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On The Edge-Balance Index Sets of Fans and Wheels

On The Edge-Balance Index Sets of Fans and Wheels. Dharam Chopra , Wichita State University Sin-Min Lee, San Jose State University Hsin-hao Su, Stonehill College MCCCC 22 At University of Nevada, Las Vegas October 23, 2008. Example : nK 2.

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On The Edge-Balance Index Sets of Fans and Wheels

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  1. On The Edge-Balance Index Sets of Fans and Wheels Dharam Chopra, Wichita State University Sin-Min Lee,San Jose State University Hsin-hao Su,Stonehill College MCCCC 22 At University of Nevada, Las Vegas October 23, 2008

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

  3. 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 and 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.

  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. Example : Pn • Lee, Tao and Lo showed that

  6. Example : Pn

  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. Notation • Let f be an edge-friendly labeling of Wn. We denote the number of edges of Cn-1 which are labeled by 0 and 1 by f+ by eC(0) and eC(1) and the number of edges in E(Wn) which are labeled by 0 and 1 by f+ by eN(0) and eN(1).

  9. Because of the nature of a wheel, we separate a wheel into a cycle and a star. Idea W6 W5

  10. Because of the nature of a wheel, we separate a wheel into a cycle and a star. Separation W5=C5+St5 W6=C6+St6

  11. Vertex of Order 2 • For a vertex of order 2 with an edge-labeling (not necessary friendly), it can only be labeled as one of the following three cases.

  12. Add an Edge to an Vertex of Order 2 • If the vertex was labeled, then the label of the vertex is not changed. • If the vertex was not labeled then the label of the vertex is labeled by the same as the adding edge.

  13. Edge-balance Index Sets of a Cycle • We focus on the cycle part and get several algebraic equations that we can use later. • We denote the number of vertices on Cn-1 which are labeled by 0, 1, and not labeled by the restricted f+ by vC(0), vC(1), and vC(x), respectively.

  14. Lemma • In a cycle Cn with a labeling f (not necessary edge friendly), we have two equations: • 2 vC(0)+ vC(x)=2eC(0), and • 2 vC(1)+ vC(x)=2eC(1).

  15. Lemmas • In a cycle Cn with a labeling f (not necessary edge friendly), eC(0) or eC(1) is zero if and only if vC(x) = 0. • In a cycle Cn with a labeling f (not necessary edge friendly), if vC(0) and vC(1) are both positive then vC(x) > 0.

  16. Lemma • In a cycle Cn with a labeling f (not necessary edge friendly), assume that eC(0) > eC(1) > 1. We have 0 < vC(0) < eC(0)-1 and 0 < vC(1) < eC(1)-1.

  17. Do we have all possible labelings of a cycle? • We know that v(x) must be even. For an even number 2k > 0, We can construct a labeling of Cn by putting k pairs of 0- and 1-edges alternatively in the middle of Pn+1 and all the rest 0s on the left side edges and all the rest 1s on the right side edges and then glue the two sided vertices together.

  18. Do we have all possible labelings of a cycle?

  19. Do we have all possible labelings of a cycle? • Since we put k pairs of 0 and 1 alternatively, these 2k edges produce exactly 2k-1 unlabeled vertices. Additionally, the glued two sided vertices become an unlabeled vertex. All the other vertices are either in the middle of 0-edges or 1-edges. Thus, we have exactly 2k unlabeled vertices.

  20. Do we have all possible labelings of a cycle? • For an edge-labeling f of Cn with v(x)=2k>0, pick an unlabeled vertex and split this vertex into two vertices. • Then, we have a Pn+1 with the very left edge is labeled by 0 and the very right edge is labeled by 1.

  21. Do we have all possible labelings of a cycle?

  22. Do we have all possible labelings of a cycle? • Let a maximal chain of 0-edges is a path with only 0-edges where its length is greater than 1 and it is not a sub chain of 0-edges with longer length. • If a maximal chain of 0-edges does not contain the very left 0-edge, we can cut it off and insert into the left side of the first from the left unlabeled vertex without altering the number of 0-vertices, 1-vertices and unlabeled vertices.

  23. Lemmas • 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 vC(1) = eC(1)-k. • 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 vC(0) = n- eC(1)-k.

  24. Proof • By the above construction and rearrangement, we can put k pairs of 0- and 1-edges in the middle of Pn+1. • Since these k pairs of edges occupy k edges labeled by 1, there are only eC(1)-k 1-edges left on the right part of Pn+1.

