Numbers of Edges in SupermagicGraphs

Svetlana Drajnova, Jaroslav Ivanco, Andrea Semanicova∗

Institute of Mathematics, P.J.Safarik University, Jesenna 5, 04154 Kosice,Slovakia

Abstract

A graph is called supermagic if it admits a labelling of the edges bypairwise different consecutive integers such that the sum of the labelsof the edges incident with a vertex is independent of the particularvertex. In the paper we establish some bounds for the number ofedges in supermagic graphs.

Keywords: magic graph, supermagic graph, size of graph

1 Introduction

We consider finite undirected graphs without loops, multiple edges and iso-lated vertices. If G is a graph, then V (G) and E(G) stand for the vertex setand edge set of G, respectively.

Let a graph G and a mapping f from E(G) into positive integers be given.The index-mapping of f is the mapping f ? from V (G) into positive integersdefined by

f ?(v) =∑

e∈E(G)

η(v, e)f(e) for every v ∈ V (G),

where η(v, e) is equal to 1 when e is an edge incident with a vertex v, and 0otherwise. An injective mapping f from E(G) into positive integers is calleda magic labelling of G for an index λ if its index-mapping f ? satisfies

f ?(v) = λ for all v ∈ V (G).

∗E-mail address: ivanco@science.upjs.sk, semanic@science.upjs.skThis is a preprint of an article published in J Graph Theory 52: 15-26, 2006 c© Copyright

2005 Wiley Periodicals, Inc., A Wiley Company, http://www.interscience.Wiley.com/.

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2 IM Preprint series A, No. 2/2004

A magic labelling f of G is called a supermagic labelling of G if the set{f(e) : e ∈ E(G)} consists of consecutive positive integers. We say thata graph G is supermagic (magic) if and only if there exists a supermagic(magic) labelling of G. Note that any supermagic regular graph G admitsa supermagic labelling into the set {1, . . . , |E(G)|}.

The concept of magic graphs was introduced by Sedlacek [11]. The regularmagic graphs are characterized in [2]. Two different characterizations of allmagic graphs are given in [9] and [8].

Supermagic graphs were introduced by M. B. Stewart [12]. It is easyto see that the classical concept of a magic square of n2 boxes correspondsto the fact that the complete bipartite graph Kn,n is supermagic for everypositive integer n 6= 2 (see also [12]). Stewart [13] characterized supermagiccomplete graphs. In [6] supermagic regular complete multipartite graphs andsupermagic cubes are characterized. In [7] there are given characterizationsof magic line graphs of general graphs and supermagic line graphs of re-gular bipartite graphs. In [10] and [1] supermagic labellings of the Mobiusladders and two special classes of 4-regular graphs are constructed. Someconstructions of supermagic labellings of various classes of regular graphsare described in [5] and [6]. More comprehensive information on magic andsupermagic graphs can be found in [3].

In this paper we deal with the number of edges in supermagic graphs.

2 Number of edges

For the number of edges in magic graphs it holds

Proposition 1. [14] A connected magic graph with n vertices and ε edges

exists if and only if n = 2 and ε = 1 or n ≥ 5 and 5n4

< ε ≤ n(n−1)2

.

It is easy to see that all components of a magic graph are magic graphsand at most one of them is isomorphic to the complete graph K2. Thus wehave immediately

Corollary 1. A magic graph of order n and size ε exists if and only if n = 2and ε = 1 or n ∈ {5, 6} and 5n

4< ε ≤ n(n−1)

2or n ≥ 7 and 5n−6

4< ε ≤ n(n−1)

2.

Moreover, any magic graph with at most 54n edges contains a component

isomorphic to K2.

The previous assertions imply the following interpolation theorem: LetG1 and G2 be magic graphs of order n. Then there exists a magic graphof order n and size ε for each integer ε satisfying |E(G1)| ≤ ε ≤ |E(G2)|.A similar result for supermagic graphs is not valid.

S. Drajnova, J. Ivanco, A. Semanicova: Numbers of Edges ... 3

Theorem 1. Let d be the greatest common divisor of integers n and ε, andlet n1 = n

d. If n1 and ε are both even, then there exists no supermagic graph

of order n and size ε.

Proof. Let d denote the greatest common divisor of n and ε and let n1 = nd,

ε1 = εd. Suppose that G is a supermagic graph of order n and size ε. Then it

admits a supermagic labelling f : E(G) −→ {a, . . . , a + ε − 1} for an indexλ. It holds

nλ =∑

v∈V (G)

∑e∈E(G)

η(v, e)f(e) = 2∑

e∈E(G)

f(e)

=2[a + . . . + (a + ε− 1)] = (2a + ε− 1)ε.

As ε and n1 are both even, the index λ = (2a+ε−1)εn

= (2a+ε−1)ε1

n1is not

an integer, a contradiction.

For example there exists no supermagic graph of order 8 and size ε ≡2, 4, 6 (mod 8) (i.e., with 10, 12, 14, 18, 20, 22, 26, 28 edges). Thus theproblem to characterize the numbers of edges in supermagic graphs seems tobe difficult.

Let M(n) (m(n)) denote the maximal (minimal) number of edges in asupermagic graph of order n. Evidently, M(n) and m(n) are not defined forn = 1, 3, 4 and M(2) = m(2) = 1. In the next sections we determine M(n)and establish some bounds of m(n) for n ≥ 5.

3 Upper bound

Finding the maximum number M(n) of edges in a supermagic graph is closelyassociated with the characterization of supermagic complete graphs.

Proposition 2. [13] A complete graph of order n is supermagic if and onlyif n = 2 or 5 < n 6≡ 0 (mod 4).

Next we prove that by deleting an edge from a complete graph Kn, n ≥ 6,we obtain a supermagic graph.

Theorem 2. For every positive integer n ≥ 6, the complete graph Kn withoutan edge is supermagic.

Proof. We will consider the following casesA. 6 ≤ n 6≡ 0 (mod 4). By Proposition 2 the complete graph Kn

is supermagic. Thus there exists a supermagic labelling f : E(Kn) −→

4 IM Preprint series A, No. 2/2004

{1, . . . , n(n−1)2

} for an index λ. Let e be an edge of Kn such that f(e) = 1.

Define a labelling g : E(Kn − e) −→ {1, . . . , n(n−1)2

− 1} by

g(e) = f(e)− 1 for all e ∈ E(Kn − e).

Since Kn is an (n− 1)-regular graph we have

g?(v) = f ?(v)− (n− 1) for all v ∈ V (Kn).

Therefore, g is a supermagic labelling of Kn − e.B. 8 ≤ n ≡ 0 (mod 4). Let e be an arbitrary edge of the complete graph

Kn. Denote the vertices of Kn by v1, . . . , vn in such a way that e = vn−1vn.Let G be a subgraph of Kn induced by the set {v1, . . . , vn−2}. The graph

G is isomorphic to Kn−2 and by Proposition 2 there exists a supermagiclabelling f from E(G) into {1, . . . ,

(n−2

2

)}. Clearly, f ?(vi) = (

(n−2

2

)+ 1)n−3

2

for all i, 1 ≤ i ≤ n− 2.Put a positive integer a := 1

4[n3 − 6n2 + 7n + 4] and define a mapping

g : E(Kn − e) −→ {a, . . . , a + n(n−1)2

− 2} by

g(vivj) =

a− 1 + f(vivj) for 1 ≤ j ≤ n− 2 and 1 ≤ i ≤ n− 2,

a +(

n−22

)− 1 + i for j = n− 1 and n− 8 ≥ i ≡ 0, 3 (mod 4),

or j = n− 1 and i = n− 7, n− 5,

or j = n and n− 8 ≥ i ≡ 1, 2 (mod 4),

or j = n and i = n− 6, n− 4, n− 3, n− 2,

a + (n−2)(n+1)2

− i otherwise.

A part of definition of g was inspired by [4]. It is easy to see that themapping g is a bijection and for its index-mapping we get

g?(vi) =1

4(n4 − 6n3 + 9n2 + 4n− 12) for 1 ≤ i ≤ n.

Thus, g is a supermagic labelling and Kn − e is a supermagic graph.

It is not difficult to check that M(5) = 8. Therefore, using previousresults we get the following theorem

Theorem 3. Let n ≥ 5 be a positive integer. Then

M(n) =

8 for n = 5,n(n−1)

2for 6 ≤ n 6≡ 0 (mod 4),

n(n−1)2

− 1 for 8 ≤ n ≡ 0 (mod 4).

S. Drajnova, J. Ivanco, A. Semanicova: Numbers of Edges ... 5

4 Lower bound

In this section we establish some bounds for m(n). The main result is

Theorem 4. Let n ≥ 5 be a positive integer. Then

m(n) ≥ 3n− 1

2−

√3n2 − 2n +

1

4.

Proof. Suppose that G is a supermagic graph of order n with ε = m(n) edges.It admits a supermagic labelling f : E(G) −→ {a, . . . , a+ε−1} for an index

λ =(2a + ε− 1)ε

n. (1)

Let V3 denote the set of vertices of degree at least 3, the cardinality ofthis set is denoted by n3. By n2 denote the number of 2-vertices (i.e., verticesof degree 2). As every vertex of a supermagic graph G has degree at least 2,n = n2 + n3. For the number of edges we have

2ε =∑

v∈V (G)

deg(v) = 2n2 +∑v∈V3

deg(v) ≥ 2n2 + 3n3 = 3n− n2,

thus

ε ≥ 3n

2− n2

2. (2)

If G contains no 2-vertex, then ε ≥ 3n2

and the assertion is satisfied. Sowe can assume that n2 ≥ 1.

