On the integer {k}-domination number of circulant graphs

TL;DR

This paper determines the exact integer {k}-domination number for a class of circulant graphs with specific difference sets, providing a precise formula based on graph parameters.

Contribution

It introduces a closed-form expression for the {k}-domination number of circulant graphs with difference set D={1,2,...,t}, extending understanding of domination in these graphs.

Findings

Derived the {k}-domination number formula for D={1,2,...,t}

Established the exact value as a ceiling function involving n, k, and t

Enhanced the theoretical framework for domination in circulant graphs.

Abstract

Let $G=(V,E)$ be a simple undirected graph. $G$ is a circulant graph defined on $V=\mathbb{Z}_n$ with difference set $D\subseteq \{1,2,\ldots,\lfloor\frac{n}{2}\rfloor\}$ provided two vertices $i$ and $j$ in $\mathbb{Z}_n$ are adjacent if and only if $\min\{|i-j|, n-|i-j|\}\in D$. For convenience, we use $G(n;D)$ to denote such a circulant graph. A function $f:V(G)\rightarrow\mathbb{N}\cup\{0\}$ is an integer $\{k\}$-domination function if for each $v\in V(G)$, $\sum_{u\in N_G[v]}f(u)\geq k.$ By considering all $\{k\}$-domination functions $f$, the minimum value of $\sum_{v\in V(G)}f(v)$ is the $\{k\}$-domination number of $G$, denoted by $\gamma_k(G)$. In this paper, we prove that if $D=\{1,2,\ldots,t\}$, $1\leq t\leq \frac{n-1}{2}$, then the integer $\{k\}$-domination number of $G(n;D)$ is $\lceil\frac{kn}{2t+1}\rceil$.

Equations

$$(2t+1)\sum _{v\in V(G)}f(v)=(f(v_1),f(v_2),\dots ,f(v_n))(A+I_n)1_n\ge 1_n^T\cdot k\cdot 1_n=nk,$$

$$\lceil \frac{ak}{b}\rceil$$

$$\lceil \frac{ak}{b}\rceil =\sum _{i=0}^{a-1}\lfloor \frac{k+s_i}{b}\rfloor$$

$$S_i=\left\{\lceil \frac{bi}{a}\rceil ,\lceil \frac{bi}{a}\rceil +1,\dots ,\lceil \frac{b(i+1)}{a}\rceil -1\right\}$$

$$\lceil \frac{3k}{8}\rceil =\sum _{i=1}^3\lfloor \frac{k+\lceil 8i/3\rceil -1}{8}\rfloor =\lfloor \frac{k+2}{8}\rfloor +\lfloor \frac{k+5}{8}\rfloor +\lfloor \frac{k+7}{8}\rfloor ,$$

$$S_0=\{0,1,2\},S_1=\{3,4,5\}\text{and}{S}_2=\{6,7\},$$

$$B\setminus A^{\prime }\le B\setminus \tilde{A}\le B\setminus A,$$

$$A(i)<A(i+r)\text{for}0\le i\le a-r-1$$

$$A(j)<A(j-a+r)\text{for}a-r\le j\le a-d-1,$$

$$\lceil \frac{rk}{a}\rceil =\sum _{i=a-r}^{a-1}\lfloor \frac{k+A(i)}{a}\rfloor ,$$

$$\lceil \frac{ak}{b}\rceil =\sum _{i=b-a}^{b-1}\lfloor \frac{k+B(i)}{b}\rfloor$$

$$\left\{A(i)\mid a-r\le i\le a-1\right\}=\left\{\lceil \frac{aj}{r}\rceil -1\mid 1\le j\le r\right\}.$$

$$A(i)=a-1-A^{\prime }(a-s+i)\text{for}i=0,1,\dots ,s-1,$$

$$a-1-(\lceil \frac{ai}{r}\rceil -1)=a+\lfloor \frac{-ai}{r}\rfloor =\lfloor \frac{a(r-i)}{r}\rfloor$$

$$\left\{A^{\prime }(i)\mid 0\le i\le r-1\right\}=\left\{\lfloor \frac{aj}{r}\rfloor \mid 0\le j\le r-1\right\}$$

