on edge-balance index sets of l-product of cycles by cycles
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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, 2010. - PowerPoint PPT PresentationTRANSCRIPT
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