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Vladikavkaz Mathematical Journal 2018, Volume 20, Issue 4, P. 59 66
УДК 519.17
DOI 10.23671/VNC.2018.4.23388
TRANSVERSAL DOMINATION IN DOUBLE GRAPHS
S. R. Nayaka1, Puttaswamy1 and K. N. Prakash2
1 P.E.S. College of Engineering, Mandya, Karnataka 571401, India;
2 Vidyavardhaka College of Engineering, P.B. No.206, Gokulam III Stage, Mysuru 570002, Karnataka, India E-mail: [email protected], prof.puttaswamy@gmail. com, [email protected]
Abstract. Let G be any graph. A subset S of vertices in G is called a dominating set if each vertex not in S is adjacent to at least one vertex in S. A dominating set S is called a transversal dominating set if S has
G
of a transversal dominating set is called the transversal domination number denoted by ytd(G). In this paper, we are considering special types of graphs called double graphs obtained through a graph operation. We study the new domination parameter for these graphs. We calculate the exact value of domination and transversal domination number in double graphs of some standard class of graphs. Further, we also estimate some simple bounds for these parameters in terms of order of a graph.
Key words: transversal dominating set, transversal domination number, direct product, double graph. Mathematical Subject Classification (2010): 05C69.
For citation: Nayaka, S. R., Puttaswamy and Prakash, K. N. Transversal Domination in Double Graphs, Vladikavkaz Math. J., 2018, vol. 20, no. 4, pp. 59-66. DOI: 10.23671/VNC.2018.4.23388.
1. Introduction
Let G be a graph. A subset S of vertices is called a dominating set of G if every vertex not in S is adjacent to at least one vertex in S. The minimum cardinality of a dominating set is called the domination number, denoted bv 7(G). For any graph G, there may be many dominating sets of different cardinalities between 7(G) and the order of G. The concept of transversal domination in graphs is defined and studied in [1]. A dominating set is called the transversal
G
of a transversal dominating set is called the transversal domination number, denoted by 7td(G). In fl], authors have obtained fundamental results related to transversal domination parameter including exact values for standard graphs and bounds in terms of order and domination number.
Let G and H be any two graphs. The direct product of G and H is a graph denoted by G x H with the vertex set V(G) x V(H) such that two vertices (v1,w1) and (v2,w2) are adjacent in G x H if and only if v1 and v2 are adjacent in G and w1 and w2 are adjacent in H. The total graph Tn of order n is the graph associated to the total relation (where every vertex is adjacent to each vertex). In fact, Tn can be obtained from the complete graph Kn
GG
© 2018 Nayaka, S. R., Puttaswamy, Prakash, K. N.
denoted by D(G) and is defined by D(G) = G x T2. In the double graph D(G), two vertices (vi,wi) and (v2, w2) are adjacent if and only if v1 and v2 are adjacent in G.
From the definition of a double graph [2], it follows that if G is a graph of order n and size m then D(G) is a graph of order 2n and size 4m. In particular, the degree of a vertex (v, k) will be 2degG v. The pentagonal prism with modified lateral edges and its double graph are as shown in Figure 1. The double graph D(G) always decomposes into two subgraphs Go and G1 such that G0 n G1 = 0 and G0 U G1 is a spanning subgraph of D(G). Then {G0, G1} is called the decomposition of D(G). The double graph operation is defined for any graph G,
G
The multi-star graph Km(a1, a2,..., am) is a graph of order a1 + a2 + ■ ■ ■ + am + m formed by joining a1, a2,..., am end-edges to m vertices of Km. For example, K2(a1, a2) is a double star.
Fig. 1. Double graph of a Pentagonal prism.
Gn Y (D(G)) =
2[fJ, ifn = 0 or l(mod 3);
2[fJ+l, ifn = 2 (mod 3).
< Let G be a path of order n. Then D(G) is a {2,4}-regular graph of order 2n. First, suppose n = ^r 1 ^^od 3). Let S' ^e a ^^^^^^m dominating set in For each vertex ui of S', attach a vertex vi+1 from another copy of G, which is adjacent to u' in D(G). The resulting set S of cardinalitv 2y(G) ^^rnnates D(G) and minimality holds since each vertex in S has a private neighbor. Thus, 7(35(G)) = 2|_f_|. Finally, assume n = 2(mod 3). Let v be a pendant vertex of G and let G' be a graph obtained by removing the vertex v. Then, clearly, y(D(G)) = y(D(G')) +1. Since, G' will be isomorphic to a path of order 3n or 3n +1, it follows that 7(D(G)) =2[fJ + 1. >
Theorem 1.1. Let G he a path of order n ^ 3. Then Ytd(D(G)) = y(D(G)) + 1.