In any supermagic graph there exists no edge joining the vertices of degree2, i.e. every vertex of degree 2 is adjacent to two distinct vertices of degreeat least 3. It means all edges incident with the n2 vertices of degree 2 aremutually distinct and their number is 2n2. The sum of the labels of edgesincident with 2-vertices has to be less or equal to the sum of maximal valueswhich can be assigned to any 2n2 edges in the supermagic labelling f

n2λ ≤ (a + ε− 1) + . . . + (a + ε− 2n2) = (2a + 2ε− 2n2 − 1)n2.

As n2 6= 0, by (1) we get

(2a + ε− 1)ε

n= λ ≤ 2a + 2ε− 2n2 − 1.

From this inequality we yield

2n2 ≤ 2a(1− ε

n

)− ε2

n+

ε

n+ 2ε− 1. (3)

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Any supermagic graph of order n > 2 has more edges than vertices and so1− ε

n< 0. Since a ≥ 1,

2a(1− ε

n

)≤ 2

(1− ε

n

).

Appointing this in (3) we have

2n2 ≤ 1− ε2

n− ε

n+ 2ε.

From (2) we get n2 ≥ 3n− 2ε, and combining it we get

0 ≥ ε2 + ε(1− 6n) + 6n2 − n.

This inequality immediately implies

ε ≥ 3n− 1

2−

√3n2 − 2n +

1

4,

which is the desired lower bound for m(n).

The previous theorem implies m(n) ≥ d(3 −√

3 )ne. It seems that it isnot possible to reach this bound. The best bound we know is 9n

7for n = 14,

42, 70. (We suppose that there exists an infinite family of supermagic graphsof size 9n

7.) The corresponding supermagic graph for n = 14 is illustrated on

Figure 1.

Figure 1. Figure 2.

Now we deal with an upper bound for m(n).Let d and p be non-negative integers such that k := d + p ≥ 2. By Md,p

denote the graph with the vertex set {u1, . . . , u2k, v1, . . . , vp} and the edgeset consisting of edges

e1 = u1u2, e2 = u2u3, . . . , e2k−1 = u2k−1u2k, e2k = u2ku1,

f1 = u1uk+1, f2 = u2uk+2, . . . , fd = uduk+d,

l1 = v1ud+1, l2 = v2ud+2, . . . , lp = vpud+p,

r1 = v1u2k, r2 = v2u2k−1, . . . , rp = vpu2k−p+1.

S. Drajnova, J. Ivanco, A. Semanicova: Numbers of Edges ... 7

Lemma 1. The graph Md,p is supermagic for every odd positive integer d.

Proof. For every odd positive integer d there exists a positive integer s suchthat d = 2s− 1. Put a = p + s. By g denote the mapping from the edge setof Md,p to the set {a, . . . , a + |E(Md,p)| − 1} defined by

g(e) =

a + i−12

for e = ei and i = 1, 3, . . . , 2k − 1,

a + k + i−1−d2

for e = ei and i = d + 1, d + 3, . . . , 2k,

a + k + 2k+i−1−d2

for e = ei and i = 2, 4, . . . , d− 1,

a + 2k − 1 + i for e = ri and i = 1, 2, . . . , p,

a + 3k − i for e = fi and i = 1, 2, . . . , d,

a + 3k + p− i for e = li and i = 1, 2, . . . , p.

It is easy to see that g is a bijection and its index-mapping g? satisfies

g?(v) = 8p + 12s− 6 for every v ∈ V (Md,p).

Thus g is a supermagic labelling and Md,p is a supermagic graph.

Lemma 2. For every positive integer k ≥ 2 there exists a supermagic graphof order 3k and size 4k.

Proof. Consider a cycle C2k with vertices u1, u2, . . . , u2k and edges e1 = u1u2,. . . , e2k−1 = u2k−1u2k, e2k = u2ku1. Let f be a mapping from E(C2k) to theset of positive integers defined by

f(e) =

k − 1 + i−12

for i = 1, 3, . . . , k, k + 4, . . . , 2k − 1,

4k − 2 for i = 2,

k − 1 + i2

for i = k + 1,

2k − 2 + i−32

for i = k + 2,

2k − 2 + i−22

for i = 4, 6, . . . , k − 1, k + 3, . . . , 2k,

for k odd, and

f(e) =

k − 1 + i−12

for i = 1, 3, k + 1,

4k − 2 for i = 2,

2k − 2 + i−22

for i = k + 2, 2k,

2k − i2

for i = 4, 6, . . . , k, k + 4, . . . , 2k − 2,

3k − 3− i−32

for i = 5, 7 . . . , k − 1, k + 3, . . . , 2k − 1,

for k even.

