# L(3, 2, 1)-Labeling of Simple Graphs

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1 L(3,, 1)-Labeling of Simple Graphs Jean Clipperton Simpson College, IA Jessica Gehrtz Concordia College-Moorhead, MN Zsuzsanna Szaniszlo Valparaiso University, IN Desmond Torkornoo University of Richmond, VA July 11, 005 Abstract An L(3,, 1)-labeling is a simplified model for the channel assignment problem. It is a natural generalization of the widely studied L(, 1)- labeling. An L(3,, 1)-labeling of a graph G is a function f from the vertex set V (G) to the set of positive integers such that for any two vertices x, y, if d(x, y) = 1, then f(x) f(y) 3; if d(x, y) =, then f(x) f(y) ; and if d(x, y) = 3, then f(x) f(y) 1. The L(3,, 1)-labeling number k(g) of G is the smallest positive integer k such that G has an L(3,, 1)-labeling with k as the maximum label. In this paper we determine the L(3,, 1)-labeling number for paths, cycles, caterpillars, n- ary trees, complete graphs and complete bipartite graphs. We also present an upper bound for k(g) in terms of the maximum degree of G. 1 Introduction The assignment of FM frequencies to stations became a problem as technology advanced in the early 0th centuy. As more and more stations requested frequencies, it became difficult to assingn frequencies without having new stations NSF Grant. 1

2 interfere with the broadcast of other stations nearby. The channel assignment problem is an engineering problem in which the task is to assign a channel (nonnegative integer) to each FM radio station in a set of given stations such that there is no interference between stations and the span of the assigned channels is minimized. The level of interference between any two FM radio stations correlates with the geographic locations of the stations. Closer stations have a stronger interference, and thus there must be a greater difference between their assigned channels. In 1980, Hale introduced a graph theory model of the channel assignment problem where it was represented as a vertex coloring problem [3]. Vertices on the graph correspond to the radio stations and the edges show the proximity of the stations. In 1991, Roberts proposed a variation of the channel assignment problem in which the FM radio stations were considered either close or very close. Close stations were vertices of distance two apart on the graph and were assigned channels that differed by two; stations that were considered very close were adjacent vertices on the graph and were assigned distinct channels [6]. More precisely, Griggs and Yeh defined the L(, 1)-labeling of a graph G = (V, E) as a function f which assigns every x, y in V a label from the set of positive integers such that f(x) f(y) if d(x, y) = 1 and f(x) f(y) 1 if d(x, y) = []. L(, 1)-labeling has been widely studied in recent years. In 001, Chartrand et al. introduced the radio-labeling of graphs; this was motivated by the regulations for the channel assignments in the channel assignment problem [1]. Radio-labeling takes into consideration the diameter of the graph, and as a result, every vertex is related. Practically, interference among channels may go beyond two levels. L(3,, 1)- labeling naturally extends from L(, 1)-labeling, taking into consideration vertices which are within a distance of three apart; however, it remains less difficult than radio-labeling. An L(3,, 1)-labeling of a graph G = (V, E) is a function f which assigns every x, y in V a label from the set of positive integers such that f(x) f(y) 3 if d(x, y) = 1, f(x) f(y) if d(x, y) =, and f(x) f(y) 1 if d(x, y) = 3. The L(3,, 1)-labeling number, k(g), of G is the smallest number k such that G has an L(3,, 1)-labeling with k as the maximum label [4]. In this paper we determine the L(3,, 1)-labeling number for paths, cycles, caterpillars, n-ary trees, complete graphs and complete bipartite graphs. We also present an upper bound for k(g) in terms of the maximum degree of G. The method for calculating the upper bound was introduced by Griggs and Yeh [] and referenced by Jonas [5].

