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On k-Edge -magic Cubic Graphs

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On k-Edge -magic Cubic Graphs. Sin-Min Lee , San Jose State University Hsin-hao Su *, Stonehill College Yung-Chin Wang , Tzu-Hui Institute of Technology 24th MCCCC At Illinois State University September 11, 2010. Supermagic Graphs. - PowerPoint PPT Presentation
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On k-Edge-magic Cubic Graphs Sin-Min Lee, San Jose State University Hsin-hao Su*, Stonehill College Yung-Chin Wang, Tzu-Hui Institute of Technology 24th MCCCC At Illinois State University September 11, 2010
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On k-Edge-magic Cubic Graphs

Sin-Min Lee, San Jose State University

Hsin-hao Su*, Stonehill College

Yung-Chin Wang, Tzu-Hui Institute of Technology

24th MCCCCAt

Illinois State University

September 11, 2010

Supermagic Graphs

For a (p,q)-graph, in 1966, Stewart defined that a graph labeling is supermagic iff the edges are labeled 1, 2, 3, …, q so that the vertex sums are a constant.

k-Edge-Magic Graphs

A (p,q)-graph G is called k-edge-magic (in short k-EM) if there is an edge labeling l: E(G) {k, k+1, …, k+q-1} such that for each vertex v, the sum of the labels of the edges incident with v are all equal to the same constant modulo p; i.e., l+(v) = c for some fixed c in Zp.

If k =1, then G is said to be edge-magic.

Examples: 1-Edge-Magic

The following maximal outerplanar graphs with 6 vertices are 1-EM.

Examples: 1-Edge-Magic

In general, G may admits more than one labeling to become a k-edge-magic graph with different vertex sums.

Examples: k-Edge-Magic

In general, G may admits more than one labeling to become a k-edge-magic graph.

Necessary Condition

A necessary condition for a (p,q)-graph G to be k-edge-magic is

Proof: The sum of all edges is Every edge is counted twice in the vertex

sums.

pkqq mod012

2

1 qkkq

k-Edge-Magic is periodic

Theorem: If a (p,q)-graph G is k-edge-magic then it is pt+k-edge-magic for all t ≥ 0 .

Cubic Graphs

Definition: 3-regular (p,q)-graph is called a cubic graph.

The relationship between p and q is

Since q is an integer, p must be even.

2

3pq

One for All

Theorem: If a cubic graph is k-edge-magic, then it is k-edge-magic for all k.

Proof: Since every vertex is of degree 3, by adding or

subtracting 1 to each adjacent edge, the vertex sum remains the same.

Examples: Complete Bipartite

The complete bipartite graph K3,3 is k-edge-magic for all k.

Not Order 4s

Theorem: A cubic graph with order 4s is not k-edge-magic for all k.

Proof: The number of edges is 6s. The necessary condition implies

It is impossible for all k.

.4mod212660 sskss

Möbius Ladders

The concept of Möbius ladder was introduced by Guy and Harry in 1967.

It is a cubic circulant graph with an even number n of vertices, formed from an n-cycle by adding edges (called “rungs”) connecting opposite pairs of vertices in the cycle.

Möbius Ladders A möbius ladder ML(2n)

with the vertices denoted by a1, a2, …, a2n. The edges are then {a1, a2}, {a2, a3}, … {a2n, a1}, {a1, an+1}, {a2, an+2}, … , {an, a2n}.

Labeling Idea

Splits all edges into two subsets. The first subset contains all the edges of C2n. The second set contains all middle edges, which forms a perfect matching.

Construct a graceful labeling for the first subset, i.e., an arithmetic progression. The rest numbers also form an arithmetic progression.

Labeling Method 1

Divides the numbers into three subsets: {0, 1, 2, 3, …, 2k = n-1}, {2n, 2n+1, 2n+2, 2n+3, …, 2n+2k = 3n-1}, {n, n+1, n+2, n+3, …, n+2k = 2n-1}.

Use the first two subsets to label C2n by the following sequence: k+1, 1, k+2, 2, k+3, 3, …, 2k, k, 0, k+1, 1, k+2, 2, k+3, 3, …, 2k, k, 0.

Example of Method 1

Labeling Method 2

We label the edges by 1, 1, 2, 2, 3, 3, …, k+1, k+1, n+k+2, k+2, n+k+3, k+3, …, 2n, 2k+1.

Label the rest numbers, k+2, k+3, …, n+k+1 to the edges in the middle.

Example of Method 2

Cylinder Graphs Theorem (Lee, Pigg, Cox; 1994): The

cylinder graph CnxP2 is a 1-edge-magic graph if n is odd.

Cylinder Graphs Examples

Generalized Petersen Graphs The generalized Petersen graphs P(n,k) were

first studied by Bannai and Coxeter. P(n,k) is the graph with vertices {vi, ui : 0 ≤ i

≤ n-1} and edges {vivi+1, viui, uiui+k}, where subscripts modulo n and k.

Theorem: The generalized Petersen graph P(n,t) is a k-edge-magic graph for all k if n is odd.

Gen. Petersen Graph Ex.

Order 6

Theorem: A cubic graph with order 6 is k-edge-magic for all k.

Order 10

Order 14: Transformation

Order 14: Transformation

Conjecture

Conjecture: A cubic graph with order 4s+2 is k-edge-magic for all k.

With the previous examples, this is a reasonable extension of a conjecture by Lee, Pigg, Cox in 1994.

Mod(m)-Edge-Magic Graphs

A (p,q)-graph G is called Mod(m)-edge-magic (in short Mod(m)-EM) if there is an edge labeling l: E(G) {1,2,…,q} such that for each vertex v, the sum of the labels of the edges incident with v are all equal to the same constant modulo m; i.e., l+(v) = c for some fixed c in Zm.

Relationship between EM

Theorem: For a graph with order p, if it is 1-edge-magic, then it is mod(m)-edge-magic for m to be a factor of p.

Proof: Since m is a factor of p, the constant sum in Zp

remains constant in Zm.

Counterexample

Proof

Since it is mod(5)-edge-magic, we have relations as followings: a + b = l + m, (1) b + c = k + l, (2) h + i = f + e, (3)

From relations (1) and (2), we have a + k = c + m. (4)

Proof (continued)

Therefore we have g = d. Then we have a new relation h + f = i + e. (5)

From relations (3) and (5), we have i = f.

Then we have h = e, and g = j = d.

Proof (continued)

Without losing generality, we say d = 0, e = 1 and f = 4.

From relation (4), we have a = 1, k = 4, c = 2, m = 3 or a = 1, k = 4, c = 3, m = 2 or a = 2, k = 3, c = 1, m = 4 or a = 2, k = 3, c = 4, m = 1.

Here, we already run out of 1 and 4 and only 2 and 3 left in the set.

Proof (continued)

For the case a = 1, we have b = 2 and n = 2. If forces that o = 3 and l = 3. But m = 3 or 2 can’t make the sum on v9 equal to 0. This is a contradiction.

With the same argument, we can show that all the possibilities can’t be true.

Therefore it is not mod(5)-edge-magic.

Future Problems

Do we have just a few counterexamples? Any better necessary condition? Possible sufficient conditions?


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