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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

Domination in 4-regular Knödel graphs

Doost Ali Mojdeh / S.R. Musawi / E. Nazari
Published Online: 2018-08-03 | DOI: https://doi.org/10.1515/math-2018-0072

Abstract

A subset D of vertices of a graph G is a dominating set if for each uV(G) ∖ D, u is adjacent to some vertex vD. The domination number, γ(G) of G, is the minimum cardinality of a dominating set of G. For an even integer n ≥ 2 and 1 ≤ Δ ≤ ⌊log2 n⌋, a Knödel graph WΔ, n is a Δ-regular bipartite graph of even order n, with vertices (i, j), for i = 1, 2 and 0 ≤ jn/2 − 1, where for every j, 0 ≤ jn/2 − 1, there is an edge between vertex (1, j) and every vertex (2, (j+2k − 1) mod (n/2)), for k = 0, 1, ⋯, Δ − 1. In this paper, we determine the domination number in 4-regular Knödel graphs W4,n.

Keywords: Knödel graph; Domination number; Pigoenhole Principal

MSC 2010: 05C69; 05C30

1 Introduction

For graph theory notation and terminology not given here, we refer to [1]. Let G = (V, E) denote a simple graph of order n = |V(G)| and size m = |E(G)|. Two vertices u, vV(G) are adjacent if uvE(G). The open neighborhood of a vertex uV(G) is denoted by N(u) = {vV(G)|uvE(G)} and for a vertex set SV(G), N(S) = uSN(u). The cardinality of N(u) is called the degree of u and is denoted by deg(u), (or degG(u) to refer it to G). The closed neighborhood of a vertex uV(G) is denoted by N[u] = N(u) ∪ {u} and for a vertex set SV(G), N[S] = uSN[u]. The maximum degree and minimum degree among all vertices in G are denoted by Δ(G) and δ(G), respectively. A graph G is a bipartite graph if its vertex set can be partitioned to two disjoint sets X and Y such that each edge in E(G) connects a vertex in X with a vertex in Y. A set DV(G) is a dominating set if for each uV(G) ∖ D, u is adjacent to some vertex vD. The domination number, γ(G) of G, is the minimum cardinality of a dominating set of G. The concept of domination theory is a widely studied concept in graph theory and for a comprehensive study see, for example [1].

An interesting family of graphs namely Knödel graphs was introduced about 1975 [2]. On a one-page note, Walter Knödel introduced a special graph as a minimum gossip graph [2]. For an even integer n ≥ 2, the graph KGn is a regular bipartite graph with degree ⌊log2 n⌋. Each vertex 2j+1 is adjacent to vertices 2j+2r, where j = 0, 1, 2, ⋯, n/2 − 1 and r = 1, 2, ⋯, ⌊ log2 n⌋. The graphs KGn are called modified Knödel graphs in the literature. In 1995, Bermond et al. presented some methods for constructing new broadcast graphs. Their constructions are based on graph compounding operation. For example, the modified Knödel graph KG2n is the compound of KGn and K2 [3]. In 1997, Bermond et al. showed that the edges of the modified Knödel graph can be grouped into dimensions which are similar to the dimensions of hypercubes. In particular, routing, broadcasting and gossiping, can be done easily in modified Knödel graphs using these dimensions [4]. The general definition of generalized Knödel Graphs were introduced in 2001 [5]. Since then, they have been widely studied by some authors.

Fraigniaud and Peters formally defined the generalized family of Knödel graphs [5].

Definition 1.1

([5]). For an even integer n ≥ 2 and 1 ≤ Δ ≤ ⌊log2 n⌋, a Knödel graph WΔ, n is a Δ-regular bipartite graph of even order n, with vertices (i, j), for i = 1, 2 and 0 ≤ jn/2 − 1, where for every j, 0 ≤ jn/2 − 1, there is an edge between vertex (1, j) and each vertex (2, (j+2k − 1) mod (n/2)), for k = 0, 1, ⋯, Δ − 1.

Knödel graphs, WΔ, n, are one of the three important families of graphs that have nice properties in terms of broadcasting and gossiping. There exist many important papers presenting graph-theoretic and communication properties of the Knödel graphs, see for example [4, 6, 7, 8, 9]. It is worth noting that any Knödel graph is a Cayley graph and so it is a vertex-transitive graph [10].

Xueliang et al. [11] studied the domination number in 3-regular Knödel graphs W3,n. They obtained exact domination number for W3,n. Mojdeh et al. [14] determined the total domination number in 3-regular Knödel graphs W3,n. In this paper, we determine the domination number in 4-regular Knödel graphs W4,n. The following is useful.

Theorem 1.2

([12, 13]). For any graph G of order n with maximum degree Δ(G),n1+Δ(G)γ(G)nΔ(G).

We need also the following simple observation from number theory.

Observation 1.3

If a, b, c, d and x are positive integers such that xaxb = xcxd ≠ 0, then a = c and b = d.