3 Definitions and Notation Definition 1. Let G = (V, E) be a graph and f be a mapping f : V N. f is an L(3,, 1)-labeling of G if, for all x, y V, 3, if d(x, y) = 1; f(x) f(y), if d(x, y) = ; 1, if d(x, y) = 3. Definition. The L(3,, 1)-number, k(g), of a graph G is the smallest natural number k such that G has an L(3,, 1)-labeling with k as the maximum label. An L(3,, 1)-labeling of a graph G is called a minimal L(3,, 1)-labeling of G if, under the labeling, the highest label of any vertex is k(g). Note. If 1 is not used as a vertex label in an L(3,, 1)-labeling of a graph, then every vertex label can be decreased by one to obtain another L(3,, 1)-labeling of the graph. Therefore in a minimal L(3,, 1)-labeling 1 will necessarily appear as a vertex label. Definition 3. Let G = (V, E) be a graph. G is called a complete graph on n vertices, K n, if for all vertices x, y V, (x, y) E. G is called a complete bipartite graph, K m,n, if: 1. The set of vertices, V, can be partitioned into two disjoint sets of vertices, A and B, such that A = m, B = n, and V = m + n. For all a i, a j A, (a i, a j ) / E and for all b i, b j B, (b i, b j ) / E 3. For all a i A and b j B, (a i, b j ) E. A star, S n, is a K 1,n complete bipartite graph. Definition 4. Let G = (V, E) be a graph. G is called a path, P n, if V = {v 1, v,..., v n } such that for 1 i < n, (v i, v i+1 ) E. G is called a cycle, C n, if V = {v 1, v,..., v n } such that for 1 i < n (v i, v i+1 ) E and (v 1, v n ) E. Definition 5. Let G = (V, E) be a graph. G is called a tree if G is connected and has no cycles. G is called a caterpillar if G is a tree such that the removal of the degree one vertices produces a path called the spine of the caterpillar. A uniform caterpillar is a caterpillar with only degree one and degree vertices. We denote a uniform caterpillar with n vertices on the spine by Cat n. G is called an n-ary tree if G is a rooted tree such that the root has degree n and all the other vertices have degree n

4 3 Theorems 3.1 Complete Graphs Theorem 1. For any complete graph on n vertices, k(k n ) = 3n. Proof. Let G = (V, E) be a complete graph with V = {v 1, v,..., v n } and let f be a minimal L(3,, 1)-labeling of G with f(v i ) < f(v j ) for all i < j. Then, for all v i, v j V and i j, (v i, v j ) E, implying that d(v i, v j ) = 1. Thus, f(v i ) f(v j ) 3 for all v i, v j V and i j. Since there exists v i in V such that f(v i ) = 1, we have f(v 1 ) = 1. Also, since f is a minimal L(3,, 1)-labeling of G, we have that for all v i with 1 < i n, f(v i ) f(v i 1 ) + 3. Recursively, we have f(v n ) f(v n 1 ) + 3 Therefore k(k n ) = 3n. f(v n ) + 3() f(v n 3 ) + 3(3). f(v n i ) + 3(n i). f(v 1 ) + 3(n 1) = 1 + 3(n 1) = 3n. 3. Complete Bipartite Graphs and Stars Theorem. For any complete bipartite graph, k(k m,n ) = (m + n). Proof. Let G = (V, E) be a complete bipartite graph, K m,n, and let A = {a 1, a,..., a m } and B = {b 1, b,..., b n } be the two sets of vertices that partition V as per definition of a complete bipartite graph. Let f be a minimal L(3,, 1)-labeling of G with f(a 1 ) f(a ) f(a m ) and f(b 1 ) f(b ) f(b n ). Note that d(a i, a j ) = for all a i, a j A with i j; the same is true for pairs of vertices in B. Let f(a 1 ) = 1, then we need each f(a i ) with i 1 to be odd since f is minimal. Thus, f(a) = {1, 3, 5,..., 1 + (m 1)}. Since each a i A is adjacent to every b i B, we need f(a i ) f(b i ) 3. Note that we need f(b 1 ) f(a m ) + 3. Then, since f(a m ) = m 1, f(b 1 ) 3 + (m 1). Since f is minimal we have f(b 1 ) = 3 + (m 1) = m +. The labeling of vertices in B follows an argument similar to the labeling of vertices in A: since f(b 1 ) = m we need each f(b i ) to be even. Thus, f(b) = {m +, m + 4,..., m + + (n 1)}. Then f(b n ) = m + + (n 1) = (m + n). Therefore k(k m,n ) = (m + n). Corollary 3. For a star, S n, k(s n ) = n +. Proof. By definition, a star, S n, is K 1,n. Therefore k(s n ) = n +. 4

