𝑥 ∈ 𝑆} (Javaid & Shokat, 2008).

Definition 2. 𝐺 is a connected graph and 𝑉(𝐺) partitioned into multiple partitions, called 𝑆₁, 𝑆₂, … , 𝑆_𝑘 and let 𝛱 is ordered set k-partition, writen 𝛱= {S1, S2,..., Sk} from 𝑉(𝐺). Next, let a vertices 𝑣 in 𝐺, so the representation is 𝑣 to 𝛱 in

from 𝑟(𝑣|∏) =

(𝑑(𝑣, 𝑆₁), 𝑑(𝑣, 𝑆₂), … , 𝑑(𝑣, 𝑆_𝑘)). Partition 𝛱 called resolving partition if k-vector r(v|𝛱) different for each 𝑣∈ 𝑉(𝐺) (Javaid & Shokat, 2008).

Definition 3. The minimum 𝑘 for which there is a resolving 𝑘-partition of 𝑉(𝐺) called partition dimension of 𝐺 which is denoted by 𝑝𝑑(𝐺) (Javaid & Shokat, 2008).

METHOD

The research method applied in this research is the literature study method by reviewing and understanding the literature and related articles from previous studies. It aims to find a supporting theory related to the problem to be studied so that a discussion of the problem is found. This research is intended to obtain the partitioning dimension of the path graph, 𝑃_𝑛. For this reason, first, we look for resolving the partition graph of the 𝑃_𝑛 path. Then determine the dimensions of the partition. So, in the last section, we will obtain the partition dimensions of the path graph 𝑃_𝑛.

RESULTANDDISCUSSION Resolving Partition of the Path Graph

In finding the resolving partition of a connected graph 𝐺, can be used lemma 1:

Lemma 1. 𝐺 is a connected graph, Let 𝛱 is a resolving partition from 𝑉(𝐺) and 𝑢, 𝑣 ∈ 𝑉(𝐺).

if 𝑑(𝑢, 𝑤) = 𝑑(𝑣, 𝑤) for each 𝑤 ∈ 𝑉(𝐺)\

{𝑢, 𝑣}, so 𝑢 𝑎𝑛𝑑 𝑣 belong to different classes from 𝛱.

Proof. Method of proof by contradiction for to Lemma 1.

Let 𝛱 = (𝑆₁, 𝑆₂, . . , 𝑆_𝑘) is the 𝑘-partition ordered set of 𝑉(𝐺) with k is a natural number bounded by the number of points in 𝐺 and let 𝑢, 𝑣 are the elements in the class 𝛱 at the same. If known 𝑑(𝑢, 𝑤) = 𝑑(𝑣, 𝑤) for each 𝑤 ∈ 𝑉(𝐺)\

{𝑢, 𝑣}, so it will be to find about 𝛱 not a resolving partition of 𝑉(𝐺).

For the first step, suppose that 𝛱 = (𝑆₁, 𝑆₂, . . , 𝑆_𝑘) with u and v included in class 𝛱 at the same, let 𝑆_𝑖. So, it can be result that 𝑑(𝑢, 𝑆_𝑖) = 𝑑(𝑣, 𝑆_𝑖) = 0 because the distance of the point to it self is zero.

Known 𝑑(𝑢, 𝑤) = 𝑑(𝑣, 𝑤) for each 𝑤 ∈ 𝑉(𝐺)-{𝑢, 𝑣}, this results in 𝑑(𝑢, 𝑆_𝑗) = 𝑑(𝑣, 𝑆_𝑗) for each j with 1≤ 𝑗 ≠ 𝑖 ≤ 𝑘. So, can be result 𝑟(𝑢| 𝛱) = 𝑟(𝑣 |𝛱) for each vertices in 𝐺.

Because there is the same representation for vertices in 𝐺 so based on Definition 2 can be 𝛱 not a resolving partition in 𝑉(𝐺). ∎ Then, the resolving partition of a path graph is presented, 𝑃_𝑛.

