Note on Distance Magic Products $G\circ C_4$

Abstract

A distance magic labeling of a graph $G=(V,E)$ of order $n$ is a bijection $l :V \rightarrow \{1, 2,\ldots , n\}$ with the property that there is a positive integer $k$ (called magic constant) such that $w(x) = k$ for every $x \in V$. If a graph $G$ admits a distance magic labeling, then we say that $G$ is a distance magic graph. In the case of non-regular graph $G$, the problem of determining whether there is a distance magic labeling of the lexicographic product $G\circ C_4$ was posted in Arumugam et al. (J Indonesian Math Soc 11–26, 2011). We give necessary and sufficient conditions for the graphs $K_{m,n}\circ C_4$ to be distance magic. We also show that the product $C^{(t)}_3\circ C_4$ of the Dutch Windmill Graph and the cycle $C_4$ is not distance magic for any $t>1$.

1 Introduction

All graphs considered in this paper are simple finite graphs. Given a graph $G$, we denote its order by $|G|=n$, its vertex set by $V(G)$ and the edge set by $E(G)$. The neighborhood $N(x)$ of a vertex $x$ is the set of vertices adjacent to $x$, and the degree $d(x)$ of $x$ is $|N(x)|$, the size of the neighborhood of $x$.

Let $w(x)=\sum _{y\in N_G(x)}l(y)$ for every $x \in V(G)$.

A distance magic labeling (also called sigma labeling) of a graph $G=(V,E)$ of order $n$ is a bijection $l :V \rightarrow \{1, 2,\ldots , n\}$ with the property that there is a positive integer $k$ (called magic constant) such that $w(x) = k$ for every $x \in V$. If a graph $G$ admits a distance magic labeling, then we say that $G$ is a distance magic graph (see [13]).

The concept of distance magic labeling has been motivated by the construction of magic squares. Finding a distance magic labeling of an $r$-regular graph turns out to be equivalent to finding equalized incomplete tournament $\mathrm {EIT}(n, r)$ [4]. In an equalized incomplete tournament $\mathrm {EIT}(n, r)$ of $n$ teams with $r$ rounds, each team plays with exactly $r$ other teams and the total strength of the opponents that team $i$ plays is $k$. Thus, it is easy to observe that finding an $\mathrm {EIT}(n, r)$ is the same as finding a distance magic labeling of any $r$-regular graph on $n$ vertices. For a survey, we refer the reader to [2].

The following observations were independently proved:

Observation 1.1

([8–10, 13]) Let $G$ be an $r$-regular distance magic graph on $n$ vertices. Then $k = \frac{r(n+1)}{2}$.

Observation 1.2

([8–10, 13]) No $r$-regular graph with $r$-odd can be a distance magic graph.

The problem of distance magic labeling of $r$-regular graphs was studied recently (see [1–4, 9, 11]). It is interesting that if you blow up an $r$-regular $G$ graph into some specific $p$-regular graph, then the obtained graph $H$ is distance magic. More formally, we have the following definition.

Definition 1.3

([7], p. 185) The lexicographic product $G\circ H$ of two graphs $G$ and $H$ is defined on $V(G\circ H)=V(G)\times V(H)$, two vertices $(u,x),(v,y)$ of $G\circ H$ being adjacent whenever $uv\in E(G)$, or $u=v$ and $xy\in E(H)$.

$G\circ H$ is also called the composition of graphs $G$ and $H$ and denoted by $G[H]$ (see [6]).

Miller at al. [9] proved the following results.

Theorem 1.4

([9]) The cycle $C_n$ of length $n$ is a distance magic graph if and only if $n = 4$.

Theorem 1.5

([9]) Let $r \ge 1$, $n \ge 3$, $G$ be an $r$-regular graph and $C_n$ the cycle of length $n$. Then $G\circ C_n$ admits a distance magic labeling if and only if $n = 4$.

Theorem 1.6

([9]) Let $G$ be an arbitrary regular graph. Then $G\circ \overline{K}_n$ is distance magic for any even $n$.

