Computing the edge irregularity strengths of chain graphs and the join of two graphs

In computer science, graphs are used in variety of applications directly or indirectly. Especially quantitative labeled graphs have played a vital role in computational linguistics, decision making software tools, coding theory and path determination in networks. For a graph G ( V, E ) with the vertex set V and the edge set E, a vertex k -labeling φ : V → { 1 , 2 , . . . , k } is deﬁned to be an edge irregular k -labeling of the graph G if for every two different edges e and f their w φ ( e ) 6 = w φ ( f ) , where the weight of an edge e = xy ∈ E ( G ) is w φ ( xy ) = φ ( x ) + φ ( y ) . The minimum k for which the graph G has an edge irregular k -labeling is called the edge irregularity strength of G , denoted by es ( G ) . In this paper, we determine the edge irregularity strengths of some chain graphs and the join of two graphs. We introduce a conjecture and open problems for researchers for further research.


Introduction
A graph G(V, E) with the vertex set V and the edge set E is connected if for any pair of vertices in G there exists a path connecting them. For a graph G, the degree of a vertex v is the number of edges incident to v and denoted by d (v). Two vertices are adjacent if and only if there is an edge between them.
A graph labeling is an assignment of integers to the vertices or edges or both with subject to certain condition(s). If the domain of the mapping is the set of vertices (or edges), then the labeling is called a vertex labeling (or an edge labeling). If the domain is V (G) ∪ E(G) then we call the labeling a total labeling. Thus, for an edge k-labeling φ : E(G) → {1, 2, . . . , k} the associated weight of a vertex x ∈ V (G) is where the sum is over all vertices y adjacent to x.
Chartrand et al. [9] introduced edge k-labeling φ of a graph G such that w φ (x) = w φ (y) for all vertices x, y ∈ V (G) with x = y. Such labelings were called irregular assignments and the irregularity strength s(G) of a graph G is known as the minimum k for which G has an irregular assignment using labels at most k. This parameter has attracted much attention [5,6,8,11].
In 2007, Bača et al. [7] investigated two modifications of the irregularity strength of graphs, namely a total edge irregularity strength, denoted by tes(G) and a total vertex irregularity strength, denoted by tvs(G). Some results on total edge irregularity strength and total vertex irregularity strength can be found in [1,2,3,12,13].
Motivated by these papers, Ahmad et al. [4] introduced the following irregular labeling: A vertex k-labeling φ : V → {1, 2, . . . , k} is defined to be an edge irregular k-labeling of the graph G if for every two different edges e and f their w φ (e) = w φ (f ), where the weight of an edge e = xy ∈ E(G) is w φ (xy) = φ(x) + φ(y). The minimum k for which the graph G has an edge irregular k-labeling is called the edge irregularity strength of G denoted by es(G).
The following theorem that is proved in [4], establishes a lower bound for the edge irregularity strength of a graph G.
In [4] it is shown that for a path P n , n ≥ 2, es(P n ) = ⌈ n 2 ⌉, for a star K 1,n , n ≥ 1, es(K 1,n ) = n, for a double star S m,n , 3 ≤ m ≤ n, es(S m,n ) = n and for the Cartesian product of two paths P n and P m , m, n ≥ 2, es(P n ✷P m ) = ⌈ 2mn−m−n+1 2 ⌉. I. Tarawneh et al. [14,15] determined the edge irregularity strength of the corona product of graphs with paths and cycle with isolated vertices.

