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

Edge Labeling A labeling f : E(G)  Z 2 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 : nK 2 EBI(nK 2 ) 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 | e f (0)  e f (1) |  1. Definition: The edge-balance index set of the graph G, EBI(G), is defined as {|v f (0) – v f (1)| : the edge labeling f is edge- friendly.}

Examples

Example : P n Lee, Tao and Lo [1] showed that [1] S-M. Lee, S.P.B. Lo, M.F. Tao, On Edge-Balance Index Sets of Some Trees, manuscript.

Wheels The wheel graph W n = N 1 +C n-1 where V(W n ) = {c 0 }  {c 1,…,c n-1 } and E(W n ) = {(c 0,c i ): i = 1, …, n-1}  E(C n-1 ). W5W5 W6W6

Edge Balance Index Set of Wheels Chopra, Lee ans Su [2] proved: Theorem: If n is even, then EBI(W n ) ={0, 2, …, 2i, …, n-2}. Theorem: If n is odd, then EBI(W n ) = {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,

EBI(W 6 ) = {0,2,4} |v(0)-v(1)|= 0 |v(0)-v(1)|= 2 |v(0)-v(1)|= 4

EBI(W 5 ) = {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(W 7 ) ={0, 1, 2, 3, 5}. EBI(W 9 ) ={0, 1, 2, 3, 4, 5, 7}. EBI(W 11 ) ={0, 1, 2, 3, 4, 5, 7, 9}. EBI(W 13 ) ={0, 1, 2, 3, 4, 5, 6, 7, 9, 11}. EBI(W 15 ) ={0, 1, 2, 3, 4, 5, 6, 7, 9, 11, 13}. EBI(W 17 ) ={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

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

Notations Let f be an edge labeling of a cycle C n. We denote the number of edges of C n which are labeled by 0 and 1 by f + by e C (0) and e C (1), respectively. We denote the number of vertices on C n which are labeled by 0, 1, and not labeled by the restricted f + by v C (0), v C (1), and v C (x), respectively

Proposition (Chopra, Lee and Su [2] ) In a cycle C n with a labeling f (not necessary edge friendly), assume that e C (0) > e C (1) > 1 and v C (x) = 2k > 0. Then we have v C (1) = e C (1) - k. and v C (0) = n - e C (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,

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.

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.

Proof Idea By the previous lemma, to maximize EBI, e C (1) has to be as small as it can be. Thus, if we label all edges in C n 1, it gives us the best chance to find the maximal EBI. Thus, might yield the maximal EBI.

Less 1 inside, Higher EBI

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.)

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 C n 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.

Degree 4 Vertices

Proof (continued) The above labeling requires n 1-edges for C n 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.

Keep Degree 4 Unchanged According to the formula we can label the rest in any way without changing EBI.

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

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.

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.

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, …, }.