On the partition dimension of two-component graphs - Indian Academy ...

Proc. Indian Acad. Sci. (Math. Sci.) Vol. 127, No. 5, November 2017, pp. 755–767. https://doi.org/10.1007/s12044-017-0363-2

On the partition dimension of two-component graphs ˇ 2 D O HARYENI1,∗ , E T BASKORO1 , S W SAPUTRO1 , M BACA 2 ˇ ˇ ˇ and A SEMANICOVÁ-FE NOV CÍKOVÁ 1 Combinatorial Mathematics Research Group, Department of Mathematics, Faculty of Mathematics and Natural Sciences, Institut Teknologi Bandung (ITB), Bandung, Indonesia 2 Department of Applied Mathematics and Informatics, Technical University, Kosice, Slovak Republic *Corresponding author. E-mail: [email protected]; [email protected]; [email protected]; [email protected]; [email protected]; [email protected]

MS received 26 August 2016; published online 17 November 2017 Abstract. In this paper, we continue investigating the partition dimension for disconnected graphs. We determine the partition dimension for some classes of disconnected graphs G consisting of two components. If G = G 1 ∪ G 2 , then we give the bounds of the partition dimension of G for G 1 = Pn or G 1 = Cn and also for pd(G 1 ) = pd(G 2 ). Keywords.

Partition dimension; disconnected graph; component.

2010 Mathematics Subject Classification.

05C12, 05C15.

1. Introduction The study of the partition dimension for graphs was initiated by Chartrand et al. [2] aimed at finding a new way to solve the problem in metric dimensions of graphs. Many results in determining the partition dimension of some connected graphs have been obtained, see [2–5,9,10]. The graphs of order n with partition dimension 2, n, n − 1 and n − 2 have been characterized, see [3,10]. The partition dimension of graphs derived from some graph operations such as corona product, Cartesian product and strong product has also been determined, see [1,8,11,12]. However, this study was only limited for connected graphs. In [6], Haryeni et al. extended the definition of partition dimension for a graph, such that it can be applied for any graph (including disconnected graphs). Let G = (V, E) be an arbitrary (connected or disconnected) graph. The distance d(u, v) between u, v ∈ V (G) is defined as the number of edges in a shortest path connecting u and v in G. If there is no u − v path in G, then define d(u, v) = ∞. A vertex x ∈ V (G) resolves G if d(w, x) = d(y, x) for any pair of distinct vertices w, y ∈ V (G). For a subset of vertices R ⊂ V (G) and v ∈ V (G), the distance d(v, R) between v and R is defined as min{d(v, x) : x ∈ R}. Let  = {R1 , R2 , . . . , Rk } be an ordered k-partition of V (G). Then we call Ri as the partition class of  for any i ∈ [1, k]. If all d(v, Ri ) are finite © Indian Academy of Sciences

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for all vertices v ∈ V (G), then the representation of v with respect to  is defined as r (v|) = (d(v, R1 ), d(v, R2 ), . . . , d(v, Rk )). The partition  is a resolving partition if any two distinct vertices in G are resolved by some Ri . The least integer k such that G has a resolving k-partition is called the partition dimension of G and denoted by pd(G) or pdd(G) for connected or disconnected G, respectively. Otherwise, pdd(G) = ∞ if no integer k satisfies the above condition. Some results in finding the partition dimension for disconnected graph have been obtained, see [6,7]. In this paper, we continue studying the partition dimension of disconnected graphs. In particular, we determine the partition dimension of G = G 1 ∪ G 2 where G 1 = Pn or Cn and G 2 is any connected graph, or for some graphs G 1 and G 2 in which pd(G 1 ) = pd(G 2 ). The following results are useful in proving our main theorem in the next section. Lemma 1.1 [3]. Let  be a resolving partition of V (G) and u, v ∈ V (G). If d(u, w) = d(v, w) for all w ∈ V (G)\{u, v}, then u and v belong to distinct partition classes of . m G i . If pdd(G) < ∞, then max{ pd(G i ) : i ∈ [1, m]} ≤ Theorem 1.2 [6]. Let G = i=1 pdd(G) ≤ min{|V (G i )| : i ∈ [1, m]}. For integers k, l1 , l2 , . . . , lk ≥ 2, define a caterpillar C(k; l1 , l2 , . . . , lk ) as a graph obtained by attaching li vertices to each vertex vi of the path Pk for i ∈ [1, k]. If l1 = l2 = . . . = lk = l, then it is called a homogeneous caterpillar and is denoted by C(k, l). Darmaji et al. [4] gave the partition dimension of a homogeneous caterpillar as follows. Theorem 1.3 [4]. Let C(k, l) be a homogeneous caterpillar with l ≥ 3 and k ≥ 2. Then,  l, if k ≤ l, pd(C(k, l)) = l + 1, otherwise. 2. Path or cycle components m G i and  = {R1 , R2 , . . . , Rk } be a partition of V (G). Following For m ≥ 1, let G = i=1 [6], for any t ≥ 1, define a vertex v of G as a t-distance vertex if d(v, R j ) = 0 or t for any R j ∈ . Such a partition  is called as a connected partition if every subgraph induced by R j ∩ V (G i ) is connected for every j ∈ [1, k] and i ∈ [1, m]. In this section we consider graph G with only two components. In particular, we will determine the partition dimension of G = G 1 ∪ G 2 where G 1 = Pn or Cn . If G 2 = K m where |V (G 1 | ≥ m and m ≥ 4, then pdd(G) = m. Otherwise, pdd(G) = ∞. For now on, we consider G 2 is not complete. The following lemma shows that any connected k-partition is a resolving partition of a path Pm or a cycle Cm . Lemma 2.1. For k ∈ [3, m], any connected k-partition of V (Pm ) or V (Cm ) is a resolving partition. Proof. Let G = Pm or Cm , where V (G) = {u i : i ∈ [1, m]}, and  = {R1 , R2 , . . . , Rk } be a connected k-partition of G where k ∈ [3, m]. Then, without loss of generality there are integers s1 < s2 < . . . < sk = m such that

