Radio Number for Square of Cycles

Radio Number for Square of Cycles Daphne Der-Fen Liu Melanie Xie Department of Mathematics California State University, Los Angeles Los Angeles, CA 9003, USA August 30, 00 Abstract Let G be a connected graph with diameter d For any two vertices u and v, let d G(u, v) denote the distance between u and v A radio labelling for G is a function f that assigns to each vertex a nonnegative integer (label) such that the in-equality f(u) f(v) d d G(u, v) + 1 holds for any two vertices u and v The span of f is the difference of the maximum and the minimum labels assigned The radio number of G is the minimum span over all radio labellings of G The radio numbers for paths and cycles were studied by Chartrand et al and Zhang, and were completely determined by Liu and Zhu We study the radio number for square cycles (cycles by adding edges between vertices of distance two apart) The radio numbers for the square of even cycles are completely determined For the square of an odd cycle, we obtain a lower bound for the radio number, and show that the bound is achieved by many cases, while not achievable by some other cases 1 Introduction Radio labelling is motivated by the AM/FM radio channel assignment problem (cf [1, 3]) Suppose we are given a set of stations (or transmitters), the task is to assign to each station (or transmitter) a channel (non-negative integer) such that the interference is avoided The interference is closely related to the geographical location of the stations, the smaller distance between two stations, the stronger interference between them might occur To Supported in part by the National Science Foundation under grant DMS 03056 Corresponding author Email: dliu@calstatelaedu 1

avoid stronger interference, the larger separation of the channels assigned to the stations must be To model this problem, we construct a graph, such that each station is represented by a vertex, and two vertices are adjacent when their corresponding stations are close The ultimate goal is to find a valid labelling for the vertices such that the span (range) of the channels used is minimized For a given simple connected graph G, the distance between any two vertices u and v, denoted by d G (u, v) (or d(u, v) when G is understood in the context), is the length (number of edges) of a shortest (u, v) path The diameter of G, denoted by diam(g) (or d if G is understood in the context), is the maximum distance between two vertices in G A radio labelling (also known as multi-level distance labeling [5]) of G is a function, f : V (G) {0, 1,, }, such that the following holds for any u and v: f(u) f(v) diam(g) d(u, v) + 1 The span of f is the difference of the largest and the smallest channels used, {f(v) f(u)} The radio number of G is defined as the minimum max u,v V (G) span of a radio labelling of G The radio numbers for different families of graphs have been studied by several authors The radio number for paths and cycles was investigated by [3, 1, 1], and was completely solved by Liu and Zhu [5] The radio number for spiders (trees with at most one vertex of degree more than ) has been investigated by Liu [6] The square of a graph G, denoted by G, has the same vertex set as G, and the edge set E(G ) = E(G) {uv : d G (u, v) = } The radio number for the square of paths is completely settled in [7] In this article, we investigate the radio number for the square of cycles (or square cycles), denoted by Cn We completely determine the radio numbers for the squares of even cycles Ck For the radio number of the square of odd cycles, we obtain a lower bound, and show that the bound is achieved by many cases, while not achievable by some other cases A lower bound Throughout the article, we denote V (C n ) = V (Cn) = {v 0, v 1, v,, v n 1 }, E(C n ) = {v i v i+1 : i = 0, 1,, n 1} and E(Cn) = {v i v i+1, v i v i+ : i = 0, 1,, n 1} All the sub-indices are taken under modulo n For instance, v n = v 0, v n+i = v i, etc For any u, v V (C n ), we denote the distance between u and v, in Cn and C n, respectively, by d(u, v) and d Cn (u, v) The diameter of Cn is denoted by d For any radio labelling f of Cn, we re-name the vertices of Cn by

V (C n) = {x 0, x 1,, x n 1 }, where 0 = f(x 0 ) < f(x 1 ) < f(x ) < < f(x n 1 ) Since we are searching for the minimum span of a radio labelling, without loss of generality, we assume f(x 0 ) = 0 Hence, the span of f is f(x n 1 ) For any i = 0, 1,,, n, we set d i = d(x i, x i+1 ); and f i = f(x i+1 ) f(x i ) A radio labelling f of a graph G is called optimal if the span of f is equal to the radio number of G Lemma 1 Let f be a radio labelling for Cn Then for any i = 0, 1,, n, k+, if n = k; k+1 f(x i+ ) f(x i ) = f i+1 + f i, if n = k + 1; k+, if n = k + ;, if n = k + 3 k+3 Proof Let n = k + r, r = 0, 1,, 3 For any i = 0, 1,,, n, let Then, we have l i = d Cn (x i, x i+1 ), and l i = d Cn (x i+, x i ) d(x i, x i+1 ) = li l, and d(x i+, x i ) = i Assume r = 0, 1 Then diam(cn) = k If l i k, then d(x i, x i+1 ) k Hence, by definition, we have f i = f(x i+1 ) f(x i ) k d(x i, x i+1 ) + 1 k k + 1 k This implies that f(x i+ ) f(x i ) = f i + f i+1 > k + 1 So the result follows Assume l i < k Then d(x i, x i+ ) k Hence, by definition, we have k + f(x i+ ) f(x i ) k d(x i, x i+ ) + 1 Thus, we assume l i, l i+1 k+1, and l i k This implies that l i+l i+1 +l i = n By definition, we have f i = f(x i+1 ) f(x i ) k li + 1, f i+1 = f(x i+ ) f(x i+1 ) k li+1 + 1, f i + f i+1 = f(x i+ ) f(x i ) k l i + 1 (1) 3

