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6URAL MATHEMATICAL JOURNAL, Vol. 7, No. 2, 2021, pp. 86-93
DOI: 10.15826/umj.2021.2.006
NOTE ON SUPER (a, 1)-P3-ANTIMAGIC TOTAL LABELING OF STAR Sn
S. Rajkumart, M. Nalliah"^ and Madhu Venkataraman^tt
Department of Mathematics, School of Advanced Sciences, Vellore Institute of Technology Vellore-632 014, India
traj26101993@gmail .com, tt nalliahklu@gmail.com, tttmadhu.v@vit.ac.in
Abstract: Let G = (V, E) be a simple graph and H be a subgraph of G. Then G admits an H-covering, if every edge in E(G) belongs to at least one subgraph of G that is isomorphic to H. An (a, d) — H-antimagic total labeling of G is bijection f : V(G) U E(G) ^ {1, 2, 3,..., |V(G)| + |E(G)|} such that for all subgraphs H' of G isomorphic to H, the H' weights w(H') = vey(h') f (v) + "12eeE(H') f (e) constitute an arithmetic progression {a, a + d, a + 2d, ...,a + (n — 1)d}, where a and d are positive integers and n is the number of subgraphs of G isomorphic to H. The labeling f is called a super (a, d) — H-antimagic total labeling if f (V(G)) = {1, 2, 3,..., | V(G)|}. In [5], David Laurence and Kathiresan posed a problem that characterizes the super (a, 1) — P3-antimagic total labeling of Star Sn, where n = 6, 7, 8, 9. In this paper, we completely solved this problem.
Keywords: H-covering, Super (a, d) — H-antimagic, Star.
1. Introduction
Let G = (V(G),E(G)) and H = (V(H),E(H)) be simple and finite graphs. Let |V(G)| = vg, |E(G)| = eG, |V(H)| = vH and |E(H)| = eH. An edge covering of G is a family of different subgraphs H1, H2, H3,..., Hk such that any edge of E(G) belongs to at least one of the subgraphs Hj, 1 < j < k. If the Hjs are isomorphic to a given graph H, then G admits an H-covering. Gutienrez and Llado [2] defined H—magic labeling, which is a generalization of Kotzig and Rosa's edge magic total labeling [4]. A bijection f : V(G) u E(G) ^ {1,2,3, ...,vG + eG} is called an H-magic labeling of G if there exists a positive integer k such that each subgraph H' of G isomorphic to H satisfies
w(H')= £ f (v)+ £ f (e) = k.
veV(H') e€E(H')
In this case, they say that G is H-magic. When f (V (G)) = {1,2,3, ...,vG}, we say that G is H—super magic. On the other hand, Inayah et al. [3] introduced (a, d) — H-antimagic total labeling of G which is defined as a bijection f : V(G) u E(G) ^ {1,2,3,..., vG + eG} such that for all subgraphs H of G isomorphic to H, the set of H -weights
w(H' )= £ f (v) + £ f (e)
veV(H') eeE(H')
constitutes an arithmetic progression a, a + d, a + 2d,..., a + (n — 1)d, where a and d are some positive integers and n is the number of subgraphs isomorphic to H. In this case, they say that G is (a, d) — H—antimagic. If f (V(G)) = {1,2,3,... , vG}, they say that f is a super (a, d) — H-antimagic total labeling and G is super (a, d) — H-antimagic. This labeling is a more general case of super (a, d)-edge-antimagic total labelings. If H = K2, then we say that super (a, d) — H-antimagic
labelings, which is also called super (a, d)-edge-antimagic total labelings and have been introduced in [6]. They studied some basic properties of such labeling and also proved the following theorem.
Theorem 1 [3]. If G has a super (a, d) — H-antimagic total labeling and t is the number of subgraphs of G isomorphic to H, then G has a super (a',d) — H-antimagic total labeling, where a' = [(vg + 1)vh + (2vg + eG + 1)e# ] — a — (t — 1)d.