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R1 = {u 1 , u 2 , . . . , u s1 }, R2 = {u s1 +1 , u s1 +2 , . . . , u s2 }, .. . Rk = {u sk−1 +1 , u sk−1 +2 , . . . , u sk }. Now, consider any two vertices x, y ∈ V (G) in Rt for some t ∈ [1, k]. If x = u i and y = u j where i < j, then d(x, Rt+1 ) > d(y, Rt+1 ) or d(x, Rt−1 ) < d(y, Rt−1 ). Therefore, r (x|) = r (y|) and so  is a resolving partition of G.  In the following theorem, we give the necessary condition such that pdd(H ∪ Pn ) = pd(H ) for any connected graph H where pd(H ) ≥ 3. Theorem 2.2. Let H be a connected graph of order at least 3 and pd(H ) ≥ 3. If n ≥ max{2 × diam(H ) + 2, pd(H )}, then pdd(H ∪ Pn ) = pd(H ). Proof. For t ≥ 3 and k ≥ 1, let pd(H ) = t, diam(H ) = k and G = H ∪ Pn . If n ≥ max{2k + 2, t}, then we assume that pdd(G) < ∞ and thus pdd(G) ≥ t by using Theorem 1.2. To prove that pdd(G) ≤ t, let  = {R1 , R2 , . . . , Rt } be any resolving partition of H and V (Pn ) = {vi : i ∈ [1, n]}. Now we consider the following two cases. Case 1. 2k + 2 > t. Let 1 = {R1 , R2 , . . . , Rt } be a partition of G, where Ri  Rt−1 Rt

= Ri ∪ {vi } for any i ∈ [1, t − 2], = Rt−1 ∪ {vi : i ∈ [t − 1, n − 1]} and = Rt ∪ {vn }.

Now, we will show that 1 is a resolving partition of G. We consider any two vertices x, y ∈ V (G) in R p for some p ∈ [1, t]. If x, y ∈ V (H ), then r (x|1 ) = r (x|) = r (y|) = r (y|1 ). If x = vi and y = v j in Pn where t −1 ≤ i < j ≤ n −1, then r (x|1 ) = r (y|1 ) since 1 is a connected partition in Pn by Lemma 2.1. Now assume for x ∈ V (H ) and y = vi ∈ V (Pn ). If i ≥ k + 2, then d(x, R1 ) ≤ k < k + 1 ≤ i − 1 = d(y, v1 ) = d(y, R1 ). Otherwise, d(x, Rn ) ≤ k = (2k +2)−(k +2) < n −i = d(y, vn ) = d(y, Rn ). Therefore, r (x|1 ) = r (y|1 ) for any two vertices x, y ∈ V (G) in R p and thus 1 is a resolving partition of G. Case 2. Let t ≥ 2k + 2. Let 2 = {R1 , R2 , . . . , Rt } be a partition of G, where Ri = Ri ∪ {vi } for any i ∈ [1, t − 1] and

Rt = Rt ∪ {vi : i ∈ [t, n]}.

We investigate any two vertices x, y ∈ V (G) in Rl for some l ∈ [1, t]. As in Case 1, if x, y ∈ V (H ) or x, y ∈ V (Pn ), then r (x|2 ) = r (y|2 ). Now we assume x ∈ V (H ) and y = vi ∈ V (Pn ) for some i ∈ [1, n]. If i ≤ 2t , then d(x, Rt ) ≤ k < k + 1 ≤ 2t ≤ t − i = d(y, vt ) = d(y, Rt ). Otherwise, d(x, R1 ) ≤ k = 2t − 1 < i − 1 = d(y, v1 ) = d(y, R1 ). Therefore, r (x|2 ) = r (y|2 ) for any two vertices x, y ∈ V (G) and so 2 is a resolving partition of G. This concludes the proof that pdd(G) = pd(H ).



