on edge-balance index sets of l-product of cycles by cycles daniel bouchard, stonehill college...

On Edge-Balance Index Sets of L-Product of Cycles by Cycles

Daniel Bouchard, Stonehill College

Patrick Clark, Stonehill College

Hsin-hao Su, Stonehill College

(Funded by Stonehill Undergraduate Research Experience)

6th IWOGL 2010University of Minnesota, Duluth

October 22, 2010

Edge Labeling

A labeling f : E(G) Z2 induces a vertex partial labeling f+ : V(G) A defined by f+(x) = 0 if the edge labeling of f(x,y) is 0 more

than 1; f+(x) = 1 if the edge labeling of f(x,y) is 1 more

than 0; f+(x) is not defined if the number of edge

labeled by 0 is equal to the number of edge labeled by 1.

Example : nK2

EBI(nK2 ) is {0} if n is even and {2}if n is odd.

Definition of Edge-balance

Definition: A labeling f of a graph G is said to be edge-friendly if | ef(0) ef(1) | 1.

Definition: The edge-balance index set of the graph G, EBI(G), is defined as

{|vf(0) – vf(1)| : the edge labeling f is edge-friendly.}

Examples

Example : Pn

Lee, Tao and Lo[1] showed that

evenisnandn

oddisnandn

n

n

n

PEBI n

6}2,1,0{

5}1,0{

4}2,1{

3}0{

2}2{

)(

[1] S-M. Lee, S.P.B. Lo, M.F. Tao, On Edge-Balance Index Sets of Some Trees, manuscript.

Wheels

The wheel graph Wn = N1 +Cn-1 where V(Wn) = {c0} {c1,…,cn-1} and E(Wn) = {(c0,ci): i = 1, …, n-1} E(Cn-1).

W5W6

Edge Balance Index Set of Wheels

Chopra, Lee ans Su[2] proved: Theorem: If n is even, then

EBI(Wn) ={0, 2, …, 2i, …, n-2}. Theorem: If n is odd, then

EBI(Wn) = {1, 3, …, 2i+1, …, n-2}

{0, 1, 2, …, (n-1)/2}.

[2] D. Chopra, S-M. Lee, H-H. Su, On Edge-Balance Index Sets of Wheels, International Journal of Contemporary Mathematical Sciences 5 (2010), no. 53, 2605-2620.

EBI(W6) = {0,2,4}

|v(0)-v(1)|= 0 |v(0)-v(1)|= 2 |v(0)-v(1)|= 4

EBI(W5) = {0,1,2,3}

|v(0)-v(1)|= 0 |v(0)-v(1)|= 1 |v(0)-v(1)|= 2 |v(0)-v(1)|= 3

A Lot of Numbers are Missing

EBI(W7) ={0, 1, 2, 3, 5}.

EBI(W9) ={0, 1, 2, 3, 4, 5, 7}.

EBI(W11) ={0, 1, 2, 3, 4, 5, 7, 9}.

EBI(W13) ={0, 1, 2, 3, 4, 5, 6, 7, 9, 11}.

EBI(W15) ={0, 1, 2, 3, 4, 5, 6, 7, 9, 11, 13}.

EBI(W17) ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 15}.

L-Product

Let H be a connected graph with a distinguished vertex s.

Construct a new graph G ×L (H,s) as follows: Take |V(G)| copies of (H,s), and identify each vertex of G with s of a single copy of H. We call the resulting graph the L-product of G and (H,s).

L-Product Example

15 StC L 25 StC L

46 StC L

Generalized L-Product

More generally, the n copies of graphs to be identified with the vertices of G need not be identical.

Generalized L-Product

Let Gph* be the family of pairs (H,s), where H is a connected graph with a distinguished vertex s. For any graph G and any mapping : V(G) Gph* we construct the generalized L-product of G and , denoted G ×L , by identifying each v V(G) with s of the respective (v).

L-Product of Cycles by Cycles

35 CC L45 CC L

Notations

Let f be an edge labeling of a cycle Cn.

We denote the number of edges of Cn which are labeled by 0 and 1 by f+ by eC(0) and eC(1), respectively.