Several authors are studied antimagic type labeling of graphs see [1]. In 2015, Laurence and Kathiresan [5] obtained an upper bound of d for any graph G, and they investigated the existence of super (a, d) — P3-antimagic total labeling of star graph Sn. First, they observed that Sn admits a Ph—covering for h = 2,3, and the star Sn contains
/ n
'=U—i
subgraphs Ph, h = 2,3, which is denoted by pj, 1 < j < h. In 2005, Sugeng et al. [7] investigated the case h = 2 using super (a, d)-edge-antimagic total labeling. In 2015, the case of h = 3 was investigated by Laurence and Kathiresan [5]. Here they observed that if the star Sn, n > 3 admits a super (a, d) — P3-antimagic total labeling then d € {0,1,2}. Now, they proved the star Sn, n > 3 has super (4n + 7,0) — P3-antimagic total labeling and Sn, n > 3 admits a super (a, 2) — P3-antimagic total labeling if and only if n = 3. Also, they proved the following theorems and posed a problem.
Theorem 2 [5]. If the star Sn, n > 3 has super (a, 1)-P3-antimagic total labeling, then 3 < n < 9. Moreover, the star Sn admits a super (a, 1)-P3-antimagic total labeling, where a = 19, n = 3 and a = 21, for n = 4.
Theorem 3 [5]. For n = 5, the star Sn has no super (a, 1)-P3-antimagic total labeling.
Problem 1. [5] For each n, 6 < n < 9 characterize the super (a, 1) — P3-antimagic total labeling for the star Sn.
In this paper, we present the complete solution to the above problem.
2. Main Results
Let Sn = Ki,n, n > 1 be the star graph and let vo be the central vertex and let Vj, 1 < i < n be its adjacent vertices. Thus Sn has n + 1 vertices and n edges.
Theorem 4. The star S6 has no super (a, 1) — P3-antimagic total labeling.
Proof. Let V(So) = {V0,V1,V2,V3,V4,V5,V6} and E(Se) = {V0V1,V0V2, V0V3, V0V4, V0V5, voVe} be the vertex set and the edge set of Star S6. Suppose there exists a super (a, 1) — P3-antimagic total labeling f : V u E ^ {1,2,3,..., 13} for S6 and let v0 be the central vertex of S6. In the computation of P3 — weights the label of the central vertex v0,f (v0) is used 15 times and label of other vertices and edges say i are used 5 times each. Therefore,
13 15 10/(Vo) + 5^(i) = y[2a + 14],
i=1
which implies a = (70 + 2f (vo))/3. Since 1 < f (vo) < 7, it follows that a = 24 if f (vo) = 1, a = 26 if f (vo) = 4 and a = 28 if f (vo) = 7.
1
2 3 4 5 6 7
Figure 1. There is no possible to obtain P3-weight 27.
Case (i): f (v0) = 1. Then a = 24 and the P3 — weights of £6 are given by W = {24,25,..., 38}. Now, the P3 — weight 24 is getting exactly two possible 5 elements sum (1,2,4,8, 9) or (1,2,3,8,10) and hence the label of edges ei = v0vi and e2 = v0v3 or v0v2 is f(ei) = 8 and ffo) = 9 or 10.
Subcase (i): f (e2 = v0v3) = 9. Then a = 24 and hence the label of the vertices and edges are f (V0) = 1,f (vi) = 2,f (V3) = 4, f (ei = V0V1) = 8 and f (e2 = V0V3) = 9. Now, the P3 — weight 25 is getting exactly one possible 5 elements sum (1, 2, 3, 8, 11) and hence the label of an edge e3 = v0v2 is f (e3) = 11. Also,the P3 — weight 26 is getting exactly one possible 5 elements sum (1,2, 5,8,10) and hence the label of an edge e4 = v0v4 is f (e4) = 10.
Let x = v0v5 and y = v0v6 be two edges of £6 (see Fig. 1). Clearly, the label of the edges x and y is f (x) = 12 or 13 and f (y) = 13 or 12. If f (x) = 12 then f (y) = 13 and hence there is no P3 — weight 27. Also, if f (x) = 13 then f (y) = 12 and hence there is no P3 — weight 27, which is a contradiction.
A similar contradiction arises, if the edges ei = v0vi and e2 = v0v2 with f (ei = 9) and f (e2) = 8 for the P3 — weight 24 is used to getting the P3 — weight 27.