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We give the condition for a graph H ∪ Cm such that pdd(H ∪ Cm ) = pd(H ) for some values of pd(H ), as follows. Theorem 2.3. Let H be a connected graph of order ≥ 3. If ( pd(H ) = 3 and m ≥ 6 × diam(H )) or ( pd(H ) = 4 and m ≥ 4 × diam(H )), then pdd(H ∪ Cm ) = pd(H ). Proof. For k ≥ 1, let diam(H ) = k and G = H ∪ Cm with V (Cm ) = {u i : i ∈ [1, m]}. Assume that pdd(G) < ∞. We consider the following two cases: Case 1. Let pd(H ) = 3 and m ≥ 6k. Then, pdd(G) ≥ 3 by Theorem 1.2. Now, we will show that pdd(G) ≤ 3. Let 1 = {R1 , R2 , R3 } be a resolving partition of H . Let us show that 1 = {R1 , R2 , R3 }, where R1 = R1 ∪ {u 1 , u 2 , . . . , u 2k }, R2 = R2 ∪ {u 2k+1 , u 2k+2 , . . . , u 4k } and R3 = R3 ∪ {u 4k+1 , u 4k+2 , . . . , u m } is a resolving partition of G. We consider two vertices x, y ∈ V (G) in R j for some j ∈ [1, 3]. Note that if x, y ∈ V (H ), then r (x|1 ) = r (x|1 ) = r (y|1 ) = r (y|1 ). For two vertices x, y ∈ V (Cm ), since 1 is a connected 3-partition of Cm , we have r (x|1 ) = r (y|1 ) by Lemma 2.1. Now assume for x ∈ V (H ) and y = u i ∈ V (Cm ). We consider three subcases. Subcase 1.1. Let x, y ∈ R1 . If i ∈ [1, k], then d(y, R2 ) = d(u i , u 2k+1 ) = 2k + 1 − i ≥ 2k + 1 − k = k + 1 > k ≥ d(x, R2 ). Otherwise, d(y, R3 ) = d(u i , u 6k ) = i ≥ k + 1 > k ≥ d(x, R3 ). Subcase 1.2. Let x, y ∈ R2 . If i ∈ [2k + 1, 3k], then d(y, R3 ) = d(u i , u 4k+1 ) = 4k + 1 − i ≥ 4k + 1 − 3k = k + 1 > k ≥ d(x, R3 ). Otherwise, d(y, R1 ) = d(u i , u 2k ) = i − 2k ≥ 3k + 1 − 2k > k ≥ d(x, R1 ). Subcase 1.3. Let x, y ∈ R3 . If i ∈ [4k + 1, 5k], then d(y, R1 ) = d(u i , u 1 ) = 6k − i + 1 ≥ 6k −5k +1 > k ≥ d(x, R1 ). Otherwise, d(y, R2 ) = d(u i , u 4k ) = i −4k ≥ 5k +1−4k > k ≥ d(x, R2 ). Hence we conclude that r (x|1 ) = r (y|1 ) for any two distinct vertices x, y ∈ V (G). This implies that 1 is a resolving partition of G = H ∪ Cm , where pd(H ) = 3 and m ≥ 6k. Case 2. Let pd(H ) = 4 and m ≥ 4k. By Theorem 1.2, we have pdd(G) ≥ 4. Let 2 = {R1 , R2 , R3 , R4 } be a resolving partition of H . Let us show that 2 = {R1 , R2 , R3 , R4 }, where Ri = Ri ∪ {u (i−1)k+1 , u (i−1)k+2 , . . . , u ik } for each i ∈ [1, 3] and the remaining vertices are in R4 , is a resolving partition of G. Consider two vertices x, y ∈ R j for some j ∈ [1, 4]. For two vertices x, y ∈ V (H ) or x, y ∈ V (Cm ), r (x|2 ) = r (y|2 ) by a similar reason to Case 1. Now, for two vertices x ∈ V (H ) and y ∈ V (Cm ) in R j , d(y, Ra ) ≥ k + 1 > d(x, Ra ) for some integer a ∈ { j + 2, j − 2}. Therefore, 2 is a  resolving partition of G = H ∪ Cm for pd(H ) = 4 and m ≥ 4k. In the next result, we give the partition dimension for a graph containing a tree. For a tree T , a vertex of degree at least 3 in T is called a major vertex of T . Any pendant vertex u of T is said to be a terminal vertex of a major vertex v if d(u, v) < d(u, w) for every other major vertex w of T . The terminal degree of a major vertex v is the number of terminal vertices of v. A major vertex v is an exterior major vertex if it has positive terminal degree. Furthermore, a fibre tree is a tree with all vertices as pendant or exterior major vertices and each exterior major vertices has terminal degree at least 5.

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Theorem 2.4. Let T be a fibre tree of order ≥ 6 and H = Pm or Cm . If m ≥ pd(T ), then pdd(H ∪ T ) = pd(T ). Proof. For t ≥ 5, let  = {R1 , R2 , . . . , Rt } be any minimum resolving partition of a fibre tree T and x ∈ V (T ). By the properties of a fibre tree T , if x is an exterior major vertex of T , then there are at least 4 distinct partition classes Ri ∈  satisfying d(x, Ri ) = 1. Otherwise, there are at least 3 distinct partition classes Ri ∈  such that d(x, Ri ) = 2. By doing any connected t-partition  in H = Pm or Cm , we always have that  is also a resolving partition of H by Lemma 2.1. Since for v ∈ V (H ), d(v, Ri ) = 1 or 2 for at most two distinct partition classes Ri ∈  in H , we immediately have r (x|) = r (y|) for any x, y ∈ V (T ) ∪ V (H ).  In the following theorem, we give the partition dimension of a disjoint union of a path Pm and a cycle Cn . Theorem 2.5. For m ≥ 3 and n ≥ 4, pdd(Pm ∪ Cn ) = 3. Proof. Let G = Pm ∪ Cn and V (G) = V (Pm ) ∪ V (Cn ) = {u i : i ∈ [1, m]} ∪ {v j : j ∈ [1, n]}. By using Theorem 1.2, we have pdd(G) ≥ 3. To prove that pdd(G) ≤ 3, we define a partition  = {R1 , R2 , R3 } of G such that R1 = {u 1 , v1 }, R2 = {u 2 } ∪ {vi : i ∈ [2, n −1]}, and R3 = {u i : i ∈ [3, m]}∪{vn }. To prove that such a partition  is a resolving partition, we consider any two distinct vertices x, y ∈ Rt for some t ∈ [1, 3]. Note that  is a connected partition of G. Hence if (x = vi and y = v j where 2 ≤ i < j ≤ n − 1), or (x = u i and y = u j where 3 ≤ i < j ≤ m), then r (vi |) = r (v j |) by Lemma 2.1. If x = u 1 and y = v1 , then d(x, R3 ) = 2 = 1 = d(y, R3 ). If x = u 2 and y = vi where i ∈ [2, n − 1] or x = vn and y = u j where j ∈ [3, m], then d(x, Ra ) = 1 < d(y, Ra ) for some a = t. Thus we can conclude that r (x|) = r (y|) for any x, y ∈ V (G) which implies that  is a resolving partition of G.  In the next theorem, we give the partition dimension of a disconnected graph consisting of a star K 1,n and a cycle Cm . Theorem 2.6. For n ≥ 3 and m ≥ 4, ⎧ if (n = 3 and m ≥ 4, m = 5), or (m ≥ n ≥ 4), ⎨ n, pdd(K 1,n ∪ Cm ) = n + 1, if (n = 3 and m = 5), ⎩ ∞, otherwise. Proof. For n ≥ 3 and m ≥ 4, let G = K 1,n ∪ Cm and V (G) = V (K 1,n ) ∪ V (Cm ) = {v, vi : i ∈ [1, n]} ∪ {u j : j ∈ [1, m]}. For (n = 3 and m ≥ 4, m = 5), or (m ≥ n ≥ 4), or (n = 3 and m = 5), we claim that pdd(G) < ∞. Hence pdd(G) ≥ n by Theorem 1.2. We consider two cases. Case 1. Let (n = 3 and m = 4, 6, 7, . . .) or (m ≥ n ≥ 4). We will prove that pdd(G) ≤ n. Let 1 = {R1 , R2 , . . . , Rn } be a partition of G, where Ri = {vi , u i } for any i ∈ [2, n] and the remaining vertices are in R1 . Let x, y be any two distinct vertices of G in Rt for some t ∈ [1, n]. If x = vi and y = u i where i ∈ [2, n], then d(x, Ra ) = 2 = 1 = d(y, Ra ) for some integer a ∈ {i + 1, i − 1}\{1}. If x = v and y ∈ {v1 , u 1 , u j : j ∈ [n + 1, m]},