5 3.3 Paths Lemma 4. For a path on n vertices, P n, with n 8, k(p n ) 8. Proof. Let f be a minimal L(3,, 1)-labeling for a path on n vertices, P n, and suppose k(p n ) < 8 for n 8. Let v 1 be a vertex with label 1. There is an induced subpath of at least 5 vertices with v 1 as an end vertex. Let V = {v 1, v, v 3, v 4, v 5 } be this path. We will continue by considering the possibilities for f(v ). Case I: f(v ) = 4: Then we need f(v 3 ) = 7, f(v 4 ) =, f(v 5 ) = 5, and f(v 6 ) 8, which contradicts our assumed k(p n ). Case II: f(v ) = 5: Then we need f(v 3 ) 8, a contradiction to our assumed k(p n ). Case III: f(v ) = 6: Then we need f(v 3 ) = 3, forcing f(v 4 ) 8, a contradiction. Case IV: f(v ) = 7: Then f(v 3 ) = 3 or 4. Either possibility for f(v ) forces f(v 4 ) 9, which is a contradiction. Therefore we can conclude that k(p n ) 8, when n 8. Theorem 5. For any path, P n, 1, if n = 1; 4, if n = ; k(p n ) = 6, if n = 3, 4; 7, if n = 5, 6, 7; 8, if n 8. Proof. Let V = {v 1, v,..., v n } be the set of vertices of P n such that v i is adjacent to v i+1 for 1 i < n. Define f such that f({v 1, v,..., v 8 }) = {1, 4, 7,, 5, 8, 3, 6} and f(v i ) = f(v j ) if i j (mod 8). By definition of f we can conclude that k(p n ) 8 for n 8. We combine this result with that of Lemma 4 to get k(p n ) = 8 for n 8. For each P n with n < 8, we proceed by cases using the labeling pattern defined by f and the observation that each P 1, P,..., P 7 is an induced subpath of P n with n 8: Case I: n = 1. This is trivially true. Case II: n =. The labeling pattern {1, 4} shows that k(p ) = 4 because we cannot do any better. 5

6 Case III: n = 3, 4. The labeling pattern {3, 6, 1, 4} shows that k(p n ) 6 for n = 3, 4. Suppose k(p n ) < 6. There is a vertex v i V such that f(v i ) = 1. If v i has degree, then vertices v i 1 and v i+1 exist such that f(v i 1 ) 4 and f(v i+1 ) 6, a contradiction. If v i has degree 1, let v i = v 1, then the possibilities for f(v ) are 4 and 5. In either case, we would need f(v 3 ) > 6, a contradiction. Case IV: n = 5, 6, 7. The labeling pattern {3, 6, 1, 4, 7,, 5} shows that k(p n ) 7 for n = 5, 6, 7. Suppose k < 7. There is a vertex v i V such that f(v i ) = 1 and vertices v i+1 and v i+ exist or v i 1 and v i exist. Without loss of generality, suppose v i+1 and v i+ exist. The possibilities for f(v i+1 ) are 4, 5, and 6. If f(v i+1 ) = 4, 5 we need f(v i+ ) 7, a contradiction. If f(v i+1 ) = 6 then f(v i+ ) = 3. Now if v i+ has degree, then v i+3 exists and f(v i+3 ) > 7. If v i+ has degree 1 then vertices v i 1 and v i exist. The only possibility for f(v i 1 ) is 4. But this forces f(v i 1 ) 7, a contradiction. 3.4 Cycles Lemma 6. Let n be an odd integer, n > 3, then k(c n ) 8. Proof. Let f be a minimal L(3,, 1)-labeling of C n where n is odd and n > 3. Suppose k(c n ) = 8. Then the only possible labelings with 8 as the maximum integer are as follows: f 1 (V ) = {8, 5,, 7, 4, 1, 6, 3, 8... }, f (V ) = {8, 5,, 7, 4, 1, 8... }, f 3 (V ) = {8, 5, 1, 7, 4, X}, f 4 (V ) = {8, 5, 1, 7, 3, X}, f 5 (V ) = {8, 4, 1, 6, 3, 8,... }, f 6 (V ) = {8, 4, 1, 7, 3, X}, f 7 (V ) = {8, 3, 6, 1, 8,... }, f 8 (V ) = {8,, 5, X}, f 9 (V ) = {8,, 6, X}, and f 10 (V ) = {8, 1, 5, X}. Observe that f 1, f, and f 7 are labelings for even cycles. Of the remaining labelings f 3, f 4, f 6, f 8, f 9, and f 10 all fail as potential labelings for a cycle because the missing X labels are forced to be greater than the assumed k(c n ). The only labeling left is f 5, which also fails to label a cycle because it does not provide a repeatable pattern. Therefore, no minimal L(3,, 1)-labeling exists for C n with odd n and n > 3 such that k(c n ) = 8. Lemma 7. k(c 4 ) = 8. Proof. The labeling pattern {1, 6, 3, 8} shows that k(c 4 ) 8. Now, let f be minimal L(3,, 1)-labeling of C 4 and suppose k(c 4 ) < 8. Let f(v 1 ) = 1 and v and v 4 be the two vertices adjacent to v 1. Then the possibilities for {f(v ), f(v 4 )} are {4, 6}, {4, 7}, and {5, 7}. Case I: {f(v ), f(v 4 )} = {4, 6}: Then we need f(v 3 ) = 9, a contradiction to our assumption that k(c 4 ) < 8. Case II: {f(v ), f(v 4 )} = {4, 7} or {5, 7}: Then we need f(v 3 ) = 10, a contradiction to the assumed k(c 4 ). Therefore k(c 4 ) = 8. 6