Subcase (ii): f (e2 = v0v2) = 10. Then a = 24 and hence the label of the vertices and edges of P3 — weight 24 is f (V0) = 1,f (vi) = 2, f (V2) = 3,f (ei = V0Vi) = 8 and f (e2 = V0V2) = 10. Now, the P3 — weight 25 is getting exactly one possible 5 elements sum (1, 2, 5, 8, 9) and hence the label of an edge e3 = v0v4 is f (e3) = 9. Also, the P3 — weight 26 is getting exactly one possible 5 elements sum (1,2,4,8,11) and hence the label of an edge e4 = v0v3 is f (e4) = 11. Let x = v0v5 and y = v0v6 be two edges of S6 (see Fig. 2). Clearly, the label of the edges x and y is f (x) = 12 or 13 and f (y) = 13 or 12. If f (x) = 12 then f (y) = 13 and hence there is no P3 — weight 27. Also, If f (x) = 13 then f (y) = 12 and hence there is no P3 — weight 27, which is a contradiction.
A similar contradiction arises, if the edges ei = v0vi and e2 = v0v2 with f (ei) = 10 and f (e2) = 8 for the P3 — weight 24 is used to getting the P3 - weight 27.
Case (ii): f (v0) = 7. Then a = 28. Now, if f is a super (28,1) — P3-antimagic total labeling of S6, then by Theorem 1 [3], f is a super (24,1) — P3-antimagic total labeling, which does not exist by Case (i).
Case (iii): f (v0) = 4. Then a = 26 and hence the P3 — weights of S6 are given by W = {26, 27, . . . , 40}. Now, the P3 — weight 26 is getting exactly four possibles 5 elements sum such as (4,1,2,8,11), (4,1,2,9,10), (4,2,3,8,9) and (4,1,3,8,10) and hence the edges ei = V0Vi or V0V2 and e2 = v0v2 or v0v3 with f (ei) = 8 or 9 and f (e2) = 9 or 10 or 11.
Subcase (i): f (ei = v0vi) = 8 and f (e2 = v0v2) = 11. Then a = 26 and hence the label of the vertices and edges of P3 — weight 26 is f (v0) = 4, f (vi) = 1, f (v2) = 2, f (ei = v0vi) = 8 and
1
2 3 4 5 6 7
Figure 2. The possible edge labels x and y are obtain P3-weight 27.
4
123567
Figure 3. There is no possible to obtain P3-weight 30.
f (e2 = v0v2) = 11. Now, the P3 — weight 27,28 and 29 are getting exactly one possible 5 elements sum (4,1, 5, 8,9),(4,1,3,8,12) and (4,1,6,8,10). Hence the label of the edges e3 = v0v3, e4 = v0v4, e5 = v0v5 and e6 = v0v6 is f (e3) = 12, f (e4) = 9, f (e5) = 10 and f (e6) = 13. From Fig. 3, there is no P3 — weight is 30, which is a contradiction.
A similar contradiction arises, if the edges ei and e2 with f(ei = v0vi) = 11 and f(e2 = v0v2) = 8 for P3 — weight 26 are used to getting the P3 — weight 33, for more details see Fig. 4.
Subcase (ii): f(e1 = v0v1) = 9 and f(e2 = v0v2) = 10. Then a = 26 and hence the label of the vertices and edges of P3 — weight 26 is f (v0) = 4, f (v1) = 1, f (v2) = 2, f (e1 = v0v1) = 9 and f(e2 = v0v2) = 10. Now, the P3 — weight 27 is getting exactly two possibles 5 elements sum such as (4,2,3,10,8), (4,1, 5,9,8) and hence the label of the edges e3 = v0v3 or v0v4 is f(e3) = 8. If an edge e3 = v0v3 with f (e3) = 8 then we get the P3 — weight as sum of 5 elements (4,1,3,9,8) is 25, which is a contradiction. If an edge e3 = v0v4 with f (e3) = 8 then we get the P3 — weights from 28 to 32 are getting exactly one possible 5 elements sum such as (4,1, 3, 9,11), (4, 2, 5,10, 8), (4, 2, 3,10,11), (4, 3, 5,11, 8) and (4,1, 6, 9,12). From Fig. 5, there is no P3 — weight 33, which is a contradiction.
