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The domination numbers of cylindrical grid graphs
Applied Mathematics and Computation 217 (2011) 4879–4889
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
The domination numbers of cylindrical grid graphs Mrinal Nandi a, Subrata Parui b, Avishek Adhikari c,⇑ a b c
Department of Statistics, West Bengal State University, Barasat, North 24-Parganas, India IMBIC, AH 317, Salt Lake City, Calcutta 700091, India Department of Pure Mathematics, University of Calcutta, 35 Ballygunge Circular Road, Calcutta 700019, India
a r t i c l e
i n f o
Keywords: Minimum dominating set Grid graph Cycle Path and Cartesian product of graphs
a b s t r a c t Let c(Pm h Cn) denote the domination number of the cylindrical grid graph formed by the Cartesian product of the graphs Pm, the path of length m, m P 2 and the graph Cn, the cycle of length n, n P 3. In this paper, methods to find the domination numbers of graphs of the form Pm h Cn with n P 3 and m = 2, 3 and 4 are proposed. Moreover, bounds on domination numbers of the graphs P5 h Cn, n P 3 are found. The methods that are used to prove that results readily lead to algorithms for finding minimum dominating sets of the above mentioned graphs. Ó 2010 Elsevier Inc. All rights reserved.
1. Introduction Graph theory, as a mathematical discipline, was created long back by Euler in his famous discussion of the Konigsberg Bridge Problem (1736). But in the current scenario, along with the remarkable developments in various areas of Applied Mathematics such as Fluid Dynamics [27,29], Game theory [8,1], Theory of solitons [2–4] etc., the field of graph theory and its applications is also a rapidly expanding area of mathematics. In this progress, demands come from fields of applications like game theory and programming [8,1], communications theory [7], electrical networks and switching circuits [24] as well as problems from biology [20,19] and psychology [21]. The problem of domination is one of the most widely studied topics in graph theory: the 1998 book by Haynes et al. [17] contains a bibliography with over 1200 papers on the subject. The domination problem was studied from the 1950s onwards, but the rate of research on domination significantly increased in the mid-1970s. Perhaps the fastest growing area within graph theory is the study of domination in graphs. The signed domination number of a graph G was defined in [9] and has been studied by several authors including [9,13]. Independent dominating sets were introduced into the theory of games by Morgenstern in [25]. For an extensive survey of domination problems and comprehensive bibliography the readers are refereed to [18]. The study of domination numbers of products of graphs was initiated by Vizing [26]. He conjectured that the domination number of the cartesian product of two graphs is always greater than or equal to the product of the domination numbers of the two factors; a conjecture which is still unproven. In [14], a link is shown between the existence of tilings in Manhattan metric with {1}-bowls and minimum total dominating sets of cartesian products of paths and cycles. Domination numbers of cartesian products were intensively investigated (see e.g., [28,12,14]). The graphs, considered here, are finite, nonempty, connected, undirected, without loops and without multiple edges. Besides these, any undefined terms in this paper may be found in Harary [15].
⇑ Corresponding author. E-mail addresses: [email protected] (M. Nandi), [email protected] (S. Parui), [email protected] (A. Adhikari). 0096-3003/$ - see front matter Ó 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2010.11.019
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Let G be a simple graph whose vertex set and edge set are V(G) and E(G), respectively. A set D # V(G) of a simple graph G is called a dominating set if every vertex v 2 V(G)nD is adjacent to some vertex u 2 D. The domination number of G is the cardinality of a smallest dominating set of the graph G and it is usually denoted by c(G). In addition, smallest such dominating set is called a minimum dominating set of G. For any two graphs G and H, the Cartesian product G h H is the graph with vertex set V(G) V(H) and with edge set E(G H) such that (u1, v1) (u2, v2) 2 E(G H), whenever v1 = v2 and u1u2 2 E(G), or u1 = u2 and v1v2 2 E(H) [23]. Throughout the paper, the following notation and terminology are used. The numbers 0, 1, 2, . . . , n 1 always denote the vertices of the path Pn or the cycle Cn. Also, let c(Pm h Pn) and c(Pm h Cn) denote the domination numbers of Cartesian product graphs Pm h Pn and Pm h Cn, respectively. The graph Pm h Cn can be termed as cylindrical grid graph as shown in Fig. 1 in two different looks. Let (G)v = G {v} where v 2 V(H) and (H)u = {u} H where u 2 V(G). (G)v and (H)u are called the layers of G and H, respectively. Moreover, layer of a dominating set means D \ (Pm)i for i 2 V(Cn). Throughout the paper, the leftmost column in all figures denotes the layer (Pm)0. Now, define the term modified concatenation, of two dominating sets of Pm C n1 and Pm C n2 . If D1 and D2 are two dominating sets of Pm C n1 and P m C n2 , respectively then the modified concatenation of D1 and D2, denoted by D1kD2, is a subset D of P m C n1 þn2 such that D \ (Pm)i = D1 \ (Pm)i, i = 0, 1, . . . , n1 1 and D \ ðPm Þn1 þi ¼ D2 \ ðPm Þi , i = 0, 1, . . . , n2 1, i.e., the ith (Pm)-layer of D is coming from the ith (Pm)-layer of D1 if 0 6 i 6 n1 1 and from the i n1th (Pm)-layer of D2 if n1 6 i 6 n1 + n2 1. The illustration is shown in the Fig. 2. One of the most challenging problems concerning the domination numbers of Cartesian products of graphs is the proof of the Vizing Conjecture, namely c(G h H) P c(G) c(H) [26]. Despite numerous results showing its validity in some special cases, till date the conjecture remains an open problem. Partial works have been made towards finding the domination numbers of some particular Cartesian product of graphs. This problem also seems to be difficult one and the authors of [6] proved that even for subgraphs of Pm h Pn, this problem is NP-complete. In [22], Jacobson and Kinch established the following results: For all n P 1, cðP2 Pn Þ ¼
nþ2
. 2 3nþ4 . 4 n þ 1; cðP4 Pn Þ ¼ n; cðP3 Pn Þ ¼
for n ¼ 1; 2; 3; 5; 6; 9; otherwise:
Fig. 1. Two different looks of the cylindrical grid graph P5 h C5.
Fig. 2. Modified concatenation of the dominating sets of P5 h C6 and P5 h C3 to get the dominating set for P5 h C9.
M. Nandi et al. / Applied Mathematics and Computation 217 (2011) 4879–4889
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In [5], Chang and Clark established the following results: 6nþ6 ; for n ¼ 2; 3; 7; 5 cðP 5 Pn Þ ¼ 6nþ8 ; otherwise: 5 In [23], the authors established the following results regarding the Cartesian product of two cycles: For n P 4, cðC 3 C n Þ ¼ n n4 . For n P 4, c(C4 h Cn) = n. n; For n P 5, cðC 5 C n Þ ¼ n þ 1;
n ¼ 5k; k P 1; n 2 f5k þ 1; 5k þ 2; 5k þ 4g;
k P 1:
Furthermore, c(C5 h C5k+3) 6 5(k + 1), k P 1. More works may be found in [10–12,16]. In this paper, the topic of interest is to find the domination numbers of cylindrical grid graphs Pm h Cn, m P 2, n P 3. Towards finding the answer, some partial results in this direction are obtained. The domination numbers as well as minimum dominating sets of the graphs Pm h Cn, for m = 2, 3, 4 and n P 3 are found. Bounds on c(Pm h Cn) for m = 5 and n P 3 are also proposed. As a brief summery, results that are proved in the subsequent sections are stated as follows. For all n P 3, nþ1 ; when n is not a multiple of 4; cðP 2 C n Þ ¼ n 2 ; when n is a multiple of 4: 2 3n cðP 3 C n Þ ¼ 4 . n þ 1; for n ¼ 3; 5; 9; cðP 4 C n Þ ¼ n; otherwise: c(P5 h C3) = 4, c(P5 h C4) = 5 and c(P5 h C5) = 7. Moreover, for n P 6, n þ 5n 6 cðP 5 C n Þ 6 n þ n4 . Throughout the paper, we use the arithmetic operations of the indices over modulo n. 2. Finding the domination numbers of some cylindrical grid graphs In this section, the domination numbers as well as minimum dominating sets of particular cylindrical grid graphs of the form Pm h Cn, for all n P 3 and for m = 2, 3 and 4 are found. To prove the results, the following Lemmas are required. Lemma 2.1. Let m P 2. Then, there exists a minimum dominating set D of Pm h Cn such that for every i 2 V(Cn), j(Pm)i \ Dj 6 m 1. Proof. Let D0 be a minimum dominating set of Pm h Cn. Suppose further that, j(Pm)i \ D0 j = m holds for k many Pm-layers (Pm)i, i.e., for k-many i’s in {0, 1, . . . , n 1}, 0 6 k 6 n 1. Let us assume that j(Pm)i \ D0 j = m for some i 2 {0, 1, 2, . . . , n 1}. In addition, if (Pm)i1 \ D0 = (Pm)i+1 \ D0 = /, then we consider D = (D0 [ {(0, i 1), (1, i + 1)})n{(0, i), (1, i)}. Next, we assume that the layer (Pm)i+1 has nonempty intersection with D0 and let (j, i + 1) 2 (Pm)i+1 \ D0 . Then, it is clear that (j, i 1) and at least one of (j + 1, i 1) and (j 1, i 1) R (Pm)i1 \ D0 , otherwise, D0 n{(j, i)} would be a dominating set, contradicting the minimality of D0 and hence j(Pm)i1 \ D0 j < m 1. Then, we consider D = (D0 [ {(j, i 1)})n{(j,i)}. Now, D is a minimum dominating set with k 1 many Pm-layers having m vertices in common with D. Repeating this construction we get the result. h Lemma 2.2. Let n P 3. Then, there exists a minimum dominating set D of Pm h Cn such that for every i 2 V(Pm), jD \ (Cn)ij 6 n 1. Proof. The proof is similar as Lemma 2.1.
h
Lemma 2.3. There cannot be two consecutive Pm-layers having empty intersection with a minimum dominating set of Pm h Cn, for m P 3 and n P 4. Proof. If possible, let there be a minimum dominating set D having empty intersection with two consecutive layers (Pm)i and (Pm)i+1 for some i 2 {0, 1, 2, . . . , n 1}. Define (Pm)1 = (Pm)n1, (Pm)2 = (Pm)n2 and (Pm)n = (Pm)0, (Pm)n+1 = (Pm)1. Thus, j(Pm)i \ Dj = j(Pm)i+1 \ Dj = 0 gives j(Pm)i1 \ Dj = j(Pm)i+2 \ Dj = m. Now, consider the following cases: Case 1. For n = 4, 5, we set D0 = D [ {(0, i)}n{(0, i 1), (1, i 1)}. Case 2. For n P 6, we set D0 = (D [ {(0, i), (1, i 2), (1, i + 3), (2, i)})n{(0, i 1), (0, i + 2), (1, i 1), (2, i 1), (2, i + 2)}. In each case we find that D0 is a dominating set and jD0 j < jDj, which is a contradiction. Therefore, we conclude that there cannot have two consecutive layers (Pm)i, (Pm)i+1 having empty intersection with a minimum dominating set. h
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Lemma 2.4. For every dominating set D of Pm h Cn the following m inequalities hold:
xi1 þ 3xi þ xiþ1 P n;
8i ¼ 0; . . . ; m 1;
where xi = j(Cn)i \ Dj for i = 0, 1, . . . , m 1 and x1 = xm = 0. Moreover, if xi1 + 3xi + xi+1 = n, then there does not exist any pair Pm1 of vertices from ((Cn)i1 [ (Cn)i [ (Cn)i+1) \ D such that they dominate a common vertex of (Cn)i. Finally, i¼0 xi ¼ jDj P cðPm C n Þ. Proof. The result follows from the fact that any vertex of D from (Cn)i dominates three vertices of (Cn)i including itself and any vertex of D from (Cn)i1 or (Cn)i+1 dominates one vertex of (Cn)i. h Remark 2.5. The similar result of Lemma 2.4 holds for yi = j(Pm)i \ Dj, where i 2 {0, 1, . . . , n 1}. Using the above Lemmas we now prove the following theorems. nþ1 ; when n is not a multiple of 4; Theorem 2.6. For n P 3, cðP2 C n Þ ¼ n 2 when n is a multiple of 4: 2; Proof. We consider a set D with
D¼
fð0; iÞ : i 2ðmod 4Þg [ fð1; iÞ : i 0ðmod 4Þg [ fð0; n 1Þg; if n 2ðmod 4Þ; fð0; iÞ : i 2ðmod 4Þg [ fð1; iÞ : i 0ðmod 4Þg;
otherwise:
nþ1
Then, D is a dominating set of P2 h Cn and that jDj ¼ 2 , whenever n is not a multiple of 4 and jDj ¼ 2n, whenever n is a , or c(P2 h Cn) 6 n/2 for different values of n. multiple of 4. Hence, cðP2 C n Þ 6 nþ1 2 n To show that cðP2 C n Þ P 2 , let D0 be a minimum dominating set. Let j(Cn)i \ D0 j = xi, for i = 0, 1, i.e., xi is the number of vertices of D0 from the layer (Cn)i. Then, using Lemma 2.4 we have
3x0 þ x1 P n;
ðiÞ
x0 þ 3x1 P n;
ðiiÞ
x0 þ x1 ¼ cðP2 C n Þ:
ðiiiÞ
Therefore, 4(x0 + x1) P 2n which gives x0 þ x1 P n2. nþ1 ; if n is odd; Consequently, cðP2 C n Þ P n 2 if n is even: 2; Now, let n = 4k + 2, k P 1. We will show that cðP 2 C n Þ P 2n þ 1, i.e., c(P2 h Cn) P 2k + 2. If possible, let cðP2 C n Þ 6 n2 ¼ 2k þ 1. Then, x0 + x1 6 2k + 1. Therefore, one of x0 or x1 must be less than or equals k. Without loss of generality,
let
x0 6 k.