2 Properties in the Knödel graphs

In this section we review some properties in the Knödel graphs that are proved in [14]. Mojdeh et al. considered a re-labeling on the vertices of a Knödel graph as follows: we label (1, i) by ui+1 for each i = 0, 1, &, n/2 − 1, and (2, j) by vj+1 for j = 0, 1, &, n/2 − 1. Let U={u1,u2,,un2} and V={v1,v2,,vn2}. From now on, the vertex set of each Knödel graph WΔ, n is UV such that U and V are the two partite sets of the graph. If S is a set of vertices of WΔ, n, then clearly, SU and SV partition S, |S| = |SU|+|SV|, N(SU) ⊆ V and N(SV) ⊆ U. Note that two vertices ui and vj are adjacent if and only if j ∈ {i+20 − 1, i+21 − 1, ⋯, i+2Δ−1 − 1}, where the addition is taken modulo n/2. For any subset {ui1, ui2, ⋯, uik} of U with 1 ≤ i1 < i2 < ⋯ < ikn2, by the indices ijs of the elements of A, we want to correspond a sequence to the set A.

Definition 2.1

For any subset A = {ui1, ui2, ⋯, uik} of U with 1 ≤ i1 < i2 < ⋯ < ikn2 we define a sequence n1, n2, ⋯, nk, called a cyclic sequence, where nj = ij+1ij for 1 ≤ jk − 1 and nk = n2+i1ik. For two vertices uij, uijA, we define index-distance of uij and uij by id(uij, uij) = min{|ijij|, n2 − |ijij|}.

Observation 2.2

Let A = {ui1, ui2, ⋯, uik} ⊆ U be a set such that 1 ≤ i1 < i2 < ⋯ < ikn2 and let n1, n2, ⋯, nk be the corresponding cyclic-sequence of A. Then,

  1. n1+n2+⋯+nk = n2.

  2. If uij, uijA, then id(uij, uij) equals sum of some consecutive elements of the cyclic-sequence of A and n2id(uij, uij) is the sum of the remaining elements of the cyclic-sequence. Furthermore, {id(uij, uij), n2id(uij, uij)} = {|ijij|, n2 − |ijij|}.

Proof

  1. By definition of cyclic-sequence, we have n1+n2+ ⋯ +nk = (i2i1)+(i3i2)+⋯+(ikik − 1)+( n2+i1ik) = n2 as desired.

  2. By vertex transitivity of Knödel graphs, without loss of generality, we assume that j′ = 1. Since 1 = j′ < jk, |ijij| = iji1 = (ijij−1)+(ij−1ij−2)+⋯+(i2i1). So, we have |iji1| = nj−1+nj−2+⋯+n1 and by (1), we have n2 − |iji1| = nj+nj+1+⋯+nk. This shows that both of |iji1| and n2 − |iji1| and consequently id(uij, ui1) are the sum of some consequent elements of cyclic-sequence. The proof of {id(uij, uij), n2id(uij, uij)} = {|ijij|, n2 − |ijij|} is straightforward. □

    Let ℳΔ = {2a − 2b : 0 ≤ b < a < Δ} for Δ ≥ 2.

Lemma 2.3

In the Knödel graph WΔ, n with vertex set UV, for every ij and 1 ≤ i, jn/2, N(ui) ∩ N(uj) ≠ ∅ if and only if id(ui, uj) ∈ ℳΔ or n2id(ui, uj) ∈ ℳΔ.

Proof

Without loss of generality, assume that i > j.

First, suppose that vtN(ui) ∩ N(uj). Therefore, there exist two integers 0 ≤ a, bΔ − 1 such that ti+2a − 1 ≡ j+2b − 1 (mod n/2). Hence, (ij) − (2b − 2a) ≡ 0 (mod n/2). Since 0 < ij < n/2 and − n/2 < 2b − 2a < n/2, we have only two different cases: (*) (ij) − (2b − 2a) = 0 and so |ij| = 2b − 2a ∈ ℳΔ and (**) (ij) − (2b − 2a) = n/2 and so n/2 − |ij| = 2a − 2b ∈ ℳΔ. Now, by Observation 2.2(2), we have id(ui, uj) ∈ ℳΔ or n2id(ui, uj) ∈ ℳΔ as desired.

To show the if part, suppose that id(ui, uj) ∈ ℳΔ or n2id(ui, uj) ∈ ℳΔ. By Observation 2.2(2), we have |ij| = 2a − 2b ∈ ℳΔ or n/2 − |ij| = 2a − 2b ∈ ℳΔ for some integers 0 ≤ b < aΔ − 1. In the first case, i+2b − 1 = j+2a − 1 and so vtN(ui) ∩ N(uj), where ti+2b − 1 = j+2a − 1 (mod n/2). In the second case, i+2a − 1 = n/2+j+2b − 1 and so vtN(ui) ∩ N(uj), where ti+2a − 1 = n/2+j+2b − 1 (mod n/2). In each case we have N(ui) ∩ N(uj) ≠ ∅ and proof is completed. □

3 4-regular Knödel graphs

In this section we determine the domination number in 4-regular Knödel graphs W4,n. Note that n ≥ 16 by the definition. For this purpose, we prove the following lemmas namely Lemma 3.1, 3.2, 3.3, 3.4 and 3.5.