Shafiq et al. [12] considered distance magic labeling for disconnected graphs and obtained the following theorems.

Theorem 1.7

([12]) Let $m \ge 1$, $n\ge 2$ and $p \ge 3$. Then $mC_p\circ K_n$ has a distance magic labeling if and only if either $n$ is even or $mnp$ is odd or $n$ is odd and $p \equiv 0 (\mathrm{mod}~4)$.

The following problem was posted in [2].

Proposition 1.8

([2]) If $G$ is non-regular graph, determine if there is a distance magic labeling of $G\circ C_4$.

The Dutch Windmill Graph $C_3^{(t)}$, also called a friendship graph, is the graph obtained by taking $t>1$ copies of the cycle graph $C_3$ with a vertex in common [5]. We show that the product $C_3^{(t)}\circ C_4$ is not distance magic for any $t>1$.

The paper is organized as follows. In the next section we focus on the products of complete bipartite graphs and cycle $C_4$. In the third section we prove that the product of the Dutch Windmill Graph and the cycle $C_4$ cannot be distance magic.

2 The Product $K_{m,n}\circ C_4$

Let $K_{m,n}$ have the vertex partite sets $A\!=\!\!\{x_0, x_1,\ldots ,x_{n-1}\}$ and $B\!\!=\!\{y_0,y_1,\ldots ,y_{m-1}\}$. Let $C_4=v^0v^1v^2v^3v^0$ and $H=K_{m,n}\circ C_4$. For $0\le i \le n-1$ and $j=0,1,2,3$, let $x_i^j$ be the vertices of $H$ that replace $x_i$ in $A$. For $0\le i \le m-1$ and $j=0,1,2,3$, let $y_i^j$ be the vertices of $H$ that replace $y_i$ in $B$. Let $A[C_4]=\{x_i^j: i=0,1,\ldots ,n-1,j=0,1,2,3, x_i\in A\}$ and $B[C_4]=\{y_i^j: i=0,1,\ldots ,m-1,j=0,1,2,3, y_i\in B\}$.

The following Lemma holds true.

Lemma 2.1

If $H=K_{m,n}\circ C_4$, where $1\le m<n$ is a distance magic graph and $k$ is the magic constant, then the following conditions hold:

• (1) $l(x_i^0)+l(x_i^2)=l(x_i^1)+l(x_i^3)=a \quad$ for some constant $a$ for all    $0 \le i\le n-1$ and $l(y_i^0)+l(y_i^2)=l(y_i^1)+l(y_i^3)=b$    for some constant $b$ for all    $0 \le i\le m-1$,

• (2) $b+2an=a+2mb=k$    and    $a<b$,

• (3) $bm+an=(m+n)(4m+4n+1)$,

Proof

(1)

Notice that: $w(x_i^j)=l(x_i^{j+1})+l(x_i^{j+3})+\sum _{i=1}^m\sum _{j=1}^4{l(y_i^j)}$ for all $0 \le i\le n-1$, $j=0,1,2,3$, where the addition in the superscripts is performed modulo $4$. Since the graph $H$ is distance magic we obtain that $l(x_i^0)+l(x_i^2)=l(x_i^1)+l(x_i^3)=a$ for some constant $a$ for all $0 \le i\le n-1$. Similarly $l(y_i^0)+l(y_i^2)=l(y_i^1)+l(y_i^3)=b$ for some constant $b$ for all $0 \le i\le m-1$.

(2) Fact (1) implies that $w(x_i^j)=a+2bm$ for all $0 \le i\le n-1$, $j=0,1,2,3$ and $w(y_i^j)=b+2an$ for all $0 \le i\le m-1$, $j=0,1,2,3$. As $m<n$, this implies that $a<b$.

(3) The labeling $l$ is a bijection, so the sum of all labels has to be equal to $\sum _{i=1}^{4m+4n}{i}$:

$$\begin{aligned} 2an+2bm=\frac{(4m+4n)(4m+4n+1)}{2}. \end{aligned}$$

$\square$

The following theorem completely characterizes the pairs $(m,n)$, for which $K_{m,n}\circ C_4$ is distance magic.