Then,
3x0 + x1 6 3x0 + (2k + 1 x0) = 2x0 + 2k + 1 6 4k + 1 < n,
cðP2 C n Þ P 2n þ 1. h Remark 2.7. For n P 3, cðP 2 C n Þ P cðP 2 Þ cðC n Þ ¼ 1 . Theorem 2.8. For n P 3, cðP3 C n Þ ¼ 3n 4
which
contradicts
(i).
Hence,
n , the equality holds only when n = 4. 3
Proof. Consider a set D with
D¼
fð0; iÞ; ð2; iÞ : i 2ðmod 4Þg [ fð1; iÞ : i 0ðmod 4Þg [ fð1; n 1Þg; if n 2ðmod 4Þ; fð0; iÞ; ð2; iÞ : i 2ðmod 4Þg [ fð1; iÞ : i 0ðmod 4Þg;
otherwise:
. For n = 3 and n = 4, it is illustrated in Fig. 3. Therefore, Then, D is a dominating set of P3 h Cn and that jDj ¼ 3n 4 cðP3 C n Þ 6 3n4 . Next, we show that cðP3 C n Þ P 3n 4 . Let n = 4k + t, where k P 0, 3 P t P 0 and let D be a minimum dominating set satisfying the property as stated in Lemma 2.1. Let s be the number of P3-layers which have empty intersection with D. Then, by Lemma 2.3, since no two empty layers are adjacent, we have s 6 2n ¼ 2k þ 2t . Now, as every empty P3-layer is s dominated by exactly two other P3-layers, there are at least 2 P3-layers with precisely two vertices from D. Hence,
2kþb2t c ¼ is maximum when s ¼ 2k þ 2t . So, jDj P n 2k þ 2t jDj P 2 2s þ n 2s s ¼ n s 2s . Also, s 2s 2
t t 2kþb2c bc ¼ 3k þ t 2t þ 22 ¼ 3k þ t (since 0 6 t 6 3Þ ¼ 3n ð4k þ tÞ 2k þ 2t 4 . h 2
Fig. 3. Vertices with bold circle form a minimum dominating set for P3 h C3 and P3 h C4.
M. Nandi et al. / Applied Mathematics and Computation 217 (2011) 4879–4889
Remark 2.9. For n P 3, cðP 3 C n Þ > cðP 3 Þ cðC n Þ ¼ 1
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n . 3
Theorem 2.10. c(P4 h Cn) = n + 1, for n = 3, 5, 9. Proof. The proof of the theorem for three different values of n are given successively. n = 3. Consider the set D1 = {(0, 2), (1, 0), (2, 2), (3, 1)}. Then, D1 is a dominating set for P4 h C3 as shown in Fig. 4. Since jD1j = 4, c(P4 h C3) 6 4. Now, if possible, let there exist a dominating set D01 such that jD01 j 6 3. Let jðC 3 Þi \ D01 j ¼ xi , where xi is the number of vertices of D01 from the layer (C3)i, for i = 0, 1, 2, 3. Therefore, 3 X
xi 6 3;
ðiÞ
3x0 þ x1 P 3;
ðiiÞ
x0 þ 3x1 þ x2 P 3;
ðiiiÞ
x1 þ 3x2 þ x3 P 3; x2 þ 3x3 P 3:
ðivÞ ðvÞ
i¼0
Also,
Now, adding (ii) and (v) and subtracting (i) we get
x0 þ x3 P 2
ðviÞ
which gives
x1 þ x2 6 1:
ðviiÞ
Therefore, at least one of x1 and x2 must be zero. Without loss of generality, let x1 = 0. Then, from (iii) we have
x0 þ x2 P 3
ðviiiÞ
and hence from (i) x3 = 0. Therefore, from (v) we get x2 = 3 and from (i) x0 = 0 which contradicts (ii). So, we have jD0 j P 4 and therefore c(P4 h C3) = 4. n = 5. Consider the set D2 = {(0, 2), (0, 4), (1, 0), (2, 3), (3, 1), (3, 4)}. Then, D2 is a dominating set of P4 h C5 with jD2j = 6 as shown in Fig. 4. Therefore, c(P4 h C5) 6 6. We now show that, c(P4 h C5) P 6. If possible, let there exist a dominating set D02
0
such that D2 6 5. Let ðC 5 Þi \ D02 ¼ xi , where xi is the number of vertices of D02 from the layer (C5)i for i = 0, 1, 2, 3. Therefore, 5 X
xi 6 5:
ðiÞ
i¼0
Also, from Lemma 2.4 we have
3x0 þ x1 P 5;
ðiiÞ
x0 þ 3x1 þ x2 P 5;
ðiiiÞ
x1 þ 3x2 þ x3 P 5;
ðivÞ
x2 þ 3x3 P 5:
ðvÞ
Now, adding (ii) and (v) and subtracting (i) we get
x0 þ x3 P 3
ðviÞ
which implies
x1 þ x2 6 2:
ðviiÞ
Now, we claim that x1 – 0. Otherwise, if x1 = 0, then from (iii) we get x0 + x2 P 5. Therefore, x0 + x2 = 5 and from (i) we have x3 = 0. Therefore, from (v) we get x2 P 5, i.e., x2 = 5. This gives x0 = 0, which contradicts (ii). Therefore, we have x1 – 0. Similarly, x2 – 0. So, we have from (vii), x1 = x2 = 1. For these values of x1 and x2 we get from (v), x3 P 2 and from (ii), x0 P 2,
contradicting (i). Therefore, we cannot have D02 6 5. So, D02 P 6 and hence c(P4 h C5) = 6.
Fig. 4. Vertices with bold circles form a minimum dominating set for P4 h C3, P4 h C5 and P4 h C9.
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n = 9. Consider the set D3 = {(0, 2), (0, 4), (0, 8), (1, 0), (1, 6), (2, 3), (2, 8), (3, 1), (3, 5), (3, 7)}. Then D3 is a dominating set of P4 h C9 with jD3j = 10 as shown in Fig. 4. Therefore, c(P4 h C9) 6 10. Now, if possible, let there exists a dominating set D03 such
0
that D3 6 9 and let ðC 9 Þi \ D03 ¼ xi , where xi is the number of vertices of D03 from the layer (C9)i for i = 0, 1, 2, 3. Therefore, 9 X
xi 6 9:
ðiÞ
i¼0
Also, from Lemma 2.4 we have
3x0 þ x1 P 9;
ðiiÞ
x0 þ 3x1 þ x2 P 9;
ðiiiÞ
x1 þ 3x2 þ x3 P 9;
ðivÞ
x2 þ 3x3 P 9:
ðvÞ
Now, adding (ii) and (v) and subtracting (i) we get
x0 þ x3 P 5
ðviÞ
which gives
x1 þ x2 6 4:
ðviiÞ
Now, we claim that x1 – 0. If x1 = 0 then from (iii) we have x0 + x2 P 9 and hence from (i), x3 = 0. Therefore, from (v), x2 = 9 and from (i), x0 = 0 which contradicts (ii). Therefore, x1 – 0. Similarly, we can show that x2 – 0. Again, we claim that, either x1 – 2 or x2 – 2. If possible, let x1 = x2 = 2. Then, from (ii) and (v), we get x0 P 3 and x3 P 3, contradicting (i). Therefore, either x1 – 2 or x2 – 2. Now, from (vii), we get one of x1 or x2 must be 1. Without loss of generality, let x1 = 1. Therefore, from (iii) we get x0 + x2 P 6 and from (i) we get x1 + x3 6 3. This gives x3 6 2. We now claim that x3 – 0. If possible, let x3 = 0, then from (v) we get x2 P 9 which contradicts (i). So, x3 – 0. Next, we show that x3 – 1. If possible, let x3 = 1. Then, from (v) we get x2 P 6. Therefore, from (i), x0 6 1 which contradicts (ii). Consequently, x3 – 1 and hence x3 = 2. This implies x0 + x2 = 6. Now, from (v), x2 P 3 and from (ii), x0 P 3. Therefore, x0 = 3, x1 = 1, x2 = 3, and x3 = 2.
Lastly, we show that x0 = 3, x1 = 1, x2 = 3, and x3 = 2 cannot give a dominating set D03 with D03 ¼ 9. Without loss of 0 generality, let us assume ð1; 0Þ 2 D3 . Then, ð1; iÞ R D03 for all i = 1, 2, . . . , 8. Again, ð0; 1Þ R D03 because if ð0; 1Þ 2 D03 then (1, 1) is dominated by (0, 1) and (1, 0) which will be a contradiction by the Lemma 2.4. Similarly, ð0; 0Þ; ð0; 8Þ R D03 . Therefore, to dominate (0, 1) and (0, 8) we must take ð0; 2Þ; ð0; 7Þ 2 D03 . Till now, the vertices (0, 4) and (0, 5) in the layer (C9)0 are not dominated by any vertices of D03 . Therefore, either ð0; 4Þ 2 D03 or ð0; 5Þ 2 D03 . Without loss of generality, let ð0; 4Þ 2 D03 . Again, the vertices (1, 3), (1, 5), (1, 6) in the layer (C9)1 are not dominated by any vertices of D03 . To dominate these vertices we must take the vertices (2, 3), (2, 5), (2, 6) in D03 . In a similar way, to dominate the vertices (2, 1), (2, 8) of the layer (C9)2 we must take the vertices (3, 1), (3, 8) in D03 . Still (3, 4) is not dominated by any vertex of D03 as shown in Fig. 5. Hence, D03 can not be a dominating set. This contradiction shows that c(P4 h C9) P 10. h Theorem 2.11. For n P 3, c(P4 h Cn) = n, for n – 3, 5, 9. Proof. The three diagrams (i), (ii) and (iii) shown in Fig. 6 give dominating sets for P4 h C4, P4 h C6 and P4 h C7, respectively. Note that, a dominating set for P4 h C8 can be obtained by the modified concatenation of the two copies of the diagram (i). A dominating set for P4 h C10 can be constructed by the modified concatenation of the diagrams (i) and (ii), a dominating set for P4 h C11 can be constructed by the modified concatenation of the diagrams (i) and (iii), a dominating set for P4 h C12 can be constructed by the modified concatenation of the two copies of the diagram (ii) and a dominating set for P4 h C13 can be constructed by modified concatenation of the diagrams (ii) and (iii). Finally, the dominating set for n P 14 can be constructed by modified concatenation of the diagram for n = 4 with the diagrams for n = 10, 11, 12, 13. Hence, we have c(P4 h Cn) 6 n.
Fig. 5. Non admissibility of c(P4 h C9) = 9.