Summing the three inequalities in the above, ( li (f(x i+ ) f(x i )) 3k + 3 + li+1 ) l + i () If n = k, then at most two of the elements in {l i, l i+1, li } are odd, so li + li+1 + l i k+ = k + 1 If n = k + 1, then li + li+1 l + i k + = k + (3) Therefore, the result follows by combining () and (3) Similarly, one can show the cases for r =, 3 (in which d = k + 1, and one can first prove that the results follow if l i < k + 1 or li < k, then verify the rest by some calculation similar to the above) We leave the details to the reader Theorem Let n = k + r, r = 0, 1,, 3 Then k +5k 1, if r = 0 and k is odd; k +3k, if r = 0 and k is even; k + k, if r = 1 and k is odd; rn(cn) k + k, if r = 1 and k is even; k + 5k + 1, if r = and k is odd; k + k + 1, if r = and k is even; k +7k+3, if r = 3 and k is odd; k +9k+, if r = 3 and k is even Proof Let f be a radio labelling for Cn Assume r = 0 (so n 1 is odd) By Lemma 1, the span of f has: ( ) n k + f(x n 1 ) = f 0 + f 1 + f + + f n + 1 Considering the parity of k separately to the above, the results for r = 0 follow Assume r = 1 (so n 1 is even) By Lemma 1, we get ( ) n 1 f(x n 1 ) = f 0 + f 1 + f + + f n k + 1 Considering the parity of k separately, the results for r = 1 follow The cases for r =, 3 can be proved by the same method We leave the details to the reader

3 Square of Even Cycles We prove that for any even cycle, the radio number is equal to the lower bound proved in Theorem It suffices to give radio labellings with spans equal to the lower bound Throughout this article, we express each radio labelling f, with 0 = f(x 0 ) < f(x 1 ) < f(x ) < < f(x n 1 ), by the following two: A permutation τ on {0, 1,,, n 1}, with τ(0) = 0 and x i = v τ(i) Unless indicated, we let f(x 0 ) = f(v 0 ) = 0, and f(x i+1 ) = f(x i ) + d + 1 d(x i+1, x i ) for i = 0, 1,, n To show that each f, expressed in the above format, is a radio labelling, we shall (and it suffices to) verify that all the following hold, for any i: (1) τ is a permutation () f(x i+ ) f(x i ) d d(x i, x i+ ) + 1 (3) f(x i+3 ) f(x i ) d d(x i, x i+3 ) + 1 () f(x i+ ) f(x i ) d + 1 Theorem 3 Let C n be an n-vertex square cycle where n is even Then k +5k 1, if n = k and k is odd; rn(cn) k +3k =, if n = k and k is even; k + 5k + 1, if n = k + and k is odd; k + k + 1, if n = k + and k is even Proof By Theorem, it suffices to find radio labellings with spans achieving the desired numbers We consider cases Case 1: n = k and k is odd Define τ(0) = 0 and: τ(i) + k (mod n), if i 0, (mod ); τ(i + 1) = τ(i) + k (mod n), if i 1 (mod ); τ(i) + k + 1 (mod n), if i 3 (mod ) By the definition of τ, and the fact that k is odd, we have { k, if i is even; d i = d(x i, x i+1 ) = d(v τ(i), v τ(i+1) ) = k+1, if i is odd Now we show that f is a radio labelling 5

Claim (1): τ is a permutation For i = 0, 1,, k 1, we re-write τ by: τ(i) = (k + k + k + k + 1)i = (i + 0)k + i (mod n); τ(i + 1) = ki + i + k = (i + )k + i (mod n); τ(i + ) = ki + i + k + k = (i + 3)k + i (mod n); τ(i + 3) = ki + i + 3k + k = (i + 1)k + i (mod n) Assume τ(i + j) = τ(i + j ) Then, (i i + t t )k i i (mod n) for some t, t {0, 1,, 3} This implies, i = i, as 0 i i < k Therefore, it must be t = t, because t t 3 So, τ is a permutation Claim (): f(x i+ ) f(x i ) d d(x i, x i+ ) + 1 By the definition of τ, we have d i + d i+1 = k + k+1 = 3k+1, and d(x i, x i+ ) { k 1, k+1 } So, d(x i, x i+ ) k 1 Therefore, f(x i+ ) f(x i ) = f i + f i+1 = (d d i + 1) + (d d i+1 + 1) = k + 3k+1 k d(x i, x i+ ) + 1 Claim (3): f(x i+3 ) f(x i ) d d(x i, x i+3 ) + 1 By the definition of τ, we have { k+1 d(x i, x i+3 ) =, if d i + d i+1 + d i+ = k + k+1 + k; 1, if d i + d i+1 + d i+ = k+1 + k + k+1 By some calculation, we conclude that f(x i+3 ) f(x i ) = (d d i + 1) + (d d i+1 + 1) + (d d i+ + 1) d d(x i, x i+3 ) + 1 Claim (): f(x i+ ) f(x i ) k + 1 By the definition of τ, we have d i + d i+1 + d i+ + d i+3 = 3k + 1 Therefore, f(x i+ ) f(x i ) = k (d i + d i+1 + d i+ + d i+3 ) + = k + (3k + 1) k + 1 By Claims (1 ), f is a radio labelling The span of f is: f(x n 1 ) = d d 0 + 1 + d d 1 + 1 + + d d n + 1 = (n 1)(d + 1) (d 0 + d 1 + d + + d n ( ( ) ) 3k + 1 = (k 1)(k + 1) (k 1) + k Figure 1 shows an optimal radio labelling for C 1 with span 16 = k + 5k 1 6