A similar contradiction arises, if the edges e1 = v0v1 and e2 = v0v2 with f (e1 = v0v1) = 10 and f (e2 = v0v2) = 9 for the P3 — weight 26 is used to getting the P3 — weight 27, which is a contradiction.
Subcase (iii) : f (e1 = v0v2) = 8 and f (e2 = v0v3) = 9. Then a = 26 and hence the label of the vertices and edges of P3 — weight 26 is f (v0) = 4, f (v2) = 2, f (v3) = 3, f (e1 = v0v2) = 8 and f (e2 = v0v3) = 9. Now, the P3 — weight 27 is getting exactly one possible 5 elements sum (4,1,3,9,10) and hence the label of an edge e3 = v0v1 is f (e3) = 10. Thus, we get a P3 — weight
4
1 2 3 5 6 7
Figure 4. The possible edge label is obtain to P3-weight 33.
4
123567
Figure 5. There is no possible to obtain P3-weight 33.
as sum of 5 elements (4,1,2,10,8) is 25, which is a contradiction.
A similar contradiction arises, if the edges e1 = v0v2 and e2 = v0v3 with f (ei = v0v2) = 9 and f (e2 = v0v3) = 8 for the P3 — weight 26. The P3 — weight 27 is getting exactly one possible 5 elements sum (4,1,2,11,9) and hence the label of an edge f (e3 = v0v1) = 11. Thus, we get the P3 = (v0,v1,v3,e3 = v0v1,e2 = v0v3) with weight (4 + 1 + 3 + 11 + 8) is 27, which is a contradiction.
Subcase (iv): f (e1 = v0v1) = 8 and f (e2 = v0v3) = 10. Then a = 26 and hence the label of the vertices and edges of P3 — weight 26 is f (v0) = 4, f (v1) = 1, f (v3) = 3, f (e1 = v0v1) = 8 and f (e2 = v0v3) = 10. Now, the P3 — weight 27 is getting exactly two possibles 5 elements sum such as (4,1,2,8,12), (4,1, 5,8,9) and hence the label of the edges e3 = v0v2 or v0v4 is f (e3) = 12 or 9. If an edge e3 = v0v2 with f (e3) = 12 then the P3 — weights 28 and 29 are getting exactly one possible 5 elements sum (4,1,6,8,9) and (4,1, 5,8,11). From Fig. 6, there is no P3 — weight 30, which is a contradiction. If an edge e4 = v0v4 with f (e4) = 9 then the P3 — weight 28 is getting exactly one possible 5 elements sum (4,1,2,8,13) and hence the label of an edge e5 = v0v2 is f (e5) = 13. From Fig. 7, there is no P3 — weight 29 when x = 11 or 12 and y = 12 or 11, which is a contradiction.
A similar contradiction arises, if the edges e1 = v0v1 and e2 = v0v3 with f (e1 = v0v1) = 10 and f(e2 = v0v3) = 8 for the P3 — weight 26 are used to getting the P3 — weight 27, which is a contradiction. □
Theorem 5. The star has no super (a, 1) — P3-antimagic total labeling.
Proof. Let V(S7) = {v0, v1, v2, v3, v4, v5, ,v7} and E(S7) = {v0v1 ,v0v2,v0v3,v0v4,v0v5, v0v6,v0v7} be the vertex and edge set of star S7. Suppose there exists a super (a, 1) — P3-antimagic total labeling f : V u E ^ {1,2,3,..., 15} for S7 and let v0 be the central vertex of S7. In the
y 12/ 10/ 11 \ 9\ X 1 2 3 5 6 7
Figure 6. There is no possible to obtain P3-weight 30.
Y 1y 10/ 9\ x 1 2 3 5 6
Figure 7. There is no possible to obtain P3-weight 29.
computation of P3 — weights the label of the central vertex vo, f (vo) is used 21 times and label of other vertices and edges say i are used 6 times each. Therefore,
15
21,
15/(vo)+6^Ci) = —[2a + 20],
i=1
which implies that we get
15/(vo) + 510
21
Since 1 < /(v0) < 8, we have only two values a such as a = 25 if /(v0) = 1 and a = 30 if /(v0) = 8.