$$C_1(a,b):=[0,1,\dots ,a-1].$$

$$C_1(a,b):=\text{the extension sequence of}(\overline{C_1(r,a)},a)$$

$$C:=B[b-a:b]\circ \underset{q}{\underbrace{B\circ B\circ \cdots \circ B}}$$

$${\gamma }_k(G)=\lceil \frac{kn}{2t+1}\rceil .$$

$$B[b-a:b]=[\ell -1,2\ell -1,\dots ,a\ell -1]$$

$$\sum _{i=0}^{b-1}\lfloor \frac{k+i}{b}\rfloor =k,$$

$$C[i:i+a]\le B[b-a:b]\text{for}0\le i\le a-d-1,$$

$$C_2(3,5)=[2,0,4,3,1],$$

$$(\lfloor \frac{k+4}{5}\rfloor ,\lfloor \frac{k+3}{5}\rfloor ,\lfloor \frac{k+1}{5}\rfloor ,\lfloor \frac{k+2}{5}\rfloor ,\lfloor \frac{k}{5}\rfloor ,\lfloor \frac{k+4}{5}\rfloor ,\lfloor \frac{k+3}{5}\rfloor ,\lfloor \frac{k+1}{5}\rfloor )$$

Full text

A function $f:V(G)\rightarrow\mathbb{N}\cup\{0\}$ is an integer $\{k\}$-domination function if for each $v\in V(G)$, $\sum_{u\in N_{G}[v]}f(u)\geq k.$ By considering all $\{k\}$-domination functions $f$, the minimum value of $\sum_{v\in V(G)}f(v)$ is the $\{k\}$-domination number of $G$, denoted by $\gamma_{k}(G)$. In this paper, we prove that if $D=\{1,2,\ldots,t\}$, $1\leq t\leq\frac{n-1}{2}$, then the integer $\{k\}$-domination number of $G(n;D)$ is $\lceil\frac{kn}{2t+1}\rceil$.

MSC 2010: 05C69, 11A05.

Keywords: Circulant graph, integer $\{k\}$-domination number, Euclidean algorithm.

1 Introduction and preliminaries

The study of domination number of a graph $G$ has been around for quite a long time. Due to its importance in applications, there are various versions of extension study, see [6] for reference.

The idea of integer $\{k\}$-domination was proposed by Domke et al. in [3]. It can be dealt as a labeling problem. The vertices of the graph $G$ are labeled by integers in $\mathbb{N}\cup\{0\}$ such that for each vertex $v$, the total (sum) values in its closed neighborhood $N_{G}[v]$ must be at least $k$. The problem is asking for finding the minimum total value labeled on $G$. Finally, we say that $f:V(G)\rightarrow\mathbb{N}\cup\{0\}$ is an integer $\{k\}$-domination function if for each $v\in V(G)$, $\sum_{u\in N_{G}[v]}f(u)\geq k$. Among all such functions $f$, the minimum value of $\sum_{v\in V(G)}f(v)$ is called the integer $\{k\}$-domination number of $G$, denoted by $\gamma_{k}(G)$.

It is not difficult to see that the original domination number of a graph $G$, $\gamma(G)$, can be recognized as $\gamma_{1}(G)$ since the vertices with label "1" gives a dominating set. For more information about domination problem, the readers may refer to [2, 4, 5, 6, 8]. Hence, the integer $\{k\}$-domination problem is also an NP-hard problem. So far, results obtained are all on special classes of graphs, see [1, 3, 7, 9].

In this paper, we shall consider the class of circulant graph $G=G(n;D)$ where $D=\{1,2,\ldots,t\}$, $1\leq t\leq\frac{n-1}{2}$, i.e., $V(G)=\mathbb{Z}_{n}$ and two vertices $i$ and $j$ are adjacent if and only if $d(i,j):=\min\{|i-j|,n-|i-j|\}\in D$. Since $D=\{1,2,\ldots,t\}$, $G(n;D)$ is exactly the power graph $C_{n}^{t}$ where $C_{n}$ is a cycle of order $n$.

The following results are obtained by Lin [10]. For clearness, we also outline its proof in which basic linear algebra is applied.

