Distance magic circulant graphs

Transcription

1 Distace magic circulat graphs Sylwia Cichacz 1, Dalibor Frocek 2 2 correspodig author 1 cichacz@agh.edu.pl 2 dalibor@d.um.edu May 5, 2014 Abstract Let G = (V, E) be a graph of order. A distace magic labelig of G is a bijectio l: V {1, 2,..., } for which there exists a positive iteger k such that x N(v) l(x) = k for all v V, where N(v) is the ope eighborhood of v. I this paper we deal with circulat graphs C(1, p). The circulat graph C (1, p) is the graph o the vertex set V = {x 0, x 1,..., x 1 } with edges (x i, x i+p ) for i = 0,..., 1 where i + p is take modulo. We completely characterize distace magic graphs C (1, p) for p odd. We also give some sufficiet coditios for p eve. Moreover, we also cosider a group distace magic labelig of C (1, p). 1 Itroductio All graphs cosidered i this paper are simple fiite graphs. Cosider a simple graph G whose order we deote by = G. Write V (G) for the vertex set ad E(G) for the edge set of a graph G. The ope eighborhood N(x) of a vertex x is the set of vertices adjacet to x, ad the degree d(x) of x is N(x), the size of the eighborhood of x. By C we deote a cycle o vertices. The author was supported by Natioal Sciece Cetre grat r 2011/01/D/ST/

2 Distace magic circulat graphs 2 I this paper we ivestigate distace magic labeligs, which belog to a large family of magic type labeligs. Geerally speakig, a magic type labelig of a graph G(V, E) is a mappig from V, E, or V E to a set of labels which most ofte is a set of itegers or group elemets. The the weight of a graph elemet is typically the sum of labels of the eighborig elemets of oe or both types. If the weight of each elemet is required to be equal, the we speak about magic-type labelig; whe the weights are all differet (or eve form a arithmetic progressio), the we speak about a ati-magic-type labelig. Probably the best kow problem i this area is the ati-magic cojecture by Hartsfield ad Rigel [10], which claims that the edges of every graph except K 2 ca be labeled by itegers 1, 2,..., E so that the weight of each vertex is differet. A comprehesive dyamic survey of graph labeligs is maitaied by Gallia [9]. A more detailed survey related to our topic by Arumugam at al. [1] was published recetly. A distace magic labelig (also called sigma labelig) of a graph G = (V, E) of order is a bijectio l: V {1, 2,..., } with the property that there is a positive iteger k (called the magic costat) such that w(x) = l(y) = k for every x V (G), y N G (x) where w(x) is the weight of vertex x. If a graph G admits a distace magic labelig, the we say that G is a distace magic graph. It is worth metioig that fidig a r-regular distace magic labelig turs out equivalet to fidig equalized icomplete touramet EIT(, r) [8]. I a equalized icomplete touramet EIT(, r) of teams with r rouds, every team plays exactly r other teams ad the total stregth of the oppoets that team i plays is k. Thus, it is easy to otice that fidig a EIT(, r) is the same as fidig a distace magic labelig of ay r-regular graph o vertices. The followig observatios were idepedetly proved: Observatio 1 ([11], [12], [14], [15]) Let G be a r-regular distace magic graph o vertices. The k = r(+1) 2. Observatio 2 ([11], [12], [14], [15]) There is o distace magic r-regular graph with r odd.

3 Distace magic circulat graphs 3 The followig cycle-related results were proved by Miller, Rodger, ad Simajutak, ad by Rao, Sigh, ad Parameswara, respectively. Theorem 3 ([12]) The cycle C of legth is distace magic if ad oly if = 4. Theorem 4 ([13]) The cartesia product C C m,, m 3 is distace magic if ad oly if = m 2 (mod 4). Circulat graphs are aother iterestig family of vertex-trasitive graphs. These graphs arise i various settigs; for istace, they are the Cayley graphs over the cyclic group of order. The circulat graph C (s 1, s 2,..., s k ) is the graph o the vertex set V = {x 0, x 1,..., x 1 } with edges (x i, x i+sj ) for i = 0,..., 1, j = 1,..., k where i + s j is take modulo. Moreover, sice there is o distace magic r-regular graph for r odd by Observatio 2, we ca assume that > 2s k + 1. It was show i [2] that a graph C (1, 2) is ot distace magic uless = 6. For p odd the followig theorem was proved: Theorem 5 ([2]) If p is odd, the C (1, 2,..., p) is a distace magic graph if ad oly if 2p(p + 1) 0 (mod ), 2p + 2 ad 0 (mod 2). gcd(,p+1) I this paper we cosider the correspodig problem for circulat graphs C (1, p). The motivatio for cosiderig circulats is a problem stated i [14]. Problem 6 ([14]) Characterize 4-regular distace magic graphs. We will also cosider the otio of group distace magic labelig of graphs that was itroduced i [7]. A Γ-distace magic labelig of a graph G(V, E) with V = is a ijectio from V to a Abelia group Γ of order such that the weight of every vertex x V is equal to the same elemet µ Γ, called the magic costat. Some families of graphs that are Γ-distace magic were studied i [4, 3, 5, 7]. The followig result was proved i [7]: Theorem 7 ([7]) The cartesia product C C m,, m 3 is a Zm-distace magic graph if ad oly if m is eve. The paper is orgaized as follows. I the ext three sectios we cosider distace magic circulat graphs C (1, p). I the last sectio we cosider a Γ-distace magic labelig.

4 Distace magic circulat graphs 4 2 Distace magic graphs C (1, p) We start with some observatios: Observatio 8 Let C (1, p) be a distace magic graph with magic costat k. The for ay i {0, 1,..., 1} ad ay γ N l(x i ) + l(x i+p 1 ) = l(x i+γ(2p+2) ) + l(x i+p 1+γ(2p+2) ). Proof. Sice C (1, p) is distace magic we obtai w(x 0 ) w(x 1 ) = w(x 1 ) w(x 2 ) =... = w(x 1 ) w(x 0 ) = 0. Hece w(x i ) w(x i+p+1 ) = l(x i 1 ) + l(x i p ) (l(x i+p+2 ) + l(x i+2p+1 )) = 0 ad w(x i+2p+2 ) w(x i+3p+3 ) = l(x i+p+2 )+l(x i+2p+1 ) (l(x i+3p+4 )+l(x i+4p+3 ) = 0. for i {0, 1,..., }. So we obtai l(x i ) + l(x i+p 1 ) = l(x i+γ(2p+2) ) + l(x i+p 1+γ(2p+2) ). Observatio 9 Let C (1, p) be a distace magic graph with magic costat k, the for ay i {0, 1,..., 1} ad ay γ N l(x i ) + l(x i+p+1 ) = l(x i+γ(2p 2) ) + l(x i+p+1+γ(2p 2) ). Proof. Sice C (1, p) is distace magic we obtai Hece w(x 0 ) w(x 1 ) = w(x 1 ) w(x 2 ) =... = w(x 1 ) w(x 0 ) = 0. w(x i ) w(x i+p 1 ) = l(x i+1 ) + l(x i p ) (l(x i+p 2 ) + l(x i+2p 1 )) = 0.

5 Distace magic circulat graphs 5 Also w(x i+2p 2 ) w(x i+3p 3 ) = l(x i+p 2 )+l(x i+2p 1 ) (l(x i+3p 4 )+l(x i+4p 3 )) = 0 ad so o for i {0, 1,..., }. So we obtai l(x i ) + l(x i+p+1 ) = l(x i+γ(2p 2) ) + l(x i+p+1+γ(2p 2) ). Observatios 8 ad 9 imply the followig corollary. Corollary 10 If C (1, p) is a distace magic graph, the for ay i {0, 1,..., 1} ad for ay α, γ N l(x i ) + l(x i+(2α+1)(p 1) ) = l(x i+γ(2p+2) ) + l(x i+(2α+1)(p 1)+γ(2p+2) ), l(x i ) + l(x i+(2α+1)(p+1) ) = l(x i+γ(2p 2) ) + l(x i+(2α+1)(p+1)+γ(2p 2) ). Proof. Sice C (1, p) is distace magic we obtai by Observatio 8 l(x i ) + l(x i+p 1 ) = l(x i+γ(2p+2) ) + l(x i+p 1+γ(2p+2) ), l(x i+p 1 ) + l(x i+2(p 1) ) = l(x i+p 1+γ(2p+2) ) + l(x i+2(p 1)+γ(2p+2) ), l(x i+2(p 1) ) + l(x i+3(p 1) ) = l(x i+2(p 1)+γ(2p+2) ) + l(x i+3(p 1)+γ(2p+2) ),. l(x i+2α(p 1) ) + l(x i+(2α+1)(p 1) ) = l(x i+2α(p 1)+γ(2p+2) ) + l(x i+(2α+1)(p 1)+γ(2p+2) ). Alteratively subtractig ad summarizig the above equatios we obtai l(x i+2(p 1) ) + l(x i+3(p 1) ) = l(x i+2(p 1)+γ(2p+2) ) + l(x i+3(p 1)+γ(2p+2) ). Usig Observatio 9 by similar argumets we obtai l(x i ) + l(x i+(2α+1)(p+1) ) = l(x i+γ(2p 2) ) + l(x i+(2α+1)(p+1)+γ(2p 2) ). Theorem 11 If C (1, p) is distace magic the 0 (mod 2). gcd(,p 1) gcd(,p+1) 0 (mod 2) ad

6 Distace magic circulat graphs 6 Proof. Let k be a magic costat for C (1, p). It is well kow that if a, b Z ad gcd(a, ) = gcd(b, ), the a ad b geerate the same subgroup of Z, that is, a = b. Suppose that 1 (mod 2), the we have gcd(,p+1) gcd(, p+1) = gcd(, 2p+2) ad 2(p+1) = p+1. Hece, p+1 = 2c(p+1) for some c 1. The we use Lemma 8, set γ = c, i = 0, p 1 ad obtai respectively: l(x 0 ) + l(x p 1 ) = l(x 2c(p+1) ) + l(x p 1+2c(p+1) ) = l(x p+1 ) + l(x 2p ), l(x p 1 ) + l(x 2p 2 ) = l(x p 1+2c(p+1) ) + l(x 2p 2+2c(p+1) ) = l(x 2p ) + l(x 3p 1 ). Sice N(x i ) = {x i p, x i 1, x i+1, x i+p } ad C (1, p) is distace magic, we obtai: l(x 0 ) + l(x p 1 ) = l(x p+1 ) + l(x 2p ) = k 2, l(x p 1 ) + l(x 2p 2 ) = l(x 2p ) + l(x 3p 1 ) = k 2. Therefore l(x 0 ) = l(x 2p 2 ) ad we have a cotradictio, because > 2p + 1. Suppose ow 1 (mod 2), the gcd(, p 1) = gcd(, 2p 2). gcd(,p 1) Thus, p 1 = c2(p 1) for some c 1. The we use Lemma 9, set γ = c, i = 0, p + 1 ad obtai respectively: l(x 0 ) + l(x p+1 ) = l(x 2c(p 1) ) + l(x p+1+2c(p 1) ) = l(x p 1 ) + l(x 2p ), l(x p+1 ) + l(x 2p+2 ) = l(x p+1+2c(p 1) ) + l(x 2p+2+2c(p 1) ) = l(x 2p ) + l(x 3p+1 ). Sice N(x i ) = {x i p, x i 1, x i+1, x i+p } ad C (1, p) is distace magic, we obtai: l(x 0 ) + l(x p+1 ) = l(x p 1 ) + l(x 2p ) = k 2, l(x p 1 ) + l(x 2p+2 ) = l(x 2p ) + l(x 3p+1 ) = k 2. Therefore l(x 0 ) = l(x 2p+2 ). For 2p + 2 we obtai a cotradictio, sice the labelig l. For = 2p + 2 we have 0 (mod 2), a cotradictio. gcd(,p+1) From the above theorem the below observatio easily follows:

7 Distace magic circulat graphs 7 Observatio 12 If C (1, p) is a distace magic circulat graph, the 0 (mod 2). Moreover, whe p is odd, the 0 (mod 8). Proof. If is odd, the is odd ad C gcd(,p+1) (1, p) caot be distace magic by Theorem 11. Whe p is odd, oe of p 1, p + 1 is cogruet to 2 modulo 4 ad we ca write {p 1, p + 1} = {2q 1, 2 t q 2 }, where t 2 ad q 1, q 2 are both odd. Let = 2 s q 3, where q 3 is odd. Because 0 (mod 2) ad gcd(,p+1) 0 (mod 2), we must have s > t 2. Hece gcd(,p 1) 2s 8 ad 0 (mod 8). Observatio 13 A graph C 2p+2 (1, p) is distace magic. Proof. Let l(x i ) = i + 1, l(x i+p+1 ) = 2p + 2 i for i = 0, 1,..., p. Notice that w(x i ) = l(x i p ) + l(x i+1 ) + l(x i 1 ) + l(x i+p ) = 4p + 6 for every x i V (C 2p+2 (1, p)). 2.1 C (1, 2p + 1) distace magic graphs Theorem 14 If p is odd ad C (1, p) is distace magic, the p 2 1 (0 mod ). Proof. Let k be a magic costat for C (1, p). Assume first that p 3 (mod 4). By Observatio 9 we obtai l(x 0 ) + l(x p+1 ) = l(x γ(2p 2) ) + l(x p+1+γ(2p 2) ) = k 0, l(x 1 ) + l(x p+2 ) = l(x 1+γ(2p 2) ) + l(x p+2+γ(2p 2) ) = k 1,. l(x 2p 3 ) + l(x 3p 2 ) = l(x 2p 3+γ(2p 2) ) + l(x 3p 2+γ(2p 2) ) = k 2p 3, for ay γ. It implies that l(x (p+1) ) = k 0 l(x 0 ), l(x 2(p+1) ) = k p+1 k 0 + l(x 0 ), l(x 3(p+1) ) = k 2(p+1) mod(2p 2) + k 0 k p+1 l(x 0 ) = k 4 + k 0 k p+1 l(x 0 ). Repeatig the argumet, we get j 1 l(x j(p+1) ) = ( 1) j 1 i k i(p+1) mod(2p 2) + ( 1) j l(x 0 ). i=0

8 Distace magic circulat graphs 8 So i particular for j = p 1, (p 1)/2 1 i=0 p 2 l(x p 2 1) = ( 1) p 2 i k i(p+1) mod(2p 2) + l(x 0 ) = i=0 ( 1) p 2 i k i(p+1) mod(2p 2) + p 2 i=(p 1)/2 ( 1) p 2 i k i(p+1) mod(2p 2) + l(x 0 ) Hece if p 3 (mod 4), the gcd(p + 1, 2(p 1)) = 4 ad p 1 is odd. It is 2 kow fact that the order of a subgroup geerated by p + 1 i Z2p 2 is p 1, 2 amely p+1 = p 1 p 1. Moreover, otice that (p+1) 0 (mod (2p 2)). 2 2 It implies that (p 1)/2 1 i=0 p i=(p 1)/2 ( 1) p 2 i k i(p+1) mod(2p 2) = k 0 +k p+1 k 2(p+1) +k 3(p+1)... k ((p 1)/2 1)(p+1), ( 1) p 2 i k i(p+1) mod(2p 2) = k 0 k p+1 +k 2(p+1) k 3(p+1) +...+k ((p 1)/2 1)(p+1). Hece l(x p 2 1) = l(x 0 ). Assume ow that p 1 (mod 4). By Observatio 8 we obtai l(x 0 ) + l(x p 1 ) = l(x γ(2p+2) ) + l(x p 1+γ(2p+2) ) = k 0, l(x 1 ) + l(x p ) = l(x 1+γ(2p+2) ) + l(x p+γ(2p+2) ) = k 1,. l(x 2p+1 ) + l(x 3p ) = l(x 2p+1+γ(2p+2) ) + l(x 3p+γ(2p+2) ) = k 2p+1, for ay γ. It implies that l(x (p 1) ) = k 0 l(x 0 ), l(x 2(p 1) ) = k p 1 k 0 + l(x 0 ), l(x 3(p 1) ) = k 2(p 1) mod(2p+2) + k 0 k p+1 l(x 0 ). Repeatig the argumet, we get j 1 l(x j(p 1) ) = ( 1) j 1 i k i(p 1) mod(2p+2) + ( 1) j l(x 0 ). i=0

9 Distace magic circulat graphs 9 So i particular for j = p + 1: l(x p 2 1) = p ( 1) p i k i(p 1) mod(2p+2) + l(x 0 ) = i=0 (p+1)/2 1 i=0 ( 1) p i k i(p 1) mod(2p+2) + p i=(p+1)/2 ( 1) p i k i(p 1) mod(2p+2) + l(x 0 ) Hece if p 1 (mod 4) the gcd(p 1, 2(p + 1)) = 4 ad p+1 2 is odd. As above, we obtai (p+1)/2 1 i=0 p i=(p+1)/2 ( 1) p i k i(p 1) mod(2p+2) = k 0 +k p 1 k 2(p 1) +k 3(p 1)... k ((p+1)/2 1)(p 1), ( 1) p i k i(p 1) mod(2p+2) = k 0 k p 1 +k 2(p 1) k 3(p 1) +...+k ((p+1)/2 1)(p 1). Hece l(x p 2 1) = l(x 0 ). Observatio 15 If p is odd, p (mod ), 0 (mod 2) gcd(,p+1) ad 0 (mod 2), the C gcd(,p 1) (1, p) is a distace magic graph. Proof. Because we always suppose that > 2p ad the case = 2p + 2 was treated i Observatio 13, we will assume that > 2p + 2. By the assumptio p 1 ad p + 1 are eve ad hece oe of them is cogruet to 0 modulo 4. Moreover, by Observatio 12 we kow that 0 (mod 8). We will further assume that p (mod 4) ad leave the other case to the reader, sice it is essetially similar. It is well kow that whe a, b Z ad gcd(, a) = gcd(, b), the a = b. Obviously, gcd(, gcd(, p+1)) = gcd(, p+1). Hece, gcd(, p+1) = p + 1 ad gcd(, p + 1) = p + 1. Because 0 (mod 2) ad gcd(,p+1) we assumed that p (mod 4), we observe that gcd(, p + 1) = 4s for some s ad that the subgroup H = p + 1 = 4s of Z is of order 2k for some k. Let us deote by X j for j = 0, 1, 2,..., 4s 1 the set of all vertices whose subscripts belog to the coset H + j. First we label vertices of X 0, X 2,..., X 4s 2. Notice that there are 2s of them.

10 Distace magic circulat graphs 10 Case 1: k 0 (mod 2). Label the vertices of X 0 as follows: If k = 2, the l(x 0 ) = 1, l(x p+1 ) = 1, l(x 2(p+1) ) = 2, l(x 3(p+1) ) =. If k = 4, the l(x 0 ) = 1, l(x p+1 ) = 1, l(x 2(p+1) ) = 3, l(x 3(p+1) ) = 3, l(x 4(p+1) ) = 4, l(x 5(p+1) ) = 2, l(x 6(p+1) ) = 2, l(x 7(p+1) ) =. For k 6 let: l(x 0 ) = 1, l(x 2(p+1) ) = 3, l(x 4(p+1) ) = 5,..., l(x 2i(p+1) ) = 2i + 1,..., l(x (k 4)(p+1) ) = k 3, l(x (k 2)(p+1) ) = k 1, l(x k(p+1) ) = k, l(x (k+2)(p+1) ) = k 2, l(x (k+4)(p+1) ) = k 4,..., l(x (2k 4)(p+1) ) = 4, l(x (2k 2)(p+1) ) = 2, ad l(x p+1 ) = 1, l(x 3(p+1) ) = 3, l(x 5(p+1) ) = 5,..., l(x (2i+1)(p+1) ) = 2i 1,..., l(x (k 3)(p+1) ) = k + 3, l(x (k 1)(p+1) ) = k + 1, l(x (k+1)(p+1) ) = k + 2, l(x (k+3)(p+1) ) = k , l(x (2k 3)(p+1) ) = 2, l(x (2k 1)(p+1) ) =. Notice that a vertex x m belogs to X j if the vertex x m 2+i(p+1) for ay i belogs to X j 2. The vertices i X 2, X 4..., X 4s 2 will be labeled recursively as follows: l(x m ) = l(x m 2+(k+1)(p+1) ) + k whe l(x m 2+(k+1)(p+1) ) < 2 ad l(x m ) = l(x m 2+(k+1)(p+1) ) k whe l(x m 2+(k+1)(p+1) ) > 2. I particular, we have l(x v(p+1)+2z ) = l(x v(p+1)+z(k+1)(p+1) ) + zk whe l(x v(p+1)+z(k+1)(p+1) ) < 2 ad l(x v(p+1)+2z ) = l(x v(p+1)+z(k+1)(p+1) ) zk whe l(x v(p+1)+z(k+1)(p+1) ) > 2. We otice that the sum of two cosecutive labels i each X j falls ito oe of three cases. For istace, i X 0 we have l(x 0 ) + l(x (p+1) ) = l(x 2(p+1) ) + l(x 3(p+1) ) =... ad also = l(x (k 2)(p+1) ) + l(x (k 1)(p+1) ) = l(x (k+1)(p+1) ) + l(x (k+2)(p+1) ) = = l(x (2k 3)(p+1) ) + l(x (2k 2)(p+1) ) =. The we have l(x (p+1) ) + l(x 2(p+1) ) = l(x 3(p+1) ) + l(x 4(p+1) ) =... = l(x (k 3)(p+1) ) + l(x (k 2)(p+1) ) = + 2

11 Distace magic circulat graphs 11 ad also l(x k(p+1) ) + l(x (k+1)(p+1) ) = = l(x (2k 2)(p+1) ) + l(x (2k 1)(p+1) ) = + 2. Fially, we have l(x (2k 1)(p+1) ) + l(x 0 ) = l(x (k 1)(p+1) ) + l(x k(p+1) ) = + 1. I X 2 we have l(x 2 ) + l(x 2+(p+1) ) = l(x 2+2(p+1) ) + l(x 2+3(p+1) ) =... = l(x 2+(k 4)(p+1) ) + l(x 2+(k 3)(p+1) ) =, l(x 2+(k+1)(p+1) ) + l(x 2+(k+2)(p+1) ) =... = l(x 2+(2k 3)(p+1) ) + l(x 2+(2k 2)(p+1) ) =, l(x 2+(2k 1)(p+1) ) + l(x 2 ) = l(x 2+(p+1) ) + l(x 2+2(p+1) ) =... = l(x 2+(k 1)(p+1) ) + l(x 2+k(p+1) ) = + 2, l(x 2+k(p+1) ) + l(x 2+(k+1)(p+1) ) =... = l(x 2+(2k 2)(p+1) ) + l(x 2+(2k 1)(p+1) ) = + 2. Ad we agai have l(x 2+(2k 2)(p+1) )+l(x 2+(2k 1)(p+1) ) = l(x 2+(k 2)(p+1) )+l(x 2+(k 1)(p+1) ) = +1. Now we look at the weights of vertices i X 1. We have w(x 1 ) = l(x 0 )+l(x (p+1) )+l(x 2+(2k 2)(p+1) )+l(x 2 ) = +(+2) = 2+2. Similarly, we have w(x 1+2(p+1) ) = = w(x 1+(k 2)(p+1) ) = + ( + 2) = ad also w(x 1+(k+1)(p+1) ) = = w(x 1+(2k 3)(p+1) ) = + ( + 2) =

12 Distace magic circulat graphs 12 The we have w(x 1+(p+1) ) = l(x (p+1) ) + l(x 2(p+1) ) + l(x 2 ) + l(x 2+(p+1) ) Similarly, we have = ( + 2) + = w(x 1+3(p+1) ) = = w(x 1+(k 3)(p+1) ) = ( + 2) + = ad also w(x 1+k(p+1) ) = = w(x 1+(2k 2)(p+1) ) = ( + 2) + = Fially, the two special vertices are x 1+(2k 1)(p+1) ad x 1+(k 1)(p+1) with ad w(x 1+(2k 1)(p+1) ) = l(x (2k 1)(p+1) ) + l(x (p+1) ) + l(x 2+(2k 2)(p+1) ) + l(x 2+(2k 1)(p+1) ) = ( + 1) + ( + 1) = w(x 1+(k 1)(p+1) ) = l(x (k 1)(p+1) )+l(x k(p+1) )+l(x 2+(k 2)(p+1) )+l(x 2+(k 1)(p+1) ) = ( + 1) + ( + 1) = Usig similar reasoig, oe ca verify that the weights of all vertices with odd subscripts will be equal to I particular, usig the recursive ature of the labelig, we have w(x (2i+1)+r(p+1) ) =l(x 2i+r(p+1) ) + l(x 2i+(r+1)(p+1) ) + l(x (2i+2)+(r 1)(p+1) ) + l(x (2i+2)+r(p+1) ) =l(x 2i 2+(r+k+1)(p+1) ) + k + l(x 2i 2+(r+k+2)(p+1) ) k + l(x 2i+(r+k)(p+1) ) + k + l(x 2i+(r+k+1)(p+1) ) k =l(x 2i 2+(r+k+1)(p+1) ) + l(x 2i 2+(r+k+2)(p+1) ) + l(x 2i+(r+k)(p+1) ) + l(x 2i+(r+k+1)(p+1) ) =w(x 2i 1+(r+k+1)(p+1) ) =

13 Distace magic circulat graphs 13 The roles of k ad k may be iterchaged, depedig o the value of r. Fially, the vertices i X 1, X 3,..., X 4s 1 will be labeled as l(x m ) = l(x m 1 ) + 2sk whe l(x m 1 ) < 2 ad l(x m ) = l(x m 1 ) 2sk whe l(x m 1 ) > 2. Obviously, l is a bijectio. Usig similar argumets as above oe ca check that w(x i ) = 2( + 1) for x i X 0, X 2,..., X 4s 2. The case of p 1 0 (mod 4) is essetially the same ad is left to the reader. Case 2: k 1 (mod 2). Label the vertices of X 0 as follows: If k = 1, the l(x 0 ) = 1, l(x p+1 ) =. If k = 3, the l(x 0 ) = 1, l(x p+1 ) = 1, l(x 2(p+1) ) = 3, l(x 3(p+1) ) = 2, l(x 4(p+1) ) = 2, l(x 20 ) =. For k 5 let: l(x 0 ) = 1, l(x 2(p+1) ) = 3, l(x 4(p+1) ) = 5,..., l(x 2i(p+1) ) = 2i + 1,..., l(x (k 3)(p+1) ) = k 2, l(x (k 1)(p+1) ) = k, l(x (k+1)(p+1) ) = k 1, l(x (k+3)(p+1) ) = k 3, l(x (k+5)(p+1) ) = k 5,..., l(x (2k 4)(p+1) ) = 4, l(x (2k 2)(p+1) ) = 2, ad l(x p+1 ) = 1, l(x 3(p+1) ) = 3, l(x 5(p+1) ) = 5,..., l(x (2i+1)(p+1) ) = 2i 1,..., l(x (k 4)(p+1) ) = k + 4, l(x (k 2)(p+1) ) = k + 2, l(x k(p+1) ) = k + 1, l(x (k+2)(p+1) ) = k , l(x (2k 3)(p+1) ) = 2, l(x (2k 1)(p+1) ) =. The vertices i X 2 will be labeled as l(x m ) = l(x m+2 2(p+1) ) + k whe l(x m+2 2(p+1) ) < 2 ad l(x m ) = l(x m+2 2(p+1) ) k whe l(x m+2 2(p+1) ) > 2. Notice that a vertex x m belogs to X k if the vertex x m+2p 2 belogs to X k 4. The vertices i X 4, X 6,..., X p 3 will be labeled recursively as follows. l(x m ) = l(x m+2p 2 ) + k whe l(x m+2p 2 ) < 2 ad l(x m ) = l(x m+2p 2 ) k whe l(x m+2p 2 ) > 2.