Theorem 2.2

Let $m$ and $n$ be integers such that $1\le m<n$. Then $K_{m,n}\circ C_4$ is distance magic if and only if the following conditions hold.

• (1) The numbers

  $$\begin{aligned} a=\frac{(m+n)(4m+4n+1)(2m-1)}{4mn-m-n} \end{aligned}$$

  and

  $$\begin{aligned} b=\frac{(m+n)(4m+4n+1)(2n-1)}{4mn-m-n} \end{aligned}$$

  are integers.

• (2) There exist integers $p,q,t\ge 1$, such that

  $$\begin{aligned}&p+q=b-a,\\&4n=pt,\\&4m=qt. \end{aligned}$$

Proof

First, let us assume that for given $m$ and $n$, $1\le m<n$ there exist $a$, $b$, $p$, $q$ and $t$ with desired properties. Then the following labeling is distance-magic: If $t=4s$ for some integer $s$, then let

$$\begin{aligned} l(x_{kp+i}^j)=\left\{ \begin{array}{lll} k(2p+2q)+i+1 &{}\quad \mathrm {for}\quad &{} j=0 \\ k(2p+2q)+p+q+i+1 &{}\quad \mathrm {for}\quad &{} j=1, \\ a-f(x_{kp+i}^{j-2}) &{}\quad \mathrm {for}\quad &{} j=2,3, \end{array}\right. \end{aligned}$$

for $0\le k\le t/4-1,i=0,1,\ldots ,p-1$,

$$\begin{aligned} l(y_{kq+i}^j)=\left\{ \begin{array}{lll} k(2p+2q)+p+i+1 &{}\quad \mathrm {for}\quad &{} j=0, \\ k(2p+2q)+2p+q+i+1 &{}\quad \mathrm {for}\quad &{} j=1, \\ b-f(y_{kq+i}^{j-2}) &{}\quad \mathrm {for}\quad &{} j=2,3. \end{array}\right. \end{aligned}$$

for $0\le k\le t/4-1,i=0,1,\ldots ,q-1$. Observe that the sets of labels of vertices $x_i^j$ and $y_i^j$ for $j=0,1$ do not intersect and their elements are consecutive numbers from the set $\{1,\dots ,2(m+n)\}$. And as $b-a=p+q$, also the sets of labels for $j=2,3$ do not intersect and they are consecutive numbers from the set $\{a-(2m+2n)+q,\dots , a+q-1=b-p-1\}$. In order to prove that $l$ is a bijection it is enough to show that $a+q-1=4m+4n$. It is true, as we have:

$$\begin{aligned} a(n+m)+(b-a)m=an+bm=(m+n)(4m+4n+1). \end{aligned}$$

But on the other hand,

$$\begin{aligned} (b-a)m=(p+q)m=\frac{4(m+n)m}{t}=(m+n)q, \end{aligned}$$

so finally

$$\begin{aligned} a(n+m)+(m+n)q=(m+n)(4m+4n+1) \end{aligned}$$

and $a+q=4m+4n+1$.

If $t\not \equiv 0(\mathrm{mod}~4)$, then $p$ and $q$ are even. In such a situation, let

$$\begin{aligned} l(x_{kp/2+i}^j)=\left\{ \begin{array}{lll} k(p+q)+2i+j+1 &{}\quad \mathrm {for}\quad &{} j=0,1, \\ a-f(x_{kp/2+i}^{j-2}) &{}\quad \mathrm {for}\quad &{} j=2,3, \end{array}\right. \end{aligned}$$

for $0\le k\le \lfloor {t/2}\rfloor -1,i=0,1,\ldots ,p/2-1$, and for $k=\lfloor {t/2}\rfloor =\lceil {t/2}\rceil -1,i=0,1,\ldots ,p/4-1$,

$$\begin{aligned} l(y_{kq/2+i}^j)=\left\{ \begin{array}{lll} k(p+q)+p+2i+j+1 &{}\quad \mathrm {for}\quad &{} j=0,1, \\ b-f(y_{kq/2+i}^{j-2}) &{}\quad \mathrm {for}\quad &{} j=2,3, \end{array}\right. \end{aligned}$$

