1. A Study of IC-coloring of Graphs 研 究 生：林耀仁 指導教授：江南波

2. Sum-saturable  Let G = (V, E) be an undirected graph with p vertices and let K = p(p+1)/2. Let f be a bijective function from V to {1,2,...,p}. Then f is said to be a saturating labelling of G if, given any k (1 k K), there exists a connected subgraph H of G such that.  If a saturating labelling of G exists, then G is said to be sum-saturable.

3. IC-coloring  let G=(V, E) be an undirected graph and let f be a function from V to N. For each subgraph H of G, we define f s (H) =.Then f is said to be a IC-coloring of G if, given any k ( 1 k f s (G)) there exists a connected subgraph H of G such that f s (H)=k.  The IC-index of G is defined to be M(G) = max{f s (G) | f is an IC-coloring of G}

4. 1995, Penrice[5] Theorem 1.3.1. For the complete graph K n, M(K n )=2 n -1. Theorem 1.3.2. For every n 4, M(K n -e)=2 n -3. Theorem 1.3.3. For all positive integers n 2, M(K 1,n )=2 n +2.

5. 2005, E. Salehi et.al.[6] Observation 1.3.5. If H is a subgraph of G, then M(H) M(G). Observation 1.3.6. If c(G) is the number of connected induced subgraph of G, then M(G) c(G).

6. 2007, Chin-Lin Shiue[7] Theorem 1.3.8. For any complete bipartite graph K m, n, 2 m n, M(K m, n )=3 ． 2 m+n-2 -2 m-2 +2.

7.  We display the results of the sum-saturability of all non-isomorphic trees with at most p=8 vertices p=1 p=2 p=3 p=4 1 12 132 143 2

8. p=5 p=6 2431 5 3 452 1 16453 2 1456 23 1 2456 3 234 1 6 5

9. p=7 14567 23 17356 24 267354 1 1 2 4 567 3 1 24567 3 25 6 7 1 34 3 567 4 21 7 1 2 34 5 6

10. p=8 158 4 7362 1475 8 2 63 145678 23

13.  A T-graph be an undirected graph with p vertices consisting of vertex set V(T) = {V 1, V 2, …, V p } and edge set E(T) = {V 1 V 2, V 2 V 3, V 3 V 4, …, V p-2 V p-1, V p V 2 }.

14. Theorem 2.1.1. The T-graph with order p=6 is not Sum- saturable. proof.  Assume T-graph with order p=6 is sum-saturable, then there is a saturating labeling f of T-graph.  Given any k (1 k 21=K), there exists a connected subgraph H of T such that.  Because K-1 and K-2 exists a connected subgraph H of T. So, number 1 and number 2 are labeling in end-vertex.  Hence we have following three cases :

15. Case 1.

16. Case 2.

17. Case 3.

18. Theorem 2.1.2. The T-graph with order p=7 is not Sum- saturable. proof.  Assume T-graph with order p=7 is sum-saturable, then there is a saturating labeling f of T-graph.  we given any k (1 k 28=K), there exists a connected subgraph H of T such that.  Because K-1 and K-2 exists a connected subgraph H of T. So, number 1 and number 2 are labeling in end-vertex.  Hence we have following three cases :

19. Case 1.

20. Case 2.

21. Case 3.

22. Theorem 2.1.3. The T-graph with order p=8 is not Sum- saturable. proof.  Assume T-graph with order p=8 is sum-saturable, then there is a saturating labeling f of T-graph.  we given any k (1 k 36=K), there exists a connected subgraph H of T such that.  Because K-1 and K-2 exists a connected subgraph H of T. So, number 1 and number 2 are labeling in end-vertex.  Hence we have following three cases :

26. Remark. We use the same way in Theorem 2.1.3, and we got the T-graph of order p 9 is not sum- saturable.

27. Conjecture 2.1.4. Suppose that T is a tree of order p. If Δ(T) >,then T is sum-saturable.

28.  A rooted tree T is a complete n-ary tree, if each vertex in T except the leaves has exactly n children.  For each vertex v in T, if the length of the path from the root to v is L, then we say that v is on the level L.  If all leaves are on the same level h, then we call T a perfect complete n-ary tree with the hight h.

