On Edge-Balance Index Sets of L-Product of Cycles by Cycle - PowerPoint Presentation

Slide1

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 2010

University of Minnesota, Duluth

October 22, 2010Slide2

Edge

Labeling

A labeling

f

: E(G)



Z

2

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 label

ed by

0 is equal to the number of edge labe

led by 1.Slide3

Example : n

K2

EBI(

n

K

2

) is {0} if n is even and {2}if n is odd. Slide4

Definition of E

dge-balance

Definition

:

A labeling

f of a graph G is said to be edge-friendly if | ef(0)



e

f

(1) |



1.

Definition

:

The

edge-balance index set

of the graph G, EBI(G), is defined as

{|v

f

(0) – v

f

(1)| : the edge labeling

f

is edge-friendly.}Slide5

ExamplesSlide6

Example : Pn

Lee, Tao and Lo

[1]

showed that

[1] S-M. Lee, S.P.B. Lo, M.F. Tao,

On Edge-Balance Index Sets of Some Trees

, manuscript.Slide7

Wheels

The wheel graph

W

n

= N

1

+Cn-1 where V(

W

n

) = {c

0

}



{c

1

,…,c

n

-1

} and E(

Wn) = {(c0,ci): i = 1, …, n-1} E(Cn-1).

W5

W

6Slide8

Edge Balance Index Set of Wheels

Chopra, Lee ans Su

[2]

proved:

Theorem

:

If n is even, thenEBI(Wn

) ={0, 2, …, 2

i

, …,

n

-2}.

Theorem

:

If

n

is odd, then

EBI(W

n) = {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.Slide9

EBI(W6

) = {0,2,4}

|v(0)-v(1)|= 0 |v(0)-v(1)|= 2 |v(0)-v(1)|= 4Slide10

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)|= 3Slide11

A Lot of Numbers are Missing

EBI(W7

) ={0, 1, 2, 3, 5}.

EBI(W

9

) ={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(W

15

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

EBI(W

17

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

Slide12

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

). Slide13

L-Product ExampleSlide14

Generalized L-Product

More generally, the

n

copies of graphs to be identified with the vertices of G need not be identical.Slide15

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).Slide16

L-Product of Cycles by CyclesSlide17

Notations

Let f

be an edge labeling of a cycle C

n

.

We denote the number of edges of C

n which are labeled by 0 and 1 by f+ by eC

(0) and

e

C

(1), respectively.

We denote the number of vertices on C

n

which are labeled by 0, 1, and not labeled by the restricted

f

+

by

v

C

(0), vC(1), and vC(x), respectivelySlide18

Proposition

(Chopra

, Lee and Su

[2]

) In a cycle

C

n with a labeling f (not necessary edge friendly), assume that eC(0)

>

e

C

(1)

>

1 and

v

C

(x) = 2

k

> 0. Then we have

vC(1) = eC(1) - k. andvC(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.Slide19

EBI of Cycles

Lemma:

For an e

dge 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.Slide20

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.Slide21

Proof Idea

By the previous lemma, to maximize EBI, e

C

(1) has to be as small as it can be.

Thus, if we label all edges in C

n

1, it gives us the best chance to find the maximal EBI.Thus, might yield the maximal EBI.Slide22

Less 1 inside, Higher EBISlide23

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 .)Slide24

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 C

n

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.Slide25

Degree 4 VerticesSlide26

Proof (continued)

The above labeling requires n

1-edges for

C

n

and 2

n 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. Slide27

Keep Degree 4 Unchanged

According to the formula

we can label the rest in any way without changing EBI.Slide28

Proof (continued)

The highest EBI of is If

n

is even or

m

is odd, then

If n is odd and m

is even, thenSlide29

Switching Edges

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

Slide30

Switching Edges

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

Slide31

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.Slide32

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.Slide33

Special Edge-labeling to Yield the Highest EBI

According to the formula

we can label the rest in any way without changing EBI.Slide34

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.Slide35

Proof (continued)

When n

is odd and

m

is even, a special labeling like the one on the right produces an EBI 0.Slide36

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, …,

}.