We denote the number of vertices on Cn which are labeled by 0, 1, and not labeled by the restricted f+ by vC(0), vC(1), and vC(x), respectively

Proposition

(Chopra, Lee and Su[2]) In a cycle Cn with a labeling f (not necessary edge friendly), assume that eC(0) > eC(1) > 1 and vC(x) = 2k > 0. Then we have

vC(1) = eC(1) - k.

and

vC(0) = n - eC(1) - k. [2] D. Chopra, S-M. Lee, H-H. Su, On Edge-Balance Index Sets of Wheels, International Journal of Contemporary Mathematical Sciences 5 (2010), no. 53, 2605-2620.

EBI of Cycles

Lemma: For an edge labeling f (not necessary edge friendly) of a finite disjoint union of cycles , we have

Note that this EBI of the disjoint union of cycles depends on the number of 1-edges only, not how you label them.

i

ini

C

.1210 Ci

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Maximal Edge-balance Index

Theorem: The highest edge-balance index of when m ≥ 5 is n if m is odd or n is even; n+1 if n is odd and m is even.

mn CC

Proof Idea

By the previous lemma, to maximize EBI, eC(1) has to be as small as it can be.

Thus, if we label all edges in Cn 1, it gives us the best chance to find the maximal EBI.

Thus, might yield the maximal EBI.

neeC 11

Less 1 inside, Higher EBI

.12

10

Ci

i

CC

en

vv

Proof

The number of edges of is

If n is even or m is odd, then

If n is odd and m is even, then

(Note that w.l.o.g we assume that .)

mn CC .1 mnmnn

.

2

11

mne

.

2

111

mne

10 ee

Proof (continued)

Since the outer cycles of contain all vertices, the EBI calculated by the previous lemma could be our highest EBI.

We already label all edges in Cn by 1. Thus, to not alter the label of the vertex adjuncts to a outer cycle, we have to have all two edges of outer cycle labeled by 1 too.

mn CC

Degree 4 Vertices

Proof (continued)

The above labeling requires n 1-edges for Cn and 2n 1-edges for outer cycles.

In order to have at least 3n 1-edges, the number of edges of must be greater or equal to 6n.

Thus, implies m must be greater or equal to 5.

mn CC

nmnmnn 61

Keep Degree 4 Unchanged

According to the formula

we can label the rest in any way without changing EBI.

,12

10

Ci

i

CC

en

vv

Proof (continued)

The highest EBI of is If n is even or m is odd, then

If n is odd and m is even, then

mn CC

.

2

1210 nn

mnnmvv

.1

2

11210

nn

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Switching Edges

By switching a 0-edge with an 1-edge adjacent to the inner cycle, we reduces the EBI by 1.

Switching Edges

By switching a 0-edge with an 1-edge adjacent to the inner cycle, we reduces the EBI by 1.

Main Results

Theorem: EBI( ) when m ≥ 5 is {0,1,2,…, n} if m is odd or n is even; {0,1,2,…, n+1} if n is odd and m is even.

mn CC

Proof

While creating an edge-labeling to yield the highest EBI, we label all edges adjacent to the inner cycle vertex 1.

Since the formula in the lemma says that the EBI of all outer cycles depends only on the number of 1-edges, we can label the edges adjacent to the edges adjacent the inner cycle vertex 0 without alter the EBI.

Special Edge-labeling to Yield the Highest EBI

According to the formula

we can label the rest in any way without changing EBI.

,12

10

Ci

i

CC

en

vv

Proof (continued)

Each outer cycle can reduce the EBI by 1 by switching edges.

Since there are n outer cycles, we can reduce the EBI by 1 n times.

Therefore, we have the EBI set contains {0,1,2,…, n} if m is odd or n is even; {1,2,…, n+1} if n is odd and m is even.

Proof (continued)

When n is odd and m is even, a special labeling like the one on the right produces an EBI 0.

When m = 3 or 4

Theorem: EBI( ) is {0, 1, 2, …, } if n is even.

{0, 1, 2, …, } if n is odd.

Theorem: EBI( ) is

{0, 1, 2, …, }.

4CCn

4

3n

3CCn

2n

4

13n