15 10 3 13 6 1 0 7 1 9 16 Figure 1: An optimal radio labelling for C 1 Case : n = k and k is even If k =, then (0, 3, 6, 1,, 7,, 5) gives an optimal radio labelling for V (C8) = (v 0, v 1, v, v 3,, v 7 ) Assume k Define τ(0) = 0 and: { τ(i) + k (mod n), if i is even; τ(i + 1) = τ(i) + k + 1 (mod n), if i is odd By the definition of τ, we obtain { k, if i is even; d i = d(x i, x i+1 ) = d(v τ(i), v τ(i+1) ) = k+, if i is odd Now we prove that f is a radio labelling Claim (1): τ is a permutation For i = 0, 1,, k 1, we re-write τ by: τ(i) = (3k + 1)i (mod n), τ(i + 1) = (3k + 1)i + k (mod n) Assume τ(i) = τ(j) or τ(i + 1) = τ(j + 1) for some 0 i, j k 1 Then we have (3k + 1)(i j) 0 (mod k) This implies that i = j, because j i k 1, and gcd(k, 3k + 1) = 1 (as k is even) Assume τ(i + 1) = τ(j) for some i, j k 1 Because (3k + 1) (1 k) (mod k), we have (i j)(3k+1) (j i)(k 1) k (mod k) This implies that (j i+)k (j i) (mod k) Because j i k 1, it must be either j i = 0 or j i = k The former case implies that (k) 0 (mod k), a contradiction; and the latter case implies that (k + ) 1 (mod ), contradicting that k is even Therefore, τ is a permutation Claim (): f(x i+ ) f(x i ) d d(x i, x i+ ) + 1 By the definition of τ, we have, for any i: d(x i, x i+ ) = k, and d i + d i+1 = k + k + 7 = 3k +

Hence f(x i+ ) f(x i ) = (d d i + 1) + (d d i+1 + 1) = k + 3k + k d(x i, x i+ ) + 1 Claim (3): f(x i+3 ) f(x i ) d d(x i, x i+3 ) + 1 d(x i, x i+3 ) = By some calculation, we get By τ, we have { k+, if d i + d i+1 + d i+ = k + k+ + k; 1, if d i + d i+1 + d i+ = k+ + k + k+ f(x i+3 ) f(x i ) = (d d i + 1) + (d d i+1 + 1) + (d d i+ + 1) d d(x i, x i+3 ) + 1 Claim (): f(x i+ ) f(x i ) k + 1 As d i + d i+1 + d i+ + d i+3 = 3k +, so f(x i+ ) f(x i ) = d (d i + d i+1 + d i+ + d i+3 ) + = k + (3k + ) k + 1 Therefore, f is a radio labelling with span k +3k, as f(x n 1 ) = (d d 0 + 1) + (d d 1 + 1) + + (d d n + 1) = (k 1)k (k 1)(3k + )/ k + (k 1) = k + 3k Figure shows an optimal radio labelling for C 16 with span 10 18 3 13 1 6 16 0 15 7 1 19 1 9 Figure : An optimal radio labelling for C 16 Case 3: n = k + Define τ(0) = 0 and: τ(i + 1) = { τ(i) + k + 1 (mod n), if i is even; τ(i) + k + 1 (mod n), if i is odd 8

By τ, we obtain { k + 1, if i is even; d i = d(x i, x i+1 ) = k+1, if i is odd We now verify that f is a radio labelling Claim (1): τ is a permutation For i = 0, 1,, k, we re-write τ by the following: τ(i) = (3k + )i (mod n), τ(i + 1) = (3k + )i + k + 1 (mod n) Assume τ(i) = τ(j) or τ(i + 1) = τ(j + 1), for some i, j k Then, (3k + )(i j) 0 (mod k + ) Moreover, because 3k + = k (mod n), we have (j i)k 0 (mod n) This implies that i = j, as j i k < n and gcd(k, k + ) Assume τ(i) = τ(j +1), for some i, j k Then, we have (3k+)(i j) k(j i) k + 1 (mod k + ) This implies that (j i )k 1 (mod k + ) Hence, gcd(k, k + ) = 1, and so k must be odd Note that, (k + 1)(k 1) = (k + ) ( ) k 1 0 (mod k + ), so (k 1)k 1 (mod k+) Therefore, j i = k 1, which is impossible, as j i k Therefore, τ is a permutation Claim (): f(x i+ ) f(x i ) k + 1 d(x i, x i+ ) + 1 By τ, we have d i + d i+1 = k + 1 + k+1 and d(x i, x i+ ) = k Hence, f(x i+ ) f(x i ) = (k + ) (k + 1 + k + 1 ) = k + 1 d(x i, x i+ ) + 1 Claim (3): f(x i+3 ) f(x i ) k + 1 d(x i, x i+3 ) + 1 By τ, we have { 1, if di + d d(x i, x i+3 ) = i+1 + d i+ = k + 1 + k+1 ; k+1, if d i + d i+1 + d i+ = k + + k+1 By some calculation, we conclude that f(x i+3 ) f(x i ) = (d d i +1)+(d d i+1 +1)+(d d i+ +1) k+ d(x i, x i+3 ) Claim (): f(x i+ ) f(x i ) k + This follows by the fact that d i + d i+1 + d i+ + d i+3 (k + 1 + k+ ) Hence, f is a radio labelling The span of f has two possibilities: If k is odd, then d i + d i+1 = k + 1 + k+1 = 3k+3 for all i Therefore, f(x n 1 ) = (n 1)d + (n 1) (d 0 + d 1 + d + + d n ) = (k + 1)(k + ) (k(3k + 3) + k + 1) = k + 5k + 1 9