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then d(x, Ra ) = 1 < d(y, Ra ) for some a ∈ {2, 3}. If x = v1 and y ∈ {u 1 , u j : j ∈ [n + 1, m]}, then d(x, Ra ) = 2 = d(y, Ra ) for some a = t. If x = u j and y = u k where n + 1 ≤ j < k ≤ m, by Lemma 2.1 since 1 is connected, r (x|1 ) = r (y|1 ). Therefore any two distinct vertices x, y ∈ V (G) have distinct representation under 1 and thus 1 is a resolving partition of G. Case 2. Let n = 3 and m = 5 and so G = K 1,3 ∪ C5 . It is easy to check that for any resolving 3-partition  of K 1,3 and C5 , there always exist x ∈ V (K 1,3 ) and y ∈ V (C5 ) such that r (x|) = r (y|). Therefore, pdd(G) ≥ 4. To prove that pdd(G) ≤ 4, it is routine to verify that a partition 2 = {R1 , R2 , R3 , R4 } of G where for i ∈ [2, 4], Ri = {vi−1 , u i } and the remaining vertices are in R1 is a resolving partition of G. Furthermore, for m < n, we suppose for the contrary that pdd(G) < ∞. Then, n ≤ pdd(G) ≤ m by Theorem 1.2, a contradiction.  In the following result, we give partition dimension of a graph containing a path Pn and a homogeneous caterpillar C(k, l). Theorem 2.7. For n, l ≥ 3 and k ≥ 2, ⎧ l, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ pdd(Pn ∪ C(k, l)) = l + 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ ∞,

if (n ≥ l = 3 and k = 2), or (n ≥ 6 and l = k = 3), or (n ≥ l ≥ 4 and k ≤ l), if (n = 4, 5 and l = k = 3), or (n > l ≥ 3 and k ≥ l + 1), otherwise.

Proof. Let G = Pn ∪ C(k, l) where n, l ≥ 3, k ≥ 2 and V (G) = V (Pn ) ∪ V (C(k, l)) = {vi : i ∈ [1, n]}∪ {u i , u i, j : i ∈ [1, k], j ∈ [1, l]}. In the following Cases 1–4, we claim that pdd(G) < ∞. Case 1. Let n ≥ l = 3 and k = 2. Then, pdd(G) ≥ pd(C(k, l)) ≥ 3 by Theorem 1.2. Now we will show that pdd(G) ≤ 3. Define a partition 1 = {R1 , R2 , R3 } of G where R1 = {v1 , u 1 , u 1,1 , u 2,1 }, R2 = {v2 , u 1,2 , u 2,2 } and the remaining vertices are in R3 . It is routine to verify that any two vertices x, y ∈ V (G) in Rt for some t ∈ [1, 3] have distinct representation under 1 . Therefore, 1 is a resolving partition of G. Case 2. Let n ≥ 6 and l = k = 3. By Theorem 1.2, pdd(G) ≥ 3. Let 2 = {R1 , R2 , R3 } be a partition of V (G) where for each j ∈ [1, 3], R1 = {v1 , u 1 , u j,1 }, R2 = {v2 , v3 , v4 , v5 , u 2 , u j,2 } and the remaining vertices are in R3 . Now, consider any two distinct vertices x, y ∈ V (G) in Rt for some t ∈ [1, 3]. Clearly, if x = va and y = vb where a < b, then r (x|2 ) = r (y|2 ) by Lemma 2.1, since 2 is a connected partition of Pn . If x = u i and y ∈ {u j,i , va } for some a, then d(x, R p ) = 1 < d(y, R p ) for some p = t. If x = u i, j and y ∈ {u k, j , va }, then d(x, R p ) = 2 = d(y, R p ) for some p = t. Therefore, r (x|2 ) = r (y|2 ) and so 2 is a resolving partition of G. Case 3. Let n ≥ l ≥ 4 and k ≤ l. Then, pdd(G) ≥ l by Theorem 1.2. Let 3 = {R1 , R2 , . . . , Rl } be a partition of G induced by the function f 3 : V (G) → {1, 2, . . . , l} as follows.