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Let Sk be a graph with vertex set V (C2k) ∪ {v1, . . . , vk} and edge setE(C2k) ∪

⋃ki=1{viui1, viui2}, where ui1, ui2 are vertices of C2k such that

f ?(ui1) = 3k − 3 + i and f ?(ui2) = 5k − 1− i.

Consider a mapping g : E(Sk) −→ {k − 1, . . . , 5k − 2} defined by

g(e) =

f(e) for e ∈ E(C2k),

5k − 1− i for e = viui1,

3k − 3 + i for e = viui2.

It is easy to check that g is a bijection. Moreover its index-mapping g?

satisfiesg?(v) = 8k − 4 for every v ∈ V (Sk).

Thus g is a supermagic labelling and Sk is a desired supermagic graph.

On Figure 3 there are illustrated the graphs M3,0, M1,3 and S4 and theirsupermagic labellings.

Figure 3.

From previous lemmas we immediately obtain

Theorem 5. Let n ≥ 5 be a positive integer. Then

m(n) ≤

4n3

for n ≡ 0 (mod 3),4n3

+ 53

for n ≡ 1 (mod 3),4n3

+ 13

for n ≡ 2 (mod 3).

Proof. It is obvious that the graphs Sn/3 (for n ≡ 0 (mod 3)), the graph onFigure 2 and the graphs M5,(n−10)/3 (for n ≡ 1 (mod 3)) and M1,(n−2)/3 (forn ≡ 2 (mod 3)) are supermagic graphs of order n with the required numberof edges.

S. Drajnova, J. Ivanco, A. Semanicova: Numbers of Edges ... 9

We conclude this paper with a determination of m(n) for prime numbern, but first we prove the following auxiliary result.

Lemma 3. Let G be a supermagic graph of order n ≥ 5 and size ε. If thegreatest common divisor of the numbers n and ε is 1, then ε > 4n

3. Moreover,

if ε is an even integer, then ε > 4n+23

.

Proof. Consider a supermagic labelling f : E(G) −→ {a, . . . , a + ε − 1} for

an index λ = (2a+ε−1)εn

. As n and ε are coprime and λ is a positive integer,then γ := 2a+ε−1

nis also a positive integer. From this we can express

λ = γε (4)

a = 12(γn− ε + 1) (5)

Let n2 denote the number of 2-vertices in G. Values of the edges (mutuallydistinct) incident with the 2-vertices are at most a + ε − 1, a + ε − 2, . . . ,a + ε− 2n2. Thus

λ ≤ (a + ε− 1) + (a + ε− 2n2) = 2a + 2ε− 2n2 − 1.

Appointing (4), (5) in this inequality we get

n2 ≤ 12((1− γ)ε + γn).

As in the proof of Theorem 4 we get (2), and then n2 ≥ 3n − 2ε. Com-bining it we have

(5− γ)ε ≥ (6− γ)n. (6)

Since γ is a positive integer it is sufficient to consider the following cases.A. γ ≥ 5. According to Corollary 1 it yields

ε >5n

4=

(1 +

1

4

)n ≥

(1 +

1

γ − 1

)n =

γ

γ − 1n.

Therefore, ε(γ − 1) > γn > γn− 2. Hence ε− 2 < γ(ε− n) = 2a+ε−1n

(ε− n).After some manipulation we have

(a + ε− 1) + (a + ε− 2) <(2a + ε− 1)ε

n= λ.

It means n2 = 0, and then ε ≥ 3n2

> 4n+23

.

B. γ ∈ {3, 4}. By (6) we get ε ≥ 6−γ5−γ

n ≥ 3n2

> 4n+23

.

C. γ = 2. According to (6) we have ε ≥ 43n. Moreover, ε 6= 4

3n. In

opposite case we get ε = 4k and n = 3k for some integer k > 1. It meansthe greatest common divisor of n and ε is k, a contradiction. Note also that(5) implies ε = 2(n− a) + 1, therefore ε is an odd integer in this case.

D. γ = 1. From (5) we get ε = n− 2a + 1 < n, contrary to Corollary 1.Thus, this case is impossible.

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If n is the prime number, then using Theorem 5 and Lemma 3 we imme-diately obtain

Theorem 6. Let n ≥ 5 be a prime number. Then

m(n) =

{4n3

+ 53

for n ≡ 1 (mod 3),4n3

+ 13

for n ≡ 2 (mod 3).

Acknowledgement. Support of Slovak VEGA Grant 1/0424/03 is a-cknowledged.

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