Lemma 3.1

For each even integer n ≥ 16, we have γ(W4,n)=2n10+0n0 (mod 10)2n2,4 (mod 10).

Proof

First assume that n ≡ 0 (mod 10). Let n = 10t, where t ≥ 2. By Theorem 1.2, γ(W4,n) ≥ n5 = 2t. On the other hand, we can see that the set D = {u1, u6, ⋯, u5t−4} ∪ {v5, v10}, ⋯, v5t} is a dominating set with 2t elements, and we have γ (W4,n) = 2t = 2⌊ n10⌋, as desired.

Next assume that n ≡ 2 (mod 10). Let n = 10t+2, where t ≥ 2. By Theorem 1.2, we have γ (W4,n) ≥ n5 > 2t. Suppose that γ(W4,n) = 2t+1. Let D be a minimum dominating set of W4,n. Then by the Pigeonhole Principal either |DU| ≤ t or |DV| ≤ t. Without loss of generality, assume that |DU| ≤ t. Let |DU| = ta, where a ≥ 0. Then |DV| = t+1+a. Observe that DU dominates at most 4t − 4a vertices of V and therefore D dominates at most (4t − 4a) + (t+1+a) = 5t − 3a+1 vertices of V. Since D dominates all vertices of V, we have 5t − 3a+1 ≥ 5t+1 and so a = 0, |DU| = t and |DV| = t+1. Let DU = {ui1, ui2, ⋯, uit} and n1, n2, ⋯, nt be the cyclic-sequence of DU. By Observation 2.2, we have k=1tnk=5t+1 and, therefore, there exists some k such that nk ∈ ℳ4 = {1, 2, 3, 4, 6, 7}. Then by Lemma 2.3, |N(uk) ∩ N(uk+1) ≥ 1. Hence, DU dominates at most 4t − 1 vertices of V, that is, D dominates at most (4t − 1)+(t+1) = 5t vertices of V, a contradiction. Now we deduce that γ(W4,n) ≥ 2t+2. On the other hand the set D = {u1, u6, ⋯, u5t+1} ∪ {v5, v10}, ⋯, v5t} ∪ {v5t+1} is a dominating set for W4,n with 2t+2 elements. Consequently, γ(W4,n) = 2t+2.

It remains to assume that n ≡ 4 (mod 10). Let n = 10t+4, where t ≥ 2. By Theorem 1.2, we have γ (W4,n) ≥ n5 > 2t. Suppose that γ(W4,n) = 2t+1. Let D be a minimum dominating set of γ (W4,n). Then by the Pigeonhole Principal either |DU| ≤ t or |DV| ≤ t. Without loss of generality, assume that |DU| = ta and a ≥ 0. Then |DV| = t+1+a. Observe that DU dominates at most 4(ta) elements of V and therefore D dominates at most 4(ta)+(t+1+a) = 5t − 3a+1 vertices of V. Since D dominates all vertices of V, we have 5t − 3a+1 ≥ |V| = 5t+2 and so −3a ≥ 1, a contradiction. Thus γ (W4,n) > 2t+1. On the other hand, the set {u1, u6, ⋯, u5t+1} ∪ {v5, v10, ⋯, v5t} ∪ {v3} is a dominating set with 2t+2 elements. Consequently, γ (W4,n) = 2t+2. □

Lemma 3.2

For each even integer n ≥ 46 with n ≡6 (mod 10), we have γ (W4,n) = 2⌊ n10⌋+3.

Proof

Let n = 10t+6 and t ≥ 4. By Theorem 1.2, we have γ (W4,n) ≥ n5 > 2t+1. The set D = {u1, u6, ⋯, u5t+1} ∪ {v5, v10, ⋯, v5t} ∪ {v2, v3} is a dominating set with 2t+3 elements. Thus, 2t+2 ≤ γ (W4,n) ≤ 2t+3. We show that γ(W4,n) = 2t+3. Suppose to the contrary that γ(W4,n) = 2t+2. Let D be a minimum dominating set of W4,n. Then by the Pigeonhole Principal either |DU| ≤ t+1 or |DV| ≤ t+1. Without loss of generality, assume that |DU| = t+1 − a, where a ≥ 0. Then |DV| = t+1+a. Note that DU dominates at most 4t+4 − 4a vertices and therefore D dominates at most (4t+4 − 4a)+(t+1+a) = 5t − 3a+5 vertices of V. Since D dominates all vertices of V, we have 5t − 3a+5 ≥ 5t+3 and so a = 0 and |DU| = |DV| = t+1. Also we have |VD| = 4t+2 ≤ |N(DU)| ≤ 4t+4, and similarly, |DU| = 4t+2 ≤ |N(DV)| ≤ 4t+4.

Let DU = {ui1, ui2, ⋯, uit+1} and n1, n2, ⋯, nt+1 be the cyclic-sequence of DU. By Observation 2.2, we have k=1t+1nk = 5t+3 and, therefore, there exists k′ such that nk < 5. Then nk ∈ ℳ4 and by Lemma 2.3, |N(uik) ∩ N(uik′+1)| ≥ 1. Hence, DU dominates at most 4t+3 vertices from V and therefore 4t+2 ≤ N(DU) ≤ 4t+3.