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f 3 (v j ) = f 3 (u i, j ) = j, for each i ∈ [1, k], j ∈ [1, l − 1], f 3 (vi ) = l, for each i ∈ [l, n], f 3 (u i ) = i, for each i ∈ [1, k], where f 3 (x) = i means x ∈ Ri . By a similar reason to Case 2, for any two vertices x, y ∈ V (G) in Rt , we have r (x|3 ) = r (y|3 ). Hence 3 is a resolving partition of G. Case 4. Let (n = 4, 5 and l = k = 3) or (n > l ≥ 3 and k ≥ l + 1). For n = 4, 5 and l = k = 3, we can check easily that for any 3-partition of G there always exists at least a vertex x ∈ V (Pn ) and y ∈ V (C(k, l)) with the same representation under such 3-partition. Thus, pdd(G) ≥ 4. For n > l ≥ 3 and k ≥ l + 1, we have pdd(G) ≥ l + 1 by Theorem 1.3. For these two conditions, we will show that pdd(G) ≤ l + 1. Let 4 = {R0 , R1 , R2 , . . . , Rl } be a partition of G induced by the function f 4 as follows. f 4 (vi ) = i, for any i ∈ [1, l], f 4 (v j ) = 0, for any j ∈ [l + 1, n] f 4 (u 1 ) = l, f 4 (u i ) = (i − 1)

mod l, for any i ∈ [2, k],

f 4 (u i, j ) = j − 1, for any i ∈ [1, k], j ∈ [1, l]. We consider any two vertices x, y ∈ V (G) in Rt for some t ∈ [0, l]. If x = vi and y = v j where i, j ∈ [l + 1, n], then clearly r (x|4 ) = r (4 ) by Lemma 2.1. If x ∈ {u 1 , u 2 } and y ∈ {vi , u j , u k,l }, then d(x, R p ) = 1 < d(y, R p ) for some p = t. If x = u i and y = u j for some i, j ∈ [3, k], then d(x, Rl ) < d(y, Rl ) for i < j. If x = u i and y ∈ {u a,b , vc } / {t, a − 1( mod l), c − 1, c + 1}. for i ≥ 3, then d(x, R p ) = 1 < d(y, R p ) for some p ∈ If x = u i,a and y ∈ {u j,a , vb }, then d(x, R p ) = d(y, R p ) for some p ∈ {l, 0}. Thus r (x|4 ) = r (y|4 ) for any two distinct vertices x, y ∈ V (G). Therefore, 4 is a resolving partition of G. Case 5. Let n < l or n = l = k = 3 or n = l < k. For n < l or n = l = k = 3, it is easy to see that pdd(G) = ∞. For n = l < k, we suppose for the contrary that pdd(G) < ∞. By using Theorem 1.3, we have pd(C(k, l)) = l + 1. By Theorem 1.2, we have l + 1 ≤ pdd(G) ≤ n = l, a contradiction.  By observing the previous results, we have the following conjecture. Conjecture 2.8. Let G = F ∪ H where F = Pn or Cn , pd(H ) ≥ pd(F) and H = K m . If pdd(G) < ∞, then pd(H ) ≤ pdd(G) ≤ pd(H ) + 1. The upper bound of Conjecture 2.8 is tight, since pdd(P4 ∪ C(3, 3)) = pd(C(3, 3)) + 1 = 4 and pdd(C5 ∪ K 1,3 ) = pd(K 1,3 ) + 1 = 4. 3. Components with the same partition dimension We know that the only graph of order n having the partition dimension n − 1 are K 1,n−1 , K n − e, or K 1 + (K 1 ∪ K n−2 ) [3]. In this section, we determine pdd(G 1 ∪ G 2 ), where |V (G 1 )| = |V (G 2 )| = n and pd(G 1 ) = pd(G 2 ) = n − 1. We also derive the upper bound of pdd(2T ) where T is a tree. We begin this part with the following lemma.

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Lemma 3.1. For n ≥ 4, let  be a resolving partition of K n − e and α() be the number of 1-distance vertex of K n − e with respect to . Then,  n − 1, if  is a minimum resolving partition, α() = n − 2, otherwise. Proof. For n ≥ 4, let G = K n − e, where V (G) = {vi : 1 ≤ i ≤ n} and E(G) = {vi v j : i = j}\{v1 vn }. Since pd(G) = n − 1, let 1 = {R1 , R2 , . . . , Rn−1 } be any minimum resolving partition of G. Note that d(vi , vk ) = d(v j , vk ) for all vk ∈ V (G)\{vi , v j } for all i, j ∈ [2, n − 1]. Therefore, vi and v j for each i, j ∈ [2, n − 1] are belong to distinct partition class of  by Lemma 1.1. Without loss of generality, assume that vi ∈ Ri−1 for all i ∈ [2, n − 1]. Since v1 and vn have the same distance to the other vertices, these two vertices belong to distinct partition classes of 1 . Thus, assume that v1 ∈ Rn−1 and vn ∈ Rk for some k ∈ [1, n −2]. Therefore, v j are 1-distance vertices for all j ∈ [1, n −1]. On the other hand, if we have a resolving n-partition 2 = {R1 , R2 , . . . , Rn } of K n − e, then Ri is a singleton partition for all i ∈ [1, n]. So, we could assume that Ri = {vi } for each i ∈ [1, n]. It follows that vk are 1-distance vertices for all k ∈ [2, n − 1]. This concludes the proof.  Theorem 3.2. Let G 1 and G 2 be two graphs of order n ≥ 4 and pd(G 1 ) = pd(G 2 ) = n − 1. Then, ⎧ n − 1, if (G 1 = K n − e and G 2 = K n − e), ⎪ ⎪ ⎨ n, if (G 1 = K n − e, G 2 = K n − e), or pdd(G 1 ∪ G 2 ) = (n = 4 and G 1 = G 2 = K n − e), ⎪ ⎪ ⎩ ∞, if (n ≥ 5 and G 1 = G 2 = K n − e). Proof. Let G = G 1 ∪ G 2 where pd(G 1 ) = pd(G 2 ) = n − 1. Then, G 1 and G 2 are one of these graphs: K 1,n−1 , K n − e or K 1 + (K 1 ∪ K n−2 ). Therefore, to determine the partition dimension of G, we only need to consider two disjoint union of these graphs. Now, let V (K 1,n−1 ) = {v, vi : i ∈ [1, n − 1]}, E(K 1,n−1 ) = {vvi : i ∈ [1, n − 1]}, V (K n − e) = {wi : i ∈ [1, n]}, E(K n − e) = {wi w j : i, j ∈ [1, n]}\{w1 wn }, V (K 1 + (K 1 ∪ K n−2 )) = {x, y, u i : i ∈ [1, n − 2]} and E(K 1 + (K 1 ∪ K n−2 )) = {x y, xu i , u i u j : i, j ∈ [1, n − 2]}. Notice that for G i = K n − e for some i ∈ [1, 2], we claim that pdd(G) < ∞ so that pdd(G) ≥ n − 1 by Theorem 1.2. Now, we investigate the following three cases. Case 1. Let G 1 = K n − e and G 2 = K n − e. We will show that pdd(G) ≤ n − 1. We consider three subcases. Subcase 1.1. Let G = K 1,n−1 ∪ (K 1 + (K 1 ∪ K n−2 )). Let 1 = {R1 , R2 , . . . , Rn−1 } be a partition of V (G) where R1 = {v, v1 , u 1 }, R2 = {x, v2 , u 2 }, R j = {v j , u j : j ∈ [3, n − 2]} and Rn−1 = {y, vn−1 }. Let us consider any two vertices p, q ∈ V (G) in Rt for some t ∈ [1, n − 1]. If p = v and q ∈ {v1 , u 1 } or p = x and q ∈ {v2 , u 2 }, then d( p, Rn−1 ) = 1 = 2 = d(q, Rn−1 ). If p = v j and q = u j for some j ∈ [1, n − 2], then / { j, n − 1}. If p = y and q = vn−1 , then d( p, Rl ) = 2 = 1 = d(q, Rl ) for some l ∈ d( p, R1 ) = 1 = 2 = d(y, R1 ). Therefore, r ( p|1 ) = r (q|1 ) for any two distinct vertices p, q ∈ V (G), and thus 1 is a resolving partition of G = K 1,n−1 ∪ (K 1 + (K 1 ∪ K n−2 )). Subcase 1.2. Let G = 2K 1,n−1 . Define a partition 2 = {R1 , R2 , . . . , Rn−1 } of G by taking the center vertex v of the first copy of G in R1 and the center vertex v of the second