7 Lemma 8. k(c 5 ) = 9. Proof. By Theorem 5 we know that k(c 5 ) 7 since k(p 5 ) = 7. Let f be minimal L(3,, 1)-labeling of C 5 and suppose k(c 5 ) = 7. Let f(v 1 ) = 1 and let v and v 5 be the two vertices adjacent to v 1. The possibilities for {f(v ), f(v 5 )} are {4, 6}, {4, 7}, and {5, 7}. Case I: {f(v ), f(v 5 )} = {4, 6}: Then we need f(v 3 ) = 8, a contradiction to our assumption that k(c 5 ) < 8. Case II: {f(v ), f(v 5 )} = {4, 7} or {5, 7}: Then we need f(v 3 ) = 9, a contradiction to the assumed k(c 5 ). Since k(c 5 ) = 7 is impossible, we have k(c 5 ) 8. By Lemma 6 for odd cycles, we know that k(c 5 ) 8. So we have k(c 5 ) 9. The labeling pattern {5, 1, 7, 3, 9} shows that k(c 5 ) 9. Therefore k(c 5 ) = 9. Lemma 9. k(c 6 ) = 8. Proof. The labeling pattern {1, 4, 7,, 5, 8} shows that k(c 6 ) 8. Now, let f be minimal L(3,, 1)-labeling of C 6 and suppose k(c 4 ) < 8. Let f(v 1 ) = 1 and let v and v 6 be the two vertices adjacent to v 1. The possibilities for {f(v ), f(v 6 )} are {4, 6}, {4, 7}, and {5, 7}. Case I: {f(v ), f(v 6 )} = {4, 6}: Then we need f(v 3 ) = 7 and f(v 4 ) =. contradiction to k(c 6 ) < 8. Case II: {f(v ), f(v 6 )} = {4, 7} or {5, 7}: Then we need f(v 3 ) = 8, a contradiction. Therefore k(c 6 ) = 8. For v 5, we need f(v 5 ) = 9, a Lemma 10. k(c 7 ) = 10. Proof. Let V = {v 1, v,..., v 7 } be the set the vertices of C 7. Suppose k(c 7 ) < 10 and let f be a minimal L(3,, 1)-labeling of C 7, then the possible values for f(v ) are in S = {1,,..., 9}. Since the greatest distance between any two vertices in V is 3, none of the possible values for f(v ) can be repeated. Also, only up to two consecutive labels may be used. Using three consecutive labels is not possible due to the distance constraints of labeling V. Then at most 6 labels can be used from S, forcing the unlabeled vertex to be labeled with a value greater than 10, a contradiction to the assumed k(c 7 ). Thus, k(c 7 ) 10. The labeling pattern {1, 5, 8,, 10, 7, 4} shows that k(c 7 ) 10. Therefore, k(c 7 ) = 10. Lemma 11. Let n be an even integer. If n 4, then n = 4a + 6b; where a and b are non-negative integers. 7