If k is even, then d i + d i+1 = k + 1 + k+ = 3k+ for all i So, we get f(x n 1 ) = k + k + 1 Figure 3 shows optimal radio labellings for C and C10 with spans 50 and 13, respectively 0 11 0 31 0 51 6 15 6 35 6 5 36 5 16 5 50 1 30 1 10 1 10 0 3 7 13 1 6 1 9 Figure 3: An optimal radio labelling for C and C 10, respectively Square of Odd Cycles The radio number for the square of odd cycles turned out to be more complicated than the even cycles We show that the lower bound proved in Theorem is sharp for many square of odd cycles, while this is not the case for some others Theorem Let Cn be an n-vertex square cycle with n = k + 1 Then { rn(cn) k = + k, if k is even; k + k, if k 3 (mod ) Moreover, if k 1 (mod ), then k + k + 1 rn(c k+1) k + k + 10

Proof If k 0,, 3, by Lemma, it suffices to find a radio labelling f with span equal to the desired number In each of the following labelling f, we use the same notations as in the previous section Case 1: k 0 (mod ) Define τ by: τ(i) = i(3k + 1) (mod n), 0 i k; τ(i + 1) = i(3k + 1) + k i (mod n), 0 i k 1; τ(i + ) = (i + 1)(3k + 1) (mod n), 0 i k 1; τ(i + 3) = (i + 1)(3k + 1) + k i (mod n), 0 i k 1 By the definition of τ, we obtain k i k + 1 + i d i = d i+ =, and d i+1 = d i+3 = We verify that f is a radio labelling with span equal to the desired number Claim (1): τ is a permutation By some calculation, using the facts that (3k + 1) k (mod k + 1) and (3k + 1) k + 1 (mod k + 1), we re-write τ as: τ(i) = ik (mod n), 0 i k; τ(i + 1) = (i + 1)k (mod n), 0 i k 1; τ(i + ) = (i + 1)k (mod n), 0 i k 1; τ(i + 3) = (i + 1)k (mod n), 0 i k 1 Thus, for any 0 j n 1, τ(j) = ak (mod n) for some k a k Moreover, because there are k + 1 = n integers in the interval [ k, k], we conclude that if τ(j) = ak, τ(j ) = a k and a = a, then j = j Assume that τ(j) = τ(j ) for some j, j Let τ(j) = ak and τ(j ) = a k, for some k a, a k Then (a a )k 0 (mod k +1) This implies that a = a, since gcd(k, k + 1) = 1 and a a k By the discussion in the previous paragraph, it must be that j = j So, τ is a permutation Claim (): f(x i+ ) f(x i ) k d(x i, x i+ ) + 1 It suffices to show that d i + d i+1 d(x i, x i+ ) + k + 1 Let i = τ(j + t), and i + = τ(j + t ) for some 0 t, t 3 By the definition of τ, it suffices to consider the following two cases: Assume j = j Then d(x i, x i+ ) = k, and d i + d i+1 = k j + k+1+j 3k+ So, the claim follows Assume j = j + 1 Then d(x i, x i+ ) = k+, and d i + d i+1 = k+1+j + k (j+1) 3k+ So, the claim follows Claim (3): f(x i+3 ) f(x i ) k+1 d(x i, x i+3 )+1 It suffices to show that d i +d i+1 +d i+ d(x i, x i+3 )+k+ Let i = τ(j+t) and i+3 = τ(j +t ), 11

for some 0 t, t 3 By the definition of τ, we consider the following two cases: Assume j = j Then, d(x i, x i+3 ) { k j j+1, }, and { k j d i +d i+1 +d i+ = + k+1+j + k j, if d(x i, x i+3 ) = k j ; k+1+j + k j + k+1+j, if d(x i, x i+3 ) = j+1 By some calculation, we get d i + d i+1 + d i+ d(x i, x i+3 ) + k + So, the claim follows Assume j = j + 1 Then, d(x i, x i+3 ) { k 1 j, j+1 }, and { k j d i +d i+1 +d i+ = + k+1+j + k j 1, d(x i, x i+3 ) = k j 1 ; k+1+j + k j 1 + k+j+, d(x i, x i+3 ) = j+1 By some calculation, we get d i + d i+1 + d i+ d(x i, x i+3 ) + k + Claim (): f(x i+ ) f(x i ) k + 1 By the definition of τ, we have d i + d i+1 + d i+ + d i+3 3k So the result follows Therefore, f is a radio labelling Notice that d i + d i+1 = 3k+ for all i So the span of f is k + k, because f(x n 1 ) = (d d 0 + 1) + (d d 1 + 1) + + (d d n + 1) = (k)(k + 1) (d 0 + d 1 + d + + d n ) = k + k Figure shows an optimal radio labelling for C 17 with span 0 15 3 10 3 18 6 1 Figure : An optimal radio labelling for C 17 Case : k 3 (mod ) For 0 i n 1, let: ( ) 3k + 1 τ(i) = i 7 0 1 17 9 1 1 (mod n) 1

Hence, for 0 i n, 3k + 1 d i = = 3k + 3 We now show that the labelling f is a radio labelling Claim (1): τ is a permutation Assume, to the contrary, that ( ) 3k+1 i ( 3k+1 ) i (mod k + 1) for some 0 i, i k and i i This is impossible, since by the Euclidean Algorithm (assuming k = m + 3 for some m), one can easily show that gcd(k + 1, 3k+1 ) = gcd(16m + 13, 6m + 5) = 1 Claim (): f(x i+ ) f(x i ) k d(x i, x i+ )+1 By the definition of τ, we have d(x i, x i+ ) = k+1, and d i + d i+1 = ( 3k+3 ) So, the result follows Claim (3): f(x i+3 ) f(x i ) k d(x i, x i+3 )+1 By τ, we have d(x i, x i+3 ) = k+1, and d i + d i+1 + d i+ = 3 ( 3k+3 ) So the result follows Claim (): f(x i+ ) f(x i ) k + 1 This follows directly by d i + d i+1 + d i+ + d i+3 = 3k + 3 The span of f is k + k, because d 0 + d 1 + d + + d n = k(3k + 3) Figure 5 shows an optimal radio labelling for C9 with span 56 0 16 3 8 6 38 5 1 8 18 3 6 10 5 36 0 6 30 1 56 0 8 50 Figure 5: An optimal radio labelling for C 9 Case 3: k (mod ) To define the permutation τ, we first partition the vertices of C n into eight sets, in the following order: A, D, G, B, E, H, C, F, such that A consists of the first (k/)+1 vertices, A = {v 0, v 1, v,, v k/ }, and each other set consists of k/ consecutive numbers For instance, D = {v k +1,, v k 1, v k } and F = {v 7k +1,, v k} Secondly, we arrange the vertices in the above sets into Table 1: The permutation of τ is defined by the ordering given by the arrows Table 1 That is, x 0 = v 0, x 1 = v 3k +1,, x k = v k/, etc 13