For any path graph, 𝑃_𝑛 with 𝑛 is order of the path graph, it will obtain some resolving partitions. However, in general if it is known 𝑃_𝑛: 𝑣₁, 𝑣₂, ⋯ , 𝑣_𝑘−1, 𝑣_𝑘, 𝑣_𝑘+1, ⋯ , 𝑣_𝑛 and 𝛱 is ordered set from 2-partition to 𝑉(𝑃𝑛), writen 𝛱 = {𝑆₁, 𝑆₂}, with 𝑆₁= {𝑣₁} and 𝑆₂ = {𝑣₂, … , 𝑣_𝑘−1, 𝑣_𝑘, 𝑣_𝑘+1, … , 𝑣_𝑛}, so 𝛱 = {𝑆₁, 𝑆₂} will be resolving partition in 𝑃𝑛. This means r(𝑣_𝑖| Π) different for each 1 ≤ 𝑖 ≤ 𝑛 with r(𝑣₁| 𝛱) = (0,1) and r(𝑣_𝑖| Π) = (i − 1,0) for 2 ≤ 𝑖 ≤ 𝑛 . Let see Figure 3.

Figure 3. Path Graph, 𝑃_𝑛

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Based on the explanation in the previous paragraph, it can be seen that:

r(𝑣₁| 𝛱) = (𝑑(𝑣₁, 𝑆₁), 𝑑(𝑣₁, 𝑆₂)).

Based on the Definition 1, in results:

𝑑(𝑣₁, 𝑆₁) = 𝑚𝑖𝑛 {𝑑(𝑣₁, 𝑥)|𝑥 ∈ 𝑆₁} and

𝑑(𝑣₁, 𝑆₂) = 𝑚𝑖𝑛 {𝑑(𝑣₁, 𝑥)|𝑥 ∈ 𝑆₂}, so:

(𝑑(𝑣₁, 𝑆₁), 𝑑(𝑣₁, 𝑆₂)) = (𝑑(𝑣₁, 𝑣₁), 𝑑(𝑣₁, 𝑣₂)).

Since the distance of a point to itself is zero, we get:

(𝑑(𝑣₁, 𝑣₁), 𝑑(𝑣₁, 𝑣₂)) = (0,1).

 So, we get 𝑟(𝑣₁| 𝛱) = (0,1).

Next, representation 𝑣_𝑘 to 𝛱 can be written:

𝑟(𝑣_𝑘| 𝛱) = (𝑑(𝑣_𝑘, 𝑆₁), 𝑑(𝑣_𝑘, 𝑆₂)).

Based on Definition 2, we results:

(𝑣_𝑘, 𝑆₁) = 𝑚𝑖𝑛 {𝑑(𝑣_𝑘, 𝑥)|𝑥 ∈ 𝑆₁} and

𝑑(𝑣_𝑘, 𝑆₂) = 𝑚𝑖𝑛 {𝑑(𝑣_𝑘, 𝑥)|𝑥 ∈ 𝑆₂}, so:

(𝑑(𝑣_𝑘, 𝑆₁), 𝑑(𝑣_𝑘, 𝑆₂)) = (𝑑(𝑣_𝑘, 𝑣₁), 𝑑(𝑣_𝑘, 𝑣_𝑘)) Because of the length path graph connected 𝐺 is 𝑚𝑖𝑛{𝑝 − 1, 𝑞} with 𝑞 is size of the graph and 𝑝 is the order. So, the distance for each point on the path graph, 𝑃_𝑛. Let see again Figure 3, known about 𝑣_𝑘 is vertices-k in 𝑃_𝑛, means for a 𝑣₁𝑣_𝑘−path there are 𝑘 vertices and 𝑘 − 1 edge.

The results:

𝑑(𝑣₁, 𝑣_𝑘) = 𝑑(𝑣_𝑘, 𝑣₁)

= min{𝑘 − 1, 𝑘 − 1}

= 𝑘 − 1.

Using this effect and knowing that the distance of a point from itself is zero, then:

(𝑑(𝑣_𝑘, 𝑣₁), 𝑑(𝑣_𝑘, 𝑣_𝑘)) = (𝑘 − 1,0).

 So, the results is 𝑟(𝑣_𝑘| 𝛱) = (𝑘 − 1,0).

