On Balance Index Sets of Trees of Diameter Four

Sin-Min Lee, San Jose State University

Hsin-hao Su, Stonehill College

Yung-Chin Wang, Tzu-Hui Institute of Technology

MCCCC 23At

Rochester Institute of Technology

October 4, 2009

Example : nK2

BI(G) is {0,1,2}.

Vertex Labeling

A labeling f : V(G) {0,1} induces an edge partial labeling f*: E(G) {0,1} defined by

f*(uv) = 0 if f(u) = f(v) = 0,

f*(uv) = 1 if f(u) = f(v) = 1. Note that if f(u) ≠ f(v), then the edge uv

is not labeled by f*.

Definition of Balance Index Set

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

Definition: The balance index set of the graph G, EBI(G), is defined as {|ef(0) – ef(1)| : the vertex labeling f is friendly.}

Definition: A labeling f of a graph G is said to be balanced if | vf(0) vf(1) | 1 and | ef(0) ef(1) | 1

Example : Cycles

The balance index set of cycles is

.1

.0)(

1

1

1 oddisnif

evenisnifCBI k

ii

k

iik

i

ini

Example : Stars

The balance index set of stars St(n) is

.2,1

.12)(

evenisknifkk

oddisknifknStBI

Example : Cn(t)

The balance index set of Cn(t) is

.2,1,0

.1,0)(

oddisnif

evenisniftCBI n

Example : Cn(t)

The balance index set of C4(3) is

Notations For i = 0 or 1, let

vf(i) = |{v V(G) | f(v) = i}|and

ef(i) = |{e E(G) | f*(e) = i}|. We also denote ef(x) to be the subset of

E(G) containing all the unlabeled edges. When the text is clear, we omit the

subscript for convenience

Lemma

For a vertex labeling f (not necessary friendly), we have three equations:

0

deg02vv

vxee

1

deg12vv

vxee

10

degdegdeg)(2vvvvGVv

vvvGE

Corollary

For any friendly vertex labeling f,

10

degdeg2

110

vvvv

vvee

Caterpillar CT(a,b,c)

For CT(a,b,c), where a+b+c is odd, the balance index set is { |½(a+b+c+1)|, |½(a+b-c+1)|, |½(a-b+c-1)|, |½(-a+b+c+1)| }.

For CT(a,b,c), where a+b+c is even, the balance indexes is { |½(a+b+c+2)|, |½(a+b-c+2)|, |½(a-b+c)|, |½(-a+b+c+2)|, |½(a-b-c)|, |½(-a+b-c+2)|, |½(-a-b+c)|, |½(-a-b-c)| }

Proof (Odd) When a+b+c is odd, the number of vertices

of CT(a,b,c) is equal to a+b+c+3 which is even. Let a+b+c+3=2M. For a friendly labeling, there are M vertices labeled 0 and M vertices labeled 1.

We name the three vertices on the spine, va, vb, and vc. In CT(a,b,c), we have a+b+c degree 1 vertices. The degrees of va, vb, and vc are a+1, b+2, and c+1, respectively.

Calculation (Odd)

We first consider the case that va, vb, and vc are all labeled 0. Then there are M-3 end-vertices labeled 0 and M end-vertices labeled 1.

12

1

12132

1

degdeg2

1

10

10

cba

McbaM

vv

ee

vvvv

Rest Cases (Odd)

Label va Label vb Label vc # of deg 1 0-vertices# of deg 1 1-

vertices Balance Index

0 0 1 M-2 M-1 ½(a+b-c+1)

0 1 0 M-2 M-1 ½(a-b+c-1)

1 0 0 M-2 M-1 ½(-a+b+c+1)

0 1 1 M-1 M-2 ½(a-b-c-1)

1 0 1 M-1 M-2 ½(-a+b-c+1)

1 1 0 M-1 M-2 ½(-a-b+c-1)

1 1 1 M M-3 ½(-a-b-c-1)

Example: CT(1,1,3)

Example: CT(1,1,3)

Proof (Even) When a+b+c is even, the number of

vertices of CT(a,b,c) is equal to a+b+c+3 which is odd. Let a+b+c+3=2M+1. For a friendly labeling, without loss of generality, there are M+1 vertices labeled 0 and M vertices labeled 1.

We name the three vertices on the spine, va, vb, and vc. In CT(a,b,c), we have a+b+c degree 1 vertices. The degrees of va, vb, and vc are a+1, b+2, and c+1, respectively.

Calculation (Even)

We first consider the case that va, vb, and vc are all labeled 0. Then there are M-2 end-vertices labeled 0 and M end-vertices labeled 1.

22

1

12122

1

degdeg2

1

10

10

cba

McbaM

vv

ee

vvvv

Rest Cases (Even)

Label va Label vb Label vc # of deg 1 0-vertices# of deg 1 1-vertices

Balance Index

0 0 1 M-1 M-1 ½(a+b-c+2)

0 1 0 M-1 M-1 ½(a-b+c)

1 0 0 M-1 M-1 ½(-a+b+c+2)

0 1 1 M M-2 ½(a-b-c)

1 0 1 M M-2 ½(-a+b-c+2)

1 1 0 M M-2 ½(-a-b+c)

1 1 1 M+1 M-3 ½(-a-b-c)

Example: CT(3,0,3)

Example: CT(3,0,3)

Corollary

BI(CT(a,1,a))={a+1,a-1,1}.

Example: CT(3,1,3)

BI(CT(3,1,3))={4,2,1}.

Example: CT(a,b,c),ub(t1, t2,…, tb))

For notational convenience, we rename CT(a,b,c),ub(t1, t2,…, tb)) as

CT(d1, d0, d2)(ub(d3,d4,…,dd0+2))

CT(1,3,1)(ub(2,2,2))

Balance Indexes of Trees of Diameter Four

Theorem: For CT(d1, d0, d2)(ub(d3,d4,…,dd0+2)), If the sum of all d’s is even, then

If the sum of all d’s is odd, then

00

112

110

2

0

vfd

ii

vf dee i

00

1112

110

2

0

vfd

ii

vf dee i

Example: CT(1,3,2)(ub(0,0,3))

BI(CT(1,3,2)(ub(0,0,3)))={0,1,2,3,4,5}.

Example: CT(1,3,2)(ub(0,0,3))

Example: CT(1,2,3)(ub(2,2))

BI(CT(1,2,3)(ub(2,2)))={0,1,2,3,4,5,6}.

Example: CT(1,2,3)(ub(2,2))

Proof (Sum is odd)

NvPv

NvPv

NvPv

NvPv

vvvv

vv

kdvkvkd

vvkdMkM

vkdMvkM

vv

ee

1deg1deg2

1

31deg1deg232

1

11deg11deg32

1

deg3deg2

1

degdeg2

1

10

00

0

0

10

Proof (Sum is even)

NvPv

NvPv

NvPv

NvPv

vvvv

vv

kdvkvkd

vvkdMkM

vkdMvkM

vv

ee

1deg1deg12

1

31deg1deg2312

1

11deg11deg312

1

deg3deg12

1

degdeg2

1

10

00

0

0

10

Future?

A computer program to calculate? A better way to reduce the

computational complexity of using degrees sequence to find the balance index sets?