Proposition 1.1** ([10]).**

Let $G$ be the circulant graph $G(n;D)$ where $D=\{1,2,\ldots,t\}$. Then, $\gamma_{k}(G)\geq\lceil\frac{kn}{2t+1}\rceil.$

Proof.

Let $A$ be the adjacency matrix of $G$ and $f$ be an $\{k\}$-domination function of $G$. Let ${\bf 1}_{n}$ denote the all 1 column vector of length $n$. Then, we have

[TABLE]

which implies the inequality. ∎

By the aid of an algorithm, Lin was able to show the following.

Proposition 1.2** ([10]).**

For $t\leq 5$, $\gamma_{k}(G(n;\{1,2,\ldots,t\}))=\lceil\frac{kn}{2t+1}\rceil$.

But, for larger $t$, it remains unsettled. Our main result of this paper shows that the equality holds for all $1\leq t\leq\frac{n-1}{2}$.

2 The main result

By Proposition 1.1, in order to determine $\gamma_{k}(G)$, it suffices to show that $\gamma_{k}(G)\leq\lceil\frac{nk}{2t+1}\rceil$. That is, we need a proper distribution of values for $f(v_{1}),f(v_{2}),\ldots,f(v_{n})$ such that for each $v_{i}$, $\sum_{u\in N_{G}[v_{i}]}f(u)\geq k$ and $\sum_{i=1}^{n}f(v_{i})\leq\lceil\frac{nk}{2t+1}\rceil$. Since we are dealing with circulant graphs, $\sum_{u\in N_{G}[v_{i}]}f(u)$ is in fact the sum of $2t+1$ consecutive labels assigned to the circle $C_{n}=(v_{1},v_{2},\ldots,v_{n})$. Therefore, we turn our focus on providing suitable labels to meet the condition.

For example, let $n=8$ and $t=2$. Then, the following labeling of $(v_{1},v_{2},\ldots,v_{8})$, $(x_{4},x_{3},x_{1},x_{2},x_{0},x_{4},x_{3},x_{1})$ will satisfy the requirement, where $x_{i}=\lfloor\frac{k+i}{5}\rfloor$, $i=0,1,2,3,4.$

We are considering $1\leq t\leq\frac{n-1}{2}$ in what follows. First, we need an estimaton of the sum of rational numbers which take its floor or ceiling values.

Lemma 2.1**.**

For positive integers $a,b$ and nonnegative integer $k$, we have the following.

1. (1)

$\lfloor\frac{k+\lfloor x\rfloor}{a}\rfloor=\lfloor\frac{k+x}{a}\rfloor$* for any real number $x,$* 2. (2)

$k=\sum_{i=0}^{a-1}\lfloor\frac{k+i}{a}\rfloor$, and 3. (3)

$\lceil\frac{ak}{b}\rceil=\sum_{i=1}^{a}\lfloor\frac{k+\lceil ib/a\rceil-1}{b}\rfloor$.

Proof.

(1) and (2) are easy to check, we prove (3).

[TABLE]

∎

Since the variables $a,b$ and $k$ are all integers, the uniqueness of the formula in Lemma 2.1(3) can be confirmed.

Corollary 2.2**.**

If integers $0\leq s_{0}\leq s_{1}\leq\cdots\leq s_{a-1}<b$ satisfy

[TABLE]

*for positive integers $a,b$ and nonnegative integer $k$, then $s_{i}=\lceil(i+1)b/a\rceil-1$ for $i=0,1,\ldots,a-1.$

∎*

According to the $s_{i}$’s given above, we split $[b]:=\{0,1,\ldots,b-1\}$ into subintervals with maximal elements $s_{i}$’s. For positive integers $a<b,$ let $[b]$ be partitioned into $a$ subsets such that

[TABLE]

for $i=0,1,\ldots,a-1.$ It is clear that ||S_{i}|-|S_{j}||$$\leq 1 for all $i,j.$ We analyze the subsets containing more elements in the following.

Lemma 2.3**.**

Let $q$ and $r$ be the quotient and remainder of $b$ divided by $a,$ respectively. Then the cardinality $|S_{i}|=q+1$ if $i=\lfloor\frac{aj}{r}\rfloor$ for $j=0,1,\ldots,r-1.$

Proof.

By definition, $|S_{i}|=\lceil(i+1)b/a\rceil-\lceil ib/a\rceil=q+\lceil(i+1)r/a\rceil-\lceil ir/a\rceil.$ Therefore, $|S_{i}|=q+1$ if and only if there exists some integer $0\leq j\leq r-1$ such that $ir/a\leq j<(i+1)r/a.$ The above inequality can be rewritten as $i\leq ja/r{<}i+1,$ and hence $i=\lfloor ja/r\rfloor.$ ∎

Example 2.4**.**

Let $a=3,$ $b=8,$ and $r=2$ be the remainder of $b$ divided by $a.$ Then

[TABLE]

and $[8]=\{0,1,\ldots,7\}$ can be partitioned into 3 subsets such that

[TABLE]

where the subsets numbered with $\lfloor\frac{aj}{r}\rfloor=0$ and $1$ as $j=0$ and $1,$ respectively, have more than $1$ elements. Note that the maximal elements $2,5,7$ of subsets $S_{i}$’s are the integers in (1) that construct $\lceil\frac{3k}{8}\rceil$.

Additionally, we need a result of the comparison between two sequences. For two real finite non-decreasing sequences $A=(a_{i}),A^{\prime}=(a^{\prime}_{i})$ of the same length $n$, we say that $A\leq A^{\prime}$ if $a_{i}\leq a^{\prime}_{i}$ for $i=0,1,\ldots,n-1.$

Lemma 2.5**.**

Let $A$ and $A^{\prime}$ be two subsequences of a real finite non-decreasing sequence $B$ which have equal length $0<|A|=|A^{\prime}|<|B|$. Then $A\leq A^{\prime}$ if and only if $B\setminus A^{\prime}\leq B\setminus A.$

Proof.

Because of the symmetry, we prove $A\leq A^{\prime}$ implies $B\setminus A^{\prime}\leq B\setminus A$ by induction on $|A|$ in the following. It is clearly true when $|A|=1.$ Suppose the statement is correct for $|A|<m<|B|.$ Assume that $A=(a_{i})_{i=0}^{m-1}$ and $A^{\prime}=(a^{\prime}_{i})_{i=0}^{m-1}$ satisfying $A\leq A^{\prime}.$ From induction hypothesis, $B\setminus(a^{\prime}_{i})_{i=1}^{m-1}\leq B\setminus(a_{i})_{i=1}^{m-1}.$ It is clear that the non-decreasing sequence obtained by exchanging an entry $a$ of the original sequence into $\tilde{a}\geq a$ (and inserting $\tilde{a}$ to the appropriate position) is not less than the original sequence. Thus, we have

[TABLE]

where $\widetilde{A}$ is obtained from $A$ by deleting $a_{0}$ and adding $a^{\prime}_{0}.$ The result follows. ∎

Now, we are ready to find the desired integer $\{k\}$-domination function $f$. Let $[a]:=\{0,1,\ldots,a-1\}$ for each positive integer $a.$ For a sequence $A$ of length $a,$ let the entries of $A$ indexed by $[a]$ and $A(i)$ be the $i$-th entry of $A.$ For $0\leq i<j\leq a,$ the subsequence $A[i:j]:=[A(i),A(i+1),\ldots,A(j-1)].$ If $A$ is a permutation of $[a],$ then the complement of $A$ is a sequence $\overline{A}$ of length $a$ defined as $\overline{A}(i)=a-1-A(i)$ for $0\leq i\leq a-1.$ The concatenation $A\circ B$ of two sequences $A$ and $B$ of lengths $a$ and $b,$ respectively, is a sequence of length $a+b$ obtained by attaching $B$ to $A$ defined as $(A\circ B)(i)=A(i)$ if $0\leq i\leq a-1$ and $(A\circ B)(j)=B(j-a)$ if $a\leq j\leq a+b-1.$

Let $A$ be a permutation of $[a]$ and thus a sequence of length $a$. For positive integers $a<b,$ we call $B$ the extension sequence of the pair $(A,b)$ if $B$ is a permutation of $[b]$ satisfying $B(i)<B(j)$ if and only if $A(i_{0})<A(j_{0})$ or $A(i_{0})=A(j_{0})$ with $i<j,$ where $i_{0}$ and $j_{0}$ are the remainders of $i$ and $j$ divided by $a,$ respectively. For example, when $(a,b)=(3,7)$ and $A=[0,1,2],$ the extension sequence of $(A,b)$ is $B=[0,3,5,1,4,6,2],$ which is attained by extending $A$ to the sequence $[0,1,2,0,1,2,0]$ of length $7$ and renumbering it with $0,1,\ldots,6.$