A B C D E F G H v 0 v 3k +1 v 3k+1 v k +1 v k+1 v 7k +1 v k+1 v 5k +1 v 1 v 3k + v 3k+ v k + v k+ v 7k + v k+ v 5k + v k 1 v 3k + v k 3k+ k v k + v k k+ k v 7k + v k k+ k v 5k + k v k Table 1: By the arrangement of the vertices in Table 1, we observe that, for any i: d i = d(x i, x i+1 ) = 3k + ; d(x i, x i+ ) = k k ; and d(x i, x i+3 ) = = k + To verify that f is a radio labelling, it suffices to prove Claims ( ) Claim (): f(x i+ ) f(x i ) k d(x i, x i+ ) + 1 By the observation above, we have f(x i+ ) f(x i )=k + (d i + d i+1 ) =k + ( 3k + ) k d(x i, x i+ ) + 1 Claim (3): f(x i+3 ) f(x i ) k d(x i, x i+3 ) + 1 Similar to Claim (), as d i + d i+1 + d i+ = 3 ( 3k+ ), we get f(x i+3 ) f(x i ) = 3k + 3 (d i + d i+1 + d i+ ) k d(x i, x i+3 ) + 1 Claim (): f(x i+ ) f(x i ) k + 1 This can be proved similar to Claim (3) Therefore, f is a radio labelling The span of f is k + k, as d 0 + d 1 + d + + d n = k(3k + ) Figure 6 shows an optimal radio labelling for C5 with span 8 Case : k 1 (mod ) We first show that rn(cn) > k + k Suppose, to the contrary, there exists a radio labelling f with span k + k Since k is odd, by Lemma 1 and Theorem, we have f(x i+ ) f(x i ) = k + 1, for all i (1) Referring to the proof of Lemma 1, we get l i, l i+1, l i are all odd for any i, () 1

0 16 3 8 6 38 1 8 18 6 10 36 0 6 30 1 0 8 3 Figure 6: An optimal radio labelling for C 5 and all the in-equalities in (1), () and (3) are equalities (note, because the span of f is k +k and because k is odd, it must be that l i, l i+1 k +1, and li k) In particular, f(x i+) f(x i ) = k + 1 l i +1 for any 0 i n Combining this with (1), we get l i = d Cn (x i+, x i ) = k, for all i (3) By definition, for any i, 1 f(x i+1 ) f(x i ) f(x i+ ) f(x i ) = k 1 So, we have 1 k + 1 li+1, which implies k 1 k + l i = d Cn (x i+1, x i ) k 1 () Because f(x 0 ) = 0, by (1) and (3), we have f(x i ) = f(v n ki ) = (k + 1)i (mod n), for all 0 i k Because n = k + 1, the above implies that the following four blocks of vertices are occupied by f(x i ), i = 0, 1,,, k: A=[v 0,, v k+1 1], B=[v k+1,, v k+ k+1 1], ], ] C=[v k+1,, v k+ k+1 D=[v 3k+1,, v 3k+ k+1 Consider the remaining vertices (which are x i+1, i = 0, 1,, k 1): W =[v k+1,, v k ], X=[v k+ k+1,, v k ], Y =[v k+ k+1 +1,, v 3k], Z=[v 3k+ k+1 +1,, v k] 15

By (), if x i A, then x i+1 X, for some i = 0,, 8, 1,, k If x i = v j A for some even j, by (3), then x i+1 = v j X for some odd j Notice that there are ( k 1 + 1) vertices with even sub-indices t of v t in block A, while there are ( k 1 ) vertices with odd sub-indices t of v t in block X By pigeonhole principle, this is impossible Now, it suffices to find a labelling with span k + k + Define τ by: τ(i) =(3k + 1)i, i = 0, 1, k; τ(i + 1)=(3k + 1)i + k, i = 0, 1; τ(i + 1)=(3k + 1)i + k 1, i =, 3; τ(i + 1)=(3k + 1)(i ) + 3k+1,i =, 5,, k 1 We verify that f is a radio labelling Claim (1): τ is a permutation Assume τ(i) = τ(i ) for some 0 i, i k Then (3k + 1)(i i ) 0 (mod n) Because (3k + 1) k (mod n), we have k(i i) 0 (mod n) This implies i = i, since i i k Similarly, one can show that if τ(i+1) = τ(i +1) for some i, i =0, 1,, 3, or τ(i + 1) = τ(i + 1) for some i, i {, 5,, k}, then i = i Assume τ(i) = τ(i +1) for some i = 0, 1,,, k and i =, 5,, k 1 Because (3k + 1) 1 (mod n), so we have (3k + 1)(i i ) 3k 1 (mod n) Multiplying by to both sides, we get (3k + 1)(i i ) (i i ) 6k k 3 (mod n) Then, it must be i = i, because 0 i i k (if i > i ), and i i k 1 (if i > i) Note that, (k 3) k + (mod n) Assume τ(i) = τ(i +1) for some i = 0, 1,,, k and i = 0, 1 Then we have (3k + 1)(i i ) k (mod n), implying (i i) (mod n) This is impossible as 1 i i k Similarly, one can show that it is impossible that τ(i) = τ(i + 1) for some i = 0, 1,,, k and i =, 3 Assume τ(i + 1) = τ(i + 1) for some i =, 5,, k 1 and i = 0, 1 Then, we obtain (3k + 1)(i i ) k+1 (mod n) Multiplying both sides by, we get i i k + (mod n), which is impossible, because 3 i i k 1 Similarly, one can show that it is impossible that τ(i + 1) = τ(i + 1) for some i =, 5,, k 1 and i =, 3 Claim (): f(x i+ ) f(x i ) k + 1 d(x i, x i+ ) for all i It suffices to show that (d i + d i+1 ) d(x i, x i+ ) k + 1 By the definition of τ, we have d(x i, x i+ ) k+1 and d i + d i+1 6k+6 Hence, the result follows 16