If |N(DU)| = 4t+3, then for each kk′ we have nk ∉ ℳ4. If there exists k″ ≠ k′ such that nk ≥ 8, then 5t+3 = Σk=1t+1nk ≥ 1+8+5(t − 1) = 5t+4, a contradiction. By symmetry we have n1 = n2 = ⋯ = nt = 5, nt+1 = 3 and DU = {u1, u6, ⋯, u5t+1}. Observe that DU doesn’t dominate vertices v3, v5, v10, ⋯, v5t and so {v3, v5, v10, ⋯, v5t} ⊆ D. Thus D = {u1, u6, ⋯, u5t+1, v3, v5, v10, ⋯, v5t}. But the vertices u4, u5, u5t−2, u5t+2 are not dominated by D, a contradiction.

Thus, |N(DU)| = 4t + 2. Then there exists precisely two pairs of vertices in DU with index-distances belonging to ℳ4. If there exists an integer 1 ≤ i′ ≤ t+1 such that ni+ni′+1 ∈ ℳ4, then min{ni, ni′+1} ≤ 3. Then min{ni, ni′+1} ∈ ℳ4 and \max{ni, ni′+1} ∉ ℳ4. Now we have max{ni, ni′+1} = 5, min{ni, ni′+1} ∈ {1, 2}, and ni ∉ ℳ4, for each i∉{i′, i′+1}. Now a simple calculation shows that the equality 5t+3 = k=1t+1nk does not hold. (Note that if each ni is less than 8, then we have k=1t+1nk ≤ 2+5+5(t − 1) = 5t+2; otherwise we have k=1t+1nk ≥ 1+5+8+5(t − 2) = 5t+4.) Thus there exist exactly two indices j and k such that nj, nk ∈ ℳ4 and {n1+n2, n2+n3, ⋯, nt+nt+1, nt+1+n1} ∩ ℳ4 = ∅. By this hypothesis, the only possible cases for the cyclic-sequence of DU are those demonstrated in Table 1.

Table 1

n = 10t+6

Note that each column of Table 1 shows the cyclic sequence of DU. We show that each case is impossible. For this purpose, we show that the cyclic-sequence of DU posed in the column i, for i ≥ 1 is impossible.

i = 1). If n1 = 8, n2 = 1, n3 = 4, n4 = ⋯ = nt+1 = 5 and DU = {u1, u9, u10, u14, u19, ⋯, u5t − 1}, then DU does not dominate the vertices v5, v6, v7, v18, v23, v28, …, v5t+3.

Thus we have DV = {v5, v6, v7, v18, v23, v28, …, v5t+3}. But D does not dominate three vertices u8, u12, u13, a contradiction.

i = 2). If n1 = 4, n2 = 1, n3 = 8, n4 = ⋯ = nt+1 = 5 and DU = {u1, u5, u6, u14, u19, ⋯, u5t − 1}. Then DU does not dominate the vertices v10, v11, v16, v18, v23, v28, …, v5t+3. Thus we have DV = {v10, v11, v16, v18, v23, v28, …, v5t+3}. But D does not dominate three vertices u2, u12, u5t+1, a contradiction.

i ∈ {3, 4}). As before, we obtain that u8D. But N(u8) = {v8, v9, v11, v15} ⊆ N(DU) and therefore N(u8) ∩ D = ∅. Hence, N[u8] ∩ D = ∅ and D does not dominate u8, a contradiction.

i ≥ 5). As before, we obtain that u3D. But N(u3) = {v3, v4, v6, v10} ⊆ N(DU) and therefore N(u3) ∩ D = ∅. Hence, N[u3] ∩ D = ∅ and D does not dominate u3, a contradiction.

Consequently, γ(W4,n) = 2t+3, as desired. □

Lemma 3.2 determines the domination number of W4,n when n ≡ 6 (mod 10) and n ≥ 46. The only values of n for n ≡ 6 (mod 10) are thus 16, 26 and 36. We study these cases in the following lemma.

Lemma 3.3

For n ∈ {16, 26, 36}, we have:

Proof

For n = 16, by Theorem 1.2 we have γ (W4,16) ≥ 165 > 3. On the other hand, the set D = {u1, u2, v6, v7} is a dominating set for W4,16, and therefore γ (W4,16) = 4.

For n = 26, by Theorem 1.2 we have γ (W4,26) ≥ 265 > 5. On the other hand, the set D = {u1, u4, u9, u10, v1, v2, v6} is a dominating set for W4,26 and therefore 6 ≤ γ (W4,26) ≤ 7. We show that γ (W4,26) = 7. Suppose, on the contrary, that γ (W4,26) = 6. Let D be a minimum dominating set for W4,26. Then by the Pigeonhole Principal either |DU| ≤ 3 or |DV| ≤ 3. If |DU| = 3 − a, where a ≥ 0, then |DV| = 3+a. Now, the elements of DU dominate at most 4(3 − a) elements of V and D dominates at most 4(3 − a) + (3+a) = 15 − 3a vertices of V. Thus 15 − 3a ≥ |V| = 13, which implies a = 0 and |DU| = |DV| = 3. Let |DU| = {u1, ui, uj}, where 1 < i < j ≤ 13 and n1 = i − 1, n2 = ji, n3 = 13+1 − j. Since n1+n2+n3 = 13, we have {n1, n2, n3} ∩ ℳ4 ≠ ∅.