Case (i): /(v0) = 1. Then a = 25 and the P3 — weights of is given by W = {25, 26,... , 45}. Now, the P3 — weight 25 is getting exactly one possible 5 elements sum (1,2,3,9,10) and hence the label of edges e1 = v0v1 and e2 = v0v2 is /(e1) = 9 and /(e2) = 10. Since the minimum possible sum of vertices labels for P3 — weight is 7, it follows that there is no P3 — weight 26, which is a contradiction. A similar contradiction arises, if the edges e1 = v0v1 and e2 = v0v2 with /(e1) = 10 and /(e2) = 9 for the P3 — weight 25 is used to getting the P3 — weight 27.
4
4
7
a
Case (ii): f (v0) = 8. Then a = 30. Now, if f is a super (30,1) — P3-antimagic total labeling of then by Theorem 1 [3], f is a super (25,1) — P3-antimagic total labeling, which does not exist by Case (i). □
Theorem 6. The star has no super (a, 1) — P3-antimagic total labeling.
Proof. Let V(Sg) = {vo,vi, v2, V3, V4, V5, V6, V7, vs} and E(Sg) = {vovi,vov2,voV3,vov4,voV5, v0v6,v0v7,vov8} be the vertex and edge set of star S8. Suppose there exists a super (a, 1) — P3-antimagic total labeling f : V u E ^ {1,2,3,..., 17} for S8 and let vo be the central vertex of S8.
In the computation of P3 — weights the label of the central vertex v0, f (v0) is used 28 times and
label of other vertices and edges say i are used 7 times each. Therefore,
17 28
2№o) + 7^(i) = y[2a + 27],
i=1
which implies that we get
= 21/fco) + 693 a 28
Since 1 < f (v0) < 9, we have only two values a such as a = 27, if f (v0) = 3 and a = 30, if f (v0) = 7.
Case (i): f (vo) = 3. Then a = 27 and the P3 — weights of £g is given by W = {27, 28,... , 54}. Now, the P3 — weight 27 is getting exactly one possible 5 elements sum (3,1,2,10,11) and hence the label of edges e1 = v0v1 and e2 = v0v2 is f (e1) = 10 and f (e2) = 11. Since the minimum possible sum of vertices labels for P3 — weight is 8, it follows that there is no P3 — weight 29, which is a contradiction. A similar contradiction arises, if the edges e1 = v0v1 and e2 = v0v2 with f (e1) = 11 and f (e2) = 10 for the P3 — weight 27 is used to getting the P3 — weight 29.
Case (ii) f (v0) = 7 Then a = 30. Now, if f is a super (30,1) — P3-antimagic total labeling of £6, then by Theorem 1 [3], f is a super (27,1) — P3-antimagic total labeling, which does not exist by Case (i). □
Theorem 7. The star £9 has no super (a, 1) — P3-antimagic total labeling.
Proof. Let V(£9) = {v0,v1 ,v2,v3, v4, v5, v6,v7,vs,v9} be the vertex set of star £9. Suppose there exists a super (a, 1) — P3-antimagic total labeling f : V u E ^ {1,2,3,..., 19} for £9 and let v0 be the central vertex of £9. In the computation of P3 — weights the label of the central vertex v0, f (v0) is used 36 times and label of other vertices and edges say i are used 8 times each. Therefore,
19
19 36
28/(Vo)+8^C0 = y[2a + 35],
i=1
which implies that we get
_ Ufjvo) + 445 a~ 18
Since 1 < f (v0) < 10, we have that a is not an integer, which is a contradiction. □
From Theorem 2-3 [5], Theorem 4-7, we get the following result.
Theorem 8. The star £n, n > 3 admits a super (a, 1) — P3-antimagic total labeling if and only if n = 3 and 4.
3. Conclusion and Scope
In [5], they investigated the existence of super (a, d)-P3-antimagic total labeling of star £n and posed the Problem 1 [5]. This paper proved the star £n has no super (a, 1)-P3-antimagic total labeling, where n = 6, 7,8,9. Therefore, we have entirely solved Problem 1 [5].
Acknowledgements
The authors are thankful to the reviewers for helpful suggestions which led to substantial improvement in the presentation of the paper.
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