M. Nandi et al. / Applied Mathematics and Computation 217 (2011) 4879–4889
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Fig. 6. Vertices with bold circle form a minimum dominating set for P4 h C4, P4 h C6 and P4 h C7.
It now remains to show that c(P4 h Cn) P n. To prove this, consider a minimum dominating set D which has no empty (P4)i layer. Then, jDj P n and hence the theorem follows. Now, consider a minimum dominating set D0 with s empty (P4)i layers. Further, let there be t nonempty (P4)i layers each adjacent to exactly one empty (P4)i layer. Then, there are k ¼ 2st nonempty (P4)i layers each adjacent to two empty layers. Since there do not exist two consecutive 2 empty (P4)i layers, the number of nonempty (P4)i layers which have no adjacent empty layer is p ¼ n t 2st s¼ 2 n 2s 2t . Now, let x01 ; x02 ; . . . ; x0t be the numbers of elements of D0 which are contained in t nonempty (P4)i layers, respectively such that each of these (P4)i layers is adjacent with exactly one empty (P4)i layer. Further, we define x001 ; x002 ; . . . ; x00k and x01 ; x02 ; . . . ; x0p in an analogous way. Now, by Remark 2.5, 2 x001 þ x002 þ þ x00k þ x01 þ x02 þ þ x0t P 4s. 00 00 00 0 0 0 Therefore, 2 x 1 þ x2 þ þ xk þ x1 þ x2 þ þ xt P 4s þ t. This implies 2 x01 þ x02 þ þ x0p þ x001 þ x002 þ þ x00k þ x01 þ x02 þ þ x0t P 4s þ t þ 2p ¼ 4s þ t þ 2n 4s t ¼ 2n. Consequently, x1 + x2 + + xn P n where xi = j(P4)i \ D0 j. h Remark 2.12. For n P 3, cðP 4 C n Þ > cðP 4 Þ cðC n Þ ¼ 2 3n . 3. Bounds on the domination numbers of P5 h Cn, n > 3 In this section, the exact domination numbers are given for P5 h C3, P5 h C4 and P5 h C5. Furthermore, bounds on the domination numbers are also proposed for P5 h Cn, n P 6. The following Lemmas are essential for proving the subsequent theorems. Lemma 3.1. For n P 3, there exists a minimum dominating set D of P5 h Cn such that for every i 2 V(Cn), j(P5)i \ Dj 6 3. Proof. Let D be a minimum dominating set of P5 h Cn such that for every i 2 V(Cn), j(P5)i \ Dj 6 4. Such a D exists by Lemma 2.1. Suppose further that j(P5)i \ Dj = 4 holds for k many P5-layers. We now construct a dominating set D0 with jD0 j = jDj such that only k 1 many P5-layers have 4 vertices in common with D. Assume that j(P5)i \ Dj = 4 for some i 2 V(Cn). We claim that j(P5)i1 \ Dj 6 2, and j(P5)i+1 \ Dj 6 2. If possible, let j(P5)i+1 \ Dj P 3. Then, D00 = (Dn((P5)i [ (P5)i+1)) [ {(2, i 1), (0, i), (4, i), (1, i + 1), (0, i + 2), (3, i + 2)} is a dominating set with jD00 j < jDj which contradicts the minimality of D. Therefore, j(P5)i+1 \ Dj 6 2. Similarly, j(P5)i1 \ Dj 6 2. Now, we set D0 = (Dn(P5)i) [ {(2, i 1), (0, i), (4, i), (2, i + 1)}. Then, D0 is a dominating set with jD0 j 6 jDj but since D is a minimum dominating set we have jD0 j = jDj. Also, j(P5)j \ D0 j < 4 for j = i 1, i, i + 1 and j(P5)j \ D0 j = j(P5)j \ Dj for other values of j. Repeating this process we get the desired minimum dominating set. h Lemma 3.2. For n P 5, there exists a minimum dominating set D of P5 h Cn such that for every i 2 V(Cn) either (a) j(P5)i \ Dj 6 2 or (b) j(P5)i \ Dj = 3 with (P5)i1 \ D = / and (P5)i+1 \ D = / for i 2 V0 and j(P5)i \ Dj 6 2 for all i R V0 , for some V0 # V(Cn). Proof. Lemma 3.1 shows there exists a minimum dominating set D such that j(P5)i \ Dj 6 3 for all i 2 V(Cn). Now, let j(P5)i \ Dj = 3 hold for k many P5-layers and among these k many P5-layers there exist k0 many P5-layers with the property that both the adjacent P5-layers of such a layer have empty intersection with D. Now, if k0 = k then D has the desired property. When k0 < k we construct a dominating set D0 with the property that D0 has k 1 or less P5-layers having three vertices in common with D and jD0 j = jDj. Let j(P5)i \ Dj = 3 for some i 2 V(Cn). We claim that, both of j(P5)i1 \ Dj and j(P5)i+1 \ Dj cannot be simultaneously equal to 2. On contrary, if possible, let j(P5)i1 \ Dj = j(P5)i+1 \ Dj = 2. Now, we consider the following sets
A1 ¼ fð1; i 2Þ; ð3; i 1Þ; ð4; i 1Þ; ð0; iÞ; ð2; i þ 1Þ; ð4; i þ 2Þg; A2 ¼ fð3; i 2Þ; ð0; i 1Þ; ð1; i 1Þ; ð4; iÞ; ð2; i þ 1Þ; ð0; i þ 2Þg; A3 ¼ fð2; i 2Þ; ð4; i 2Þ; ð0; i 1Þ; ð3; iÞ; ð1; i þ 1Þ; ð4; i þ 2Þg; A4 ¼ fð2; i 2Þ; ð0; i 2Þ; ð4; i 1Þ; ð1; iÞ; ð3; i þ 1Þ; ð0; i þ 2Þg and A5 ¼ fð2; i 2Þ; ð0; i 1Þ; ð4; i 1Þ; ð2; iÞ; ð0; i þ 1Þ; ð4; i þ 1Þg:
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Let Bkl = {(k, i + 1), (l, i + 1)}, where k, l = 0, 1, 2, 3, 4. Now, if (P5)i+1 \ D = B01 or B02 or B12 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A2, if (P5)i+1 \ D = B03 or B13 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A4, if (P5)i+1 \ D = B04 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A5, if (P5)i+1 \ D = B14 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A3, and if (P5)i+1 \ D = B23 or B24 or B34 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A1. Then, in every case D0 will be a dominating set with jD0 j < jDj, which is a contradiction. Hence, the claim follows. Now, we consider the all other possible cases: Case 1. Let j(P5)i+1 \ Dj = 3. Then, we claim that j(P5)i+2 \ Dj < 2. To show this if possible, let j((P5)i [ (P5)i+1 [ (P5)i+2) \ Dj P 8. Then, D00 = (Dn{(P5)i [ (P5)i+1 [ (P5)i+2)} [ {(0, i 1), (4, i 1), (2, i), (0, i + 1), (3, i + 2), (4, i + 2), (1, i + 3)} is a dominating set with jD00 j < jDj, contradicting the minimality of D. Therefore, j(P5)i+2 \ Dj < 2. Similarly, j(P5)i1 \ Dj < 2. Now, we construct D0 = (Dn{(P5)i [ (P5)i+1}) [ {(2, i 1), (0, i), (4, i), (0, i + 1), (4, i + 1), (2, i + 2)}. Then jD0 j 6 jDj and D0 is a dominating set. Hence, D0 is a minimum dominating set having (k 2) many P5-layers, each of which has three vertices in common with D. Similar argument for j(P5)i1 \ Dj = 3. Case 2. Let j(P5)i1 \ Dj = 2 and j(P5)i+1 \ Dj 6 1. Then, we can construct D0 such that j(P5)i1 \ D0 j = 2, j(P5)i \ D0 j = 2 and j(P5)i+1 \ D0 j = j(P5)i+1 \ Dj + 1. Similarly, we can construct D0 for j(P5)i1 \ Dj 6 1 and j(P5)i+1 \ Dj = 2. Case 3. Let j(P5)i1 \ Dj = 1 and j(P5)i+1 \ Dj 6 1. subcase 3.1. When j(P5)i \ Dj = {(0, i), (1, i), (2, i)} or {(2, i), (3, i), (4, i)}, then construct D0 = (Dn{(0, i), (1, i)}) [ {(0, i 1), (1, i + 1)} for the first one. Similarly, we can construct D0 for the other one. subcase 3.2. Other than the above case we can construct D0 such that j(P5)i1 \ D0 j = 1, j(P5)i \ D0 j = 2, j(P5)i+1 \ D0 j = j(P5)i+1 \ Dj + 1. Similar construction will be made when j(P5)i+1 \ Dj = 1 and j(P5)i1 \ Dj 6 1. Then, D0 is a minimum dominating set having k 1 many or less P5-layers each of which has three vertices in common with D. Repeating this replacement, we will get the minimum dominating set with the desired property as stated in this Lemma. h Lemma 3.3. For n P 5, there cannot be a dominating set D with five consecutive P5-layers having exactly one vertex in common with D. Proof. On contrary, let there exists a dominating set D and there exists i 2 V(Cn) such that j(P5)j \ Dj = 1 for all j = i, i + 1, i + 2, i + 3, i + 4. Note that, (0, j) R D and (4, j) R D for all j = i + 1, i + 2, i + 3. Therefore, (1, i + 2) 2 D and (3, i + 2) 2 D, a contradiction. h Lemma 3.4. Let D be a minimum dominating set with the property as stated in Lemma 3.2 . Again, let (P5)i and (P5)j be two layers having two vertices in common with D and j(P5)i+1 \ Dj – 2, j(P5)i+2 \ Dj – 2, . . . , j(P5)j1 \ Dj – 2. Then either (a) j(P5)i+1 \ Dj = 0, j(P5)i+2 \ Dj = 3, j(P5)i+3 \ Dj = 0, j(P5)i+4 \ Dj = 3, . . . , j(P5)j1 \ Dj = 0, or (b) j i 6 5 and j(P5)l \ Dj = 1 for all l = (i + 1), (i + 2), . . . , (j 1). Proof. If j(P5)l \ Dj = 0, then j(P5)l1 \ Dj > 1. Otherwise, j(P5)l+1 \ Dj P 4. Hence, j(P5)l1 \ Dj = 2 or 3. Similarly, j(P5)l+1 \ Dj = 2 or 3. Therefore, using Lemmas 3.2 and 3.3 we get the desired result. h Theorem 3.5. For n P 6, cðP5 C n Þ P n þ n5 . Proof. Let D be a minimum dominating set with the property as stated in Lemma 3.2. Let us call the collection of P5-layers {(P5)k: k = i + 1, i + 2, . . . , j 1} as a block where i and j as in the above Lemma 3.4. Let x0, x1, x3 be the number of P5-layers having 0, 1, 3 vertices, respectively in common with D. Let x02 be the number of blocks in which every P5-layer has exactly one vertex in common with D and x002 be the number of blocks in which every P5-layer has either 0 or 3 vertices in common with D and x02 be the number of blocks where j = i + 1, i.e., when the block contains no P5-layer. Then, clearly, x2 ¼ x02 þ x02 þ x002 be the number of P5-layers having 2 vertices in common with D and we have therefore x3 P x002 ; x0 ¼ x002 þ x3 ; x1 6 4x02 and 6ðx þx þx0 þx0 þx00 þx Þ ¼ 3x3 þ 2x2 þ x1 6n ¼ 3x3 þ 2x02 þ 2x02 þ 2x002 þ x1 0 1 25 2 2 3 ¼ 15 ð9x3 þ x0 þ x1 þ x02 þ x02 þ x002 þ x3 ¼ n. Now, jDj 6n 5 5 4x0 þ 4x02 þ 4x002 x1 6x0 Þ ¼ 15 9x3 þ 4x02 þ 4x02 þ 4x002 x1 6x002 6x3 ¼ 15 3x3 2x002 þ 4x02 þ 4x02 x1 ¼ 15 x3 þ 2 x3 x002 þ 20 n 0 6n 4x2 x1 þ 4x2 g P 0. Therefore, we have jDj P 5 . Thus, cðP5 C n Þ P n þ 5 . Note 3.6 jDj ¼ 6n 5 þ
x3 þ2ðx3 x002 Þþð4x02 x1 Þþ4x02 . 5