A permutation $A$ of $[a]$ is said to be nice corresponding to some $b>a$ with $a\nmid b$ if

[TABLE]

and

[TABLE]

where $r$ is the remainder of $b$ divided by $a$ and $d=\gcd(a,b).$ Note that if $r=d$ then the condition (3) can be ignored. For example, $[1,3,0,2,4]$ is nice corresponding to $8$ (or any larger integer congruent to $3$ modulo $5$) and $[4,1,6,3,0,5,2,7]$ is nice corresponding to $13.$

The following properties will carry out the recursive constructions.

Proposition 2.6**.**

Suppose that $R$ is a nice permutation of $[r]$ corresponding to some $a>r$ with $r\nmid a.$ Let $\overline{R}$ be the complement of $R.$ Then the extension sequence of $(\overline{R},a)$ is also nice corresponding to some $b>a$ with $b\equiv r~{}(\text{mod}~{}a).$

Proof.

Let $A$ be the extension sequence of $(\overline{R},a).$ Note that $A(i)<A(i+r)$ for $0\leq i\leq a-r-1$ can be verified directly by the definition of extension sequences. Assume that $s$ is the remainder of $a$ divided by $r.$ It’s left to consider the case $a-r\leq j\leq a-d-1$ where $r$ is the remainder of $b$ divided by $a$ and $d=\gcd(a,b).$ Assume that $s$ is the remainder of $a$ divided by $r.$ By Euclidean algorithm, $s\geq d.$

**Case 1: $a-r\leq j\leq a-s-1.$

**In order to show $A(j)<A(j-a+r),$ we observe that the remainders of $j$ and $j-a+r$ divided by $r$ are $j^{\prime}+s$ and $j^{\prime}$, respectively, where $j^{\prime}=j-a+r.$ Moreover, since $R$ is nice, we have $\overline{R}(i+s)<\overline{R}(i)$ for $0\leq i\leq r-s-1.$ The result is straightforward by the definition of extension sequences.

**Case 2: $a-s\leq j\leq a-d-1.$

**In this case, the remainders of $j$ and $j-a+r$ divided by $r$ become $j^{\prime}-r+s$ and $j^{\prime}$, respectively, where $j^{\prime}=j-a+r.$ Once again, since $R$ is nice, $\overline{R}(i-r+s)<\overline{R}(i)$ for $r-s\leq i\leq r-d-1.$ We have the proof. ∎

Proposition 2.7**.**

Let positive integers $a<b$ with $r>0$ the remainder of $b$ divided by $a$ and $R$ a permutation of $[r]$ with complement $\overline{R}.$ If the extension sequence $A$ of $(R,a)$ satisfies

[TABLE]

then the extension sequence $B$ of $(A^{\prime},b)$ satisfies

[TABLE]

where $A^{\prime}$ is the extension sequence of $(\overline{R},a).$

Proof.

By Corollary 2.2,

[TABLE]

Let $q$ and $s$ be the quotient and remainder of $a$ divided by $r,$ respectively. Claim that the set of ${A^{\prime}}[0:r]$ equals the set of $a-1-A[a-r:a-1].$ It is clear for $s=0.$ If $s>0,$ we have

[TABLE]

since entries in $A$ that larger than $A(i)$ become smaller than ${A^{\prime}}(a-s+i)$ in ${A^{\prime}},$ and vice versa. Therefore, the set of $A[s+jr:s+(j+1)r]$ equals to the set of $a-1-{A^{\prime}}[a-s-(j+1)r:a-s-jr]$ for $j=0,1,\ldots,q.$ The claim follows by taking $j=q-1$. Moreover, since

[TABLE]

for $1\leq i\leq r,$ we have

[TABLE]

which exactly indicates the indices of subsets defined in Lemma 2.3. Hence the set of $B[b-a:b]$ gives the maximal elements in each of the subsets $S_{0},S_{1},\ldots,S_{a-1},$ and this fact completes the proof. ∎

