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On Balance Index Sets of Trees of Diameter Four Sin-Min Lee, San Jose State University Hsin-hao Su,...
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Transcript of On Balance Index Sets of Trees of Diameter Four Sin-Min Lee, San Jose State University Hsin-hao Su,...
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?