Zero-Sum Mgic Grphs nd Their Null Sets Ebrhim Slehi Deprtment of Mthemticl Sciences University of Nevd Ls Vegs Ls Vegs, NV 89154-4020. ebrhim.slehi@unlv.edu Abstrct For ny h N, grph G = (V, E) is sid to be h-mgic if there eists lbeling l : E(G) Z h {0} such tht the induced verte set lbeling l + : V (G) Z h defined by l + (v) = l(uv) uv E(G) is constnt mp. When this constnt is 0 we cll G zero-sum h-mgic grph. The null set of G is the set of ll nturl numbers h N for which G dmits zero-sum h-mgic lbeling. In this pper we will identify severl clsses of zero sum mgic grphs nd will determine their null sets. Key Words: mgic, non-mgic, zero-sum, null set. AMS Subject Clssifiction: 05C15 05C78 1 Introduction For n belin group A, written dditively, ny mpping l : E(G) A {0} is clled lbeling. Given lbeling on the edge set of G one cn introduce verte set lbeling l + : V (G) A by l + (v) = l(uv). uv E(G) A grph G is sid to be A-mgic if there is lbeling l : E(G) A {0} such tht for ech verte v, the sum of the lbels of the edges incident with v re ll equl to the sme constnt; tht is, l + (v) = c for some fied c A. In generl, grph G my dmit more thn one lbeling to become A-mgic; for emple, if A > 2 nd l : E(G) A {0} is mgic lbeling of G with sum c, Ars Combintori 82 (2007), 41-53. 41
then λ : E(G) A {0}, the inverse lbeling of l, defined by λ(uv) = l(uv) will provide nother mgic lbeling of G with sum c. A grph G = (V, E) is clled fully mgic if it is A-mgic for every belin group A. For emple, every regulr grph is fully mgic. A grph G = (V, E) is clled non-mgic if for every belin group A, the grph is not A-mgic. The most obvious emple of non-mgic grph is P n (n 3), the pth of order n. As result, ny grph with pth pendnt of length n 3 would be non-mgic. Here is nother emple of non-mgic grph: Consider the grph H Figure 1. Given ny belin group A, typicl mgic lbeling of H is illustrted in tht figure. Since l + (u) = 0 nd l + (v) = 0, H is not A-mgic. This fct cn be generlized s follows: Lemm 1.1. Every even cycle C n with 2k + 1 (< n) consecutive pendnts is non-mgic. Lemm 1.2. Every odd cycle C n with 2k (< n) consecutive pendnts is non-mgic. H: u - - v Figure 1: An emple of non-mgic grph. Certin clsses of non-mgic grphs re presented in [1]. The originl concept of A-mgic grph is due to J. Sedlcek [11, 12], who defined it to be grph with rel-vlued edge lbeling such tht 1. distinct edges hve distinct nonnegtive lbels; nd 2. the sum of the lbels of the edges incident to prticulr verte is the sme for ll vertices. Jenzy nd Trenkler [4] proved tht grph G is mgic if nd only if every edge of G is contined in (1-2)-fctor. Z-mgic grphs were considered by Stnley [13, 14], who pointed out tht the theory of mgic lbeling cn be put into the more generl contet of liner homogeneous diophntine equtions. Recently, there hs been considerble reserch rticles in grph lbeling, interested reders re directed to [3, 15]. For convenience, the nottion 1-mgic will be used to indicte Z-mgic nd Z h -mgic grphs will be referred to s h-mgic grphs. Clerly, if grph is h-mgic, it is not necessrily k-mgic (h k). Definition 1.3. For given grph G the set of ll positive integers h for which G is h-mgic is clled the integer-mgic spectrum of G nd is denoted by IM (G). 42
Since ny regulr grph is fully mgic, then it is h-mgic for ll positive integers h 2; therefore, IM (G) = N. On the other hnd, the grph H, Figure 1, is non-mgic, hence IM (H) =. The integer-mgic spectr of certin clsses of grphs resulted by the mlgmtion of cycles nd strs hve lredy been identified [5], nd in [6] the integer-mgic spectr of the trees of dimeter t most four hve been completely chrcterized. Also, the integer-mgic spectr of some other grphs hve been studied in [7, 8, 9, 10]. 1 1 2-2 Figure 2: The grph K(2, 3) is h-mgic ( h 3). Definition 1.4. An h-mgic grph G is sid to be h-zero-sum (or just zero-sum) if there is mgic lbeling of G in Z h tht induces verte lbeling with sum 0. The grph G is sid to be uniformly null if every mgic lbeling of G induces 0 sum. Clerly, grph tht hs n edge pendnt is not zero-sum. Here is n emple of uniformly null grph: Lemm 1.5. The complete biprtite grph K(2, 3) is uniformly null mgic grph. Proof. Since the degree set of K(2, 3) is {2, 3}, it is not 2-mgic. On the other hnd, the lbeling presented in Figure 2 indictes tht the integer-mgic spectrum of K(2, 3) is N {2} with sum being 0. Now we wish to show tht 0 is the only possible sum. