Supermagic Graphs with Many Odd Degrees

Abstract

A graph $G=(V,E)$ is called supermagic if there exists a bijection $f: E\rightarrow \{1,2,\dots ,|E|\}$ such that the weight of every vertex $x\in V$ defined as the sum of labels f(xy) of all edges xy incident with x is equal to the same number m, called the supermagic constant.

Recently, Kovář et al. affirmatively answered a question by Madaras about existence of supermagic graphs with arbitrarily many different degrees. Their construction provided graphs with all degrees even. Therefore, they asked if there exists a supermagic graph with d different odd degrees for any positive integer d.

We answer this question in the affirmative by providing a construction based on the use of 3-dimensional magic rectangles.

1 Introduction

A finite simple graph $G=(V,E)$ is called supermagic if there exists a bijection $f: E\rightarrow \{1,2,\dots ,|E|\}$ called supermagic labeling such that the weight of every vertex $x\in V$ defined as the sum of labels f(xy) of all edges xy incident with x is equal to the same number m, called the supermagic constant. That is,

$$ \exists m\in \mathbb {N} \ \forall x\in V: w(x) = \sum _{xy\in E} f(xy) = m. $$

Most often, graphs studied in this context are vertex-regular or even vertex-transitive.

Recently, Kovář, Kravčenko, Krbeček, and Silber, [2] affirmatively answered a question by Madaras: Does there exist a supermagic graph with arbitrarily many different degrees?

Because their construction provided only graphs where all degrees were even, they asked the following more specific question: Does there exist a supermagic graph with d different odd degrees for any positive integer d?

We construct a class of disconnected graphs based on lexicographic products (also called compositions) of graphs with many different odd degrees and complements of complete graphs. We also present a modification of this construction that provides connected graphs with the same required properties.

The labeling used in the construction is based on the existence of three-dimensional magic rectangles.

2 Tools

To construct the graphs, we will use one of the standard graph products, called the lexicographic product or also composition.

A graph $G\circ H$ called the lexicographic product or composition of graphs G and H arises from G by replacing each vertex of G by a copy of H, and every edge of G by the complete bipartite graph $K_{t,t}$, where t is the number of vertices of H.

More formally, let $V(G)=\{g_1,g_2,\dots ,g_s\}$ and $V(H)=\{h_1,h_2,\dots ,h_t\}$. Then $V(G\circ H) = V(G) \times V(H)$ and $(g_a,h_b)(g_c,h_d)\in E(G\circ H)$ if and only if $g_ag_c\in E(G)$ or $g_a=g_c$ and $h_bh_d\in E(H)$.

If the graph H is isomorphic to $\overline{K_t}=tK_1$, that is, consists of t independent vertices, then we say that $G\circ tK_1$ is a blown up G, or that we have blown up G by $tK_1$. In this case, the graph $G\circ tK_1$ will be denoted simply by G[t].

An important ingredient of our construction is a 3-dimensional magic rectangle. We start with a more general definition introduced by Hagedorn [1].

Definition 1

An n-dimensional magic rectangle n-$\mathrm {MR}(a_1,a_2,\dots ,a_n)$ is an $a_1\times a_2\times \dots \times a_n$ array with entries $d_{i_1,i_2,\dots ,i_n}$ which are elements of the set $\{1,2,\ldots ,a_1a_2\dots a_n\},$ each appearing once, such that all sums in the k-th direction are equal to a constant $\sigma _k$. That is, for every $k,~1\le k\le n$, and every selection of indices $i_1,i_2,\dots ,i_{k-1},i_{k+1},\dots ,i_n,$ we have

$$\sum _{j=1}^{a_k} d_{i_1,i_2,\dots ,i_{k-1},j,i_{k+1},\dots ,i_n}=\sigma _k, $$

where $\sigma _k=a_k(a_1a_2\dots a_n +1)/2$.

The following existence results were also proved by Hagedorn in [1].

Theorem 1

[1]. If there exists an n-dimensional magic rectangle n-$\mathrm {MR}(a_1,a_2,\dots ,a_n)$, then $a_1\equiv a_2\equiv \dots \equiv a_n\pmod 2$.

For $a_1\equiv a_2\equiv \dots \equiv a_n\equiv 0\pmod 2$, Hagedorn found a complete existence characterization.

Theorem 2

[1]. An n-dimensional magic rectangle n-$\mathrm {MR}(a_1,a_2,\dots ,a_n)$ with $a_1\le a_2\le \dots \le a_n$ and all $a_i$ even exists if and only if $2\le a_1$ and $4\le a_2\le \dots \le a_n$.

For odd dimensions, Hagedorn proved the following.

Theorem 3

[1]. A 3-dimensional magic rectangle 3-$\mathrm {MR}(a_1,a_2,a_3)$ with $3\le a_1\le a_2\le a_3$ exists whenever $\gcd (a_i,a_j)>1$ for some $i,j\in \{1,2,3\}$.

We will use a special case of his result in our construction.

Corollary 1

[1]. A 3-dimensional magic rectangle 3-$\mathrm {MR}(3,3,a)$ exists for every odd a, $a\ge 3$.

Recently, Zhou, Li, Zhang, and Su [3] proved that for 3-dimensional magic rectangles the necessary conditions are also sufficient.

Theorem 4

[3]. A 3-dimensional magic rectangle 3-$\mathrm {MR}(a_1,a_2,a_3)$ with $3\le a_1\le a_2\le a_3$ exists whenever $a_1\equiv a_2\equiv a_3\equiv 1\pmod 2$.

3 Construction

We build our graphs in two steps. First we build a graph G with the required number of different odd degrees, and label the edges with just two different labels so that the sum of labels at every vertex is constant. Then we blow up G into G[3] and label the edges of each $K_{3,3}$ using entries of one $3\times 3$ rectangle from a 3-$\mathrm {MR}(3,3,a)$, or from a slightly modified 3-$\mathrm {MR}(3,3,b)$ with $b=8a+1$.

We call a graph G 2-pseudomagic if there are positive integers l and h and a mapping $\tilde{g}:E(G)\rightarrow \{l,h\}$ called 2-pseudomagic labeling such that the weight of every vertex x, defined as the sum of labels of all edges incident with x, is equal to the same constant $\widetilde{m}$. The edges labeled l will be called light, and those labeled h will be called heavy.

Let 3-$\mathrm {MR}(a_1,a_2,a_3)$ be a 3-dimensional magic rectangle as defined above, with entries $d_{j_1,j_2,j_3}$ and magic constants $\sigma _i, i=1,2,3$. For 3-$\mathrm {MR}(3,3,a)$ we have $\sigma _1=\sigma _2=3(9a+1)/2$. Recall that a must be odd.

Then the 3-dimensional $a_1\times a_2\times a_3$ array with entries $d^+_{j_1,j_2,j_3}=c+d_{j_1,j_2,j_3}$ is called a 3-dimensional c-lifted magic rectangle. It should be obvious that the magic constants here are equal to $a_ic+\sigma _i$ for each $i=1,2,3$. Such a rectangle will be denoted 3-$\mathrm {MR}^+(a_1,a_2,a_3;c)$ with magic constants $\sigma ^+_i$, where by the reasoning above we have $\sigma ^+_i = a_ic+\sigma _i$.

In order to use consecutive integers in the labeling of G[3], we want to find 3-$\mathrm {MR}(3,3,a)$ and 3-$\mathrm {MR}^+(3,3,b;c)$ such that $c=9a$ and $l\sigma ^+_i = h\sigma _i$ for some positive integers l and h. Note that because 3-$\mathrm {MR}^+(3,3,b;c)$ is constructed from a 3-dimensional magic rectangle 3-$\mathrm {MR}(3,3,b)$, we must have b odd.

We present one such pair in the following. For simplicity, we choose $l=1$.

Lemma 1

Let $c=9a, \ b=8a+1$, and $\sigma _1=\sigma _2$ and $\sigma ^+_1=\sigma ^+_2$ be magic constants of 3-$\mathrm {MR}(3,3,a)$ and 3-$\mathrm {MR}^+(3,3,b;c)$, respectively. Then $\sigma ^+_1=10\sigma _1$.

Proof

We have

$$\begin{aligned} \sigma _1 = \frac{9a(9a+1)}{2\cdot 3a} = \frac{3(9a+1)}{2} \end{aligned}$$

Since the labels in 3-$\mathrm {MR}^+(3,3,b;c)$ start with $9a+1$, we have

$$\begin{aligned} \sigma ^+_1 = \frac{(9a+1+9(a+b))9b}{2\cdot 3b} = \frac{3(9a+1+9(a+b))}{2} \end{aligned}$$

Substitute $b=8a+1$ to get

$$\begin{aligned} \sigma ^+_1 = \frac{3(9a+1+9(a+8a+1))}{2} = \frac{3(9a+1)(1+9))}{2} = 10\frac{3(9a+1)}{2} =10\sigma _1 \end{aligned}$$

as desired.

Now we construct a 2-pseudomagic host graph G with d different odd degrees. The labels we use are $l=1$ and $h=10$. We will say that a vertex x has a light degree $\deg _l(x)$ if it is incident with $\deg _l(x)$ light edges, that is, edges labeled 1. Similarly, vertex x has a heavy degree $\deg _h(x)$ if it is incident with $\deg _h(x)$ heavy edges, that is, edges labeled 10. The weight of a vertex x in G will be denoted $\widetilde{w}(x)$. From the above it follows that

$$\begin{aligned} \widetilde{w}(x)=\deg _l(x) + 10\deg _h(x). \end{aligned}$$

(1)

Construction 1. Let p be a prime, $p\ge 2d+1$, and $d>1$. Our host graph G will consist of $d+2$ components. A light component $G_0$, mixed components $G_1,G_2,\dots ,G_d$, and heavy component $G_{d+1}$. To use the magic rectangles from Lemma 1 for blowing up G to G[3], we will need a light and $b=8a+1$ heavy edges. We denote by $a_i$ and $b_i$ the number of light and heavy edges in $G_i$, respectively.

We start with $G_0 \cong K_{10p+1}$ with all light edges. Notice that the number of edges in $G_0$ is $(10p+1)5p$, which is odd, since p is a prime. We have $\deg (x_{0,j})=\deg _l(x_{0,j})=10p$ and from (1) it follows that $\widetilde{w}(x_{0,j})=10p$ for every $x_{0,j}\in G_0$.

Then for $i=1,2,\dots ,d-1$ we first take $G_i' \cong K_{10p+2} - M_{10p+2}$, where $M_{10p+2 }$ is a perfect matching. This is a 10p-regular graph, so we have used an even number of light edges in each $G'_i, i>0$.

Next we build $G_i$, $i=1,2,\dots ,d-1$ by removing $10(2i-1)$ light one-factors and adding back $(2i-1)$ heavy one-factors. The number of light edges in $G_i$ is still even, since for every $x_{i,j}\in G_i$ we have

$$\begin{aligned} \deg _l(x_{i,j})=10p-10(2i-1) \end{aligned}$$

(2)

and the order of $G_i$ is $10p+2$. Because we added back $(2i-1)$ heavy one-factors, we have

$$\begin{aligned} \deg _h(x_{i,j})=2i-1. \end{aligned}$$

(3)

Therefore, by adding (2) and (3), the degree of $x_{i,j}$ is

$$\begin{aligned} \deg (x_{i,j})=\deg _l(x_{i,j}) + \deg _h(x_{i,j}) = 10p-9(2i-1)=10p-18i+9, \end{aligned}$$

(4)

which is indeed odd. From (1) it follows that for every $x_{i,j}\in G_i$ we have

$$\begin{aligned} \widetilde{w}(x_{i,j})=\deg _l(x_{i,j})+10\deg _h(x_{i,j})=10p-10(2i-1)+ 10(2i-1)=10p. \end{aligned}$$

(5)

All components except $G_0$ contain an even number of light edges, making a odd as needed. It should be obvious that

$$\begin{aligned} b_1+b_2+\dots +b_d < 8(a_1+a_2+\dots +a_d). \end{aligned}$$

(6)

Since $b = 8a+1$ and $b_0=0$, we must add some heavy edges. We denote the number of lacking heavy edges by $b^*$. If $b^*$ is a multiple of p, say $b^*=ps$, we can build a p-regular component $G_{d}$ with 2s vertices, and we are done. Notice that $2s>p$, because we still need to compensate for $a_0=(10p+1)5p$ light edges in $G_0$, that is, we need to add at least $8a_0=40(10p+1)p$ heavy edges.

If $b^*$ is not a multiple of p, we keep adding copies of $G_1$ to G until the number of lacking heavy edges is divisible by p.

For each copy of $G_1$, we will add

$$ b_1=\frac{10p+2}{2} = 5p+1 $$

heavy edges and

$$ a_1=\frac{(10p-10)(10p+2)}{2} = 10(p-1)(5p+1) $$

light edges. Suppose we are adding q copies, and recall that the number of lacking heavy edges is $b^*$. Denote the number of light and heavy edges in G by $\bar{a}$ and $\bar{b}$, respectively. Hence, we have

$$ b=\bar{b}+qb_1 +b^*, $$

and

$$ a=\bar{a} + qa_1. $$

Because $b=8a+1$, we obtain

$$ \bar{b}+qb_1+b^*=8(\bar{a}+qa_1)+1, $$

or

$$\begin{aligned} b^*=(8\bar{a}+1-\bar{b}) + q(8a_1-b_1), \end{aligned}$$

(7)

where

$$\begin{aligned} 8a_1-b_1&= 8(10(p-1)(5p+1)) - (5p+1) \nonumber \\&= (5p+1)(80(p-1)-1) \nonumber \\&= (5p+1)(80p-81) \end{aligned}$$

(8)

We assumed that $8\bar{a}+1-\bar{b}$ is not divisible by p, so suppose

$$ 8\bar{a}+1-\bar{b} \equiv k\pmod {p}. $$

If we want to have $b^*$ a multiple of p, it follows from (7) that we must have

$$\begin{aligned} q(8a_1-b_1) \equiv -k \pmod {p}. \end{aligned}$$

(9)

Note that the congruence has a solution for q if and only if

$$ \gcd (8a_1-b_1,p)\mid k $$

and moreover, whenever there is a solution, then there is always one such that $1\le q\le p-1$.

Recall that we assumed $d>1$ and $p\ge 2d+1$. Hence, $p>3$ and we clearly have $\gcd (81,p) = 1$. Then from Eq. (8) we will have $\gcd (8a_1-b_1,p)=1$ because $p\not \mid 5p+1$ and $p\not \mid (80p-81)$ when $\gcd (81,p) = 1$. Therefore, we can always solve congruence (9) for q.

Hence, we can always construct a 2-pseudomagic graph G with edge labels 1 and 10 and d different odd degrees, where the number of light edges is a and the number of heavy edges is $b=8a+1$.

4 Main Result

We start with an easy observation, tying magic rectangles to supermagic labelings of complete bipartite graphs.

Lemma 2

Let 3-$\mathrm {MR}(a_1,a_1,a_3)$ be a 3-dimensional magic rectangle and $H=a_3 K_{a_1,a_1}$ a disjoint union of $a_3$ copies of the complete bipartite graph $K_{a_1,a_1}$. Then there exists a supermagic labeling of H.

Proof

Denote the k-th copy of $K_{a_1,a_1}$ by $H^k$ and vertices in its respective partite sets by $u^k_1,u^k_2,\dots ,u^k_{a_1}$ and $v^k_1,v^k_2,\dots ,v^k_{a_1}$. Let

$$f(u^k_i v^k_j)=d_{i,j,k}$$

for every $i,j=1,2,\dots ,a_1$ and every $k=1,2,\dots ,a_3.$ Then

$$w(u^k_i)=\sum _{j=1}^{a_1}f(u^k_i v^k_j)=\sum _{j=1}^{a_1} d_{i,j,k}$$

and

$$w(v^k_j)=\sum _{i=1}^{a_1}f(u^k_i v^k_j)=\sum _{i=1}^{a_1} d_{i,j,k}.$$

Because we have $a_1=a_2$, from Definition 1 it follows that

$$\sum _{j=1}^{a_1} d_{i,j,k}=\sum _{i=1}^{a_1} d_{i,j,k}= \frac{a_1(a^2_1a_3 + 1)}{2} =\sigma _1,$$

which implies

$$w(u^k_i)= w(v^k_j)=\sigma _1 $$

for every $i,j=1,2,\dots ,a_1$ and every $k=1,2,\dots ,a_3,$ and f is the desired supermagic labeling.

Now we are ready to prove our main result.

Theorem 5

For every positive integer d there exists a supermagic graph with d different odd vertex degrees.

Proof

For $d=1$ we take $G\cong K_2$ and label the only edge by 1. For $d>1$, we take the graph G with d different odd degrees from Construction 1 and blow it up to G[3]. Each vertex with an odd degree $\deg _G(x)$ in G is now of odd degree $3\deg _G(x)$. Then we label each $K_{3,3}$ arising from a light edge by entries of a 3-$\mathrm {MR}(3,3,a)$ and each $K_{3,3}$ arising from a heavy edge by entries of a 3-$\mathrm {MR}^+(3,3,b;9a)$. We will call the graphs $K_{3,3}$ in G[3] arising from light and heavy edges in G also light and heavy graphs, respectively. The triple of vertices arising from a vertex x will be denoted by $x^{[1]},x^{[2]},x^{[3]}$.

In particular, for $a_1=a_2=3$ and $a_3=a$ in Lemma 2 we have $\sigma _1=3(9a+1)/2$. From Lemma 1 it follows that $\sigma ^+_1=10\sigma _1 = 15(9a+1)$.

We observe that each light $K_{3,3}$ contributes to every $w(x^{[i]})$ for $i=1,2,3$ by $\sigma _1$ and a heavy $K_{3,3}$ contributes by $\sigma ^+_1=10\sigma _1$. Therefore, the total contribution of all light graphs at a vertex $x^{[i]}$ is $\sigma _1 \deg _l(x)$ and the contribution of all heavy graphs is $\sigma ^+_1 \deg _h(x)=10\sigma _1 deg_h(x)$. It follows that

$$\begin{aligned} w(x^{[i]})=\sigma _1 \deg _l(x) + 10\sigma _1 deg_h(x) = \sigma _1 \left( \deg _l(x) + 10 deg_h(x)\right) . \end{aligned}$$

(10)

But from (5) we have $\widetilde{w}(x)= \deg _l(x) + 10 \deg _h(x) = 10p$, and thus we immediately obtain from (10) that

$$\begin{aligned} w(x)= \sigma _1 \widetilde{w}(x) =10p\sigma _1 \end{aligned}$$

(11)

for every vertex $x^{[i]}$ in G[3], which concludes our proof.

It is not difficult to modify the graphs constructed above to make them connected.

Corollary 2

For every positive integer d there exists a connected supermagic graph with d different odd vertex degrees.

Proof

For $d=1$, we again have $G\cong K_2$. For $d>1$, we construct a connected graph H from G first, and find the blown-up graph H[3] exactly the same way as for G[3] in Theorem 5.

Recall that the graph G may contain more than one copy of $G_1$. We denoted the number of these components by $q+1$ and observed that $q+1\le p$. Denote the copies of $G_1$ by $G^k_1$ for $k=0,1,\dots ,q$. First we observe that each component $G^k_1,G_2,\dots ,G_{d}$ is large enough to contain two independent heavy edges, say $x_{i,1}y_{i,1}$ and $x_{i,2}y_{i,2}$ in component $G_i$.

For $i=1,2,\dots ,d$ we replace edges $x_{i,1}y_{i,1}$ and $x_{i+1,2}y_{i+1,2}$ by $x_{i,1}y_{i+1, 2}$ and $x_{i+1, 2}y_{i, 1}$ (here $x_{1,1}y_{1,1}$ and $x_{1,2}y_{1,2}$ belong to $G^0_1$). This connects all graphs $G^0_1,G_2,\dots ,G_{d}$ into one component without changing any vertex degrees.

Now we connect $G_0$ to each $G^k_1$ for $k=0,1,\dots ,q$ in a similar manner. Clearly, $G_0\cong K_{10p+1}$ contains $p\ge q+1$ independent light edges. Select $q+1$ and call them $u_iv_i$ for $i=0,1,\dots ,q$. Then select a light edge $t_iz_i$ in the i-th copy of $G_1$ and replace the pair of edges $u_iv_i$ and $t_iz_i$ by edges $u_it_i$ and $v_iz_i$.

As before, this connects all copies of $G_1$ to $G_0$ without changing the degree of any vertex.

We have thus constructed a connected graph H with d different odd degrees, a light edges, $b=8a+1$ heavy edges, and a constant weight $\widetilde{w}(x)=10p$ for every vertex x. Blowing it up the same way as in the proof of Theorem 5 concludes this proof.