For each pair of positive integers $(a,b)$ with $a<b,$ define two codes ${C_{1}}$ and ${C_{2}}$ as follows. If $a$ divides $b,$ then

[TABLE]

If the remainder $r$ of $b$ divided by $a$ is positive, then

[TABLE]

where $\overline{C_{1}}(r,a)$ is the complement of ${C_{1}}(r,a).$ Now $C_{2}$ can be constructed subsequently. Let $C_{2}(a,b)$ be the extension sequence of $(C_{1}(a,b),b).$ It is clear that $C_{1}(a,b)$ and $C_{2}(a,b)$ are permutations of $[a]$ and $[b],$ respectively. Suppose that each entry $\alpha$ in $C_{2}(a,b)$ is corresponding to $\lfloor\frac{k+\alpha}{b}\rfloor.$ Then the following result can be obtained by proving that

[TABLE]

is a feasible distribution of the circulant graph $G,$ where $b=2t+1,$ $n=qb+a,$ and $B=C_{2}(a,b).$

Theorem 2.8**.**

[TABLE]

Proof.

By Proposition 1.1, it suffices to show that $\gamma_{k}\leq\lceil\frac{kn}{2t+1}\rceil$. Let $G=G(n;\{1,2,\ldots,t\})$, $n=qb+a$ and $b=2t+1$. First, we construct $B[b-a:b]$. If $a$ divides $b$ such that $b=\ell a,$ then

[TABLE]

which collects the numbers $\lceil ib/a\rceil-1$ for $1\leq i\leq a$ given in Lemma 2.1. Therefore, any substring of $C$ of length $a$ is not larger than $B[b-a:b].$ By Lemma 2.5, every length $b$ string of $B\circ B[b-a:b]$ or $B[b-a:b]\circ B$ is not less than $[0,1,\ldots,b-1],$ so does $C.$ Furthermore, the sequence $[0,1,\ldots,b-1]$ is of sum

[TABLE]

which confirms the case for $a$ divides $b.$

On the other hand, let $a>\gcd(a,b)$ and $r$ be the remainder of $b$ divided by $a.$ Since the initial case is examined above, by Proposition 2.7, $B[b-a:b]$ collects the elements $\{\lceil ib/a\rceil-1\}_{i=1}^{a}.$ For the initial case, if $r$ divides $a$ then $C_{1}(r,a)=[0,1,\ldots,r-1]$ and it is easy to check that $C_{1}(a,b)$ is nice. Moreover, by Proposition 2.6, $C_{1}(a,b)$ is always nice, and hence

[TABLE]

where $d=\gcd(a,b).$ Moreover, we also have $C[i:i+a]\leq B[b-a:b]$ for $a-d\leq i\leq qb-1$ immediately from the construction of $C.$ The result follows. ∎

Example 2.9**.**

Assume that $G(n;D)$ is a circulant graph on $n=8$ vertices with $D=\{1,2\}$ (i.e., $t=2$). Let $b=2t+1=5$ and $a=3$ be the remainder of $n$ divided by $b$. First of all, we obtain $C_{1}(3,5)$ by the process of Euclidean algorithm. Since the initial condition $C_{1}(1,2)=[0],$ $C_{1}(2,3)=[0,1]$ is directly the extension code of $[0].$ Next, the complement of $C_{1}(2,3)$ is $\overline{C_{1}(2,3)}=[1,0]$. Thus, $C_{1}(3,5)=[1,0,2],$ the extension code of $([1,0],3).$ Hence,

[TABLE]

the extension code of $(C_{1}(3,5),5).$ Attach the last $3$ entries in front of $C_{2}(3,5),$ we attain the distribution $[4,3,1,2,0,4,3,1].$ Then the circular sequence $f(v):~{}v\in V(G)$ is given by

[TABLE]

which satisfies $\sum_{v\in V(G)}f(v)=\lceil 8k/5\rceil$ and $\sum_{u\in N_{G}[v]}f(u)\geq k$ for each $v\in V(G).$

3 Concluding remark

We remark finally that the construction of code $C_{2}$ can be obtained by giving an algorithm with inputs $a$ and $b$.