  25. Proof (continues) • Since the very right edge of the k pairs of 0- and 1-edges is an edge labeled by 1, we have a chain of 1-edges with eC(1)-k+1 edges and eC(1)-k vertices in between edges. Thus, vC(1) = eC(1)-k. • In Cn, n= vC(0)+ vC(1)+ vC(x). Thus, vC(0) = n- vC(1)- vC(x) = n- (eC(1)-k) -2k = n- eC(1)-k.

  26. Assumption • By the symmetry of the role of 0 and 1 in the labeling to calculate the edge balance index, without loss of generosity, we may assume that eC(0) > eC(1) > 0. Since eC(0) > eC(1) and eC(0)+eC(1)=n-1. We have • 0 <eC(1) < (n-1)/2 if n is odd, or, • 0 <eC(1) < n/2 if n is even.

  27. Three Cases on Cn-1 • It’s easy to see that there are three possible cases: • eC(0) > eC (1) > 1, or, • eC(0) = eC(1) > 1 only when n is odd (which implies eC(0) = eC(1) = (n-1)/2,) or, • eC(1) = 0 (which implies that eC(0)=n-1.)

  28. Examples |v(0)-v(1)|= 4 |v(0)-v(1)|= 2 |v(0)-v(1)|= 9 W8 W7 W11

  29. h • Note that when we put all E(Wn) edges and c0 into Cn-1 to get our wheel back, all the vertices of Cn-1 become of order 3. • Thus, all the unlabeled vertices in Cn-1 become labeled by either 0 or 1. • All other labeled vertices remain labeled by the same value. • Assume that there are h 0-edges connected to an unlabeled vertex of Cn-1. Obviously, we have 0 < h < eN(0)=eC(1).

  30. Edge Balance Index of a Wheel • Theorem:Let Wn be a wheel. • If n is even, there are two types of edge-balance indexes: • v(0)-v(1) = n-2 – 2(eC(1)+k-h) where 0 < h < eC(1), 0 < k < eC(1) and 0 < eC(1) < n/2; • v(0)-v(1) = n-2.

  31. Edge Balance Index of a Wheel • If n is odd, • v(0)-v(1) = n-2 – 2(eC(1)+k-h) where 0 < h < eC (1), 0 < k < eC (1) and 0 < eC (1) < (n-1)/2; • v(0)-v(1) = 2(h-k) where 0 < k < (n-1)/2 and 0 < h < (n-1)/2 if 0 < k < (n-1)/4 and k-(n-1)/4 < h < (n-1)/2 if (n-1)/4 < k < (n-1)/2. • v(0)-v(1) = n-2.

  32. Proof (Case 1) • By lemma, vC(x) is not zero. Since vC(x) must be even, we assume that vC(x)=2k>0 where k is an integer. By lemmas, we have vC(1) = eC(1)-k and vC(0) = n- eC(1)-k. • Since the number of edges of our wheel Wn is even, (actually, it is 2n-2,) and f is an edge-friendly labeling of Wn, we have eN(0) = eC(1) and eN(1) = eC(0). Therefore, by the assumption that eC(0) > eC(1), we have eN(1) > eN(0). This implies that c0 must be labeled by 1.

  33. Case 1 Examples |v(0)-v(1)|= 0 |v(0)-v(1)|= 2 |v(0)-v(1)|= 1 W8 W8 W7

  34. Proof (Case 1 Continues) • If eC(1) < vC(x), then we have h 0-vertices converted from unlabeled vertices and the rest vC(x)-h unlabeled vertices are converted into vertices labeled by 1. Therefore, the number of vertices labeled by 0 v(0) = vC(0) + h and the number of vertices labeled by 1 v(1) = vC(1) + vC(x)-h + 1.

  35. Proof (Case 1 Continues) • So, the edge-balance index v(0)-v(1) = (vC(0) + h) – (vC(1) + vC(x)-h + 1) = vC(0) + 2h – vC(1) – 2k – 1 = (n-1-eC(1)-k) + 2h – (eC(1)-k) – 2k – 1 = n-2 – 2(eC(1)+k-h). • Note here that we have 0 < h < eC(1) and 0 < k < eC(1). Also, since eC(0) > eC(1) > 1, we have 0 < eC(1) < (n-1)/2 if n is odd, or, 0 < eC(1) < n/2 if n is even.

  36. Proof (Case 1 Continues) • If vC(x) < eN(0)=eC(1), then h must be less than vC(x) since those are all unlabeled vertices available to be connected by an 0-edge. We have a similar inequality 0 < h < vC(x) < eC(1). In the calculation of edge-balance index set, this case will be override by the previous one since we have to calculate the balance index for all possible edge friendly labelings. Thus, it will not generate any new edge-balance index.

