Research Article | Open Access

# The Larger Bound on the Domination Number of Fibonacci Cubes and Lucas Cubes

**Academic Editor:**Yuantong Gu

#### Abstract

Let and be the -dimensional Fibonacci cube and Lucas cube, respectively. Denote by the subgraph of induced by the end-vertex that has no up-neighbor. In this paper, the number of end-vertices and domination number of and are studied. The formula of calculating the number of end-vertices is given and it is proved that . Using these results, the larger bound on the domination number of and is determined.

#### 1. Introduction

The Fibonacci cube and Lucas cube were presented in [1, 2], respectively. Because their many properties (see [1–7]) such as domination number, 2-packing number, and observability can be applied to interconnection networks [1].

However, the number of vertices of Fibonacci cube and Lucas cube grows rapidly as increases. So it is hard to calculate exactly the number of domination number of Fibonacci cubes and Lucas cubes. The lower bound on the domination number of Fibonacci cubes and Lucas cubes is determined in [3, 6], respectively. In this paper, we will give a larger bound on the domination number of Fibonacci cubes and Lucas cubes using construction method. We begin with some basic definitions.

Graphs considered in this paper are finite, simple, connected, and undirected. Let be the dimensional hypercube with . A Fibonacci string of order is a binary string of length without two consecutive ones. The Fibonacci cube (see Figure 1) is the subgraph of induced by the Fibonacci strings of length , whose vertices are the Fibonacci strings of length , and two vertices are joined by an edge if their Hamming distance is exactly 1. A Fibonacci string is a Lucas string if . The Lucas cube is the subgraph of induced by the Lucas strings of length . It is well known that and , where and are Fibonacci numbers and Lucas numbers, respectively. Recall that the Fibonacci numbers and Lucas numbers form a sequence of positive integers and , respectively, where , and and , and for .

For a connected graph , the distance between vertices and is the usual shortest path distance. For , , let be the set of vertices of that contain ones. Hence is the set of vertices of at distance from . is defined analogously. If , where and for , then we call a down-neighbor of and an up-neighbor of .

If a vertex has no up-neighbor, we call it an end-vertex and denote by the number of end-vertices of . Let be an end-vertex with string length , where are the number of ones and consecutive , respectively. We denote by the subgraph of induced by the end-vertex , whose strings of vertices were obtained from string of the vertex by changing ones into zeroes and any two vertices have an edge if their Hamming distance is exactly 1 (see Figure 2).

Let be a graph. Then is a dominating set if every vertex from is adjacent to some vertex from . The domination number is the minimum cardinality of a dominating set of .

All graph-theoretical terms and concepts used but unexplained in this paper are standard, and that can be found in many textbooks such as [8].

#### 2. The Larger Bound on the Domination Number of Fibonacci Cubes

In this section, we will determine the larger bound on the domination number of Fibonacci cubes. Firstly, we mention the following properties of Fibonacci cubes which will be used later.

Lemma 1 (see [3]). *Let and . Then any different have different sets of down-neighbors.*

Lemma 2 (see [6]). *For any , .*

Lemma 3. *Let be an end-vertex with string length , where and are the number of ones and consecutive , respectively. Then (see Figure 2).*

*Proof. *Let , and it will be frequently used in latter. Let , and be a vertex with , and
whose subset contains all vertices such that satisfy the condition for . We will prove that is a dominating set. In order to prove that is a dominating set. It suffices to prove that any vertices in can dominate vertices in . If , the result is obviously correct. We assume that . From Lemma 1, we know that any different have different sets of down-neighbors and have at most one common down-neighbor vertex. Since each vertex has exactly ones. If , then the number of vertices dominated by vertices set is at least as follows:

If , there must exist integer such that and , and then the number of vertices dominated by vertices set is at least as follows:
Therefore the set is a dominating set, and
This completes our proof.

Theorem 4. *Let be the number of end-vertices in . Then the followings hold.*(i)*If is odd, then
*(ii)*If is even, then
*

*Proof. *We prove only that the theorem is correct if is odd and . And the proofs of the others cases are similar. End-vertices of can be divided into two cases as follows.*Case 1*. The end-vertices set is composed of end-vertices with strings form .*Case 2*. The end-vertices set is composed of end-vertices with strings form .

In Case 1, since and , then end-vertices are divided into cases with . Therefore the number of end-vertices with strings form is

In Case 2, as similar as Case 1, end-vertices are divided into cases with . Then the number of end-vertices with strings form is
Therefore
This completes our proof.

Now we give the larger bound on the domination number of Fibonacci cubes as follows.

Theorem 5. *Let . Then for the Fibonacci cube the followings hold.*(i)*If is odd, then
*(ii)*If is even, then
*

*Proof. *According to the process of proof of Lemma 3, there must exist a dominating set of with the string form (or ) such that it contains the vertex with the string form (or ). So all dominating sets of of with strings form (or ) have at least two common vertices with strings (or ). Then the theorem follows directly from Lemmas 2 and 3 and Theorem 4.

#### 3. The Larger Bound on the Domination Number of Lucas Cubes

In this section, we will determine the larger bound on the domination number of Lucas cubes as follows.

Lemma 6 (see [3]). *For any , .*

Theorem 7. *Let be the number of end-vertices in . Then the followings hold.*(i)*If is odd, then
*(ii)*If is even, then
*

*Proof. *The proof is similar to Theorem 4.

Theorem 8. *Let . Then for the Lucas cube the followings hold.*(i)*If is odd, then
*(ii)*If is even, then
*

*Proof. *That follows directly from Lemmas 6 and Theorem 7.

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

The author is very grateful to Professor Wenpeng Zhang for useful suggestions and the two referees for valuable comments. This work is supported by TSNC of China (TSA1013).

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#### Copyright

Copyright © 2014 Shengzhang Ren. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.