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Theorem 3.7. (a) c(P5 h C3) = 4, (b) c(P5 h C4) = 5, and (c) c(P5 h C5) = 7. Proof. (a) Let D be a minimum dominating set of P5 h C3. Further, let j(Cn)i \ Dj = xi, for i = 0, 1, 2, 3, 4. Then, by Lemma 2.4 we have
3x0 þ x1 P 3;
ðiÞ
x0 þ 3x1 þ x2 P 3;
ðiiÞ
x1 þ 3x2 þ x3 P 3;
ðiiiÞ
x2 þ 3x3 þ x4 P 3;
ðivÞ
x3 þ 3x4 P 3:
ðvÞ
If possible, let
jDj ¼ x0 þ x1 þ x2 þ x3 þ x4 6 3:
ðviÞ
Now, x0 – 0. Otherwise, x1 = 3 would imply x2 = x3 = x4 = 0, contradicting (iv). Similarly, x4 – 0. Next, if possible, let x0 = 2 and x4 = 1. Then, we have x1 = x2 = x3 = 0, contradicting (ii). Again, taking x0 = 1 and x4 = 2 we arrived at the same contradiction. Now, if possible, let x0 = x4 = 1. Adding (ii), (iii), and (iv) we get x0 + 4x1 + 5x2 + 4x3 + x4 P 9. Therefore, 4x1 + 5x2 + 4x3 P 7. This implies 5x1 + 5x2 + 5x3 P 7 and hence x1 + x2 + x3 P 2, contradicting (vi). Consequently, jDj P 4 and therefore c(P5 h C3) P 4. Now, D = {(0, 1), (2, 0), (2, 2), (4, 1)} is a dominating set of P5 h C3 as shown in Fig. 7. Therefore, c(P5 h C3) 6 4 and hence c(P5 h C3) = 4. This completes the proof of (a). (b) Let D be a minimum dominating set of c(P5 h C4). Further, let j(Cn)i \ Dj = xi for i = 0, 1, 2, 3, 4. Then, by Lemma 2.4,
3x0 þ x1 P 4;
ðiÞ
x0 þ 3x1 þ x2 P 4;
ðiiÞ
x1 þ 3x2 þ x3 P 4; x2 þ 3x3 þ x4 P 4;
ðiiiÞ ðivÞ
x3 þ 3x4 P 4:
ðvÞ
If possible, let
jDj ¼ x0 þ x1 þ x2 þ x3 þ x4 6 4:
ðviÞ
Then, as in the proof of (a) we have x0 – 0 and x4 – 0. From (i) and (v) we get 3x0 + x1 + x3 + 3x4 P 8. Therefore, 2x0 + 2x4 P 4 + x2. Now, if x2 – 0 then x0 + x4 P 3. Hence, x1 + x2 + x3 6 1 which implies x2 = 1 and x1 = x3 = 0. Therefore, from (ii) and (iv) x0 + x2 P 4 and x2 + x4 P 4. Hence, x0 P 3 and x4 P 3, contradicting (vi). If x2 = 0 then from (iii), x1 + x3 P 4. Now, from (vi) we have x0 = x4 = 0. From (i) we have x1 P 4 and hence x3 = 0, contradicting (iv). Therefore, we have c(P5 h C4) P 5. Now, D = {(0, 2), (1, 0), (2, 3), (3, 1), (4, 3)} is a dominating set of P5 h C4 as shown in Fig. 8. Therefore, c(P5 h C4) 6 5 and hence c(P5 h C4) = 5. h (c) Let D be a minimum dominating set of c(P5 h C5). Further, let j(Cn)i \ Dj = xi where i = 0, 1, 2, 3, 4. Then, from Lemma 2.4 we have
3x0 þ x1 P 5;
ðiÞ
x0 þ 3x1 þ x2 P 5;
ðiiÞ
x1 þ 3x2 þ x3 P 5;
ðiiiÞ
x2 þ 3x3 þ x4 P 5;
ðivÞ
x3 þ 3x4 P 5:
ðvÞ
If possible, let
jDj ¼ x0 þ x1 þ x2 þ x3 þ x4 6 6:
ðviÞ
From (i) and (v) we get 3x0 + x1 + x3 + 3x4 P 10. Therefore, 2(x0 + x4) P 4 + x2. Now, if possible, let x2 = 0. Then, x0 + x4 P 2 and x1 + x3 P 5(by (iii)), contradicting (iv). Therefore, x2 P 1 and hence x0 + x4 P 3.
Fig. 7. Vertices with bold circle form a minimum dominating set for P5 h C3.
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Fig. 8. Vertices with bold circle form a minimum dominating set for P5 h C4.
Again, x0 – 0 and x4 – 0 as in (a). If possible, let x0 + x4 P 4. Then, x1 + x3 6 1 implies either x1 = 0 or x3 = 0. But this is a contradiction (since x1 = 0 implies x0 + x2 P 5 implies x0 = 4 implies x2 = 1 implies x3 P 2, contradicting (iv). Similar contradiction for x3 = 0). Therefore, x0 + x4 = 3, x2 P 1 and x1 + x3 6 2. Again, if x1 = x3 = 0 then the only possibility is x0 = 1, x1 = 0, x2 = 4, x3 = 0 and x4 = 1 which contradicts x0 + x4 = 3. If x1 = 1, x2 = 1,x3 = 1 and x0 + x4 = 3 then it will contradict either (i) or (v). Finally, if x1 = 1,x3 = 0 then x2 = 2 and x0 + x4 = 3, contradicting (i) or (v). Similar contradiction for x1 = 0,x3 = 1. Hence, we have c(P5 h C5) P 7. Now, D = {(0, 0), (0, 4), (1, 2), (2, 0), (3, 3), (3, 4), (4, 1)} is a dominating set of P5 h C5 as shown in Fig. 9. Therefore, c(P5 h C5) 6 7 and hence c(P5 h C5) = 7. h Lemma 3.8. 9 þ 95 6 cðP 5 C 9 Þ 6 9 þ 94 . Proof. The result follows from the Theorem 3.5 and the Fig. 10. h Theorem 3.9. For n P 6, n þ n5 6 cðP5 C n Þ 6 n þ n4 . Proof. For n = 9, the result follows from Lemma 3.8. For n P 6, the inequality in the left side is already being proved in Theorem 3.5. For the other part of the inequality, let us consider the minimum dominating sets for P4 h C6 and P4 h C7 as shown in Fig. 6. Now adding the vertices {(4, 3), (4, 5)} and {(4, 3), (4, 6)} to the dominating sets of P4 h C6 and P4 h C7, respectively, the dominating sets for P5 h C6 and P5 h C7 can be obtained as shown in Fig. 11. Finally, using the modified concatenation suitably as in the case of P4 h Cn, n P 3, n – 3, 5, 9, (Theorem 2.11), among these two dominating sets and the minimum dominating set of P5 h C4, as shown in Fig. 8, we get the dominating sets for P5 h Cn, n P 6, n – 9, with required cardinality. h
Fig. 9. Vertices with bold circle form a minimum dominating set for P5 h C5.
Fig. 10. Vertices with bold circle form a dominating set for P5 h C9.
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Fig. 11. Vertices with bold circle form a dominating set for P5 h C6 and P5 h C7.
4. Conclusion In this paper, we find the domination numbers of the graphs Pm h Cn, m = 2, 3 and 4, n P 3. Minimum dominating sets corresponding to the above mentioned graphs are also constructed. Finally, we give bounds on c(P5 h Cn), n P 3. As a future work, we would like to find the exact values of c(P5 h Cn), n P 3 as well as minimum dominating sets for P5 h Cn, n P 3. Moreover, we would also like to work on the bounds of c(Pm h Cn), m P 6 and n P 3. References [1] C.L. Barrett, C.M. Reidys, Elements of a theory of computer simulation I: sequential CA over random graphs, Appl. Math. Comput. 98 (2–3) (1999) 241– 259. [2] Anjan Biswas, Daniela Milovic, Optical solitons in a parabolic law media with fourth order dispersion, Appl. Math. Comput. 208 (1) (2009) 299–302. [3] Anjan Biswas, Arjuna Ranasinghe, 1-Soliton solution of Kadomtsev–Petviashvili equation with power law nonlinearity, Appl. Math. Comput. 214 (2) (2009) 645–647. [4] Anjan Biswas, Soliton-soliton interaction with dual-power law nonlinearity, Appl. Math. Comput. 198 (2) (2008) 605–612. [5] T.Y. Chang, W.E. Clark, The Domination numbers of the 5 n and 6 n grid graphs, J. Graph Theor. 17 (1993) 81–107. [6] B.N. Clark, C.J. Colbourn, D.S. Johnson, Unit disk graphs, Discrete Math. 86 (1990) 165–177. [7] Matthias Dehmer, Information processing in complex networks: graph entropy and information functionals, Appl. Math. Comput. 201 (1–2) (2008) 82– 94. [8] Matthias Dehmer, Frank Emmert-Streib, Jrgen Kilian, A similarity measure for graphs with low computational complexity, Appl. Math. Comput. 182 (1) (2006) 447–459. [9] J.E. Dunbar, S.T. Hedetniemi, M.A. Henning, P.J. Slater, Signed domination in graphs, graph theory, combinatorics, and applications, John Wiley & Sons, 1995. pp. 311–322. [10] M.H. El-Zahar, C.M. Pareek, Domination number of products of graphs, Ars Combin. 31 (1991) 223–227. [11] M.H. El-Zahar, S.M. Khamis, Kh. M. Nazzal, On the domination number of the Cartesian product of the cycle of length n and any graph, Discrete Appl. Math. 155 (2007) 515–522. [12] R.J. Faudree, R.H. Schelp, The domination number for the product of graphs, Congr. Numer. 79 (1990) 29–33. [13] O. Favaron, Signed domination in regular graphs, Discrete Math. 158 (1996) 287–293. [14] S. Gravier, Total domination number of grid graphs, Discrete Appl. Math. 121 (2002) 119–128. [15] F. Harary, Graph Theory, Addison Weslay, Reading Mass, 1969. [16] B.L. Hartnell, D.F. Rall, On Vizing’s conjecture, Congr. Numer. 82 (1991) 87–96. [17] T. Haynes, S. Hedetniemi, P. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, New York, 1998. [18] S.T. Hedetniemi, R.C. Laskar, Bibliography on domination in graphs and some basic definitions of domination parameters, Discrete Math. 86 (1–3) (1990) 257–277. [19] Sun-Yuan Hsieh, Chao-Wen Huang, Hsin-Hung Chou, A DNA-based graph encoding scheme with its applications to graph isomorphism problems, Appl. Math. Comput. 203 (2) (2008) 502–512. [20] Sun-Yuan Hsieh, Ming-Yu Chen, A DNA-based solution to the graph isomorphism problem using AdlemanLipton model with stickers, Appl. Math. Comput. 197 (2) (2008) 672–686. [21] Takehiro Inohara, Characterization of clusterability of signed graph in terms of Newcomb’s balance of sentiments, Appl. Math. Comput. 133 (1) (2002) 93–104. [22] M.S. Jacobson, L.F. Kinch, On the domination of the products of graphs I, Ars Combin. 18 (1983) 33–44. [23] S. Klavzar, N. Seifter, Dominating Cartesian products of cycles, Discrete Appl. Math. 59 (1995) 129–136. [24] W. Mayeda, Synthesis of switching functions by linear graph theory, IBM J. Res. Develop. 4 (3) (1960) 321–328. [25] O. Morgenstern, The collaboration between Oskar Morgenstern and John von Neumann on the theory of games, J. Eco. Literat. 14 (1976) 805–816. [26] V.G. Vizing, The Cartesian product of graphs, Vychisl Sistemy 9 (1963) 30–43. [27] Xiao-Ge Xu, Xiang-Hua Meng, Yi-Tian Gao, Xiao-Yong Wen, Analytic N-solitary-wave solution of a variable-coefficient Gardner equation from fluid dynamics and plasma physics, Appl. Math. Comput. 210 (2) (2009) 313–320. [28] M. El-Zahar, C.M. Pareek, Domination number of products of graphs, Ars Combinatoria 31 (1991) 223–227. [29] Gholam-Ali Zakeri, Analysis of ray tube area in geometrical shock dynamics, Appl. Math. Comput. 217 (2) (2010) 830–836.