Claim (3): f(x i+3 ) f(x i ) k + 1 d(x i, x i+3 ) for all i It suffices to show that (d i + d i+1 + d i+ ) d(x i, x i+3 ) k + By τ, we obtain: k+1, if d i + d i+1 + d i+ { 5k+1, 5k+3 }; 1, if d d(x i, x i+3 ) = i + d i+1 + d i+ { k+, k+, k+6 }, k 1, if d i + d i+1 + d i+ = 9k+7 ; k+3, if d i + d i+1 + d i+ = 9k+11 By some calculations, we conclude that the claim follows Claim (): f(x i+ ) f(x i ) k + 1 for any i By τ, we have d i + d i+1 + d i+ + d i+3 3k + 3 for any i So, the result follows Therefore, f is a radio labelling The span of f is k + k +, because f(x n 1 )=(n 1)(k + 1) (d 0 + d 1 + d + + d n ) ( =k + k k + k + 1 + k + k + 1 + k + k + 3 ( ) ( ) ( ) ( )) 3k + 1 k 8 3k + 5 k 8 + =k + k + + k + k + 3 + Figure 7 shows a radio labelling for C 1 with span 3 1 6 19 31 5 11 3 16 8 1 0 9 9 17 1 5 3 0 8 Figure 7: A radio labelling for C 1 Conjecture 1 If k 3 (mod ), then rn(c k+1 ) = k + k + For the radio numbers of Ck+3, we first prove a lemma Lemma 5 If k is odd and gcd(n, k) > 1, then rn(c k+3 ) > k +7k+3 17

Proof By Theorem, rn(cn) k +7k+3 Assume to the contrary, there exists a radio labelling f with f(x 0 ) = 0 and span f(x n 1 ) = k +7k+3 Then by the proof of Lemma 1, we have f i + f i+1 = k+3 for all i Similar to the proof of Case for n = k + 1 case (see (1) ()), by some calculation, we get for any i, This implies that, f(x i+ ) f(x i ) = k + 3, and li = d Cn (x i+, x i ) = k f(x i ) = f(v n ki ) = (k + 3)i (mod n), for all i This is impossible, since the function defined by τ(0) = 0 and τ(i) = n ki (mod n) can not be a permutation on {0, 1,, n 1}, as gcd(n, k) 3 (since k is odd) Theorem 6 Let C n be an n-vertex square cycle where n = k + 3 If k = m for some m Z +, or k = m + for some m 5 (mod 7), then rn(cn) = k + 9k + If k = m + for some m 5 (mod 7), then k + 9k + rn(cn) k + 9k + 10 If k = m + 1 for some m Z +, then { k rn(cn) +7k+3 =, if m 0, 1 (mod 3); k +7k+5, if m (mod 3) If k = m + 3 for some m Z +, then { k rn(cn) +7k+5, if m 0 (mod 3); k +7k+3, if m 1, (mod 3) Proof Let n = k + 3 Then d = k + 1 By Lemma 5, the bound proved in Theorem can not be achieved for the following two cases for n = k + 3: 1) k = m + 1 and m (mod 3) Let m = 3p + for some p Z + Then, gcd(n, k) = gcd(8p + 39, 1p + 9) 3 18

) k = m + 3 and m 0 (mod 3) Let m = 3p for some p Z + Then gcd(n, k) = gcd(8p + 15, 1p + 3) 3 If 1) or ) holds, then by Lemma 5, rn(ck+3 ) k +7k+5 Hence, by Theorem and Lemma 5, it suffices to give radio labellings for k 0, 1, (mod ), with span achieving the desired numbers We consider cases, and in each of the labelling f given, we use the same notations used in the previous section Case 1: k 0 (mod ) Let k = m For 0 i n 1, define τ by: ( ) 3k + τ(i) = i Hence, for any i = 0, 1,, n, 3k + d i = (mod n) = 3k + We now claim that f is a radio labelling Claim (1): τ is a permutation By the Euclidean Algorithm, gcd(k + 3, 3k+ ) = gcd(16m + 3, 6m + 1) = 1 Therefore, τ is a permutation Claim (): f(x i+ ) f(x i ) k + 1 d(x i, x i+ ) + 1 By the definition of τ, we have d(x i, x i+ ) = k+ and d i + d i+1 = 3k+ So, the result follows Claim (3): f(x i+3 ) f(x i ) k + 1 d(x i, x i+3 ) + 1 By τ, d(x i, x i+3 ) = k and d i + d i+1 + d i+ = 9k+1 Hence, the result follows Claim (): f(x i+ ) f(x i ) k + This is true since d i + d i+1 + d i+ + d i+3 = 3k + So, f is a radio labelling The span of f is k +9k+, by some easy calculation Figure 8 shows an optimal radio labelling for C19 with span 36 Case : k (mod ) Let k = m+ for m Z + So, n = 16m+11 Subcase 1: m 0, 1,, 3,, 6 (mod 7) For 0 i n 1, define τ by: ( ) 3k + τ(i) = i By the definition of τ, we have 3k + d i = (mod n) = 3k + 6 The labelling f is defined by f(x 0 ) = f(v 0 ) = 0, and for 0 i n 1, { f(xi ) + k + 1 d f(x i+1 ) = i + 1, if i is even; f(x i ) + k + 1 d i +, if i is odd 19

