The total disjoint irregularity strength of some certain graphs

Under a totally irregular total k-labeling of a graph G = (V,E), we found that for some certain graphs, the edge-weight set W (E) and the vertex-weight set W (V ) of G which are induced by k = ts(G), W (E) ∩W (V ) is a nonempty set. For which k, a graph G has a totally irregular total labeling if W (E) ∩W (V ) = ∅? We introduce the total disjoint irregularity strength, denoted by ds(G), as the minimum value k where this condition satisfied. We provide the lower bound of ds(G) and determine the total disjoint irregularity strength of cycles, paths, stars, and complete graphs.


Introduction
Let G be a finite, simple, and undirected graph with the vertex set V and the edge set E. Let f : V ∪ E → {1, 2, · · · , k} be a total k-labeling. Under f , the weight of a vertex v ∈ V is w(v) = f (v) + uv∈E f (uv) and the weight of an edge uv ∈ E is w(uv) = f (u) + f (uv) + f (v). If all the vertex (or edge)-weights are distinct then f is called a vertex (or edge) irregular total www.ijc.or.id The total disjoint irregularity strength of some certain graphs | M.I. Tilukay and A.N.M. Salman Figure 1. Totally disjoint irregular total labeling of (a) P 3 , (b) C 3 , (c) K 4 , and (d) K 5 k-labeling and the minimum value k such that G has a vertex (or edge) irregular total k-labeling is called the total vertex (or edge) irregularity strength, denoted by tvs(G) (or tes(G)), respectively. This parameters were introduced by Baca et al. [2]. They gave the boundary of tes(G) and tvs(G) and determined that for n vertices, tvs(C n ) = tes(C n ) = n+2 3 , tes(P n ) = n+1 3 , tvs(S n ) = tes(S n ) = n+1 2 , and tvs(K n ) = 2. Later, Jendrol et al. [7] provided a better lower bound of tes(G) and determined that tes(K 5 ) = 5 and tes(K n ) = n 2 −n+4 6 , for n = 5. For any tree T , Ivanco and Jendrol [6] proved that tes(T ) is equal to the lower bound. Nurdin et al. [9] gave the lower bound for tvs for any graph G.
Recently, Marzuki et al. [8] combined the properties of tes(G) and tvs(G) and gave new parameter called the total irregularity strength, denoted by ts(G). It is the minimum value k for which G has a totally irregular total k-labeling. They proved that the lower bound ts(G) ≥ max{tes(G), tvs(G)} is sharp for C n and P n except for P 2 or P 5 . In [14], we proved that for n = 2, ts(K n ) = tes(K n ). In [5], Indriati et al. proved that for n ≥ 3, ts(S n ) = tvs(S n ). For further reading, one can see [1], [3], [4], [5], and [10] - [13]. All these results showed that the lower bound is sharp.
Observing ts(G), for the vertex weight-set W (V ) and the edge weight-set W (E) under a totally irregular total labeling on G, W (V ) ∩ W (E) = ∅. Considering this condition, we define a new parameter called the total disjoint irregularity strength. A totally disjoint irregular total k-labeling of a graph G as a total labeling f : V ∪ E → {1, 2, · · · , k} which satisfies: (i) for any two vertices x = y ∈ V , w(x) = w(y); (ii) for any two edges x 1 y 1 = x 2 y 2 ∈ E, w(x 1 y 1 ) = w(x 2 y 2 ); (iii) W (V ) ∩ W (E) = ∅; where W (V ) (and W (E)) is the vertex (and edge) weight-set, respectively. The minimum value k such that a graph G has a totally disjoint irregular total labeling is called the total disjoint irregularity strength of a graph G, denoted by ds(G). Thus, for any graph G, For instance, Fig. 1 shows a totally disjoint irregular total labeling of P 3 , C 3 , K 4 , and K 5 .
In this paper, we determine ds of cycles, paths, stars, and complete graphs.

Main Results
Let G = (V, E) be a connected graph. For G has a totally irregular total k-labeling f : V ∪E → {1, 2, · · · , k}, we need |V | + |E| distinct weights. Let δ = δ(G) (or ∆ = ∆(G)) be the minimum (or maximum) degree of vertex in G, respectively. Let n i be the number of vertices of degree i, where i = δ, δ + 1, · · · , ∆. Then |V | = ∆ i=δ n i . Now, assume that δ = 1. Let f be a optimal labeling with respect to ds(G). Then the maximum weight has to be at least |E| + |V | + 1. The maximum vertex weight is the sum of ∆ + 1 labels and every edge weight is the sum of three labels imply that k ≥ |E|+|V |+1
Let v be a pendant vertex and n i (i = 1, 2) be the number of vertices of degree i. Then Our next results show that this lower bound is sharp.
Let C m 1 be a cycle with m 1 vertices and P m 2 be a path with m 2 vertices. Then