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copy in R2 , while the other remaining vertices vi ∈ Ri for each i ∈ [1, n − 1]. We can easily show that 2 is a resolving partition of G = 2K 1,n−1 . Subcase 1.3. Let G = 2(K 1 + (K 1 ∪ K n−2 )). Let 3 = {R1 , R2 , . . . , Rn−1 } be a partition of G in which x of the first copy of G in R1 , x of the second copy in Rn−1 , y of the first copy in Rn−1 while y of the second copy in R1 , and the remaining vertices u i ∈ Ri for all i ∈ [1, n − 2]. We also can verify directly that 3 is a resolving partition of G = 2(K 1 + (K 1 ∪ K n−2 )). This concludes the proof that pdd(G 1 ∪G 2 ) = n−1 for G 1 = K n −e and G 2 = K n −e. Case 2. Let (G 1 = K n − e and G 2 = K n − e) or (G 1 = G 2 = K 4 − e). Note that for any resolving (n − 1)-partition  of each K 1,n−1 or K 1 + (K 1 ∪ K n−2 ), v ∈ V (K 1,n−1 ) or x ∈ V (K 1 + (K 1 ∪ K n−2 )) must be a 1-distance vertex with respect to such partition . For any minimum resolving (n − 1)-partition of K n − e, we have that K n − e has (n − 1) 1-distance vertices under such partition, by Lemma 3.1. Therefore, if G 1 = K n −e and (G 2 = K 1,n−1 or K 1 + (K 1 ∪ K n−2 )), then pdd(G) ≥ n. Now for G = K n − e ∪ K 1,n−1 , let 4 = {R1 , R2 , . . . , Rn } be a partition of V (G) where R1 = {w1 , v} and Ri = {wi , vi−1 } for every i ∈ [2, n], while for G = K n − e ∪ (K 1 + (K 1 ∪ K n−2 )), let 5 = {R1 , R2 , . . . , Rn } be a partition of V (G) where Ri = {wi , u i } for all i ∈ [1, n − 2], Rn−1 = {wn−1 , y} and Rn = {wn , x}. It is routine to show that 4 or 5 are resolving partitions of G = K n − e ∪ K 1,n−1 or G = K n − e ∪ (K 1 + (K 1 ∪ K n−2 )), respectively. Now, assume G = 2(K 4 − e). By Lemma 3.1, for any minimum resolving 3-partition of K 4 − e, the number of 1-distance vertex of K 4 − e is 3. Thus, pdd(G) ≥ 4. Again by Lemma 3.1, the number of 1-distance vertex of K 4 − e with resolving 4-partition is 2. This is easy to define a resolving 4-partition  of each component of G in which the two 1-distance vertices in the first copy and the two 1-distance vertices in the second copy of G are in distinct partition classes. Since each partition class of each component of G are singleton,  is a resolving partition of G. It follows that for (G 1 = K n − e and G 2 = K n − e) or (n = 4 and G 1 = G 2 = K n − e), we have pdd(G) = n. Case 3. Let G = 2(K n − e) where n ≥ 5. We suppose for the contrary that there exists a resolving partition  of G. By considering Lemma 3.1, G has at least 2(n − 2) 1-distance vertices under . This is a desired contradiction since || ≤ n < n + 1 ≤ 2(n − 2).  For a tree T , the maximum distance between a vertex x ∈ V (T ) to all other vertices in T is considered as eccentricity of x, and denoted by ecc(x). Furthermore, a vertex x ∈ V (T ) is called a peripheral vertex if ecc(x) = diam(T ). Now we are going to derive the upper bound of the partition dimension of 2T . We begin with the following lemma. Lemma 3.3. Let T be a tree of order ≥ 3 and  = {R1 , R2 , . . . , Rt } be a minimum resolving partition of T . Let x y ∈ E(T ) and x be a peripheral vertex in Ri for some i ∈ [1, t]. If |Ri | = 1, then a new partition   = {R1 , R2 , . . . , Rt } where R j = R j \{y} for all j = i and Ri = {x, y} is also a resolving partition of T . Proof. Assume that y ∈ R j where j = i. If |R j | = 1, then a partition  ∗ = {R1∗ , R2∗ , . . . , Rt∗ }\{R ∗j } where Rl∗ = Rl for all l = i and Ri∗ = {x, y} is also a resolving partition of T . Contradict to the fact that  is a minimum resolving partition of T . Therefore, we have |R j | ≥ 2. Now, let us show that a partition   = {R1 , R2 , . . . , Rt } of T where R j = R j \{y} for all j = i and Ri = {x, y}, is a resolving partition of T .