16 3 10 6 0 36 1 30 8 0 6 8 1 3 18 Figure 8: An optimal radio labelling for C 19 We show that f is a radio labelling Claim (1): τ is a permutation This is true, because gcd(n, 3k+ ) = gcd(16m + 11, 6m + 5) = 1 Claim (): f(x i+ ) f(x i ) k + 1 d(x i, x i+ ) + 1 By the definition of τ, we have d(x i, x i+ ) = k and d i + d i+1 = 3k+6 Hence, f(x i+ ) f(x i ) = k + 5 (d i + d i+1 ) = k + 1 d(x i, x i+ ) + 1 Claim (3): f(x i+3 ) f(x i ) k + 1 d(x i, x i+3 ) + 1 Because d(x i, x i+3 ) = k+6 and d i + d i+1 + d i+ = 9k+18, we have f(x i+ ) f(x i ) (d d i +1)+(d d i+1 +)+(d d i+ +1) k+ d(x i, x i+3 ) Claim (): f(x i+ ) f(x i ) k+ Because d i +d i+1 +d i+ +d i+3 = 3k+6, so f(x i+ ) f(x i ) (k + ) + (d i + d i+1 + d i+ + d i+3 ) > k + The span of f is k +9k+, because f(x n 1 ) = (d d i + 1) + (d d i+1 + ) + + (d d n + ) = ( ) n n 1 (n 1)d + (1 + ) d i = k + 9k + Subcase : m 5 (mod 7) It suffices to find a radio labelling with span k +9k+10 Let m = 7p + 5 for some p Z + Since n = k + 3 and k = m +, we have gcd(n, 3k+ ) = 7, and that n 7 is odd For i = 0, 1,,, n 1 and j = 0, 1,, 6, define τ by: ( ) 3k + jn (j + 1)n τ(i) = i j (mod n), if i < 7 7 i=0 0

By the definition of τ, and since k is even, we obtain d i = { 3k+, for i = jn 7 1 and j = 1,, 6;, otherwise 3k+6 The labelling f is defined by f(x 0 ) = f(v 0 ) = 0, and (5) For j = 0,,, 6 and jn 7 i < (j+1)n 7 : { f(xi ) + k + 1 d f(x i+1 ) = i + 1, if i is even; f(x i ) + k + 1 d i +, if i is odd (6) For j = 1, 3, 5 and jn 7 i < (j+1)n 7 : { f(xi ) + k + 1 d f(x i+1 ) = i + 1, if i is odd; f(x i ) + k + 1 d i +, if i is even (7) We verify that f is a radio labelling Claim(1): τ is a permutation Assume, to the contrary, that τ(i) = τ(i ) for some 0 i, i n 1 Then we have ( ) ( 3k+ i j 3k+ ) i j (mod n) for some 0 j, j 6 This implies (i i ) ( ) 3k+ j j (mod n) Because 3k+ n 0 (mod 7) and 0 j j 6, it must be that j = j Hence, i i < n 3k+ 7 (Recall that gcd(n, ) = 7) Let n = 7t and 3k+ = 7s for some s, t Z + with gcd(s, t) = 1 Then we have (i i )7s 0 (mod 7t), implying (i i )s 0 (mod t) This is impossible because i i < n 7 = t and gcd(s, t) = 1 Therefore, τ is a permutation Claim(): f(x i+ ) f(x i ) k + 1 d(x i, x i+ ) + 1 for any i By the definition of τ, we have d(x i, x i+ ) = k By (5), (6) and (7), we have d i + d i+1 = { 3k+, if f i = d d i + 1 and f i+1 = d d i+1 + 1; 3k+6, otherwise Now we consider the two cases in the above separately If d i + d i+1 = 3k+, then f(x i+ ) f(x i ) = f i + f i+1 = (d d i + 1) + (d d i+1 + 1) If d i + d i+1 = 3k+6, then we obtain = k + d(x i, x i+ ) f(x i+ ) f(x i ) = (d d i + 1) + (d d i+1 + ) = k + d(x i, x i+ ) 1

Claim(3): f(x i+3 ) f(x i ) k + 1 d(x i, x i+3 ) + 1 for any i By the definition of τ, we have d(x i, x i+3 ) = k+6 By (5), we have d i + d i+1 + d i+ { 9k+18, 9k+1 } Hence, by (6) and (7), we have f(x i+3 ) f(x i ) = f i + f i+1 + f i+ (d d i + 1) + (d d i+1 + 1) + (d d i+ + 1) 9k + 18 3k + 6 k + d(x i, x i+3 ) Claim(): f(x i+ ) f(x i ) k + for any i By (5), d i + d i+1 + d i+ + d i+3 {3k + 6, 3k + 5} So, the result follows by a similar calculation to the above Therefore, f is a radio labelling, with span k +9k+10, because f(x n 1 ) = (d d 0 + 1) + (d d 1 + ) + + (d d n 7 1 + 1) +(d d n n 7 7 +1 + ) + + (d d n 7 1 + 1) +(d d 6n + 1) + (d d 6n 7 7 +1 + ) + + (d d n + ) = (n 1)d + 3( n 7 ( ) + 6 (n 7)( 3k + 6 ) + 6( 3k + ) ) = k + 9k + 10 Case 3: k 1 (mod ) Let k = m + 1 for some m Z + Subcase 31: m 0, 1 (mod 3) For 0 i n 1, define τ by: ( ) 3k + 3 τ(i) = i (mod n) Hence, 3k + 3 d i = = 3k + 5 The proof that f is a radio labelling with the desired span is similar to Case of Theorem We leave the details to the reader (cf [11]) Figure 9 shows an optimal radio labelling for C 3 with span Subcase 3: m (mod 3) Define τ(0) = 0 and τ(i + 1) = 3(k + 1)i + k + 1 (mod n), if 0 < i + 1 < n 3 ; τ(i) = 3(k + 1)i (mod n), if 1 i < n 3 ; τ(i) = k + + (i n 3 ) ( ) 3k+3 (mod n), if n 3 i n 1 Recall that n = k + 3, k = m + 1 and m (mod 3) So, n = 16m + 7 and n 3 is even Therefore, the above definition of τ implies that when