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
The domination numbers of cylindrical grid graphs Mrinal Nandi a, Subrata Parui b, Avishek Adhikari c,⇑ a b c
Department of Statistics, West Bengal State University, Barasat, North 24-Parganas, India IMBIC, AH 317, Salt Lake City, Calcutta 700091, India Department of Pure Mathematics, University of Calcutta, 35 Ballygunge Circular Road, Calcutta 700019, India
a r t i c l e
i n f o
Keywords: Minimum dominating set Grid graph Cycle Path and Cartesian product of graphs
a b s t r a c t Let c(Pm h Cn) denote the domination number of the cylindrical grid graph formed by the Cartesian product of the graphs Pm, the path of length m, m P 2 and the graph Cn, the cycle of length n, n P 3. In this paper, methods to find the domination numbers of graphs of the form Pm h Cn with n P 3 and m = 2, 3 and 4 are proposed. Moreover, bounds on domination numbers of the graphs P5 h Cn, n P 3 are found. The methods that are used to prove that results readily lead to algorithms for finding minimum dominating sets of the above mentioned graphs. Ó 2010 Elsevier Inc. All rights reserved.
1. Introduction Graph theory, as a mathematical discipline, was created long back by Euler in his famous discussion of the Konigsberg Bridge Problem (1736). But in the current scenario, along with the remarkable developments in various areas of Applied Mathematics such as Fluid Dynamics [27,29], Game theory [8,1], Theory of solitons [2–4] etc., the field of graph theory and its applications is also a rapidly expanding area of mathematics. In this progress, demands come from fields of applications like game theory and programming [8,1], communications theory [7], electrical networks and switching circuits [24] as well as problems from biology [20,19] and psychology [21]. The problem of domination is one of the most widely studied topics in graph theory: the 1998 book by Haynes et al. [17] contains a bibliography with over 1200 papers on the subject. The domination problem was studied from the 1950s onwards, but the rate of research on domination significantly increased in the mid-1970s. Perhaps the fastest growing area within graph theory is the study of domination in graphs. The signed domination number of a graph G was defined in [9] and has been studied by several authors including [9,13]. Independent dominating sets were introduced into the theory of games by Morgenstern in [25]. For an extensive survey of domination problems and comprehensive bibliography the readers are refereed to [18]. The study of domination numbers of products of graphs was initiated by Vizing [26]. He conjectured that the domination number of the cartesian product of two graphs is always greater than or equal to the product of the domination numbers of the two factors; a conjecture which is still unproven. In [14], a link is shown between the existence of tilings in Manhattan metric with {1}-bowls and minimum total dominating sets of cartesian products of paths and cycles. Domination numbers of cartesian products were intensively investigated (see e.g., [28,12,14]). The graphs, considered here, are finite, nonempty, connected, undirected, without loops and without multiple edges. Besides these, any undefined terms in this paper may be found in Harary [15].
⇑ Corresponding author. E-mail addresses: [email protected] (M. Nandi), [email protected] (S. Parui), [email protected] (A. Adhikari). 0096-3003/$ - see front matter Ó 2010 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2010.11.019
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Let G be a simple graph whose vertex set and edge set are V(G) and E(G), respectively. A set D # V(G) of a simple graph G is called a dominating set if every vertex v 2 V(G)nD is adjacent to some vertex u 2 D. The domination number of G is the cardinality of a smallest dominating set of the graph G and it is usually denoted by c(G). In addition, smallest such dominating set is called a minimum dominating set of G. For any two graphs G and H, the Cartesian product G h H is the graph with vertex set V(G) V(H) and with edge set E(G H) such that (u1, v1) (u2, v2) 2 E(G H), whenever v1 = v2 and u1u2 2 E(G), or u1 = u2 and v1v2 2 E(H) [23]. Throughout the paper, the following notation and terminology are used. The numbers 0, 1, 2, . . . , n 1 always denote the vertices of the path Pn or the cycle Cn. Also, let c(Pm h Pn) and c(Pm h Cn) denote the domination numbers of Cartesian product graphs Pm h Pn and Pm h Cn, respectively. The graph Pm h Cn can be termed as cylindrical grid graph as shown in Fig. 1 in two different looks. Let (G)v = G {v} where v 2 V(H) and (H)u = {u} H where u 2 V(G). (G)v and (H)u are called the layers of G and H, respectively. Moreover, layer of a dominating set means D \ (Pm)i for i 2 V(Cn). Throughout the paper, the leftmost column in all figures denotes the layer (Pm)0. Now, define the term modified concatenation, of two dominating sets of Pm C n1 and Pm C n2 . If D1 and D2 are two dominating sets of Pm C n1 and P m C n2 , respectively then the modified concatenation of D1 and D2, denoted by D1kD2, is a subset D of P m C n1 þn2 such that D \ (Pm)i = D1 \ (Pm)i, i = 0, 1, . . . , n1 1 and D \ ðPm Þn1 þi ¼ D2 \ ðPm Þi , i = 0, 1, . . . , n2 1, i.e., the ith (Pm)-layer of D is coming from the ith (Pm)-layer of D1 if 0 6 i 6 n1 1 and from the i n1th (Pm)-layer of D2 if n1 6 i 6 n1 + n2 1. The illustration is shown in the Fig. 2. One of the most challenging problems concerning the domination numbers of Cartesian products of graphs is the proof of the Vizing Conjecture, namely c(G h H) P c(G) c(H) [26]. Despite numerous results showing its validity in some special cases, till date the conjecture remains an open problem. Partial works have been made towards finding the domination numbers of some particular Cartesian product of graphs. This problem also seems to be difficult one and the authors of [6] proved that even for subgraphs of Pm h Pn, this problem is NP-complete. In [22], Jacobson and Kinch established the following results: For all n P 1, cðP2 Pn Þ ¼
nþ2
. 2 3nþ4 . 4 n þ 1; cðP4 Pn Þ ¼ n; cðP3 Pn Þ ¼
for n ¼ 1; 2; 3; 5; 6; 9; otherwise:
Fig. 1. Two different looks of the cylindrical grid graph P5 h C5.
Fig. 2. Modified concatenation of the dominating sets of P5 h C6 and P5 h C3 to get the dominating set for P5 h C9.
M. Nandi et al. / Applied Mathematics and Computation 217 (2011) 4879–4889
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In [5], Chang and Clark established the following results: 6nþ6 ; for n ¼ 2; 3; 7; 5 cðP 5 Pn Þ ¼ 6nþ8 ; otherwise: 5 In [23], the authors established the following results regarding the Cartesian product of two cycles: For n P 4, cðC 3 C n Þ ¼ n n4 . For n P 4, c(C4 h Cn) = n. n; For n P 5, cðC 5 C n Þ ¼ n þ 1;
n ¼ 5k; k P 1; n 2 f5k þ 1; 5k þ 2; 5k þ 4g;
k P 1:
Furthermore, c(C5 h C5k+3) 6 5(k + 1), k P 1. More works may be found in [10–12,16]. In this paper, the topic of interest is to find the domination numbers of cylindrical grid graphs Pm h Cn, m P 2, n P 3. Towards finding the answer, some partial results in this direction are obtained. The domination numbers as well as minimum dominating sets of the graphs Pm h Cn, for m = 2, 3, 4 and n P 3 are found. Bounds on c(Pm h Cn) for m = 5 and n P 3 are also proposed. As a brief summery, results that are proved in the subsequent sections are stated as follows. For all n P 3, nþ1 ; when n is not a multiple of 4; cðP 2 C n Þ ¼ n 2 ; when n is a multiple of 4: 2 3n cðP 3 C n Þ ¼ 4 . n þ 1; for n ¼ 3; 5; 9; cðP 4 C n Þ ¼ n; otherwise: c(P5 h C3) = 4, c(P5 h C4) = 5 and c(P5 h C5) = 7. Moreover, for n P 6, n þ 5n 6 cðP 5 C n Þ 6 n þ n4 . Throughout the paper, we use the arithmetic operations of the indices over modulo n. 2. Finding the domination numbers of some cylindrical grid graphs In this section, the domination numbers as well as minimum dominating sets of particular cylindrical grid graphs of the form Pm h Cn, for all n P 3 and for m = 2, 3 and 4 are found. To prove the results, the following Lemmas are required. Lemma 2.1. Let m P 2. Then, there exists a minimum dominating set D of Pm h Cn such that for every i 2 V(Cn), j(Pm)i \ Dj 6 m 1. Proof. Let D0 be a minimum dominating set of Pm h Cn. Suppose further that, j(Pm)i \ D0 j = m holds for k many Pm-layers (Pm)i, i.e., for k-many i’s in {0, 1, . . . , n 1}, 0 6 k 6 n 1. Let us assume that j(Pm)i \ D0 j = m for some i 2 {0, 1, 2, . . . , n 1}. In addition, if (Pm)i1 \ D0 = (Pm)i+1 \ D0 = /, then we consider D = (D0 [ {(0, i 1), (1, i + 1)})n{(0, i), (1, i)}. Next, we assume that the layer (Pm)i+1 has nonempty intersection with D0 and let (j, i + 1) 2 (Pm)i+1 \ D0 . Then, it is clear that (j, i 1) and at least one of (j + 1, i 1) and (j 1, i 1) R (Pm)i1 \ D0 , otherwise, D0 n{(j, i)} would be a dominating set, contradicting the minimality of D0 and hence j(Pm)i1 \ D0 j < m 1. Then, we consider D = (D0 [ {(j, i 1)})n{(j,i)}. Now, D is a minimum dominating set with k 1 many Pm-layers having m vertices in common with D. Repeating this construction we get the result. h Lemma 2.2. Let n P 3. Then, there exists a minimum dominating set D of Pm h Cn such that for every i 2 V(Pm), jD \ (Cn)ij 6 n 1. Proof. The proof is similar as Lemma 2.1.
h
Lemma 2.3. There cannot be two consecutive Pm-layers having empty intersection with a minimum dominating set of Pm h Cn, for m P 3 and n P 4. Proof. If possible, let there be a minimum dominating set D having empty intersection with two consecutive layers (Pm)i and (Pm)i+1 for some i 2 {0, 1, 2, . . . , n 1}. Define (Pm)1 = (Pm)n1, (Pm)2 = (Pm)n2 and (Pm)n = (Pm)0, (Pm)n+1 = (Pm)1. Thus, j(Pm)i \ Dj = j(Pm)i+1 \ Dj = 0 gives j(Pm)i1 \ Dj = j(Pm)i+2 \ Dj = m. Now, consider the following cases: Case 1. For n = 4, 5, we set D0 = D [ {(0, i)}n{(0, i 1), (1, i 1)}. Case 2. For n P 6, we set D0 = (D [ {(0, i), (1, i 2), (1, i + 3), (2, i)})n{(0, i 1), (0, i + 2), (1, i 1), (2, i 1), (2, i + 2)}. In each case we find that D0 is a dominating set and jD0 j < jDj, which is a contradiction. Therefore, we conclude that there cannot have two consecutive layers (Pm)i, (Pm)i+1 having empty intersection with a minimum dominating set. h
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Lemma 2.4. For every dominating set D of Pm h Cn the following m inequalities hold:
xi1 þ 3xi þ xiþ1 P n;
8i ¼ 0; . . . ; m 1;
where xi = j(Cn)i \ Dj for i = 0, 1, . . . , m 1 and x1 = xm = 0. Moreover, if xi1 + 3xi + xi+1 = n, then there does not exist any pair Pm1 of vertices from ((Cn)i1 [ (Cn)i [ (Cn)i+1) \ D such that they dominate a common vertex of (Cn)i. Finally, i¼0 xi ¼ jDj P cðPm C n Þ. Proof. The result follows from the fact that any vertex of D from (Cn)i dominates three vertices of (Cn)i including itself and any vertex of D from (Cn)i1 or (Cn)i+1 dominates one vertex of (Cn)i. h Remark 2.5. The similar result of Lemma 2.4 holds for yi = j(Pm)i \ Dj, where i 2 {0, 1, . . . , n 1}. Using the above Lemmas we now prove the following theorems. nþ1 ; when n is not a multiple of 4; Theorem 2.6. For n P 3, cðP2 C n Þ ¼ n 2 when n is a multiple of 4: 2; Proof. We consider a set D with
D¼
fð0; iÞ : i 2ðmod 4Þg [ fð1; iÞ : i 0ðmod 4Þg [ fð0; n 1Þg; if n 2ðmod 4Þ; fð0; iÞ : i 2ðmod 4Þg [ fð1; iÞ : i 0ðmod 4Þg;
otherwise:
nþ1
Then, D is a dominating set of P2 h Cn and that jDj ¼ 2 , whenever n is not a multiple of 4 and jDj ¼ 2n, whenever n is a , or c(P2 h Cn) 6 n/2 for different values of n. multiple of 4. Hence, cðP2 C n Þ 6 nþ1 2 n To show that cðP2 C n Þ P 2 , let D0 be a minimum dominating set. Let j(Cn)i \ D0 j = xi, for i = 0, 1, i.e., xi is the number of vertices of D0 from the layer (Cn)i. Then, using Lemma 2.4 we have
3x0 þ x1 P n;
ðiÞ
x0 þ 3x1 P n;
ðiiÞ
x0 þ x1 ¼ cðP2 C n Þ:
ðiiiÞ
Therefore, 4(x0 + x1) P 2n which gives x0 þ x1 P n2. nþ1 ; if n is odd; Consequently, cðP2 C n Þ P n 2 if n is even: 2; Now, let n = 4k + 2, k P 1. We will show that cðP 2 C n Þ P 2n þ 1, i.e., c(P2 h Cn) P 2k + 2. If possible, let cðP2 C n Þ 6 n2 ¼ 2k þ 1. Then, x0 + x1 6 2k + 1. Therefore, one of x0 or x1 must be less than or equals k. Without loss of generality,
let
x0 6 k.