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We consider any two vertices u, v ∈ V (T ) in Ra for some a ∈ [1, t]. If u = x and v = y, then d(u, Rb ) = d(v, Rb ) + 1 > d(v, Rb ) for all b = a. Now consider two vertices u, v ∈ V (T )\{x, y}. If d(u, y) = d(v, y), then d(u, Ri ) = d(v, Ri ). Otherwise, we distinguish two cases. Case 1. Assume that u and v are resolved by Rk with respect to  for some k = j. This implies that d(u, Rk ) = d(u, Rk ) = d(v, Rk ) = d(v, Rk ). Case 2. Assume that u and v are only resolved by R j with respect to . Then, there exist two vertices p, q ∈ R j such that d(u, R j ) = d(u, p) = d(v, q) = d(v, R j ). Note that at most one of p or q can be a vertex y. If neither p nor q are equal to y, then d(u, R j ) = d(u, p) = d(v, q) = d(v, R j ). If p = y, then d(u, R j ) = d(u, p) = d(u, y) = d(v, y) > d(v, q) = d(v, R j ). This implies that d(u, R j ) = d(u, y1 ) ≥ d(u, y) > d(v, q) = d(v, R j ) for other vertex y1 = y in R j . Similarly, if q = y, then d(u, R j ) = d(u, p) < d(v, y) = d(v, q) = d(v, R j ) and hence d(u, R j ) = d(u, p) < d(v, y) ≤ d(v, y2 ) = d(v, R j ) for other vertex y2 = y in R j . Therefore, u and v are also  resolved by R j . By Lemma 3.3, it is easy to verify the following result. COROLLARY 3.4 Let x be any peripheral vertex of a tree T . Then, there exists a minimum resolving partition  = {R1 , R2 , . . . , Rt } of T such that x ∈ Ri and |Ri | ≥ 2 for some i ∈ [1, t]. Furthermore, a peripheral vertex in T satisfying Corollary 3.4 is called an ideal vertex of T . Now we are ready to give the upper bound of partition dimension of two trees 2T , as follows. Theorem 3.5. Let T be a tree of order n ≥ 3. Then, pdd(2T ) ≤ pd(T ) + 1. Proof. Let V (2T ) = V (T1 ) ∪ V (T2 ) and E(2T ) = E(T1 ) ∪ E(T2 ) where for i ∈ [1, 2], V (Ti ) = {u i : u ∈ V (T )} and E(Ti ) = {u i vi : uv ∈ E(T )}. Let  = {R1 , R2 , . . . , Rt } be a minimum resolving partition of T . Thus, i = {Ri1 , Ri2 , . . . , Rit } where Ri j = {u i : u ∈ R j } for all j ∈ [1, t] and i ∈ [1, 2], is also a minimum resolving partition of each Ti . Let x ∈ V (T ) be an ideal vertex in Rs ∈  and so that |Rs | ≥ 2. Therefore, xi ∈ V (Ti ) is an ideal vertex in Ris where |Ris | ≥ 2 with respect to i for each i ∈ [1, 2]. Let xi yi ∈ E(Ti ) where yi ∈ Ril for some l ∈ [1, t].  } of T ∪ T For S j = R1 j ∪ R2 j , we define a new partition   = {S1 , S2 , . . . , St , St+1 1 2 where Si = Si \{x1 , x2 } for all i ∈ [1, t]\{l}, Sl = Sl ∪ {x2 }\{y2 },  St+1 = {x1 , y2 }. To show that   is a resolving partition, we consider any two distinct vertices u i , v j ∈ V (T1 ) ∪ V (T2 ) in Sa for some i, j ∈ [1, 2] and a ∈ [1, t + 1]. We consider three cases. Case 1. Let u 1 , v1 ∈ V (T1 )\{x1 }. We consider two subcases.