< Let G be a path of order n. Since the Y-set of D(G) is obtained by choosing vertices from the Y-set of copies of G, it will be clear that there are atmost two possibilities to select vertices from a Y-set of copies of G. Thus, to any vertex u of Y-set of G which is not in Y-set
5 of D(G), the set S U {u} will be a dominating set intersecting the minimum dominating sets in D(G). Minimality of the set Si = S U {u} since for any vertex v of Si, there always exists a 7-set of D(G) not intersecting S^. Hence, Ytd(D(G)) = y(D(G)) + 1. > Lemma 1.2. Let G be a complete graph. Then y(D(G)) = 2. Theorem 1.2. Let G be a complete graph of order n. Then Ytd(D(G)) = 2n — 1. < Let G be a complete graph of order n. Then D(G) will be a regular graph order (2n — 2). Since every pair of vertices, taken from each copies of G, is a dominating set it follows that
< Let G be a cycle of order n ^ 3. Then D(G) is a 4-regular graph of order 2n. If n = 4, then clearly y(D(G)) = 5. Assume n = 5. Then, any minimum dominating set of a copy of G, in which a vertex u of S replaced by the cor responding vertex u', will be a minimum dominating set and so y(D(G)) = 3.
Now, suppose n ^ 6. We may consider three possible cases here. First, suppose n = 3k. As the graph D(G) consists of two copies of Cn, choose a minimum dominating set S' of
one copy, which dominates D(G) except the vertices corresponding to the vertices of S'.
S S' S'
be a dominating set of D(G). Further, to any vertex v of S, the set S — {v} will not be a dominating set in G and so in D(G). Therefore, 7(3(G)) = 2|S"| = 2|_§J.
Next, suppose n = 3k + 1. As in the previous case, choose a minimum dominating set S' GG dominating set but not minimal as the vertices v1 and v'n have two neighbors in the set. Hence, S' — {v1, vn} will be a minimum dominating set in G. Therefore, y(D(G)) = |S' — {v1, vn}| = 2|_§J. Finally, if n = 3fc + 2, similar to the above case, for any set S' consists of vertices from 7-set of G and the corresponding vertices in other copy of G, the set S = (S' — {vi, vn-1}) U{vn} will be a minimum dominating set of cardinality 2|_§J. > Theorem 1.4. Let G be a cycle of order n ^ 3. Then
< Let G be a cycle of order n ^ 3 and let V(D(G)) = {vj, vj : 1 ^ i, j ^ n}. First we note that any minimum dominating set in D(G) contains a vertex from V' = {v1, v2, v2, vn, vn, }. Let H be a spanning sub-graph of G having the vertex set V — V'. Then Ytd(D(G)) = Y(H) + |V'I. Further, it can be noted that H will be isomorphic to a double graph D(Pn-3). Therefore, Ytd(D(G)) = y(D(Pn-3)) + 5 establishing the result. >
Theorem 1.5. Let G = Km(a1, a2,..., am) ^e a multi-star. Then Ytd(D(G)) = m + 1.
< Let G = Km(a1, a2,..., am) ^e a multi-star of order a1+a2 +-----ha^. Clearly y(G) = m.
Consider the double graph of G and the minimum dominating set S. As every vertex in S covers the leaves adjacent to it and the vertices adjacent to the corresponding vertices in another copy, it follows that S itself a minimum dominating set in D(G). Therefore, y(D(G)) = |S| = m. Finally, since D(G) contains exactly two vertex disjoint dominating sets, Ytd(D(G)) = m + 1. >
7id(D(G)) =2n - 1 >
Theorem 1.3. Let G be a cvcle of order n ^ 4. Then
Y (D(G)) =
>6.
Definition 1.1. For m ^ 2, Jahangir graph Jn,m is a graph of order nm + 1, consisting of a cycle of order nm with one vertex adjacent to exactly m vertices of Cnm at a distanee n to each other. Jahangir graph J2>16 is shown in figure 1.
Fig. 2. J2,i6.
Proposition 1.1 [3]. Let G = J2,m be a Jahangir graph with m ^ 3. Then
Y(G) =
2, if m = 3;
l1^] + 1, otherwise.