8 Proof. Let S = {4a + 6b a, b Z; a, b 0} and let n be an even integer. Then n 0 (mod 4) or n (mod 4). Suppose that n 4. Case I: n 0 (mod 4): Then there exists an integer q such that n = 4q. Since n 4, q 1. This implies n = 4a + 6b where a = q and b = 0. Then, n S. Case II: n (mod 4): Then there exists an integer q such that n = 4q +. This implies that n = 4(q 1) + 6. Since n 4 and n (mod 4), we get n 6. Then (q 1) 0. From this, it follows that n = 4a + 6b where a = (q 1) and b = 1. Then, n S. Therefore, if n is an even integer and n 4 then n = 4a + 6b, where a and b are non-negative integers. Lemma 1. Let n be an odd integer. If n 9 and n 11, then n = 4a + 5b; where a and b are non-negative integers. Proof. Let S = {4a + 5b a, b Z; a, b 0} and let n be an odd integer. Then n 1 (mod 4) or n 3 (mod 4). Suppose n 9 and n 11. Case I: n 1 (mod 4): Then there exists an integer q such that n = 4q + 1. This implies that n = 4(q 1) + 5. Since n 9, we need (q 1) 1. This implies that n = 4a + 5b where a = (q 1) and b = 1. Then, n S. Case II: n 3 (mod 4): Then there exists an integer q such that n = 4q + 3. This implies that n = 4(q 3) + 5(3). Since n 9, n 11, and n 3 (mod 4), we get n 15. Then we need (q 3) 0. Since n = 4a + 5b where a = (q 3) and b = 3, we get n S. Therefore, if n is an odd integer such that n 9 and n 11 then n = 4a+5b for non-negative integers a and b. Theorem 13. For any cycle, C n with n 3, 7, if n = 3; 8, if n is even; k(c n ) = 9, if n is odd and n 3, 7; 10, if n = 7. Proof. Let n 3 and V = {v 1, v,..., v n } be the vertices of C n such that for 1 i < n, (v i, v i+1 ) and (v 1, v n ) are edges in C n. The following cases describe all the possibilities for k(c n ). 8

9 Case I: n = 3: Observe that C 3 is a complete graph. Therefore, by Theorem 1 for complete graphs, k(c 3 ) = 7. Case II: n is even: We have k(c 4 ) = 8 by Lemma 7 and we have k(c 6 ) = 8 by Lemma 9. We also know from Theorem 5 for paths that k(p n ) = 8 for n 8. Then for all even cycles, k(c n ) 8. Recall the labelings used for C 4 and C 6 in Lemmas 7 and 9, respectively: for C 4 we used f(v ) = {1, 6, 3, 8} and for C 6 we used f(v ) = {1, 4, 7,, 5, 8}. Observe that the labeling used for C 4 can be repeated infinitely for all C n with n a multiple of 4. Likewise, the labeling used for C 6 can be repeated infinitely for all C n with n a multiple of 6. Moreover, the labelings of C 4 and C 6 can be joined together to label C 10 as such: f(v ) = {1, 6, 3, 8, 1, 4, 7,, 5, 8}. From Lemma 11, we know that every even integer greater than or equal to 4 can be expressed as some combination of non-negative multiples of 4 and 6. From this it becomes obvious that the labeling of every even C n can be composed of combinations of non-negative multiples of the C 4 and C 6 labeling patterns. Therefore k(c n ) = 8 for all even n. Case III: n is odd and n 3 or 7: By Lemma 8 we have k(c 5 ) = 9, by Theorem 5 for paths we know that k(c n ) 8 for n 8 since k(p n ) = 8, and by Lemma 6 we know that k(c n ) 8 for odd cycles. This implies that for all odd cycles k(c n ) 9. In Lemma 8 we labeled C 5 with {5, 1, 7, 3, 9}. Note that labeling of C 5 can be repeated infinitely for all C n with n a multiple of 5. Also, we can combine the labeling for C 5 with the labeling for C 4 used in Lemma 7 to label C 9 as such: {5, 1, 7, 3, 9, 1, 6, 3, 8}. By Lemma 8 we have that every odd integer greater than or equal to 9, with the exception of 11, can be expressed as some combination of non-negative multiples of 4 and 5. This implies that every C n, with n 9 and n 11, is composed of combinations of non-negative multiples of C 4 and C 5. Therefore, since k(c 5 ) = 9, we have k(c n ) = 9 for all C n, where n 9 and n 11. For C 11 we know that k(c 11 ) 9 (from Lemma and Theorem 3). Define f for C 11 such that f(v ) = {1, 6, 3, 8, 5, 1, 9, 6,, 8, 4}. Since max(f(v )) = 9, k(c 11 ) = 9. Case IV: n = 7: By Lemma 10, we have k(c 7 ) = Caterpillars Lemma 14. For a uniform caterpillar, Cat n with n >, k(cat n ) > +. Proof. Suppose k(cat n ) +. Begin with any vertex, v on the spine, label it i where 1 i +. Let m be the number of available labels for the vertices adjacent to v. We will show that m if i = 1 or + and m 1 otherwise. To determine m, we count the possible labels, j, for vertices adjacent to v such that i j 3. As 1 and + are the respective minimum and maximum possibilities for any label, we know that 1 j i 3 9