for $0\le k\le \lfloor {t/2}\rfloor -1,i=0,1,\ldots ,q/2-1$, and

$$\begin{aligned} l(y_{kq/2+i}^j)=\left\{ \begin{array}{lll} k(p+q)+p/2+2i+j+1 &{}\quad \mathrm {for}\quad &{} j=0,1, \\ b-f(y_{kq/2+i}^{j-2}) &{}\quad \mathrm {for}\quad &{} j=2,3, \end{array}\right. \end{aligned}$$

for $k=\lfloor {t/2}\rfloor =\lceil {t/2}\rceil -1,i=0,1,\ldots ,q/4-1$ (observe that in both cases the last range is in use if and only if $t$ is odd and thus $p$ and $q$ are divisible by $4$). Also in this case it is straightforward to see that the above labeling is bijective and that in both cases the magic constant equals to $k=a+2mb=b+2na$.

Now, let us assume that $K_{m,n}\circ C_4$ is distance magic for some integers $m$ and $n$, $1\le m<n$. Let the magic constant be $k$. From the Lemma 2.1 it follows that the following system of equations must be satisfied:

$$\begin{aligned} \left\{ \begin{array}{l} b+2an=a+2mb,\\ bm+an=(m+n)(4m+4n+1). \end{array} \right. \end{aligned}$$

The above system has only one solution with respect to $a$ and $b$:

$$\begin{aligned} \left\{ \begin{array}{l} a=\frac{(m+n)(4m+4n+1)(2m-1)}{4mn-m-n},\\ b=\frac{(m+n)(4m+4n+1)(2n-1)}{4mn-m-n}. \end{array} \right. \end{aligned}$$

Obviously, $a$ and $b$ must be integers.

Now we choose any distance magic labeling $l$ of $K_{m,n}\circ C_4$. Let $L_A=\{l(x)|x\in A[C_4]\}$ and $L_B=\{l(y)|y\in B[C_4]\}$. Let us divide the set of all labels $\{1,2,\dots , 4m+4n\}$ into intervals in the following way:

• (i) for each interval $I$, either $I\subseteq L_A$ or $I\subseteq L_B$,

• (ii) each interval is maximal, i.e., for any two neighboring intervals $I_1,I_2$ we have either $I_1\subseteq L_A$ and $I_2\subseteq L_B$ or $I_1\subseteq L_B$ and $I_2\subseteq L_A$.

In the remainder we will use the notation $I_1<I_2\Leftrightarrow \max \{l(x)|l(x)\in I_1\}<\min \{l(y)|l(y)\in I_2\}$. For any interval $I$ and integer $c$, let $c-I=\{c-l(x)|l(x)\in I\}$ (it is possible that $c-I=I$). From the Lemma 2.1 it follows that for each $x\in \{1,2,\dots , 4m+4n\}$, $l(x)\in L_A\Leftrightarrow a-l(x)\in L_A$ (in fact, $l(x_i^j)=a-l(x_i^{j+2})$, where the addition in the superscripts is performed modulo $4$). Similarly, $l(x)\in L_B\Leftrightarrow b-l(x)\in L_B$. This implies that $a-I\subseteq L_A\Leftrightarrow I\subseteq L_A$ and $b-I\subseteq L_B\Leftrightarrow I\subseteq L_B$. Also, if $I_1,I_2\subseteq L_A$ and $I_1<I_2$ then $a-I_2<a-I_1$. Similarly, if $I_1,I_2\subseteq L_B$ and $I_1<I_2$ then $b-I_2<b-I_1$.

Let the first two intervals be $I_1=\{1,\dots ,r\}$ and $I_2=\{r+1,\dots ,r+s\}$ for some $r,s\ge 1$.