With the same solution 2 ≤ 𝑖 ≤ 𝑛 with 𝑖 is the index of the sequence of points on 𝑃_𝑛 also can do that :

𝑑(𝑣₁, 𝑣_𝑖) = 𝑑(𝑣_𝑖, 𝑣₁)

= 𝑚𝑖𝑛{𝑖 − 1, 𝑖 − 1}

= 𝑖 − 1.

So, for 2 ≤ 𝑖 ≤ 𝑛 get representation 𝑣_𝑖 in 𝛱 like this:

𝑟(𝑣_𝑖| Π) = (𝑑(𝑣_𝑖, 𝑆₁), 𝑑(𝑑(𝑣_𝑖, 𝑆₂))) = (𝑑(𝑣_𝑖, 𝑣₁), 𝑑(𝑣_𝑖, 𝑣_𝑖))

= (i − 1,0).

 So, the result is 𝑟(𝑣_𝑖| Π) = (i − 1,0).

From this description it is concluded that 𝑟 (𝑣₁| 𝛱) = (0,1) and also get r(𝑣_𝑖| Π) = (i − 1,0) for 2 ≤ 𝑖 ≤ 𝑛. And then, get r(𝑣_𝑖| Π) different for each 1 ≤ 𝑖 ≤ 𝑛. So, Π is a resolving partition in 𝑃_𝑛.

In the following, one of the graphs of any paths is taken, namely graph 𝑃₅ (can be see in Figure 4) is a path graph which order 5 when 𝑉(𝑃₅) = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒}.

Let 𝑃₅ partitioned to be two partition, the 𝛱₁ is ordered set 2-partition, writen 𝛱₁ = {𝑆₁, 𝑆₂} with 𝑆₁ = {𝑎} and 𝑆₂= {𝑏, 𝑐, 𝑑, 𝑒}, make a representation for each point in 𝑃₅ be 𝛱₁.

Figure 4. Path Graph, 𝑃₅

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Based on Definition 2 the result about 𝛱₁ is a resolving partition in 𝑃₅ because representation 𝑟(𝑣| 𝛱₁) different for each 𝑣 ∈ 𝑉(𝑃₅).

Resolving partition in Graph 𝑃₅ can make another form. Let Π2= {𝑆1, 𝑆2, 𝑆3,} where 𝑆1= {𝑎}, 𝑆₂= {𝑐, 𝑑}, and 𝑆₃= {𝑏, 𝑒}. Get representation 3-vector for each vertices P5 to Π2

like this:

Based on Definition 2 can be conclusion about Π2 is a resolving partition in P5 because representation 3-vector 𝑟(𝑣| 𝛱₂) different for each v ∈ V(P5).

Now, let see another resolving, let Π3= {𝑆₁, 𝑆₂, 𝑆_3,𝑆₄, 𝑆₅} with 𝑆₁ = {𝑎}, 𝑆₂= {𝑏}, 𝑆₃= {𝑐}, 𝑆₄= {𝑑}, and 𝑆₅= {𝑒}. So, representation 5-vector, 𝑟(𝑣| 𝛱₃) for each vertices P5.

r(a|Π3)= (d(a,S1),d(a,S2),d(a,S3),d(a,S4), d(a,S5)) = (0,1,2,3,4)

r(b| Π3)=(d(b,S1),d(b,S2),d(b,S3),d(b,S4), d(b,S5)) = (1,0,1,2,3)

r(c|Π3)=(d(c,S1), d(c,S2), d(c,S3), d(c,S4), d(c,S5)) = (2,1,0,1,2)

r(d|Π3)=(d(d,S1),d(d,S2),d(d,S3), d(d,S4), d(d,S5)) = (3,2,1,0,1)

r(e|Π3)=(d(e,S1),d(e,S2), d(e,S3), d(e,S4), d(e,S5)) = (4,3,2,1,0)

Based on Definition 2 get about 𝛱₃is a resolving partition in P5 because representation 5-vector 𝑟(𝑣| 𝛱₃) different for each 𝑣 ∈ 𝑉(𝑃₅).

The conclusion, get 𝛱₁, 𝛱₂ and 𝛱₃ are some resolving partition in 𝑃₅. The means is 𝑃₅ have a some resolving partition.