14 Distace magic circulat graphs 14 As above, otice that the sum of two cosecutive labels i each X j falls ito oe of three cases. For istace, i X 0 we have l(x 0 ) + l(x (p+1) ) = l(x 2(p+1) ) + l(x 3(p+1) ) =... ad also = l(x (k 3)(p+1) ) + l(x (k 2)(p+1) ) = l(x (k)(p+1) ) + l(x (k+1)(p+1) ) = = l(x (2k 3)(p+1) ) + l(x (2k 2)(p+1) ) =. The we have l(x (p+1) ) + l(x 2(p+1) ) = l(x 3(p+1) ) + l(x 4(p+1) ) =... ad also = l(x (k 2)(p+1) ) + l(x (k 1)(p+1) ) = + 2 l(x (k+1)(p+1) ) + l(x (k+2)(p+1) ) = = l(x (2k 2)(p+1) ) + l(x (2k 1)(p+1) ) = + 2. Fially, we have I X 2 we have l(x (2k 1)(p+1) ) + l(x 0 ) = l(x (k 1)(p+1) ) + l(x k(p+1) ) = + 1. l(x 2 ) + l(x 2+(p+1) ) = l(x 2+2(p+1) ) + l(x 2+3(p+1) ) =... l(x 2+(k 3)(p+1) ) + l(x 2+(k 4)(p+1) ) =, l(x 2+k(p+1) ) + l(x 2+(k+1)(p+1) ) =... = l(x 2+(2k 3)(p+1) ) + l(x 2+(2k 2)(p+1) ) =, l(x 2+(2k 1)(p+1) ) + l(x 2 ) = l(x 2+(p+1) ) + l(x 2+2(p+1) ) =... = l(x 2+(k 4)(p+1) ) + l(x 2+(k 3)(p+1) ) = + 2, l(x 2+(k 1)(p+1) ) + l(x 2+k(p+1) ) =... = l(x 2+(2k 4)(p+1) ) + l(x 2+(2k 3)(p+1) ) = + 2.

15 Distace magic circulat graphs 15 Ad we agai have l(x 2+(2k 2)(p+1) )+l(x 2+(2k 1)(p+1) ) = l(x 2+(k 2)(p+1) )+l(x 2+(k 1)(p+1) ) = +1. Now we look at the weights of vertices i X 1. We have w(x 1 ) = l(x 0 )+l(x (p+1) )+l(x 2+(2k 2)(p+1) )+l(x 2 ) = +(+2) = 2+2. Similarly, we have w(x 1+2(p+1) ) = = w(x 1+(k 2)(p+1) ) = + ( + 2) = ad also w(x 1+(k+1)(p+1) ) = = w(x 1+(2k 3)(p+1) ) = + ( + 2) = The we have w(x 1+(p+1) ) = l(x (p+1) ) + l(x 2(p+1) ) + l(x 2 ) + l(x 2+(p+1) ) Similarly, we have = ( + 2) + = w(x 1+3(p+1) ) = = w(x 1+(k 3)(p+1) ) = ( + 2) + = ad also w(x 1+k(p+1) ) = = w(x 1+(2k 2)(p+1) ) = ( + 2) + = Fially, the two special vertices are x 1+(2k 1)(p+1) ad x 1+(k 1)(p+1) with ad w(x 1+(2k 1)(p+1) ) = l(x (2k 1)(p+1) ) + l(x (p+1) ) + l(x 2+(2k 2)(p+1) ) + l(x 2+(2k 1)(p+1) ) = ( + 1) + ( + 1) = w(x 1+(k 1)(p+1) ) = l(x (k 1)(p+1) )+l(x k(p+1) )+l(x 2+(k 2)(p+1) )+l(x 2+(k 1)(p+1) ) = ( + 1) + ( + 1) = As i Case 1, usig similar reasoig oe ca verify that the weights of all vertices with odd subscripts will be equal to By Theorems 11 ad Observatio 14 ad 15 we obtai the followig: Theorem 16 If p is odd, the C (1, p) is distace magic graph if ad oly if p (mod ), 0 (mod 2) ad 0 (mod 2). gcd(,p+1) gcd(,p 1)

16 Distace magic circulat graphs Figure 1: Distace magic labelig for C 24 (1, 5). 2.2 C (1, 2p ) distace magic graphs Observatio 17 If p is eve, the C 2(p 2 1)(1, p) is distace magic. Proof. Let = 2(p 2 1) ad H = p + 1 be the subgroup of Z of order 2(p 1). Sice it was proved that C (1, 2) is distace magic if ad oly if = 6 (see [2]) we ca assume that p 4. As before, deote for j = 1, 2,..., 2(p 1) by X j the set of all vertices whose subscripts belog to coset H + j ad by l j the set of all labels of vertices i X j. Label the vertices of X 0 as follows: l(x 0 ) = 1, l(x 2(p+1) ) = 3, l(x 4(p+1) ) = 5,..., l(x 2i(p+1) ) = 2i + 1,..., l(x (p 4)(p+1) ) = p 3, l(x (p 2)(p+1) ) = p 1, l(x p(p+1) ) = 2, l(x (p+2)(p+1) ) = 4, l(x (p+4)(p+1) ) = 6,..., l(x (2p 6)(p+1) ) = p 4, l(x (2p 4)(p+1) ) = p 2, ad for p = 4 we put l(x 5 ) = 29, l(x 15 ) = 30 ad l(x 25 ) = 28.

