1. On The Edge-Balance Index Sets of Fans and Broken Fans Dharam Chopra*, Wichita State University Sin-Min Lee, San Jose State University Hsin-hao Su, Stonehill College 40th SEICCGTC At Florida Atlantic University March 4, 2009

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

3. 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 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 0 is equal to the number of edge labeled 1.

4. 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.}

5. Example : P n Lee, Tao and Lo showed that

6. Example : P n

7. Fans A Fan graph is F 1,n = N 1 +P n where V(F 1,n ) = {c}  {v 1,…,v n } and E(F 1,n ) = {(c,v i ): i=1,…,n}  E(P n ).

8. Turn a Fan into a Wheel If we add an extra edge (v 1,v n ) into F 1,n, then it becomes W 1+n.

9. 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

10. Idea Because of the nature of a wheel, we separate a wheel into a cycle and a star. W5W5 W6W6

11. Separation Because of the nature of a wheel, we separate a wheel into a cycle and a star. W 5 =C 5 +St 5 W 6 =C 6 +St 6

12. 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.

13. 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.

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

15. Case 1 Examples |v(0)-v(1)|= 0 |v(0)-v(1)|= 2 |v(0)-v(1)|= 1 W 8 W 8 W 7

16. Case 2 Examples |v(0)-v(1)|= 0 |v(0)-v(1)|= 2 |v(0)-v(1)|= 4 W 7 W 11 W 11

17. Case 3 Examples |v(0)-v(1)|= 6 |v(0)-v(1)|= 5 |v(0)-v(1)|= 11 W 8 W 7 W 13

18. Edge-Balance Index Set of Wheels Theorem (Chopra, Lee, Su): If n is even, then EBI(W n ) ={0,2,…,2i,…,n-2}. Theorem (Chopra, Lee, Su): If n is odd, then EBI(W n ) = {1,3,…,2i+1,…,n- 2}  {0,2,4,…,(n-1)/2}.

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

20. 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

21. Equivalence of Friendly Labelings For an edge friendly labeling, since F 1,n has 2n-1 edges, to calculate the edge balance index, we may assume that v(0)=v(1)+1. Thus, if we label the extra edge 1, then we have an edge friendly labeling for W 1+n. Note that for an edge friendly labeling of W 1+n, if we remove an edge labeled 1 from its cycle part, we get an F 1,n with an edge friendly labeling. Thus, there is a 1-1 correspondence between the edge friendly labelings of F 1,n and the edge friendly labelings of W 1+n.

22. V 1 and V n There are three different pairs of labeling of v 1 and v n : v 1 is labeled by 0 and vn is lebeled by 1 and vice versa. Both v 1 and v n are labeled by 0. Both v 1 and v n are labeled by 1. Since case 2 and 3 requires at least 4 edges labeled by the same value, it only happens when 2n-1>7, that is, n>4.

23. Changes of the Edge-Balance Index Because the extra edge is labeled by 1, if v 1 or v n is labeled 0, then it must have another two 0-edges attached. Thus, it remains labeled 0 in F 1,n after removing the extra edge. If v 1 or v n is labeled 1, then it could either remain labeled 1 if there are another two 1- edges attached or become unlabeled. The first case keeps the balance index the same. But, the second one reduces the amount of 1-vertices 1.

24. Edge-Balance Index Set of Fans Theorem: EBI(F 1,n ) {0,1,2} if n=3 {0,1,…, n-2} for n> 4.

25. EBI(F 1,3 ) ={0,1,2}

26. EBI(F 1,4 ) ={0,1,2}

27. EBI(F 1,5 ) ={0,1,2,3}

28. EBI(F 1,6 ) ={0,1,2,3}

29. Proof By the 1-1 correspondence between the edge friendly labelings of W 1+n and F 1,n, the edge balance calculation tells us that in W 1+n, -(n-1) < v(0)-v(1) < n-3 where v(0)- v(1) must be odd if n is even and v(0)-v(1) must be even if n is odd. Note that v(0)-v(1) = n-3 only if k=1 and h=e C (1)>1 and v(0)-v(1) = -(n-1) only if k=0 and h=0.

30. Proof (continues) We first consider the extreme case v(0)-v(1) = n-3 and v(1) is reduced by 2. In this case, we have k=1. Thus, v1 and vn are the only two vertices which are unlabeled and both are converted to 1. But, since there are h>1 unlabeled vertices which are converted to 0, this cannot happen.

31. Proof (continues) The next extreme case is v(0)-v(1) = -(n-1) and v(1) is reduced by 0. In this case, since h=0, we know that all v C (x) unlabeled vertices are converted to 1. Also, since k=0 and e C (1)=n/2 or (n-1)/2 which depends on n is even or odd, respectively, we have v C (x)=2e C (1)=n if n is even and n-1 if n is odd. This cannot happen since we assume that e C (0)>e C (1) and the extra edge occupies the extra 1.

32. Proof (continues) Therefore, we conclude that, in W 1+n, -(n-2) < v(0)-v(1) < n-2 where v(0)-v(1) must be odd if n is even and v(0)-v(1) must be even if n is odd. For n=3, there is only case 1. So, v(1) can only be reduced by 0 or 1. Thus, the inequality becomes -(n-1) < v(0)-v(1) < n-1. Thus, the EBI= {0,1,2}.

33. Proof (continues) For n>4, since all the cases can only reduce v(1) by 0, 1 or 2, the inequality becomes - (n-2) < v(0)-v(1) < n-2. Also, it is easy to construct a friendly labeling that fits one of the above three cases. So, any integer between -(n-2) and n- 2 can be the result of v(0)-v(1). Thus, the EBI= {0,1,…, n-2}

34. Edge Balance Index Set of Broken Fans BF(a,b) Theorem: EBI(BF(a,b)) {0,1,2,…, a+b-4} for 2 <a < b

35. EBI(BF(2,2)) ={0}

36. EBI(BF(2,3)) ={0,1}

37. EBI(BF(2,4)) ={0,1,2}

38. EBI(BF(3,3)) ={0,1,2}

39. EBI(BF(3,6)) ={0,1,2,3,4,5}

40. St(n) Theorem: The edge-balance index set of the star St(n) is {0} if n is even. {2} if n is odd.