Observe first that $I_1\subseteq L_A$ and $I_2\subseteq L_B$. Otherwise we would have $b-I_1=\{b-r,\dots ,b-1\}$ and $a-I_2=\{a-r-s,\dots ,a-r-1\}<b-I_1$. Moreover, as $a-r-1<b-r-1$, this would imply that there is an interval $I\in L_A$, $a-I_2<I$ and thus $a-I<I_2$, a contradiction ($I_2$ is the first interval being subset of $L_A$).

We have $a-I_1=\{a-r,\dots ,a-1\}$ and $b-I_2=\{b-r-s,\dots ,b-r-1\}$. As $\min \{l(x)|l(x)\in a-I_1\}-1=a-r-1<b-r-1=\max \{l(y)|l(y)\in b-I_2\}$ and the intervals $a-I_1$ and $b-I_2$ are disjoint, it follows that $a-I_1<b-I_2$. Moreover there is no integer $u$ such that $a-1<u<b-r-s$, as it would mean that there is an interval $I\subseteq L_A$, $a-I_1<I$ and thus $a-I<I_1$, a contradiction. Thus the intervals $a-I_1$ and $b-I_2$ consist of $r+s$ consecutive integers. Moreover, the first entry $b-r-s$ of $b-I_2$ follows immediately after the last entry of $a-I_1$, which is $a-1$. Hence, we have $b-r-s=a$ and thus $r+s=b-a$. Observe also that the intervals $a-I_1$ and $b-I_2$ are the last ones contained in $L_A$ and $L_B$ respectively.

If $r+s=4m+4n$, this proves the hypothesis ($r=4n$, $s=4m$, so $t=1$, $p=r/2$, $q=s/2$). Otherwise let us assume that we are given $d\ge 1$ pairs of intervals $(I_1^i,I_2^i)$, $i=1,\dots ,d$, such that $I_1^i\subseteq L_A$, $I_2^i\subseteq L_B$, $a-I_1^i\subseteq L_A$, $b-I_2^i\subseteq L_B$, $|I_1^i|=r$, $|I_2^i|=s$, $I_1^i<I_2^i$, $a-I_1^i<b-I_2^i$ for $i=1,\dots ,d$ and $I_1^i<I_1^{i+1}$, $I_2^i<I_2^{i+1}$, $a-I_1^{i+1}<a-I_j^i$, $b-I_2^{i+1}<b-I_2^i$ for $i=1,\dots ,d-1$. Moreover, let us assume that there are no elements $u<\max \{l(y)|l(y)\in I_2^d\}$, $u\not \in \bigcup _{i=1}^{d}{(I_1^i\cup I_2^i)}$ and no elements $v>\max \{l(x)|l(x)\in a-I_1^d\}$, $v\not \in \bigcup _{i=1}^{d}{(a-I_1^i\cup b-I_2^i)}$. We are going to prove that we are able to extend this sequence to $d+1$ pairs of intervals.

Indeed, let us assume, that next two intervals are $I_1=\{d(r+s)+1,\dots ,d(r+s)+r_1\}$ and $I_2=\{d(r+s)+r_1+1,\dots ,d(r+s)+r_1+s_1\}$ for some $r_1,s_1\ge 1$. Obviously $I_1\subseteq L_A$ and $I_2\subseteq L_B$. Moreover we have $a-I_1=\{a-d(r+s)-r_1,\dots ,a-d(r+s)-1\}$ and $b-I_2=\{b-d(r+s)-r_1-s_1,\dots ,b-d(r+s)-r_1-1\}$. As $\min \{l(x)|l(x)\in a-I_1\}-1<\max \{l(y)|l(y)\in b-I_2\}$, it follows that $a-I_1<b-I_2$. Moreover there is no integer $u$ between $a-I_1$ and $b-I_2$, as there is no interval $I\subseteq L_A$, $a-I_1<I$, $I\not \in \{I_1^1,\dots ,I_1^d\}$. Thus the intervals $a-I_1$ and $b-I_2$ consist of $r+s$ consecutive integers. Thus $a-d(r+s)-1=b-d(r+s)-r_1-s_1-1$ and in consequence $r_1+s_1=b-a$. Similar reasoning leads us to the conclusion that there are no elements between $b-I_2$ and $a-I_1^d$, so $b-d(r+s)-r_1-1=a-(d-1)(r+s)-r-1$ and thus $b-a=r_1+s$. This means that $r_1=r$ and $s_1=s$. Obviously the intervals $a-I_1$ and $b-I_2$ are the last ones contained in $L_A$ and $L_B$, that do not belong to $\{I_1^1,\dots ,I_1^d\}$ and $\{I_2^1,\dots ,I_2^d\}$, respectively.