Then,
3x0 + x1 6 3x0 + (2k + 1 x0) = 2x0 + 2k + 1 6 4k + 1 < n,
cðP2 C n Þ P 2n þ 1. h Remark 2.7. For n P 3, cðP 2 C n Þ P cðP 2 Þ cðC n Þ ¼ 1 . Theorem 2.8. For n P 3, cðP3 C n Þ ¼ 3n 4
which
contradicts
(i).
Hence,
n , the equality holds only when n = 4. 3
Proof. Consider a set D with
D¼
fð0; iÞ; ð2; iÞ : i 2ðmod 4Þg [ fð1; iÞ : i 0ðmod 4Þg [ fð1; n 1Þg; if n 2ðmod 4Þ; fð0; iÞ; ð2; iÞ : i 2ðmod 4Þg [ fð1; iÞ : i 0ðmod 4Þg;
otherwise:
. For n = 3 and n = 4, it is illustrated in Fig. 3. Therefore, Then, D is a dominating set of P3 h Cn and that jDj ¼ 3n 4 cðP3 C n Þ 6 3n4 . Next, we show that cðP3 C n Þ P 3n 4 . Let n = 4k + t, where k P 0, 3 P t P 0 and let D be a minimum dominating set satisfying the property as stated in Lemma 2.1. Let s be the number of P3-layers which have empty intersection with D. Then, by Lemma 2.3, since no two empty layers are adjacent, we have s 6 2n ¼ 2k þ 2t . Now, as every empty P3-layer is s dominated by exactly two other P3-layers, there are at least 2 P3-layers with precisely two vertices from D. Hence,
2kþb2t c ¼ is maximum when s ¼ 2k þ 2t . So, jDj P n 2k þ 2t jDj P 2 2s þ n 2s s ¼ n s 2s . Also, s 2s 2
t t 2kþb2c bc ¼ 3k þ t 2t þ 22 ¼ 3k þ t (since 0 6 t 6 3Þ ¼ 3n ð4k þ tÞ 2k þ 2t 4 . h 2
Fig. 3. Vertices with bold circle form a minimum dominating set for P3 h C3 and P3 h C4.
M. Nandi et al. / Applied Mathematics and Computation 217 (2011) 4879–4889
Remark 2.9. For n P 3, cðP 3 C n Þ > cðP 3 Þ cðC n Þ ¼ 1
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n . 3
Theorem 2.10. c(P4 h Cn) = n + 1, for n = 3, 5, 9. Proof. The proof of the theorem for three different values of n are given successively. n = 3. Consider the set D1 = {(0, 2), (1, 0), (2, 2), (3, 1)}. Then, D1 is a dominating set for P4 h C3 as shown in Fig. 4. Since jD1j = 4, c(P4 h C3) 6 4. Now, if possible, let there exist a dominating set D01 such that jD01 j 6 3. Let jðC 3 Þi \ D01 j ¼ xi , where xi is the number of vertices of D01 from the layer (C3)i, for i = 0, 1, 2, 3. Therefore, 3 X
xi 6 3;
ðiÞ
3x0 þ x1 P 3;
ðiiÞ
x0 þ 3x1 þ x2 P 3;
ðiiiÞ
x1 þ 3x2 þ x3 P 3; x2 þ 3x3 P 3:
ðivÞ ðvÞ
i¼0
Also,
Now, adding (ii) and (v) and subtracting (i) we get
x0 þ x3 P 2
ðviÞ
which gives
x1 þ x2 6 1:
ðviiÞ
Therefore, at least one of x1 and x2 must be zero. Without loss of generality, let x1 = 0. Then, from (iii) we have
x0 þ x2 P 3
ðviiiÞ
and hence from (i) x3 = 0. Therefore, from (v) we get x2 = 3 and from (i) x0 = 0 which contradicts (ii). So, we have jD0 j P 4 and therefore c(P4 h C3) = 4. n = 5. Consider the set D2 = {(0, 2), (0, 4), (1, 0), (2, 3), (3, 1), (3, 4)}. Then, D2 is a dominating set of P4 h C5 with jD2j = 6 as shown in Fig. 4. Therefore, c(P4 h C5) 6 6. We now show that, c(P4 h C5) P 6. If possible, let there exist a dominating set D02
0
such that D2 6 5. Let ðC 5 Þi \ D02 ¼ xi , where xi is the number of vertices of D02 from the layer (C5)i for i = 0, 1, 2, 3. Therefore, 5 X
xi 6 5:
ðiÞ
i¼0
Also, from Lemma 2.4 we have
3x0 þ x1 P 5;
ðiiÞ
x0 þ 3x1 þ x2 P 5;
ðiiiÞ
x1 þ 3x2 þ x3 P 5;
ðivÞ
x2 þ 3x3 P 5:
ðvÞ
Now, adding (ii) and (v) and subtracting (i) we get
x0 þ x3 P 3
ðviÞ
which implies
x1 þ x2 6 2:
ðviiÞ
Now, we claim that x1 – 0. Otherwise, if x1 = 0, then from (iii) we get x0 + x2 P 5. Therefore, x0 + x2 = 5 and from (i) we have x3 = 0. Therefore, from (v) we get x2 P 5, i.e., x2 = 5. This gives x0 = 0, which contradicts (ii). Therefore, we have x1 – 0. Similarly, x2 – 0. So, we have from (vii), x1 = x2 = 1. For these values of x1 and x2 we get from (v), x3 P 2 and from (ii), x0 P 2,
contradicting (i). Therefore, we cannot have D02 6 5. So, D02 P 6 and hence c(P4 h C5) = 6.
Fig. 4. Vertices with bold circles form a minimum dominating set for P4 h C3, P4 h C5 and P4 h C9.
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n = 9. Consider the set D3 = {(0, 2), (0, 4), (0, 8), (1, 0), (1, 6), (2, 3), (2, 8), (3, 1), (3, 5), (3, 7)}. Then D3 is a dominating set of P4 h C9 with jD3j = 10 as shown in Fig. 4. Therefore, c(P4 h C9) 6 10. Now, if possible, let there exists a dominating set D03 such
0
that D3 6 9 and let ðC 9 Þi \ D03 ¼ xi , where xi is the number of vertices of D03 from the layer (C9)i for i = 0, 1, 2, 3. Therefore, 9 X
xi 6 9:
ðiÞ
i¼0
Also, from Lemma 2.4 we have
3x0 þ x1 P 9;
ðiiÞ
x0 þ 3x1 þ x2 P 9;
ðiiiÞ
x1 þ 3x2 þ x3 P 9;
ðivÞ
x2 þ 3x3 P 9:
ðvÞ
Now, adding (ii) and (v) and subtracting (i) we get
x0 þ x3 P 5
ðviÞ
which gives
x1 þ x2 6 4:
ðviiÞ
Now, we claim that x1 – 0. If x1 = 0 then from (iii) we have x0 + x2 P 9 and hence from (i), x3 = 0. Therefore, from (v), x2 = 9 and from (i), x0 = 0 which contradicts (ii). Therefore, x1 – 0. Similarly, we can show that x2 – 0. Again, we claim that, either x1 – 2 or x2 – 2. If possible, let x1 = x2 = 2. Then, from (ii) and (v), we get x0 P 3 and x3 P 3, contradicting (i). Therefore, either x1 – 2 or x2 – 2. Now, from (vii), we get one of x1 or x2 must be 1. Without loss of generality, let x1 = 1. Therefore, from (iii) we get x0 + x2 P 6 and from (i) we get x1 + x3 6 3. This gives x3 6 2. We now claim that x3 – 0. If possible, let x3 = 0, then from (v) we get x2 P 9 which contradicts (i). So, x3 – 0. Next, we show that x3 – 1. If possible, let x3 = 1. Then, from (v) we get x2 P 6. Therefore, from (i), x0 6 1 which contradicts (ii). Consequently, x3 – 1 and hence x3 = 2. This implies x0 + x2 = 6. Now, from (v), x2 P 3 and from (ii), x0 P 3. Therefore, x0 = 3, x1 = 1, x2 = 3, and x3 = 2.
Lastly, we show that x0 = 3, x1 = 1, x2 = 3, and x3 = 2 cannot give a dominating set D03 with D03 ¼ 9. Without loss of 0 generality, let us assume ð1; 0Þ 2 D3 . Then, ð1; iÞ R D03 for all i = 1, 2, . . . , 8. Again, ð0; 1Þ R D03 because if ð0; 1Þ 2 D03 then (1, 1) is dominated by (0, 1) and (1, 0) which will be a contradiction by the Lemma 2.4. Similarly, ð0; 0Þ; ð0; 8Þ R D03 . Therefore, to dominate (0, 1) and (0, 8) we must take ð0; 2Þ; ð0; 7Þ 2 D03 . Till now, the vertices (0, 4) and (0, 5) in the layer (C9)0 are not dominated by any vertices of D03 . Therefore, either ð0; 4Þ 2 D03 or ð0; 5Þ 2 D03 . Without loss of generality, let ð0; 4Þ 2 D03 . Again, the vertices (1, 3), (1, 5), (1, 6) in the layer (C9)1 are not dominated by any vertices of D03 . To dominate these vertices we must take the vertices (2, 3), (2, 5), (2, 6) in D03 . In a similar way, to dominate the vertices (2, 1), (2, 8) of the layer (C9)2 we must take the vertices (3, 1), (3, 8) in D03 . Still (3, 4) is not dominated by any vertex of D03 as shown in Fig. 5. Hence, D03 can not be a dominating set. This contradiction shows that c(P4 h C9) P 10. h Theorem 2.11. For n P 3, c(P4 h Cn) = n, for n – 3, 5, 9. Proof. The three diagrams (i), (ii) and (iii) shown in Fig. 6 give dominating sets for P4 h C4, P4 h C6 and P4 h C7, respectively. Note that, a dominating set for P4 h C8 can be obtained by the modified concatenation of the two copies of the diagram (i). A dominating set for P4 h C10 can be constructed by the modified concatenation of the diagrams (i) and (ii), a dominating set for P4 h C11 can be constructed by the modified concatenation of the diagrams (i) and (iii), a dominating set for P4 h C12 can be constructed by the modified concatenation of the two copies of the diagram (ii) and a dominating set for P4 h C13 can be constructed by modified concatenation of the diagrams (ii) and (iii). Finally, the dominating set for n P 14 can be constructed by modified concatenation of the diagram for n = 4 with the diagrams for n = 10, 11, 12, 13. Hence, we have c(P4 h Cn) 6 n.
Fig. 5. Non admissibility of c(P4 h C9) = 9.
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Fig. 6. Vertices with bold circle form a minimum dominating set for P4 h C4, P4 h C6 and P4 h C7.