0 36 6 16 6 3 1 38 10 0 30 0 1 3 8 18 8 Figure 9: An optimal radio labelling for C 3 i = n 3, then τ(i 1) = 3k + 1 and d(x i 1, x i ) = k+1 By some calculation, we obtain k + 1, if i is even and 0 i < n 3 1; k+3 d i =, if i is odd and 0 i < n 3 1; k+1, if i = n 3 1; 3k+5, if n 3 i < n ; We now verify that f is a radio labelling Claim (1): τ is a permutation Because m (mod 3), so n k 0 (mod 3) Therefore, we have τ(i) 0 (mod 3), 0 i < n 3, τ(i + 1) 1 (mod 3), 0 i + 1 < n 3, τ(i) (mod 3), n 3 i n 1 Hence, it suffices to show that τ is one-to-one within each case Assume τ(j) = τ(i) for some 0 i, j < n 3 Then (k + 1)(j i) 0 (mod n 3 ) Letting m = 3p +, we get gcd(k + 1, n 3 ) = gcd(1p + 10, 16p + 13) = 1 Hence, j i (mod n 3 ), implying that i = j, as i, j < n 3 By the same argument, one can show that τ(i+1) = (k+1)+(3k+3)i (mod n) is one-to-one for 0 i + 1 < n 3 Assume τ(j) = τ(i) for some i, j n k+1 3 Then (j i) 0 (mod n/3) This implies that j = i, since j i < n/3 Therefore, τ is a permutation Claim (): f(x i+ ) f(x i ) k + 1 d(x i, x i+ ) + 1 It suffices to show that d i +d i+1 d(x i, x i+ ) k+ By the definition of τ, d(x i, x i+ ) = k+1 and d i + d i+1 5k+7 Thus, the result follows Claim (3): f(x i+3 ) f(x i ) k + 1 d(x i, x i+ ) + 1 It suffices to show d i + d i+1 + d i+ d(x i, x i+3 ) k + By the definition of τ, we obtain: k+1, if d i + d i+1 + d i+ = 5k+7 ; d(x i, x i+3 ) = 1, if d i + d i+1 + d i+ {k +, k + 3}; k+3, if d i + d i+1 + d i+ { 9k+15, 9k+11 } 3

By some calculation, we conclude that d i +d i+1 +d i+ d(x i, x i+3 ) k+ Claim (): f(x i+ ) f(x i ) k + The result follows by the fact that d i + d i+1 + d i+ + d i+3 3k + 5 Therefore, f is a radio labelling with span k +7k+5, as f(x n 1 ) = (d d 0 + 1) + (d d 1 + 1) + + (d d n + 1) = (k + )(k + ) (d 0 + d + + d n 3 ) (d 1 + d 3 + + d n 3 3 ) d n 3 1 (d n 3 + + d n ) = k + 10k + ( 8k + 6 6 = k + 7k + 5 )(k + 1) ( k 3 )(k + 3 ) k + 1 ( 3k + 5 )( k 3 ) Figure 10 shows an optimal radio labelling for C 39 with span 115 91 1 5 115 36 9 100 60 73 85 6 19 109 30 3 9 5 67 79 0 1 66 106 55 8 31 18 97 7 7 11 61 8 88 37 103 13 Figure 10: An optimal radio labelling for C 39

References [1] G Chartrand, D Erwin, F Harary and P Zhang, Radio labelings of graphs, Bull Inst Combin Appl, 33 (001), 77 85 [] G Chartrand, D Erwin, and P Zhang, Radio Antipodal Colorings of Cycles, Congr Numer, 1 (000),19-11 [3] G Chartrand, D Erwin, and P Zhang, A graph labeling problem suggested by FM channel restrictions, Bull Inst Combin Appl, accepted [] G Chartrand, L Nebesky and P Zhang, Radio k-colorings of Paths, Discuss Math Graph Theory, accepted [5] D Liu, X Zhu, Multi-level Distance labelings for Path and Cycles, SIAM J Disc Math, to appear [6] D Liu, Radio Number for Spiders, manuscript, 00 [7] D Liu and M Xie, Radio Number for Square Paths, manuscript, 00 [8] W K Hale, Frequency assignment: theory and applications, Proc IEEE, 68 (1980), 197 151 [9] I Niven, H Zuckerman, and H Montgomery, An introduction to the Theory of Numbers, 5th edition, John Wiley and Sons, Inc New York, 1991 [10] J Gross and J Yellen, Graph Theory and its application, CRC Press, New York, 1999 [11] M Xie, Multiple Level Distance Labellings and Radio Number for Square Paths and Square Cycles, Master Thesis, California State University, Los Angeles, 00 [1] PZhang, Radio labellings of Cycles, Ars Combin, 65 (00), 1-3 5