10 or i + 3 j +. Since any pair of vertices adjacent to v has a distance of, no two consecutive numbers may be used as labels. Then, we can compute m by: ( ) ( ) (i 3) 1 ( + ) (i + 3) m (1) Case I: i = 1: As 1 is the lowest possible label for v, all available labels would be greater than 1. As such, equation 1 becomes: ( ) ( ) ( + ) (1 + 3) m () Thus, when v is labeled with with 1, there are labels available for the adjacent vertices. Case II: i = + : Then, since + is the highest label possible for v, equation 1 becomes: ( ) ( ) (( + ) 3) 1 m (3) Thus, when v is labeled with +, there are labels available for the adjacent vertices. Case III: 1 < i < + : As n >, at least one label must be used on the spine aside from 1 and +. For this spinal vertex, equation 1 is used and yields: m (i 3) 1 + ( + ) (i + 3) i i As m must be an integer (there can be no partial labels), for any v labeled with 1 < i < +, there are at most 1 labels available for vertices. Therefore, by the Pigeon-Hole Principle, any vertex labeled 1 < i < + on Cat n cannot have k(cat n ) +. Thus, k(cat n ) > +. Theorem 15. For a uniform caterpillar, Cat n, with > and n >, k(cat n ) = + 3. Proof. The labeling pattern below illustrates the pattern of an L(3,, 1)-labeling with k(cat n ) = + 3. To label the spine, we use the pattern 1,, + 3, repeatedly. To label the legs adjacent to a vertex labeled i on the spine use positive values from the set {i ± (3 + p) : 0 p }. As k(cat n ) > +, k(cat n ) = + 3 is the lowest possible k. + + (4) 10

11 Corollary 16. For any caterpillar Cat n with > and n vertices on the spine, either k(cat n ) = + or k(cat n ) = + 3. When there are no more than two vertices of degree within any subgraph of four spinal vertices on Cat n, k(cat n ) = +. Proof. This follows from the proof of the uniform caterpillar: only vertices labeled 1 and + had label options for vertices. All other spinal labels yielded at most 1 possible labels. Thus, all spinal vertices labeled with 1 < i < + must have a degree of at most 1, in order to keep the largest label to + or less (case III, Theorem 14). Also, cases I and II of Theorem 14 show that a non-uniform caterpillar with k(cat n ) = + may only have a degree of at vertices labeled with 1 and +. Notice that the labels 1 and + can be repeated every four labels and that using this pattern maximizes the total number of vertices on the caterpillar. The pattern below illustrates one labeling of the vertices that fulfills this requirement and maximizes the quantity of vertices. 3.6 N-ary Tree Lemma 17. For any n-ary tree, G, k(g) n + 6. Proof. This proof is by construction. Let G = (V, E) be an n-ary tree. First, observe that the induced subgraph of each vertex and all of the vertices adjacent to it form a star. With this observation, we propose the following L(3,, 1)-labeling of each star: let f be a minimal L(3,, 1)-labeling of a star with central vertex x and define f such that the adjacent vertices of x are labeled with the lowest positive integers from the set {f(x) ± (3 + i) : i Z + {0}}. In other words, adjacent vertices of x are first labeled with as many positive elements as possible from the set {f(x) 3, f(x) 5, f(x) 7... } and the remaining adjacent vertices are labeled with the lowest positive elements from the set {f(x) + 3, f(x) + 5, f(x) + 7,... }. We will show that these stars can be joined via overlapping edges to create an n-ary tree such that f is an L(3,, 1)-labeling. It follows from construction that for any two adjacent vertices x and y, f(x) f(y) 3. It also follows from construction that for any two vertices x and y that are mutually adjacent to a vertex w, f(x) f(y). As for vertices x, y with d(x, y) = 3, we observe that for each star (induced subgraph of G), with central vertex x: if f(x) is even then for all vertices y adjacent to x, f(y) is odd; and if f(x) is odd then for all vertices y adjacent to x, f(y) is even. Hence for vertices x, y with d(x, y) = 3, x and y have different parity. Thus, d(x, y) = 3 implies f(x) f(y) 1. Thus, f is an L(3,, 1)-labeling for any n-ary tree. Claim: The highest value of f used in the proposed L(3,, 1)-labeling is + 4. Note that the worst case for the proposed L(3,, 1)-labeling is when the central vertex, x, of a star has f(x) = 3. This is because none of the vertices adjacent to x can be labeled using elements of the set {f(x) 3, f(x) 5, f(x) 7,... }. 11

