1. On Balance Index Sets of Trees of Diameter Four Sin-Min Lee, San Jose State University Hsin-hao Su, Stonehill College Yung-Chin Wang, Tzu-Hui Institute of Technology MCCCC 23 At Rochester Institute of Technology October 4, 2009

2. Example : nK 2 BI(G) is {0,1,2}.

3. Vertex Labeling A labeling f : V(G)  {0,1} induces an edge partial labeling f*: E(G)  {0,1} defined by f*(uv) = 0 if f(u) = f(v) = 0, f*(uv) = 1 if f(u) = f(v) = 1. Note that if f(u) ≠ f(v), then the edge uv is not labeled by f*.

4. Definition of Balance Index Set Definition: A labeling f of a graph G is said to be friendly if | v f (0)  v f (1) |  1. Definition: The balance index set of the graph G, EBI(G), is defined as {|e f (0) – e f (1)| : the vertex labeling f is friendly.} Definition: A labeling f of a graph G is said to be balanced if | v f (0)  v f (1) |  1 and | e f (0)  e f (1) |  1

5. Example : Cycles The balance index set of cycles is

6. Example : Stars The balance index set of stars St(n) is

7. Example : C n (t) The balance index set of C n (t) is

8. Example : C n (t) The balance index set of C 4 (3) is

9. Notations For i = 0 or 1, let v f (i) = |{v  V(G) | f(v) = i}| and e f (i) = |{e  E(G) | f*(e) = i}|. We also denote e f (x) to be the subset of E(G) containing all the unlabeled edges. When the text is clear, we omit the subscript for convenience

10. Lemma For a vertex labeling f (not necessary friendly), we have three equations:

11. Corollary For any friendly vertex labeling f,

12. Caterpillar CT(a,b,c) For CT(a,b,c), where a+b+c is odd, the balance index set is { |½(a+b+c+1)|, |½(a+b-c+1)|, |½(a-b+c-1)|, |½(-a+b+c+1)| }. For CT(a,b,c), where a+b+c is even, the balance indexes is { |½(a+b+c+2)|, |½(a+b- c+2)|, |½(a-b+c)|, |½(-a+b+c+2)|, |½(a-b-c)|, |½(-a+b-c+2)|, |½(-a-b+c)|, |½(-a-b-c)| }

13. Proof (Odd) When a+b+c is odd, the number of vertices of CT(a,b,c) is equal to a+b+c+3 which is even. Let a+b+c+3=2M. For a friendly labeling, there are M vertices labeled 0 and M vertices labeled 1. We name the three vertices on the spine, v a, v b, and v c. In CT(a,b,c), we have a+b+c degree 1 vertices. The degrees of v a, v b, and v c are a+1, b+2, and c+1, respectively.

14. Calculation (Odd) We first consider the case that v a, v b, and v c are all labeled 0. Then there are M-3 end-vertices labeled 0 and M end-vertices labeled 1.

15. Rest Cases (Odd) Label v a Label v b Label v c # of deg 1 0-vertices # of deg 1 1- vertices Balance Index 001M-2 M-1 ½ (a+b-c+1) 010M-2 M-1 ½ (a-b+c-1) 100M-2 M-1 ½ (-a+b+c+1) 011M-1 M-2 ½ (a-b-c-1) 101M-1 M-2 ½ (-a+b-c+1) 110M-1 M-2 ½ (-a-b+c-1) 111M M-3 ½ (-a-b-c-1)

16. Example: CT(1,1,3)

18. Proof (Even) When a+b+c is even, the number of vertices of CT(a,b,c) is equal to a+b+c+3 which is odd. Let a+b+c+3=2M+1. For a friendly labeling, without loss of generality, there are M+1 vertices labeled 0 and M vertices labeled 1. We name the three vertices on the spine, v a, v b, and v c. In CT(a,b,c), we have a+b+c degree 1 vertices. The degrees of v a, v b, and v c are a+1, b+2, and c+1, respectively.

19. Calculation (Even) We first consider the case that v a, v b, and v c are all labeled 0. Then there are M-2 end-vertices labeled 0 and M end-vertices labeled 1.

20. Rest Cases (Even) Label v a Label v b Label v c # of deg 1 0-vertices # of deg 1 1-vertices Balance Index 001M-1 ½ (a+b-c+2) 010M-1 ½ (a-b+c) 100M-1 ½ (-a+b+c+2) 011M M-2 ½ (a-b-c) 101M M-2 ½ (-a+b-c+2) 110M M-2 ½ (-a-b+c) 111M+1 M-3 ½ (-a-b-c)

21. Example: CT(3,0,3)

23. Corollary BI(CT(a,1,a))={a+1,a-1,1}.

24. Example: CT(3,1,3) BI(CT(3,1,3))={4,2,1}.

25. Example: CT(a,b,c),ub(t 1, t 2,…, t b )) For notational convenience, we rename CT(a,b,c),ub(t 1, t 2,…, t b )) as CT(d 1, d 0, d 2 )(ub(d 3,d 4,…,d d0 +2)) CT(1,3,1)(ub(2,2,2))

26. Balance Indexes of Trees of Diameter Four Theorem: For CT(d 1, d 0, d 2 )(ub(d 3,d 4,…,d d0+2 )), If the sum of all d’s is even, then If the sum of all d’s is odd, then

27. Example: CT(1,3,2)(ub(0,0,3)) BI(CT(1,3,2)(ub(0,0,3)))={0,1,2,3,4,5}.

28. Example: CT(1,3,2)(ub(0,0,3))

29. Example: CT(1,2,3)(ub(2,2)) BI(CT(1,2,3)(ub(2,2)))={0,1,2,3,4,5,6}.

30. Example: CT(1,2,3)(ub(2,2))

31. Proof (Sum is odd)

32. Proof (Sum is even)

33. Future? A computer program to calculate? A better way to reduce the computational complexity of using degrees sequence to find the balance index sets?