  37. Case 2 Examples |v(0)-v(1)|= 0 |v(0)-v(1)|= 2 |v(0)-v(1)|= 4 W7 W11 W11

  38. Proof (Case 2) • The argument is very similar except c0 is not labeled because eC(0) = eC(1) = (n-1)/2. • Therefore, we get similar equations v(0) = vC(0) + h and v(1) = vC(1) + vC(x)-h. • Thus, v(0)-v(1) = (vC(0) + h) – (vC(1) + vC(x)-h) = 2(h-k).

  39. Proof (Case 2 continues) • Since eN(0) > 2k, it provides enough 0-edges to be used. • And k-(n-1)/4 < h < (n-1)/2 if (n-1)/4 < k < (n-1)/2 since the number of unlabeled vertices is greater or equal to eN(1). • Therefore, in this case, 0 < k < (n-1)/2 and 0 < h < (n-1)/2 if 0 < k < (n-1)/4

  40. Case 3 Examples |v(0)-v(1)|= 6 |v(0)-v(1)|= 5 |v(0)-v(1)|= 11 W8 W7 W13

  41. Proof (Case 3) • Since the number of edges of our wheel Wn is even and f is an edge-friendly labeling of Wn, we have eN(0) = eC(1) and eN(1) = eC(0). • Therefore, by the assumption that eC(1)=0, we have eN(0)=0 and eN(1)=n-1. This implies that c0 must be labeled by 1. • Since all the edges in Cn-1 are labeled by 0, we have n-1 vertices labeled by 0. Thus, the edge-balance index v(0)-v(1)=n-2.

  42. Edge Balance Index Set of Wheels • 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}.

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

  44. Proof (Even Case) • Since n is even, we have v(0)-v(1) = n-2 and v(0)-v(1) = n-2 – 2(eC(1)+k-h) where 1 < eC(1)+k-h < 2eC(1) and 0 < eC(1) < n/2. • This implies that 1 < eC(1)+k-h < n. • Let t= eC(1)+k-h. • Since k can be any integer between 1 and eC(1) including 1 and eC(1) and h can be any integer between 0 and eC(1) including 0 and eC(1), we have 1 < t < n.

  45. Proof (Even Case Continues) • Thus, v(0)-v(1) = n-2 -2t where 1<t<n-1 where t is an integer. • Hence, -(n-2) < v(0)-v(1) < n-4 and v(0)-v(1) must be even. • Therefore, EBI(Wn) ={0,2,…,2i,…,n-2}

  46. 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

  47. Proof (Odd Case) • Since n is odd, we have • v(0)-v(1) = n-2 – 2(eC(1)+k-h) where 0 < h < eC(1), 0 < k < eC(1) and 0 < eC(1) < (n-1)/2; • v(0)-v(1) = 2(h-k) where 0 < k < (n-1)/2 and 0 < h < (n-1)/2 if 0 < k < (n-1)/4 and k-(n-1)/4 < h < (n-1)/2 if (n-1)/4 < k < (n-1)/2. • v(0)-v(1) = n-2.

  48. Proof (Odd Case Continues) • For the first case, a similar argument applies. Thus, we have v(0)-v(1) = n-2 and v(0)-v(1) = n-2 – 2(eC(1)+k-h) where 1 < eC(1)+k-h < 2eC(1) and 0 < eC(1) < (n-1)/2. • This implies that 1 < eC(1)+k-h < n-1. Let t= eC(1)+k-h. • Since k can be any integer between 1 and eC(1) including 1 and eC(1) and h can be any integer between 0 and eC(1) including 0 and eC(1), we have 1 < t < n-1.

  49. Proof (Odd Case Continues) • Thus, v(0)-v(1) = n-2 -2t where 1<t<n-2 where t is an integer. • Thus, -(n-2) < v(0)-v(1) < n-4 and v(0)-v(1) must be odd since n is odd. • Therefore, EBI(Wn) contains{1,3,…,2i+1,…,n-2}.

  50. Proof (Odd Case Continues) • For the second case, v(0)-v(1) = 2(h-k) where 0 < k < (n-1)/2. • For h, 0 < h < (n-1)/2 if 0 < k < (n-1)/4 and k-(n-1)/4 < h < (n-1)/2 if (n-1)/4 < k < (n-1)/2 • We have - (n-1)/2 < v(0)-v(1) = 2(h-k) < (n-1)/2. Therefore, EBI(Wn) contains{0,1,2,…,(n-1)/2}.

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