17 Distace magic circulat graphs 17 For p 6 l(x p+1 ) = 2p 2 3, l(x 3(p+1) ) = 2p 2 5, l(x 5(p+1) ) = 2p 2 7,..., l(x (2i 1)(p+1)/2 ) = 2p 2 2i 3,..., l(x (p 5)(p+1) ) = 2p 2 p 1, l(x (p 3)(p+1) ) = 2p 2 p 3, l(x (p 1)(p+1) ) = 2p 2 2, l(x (p+1)(p+1) ) = 2p 2 4,..., l(x (2p 5)(p+1) ) = 2p 2 p + 2, l(x (2p 3)(p+1) ) = 2p 2 p. Notice that sice p is eve the i Zp+1 we have 2 = Zp+1 ad moreover a vertex x m belogs to X k if the vertex x m+(p 1)(p+2) belogs to X k+2. The vertices i X 1, X 2,..., X p will be labeled recursively as follows: l(x m ) = l(x m+(p 1)(p+2) ) + p 1 whe l(x m+(p 1)(p+2) ) < p 2 1 ad l(x m ) = l(x m+(p 1)(p+2) ) p + 1 whe l(x m+(p 1)(p+2) ) > p 2 1. As i Observatio 15 oe ca check that the weights of all vertices will be equal to Group distace magic C (1, p) The otio of group distace magic labelig of graphs was itroduced i [7]. Let G be a graph with vertices ad Γ a Abelia group with elemets. We call a bijectio g : V (G) Γ a Γ-distace magic labelig if for all x V (G) we have w(x) = µ for some µ i Γ. Obviously, every graph with vertices ad a distace magic labelig also admits a Z-distace magic labelig. The coverse is ot ecessarily true (see, e.g., Theorems 4 ad 7). Recall that ay group elemet ι Γ of order 2 (i.e., ι 0 such that 2ι = 0) is called a ivolutio, ad that a o-trivial fiite group has elemets of order 2 if ad oly if the order of the group is eve. Moreover every cyclic group of eve order has exactly oe ivolutio. The fudametal theorem of fiite Abelia groups states that the fiite Abelia group Γ ca be expressed as the direct sum of cyclic subgroups of prime-power order. This product is uique up to the order of the direct product. Whe t is the umber of these cyclic compoets whose order is a power of 2, the Γ has 2t 1 ivolutios. Moreover the sum of all the group elemets is equal to the sum of the ivolutios ad the eutral elemet. Let us deote this sum as s(γ) = g Γ g. The followig lemma was proved i [6] (see [6], Lemma 8). Lemma 18 ([6]) Let Γ be a Abelia group.

18 Distace magic circulat graphs 18 (i) If Γ has exactly oe ivolutio ι, the s(γ) = i. (ii) If Γ has o ivolutios, or more tha oe ivolutio, the s(γ) = 0. We start by provig a geeral theorem for Γ-distace magic labelig similar to Theorem 2. Theorem 19 Let G be a r-regular distace magic graph o vertices, where r is odd. There does ot exists a Abelia group Γ havig exactly oe ivolutio ι, Γ = such that G is Γ-distace magic. Proof. Sice r is odd, it implies that is eve. Let Γ be a Abelia group of order havig exactly oe ivolutio i. Suppose that G is Γ-distace magic. Recall that g = 0 for ay g Γ. Let ow w(g) = x V (G) w(x) = µ = 0. O the other had w(g) = x V (G) y N(x) w(y) = rs(γ). By Lemma 18 we obtai that w(g) = rι. Therefore sice r is odd, we have rι = ι, hece ι = 0, a cotradictio. Theorem 19 implies immediately the followig observatios: Observatio 20 Let G be a r-regular distace magic graph o 2 (mod 4) vertices, where r is odd. There does ot exists a Abelia group Γ of order such that G is Γ-distace magic. The coditio 2 (mod 4) is ecessary. For example, a graph K 3,3,3,3 has a Z3 Z2 Z2-distace magic labelig with the magic costat µ = (0, 1, 1) preseted i the table, where colums correspod to the partitio sets. (0,0,0) (0,1,0) (0,1,1) (0,0,1) (1,1,0) (1,0,1) (1,1,1) (1,0,0) (2,0,1) (2,0,0) (2,1,1) (2,1,0) Observatio 21 If G is a r-regular distace magic graph o vertices, where r is odd, the G is ot Z-distace magic. Aother observatio ca be easily proved. Observatio 22 If C (1, p) is a Γ-distace magic circulat graph for a group Γ, the is eve.

19 Distace magic circulat graphs 19 Proof. Let µ be a magic costat for C (1, p). Suppose that is odd, the 1 (mod 2). Thus gcd(, p + 1) = gcd(, 2p + 2) ad 2(p + 1) = gcd(,p+1) p + 1. Hece, p + 1 = 2c(p + 1) for some c 1. The we use Lemma 8, set γ = c, i = 0, p 1 ad obtai respectively: l(x 0 ) + l(x p 1 ) = l(x 2c(p+1) ) + l(x p 1+2c(p+1) ) = l(x p+1 ) + l(x 2p ), l(x p 1 ) + l(x 2p 2 ) = l(x p 1+2c(p+1) ) + l(x 2p 2+2c(p+1) ) = l(x 2p ) + l(x 3p 1 ). Sice N(x i ) = {x i p, x i 1, x i+1, x i+p } ad C (1, p) is Γ-distace magic, we obtai for i = p ad i = 2p 1: µ = l(x 0 ) + l(x p 1 ) + l(x p+1 ) + l(x 2p ) = 2(l(x 0 ) + l(x p 1 )), µ = l(x p 1 ) + l(x 2p 2 ) + l(x 2p ) + l(x 3p 1 ) = 2(l(x p 1 ) + l(x 2p 2 )). Therefore 2(l(x 0 ) l(x 2p 2 )) = 0. Recall that beig odd implies that there does ot exists a elemet g 0, g Γ such that 2g = 0. Thus l(x 0 ) = l(x 2p 2 ) ad we have a cotradictio, because > 2p + 1. Theorem 23 If gcd(, p + 1) = 2k + 1, p ad are both eve, ad = 2r(2k +1) the C (1, p) has a Z2α A-magic labelig for ay α 0 (mod r) ad ay Abelia group A of order r(2k + 1)/α. Proof. Let l = r(2k + 1)/α Sice Γ = Z2α A, thus if g Γ, the we ca write that g = (j, a i ) for j Z2α ad a i A for i = 0, 1,..., l 1. We ca assume that a 0 = 0 A. Let l(x) = (l 1 (x), l 2 (x)). Let X = p + 1 be the subgroup of Z of order 2r. Let us deote for j = 1, 2,..., 2k by X j the set of all vertices whose subscripts belog to coset X + j. Notice that α = rh for some h. Let H = 2h be the subgroup of Z2α of order r. Label the vertices of X 0 as follows: l(x 2i(p+1) ) = (2ih, 0), l(x (2i+1)(p+1) ) = ( 2ih 1, 0) i = 0, 1,..., k 1. If a subscript m belogs to coset X + j, the deote it by m j. Notice that a vertex x mj belogs to X j if the vertex x mj p belogs to X j 1. The vertices