Edge irregularity strength of chain graphs
A chain graph is a graph with blocks B 1 , B 2 , . . . , B n such that for every i, B i and B i+1 have a common vertex in such a way that the block-cut-vertex graph is a path. We will denote the chain graph with n blocks B 1 , B 2 , . .
According to the Theorem 1.
For the converse, we define a vertex labeling φ as follows: It is a routine matter to verify that all vertex labels are at most 2n + 1, and the edge weights form the set of different integers, namely {2, 3, 4, . . . , 4n + 1}. Thus, the labeling φ is the desired edge irregular (2n + 1)-labeling. This completes the proof.
From the Theorem 2.1, we proposed the following conjecture: We denote by mK n -path a chain graph with m blocks where each block is identical and isomorphic to the complete graph K n . We consider edge irregular k-labelings of mK n -paths for n = 2, 3 and 4. If n = 2 then mK 2 -path ∼ = P m + 1. It is well known that P n has an edge irregular ⌈ n 2 ⌉labeling. Consequently, es(P n ) = ⌈ n 2 ⌉. If n = 3, then In the next theorem, we determine the bounds for edge irregularity strength of mK 3 -path.
Proof. Let us consider the vertex set and the edge set of H m ( ∼ = mK 3 -path): Observe that the graph H m has 2m + 1 vertices, 3m edges and ∆(H m ) = 4, for m ≥ 2. According to the Theorem 1.
Since every block is a complete graph K 3 , therefore under every edge irregular labeling no couple of adjacent vertices can be assigned by the same label. This implies that the smallest edge weight 2 is not possible. So if the smallest edge weight is 3 then the largest edge weight will be at least 3m + 2. Since each edge weight is a sum of two labels, at least one label is at least 3m+2 2 , as 3m + 2 is divisible by 2, when m is even, therefore the label of both end vertices of largest edge weight will be 3m+2 2 , which is not possible because no couple of adjacent vertices can be assigned by the same label. Therefore, at least one label is at least For the upper bound, we define a vertex labeling φ 1 as follows: Proof. Let us consider the vertex set and the edge set of G( ∼ = mK 4 -path): Observe that the graph G has 3m+1 vertices, 6m edges and ∆(G) = 6. According to the Theorem 1.1, es(G) ≥ max 6m+1 2 , 6 = 3m + 1, for m ≥ 2. Since every block is a complete graph K 4 , therefore under every edge irregular labeling no two adjacent vertices can be assigned by the same label. This implies that the smallest edge weight 2 is not possible. So if the smallest edge weight is 3 then the largest edge weight will be at least 6m + 2. Since each edge weight is a sum of two labels, at least one label is at least 6m+2 2 = 3m + 1, as 6m + 2 is divisible by 2, therefore the label of both end vertices of largest edge weight will be 3m + 1, which is not possible because no two adjacent vertices can be assigned by the same label. Therefore, at least one label is at least 3m + 2. For the converse, we define the vertex labeling φ 2 as follows: It is a routine matter to verify that all vertex labels are at most 3m + 2. and the edge weights form the set of different integers, namely {3, 4, 5, . . . , 6m+2}. This implies that es(G) ≤ 3m+2, for m ≥ 2. This completes the proof.
Open Problem 2. Determine the edge irregularity strength of a mK n -path for m ≥ 2 and n ≥ 5.

Edge irregularity strength of join of two graphs
There are several ways to produce a new graph from a given pair of graphs. For two vertexdisjoint graphs G and H, G ∪ H is disconnected graph with the vertex set V (G) ∪ V (H) and the edge set E(G) ∪ E(H). The join G + H consists of G ∪ H and all the edges joining a vertex of G and a vertex of H. For detail see [10]. Proof. Let G = K 1,n + K 1 be a graph with the vertex set V (G) = {x, y} ∪ {x i : 1 ≤ i ≤ n} and the edge set E(G) = {xy, xx i , yx i : 1 ≤ i ≤ n}. Then |V (G)| = n + 2, |E(G)| = 2n + 1 and ∆(G) = n + 1. According to the Theorem 1.1 es(G) ≥ max 2n+2 2 , n + 1 = n + 1.
Since each two adjacent vertices in G are a part of a complete graph K 3 , therefore under every edge irregular labeling the all vertices in G must contain different labels. Since there are n + 2 vertices in G, then the maximum vertex label is at least n + 2. Therefore es(G) ≥ n + 2. To prove the equality, it suffices to prove the existence of an optimal edge irregular (n + 2)-labeling. Let φ : V (G) → {1, 2, . . . , n + 2} be a vertex labeling such that so the edge weights are distinct for all edges. Thus, the vertex labeling φ is an optimal edge irregular (n + 2)-labeling. This completes the proof.
When m = 1 and n ≥ 1, P m + K n is a star K 1,n , the edge irregularity strength of star is determined in [4] i. e es(K 1,n ) = n. When m = 2 and n ≥ 1, P m + K n is isomorphic to K 1,n + K 1 , the edge irregularity strength of K 1,n + K 1 is determined in Theorem 3.1. Therefore es(P m + K n ) = n + 2, for m = 2. In the next theorem, we determine the bounds of the edge irregularity strength of P m + K n for m ≤ 6 and n ≥ 3.
According to the Theorem 1.
This completes the proof.
Open Problem 3. Find the edge irregularity strength of P m + K n for any n ≥ 1 and m ≥ 7. Proof. Let us consider the vertex set V (G) = {x, y, x i , y j : 1 ≤ i ≤ m, 1 ≤ j ≤ n} and the edge set E(G) = {xy, xx i , xy j , yy j , yx i : This implies that the maximum degree ∆(G) = n + 1. According to the Theorem 1.1, es(G) ≥ max 2m+2n+2 2 , n + 1 = m + n + 1. Since each two adjacent vertices in G are a part of complete graph K 3 , in this way under every edge irregular labeling the smallest edge weight has to be at least 3 and the largest edge weight has to be at least 2m + 2n + 3. Since the edge weight 2m + 2n + 3 is the sum of two labels, so at least one label is at least 2m+2n+3 2 = m + n + 2. Therefore es(G) ≥ m + n + 2.