If ℳ4 includes at least three numbers of n1, n2, n3, n1+n2, n2+n3, n3+n1, then by Lemma 2.3, DU dominates at most 4 × 3 − 3 = 9 vertices of V and |DV| ≥ 13 − 9 = 4, a contradiction.

If ℳ4 includes exactly one number of n1, n2, n3, then we have, by symmetry, n1 = 3, n2 = 5, n3 = 5 and DU = {u1, u4, u9}, N(DU) = {v1, v2, v3, v4, v5, v7, v8, v9, v10, v11, v12}. Then {v6, v13} ⊆ D. But {u1, u4, u9, v6, v13} does not dominate the vertices {u2, u7, u8, u11} and we need at least 2 other vertices to dominate this four vertices, and hence |D| ≥ 7, a contradiction.

Thus, we assume that ℳ4 includes two numbers of n1, n2, n3 and {n1+n2, n2+n3, n3+n1} ∩ ℳ4 = ∅. We thus have five possibilities for the cyclic-sequence of DU that are demonstrated in Table 3. Note that each column of Table 3 shows the cyclic sequence of DU. We show that each case is impossible. For this purpose, we show that the cyclic sequence of DU posed in the column i, for i ≥ 1 is impossible.

Table 2

n = 16, 26, 36

Table 3

n = 26

i = 1) If n1 = 1, n2 = 4, n3 = 8, then DU = {u1, u2, u6} and N(DU) = {v1, v2, ⋯, v9, v13}. Thus {v10, v11, v12} ⊆ D and D = {u1, u2, u6, v10, v11, v12}. But D does not dominate the vertex u13, a contradiction.

i = 2) If n1 = 1, n2 = 8, n3 = 4, then DU = {u1, u2, u10} and N(DU) = {v1, v2, ⋯, v11, v13}. Thus {v6, v7, v12} ⊆ D and D = {u1, u2, u10, v6, v7, v12}. But D does not dominate the vertex u8, a contradiction.

i = 3) If n1 = 2, n2 = 3, n3 = 8, then DU = {u1, u3, u6}, N(DU) = {v1, v2, ⋯, v10, v13}.Thus {v5, v11, v12} ⊆ D and D = {u1, u3, u6, v5, v11, v12}. But D does not dominate the vertices u7 and u13.

i = 4) If n1 = 2, n2 = 8, n3 = 3, then DU = {u1, u3, u11} and N(DU) = {v1, v2, ⋯, v6, v8, v10, v11, v12}. Thus {v7, v9, v13} ⊆ D and D = {u1, u3, u11, v7, v9, v13}. But D does not dominate the vertex u5, a contradiction.

i = 5) If n1 = 4, n2 = 4, n3 = 5, then DU = {u1, u5, u9} and N(DU) = {v1, v2, ⋯, v10, v12}. Thus {v7, v11, v13} ⊆ D and D = {u1, u3, u11, v7, v9, v13}. But D does not dominate the vertices u2 and u3.

Consequently, γ(W4,26) = 7.

We now consider the case n = 36. By Theorem 1.2, γ (W4,36) ≥ 365 > 7. On the other hand, the set D = {u1, u2, u10, u11, v6, v7, v15, v16} is a dominating set for the graph W4,36 and, therefore, γ (W4,36) = 8. □

We now consider the case n ≡ 8 (mod 10). For n = 18, 28 and 38 we have the following lemma.

Lemma 3.4

For n ∈ {18, 28, 38}, we have:

Proof

For n = 18, by Theorem 1.2 we have γ (W4,18) ≥ 185 > 3. But the set D = {u1, u2, v6, v7} is a dominating set for W4,18 and, therefore, γ (W4,18) = 4.

For n = 28, by Theorem 1.2 we have γ (W4,28) ≥ 285 > 5 and the set D = {u1, u6, u11, u13, v3, v5, v9} is a dominating set for W4,28 and, therefore, 6 ≤ γ (W4,28) ≤ 7. Suppose on the contradiction that γ (W4,28) = 6 and D is a minimum dominating set for γ (W4,28). By the Pigeonhole Principal, either |DU| ≤ 3 or |DV| ≤ 3. If |DU| = 3 − a and a ≥ 0, then |DV| = 3+a. Now, the elements of DU dominate at most 4(3 − a) elements of V and D dominates at most 4(3 − a)+(3+a) = 15 − 3a vertices of V. Thus 15 − 3a ≥ |V| = 14. Therefore, a = 0 and so |DU| = |DV| = 3. Let |DU| = {u1, ui, uj} and 1 < i < j ≤ 14 and n1 = i − 1, n2 = ji, n3 = 14 + 1 − j. Since n1+n2+n3 = 14, we have {n1, n2, n3} ∩ ℳ4 ≠ ∅. If ℳ4 includes at least two numbers in {n1, n2, n3, n1+n2, n2+n3, n3+n1}, then DU dominates at most 4 × 3 − 2 = 10 vertices of V and |DV| ≥ 14 − 10 = 4, a contradiction.

