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On Edge-Balance Index Sets of L-Product of Cycles by Cycles

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.

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On Edge-Balance Index Sets of L-Product of Cycles by Cycles

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  1. 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

  2. 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.

  3. Example : nK2 • EBI(nK2 ) is {0} if n is even and {2}if n is odd.

  4. 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.}

  5. Examples

  6. Example : Pn • 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.

  7. 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). Wheels W6 W5

  8. 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.

  9. EBI(W6) = {0,2,4} |v(0)-v(1)|= 0 |v(0)-v(1)|= 2 |v(0)-v(1)|= 4

  10. 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

  11. 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}.

  12. 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).

  13. L-Product Example

  14. Generalized L-Product • More generally, the n copies of graphs to be identified with the vertices of G need not be identical.

  15. 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).

  16. L-Product of Cycles by Cycles

  17. 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

  18. 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.

  19. 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.

  20. 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.

  21. 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.

  22. Less 1 inside, Higher EBI

  23. 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 .)

  24. 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.

  25. Degree 4 Vertices

  26. 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.

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

  28. 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

  29. Switching Edges • By switching a 0-edge with an 1-edge adjacent to the inner cycle, we reduces the EBI by 1.

  30. Switching Edges • By switching a 0-edge with an 1-edge adjacent to the inner cycle, we reduces the EBI by 1.

  31. 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.

  32. 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.

  33. Special Edge-labeling to Yield the Highest EBI • According to the formula • we can label the rest in any way without changing EBI.

  34. 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.

  35. Proof (continued) • When n is odd and m is even, a special labeling like the one on the right produces an EBI 0.

  36. 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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