otherwise.
To prove Theorem 2.2, we need this lemma.
Lemma 2.1. For any integers y and Proof. Set all equations above as a linear system leads to the solution which is required.
Now, we are able to prove Theorem 2.2.
. We divide the proof into two cases as follows: Case 1. m 2 = 3 Suppose that ds(P 3 ) = 2. Since we need 5 distinct weight from 2 to 6, one endpoint (and its incidence edge) can be labeled by 1 to have smallest weight. In the other hand, maximum weight 6 only can occur when the rest vertices and edge are labeled by 2. Hence, there are one vertex and one edge of the same weight. Contrary to hypothesis. Thus, ds(P 3 ) ≥ 3. By label P 3 as in Fig.1, we have ds(P 3 ) = 3. Case 2. m 1 ≥ 3 and m 2 = 3 It is trivial for n 2 = 1. For n 1 ≥ 3 and n 2 ≥ 2, by Theorem 2.1, ds(C m 1 ) ≥ t(m 1 ) and ds(P m 2 ) ≥ t(m 2 ). For the reverse inequality, we construct f i : V ∪E → {1, 2, · · · , t(m i )}, for i = 1, 2, as follows: Let f m 1 , · · · , w(v m 2 )} be the alternating vertex (and edge) weight-sets of C m 1 and P m 2 , respectively. Let We prove by induction on m i . For the base step, it is true that for f For the inductive step, we assume that for all k 1 and k 2 , f 1 is a totally disjoint irregular total t 1 -labeling of C k 1 and f 2 is a totally disjoint irregular total t 2 -labeling of P k 2 , where and the maximum weight is w(e d(k i ) ).
. This is sufficient to apply Lemma 2.1.
d(k i )+1 } and y = w(e d(k i ) ) + 1. Then, we have v The total disjoint irregularity strength of some certain graphs | M.I. Tilukay and A.N.M. Salman Then, it can be checked that the maximum label is ds(G k i ) + 2 = ds(G (k i )+3 ). We have completed the labeling f i on G (k i )+3 and have proved that f i is a totally disjoint irregular total t(k i )-labeling. Thus, for any positive integer m 1 ≥ 3 and m 2 ∈ N, ds(C m 1 ) = 2m 1 +2
where v n+1 is the vertex of degree n. By Theorem 2.1, ds(S n ) ≥ n. To prove the reverse inequality, we construct an irregular total labeling f : Thus, f is a totally disjoint irregular total labeling and ds(S n ) = n for n ≥ 3.
Next, by using our previous result in [14], we determine the exact value of ds(K n ). For the convenient of reader, we provided the construction of totally irregular total labeling of K n for n = 5, 10, 12 given in [14]. Let n 2 −n+4 6 = t and n+1 3 = s. We divide the vertex-set into 3 mutually disjoint subsets, say A, B, and C, where , otherwise.

6
. For the reverse inequality, we divide the proof into three cases as follows: Case 1. n ≤ 5 It is obvious for n ≤ 3. Now, suppose that ds(K 4 ) = 3. We need 10 distinct weight with minimum weight 3. We can label 2 vertices and one edge by label 1. In the other hand, the maximum weight should be 12. Labeling 3 edges and one vertex by label 3 implies that there are 2 edges with the same weight 7. Contrary to hypothesis. Thus, ds(K 4 ) ≥ 4. To prove the upper bound for n = 4 or 5, we define f as in Fig. 1. Therefore, we have the exact value of ds(K n ) for n ≤ 5. Case 2. n = 77 or n ≥ 80 Consider that under the totally irregular total tlabeling of K n in (2), the maximum edge weight is w(c s−1 c s ) = n 2 + 2 and minimum vertex-weight is w(a 1 ) = s(s 2 −1) 6 + n. It follow w(c s−1 c s ) < w(a 1 ) implies vertex-weight set and edge weight set are disjoint. Thus, ds(K n ) = t for n = 77 or n ≥ 80. Consider that under the totally irregular total tlabeling of K n provided in (2), we met condition where the minimum vertex-weight w(a 1 ) is equal to the weight of an edge connecting vertices in (B, C) or (C, C). Then, we modify f . Let E(K n ) = {e i |1 ≤ i ≤ n(n − 1)/2}. Let e p ∈ E(K n ) be an edge where w(a 1 ) = w(e p ). Since t ≡ 2 mod 3, then f (e n(n−1)/2 ) = f (c s−1 c s ) = t − 1. It implies that we can change f (e i ) by f (e i + 1), for p ≤ i ≤ n(n − 1)/2 without changing the maximum label such that W (V (K n )) ∩ W (E(K n )) = ∅. It complete the proof.