13 there are enough labels in the sets {4,..., + 3} and {5, 7, 9,, + 3} for labeling each element of A, so we can use these sets in an L(3,, 1)-labeling. Now we show that no other sets are possible. Suppose the minimum label used for elements of A was at least 6. Then in order to have enough labels, a label at least 6 + ( 1) = + 4 = n + 6 would need to be used, which contradicts the assumption k(g) < n + 6. Therefore, the smallest label must be less than 6. If the smallest label is 5, or if we use only even labels, the labels are {5, 7, 9,..., + 3}, and {4, 6, 8,..., + }, respectively. Otherwise the labels must come from the set {4,..., + 3}. Theorem 0. For any n-ary tree, G, k(g) = n + 6. Proof. Let G be an n-ary tree. By Lemma 17 we know that k(g) n + 6. If k(g) < n + 6 then, by Lemma 18, we know that neither 6 nor 7 can appear as a label in A. By Lemma 19 we know that the labels of A must come from one of the following sets: {4, 6, 8,..., + }, {5, 7, 9,..., + 3} or one of the sets {4,..., + 3}. Notice that 6 is an element of the set {4, 6, 8,..., + }, 7 is an element of the set {5, 7, 9,..., + 3}, and either 6 or 7 is an element of any set of the form {4,..., + 3}. Therefore none of these sets can provide the labels for A. Therefore, k(g) = n The Upper Bound in Terms of Maximum Degree, Theorem 1. If G is a graph with maximum degree, k(g) Proof. Let N i (v) be the set of vertices of distance i from a vertex v; the members of the sets of 1 i 3 shall be referred to as neighbors of v. When constructing a greedy L(3,, 1)-labeling f on a graph G, each vertex surrounding v is given the lowest available label such that no two labels are repeated within distance 3 of v. Furthermore, each neighbor of v blocks a certain number of labels from use: N 1 (v) blocks up to 5 labels, N (v) up to 3, and N 3 (v) prohibits the use of the label given to itself. The maximum label needed for v is calculated through a sum of the quantity of vertices within distance three of v while weighting each set of neighbors according to the quantity of labels each blocks. As the maximum degree of G is at most, N 1, N ( 1), and N 3 ( 1). Thus, k(g) ( 1) + ( 1), or k(g)

14 References [1] G. Chartrand, D. Erwin, F. Harary, and P. Zhang, Radio labeling of graphs, Bull. Inst. Combin. Appl., 33 (001), [] J. R. Griggs and R. K. Yeh, Labeling graphs with a condition at distance two, SIAM J. Discrete Math., 5 (199) [3] W. K. Hale, Frequency assignment: theory and application, Proc. IEEE, 68 (1980), [4] L. Jia-zhuang and S. Zhen-dong, The L(3,, 1)-labeling problem on graphs, Mathematica Applicata, 17 (4) (004), [5] K. Jonas, Graph coloring analogues with a condition at distance two: L(,1)-labelling and list λ-labellings, Ph.D. thesis, University of South Carolina (1993), 8-9. [6] F. S. Roberts, T-colorings of graphs: recent results and open problems, Discrete Math., 93 (1991),

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