on the edge-balance index sets of fans and broken fans dharam chopra*, wichita state university...

On The Edge-Balance Index Sets of Fans and Broken Fans

Dharam Chopra*, Wichita State University

Sin-Min Lee, San Jose State University

Hsin-hao Su, Stonehill College

40th SEICCGTCAt

Florida Atlantic University

March 4, 2009

Example : nK2

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

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

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.}

Example : Pn

Lee, Tao and Lo showed that

evenisnandn

oddisnandn

n

n

n

PEBI n

6}2,1,0{

5}1,0{

4}2,1{

3}0{

2}2{

)(

Example : Pn

Fans

A Fan graph is F1,n = N1 +Pn where V(F1,n) = {c} {v1,…,vn} and E(F1,n) = {(c,vi): i=1,…,n} E(Pn).

Turn a Fan into a Wheel

If we add an extra edge (v1,vn) into F1,n, then it becomes W1+n.

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

Idea

Because of the nature of a wheel, we separate a wheel into a cycle and a star.

W5W6

Separation

Because of the nature of a wheel, we separate a wheel into a cycle and a star.

W5=C5+St5 W6=C6+St6

Vertex of Order 2

For a vertex of order 2 with an edge-labeling (not necessary friendly), it can only be labeled as one of the following three cases.

Add an Edge to an Vertex of Order 2 If the vertex was labeled, then the label of the vertex

is not changed. If the vertex was not labeled then the label of the

vertex is labeled by the same as the adding edge.

Three Cases on Cn-1

It’s easy to see that there are three possible cases: eC(0) > eC (1) > 1, or, eC(0) = eC(1) > 1 only when n is odd

(which implies eC(0) = eC(1) = (n-1)/2,) or,

eC(1) = 0 (which implies that eC(0)=n-1.)

Case 1 Examples

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

W8 W8 W7

Case 2 Examples

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

W7 W11 W11

Case 3 Examples

|v(0)-v(1)|= 6 |v(0)-v(1)|= 5 |v(0)-v(1)|= 11

W8 W7 W13

Edge-Balance Index Set of Wheels

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

Theorem (Chopra, Lee, Su): If n is odd, then EBI(Wn) = {1,3,…,2i+1,…,n-2}{0,2,4,…,(n-1)/2}.

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

Equivalence of Friendly Labelings

For an edge friendly labeling, since F1,n has 2n-1 edges, to calculate the edge balance index, we may assume that v(0)=v(1)+1.

Thus, if we label the extra edge 1, then we have an edge friendly labeling for W1+n.

Note that for an edge friendly labeling of W1+n, if we remove an edge labeled 1 from its cycle part, we get an F1,n with an edge friendly labeling.

Thus, there is a 1-1 correspondence between the edge friendly labelings of F1,n and the edge friendly labelings of W1+n.

V1 and Vn

There are three different pairs of labeling of v1 and vn:

v1 is labeled by 0 and vn is lebeled by 1 and vice versa.

Both v1 and vn are labeled by 0.

Both v1 and vn are labeled by 1.

Since case 2 and 3 requires at least 4 edges labeled by the same value, it only happens when 2n-1>7, that is, n>4.

Changes of the Edge-Balance Index

Because the extra edge is labeled by 1, if v1 or vn is labeled 0, then it must have another two 0-edges attached. Thus, it remains labeled 0 in F1,n after removing the extra edge.

If v1 or vn is labeled 1, then it could either remain labeled 1 if there are another two 1-edges attached or become unlabeled. The first case keeps the balance index the same. But, the second one reduces the amount of 1-vertices 1.

Edge-Balance Index Set of Fans

Theorem: EBI(F1,n) {0,1,2} if n=3 {0,1,…, n-2} for n> 4.

EBI(F1,3) ={0,1,2}

EBI(F1,4) ={0,1,2}

EBI(F1,5) ={0,1,2,3}

EBI(F1,6) ={0,1,2,3}

Proof

By the 1-1 correspondence between the edge friendly labelings of W1+n and F1,n, the edge balance calculation tells us that in W1+n, -(n-1) < v(0)-v(1) < n-3 where v(0)-v(1) must be odd if n is even and v(0)-v(1) must be even if n is odd.

Note that v(0)-v(1) = n-3 only if k=1 and h=eC(1)>1 and v(0)-v(1) = -(n-1) only if k=0 and h=0.

Proof (continues)

We first consider the extreme case v(0)-v(1) = n-3 and v(1) is reduced by 2. In this case, we have k=1. Thus, v1 and vn are the only two vertices which are unlabeled and both are converted to 1. But, since there are h>1 unlabeled vertices which are converted to 0, this cannot happen.

Proof (continues)

The next extreme case is v(0)-v(1) = -(n-1) and v(1) is reduced by 0. In this case, since h=0, we know that all vC(x) unlabeled vertices are converted to 1. Also, since k=0 and eC (1)=n/2 or (n-1)/2 which depends on n is even or odd, respectively, we have vC(x)=2eC (1)=n if n is even and n-1 if n is odd. This cannot happen since we assume that eC (0)>eC (1) and the extra edge occupies the extra 1.

Proof (continues)

Therefore, we conclude that, in W1+n, -(n-2) < v(0)-v(1) < n-2 where v(0)-v(1) must be odd if n is even and v(0)-v(1) must be even if n is odd.

For n=3, there is only case 1. So, v(1) can only be reduced by 0 or 1. Thus, the inequality becomes -(n-1) < v(0)-v(1) < n-1. Thus, the EBI= {0,1,2}.

Proof (continues)

For n>4, since all the cases can only reduce v(1) by 0, 1 or 2, the inequality becomes -(n-2) < v(0)-v(1) < n-2.

Also, it is easy to construct a friendly labeling that fits one of the above three cases. So, any integer between -(n-2) and n-2 can be the result of v(0)-v(1). Thus, the EBI= {0,1,…, n-2}

Edge Balance Index Set of Broken Fans BF(a,b)

Theorem: EBI(BF(a,b)) {0,1,2,…, a+b-4} for 2 <a < b

EBI(BF(2,2)) ={0}

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

EBI(BF(2,4)) ={0,1,2}

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

EBI(BF(3,6)) ={0,1,2,3,4,5}

St(n)

Theorem: The edge-balance index set of the star St(n) is {0} if n is even. {2} if n is odd.