By induction we obtain that we are able to construct such a sequence of pairs for every $d$. As the number of pairs $(I_1^i,I_2^i)$ has to be finite, after some number of steps (say $t$) we exhaust all labels. Obviously

$$\begin{aligned} rt=\sum _{i=1}^{t}{|I_1^i|}=4n, \end{aligned}$$

and

$$\begin{aligned} st=\sum _{i=1}^{t}{|I_2^i|}=4m. \end{aligned}$$

Putting $p=r$ and $q=s$, we arrive at the hypothesis. $\square$

The pairs $(m,n)$ that satisfy the assumptions of the Theorem 2.2 are very rare. We checked all the pairs where $1\le m<n\le 80000$ and only for the following ones the graphs $K_{m,n}\circ C_4$ are distance magic: $(9,21)$, $(20,32)$, $(428,548)$, $(2328,2748)$, $(6408,10368)$, $(7592,8600)$, $(10098, 24378)$, $(18860,20840)$, $(39540,42972)$, $(73808,79268)$.

3 The Product $C^{(t)}_3\circ C_4$

Let $C^{(t)}_3$ have the central vertex $x$ and vertices $x,y_i,z_i$ for $i=1,\dots ,t$ belong to $i$th copy of cycle $C_3$. Let $C_4=v^0v^1v^2v^3v^0$ and $H=C^{(t)}_3\circ C_4$. For $0\le i \le t-1$ and $j=0,1,2,3$, let $y_i^j$, $z_i^j$ be the vertices of $H$ that replace $y_i^j$, $z_i^j$ $0 \le i\le t-1$ in $C^{(t)}_3$ and $x^0,x^1,x^2,x^3$ be the vertices of $H$ that replace $x$.

Theorem 3.1

The graph $C^{(t)}_3\circ C_4$ is not distance magic for any $t>1$.

Proof

Suppose that $l$ is a distance magic labeling of the graph $H=C^{(t)}_3\circ C_4$ and $k = w(x)$, for all vertices $x\in V(H)$. It is easy to observe that there exist natural numbers $b$, $a^i_y$ and $a^i_z$, $0\le i \le t-1$, such that:

• $l(x^0)+l(x^{2})=l(x^1)+l(x{^3})=b$.

• $l(y_i^0)+l(y_i^{2})= l(y_i^1)+l(y_i^{3}) = a_y^i$ for $0 \le i\le t-1$.

• $l(z_i^0)+l(z_i^{2})= l(z_i^1)+l(z_i^{3}) = a_z^i$ for $0 \le i\le t-1$.

Since $a_y^i+2a_z^i+2b=w(y_i^j)=w(z_i^j)=a_z^i+2a_y^i+2b$, we obtain that $a_y^i=a_z^i=a^i$. This implies that for any $0 \le i,l\le t-1$ and $0 \le j,h\le 3$, $3a^i+2b=w(z_i^j)=w(z_l^h)=3a^l+2b$, hence $a^i=a^l=a$.

Since $3a+2b=b+4ta=k$ and $4ta+2b=1+2+\cdots +4(2t+1)$ we obtain that $b=\frac{(2t+1)(4t-3)(8t+5)}{6t-3}$. Recall that the biggest label we can use is $4(2t+1)$, hence $b\le 16t+7$. One can calculate that the only positive integer $t$ that satisfies the inequality

$$\begin{aligned} (2t+1)(4t-3)(8t+5)\le (16t+7)(6t-3) \end{aligned}$$

is $t=1$, a contradiction. $\square$