Improved Upper Bound on Independent Domination Number for Hypercubes

05/13/2022
by   Debabani Chowdhury, et al.
0

We revisit the problem of determining the independent domination number in hypercubes for which the known upper bound is still not tight for general dimensions. We present here a constructive method to build an independent dominating set S_n for the n-dimensional hypercube Q_n, where n=2p+1, p being a positive integer ≥ 1, provided an independent dominating set S_p for the p-dimensional hypercube Q_p, is known. The procedure also computes the minimum independent dominating set for all n=2^k-1, k>1. Finally, we establish that the independent domination number α_n≤ 3 × 2^n-k-2 for 7× 2^k-2-1≤ n<2^k+1-1, k>1. This is an improved upper bound for this range as compared to earlier work.

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1 Introduction and related work

Hypercubes are classical structures that are well-studied in many important areas such as set theory, graph theory, combinatorics and Boolean algebra [15]. Given an -dimensional hypercube graph , where be the set of vertices (nodes) and be the set of edges in , and a subset , we say that dominates its neighbourhood , i.e., the set of all nodes that are adjacent to at least one node in [1, 15]. If dominates , then is called a dominating set of . is said to be independent if no two nodes in are adjacent. An independent dominating set with minimum cardinality is called the minimum independent dominating set. The independent domination number is the cardinality of the minimum independent dominating set of .

Many results are known on independent domination for simple regular and other graphs [3]-[8]. However, there are many unsettled questions regarding minimum dominating sets in hypercubes as stated by Harary et al. [1]. They derived the minimum independent dominating sets of for dimension :. It was also reported that for and , . Yet, the upper bound on is still not tight for many other values of . Mane et al. [2] showed that for , for any . In this work, we improve this bound for certain values of . We split the range into two non-overlapping cases:

Case 1: , ;
Case 2: , .

In this paper, we show that for . This tightens the upper bound for Case 2 in comparison to the result proved earlier for both cases by Mane et al. [2]. Given an independent dominating set , our procedure constructs an independent dominating set , where , , being a positive integer . Our result follows from a procedure that iteratively constructs a minimal independent dominating set in a hypercube from that of a smaller-dimension hypercube.

2 Concepts and techniques

The -dimensional hypercube is an undirected regular graph whose vertex set where and denote the set of nodes and set of edges in , respectively. The set can be envisaged as the set of all binary -tuples of zeros and ones, i.e. . Thus, . Two nodes of are adjacent if their binary -tuples differ in exactly one place, i.e., if they are unit Hamming distance apart. Let the Hamming distance between two binary -tuples corresponding to vertices be represented as .

Lemma 1

Proof : For any , and every vertex in has adjacent vertices. That is, a vertex of can dominate at most vertices uniquely. Thus, the minimum number of vertices required to dominate all vertices of will be . As the number of vertices should be an integer, .

3 Iterative construction of an independent dominating set in a hypercube

The following procedure constructs an independent dominating set of , given an independent dominating set of , where . We use the following notation in Procedure 1.

Let be an independent dominating set of such that . Let a vertex , be represented by

-bit binary vectors

, , respectively, where , , , .

Procedure 1:
Input:
Output:

  1. Assign .

  2. For every vertex get a vertex , such that differ in LSB, i.e., , and .

  3. , .

  4. For every , and for every get respectively, such that , , and , .

  5. .

  6. Output .

  7. End.

Lemma 2

Proof : It follows from Procedure 1.

Lemma 3

as constructed by Procedure 1 is an independent dominating set of .

Proof : Follows from [9].

4 Constructing an independent dominating set for , where

Let be independent dominating sets for -, and -dimensional hypercubes and , respectively, where . Clearly, , , , . Let , be vertices of and , respectively, such that , , , . The following procedure constructs from .

Procedure 2:
Input: where every vector is represented as -bit binary , where .
Output: where every vector is represented as -bit binary , where .

  1. Assign ,

  2. Get -bit binary tuple of as , where , .

  3. Set .

  4. Set .

  5. .

  6. .

  7. If , then else .

  8. ; If then goto Step 6.

  9. If is of even parity then else .

  10. .

  11. ; If , then goto Step 4.

  12. ; If , then goto Step 2.