The only remaining case is that ℳ4 includes exactly one of the three numbers n1, n2 and n3 and also {n1+n2, n2+n3, n3+n1} ∩ ℳ4 = ∅. By symmetry we have n1 = 4, n2 = 5, n3 = 5 and DU = {u1, u5, u10}, N(DU) = {v1, v2, v3, v4, v5, v6, v8, v10, v11, v12, v13} thus {v7, v9, v14} ⊆ D and D = {u1, u5, u10, v7, v9, v14}. But D does not dominate the vertex u3. That is a contradiction and therefore γ (W4,28) = 7. For n = 38, by Theorem 1.2 we have γ (W4,38) ≥ 385 > 7 and the set D = {u1, u6, u11, u16, u18, v3, v5, v10, v13, v15} is a dominating set for W4,38 and therefore 8 ≤ γ (W4,38) ≤ 10. Let γ (W4,38) < 10 and D is a dominating set for γ (W4,38) with |D| = 9. Then by the Pigeonhole Principal either |DU| ≤ 4 or |DV| ≤ 4. If |DU| = 4 − a and a ≥ 0, then |DV| = 5+a. Now, the elements of DU dominate at most 4(4 − a) elements of V and D dominates at most 4(4 − a)+(5+a) = 21 − 3a vertices of V. Thus 21 − 3a ≥ |V| = 19 that results a = 0 and we have |DU| = 4 and |DV| = 5. Let |DU| = {u1, ui, uj, uk}, where 1 < i < j < k ≤ 19 and n1, n2, n3, n4 be the cyclic-sequence of DU. Since n1+n2+n3+n4 = 19, we have {n1, n2, n3, n4} ∩ ℳ4 ≠ ∅. If ℳ4 includes at least three numbers of {n1, n2, n3, n4, n1+n2, n2+n3, n3+n4, n4+n1}, then DU dominates at most 4 × 4 − 3 = 13 vertices of V and |DV| ≥ 19 − 13 = 6, a contradiction.

If we wish that ℳ4 includes exactly one number out of n1, n2, n3, n4, we have three cases:

i = 1) If n1 = 4, n2 = 5, n3 = 5, n4 = 5 and DU = {u1, u5, u10, u15}, then {v7, v9, v14, v19} ⊆ D but {u1, u5, u10, u15, v7, v9, v14, v19} does not dominate the vertices u3 and u17. For dominating u3 and u17, we need two vertices and therefore |D| ≥ 10, a contradiction.

i = 2) If n1 = 1, n2 = 8, n3 = 5, n4 = 5 and DU = {u1, u2, u10, u15}, then {v6, v7, v12, v14, v19} ⊆ D and D = {u1, u2, u10, u15, v6, v7, v12, v14, v19} but D does not dominate the vertices u8 and u17.

i = 3) If n1 = 8, n2 = 1, n3 = 5, n4 = 5 and DU = {u1, u9, u10, u15}, then {v5, v6, v7, v14, v19} ⊆ D and D = {u1, u9, u10, u15, v5, v6, v7, v14, v19} but D does not dominate the vertex u8.

Now we consider the cases that ℳ4 includes exactly two numbers of the cyclic-sequence n1, n2, n3, n4 and {n1+n2, n2+n3, n3+n4, n4+n1} ∩ ℳ4 = ∅. By symmetry we have ten cases:

i = 1) If n1 = 1, n2 = 9, n3 = 1, n4 = 8 and DU = {u1, u2, u11, u12}, then {v6, v7, v10, v16, v17} ⊆ D and D = {u1, u2, u11, u12, v6, v7, v10, v16, v17} but D does not dominate the vertex u8.

i = 2) If n1 = 2, n2 = 8, n3 = 1, n4 = 8 and DU = {u1, u3, u11, u12}, then {v5, v7, v9, v16, v17} ⊆ D and D = {u1, u3, u11, u12, v5, v7, v9, v16, v17} but D does not dominate the vertex u18.

i = 3) If n1 = 4, n2 = 1, n3 = 9, n4 = 5 and DU = {u1, u5, u6, u15}, then {v10, v11, v14, v17, v19} ⊆ D and D = {u1, u5, u6, u15, v10, v11, v14, v17, v19} but D does not dominate the vertex u2.

i = 4) If n1 = 9, n2 = 1, n3 = 4, n4 = 5 and DU = {u1, u10, u11, u15}, then {v5, v6, v7, v9, v19} ⊆ D and D = {u1, u10, u11, u15, v5, v6, v7, v9, v19} but D does not dominate the vertices u13 and u14.

i = 5) If n1 = 3, n2 = 2, n3 = 9, n4 = 5 and DU = {u1, u4, u6, u15}, then {v10, v12, v14, v17, v19} ⊆ D and D = {u1, u4, u6, u15, v10, v12, v14, v17, v19} but D does not dominate the vertices u2 and u8.