29. Theorem 2.2.1. A perfect complete binary tree T is sum-saturable. Proof. Let h be the hight of T. If h 2 Hence perfect complete binary trees with height h 2 are sum-saturable.

30.  If h ≧ 3, we define a labelling as follows ： 1248 3 5

31.  h = 3  h = 4 1248 7 3 5 6 9 10 1112 13 1415 124816 23 3 5 6 7 9 10 11 1213 14 15 17 18 1920 21 2224 25 26 2728 29 3031

32. Corollary 2.2.2. A perfect complete n-ary tree is sum- saturable.

33. m(h+1) = 1 + n ． m(h) < 2 ． n ． m(h) log 2 m(h+1) < log 2 n+ log 2 m(h)+1, i.e. log 2 m(h+1)- log 2 m(h) < log 2 n +1< n L= 12 p 2L2L

34. Theorem 3.1.1. For every integer n 4, we have 2 n -8 M(K n -L) 2 n -3, where L is a matching consisting of two edges. V1V1 V2V2 V n-1 VnVn

35. Proof.  Let V(K n -L)={V 1, V 2, …, V n }.  We assign the vertexV 1 is non-adjacent to the vertex V n-1 and the vertex V 2 is non-adjacent to the vertex V n.  We define f:V(K n -L) N by f(V i )=2 i-1, for all i=1, 2, …,n-2, f(V n-1 )=2 n-2 -2 and f(V n )=2 n-1 -5.  We claim that f is an IC-coloring of V(K n -L), with f s (K n -L)= +(2 n-2 -2)+(2 n-1 -5)=2 n -8.  For any integer k [1, 2 n -8], and consider the following three cases:

36. (i) k [1, 2 n-2 -1] (ii) k [2 n-2, 2 n-1 -3] Let a=k-(2 n-2 -2), then 2 a 2 n-2 -1. (iii) k [2 n-1 -2, 2 n -8] Let b=k-(2 n-1 -5), then 3 b 2 n-1 -3. We get 2 n -8 M(K n -L) 2 n -3.

37. Theorem 3.1.2. For every integer n 4, we have 2 n -5 M(K n -P 3 ) 2 n -4. V1V1 V2V2 VnVn

38. Proof.  Let V(K n -P 3 )={V 1, V 2, …, V n }.  We assign the vertexV 1 is non-adjacent to the vertex V n and the vertex V 2 is non-adjacent to the vertex V n.  We define f:V(K n -P 3 ) N by f(V i )=2 i-1, for all i=1, 2, …,n-1, and f(V n )=2 n-1 -4.  We claim that f is an IC-coloring of V(K n -P 3 ), with f s (K n -P 3 )= +(2 n-1 -4)=2 n -5.  For any integer k [1, 2 n -5], and consider the following two cases:

39. (i) k [1, 2 n-1 -1] (ii) k [2 n-1, 2 n -5] Let a=k-(2 n-1 -4), then 4 a 2 n-1 -1. We get 2 n -5 M(K n -P 3 ) 2 n -4.

40. Theorem 3.2.1. For every integer n 4, we have 2 n -9 M(K n -P 4 ) 2 n -6. V1V1 V2V2 V n-1 VnVn

41. Theorem 3.2.2. For every integer n 6, we have 2 n -20 M(K n -R) 2 n -4, where R is a matching consisting of three edges. V1V1 V2V2 V3V3 V n-2 V n-1 VnVn

42. Theorem 3.2.3. For every integer n 5, we have 2 n -12 M(K n -P 3 {e}) 2 n -5, where e E(P 3 ). V1V1 V2V2 V3V3 V n-1 VnVn

43. Theorem 3.2.4. For every integer n 4, we have 2 n -7 M(K n -C 3 ) 2 n -5. V1V1 VnVn V n-1

44. Theorem 3.2.5. For every integer n 5, we have 2 n -9 M(K n -k 1,3 ) 2 n -8. V1V1 V2V2 V3V3 VnVn

45. Corollary 3.2.6. For every integer n m+1, we have 2 n -2 m-1 M(K n -k 1,m ) 2 n -2 m.

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