on edge-balance index sets of l-product of cycles by cycles

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

evenisnandnoddisnandn

nnn

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, thenEBI(Wn) ={0, 2, …, 2i, …, n-2}.

Theorem: If n is odd, thenEBI(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

iCC envv

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 nnmnnmvv

.12

11210

nnmnnmvv

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

43n

3CCn

2n

413n