Theorem 1.6. Let G = Jn,m be a Jahangir graph with m,n ^ 3. Then
m(-",~1-> +1, if n = 1 (mod 3);
Y (G) = { 3 v 7 \Pfl,
ifn = ^r2 (mod 3).
< Let G = Jn,m be a Jahangir graph with m,n ^ 3 and let V(G) = {v1, v2,..., vnm, vnm+1}, where vnm+1 is the vertex at the center, adjacent to vertices of Cnm. First assume n = 1(mod 3), i. e., n = 3k + 1, to some positive integer k. From the definition, the vertex vnm+1 is adjacent to m vertices of Cnm at a distance 3k + 1. Removing the vertex vnm+1 from G, the graph induced by V(G) — {vnm+1} splits into m components each component isomorphic to P3&. Therefore, the minimum dominating set of G is obtained by taking dominating set from each component together with vnm+1. That is, if S = um=1Si, where Si denotes Y-set of ith component, then SU {vnm+1} will be a minimum dominating set of G. Since any vertex not in SU{vnm+1} will be adjacent to exactly one vertex in SU{vnm+1}, no proper subset will be dominating set in G. Thus, j(G) = m(-"~1'> + 1.
Next, suppose n = 2(mod 3). Here, we may consider three possible cases. First, assume m = 0(mod 3). Then {vm, v2m, Vsm,..., vnm} will be a dominating set of cardinality ^. On the other hand, let D be a dominating set in G and assume vnm+1 € D. As the vertex vnm+1 dominates m vertices, to cover the remaining vertices, at least vertices are necessary.
Thus, we must have, \D\ ^ m|~§] + 1, which is not possible. Hence, vnm+\ £ D. This shows that the domination number of G co-incides with that of a cycle. Therefore, y(G) = l"22^] • Next, suppose m = 3^. In this case {v1, v3, v6,..., vnm-1} will be a dominating set of
i. e., I"22?1]. On the other hand, as in the above case, it is easy to observe that
size
nm-\- 2 3
Vnm+i £ D, to any dominating set D of G. Therefore, y(G) = I"22?2]
Finally, assume n = 0(mod 3). For any integer m ^ 3, clearly nm will be a multiple of 3. Further, no dominating set D contains the center vertex vnm+1. Hence, y(G) = Y(Cnm), i. e., 7(G) = . >
Proposition 1.2. Suppose m ^ 3 and n = 1(mod 3), then
i^pil+2, ifm = 3 ;
I ——+ 1, otherwise.
< Let Jn,m be a Jahangir graph with m ^ 3 and n = 1(mod 3). If m = 3, then Jn,3 contains three minimum dominating sets among which two of them having a common vertex. Thus, 7td{Jn,m) = 1-> + 2. Next, Assume m ^ 4. Then, Jra;TO contains three minimum dominating sets. The dominating set {3, 7,11,... , vnm-1, vnm+1} intersects other two sets and hence itself become a transversal dominating set. Therefore, 7td(Jn,m) = 1-> + 1. >
Theorem 1.7. Suppose m ^ 3 and n = 1(mod 3) then y(D(Jn,m)) = 2y(Jn,m).
< Let Jn,m be a Jahangir graph with m ^ ^d n = 1(mod 3). Let S be any minimum dominating set of Jn,TO. Then, S dominate the double graph D(Jn,m) except the corresponding vertices of S in the other copy of Jn,m. Since none of the vertices in S have common neighbor in itself, the minimum dominating set of D(Jn,m) is obtained bv adding the corresponding vertices of S. Therefore, Y(D(Jn,m)) = 2y(Jn,m)• >
Proposition 1.3. Suppose m ^ 3 and n = 1(mod 3). Then Ytd(D(Jn,m)) = Y(D(Jn,m)).
< Let Jn,m be a Jahangir graph with m ^ ^d n = 1(mod 3) We observe that Jn,m contains a unique dominating set, the double graph D( Jn,m) also contains only one dominating set and hence, Ytd(D(Jn,m)) = Y(D(Jn,m)) >
Theorem 1.8. Let G = Jn,m be a Jahangir graph with n = 2(mod 3). Then
(Pfl+2, ifm = 0 (mod 3); ltd{G) = 1 + 1, ifm = l (mod 3);
[pf], if m = 2 (mod 3).