The Partition Dimension of a Path Graph, 𝒑𝒅(𝑷_𝒏)

Let 𝐺 is graph connected ordered 𝑛 ≥ 2 so is 2 ≤ 𝑝𝑑(𝐺) ≤ 𝑛. For each integer 𝑛 ≥ 2, only have one graph ordered n have a partition dimension 2, 𝑝𝑑(𝐺) = 2, is path graph, Pn.

For more details, let see Theorem 1.

Theorem 1. Let 𝐺 is a graph connected ordered 𝑛 ≥ 2, so 𝑝𝑑(𝐺) = 2 if and only if 𝐺 = 𝑃_𝑛. Proof. Look at a connected graph 𝐺 in order 𝑛 ≥ 2.

Let 𝐺 = 𝑃_𝑛. It will be shown that 𝑝𝑑(𝐺) = 2. Show that 𝑃_𝑛: 𝑣₁, 𝑣₂ , ⋯ , 𝑣_𝑛 and 𝛱 = {𝑆₁, 𝑆₂} be a ordered set 2-partition in 𝑉(𝑃_𝑛) with 𝑆₁= {𝑣₁} and 𝑆₂= {𝑣₂, 𝑣_3,… , 𝑣_𝑛}. Because r(𝑣₁| Π) = (0,1) and r(𝑣_𝑖| Π) = (i − 1,0) for 2 ≤ 𝑖 < 𝑛 , So, Π is a resolving partition in 𝑃_𝑛 and get 𝑝𝑑(𝐺) = 2.

In other hand, let 𝑝𝑑(𝐺) = 2 will show that 𝐺 = 𝑃_𝑛. Because 𝑝𝑑(𝐺) = 2 so writen Π = {𝑆₁, 𝑆₂} is a resolving partition in graph 𝐺 ordered 𝑛. Let see about 𝐺 is a connected graph and then there are vertices 𝑢 ∈ 𝑆₁ and 𝑣 ∈ 𝑆₂ be adjacent. Because representation from 𝑟(𝑤| 𝛱) = (0, 𝑑(𝑤|𝑆₂)) if 𝑤 ∈ 𝑆₁ different with 𝑟(w| 𝛱) = (𝑑(𝑤|S₁), 0) if 𝑤 ∈ 𝑆₂, so 𝑢 is a single vertices in 𝑆₁ which adjacent to one vertice in 𝑆₂ and 𝑣 is single vertices in 𝑆₂ which adjacent to one vertice in 𝑆₁. The conclusion we will show 〈𝑆₁〉 and 〈𝑆₂〉 is a path in 𝐺.

Because 𝐺 connected graph, if 𝑆₁− {𝑢} ≠

∅ so each vertices in 𝑆₁ be adjacent with at least one vertices in 𝑆₁. Vertice 𝑢 is vertice in 𝑆₁.

Next, let 𝑢 be adjacent with at least one vertices in 𝑆₁, let vertice 𝑢₁ and vertices 𝑢₂. Let see Figure 5.

Figure 5. Illustration u Single Vertice in 𝑆

₁

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Figure 6. 𝑆₁ Induced Subgraph from 𝐺

From Figure 5 get 𝑟(𝑢₁| 𝛱) = 𝑟(𝑢₂| 𝛱) = (0,2) with 𝑢, 𝑢₁, 𝑢₂∈ 𝑆₁ and 𝑣 ∈ 𝑆₂. This is contradiction with 𝛱 is a resolving partition in 𝑉(𝐺), then the assumption is wrong.

So, 𝑢 exactly be adjacent to one vertices in 𝑆₁..

Assume that 𝑤 is single vertices in 𝑆₁ be adjacent with 𝑢. This means that 𝑤 is be adjacent with at most one vertices in 𝑆₁ different with 𝑢, let vertice 𝑧. The same is done to get the next vertices. So, 𝑉(𝑆₁) = {𝑢, 𝑤, 𝑧, … . }. With vertices in 𝑆₁ and also in 𝐺, it means 𝑢, 𝑤, 𝑧 … . ∈ 𝑉(𝐺).

For detail can see in Figure 6.

𝑆₁ is induced subgraph from G because any two be adjacent vertices on 𝐺, the vertices also be adjacent in 𝑆₁, denoted by 〈𝑆₁〉 . Look at Figure 6 about 〈𝑆₁〉 make a path in 𝐺.