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Subcase 1.1. Assume that u and v in T are resolved by Rb where b = s. Then, u 1 and v1 in T1 are also resolved by Sb , since d(u 1 , Sb ) = d(u, Rb ) = d(v, Rb ) = d(v1 , Sb ). Subcase 1.2. Assume that u and v in T are only resolved by Rs . Thus, there exist two vertices p, q ∈ Rs such that d(u, Rs ) = d(u, p) = d(v, q) = d(v, Rs ). If d(u, y) =  ) = d(u , x ) = d(u, x)  = d(v, x) = d(v , x ) = d(v , S  ). d(v, y), then d(u 1 , St+1 1 1 1 1 1 t+1 Otherwise, we distinguish three subcases. Subcase 1.2.1. Neither p nor q are equal to x. Then, d(u 1 , Ss ) = d(u 1 , p1 ) = d(u, p) = d(v, q) = d(v1 , q1 ) = d(v1 , Ss ). Subcase 1.2.2. Let p = x. Then, d(u 1 , Ss ) = d(u 1 , s1 ) ≥ d(u 1 , p1 ) = d(u, p) = d(u, x) = d(v, x) > d(v, q) = d(v1 , q1 ) = d(v1 , Ss ) for some s1 ∈ Ss . Subcase 1.2.3. Let q = x. By a similar way to Subcase 1.2.2, then d(u 1 , Ss ) = d(u 1 , p1 ) = d(u, p) < d(v, x) = d(v, q) = d(v1 , q1 ) ≤ d(v1 , t1 ) = d(v1 , Ss ) for some t1 ∈ Ss . Therefore, if u and v are resolved by Rs and d(u, y) = d(v, y), then u 1 and v1 are resolved by Ss . Case 2. u 2 , v2 ∈ V (T2 )\{y2 }. Subcase 2.1. Assume that u and v in T are resolved by Rb where b ∈ / {l, s}. Then, u 2 and v2 in T2 are also resolved by Sb , since d(u 2 , Sb ) = d(u, Rb ) = d(v, Rb ) = d(v, Sb ). Subcase 2.2. Assume that u and v in T are only resolved by Rs . Thus, there exist two vertices p, q ∈ Rs such that d(u, Rs ) = d(u, p) = d(v, q) = d(v, Rs ). If d(u, y) =  ) = d(u , y ) = d(u, y)  = d(v, y) = d(v , y ) = d(v , S  ). d(v, y), then d(u 2 , St+1 2 2 2 2 2 t+1 Otherwise, we distinguish in three subcases. Subcase 2.2.1. Neither p nor q are equal to x. Then, d(u 2 , Ss ) = d(u 2 , p2 ) = d(u, p) = d(v, q) = d(v2 , q2 ) = d(v2 , Ss ). Subcase 2.2.2. Let p = x. Then, d(u 2 , Ss ) = d(u 2 , s2 ) ≥ d(u 2 , p2 ) = d(u, p) = d(u, x) = d(v, x) > d(v, q) = d(v2 , q2 ) = d(v2 , Ss ) for some s2 ∈ Ss . Subcase 2.2.3. Let q = x. Similarly by Subcase 2.2.2, then d(u 2 , Ss ) = d(u 2 , p2 ) = d(u, p) < d(v, x) = d(v, q) = d(v2 , q2 ) ≤ d(v2 , t2 ) = d(v2 , Ss ) for some t2 ∈ Ss . Therefore, if u and v are resolved by Rs and d(u, y) = d(v, y), then u 2 and v2 are resolved by Ss . Subcase 2.3. Assume that u and v are only resolved by Rl . Thus, there exist two vertices p, q ∈ Rl such that d(u, Rl ) = d(u, p) = d(v, q) = d(v, Rl ). If d(u, y) = d(v, y), then  ) = d(u , y ) = d(u, y)  = d(v, y) = d(v , y ) = d(v , S  ). Otherwise, we d(u 2 , St+1 2 2 2 2 2 t+1 distinguish in three subcases. Subcase 2.3.1. Neither p nor q are equal to y. Then, d(u 2 , Sl ) = d(u 2 , p2 ) = d(u, p) = d(v, q) = d(v2 , q2 ) = d(v2 , Sl ). Subcase 2.3.2. Let p = y. Then, d(u 2 , Sl ) = d(u 2 , s2 ) ≥ d(u 2 , p2 ) = d(u, p) = d(u, y) = d(v, y) > d(v, q) = d(v2 , q2 ) = d(v2 , Sl ) for some s2 ∈ Sl . Subcase 2.3.3. Let q = y. By a similar way to Subcase 2.3.2, then d(u 2 , Sl ) = d(u 2 , p2 ) = d(u, p) < d(v, y) = d(v, q) = d(v2 , q2 ) ≤ d(v2 , t2 ) = d(v2 , Sl ) for some t2 ∈ Sl . Therefore, if u and v are resolved by Rl and d(u, y) = d(v, y), then u 2 and v2 are resolved by Sl .

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Case 3. Let u 1 ∈ V (T1 ) and v2 ∈ V (T2 ). If u 1 = x1 and v2 = y2 , then d(u 1 , Sb ) = d(x1 , Sb ) = d(y1 , Sb ) + 1 = d(y2 , Sb ) + 1 > d(y2 , Sb ) = d(v2 , Sb ) for all b ∈ / {a, l}. If u 1 = y1 and v2 = x2 , then d(u 1 , Sb ) = d(y1 , Sb ) = d(y2 , Sb ) < d(y2 , Sb ) + 1 =  ) = d(u , x ) = / {a, t + 1}. Otherwise, d(u 1 , St+1 d(x2 , Sb ) = d(v2 , Sb ) for all b ∈ 1 1  d(u 1 , y1 ) + 1 = d(v2 , y2 ) + 1 > d(v2 , y2 ) = d(v2 , St+1 ). This implies that r (u i |  ) = r (v j |  ) for any two vertices u i , v j ∈ V (T1 ) ∪ V (T2 ).  By considering the previous results, we have the following conjecture. Conjecture 3.6. Let G 1 and G 2 be two graphs of order n ≥ 3 and not isomorphic to complete graphs. If pd(G 1 ) = pd(G 2 ) = k and pdd(G 1 ∪ G 2 ) < ∞, then pdd(G 1 ∪ G 2 ) ≤ k + 1. There are several graphs G with two components where pdd(G) satisfying the upper bound above, such as for two copies of caterpillar C(3, 3) where pdd(2C(3, 3)) = pd(C(3, 3))+1 = 4, and for G = K n −e∪ K 1,n−1 where pdd(G) = pd(K n −e)+1 = n.

Acknowledgements This research was supported by Research Grant: “Program Penelitian dan Pengabdian kepada Masyarakat-Institut Teknologi Bandung (P3MI-ITB)”, Ministry of Research, Technology and Higher Education, Indonesia. The research for this article was also supported by APVV-15-0116 and by VEGA 1/0233/18.

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