< Let G = Jn,m, m ^ 3 be a Jahangir graph such that n = 2(mod 3). First, we note that dominating set in Jn,m arises from dominting set of the cycle Cnm and hence any transversal dominating set in Jn,m contains at least one vertex from the set D = {v1,v2,v3}. There are three possible cases here, suppose m = 0(mod 3), then nm = 0(mod 3). Thus, Jn,m contains exactly three vertex disjoint dominating sets each of cardinality Therefore 7td(G) ^ ^ + 2. On the other hand, since any Y-set contains vertex from D, it follows that Ytd(G) = 3+y(H), where H is the graph induced by V(Jn,m) — D. Clearly, H = Pmn-5 and hence, Ytd(Jn,m) = |"rapi sUpp0se m = l(mod 3), then Jra;TO contains two vertex disjoint dominating sets. Hence, Ytd-set of Jn,m is obtained by adding one vertex to the Y~set of Jn,m- Therefore, |-?Tm-| ^ Finally suppose m = 2(mod 3). As the graph Jra;TO contains only one dominating set {v1, v4, v7,..., vmn-4, vmn-1} and hence itself a transversal dominating set. Therefore, 7td(Jn,m) = 7 (Jn,m) = l^f] ■ >
Theorem 1.9. Let G = Jn m be a Jahangir graph with n = 2(mod 3). Then Ytd(D(G)) = 2pf ] _ + 1. Further, ltd(V(G)) = 7(Q(G)).
< Let G = Jn , m be a Jahangir graph with n = 2(mod 3). Let S be a minimum dominating set in G. Then, S dominates the double graph D(G) except the vertices in the second copy of G corresponding to that of S. Also, the vertex vnm+1 at the center will be adjacent to
exactly L^J vertices of S. Hence, the minimum dominating set will be obtained by choosing \S\ - L^J vertices from the second copy of G. Therefore, 7i(J(35(G)) = 2Pf ] - [si=l\ + 1. Next, since any dominating set in D(G) contains the center vertex vnm+1, it follows that any Y-set itself a transversal dominating set in D(G). Hence, Ytd(D(G)) = y(D(G)). >
Proposition 1.4. Let G = Jn,m be a Jahangir graph with n = 0(mod 3) and m ^ 3. Then Ytd(G) = y(G).
< Let G = Jn,m be a Jahangir graph with n = 0(mod m ^ 3. For any value of m, we have nm = 0(mod 3) and so G contains unique dominating set {v1, v4, v7,..., vnm-5, vnm-2}. Therefore, Yi(i(G) = j(G) = Pf ]. >
Theorem 1.10. Let G = Jn,m be a Jahangir graph with m ^ 3. If n = 0(mod 3), then 7 (35(G)) = ^ + 1.
< Let G = Jn,m be a Jahangir graph with m ^ 3 and n = 0(mod 3). Since G contains a unique dominating set D = {v1, v4, v7,..., vnm-5, vnm-2} and the set fails to dominates the corresponding vertices in the second copy of G. Therefore, 7(G) ^ ^ + 1. On other hand, since the center vertex vnm+1 is adjacent to every vertex in D, the set D U {vnm+1} will be a dominating set and hence, 7(35(G)) = ^ + 1. >
Theorem 1.11. Let G = Jn,m be a Jahangir graph with m ^ 3. If n = 0(mod 3), then
7td(2>(G9) = + 2.
< Let G = Jn,m be a Jahangir graph with m ^ 3 and n = 0(mod 3). From the above
G
Thus adding one vertex from a dominating set to other set, the transversal dominating set will be obtained, therefore, (35(G)) = ^ + 2. >
2. Bounds for Ytd(D(G))
Theorem 2.1. Let G be any connected graph of order n. Then 1 ^ y(G) ^ Y(D(G)) ^ Ytd(D(G)) ^ 2n. Further, y(D(G)) = Ytd(D(G)) if and only if G contains a unique
dominating set of size 1.
< Let G be any connected graph of order n. Since any dominating set of double graph of G dominates G also, it follows that y(G) ^ y(D(G)). Assume y(D(G)) = Ytd(D(G)). On contrary, suppose y(G) ^ 2 and let S be a Y-set. Then S dominates the double graph D(G) except the corresponding vertices of S in other copy of G. Therefore, S cannot be a transversal dominating set in D(G), showing that y(D(G)) = Ytd(D(G)). Conversly, if G contains a unique dominating set of cardinality one, then D(G) contains unique dominating set and so y(D(G)) = Ytd(D(G)). >
Corollary 2.1. Let G be any graph. Then 2 ^ y(D(G)) ^ Ytd(®(G)). Further, Ytd(D(G)) = 2 if and only if G is a star.