20 Distace magic circulat graphs 20 i X 1, X 2, X 3..., X 2k will be labeled recursively as follows. { l1 (x l 1 (x mj ) = mj p) + 1 if l 1 (x mj p) j 1 (mod 2h) l 1 (x mj p) 1 if l 1 (x mj p) j 1 (mod 2h) { a j/r if l l 2 (x mj ) = 1 (x mj ) 0 (mod 2) a j/r if l 1 (x mj ) 1 (mod 2) Obviously l is bijectio ad l(x i p ) + l(x i+1 ) = ( 1, 0) or l(x i p ) + l(x i+1 ) = (2h 1, 0). We will cosider ow two cases: Case 1. X = p + 1 = Z. Notice that the l = 1 ad h = 1. Suppose that there exists i such that l(x i p ) + l(x i+1 ) = (1, 0) ad l(x i 1 ) + l(x i+p ) = (1, 0), or similarly l(x i p ) + l(x i+1 ) = ( 1, 0) ad l(x i 1 ) + l(x i+p ) = ( 1, 0). Recall that l(x 2i(p+1) ) + l(x (2i+1)(p+1) ) = ( 1, 0) ad l(x (2i+1)(p+1) ) + l(x (2i+2)(p+1) ) = (1, 0). It implies that there exists β Z such that i p + 2β (i 1) mod. It meas that 2β (p 1) mod. Sice is eve, p 1 odd, a cotradictio. It follows that w(x i ) = (0, 0) for every i ad the graph is Γ-distace magic with µ = (0, 0). Case 2. X = p + 1 = Z. The l(x i p ) + l(x i+1 ) = l(x i+(j 1)p) ) + l(x i+jp+1 ) for ay j = 1, 2,..., 2k. Thus for j = 2 we have l(x i p ) + l(x i+1 ) = l(x i+p ) + l(x i+2p+1 ). Sice l(x i 1 ) + l(x i+p ) l(x i+p ) + l(x i+2p+1 ) we obtai that w(x i ) = l(x i p ) + l(x i+1 ) + l(x i 1 ) + l(x i+p ) = (2h 2, 0). Thus µ = (2h 2, 0) ad the graph is Γ-distace magic. Corollary 24 If gcd(, p + 1) = 2k + 1, p is eve, = 2α(2k + 1) ad gcd(2k + 1, 2α) = 1, the C (1, p) has a Z-magic labelig. Proof. By Theorem 23 there exists Z2α Z2k+1-magic labelig. Sice gcd(2k+ 1, 2α) = 1, the group Z2α Z2k+1 is isomorphic to the group Z. Corollary 25 If = α(p 2 1) ad: α = 1 if p is odd

21 Distace magic circulat graphs 21 α = 2β ad gcd(p + 1, 2β) = 1 if p is eve the C (1, p) is Z-distace magic. Proof. If α = 1 ad p is odd, the sice C (1, p) is distace magic by Theorem 16, the C (1, p) is Z-distace magic. If α = 2β, gcd(p + 1, 2β) = 1 ad p is eve, the gcd(, p + 1) = p + 1 ad C (1, p) has Z-magic labelig by Corollary 24. Usig the same argumets as i the proof of Theorem 14 we obtai corollary: Corollary 26 If p is odd ad 2p 2 2 (0 mod ) the C (1, p) is ot Γ- distace magic for ay Abelia group Γ of order. Observatio 27 If p = 5, the C p 2 1(1, p) is Γ-distace magic for ay Abelia group Γ of order p 2 1. Proof. The fudametal theorem of fiite Abelia groups states that the fiite Abelia group Γ ca be expressed as the direct sum of cyclic subgroups of prime-power order. Sice the order of Γ is 24 we have the followig possibilities: Γ = Z8 Z3 = Z24, Γ = Z2 Z3 Z4 = Z6 Z4 = Z2 Z12 ad Γ = Z2 Z3 Z2 Z2 = Z6 Z2 Z2. If Γ = Z8 Z3 = Z24, the sice C 24 (1, 5) is distace magic by Theorem 5 it implies that it is Z24-distace magic. We will show ow that C 24 (1, 5) is Z6 A-distace magic for ay Abelia group A of order 4. If g Γ, the we ca write that g = (j, a i ) for j Z6 ad a i A for i = 0, 1, 2, 3. Defie the followig labelig. l(x 0 ) = (0, a 0 ), l(x 8 ) = (4, a 0 ), l(x 16 ) = (2, a 0 ), l(x 6 ) = (5, a 1 ), l(x 14 ) = (1, a 1 ), l(x 22 ) = (3, a 1 ), l(x 12 ) = (2, a 2 ), l(x 20 ) = (0, a 2 ), l(x 4 ) = (4, a 2 ), l(x 18 ) = (3, a 3 ), l(x 2 ) = (5, a 3 ), l(x 10 ) = (1, a 3 ), l(x 1 ) = (0, a 1 ), l(x 9 ) = (4, a 1 ), l(x 17 ) = (2, a 1 ), l(x 7 ) = (5, a 2 ), l(x 15 ) = (1, a 2 ), l(x 23 ) = (3, a 2 ), l(x 13 ) = (2, a 3 ), l(x 21 ) = (0, a 3 ), l(x 5 ) = (4, a 3 ), l(x 19 ) = (3, a 0 ), l(x 3 ) = (5, a 0 ), l(x 11 ) = (1, a 0 ). Takig ow a i = i if A = Z4 or a 0 = (0, 0), a 1 = (0, 1), a 2 = (1, 1), a 3 = (1, 0) A = Z2 Z2 oe ca check the graph C 24 (1, 5) is Z6 A-distace magic with the magic costat (4, 2a 1 ).

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