i = 6) If n1 = 9, n2 = 2, n3 = 3, n4 = 5 and DU = {u1, u10, u12, u15}, then {v5, v6, v7, v9, v14} ⊆ D and D = {u1, u10, u12, u15, v5, v6, v7, v9, v14} but D does not dominate the vertex u16.

i = 7) If n1 = 8, n2 = 3, n3 = 5, n4 = 3 and DU = {u1, u9, u12, u17}, then {v3, v6, v7, v11, v14} ⊆ D and D = {u1, u9, u12, u17, v3, v6, v7, v11, v14} but D does not dominate the vertex u16.

i = 8) If n1 = 3, n2 = 6, n3 = 5, n4 = 5 and DU = {u1, u4, u10, u15}, then {v6, v9, v12, v14, v19} ⊆ D and D = {u1, u4, u10, u15, v6, v9, v12, v14, v19} but D does not dominate the vertex u17.

i = 9) If n1 = 3, n2 = 5, n3 = 6, n4 = 5 and DU = {u1, u4, u9, u15}, then {v6, v13, v14, v17, v19} ⊆ D and D = {u1, u4, u9, u15, v6, v13, v14, v17, v19} but D does not dominate the vertices u2 and u8.

i = 10) If n1 = 3, n2 = 5, n3 = 5, n4 = 6 and DU = {u1, u4, u9, u14}, then {v3, v6, v13, v18, v19} ⊆ D and D = {u1, u4, u9, u14, v3, v6, v13, v18, v19} but D does not dominate the vertices u7 and u8.

Hence, W4,38 has not any dominating set with 9 vertices and W4,38 = 10 as desired. □

Now for n ≥ 48 with n ≡ 8 (mod 10) we determine the domination number of W4,n as follows.

Lemma 3.5

For each even integer n ≥ 48, n ≡ 8 (mod 10), we have γ (W4,n) = 2⌊ n10⌋+4.

Proof

Let n = 10t+8, where t ≥ 4. By Theorem 1.2, we have γ (W4,n) ≥ n5 > 2t+1. The set D = {u1, u6, ⋯, u5t+1} ∪ {v5, v10, ⋯, v5t} ∪ {v3, v5t − 2, v5t+3} is a dominating set with 2t+4 elements, and so, 2t+2 ≤ γ (W4,n) ≤ 2t+4. We show that γ (W4,n) = 2t+4.

First, assume that γ(W4,n) = 2t+2. Let D be a minimum dominating set of W4,n. Then by the Pigeonhole Principal either |DU| ≤ t+1 or |DV| ≤ t+1. Without loss of generality assume that |DU| = t+1 − a, where a ≥ 0. Then |DV| = t+1+a. Observe that DU dominates at most 4t+4 − 4a vertices of V, and therefore, D dominates at most (4t+4 − 4a)+(t+1+a) = 5t − 3a+5 vertices of V. Since D dominates all vertices in V, we have 5t − 3a+5 ≥ 5t+4 and so a = 0. Then |DU| = |DV| = t+1. Also we have |VD| = 4t+3 ≤ |N(DU)| ≤ 4t+4 and |UD| = 4t+3 ≤ |N(DV)| ≤ 4t+4. Let DU = {ui1, ui2, ⋯, uit+1} and n1, n2, ⋯, nt+1. By Observation 2.2, we have Σk=1t+1nk = 5t+4 and therefore there exist k′ such that nk ∈ ℳ4. By Lemma 2.3, DU dominates at most 4t+3 vertices from V. Then |N(DU)| = |N(DV)| = 4t+3 and k′ is unique. If there exists 1 ≤ k″ ≤ t+1 such that nk ≥ 8, then 5t+4 = Σk=1t+1nknk+nk+5(t − 1) ≥ 1+8+5(t − 1) = 5t+4 which implies nk = 1, nk = 8 and for each k∉{k′, k″} we have nk = 5. Now in each arrangement of the cyclic sequence of DU, we have one adjacency between 1 and 5. Then we have two vertices in DU with index-distance equal to 6, a contradiction. Thus for kk′ we have nk = 5 and nk = 4. We have (by symmetry) n1 = n2 = ⋯ = nt = 5 and nt+1 = 4 and DU = {u1, u6, ⋯, u5t+1}. Now DU doesn’t dominate the vertices v3, v5, v10, ⋯, v5t and so {v3, v5, v10, ⋯, v5t} ⊆ D. Thus D = {u1, u6, ⋯, u5t+1, v3, v5, v10, ⋯, v5t}, but D does not dominate two vertices u5t−2 and u5t+3, a contradiction.