< Let G be any graph. First, assume Ytd(D(G)) = 2. From the above theorem it follows that y(D(G)) = Ytd(D(G)) and so G must contain exactly one vertex of degree n — 1, proving that G is a star. Converse is obvious. >
There is no exact relation between Ytd(G) and y(D(G)). For example, if G is a star, then y(D(G)) = Ytd(D(G)). Let G be a complete graph of order n ^ 4, then Ytd(G) = n — 1 > 2 = y(d(g)). Finally, let G be a path of Pfe- Then Ytd(G) = 5 but Y(D(G)) = 6.
Proposition 2.1. Let G be a connected graph of order n ^ 2. Then Ytd(D(G)) ^ Y(D(G)) + 5(D(G)).
< Let G be a connected graph of order n ^ 2. Then, á(G) ^ 1 and let v be a vertex of degree ¿(G). Clearly, any dominating set in D(G) must contain either v от a vertex from N(v). Thus, Ytd(D(G)) < y(D(G)) + |N(v)|. This proves that, Ytd(D(G)) < y(D(G)) + 5(D(G)). >
Theorem 2.2. Let G be any graph. Then Ytd(D(G)) = 2n — 1 if and only if G is a complete graph.
< Let G be a connected graph of order n. Assume that Ytd(D(G)) = 2n — 1. Then, any subset S' of vertices of order atmost 2n — 2 is not a transversal dominating set in D(G). From the minimality of Ytd(D(G)), it follows that, V — S' = {u, v} is a dominating set in D(G). Further, V — S' must contains at least one vertex from each copy of G. Thus, y(G) = 1. As the vertices u, v are chosen arbitrarily, each vertex in G must have degree n — 1, proving
G>
References
1. Nayaka, S. R., Alwardi A. and Puttaswamy. Transversal Domination in Graphs, Gulf Journal of Mathematics, 2018, vol. 6, no. 2, pp. 41-49.
2. Munarini, E., Perelli Cippo, C., Scagliola, A. and Zagaglia Salvi, N. Double Graphs, Discrete Mathematics, 2008, vol. 308, pp. 242-254.
3. Mojdeh, D. A. and Ghameshlou, A. N. Domination in Jahangir Graph J2,m, Int. J. Contemp. Math. Sciences, 2007, vol. 2, no. 24, pp. 1193-1199.
Received November 29, 2017
SOMESHWARAPURA R. nayaka P.E.S. College of Engineering, Mandya, Karnataka 571401, India, Assistant Professor E-mail: [email protected]
Puttaswamy
P.E.S. College of Engineering, Mandya, Karnataka 571401, India, Professor
E-mail: prof. puttaswamy@gmail. com
K. N. Prakasha
Vidyavardhaka College of Engineering, P.B. No.206, Gokulam III Stage, Mysuru 570002, Karnataka, India Assistant Professor E-mail: [email protected]
Владикавказский математический журнал 2018, Том 20, Выпуск 4, С. 59^66
ТРАНСВЕРСАЛЬНОЕ ДОМИНИРОВАНИЕ В ДВОЙНЫХ ГРАФАХ
Наяка С. Р.1, Путтасвами1, Пракаша К. Н.2 1 Инженерный колледж, Мандья, Карнатака 571401, Индия;
2
E-mail: [email protected], prof.puttaswamy@gmail. com, [email protected]
Аннотация. Пусть G — произвольный граф. Подмножество S множества всех вершин G называется доминирующим множеством, если каждая вершина, не входящая в S, примыкает, по меньшей мере, SS
множеством, если S имеет непустое пересечение с каждым доминирующим множеством минимальной мощности в G. Минимальная мощность трансверсального доминирующего множества называется числом трансверсального доминирования, обозначаемым jtd(G). В данной статье рассматриваются специальные типы графов, называемые двойными графами, получаемыми с помощью операций над графами. Мы изучаем новый параметр доминирования для этих графов. Вычисляется точное значение числа доминирования и числа поперечного доминирования в двойных графах некоторого стандартного класса графов. Кроме того, получены некоторые простые оценки для этих параметров в терминах порядка графа.
Ключевые слова: поперечное доминирующее множество, число поперечного доминирования, прямое произведение, двойной граф.
Mathematical Subject Classification (2010): 05С69.
Образец цитирования: Nayaka S. R., Puttaswamy, Prakash K. N. Transversal Domination in Double Graphs // Владикавк. мат. журн— 2018.^Т. 20, № 4.—С. 59-66 (in English). DOI: 10.23671/VNC.2018.4.23388.
