on edge-balance index sets of l-product of cycles by cycles daniel bouchard, stonehill college...
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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 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
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PEBI n
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[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, then
EBI(Wn) ={0, 2, …, 2i, …, n-2}. Theorem: If n is odd, then
EBI(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.
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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.
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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.
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Less 1 inside, Higher EBI
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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 .)
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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.
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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.
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Keep Degree 4 Unchanged
According to the formula
we can label the rest in any way without changing EBI.
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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
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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.
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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, …, }.
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