Now, assume that γ(W4,n) = 2t+3. Let D be a minimum dominating set of W4,n. Then by the Pigeonhole Principal either |DU| ≤ t+1 or |DV| ≤ t+1. Without loss of generality, suppose |DU| = t+1 − a, where a ≥ 0. Then |DV| = t+2+a. Observe that DU dominates at most 4t+4 − 4a vertices of V and, therefore, D dominates at most (4t+4 − 4a)+(t+2+a) = 5t − 3a+6 vertices of V. Since D dominates all vertices in V, we have 5t − 3a+6 ≥ 5t+4 and a = 0, |DU| = t+1 and |DV| = t+2. Also we have 4t+2 ≤ |N(DU)| ≤ 4t+4. Since 4t+2 ≤ |N(DU)| ≤ 4t+4, at most two elements of n1, n2, ⋯, nt+1 can be in ℳ4. If x is the number of 5′s in the cyclic-sequence of DU, then by Observation 2.2, we have Σk=1t+1nk = 5t+4 ≥ 1+1+8(t − x − 1)+5x and, therefore, 3x ≥ 3t − 10 which implies xt − 3. Thus t − 3 elements of the cyclic sequence are equal to 5. The sum of the remaining four values of the cyclic sequence is 19, and at most two of them are in ℳ4. In the last case of Lemma 3.3, for n = 38, we identified all such cyclic sequences and placed them in two tables, Table 5 and Table 6. We now continue according to Table 5 and Table 6.

Table 4

n = 18, 28, 38

Table 5

n = 38 with one ni in ℳΔ

Table 6

n = 38 with two ni in ℳΔ

In the case (i = 1) in Table 5 we have n1 = 4, n2 = ⋯ = nt+1 = 5 and DU = {u1, u5, u10, ⋯, u5t}. Thus {v7, v9, v14, ⋯, v5t+4} ⊆ D. But {u1, u5, u10, ⋯, u5t, v7, v9, v14, ⋯, v5t+4} does not dominate the vertices u3 and u5t+2. For dominating u3 and u5t+2, we need two vertices and therefore |D| ≥ 2t+4, a contradiction. In the cases (i ∈{2, 3}) in Table 5 we have to add 5′s to the end of the cyclic sequence and construct the corresponding set D with 2t+3 elements. In both cases we obtain that N[u8] ∩ D = ∅. Then D is not a dominating set, a contradiction.

In the case (i = 1) in Table 6 we cannot add a 5 to the cyclic sequence, since by adding a 5 to the cyclic-sequence we obtain two consecutive values of the cyclic sequence which one is 5 and the other is 1 and their sum is 6 which belongs to ℳ4, a contradiction.

In the case (i = 2) in Table 6 we cannot add a 5 to the cyclic sequence, since by adding a 5 to the cyclic sequence we obtain two consecutive values of the cyclic sequence which one is 5 and the other is 1 or 2, and their sum is 6 or 7, which belongs to ℳ4, a contradiction.

In the cases (i ∈ {3, 4, 5, 6}) in Table 6 we have to add 5′s to the end of the cyclic sequence and construct the corresponding set D with 2t+3 elements. In (i = 3), we obtain that N[u2] ∩ D = ∅, in (i = 4), we obtain that N[{u13, u14}] ∩ D = ∅, in (i = 5), we obtain that N[{u2, u8}] ∩ D = ∅, and in (i = 6), we obtain that N[u16] ∩ D = ∅. In all four cases, D is not a dominating set, a contradiction.

In the case (i = 7) in Table 6, by adding 5′s to the cyclic-sequence, we obtain some different new cyclic-sequences. We divide them into three categories.

  • c1)

    n1 = 8, n2 = 3 and n3 = 5. In this category, the constructed set, D, does not dominate u16, a contradiction.

  • c2)

    n1 = 8, n2 = 5, n3 = 3 and n4 = 5. In this category, the constructed set, D, does not dominate u15 a contradiction.

  • c3)

    n1 = 8, n2 = n3 = 5 and if ni = nj = 3, then |ij| ≥ 2. In this category, the constructed set, D, does not dominate u5i+1, a contradiction. (Notice that this category does not appear for n ≤ 5.) In the cases (i ∈{8, 9, 10}) in Table 6, by adding 5’s to the cyclic-sequences, we obtain some different new cyclic-sequences. we divide them into two categories.

  • c1)

    n1 = 3 and nt+1 = 5. In this category, the constructed set, D, does not dominate u2, a contradiction.

  • c2)

    n1 = 3 and nt+1 = 6. In this category, the constructed set, D, does not dominate u7 and u8. Hence, D is not a dominating set, a contradiction.

Hence, γ (W4,n) = 2t+4 = 2⌊ n10⌋+4 □

Now a consequence of Lemmas 3.1, 3.2, 3.33, 3.4 and 3.5 implies the following theorem which is the main result of this section.

Theorem 3.6

For each integer n ≥ 16, we have: γ(W4,n)=2n10+0n0 (mod 10)2n=16,18,36;n2,4 (mod 10)3n=28;n6 (mod 10),n16,364n8 (mod 10),n18,28.

4 Conclusion and Suggestion

In this manuscript we studied the domination number of 4-regular Knödel graphs. The following are some open related problems.

Problem 1

Obtain the domination number of k-regular Knödel graphs for k ≥ 5.

Problem 2

Obtain the total domination number of k-regular Knödel graphs for k ≥ 4.

Problem 3

Obtain the connected domination number of k-regular Knödel graphs for k ≥ 3.

Problem 4

Obtain the independent domination number of k-regular Knödel graphs for k ≥ 3.

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About the article

Received: 2017-11-29

Accepted: 2018-05-08

Published Online: 2018-08-03


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 816–825, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0072.

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© 2018 Mojdeh et al., published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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