Now, will be shown about

〈𝑆₂〉 also form a path di 𝐺.

Because

𝐺 is connected graph so there are

vertices 𝑢 ∈ 𝑆

₁

and 𝑣 ∈ 𝑆

₂

will be adjacent.

Because different representation in

𝑟(𝑡| 𝛱) = (0, 𝑑(𝑡|𝑆₂))

if

𝑡 ∈ 𝑆₁

and

r(𝑡| 𝛱) = (d(t|S₁), 0) if 𝑡 ∈ 𝑆₂, so 𝑢 is

single vertices in 𝑆

₁ be adjacent to a vertices

in 𝑆

₂

and 𝑣 is single vertices in 𝑆

₂

. If 𝑆

₂− {𝑣} ≠ ∅ so each vertices in 𝑆₂

be adjacent with at least one vertices in 𝑆

₂

.

Next, suppose that 𝑣 is adjacent to more than one vertice in 𝑆₂, for example, 𝑣₁, 𝑣₂∈ 𝑆2. For detail can see Figure 7. From Figure 7 can see 𝑟(𝑣₁| 𝛱) = 𝑟(𝑣₂| 𝛱) = (0,2) with 𝑣, 𝑣₁, 𝑣₂∈ 𝑆₂ and 𝑢 ∈ 𝑆₁and 𝛱 = {𝑆₁, 𝑆₂}. This contradicts the statement that 𝛱 is resolving partition in 𝑉(𝐺), then the assumption is wrong. Jadi, 𝑣 exactly be adjacent to one vertices in 𝑆₁ 𝑆₂..

Assume that 𝑡 is single vertices in 𝑆₂ be adjacent with 𝑣. This means that 𝑡 is be adjacent with at most one vertices in 𝑆₂ different with 𝑣, let vertices 𝑥. The same is done to get the next vertices. So, 𝑉(𝑆₂) = {𝑣, 𝑡, 𝑥, … . } With vertices in 𝑆2 and also in 𝐺, it means 𝑣, 𝑡, 𝑥 … . ∈ 𝑉(𝐺).

For detail can see in Figure 8.

Figure 7. Illustration 𝑣 Single Vertices in 𝑆₂

Figure 8. 𝑆₂ Induced Subgraph 𝐺

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𝑆₂ is induced subgraph from 𝐺, denoted by 〈𝑆₂〉.

Look at Figure 6 about 〈𝑆₂〉 make a path in 𝐺.

Based on the description, it has been result that 〈𝑆₁〉 and 〈𝑆₂〉 form a path in the connected graph 𝐺. So, it can be concluded that G is a path. ∎ For example aplication of Theorem 1, let see 𝑃₅ in Figure 4. In the previous explanation it has been obtained about resolving partition of path graph 𝑃₅ where 𝛱₁, 𝛱₂, 𝛱₃ are some resolving partition 𝑃₅ with 𝛱₃= {𝑆₁, 𝑆₂, 𝑆_3,𝑆₄, 𝑆₅}, 𝛱₂= {𝑆₁, 𝑆₂, 𝑆_3,}, and 𝛱₁= {𝑆₁, 𝑆₂}. Based on Definition 3 get about 𝑝𝑑(𝑃₅) = 2 because two is the minimum cardinality of resolving k-partition to 𝑉(𝑃₅).

CONCLUSION

Based on the results in the discussion, it can be concluded that:

1. Path graph, 𝑃_𝑛 with 𝑛 ≥ 2, have some resolving partition, is 𝛱₁, 𝛱₂, 𝛱₃with 𝛱₃= {𝑆₁, 𝑆₂, 𝑆_3,𝑆₄, 𝑆₅}, 𝛱₂= {𝑆₁, 𝑆₂,

𝑆_3,} and 𝛱₁= {𝑆₁, 𝑆₂}.

2.

Let

𝑛

is order in a connected graph with

𝑛 ≥ 2,

then there is exactly one graph with

pd(𝐺)= 2,

is a path graph. So, get

pd(𝑃_𝑛) = 2.

After writing this article, the author suggests examining the partition dimensions of other graph because there are still many of these graph that have not revealed the cardinality of the partition dimensions.

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