  13. Output .

  14. End.

Example 1

Table 1 shows how Procedure 2 is executed, given a minimum independent dominating set for the 3-dimensional hypercube to obtain an independent dominating set for the 7-dimensional hypercube . Also, set produced by Procedure 2 is a minimum independent dominating set in . Figure 1 shows how a vector of is modified to produce a vector of . It also includes, in general, the concatenation process that leads to an -dimensional vector of the dominating set starting from a -dimensional vector of the dominating set . Given, , , , i.e., , and .

Lemma 4

as produced by Procedure 2 has cardinality .

Proof : Immediate from Procedure 2.

Theorem 1

as obtained by Procedure 2, is an independent dominating set of .

Proof : First, we prove that is an independent set of . Next, we will show that is also a dominating set of .
Let be two vertices of obtained by Procedure 2. Clearly,

and

For any two vetices , there are three possible cases;
Case 1: ;
Case 2: , , i.e., both

have the same parity, either both even or both odd;


Case 3: , i.e., if has even (odd) parity, then has odd (even) parity.
For Case 1, the Hamming distance between -bit binary vectors of is

For Case 2,

For Case 3,

Thus, for any two vertices , we have . That is, no two vertices in are adjacent. Hence, is an independent set of .

To prove that is a dominating set, we argue as follows. We have proved that is an independent set, i.e., its no two vertices are adjacent, i.e., , . Thus, for every vertex , its all adjacent vertices are in . Thus, we have the following four cases:
Case 1: contains only ;
Case 2: ;
Case 3: , a , such that is adjacent to .

Let be the neighbourhood of , respectively. Now, , if every , then

So,

Cace 4: .
By combining Cases (1), (2) and (4), we have , i.e., contains all those vertices of which are not in . Again, from Case 3, a , such that is adjacent to . Thus, is a dominating set of . Hence, is an independent dominating set of .

Procedure 2 leads to the following known result [1].

Lemma 5

for .

Proof: From Lemma 1, . From Lemma 1, it follows that . When , is an integer, and hence, .

(parity
of
0 000 1 1 0 000 000 0000000
0 000 2 2 0 111 111 0000111
1 001 1 3 1 000 001 1001001
1 001 2 4 1 111 110 1001110
2 010 1 5 1 000 010 1010010
2 010 2 6 1 111 101 1010101
3 011 1 7 0 000 011 0011011
3 011 2 8 0 111 100 0011100
4 100 1 9 1 000 100 1100100
4 100 2 10 1 111 011 1100011
5 101 1 11 0 000 101 0101101
5 101 2 12 0 111 010 0101010
6 110 1 13 0 000 110 0110110
6 110 2 14 0 111 001 0110001
7 111 1 15 1 000 111 1111111
7 111 2 16 1 111 000 1111000
Table 1: Construction of from

Parity of

Parity of

Figure 1: Construction of an -dimensional vector of the independent dominating set from a -dimensional vector of the dominating set , where . It illustatrates the construction of a vector for , , i.e., the construction of a vector of from a vector of .

5 Upper bound on the independent domination number for hypercubes

Theorem 2

For , where ,

Proof : From [1], we know that . In this case, . Following Lemma 4 and Theorem 1, for , we obtain , where . By iteration, we can then derive results for corresponding to , respectively. Following Procedure 2, we obtain for , respectively. The term in the series can be evaluated as . Hence, the proof.

Theorem 3

For , where ,

Proof : If is given, we can construct following Procedure 1. Thus, when , . Hence, the proof.

Thus it improves the previous upper bound [2] for the range of .

Theorem 4

For , , .

Proof : If is given, we can construct following Procedure 1. Thus, when , . Hence, the proof.

6 Independent domination number for hypercubes

Here, we summarize our results and compare them with previously known bounds.

6.1 Previous results

For , , [1, 10, 11, 12]
For , , [1, 2, 13]
For , , [2]

6.2 Proposed work

For , ,
For , ,
For , ,

6.3 Numerical comparison for some values of

We report the values of for different values of in Table 2. Results in Table 3 are obtained from Table 2, which show improved upper bound for some values on .

[1, 2, 10, 11, 12, 13]
[2]
(this work)
Table 2: of for
[1, 2, 10] [2] (this work)
[11, 12]
[13, 14]
1 2 2
2 3 2
4 4
5 8