It now remains to show that c(P4 h Cn) P n. To prove this, consider a minimum dominating set D which has no empty (P4)i layer. Then, jDj P n and hence the theorem follows. Now, consider a minimum dominating set D0 with s empty (P4)i layers. Further, let there be t nonempty (P4)i layers each adjacent to exactly one empty (P4)i layer. Then, there are k ¼ 2st nonempty (P4)i layers each adjacent to two empty layers. Since there do not exist two consecutive 2 empty (P4)i layers, the number of nonempty (P4)i layers which have no adjacent empty layer is p ¼ n t 2st s¼ 2 n 2s 2t . Now, let x01 ; x02 ; . . . ; x0t be the numbers of elements of D0 which are contained in t nonempty (P4)i layers, respectively such that each of these (P4)i layers is adjacent with exactly one empty (P4)i layer. Further, we define x001 ; x002 ; . . . ; x00k and x01 ; x02 ; . . . ; x0p in an analogous way. Now, by Remark 2.5, 2 x001 þ x002 þ þ x00k þ x01 þ x02 þ þ x0t P 4s. 00 00 00 0 0 0 Therefore, 2 x 1 þ x2 þ þ xk þ x1 þ x2 þ þ xt P 4s þ t. This implies 2 x01 þ x02 þ þ x0p þ x001 þ x002 þ þ x00k þ x01 þ x02 þ þ x0t P 4s þ t þ 2p ¼ 4s þ t þ 2n 4s t ¼ 2n. Consequently, x1 + x2 + + xn P n where xi = j(P4)i \ D0 j. h Remark 2.12. For n P 3, cðP 4 C n Þ > cðP 4 Þ cðC n Þ ¼ 2 3n . 3. Bounds on the domination numbers of P5 h Cn, n > 3 In this section, the exact domination numbers are given for P5 h C3, P5 h C4 and P5 h C5. Furthermore, bounds on the domination numbers are also proposed for P5 h Cn, n P 6. The following Lemmas are essential for proving the subsequent theorems. Lemma 3.1. For n P 3, there exists a minimum dominating set D of P5 h Cn such that for every i 2 V(Cn), j(P5)i \ Dj 6 3. Proof. Let D be a minimum dominating set of P5 h Cn such that for every i 2 V(Cn), j(P5)i \ Dj 6 4. Such a D exists by Lemma 2.1. Suppose further that j(P5)i \ Dj = 4 holds for k many P5-layers. We now construct a dominating set D0 with jD0 j = jDj such that only k 1 many P5-layers have 4 vertices in common with D. Assume that j(P5)i \ Dj = 4 for some i 2 V(Cn). We claim that j(P5)i1 \ Dj 6 2, and j(P5)i+1 \ Dj 6 2. If possible, let j(P5)i+1 \ Dj P 3. Then, D00 = (Dn((P5)i [ (P5)i+1)) [ {(2, i 1), (0, i), (4, i), (1, i + 1), (0, i + 2), (3, i + 2)} is a dominating set with jD00 j < jDj which contradicts the minimality of D. Therefore, j(P5)i+1 \ Dj 6 2. Similarly, j(P5)i1 \ Dj 6 2. Now, we set D0 = (Dn(P5)i) [ {(2, i 1), (0, i), (4, i), (2, i + 1)}. Then, D0 is a dominating set with jD0 j 6 jDj but since D is a minimum dominating set we have jD0 j = jDj. Also, j(P5)j \ D0 j < 4 for j = i 1, i, i + 1 and j(P5)j \ D0 j = j(P5)j \ Dj for other values of j. Repeating this process we get the desired minimum dominating set. h Lemma 3.2. For n P 5, there exists a minimum dominating set D of P5 h Cn such that for every i 2 V(Cn) either (a) j(P5)i \ Dj 6 2 or (b) j(P5)i \ Dj = 3 with (P5)i1 \ D = / and (P5)i+1 \ D = / for i 2 V0 and j(P5)i \ Dj 6 2 for all i R V0 , for some V0 # V(Cn). Proof. Lemma 3.1 shows there exists a minimum dominating set D such that j(P5)i \ Dj 6 3 for all i 2 V(Cn). Now, let j(P5)i \ Dj = 3 hold for k many P5-layers and among these k many P5-layers there exist k0 many P5-layers with the property that both the adjacent P5-layers of such a layer have empty intersection with D. Now, if k0 = k then D has the desired property. When k0 < k we construct a dominating set D0 with the property that D0 has k 1 or less P5-layers having three vertices in common with D and jD0 j = jDj. Let j(P5)i \ Dj = 3 for some i 2 V(Cn). We claim that, both of j(P5)i1 \ Dj and j(P5)i+1 \ Dj cannot be simultaneously equal to 2. On contrary, if possible, let j(P5)i1 \ Dj = j(P5)i+1 \ Dj = 2. Now, we consider the following sets
A1 ¼ fð1; i 2Þ; ð3; i 1Þ; ð4; i 1Þ; ð0; iÞ; ð2; i þ 1Þ; ð4; i þ 2Þg; A2 ¼ fð3; i 2Þ; ð0; i 1Þ; ð1; i 1Þ; ð4; iÞ; ð2; i þ 1Þ; ð0; i þ 2Þg; A3 ¼ fð2; i 2Þ; ð4; i 2Þ; ð0; i 1Þ; ð3; iÞ; ð1; i þ 1Þ; ð4; i þ 2Þg; A4 ¼ fð2; i 2Þ; ð0; i 2Þ; ð4; i 1Þ; ð1; iÞ; ð3; i þ 1Þ; ð0; i þ 2Þg and A5 ¼ fð2; i 2Þ; ð0; i 1Þ; ð4; i 1Þ; ð2; iÞ; ð0; i þ 1Þ; ð4; i þ 1Þg:
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Let Bkl = {(k, i + 1), (l, i + 1)}, where k, l = 0, 1, 2, 3, 4. Now, if (P5)i+1 \ D = B01 or B02 or B12 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A2, if (P5)i+1 \ D = B03 or B13 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A4, if (P5)i+1 \ D = B04 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A5, if (P5)i+1 \ D = B14 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A3, and if (P5)i+1 \ D = B23 or B24 or B34 then let D0 = (Dn{(P5)i1 [ (P5)i [ (P5)i+1}) [ A1. Then, in every case D0 will be a dominating set with jD0 j < jDj, which is a contradiction. Hence, the claim follows. Now, we consider the all other possible cases: Case 1. Let j(P5)i+1 \ Dj = 3. Then, we claim that j(P5)i+2 \ Dj < 2. To show this if possible, let j((P5)i [ (P5)i+1 [ (P5)i+2) \ Dj P 8. Then, D00 = (Dn{(P5)i [ (P5)i+1 [ (P5)i+2)} [ {(0, i 1), (4, i 1), (2, i), (0, i + 1), (3, i + 2), (4, i + 2), (1, i + 3)} is a dominating set with jD00 j < jDj, contradicting the minimality of D. Therefore, j(P5)i+2 \ Dj < 2. Similarly, j(P5)i1 \ Dj < 2. Now, we construct D0 = (Dn{(P5)i [ (P5)i+1}) [ {(2, i 1), (0, i), (4, i), (0, i + 1), (4, i + 1), (2, i + 2)}. Then jD0 j 6 jDj and D0 is a dominating set. Hence, D0 is a minimum dominating set having (k 2) many P5-layers, each of which has three vertices in common with D. Similar argument for j(P5)i1 \ Dj = 3. Case 2. Let j(P5)i1 \ Dj = 2 and j(P5)i+1 \ Dj 6 1. Then, we can construct D0 such that j(P5)i1 \ D0 j = 2, j(P5)i \ D0 j = 2 and j(P5)i+1 \ D0 j = j(P5)i+1 \ Dj + 1. Similarly, we can construct D0 for j(P5)i1 \ Dj 6 1 and j(P5)i+1 \ Dj = 2. Case 3. Let j(P5)i1 \ Dj = 1 and j(P5)i+1 \ Dj 6 1. subcase 3.1. When j(P5)i \ Dj = {(0, i), (1, i), (2, i)} or {(2, i), (3, i), (4, i)}, then construct D0 = (Dn{(0, i), (1, i)}) [ {(0, i 1), (1, i + 1)} for the first one. Similarly, we can construct D0 for the other one. subcase 3.2. Other than the above case we can construct D0 such that j(P5)i1 \ D0 j = 1, j(P5)i \ D0 j = 2, j(P5)i+1 \ D0 j = j(P5)i+1 \ Dj + 1. Similar construction will be made when j(P5)i+1 \ Dj = 1 and j(P5)i1 \ Dj 6 1. Then, D0 is a minimum dominating set having k 1 many or less P5-layers each of which has three vertices in common with D. Repeating this replacement, we will get the minimum dominating set with the desired property as stated in this Lemma. h Lemma 3.3. For n P 5, there cannot be a dominating set D with five consecutive P5-layers having exactly one vertex in common with D. Proof. On contrary, let there exists a dominating set D and there exists i 2 V(Cn) such that j(P5)j \ Dj = 1 for all j = i, i + 1, i + 2, i + 3, i + 4. Note that, (0, j) R D and (4, j) R D for all j = i + 1, i + 2, i + 3. Therefore, (1, i + 2) 2 D and (3, i + 2) 2 D, a contradiction. h Lemma 3.4. Let D be a minimum dominating set with the property as stated in Lemma 3.2 . Again, let (P5)i and (P5)j be two layers having two vertices in common with D and j(P5)i+1 \ Dj – 2, j(P5)i+2 \ Dj – 2, . . . , j(P5)j1 \ Dj – 2. Then either (a) j(P5)i+1 \ Dj = 0, j(P5)i+2 \ Dj = 3, j(P5)i+3 \ Dj = 0, j(P5)i+4 \ Dj = 3, . . . , j(P5)j1 \ Dj = 0, or (b) j i 6 5 and j(P5)l \ Dj = 1 for all l = (i + 1), (i + 2), . . . , (j 1). Proof. If j(P5)l \ Dj = 0, then j(P5)l1 \ Dj > 1. Otherwise, j(P5)l+1 \ Dj P 4. Hence, j(P5)l1 \ Dj = 2 or 3. Similarly, j(P5)l+1 \ Dj = 2 or 3. Therefore, using Lemmas 3.2 and 3.3 we get the desired result. h Theorem 3.5. For n P 6, cðP5 C n Þ P n þ n5 . Proof. Let D be a minimum dominating set with the property as stated in Lemma 3.2. Let us call the collection of P5-layers {(P5)k: k = i + 1, i + 2, . . . , j 1} as a block where i and j as in the above Lemma 3.4. Let x0, x1, x3 be the number of P5-layers having 0, 1, 3 vertices, respectively in common with D. Let x02 be the number of blocks in which every P5-layer has exactly one vertex in common with D and x002 be the number of blocks in which every P5-layer has either 0 or 3 vertices in common with D and x02 be the number of blocks where j = i + 1, i.e., when the block contains no P5-layer. Then, clearly, x2 ¼ x02 þ x02 þ x002 be the number of P5-layers having 2 vertices in common with D and we have therefore x3 P x002 ; x0 ¼ x002 þ x3 ; x1 6 4x02 and 6ðx þx þx0 þx0 þx00 þx Þ ¼ 3x3 þ 2x2 þ x1 6n ¼ 3x3 þ 2x02 þ 2x02 þ 2x002 þ x1 0 1 25 2 2 3 ¼ 15 ð9x3 þ x0 þ x1 þ x02 þ x02 þ x002 þ x3 ¼ n. Now, jDj 6n 5 5 4x0 þ 4x02 þ 4x002 x1 6x0 Þ ¼ 15 9x3 þ 4x02 þ 4x02 þ 4x002 x1 6x002 6x3 ¼ 15 3x3 2x002 þ 4x02 þ 4x02 x1 ¼ 15 x3 þ 2 x3 x002 þ 20 n 0 6n 4x2 x1 þ 4x2 g P 0. Therefore, we have jDj P 5 . Thus, cðP5 C n Þ P n þ 5 . Note 3.6 jDj ¼ 6n 5 þ
x3 þ2ðx3 x002 Þþð4x02 x1 Þþ4x02 . 5
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Theorem 3.7. (a) c(P5 h C3) = 4, (b) c(P5 h C4) = 5, and (c) c(P5 h C5) = 7. Proof. (a) Let D be a minimum dominating set of P5 h C3. Further, let j(Cn)i \ Dj = xi, for i = 0, 1, 2, 3, 4. Then, by Lemma 2.4 we have
3x0 þ x1 P 3;
ðiÞ
x0 þ 3x1 þ x2 P 3;
ðiiÞ
x1 þ 3x2 þ x3 P 3;
ðiiiÞ
x2 þ 3x3 þ x4 P 3;
ðivÞ
x3 þ 3x4 P 3:
ðvÞ
If possible, let
jDj ¼ x0 þ x1 þ x2 þ x3 þ x4 6 3:
ðviÞ
Now, x0 – 0. Otherwise, x1 = 3 would imply x2 = x3 = x4 = 0, contradicting (iv). Similarly, x4 – 0. Next, if possible, let x0 = 2 and x4 = 1. Then, we have x1 = x2 = x3 = 0, contradicting (ii). Again, taking x0 = 1 and x4 = 2 we arrived at the same contradiction. Now, if possible, let x0 = x4 = 1. Adding (ii), (iii), and (iv) we get x0 + 4x1 + 5x2 + 4x3 + x4 P 9. Therefore, 4x1 + 5x2 + 4x3 P 7. This implies 5x1 + 5x2 + 5x3 P 7 and hence x1 + x2 + x3 P 2, contradicting (vi). Consequently, jDj P 4 and therefore c(P5 h C3) P 4. Now, D = {(0, 1), (2, 0), (2, 2), (4, 1)} is a dominating set of P5 h C3 as shown in Fig. 7. Therefore, c(P5 h C3) 6 4 and hence c(P5 h C3) = 4. This completes the proof of (a). (b) Let D be a minimum dominating set of c(P5 h C4). Further, let j(Cn)i \ Dj = xi for i = 0, 1, 2, 3, 4. Then, by Lemma 2.4,
3x0 þ x1 P 4;
ðiÞ
x0 þ 3x1 þ x2 P 4;
ðiiÞ
x1 þ 3x2 þ x3 P 4; x2 þ 3x3 þ x4 P 4;
ðiiiÞ ðivÞ
x3 þ 3x4 P 4:
ðvÞ
If possible, let
jDj ¼ x0 þ x1 þ x2 þ x3 þ x4 6 4:
ðviÞ
Then, as in the proof of (a) we have x0 – 0 and x4 – 0. From (i) and (v) we get 3x0 + x1 + x3 + 3x4 P 8. Therefore, 2x0 + 2x4 P 4 + x2. Now, if x2 – 0 then x0 + x4 P 3. Hence, x1 + x2 + x3 6 1 which implies x2 = 1 and x1 = x3 = 0. Therefore, from (ii) and (iv) x0 + x2 P 4 and x2 + x4 P 4. Hence, x0 P 3 and x4 P 3, contradicting (vi). If x2 = 0 then from (iii), x1 + x3 P 4. Now, from (vi) we have x0 = x4 = 0. From (i) we have x1 P 4 and hence x3 = 0, contradicting (iv). Therefore, we have c(P5 h C4) P 5. Now, D = {(0, 2), (1, 0), (2, 3), (3, 1), (4, 3)} is a dominating set of P5 h C4 as shown in Fig. 8. Therefore, c(P5 h C4) 6 5 and hence c(P5 h C4) = 5. h (c) Let D be a minimum dominating set of c(P5 h C5). Further, let j(Cn)i \ Dj = xi where i = 0, 1, 2, 3, 4. Then, from Lemma 2.4 we have
3x0 þ x1 P 5;
ðiÞ
x0 þ 3x1 þ x2 P 5;
ðiiÞ
x1 þ 3x2 þ x3 P 5;
ðiiiÞ
x2 þ 3x3 þ x4 P 5;
ðivÞ
x3 þ 3x4 P 5:
ðvÞ
If possible, let
jDj ¼ x0 þ x1 þ x2 þ x3 þ x4 6 6:
ðviÞ
From (i) and (v) we get 3x0 + x1 + x3 + 3x4 P 10. Therefore, 2(x0 + x4) P 4 + x2. Now, if possible, let x2 = 0. Then, x0 + x4 P 2 and x1 + x3 P 5(by (iii)), contradicting (iv). Therefore, x2 P 1 and hence x0 + x4 P 3.
Fig. 7. Vertices with bold circle form a minimum dominating set for P5 h C3.
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Fig. 8. Vertices with bold circle form a minimum dominating set for P5 h C4.
Again, x0 – 0 and x4 – 0 as in (a). If possible, let x0 + x4 P 4. Then, x1 + x3 6 1 implies either x1 = 0 or x3 = 0. But this is a contradiction (since x1 = 0 implies x0 + x2 P 5 implies x0 = 4 implies x2 = 1 implies x3 P 2, contradicting (iv). Similar contradiction for x3 = 0). Therefore, x0 + x4 = 3, x2 P 1 and x1 + x3 6 2. Again, if x1 = x3 = 0 then the only possibility is x0 = 1, x1 = 0, x2 = 4, x3 = 0 and x4 = 1 which contradicts x0 + x4 = 3. If x1 = 1, x2 = 1,x3 = 1 and x0 + x4 = 3 then it will contradict either (i) or (v). Finally, if x1 = 1,x3 = 0 then x2 = 2 and x0 + x4 = 3, contradicting (i) or (v). Similar contradiction for x1 = 0,x3 = 1. Hence, we have c(P5 h C5) P 7. Now, D = {(0, 0), (0, 4), (1, 2), (2, 0), (3, 3), (3, 4), (4, 1)} is a dominating set of P5 h C5 as shown in Fig. 9. Therefore, c(P5 h C5) 6 7 and hence c(P5 h C5) = 7. h Lemma 3.8. 9 þ 95 6 cðP 5 C 9 Þ 6 9 þ 94 . Proof. The result follows from the Theorem 3.5 and the Fig. 10. h Theorem 3.9. For n P 6, n þ n5 6 cðP5 C n Þ 6 n þ n4 . Proof. For n = 9, the result follows from Lemma 3.8. For n P 6, the inequality in the left side is already being proved in Theorem 3.5. For the other part of the inequality, let us consider the minimum dominating sets for P4 h C6 and P4 h C7 as shown in Fig. 6. Now adding the vertices {(4, 3), (4, 5)} and {(4, 3), (4, 6)} to the dominating sets of P4 h C6 and P4 h C7, respectively, the dominating sets for P5 h C6 and P5 h C7 can be obtained as shown in Fig. 11. Finally, using the modified concatenation suitably as in the case of P4 h Cn, n P 3, n – 3, 5, 9, (Theorem 2.11), among these two dominating sets and the minimum dominating set of P5 h C4, as shown in Fig. 8, we get the dominating sets for P5 h Cn, n P 6, n – 9, with required cardinality. h
Fig. 9. Vertices with bold circle form a minimum dominating set for P5 h C5.
Fig. 10. Vertices with bold circle form a dominating set for P5 h C9.
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Fig. 11. Vertices with bold circle form a dominating set for P5 h C6 and P5 h C7.
4. Conclusion In this paper, we find the domination numbers of the graphs Pm h Cn, m = 2, 3 and 4, n P 3. Minimum dominating sets corresponding to the above mentioned graphs are also constructed. Finally, we give bounds on c(P5 h Cn), n P 3. As a future work, we would like to find the exact values of c(P5 h Cn), n P 3 as well as minimum dominating sets for P5 h Cn, n P 3. Moreover, we would also like to work on the bounds of c(Pm h Cn), m P 6 and n P 3. References [1] C.L. Barrett, C.M. Reidys, Elements of a theory of computer simulation I: sequential CA over random graphs, Appl. Math. Comput. 98 (2–3) (1999) 241– 259. [2] Anjan Biswas, Daniela Milovic, Optical solitons in a parabolic law media with fourth order dispersion, Appl. Math. Comput. 208 (1) (2009) 299–302. [3] Anjan Biswas, Arjuna Ranasinghe, 1-Soliton solution of Kadomtsev–Petviashvili equation with power law nonlinearity, Appl. Math. Comput. 214 (2) (2009) 645–647. [4] Anjan Biswas, Soliton-soliton interaction with dual-power law nonlinearity, Appl. Math. Comput. 198 (2) (2008) 605–612. [5] T.Y. Chang, W.E. Clark, The Domination numbers of the 5 n and 6 n grid graphs, J. Graph Theor. 17 (1993) 81–107. [6] B.N. Clark, C.J. Colbourn, D.S. Johnson, Unit disk graphs, Discrete Math. 86 (1990) 165–177. [7] Matthias Dehmer, Information processing in complex networks: graph entropy and information functionals, Appl. Math. Comput. 201 (1–2) (2008) 82– 94. [8] Matthias Dehmer, Frank Emmert-Streib, Jrgen Kilian, A similarity measure for graphs with low computational complexity, Appl. Math. Comput. 182 (1) (2006) 447–459. [9] J.E. Dunbar, S.T. Hedetniemi, M.A. Henning, P.J. Slater, Signed domination in graphs, graph theory, combinatorics, and applications, John Wiley & Sons, 1995. pp. 311–322. [10] M.H. El-Zahar, C.M. Pareek, Domination number of products of graphs, Ars Combin. 31 (1991) 223–227. [11] M.H. El-Zahar, S.M. Khamis, Kh. M. Nazzal, On the domination number of the Cartesian product of the cycle of length n and any graph, Discrete Appl. Math. 155 (2007) 515–522. [12] R.J. Faudree, R.H. Schelp, The domination number for the product of graphs, Congr. Numer. 79 (1990) 29–33. [13] O. Favaron, Signed domination in regular graphs, Discrete Math. 158 (1996) 287–293. [14] S. Gravier, Total domination number of grid graphs, Discrete Appl. Math. 121 (2002) 119–128. [15] F. Harary, Graph Theory, Addison Weslay, Reading Mass, 1969. [16] B.L. Hartnell, D.F. Rall, On Vizing’s conjecture, Congr. Numer. 82 (1991) 87–96. [17] T. Haynes, S. Hedetniemi, P. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, New York, 1998. [18] S.T. Hedetniemi, R.C. Laskar, Bibliography on domination in graphs and some basic definitions of domination parameters, Discrete Math. 86 (1–3) (1990) 257–277. [19] Sun-Yuan Hsieh, Chao-Wen Huang, Hsin-Hung Chou, A DNA-based graph encoding scheme with its applications to graph isomorphism problems, Appl. Math. Comput. 203 (2) (2008) 502–512. [20] Sun-Yuan Hsieh, Ming-Yu Chen, A DNA-based solution to the graph isomorphism problem using AdlemanLipton model with stickers, Appl. Math. Comput. 197 (2) (2008) 672–686. [21] Takehiro Inohara, Characterization of clusterability of signed graph in terms of Newcomb’s balance of sentiments, Appl. Math. Comput. 133 (1) (2002) 93–104. [22] M.S. Jacobson, L.F. Kinch, On the domination of the products of graphs I, Ars Combin. 18 (1983) 33–44. [23] S. Klavzar, N. Seifter, Dominating Cartesian products of cycles, Discrete Appl. Math. 59 (1995) 129–136. [24] W. Mayeda, Synthesis of switching functions by linear graph theory, IBM J. Res. Develop. 4 (3) (1960) 321–328. [25] O. Morgenstern, The collaboration between Oskar Morgenstern and John von Neumann on the theory of games, J. Eco. Literat. 14 (1976) 805–816. [26] V.G. Vizing, The Cartesian product of graphs, Vychisl Sistemy 9 (1963) 30–43. [27] Xiao-Ge Xu, Xiang-Hua Meng, Yi-Tian Gao, Xiao-Yong Wen, Analytic N-solitary-wave solution of a variable-coefficient Gardner equation from fluid dynamics and plasma physics, Appl. Math. Comput. 210 (2) (2009) 313–320. [28] M. El-Zahar, C.M. Pareek, Domination number of products of graphs, Ars Combinatoria 31 (1991) 223–227. [29] Gholam-Ali Zakeri, Analysis of ray tube area in geometrical shock dynamics, Appl. Math. Comput. 217 (2) (2010) 830–836.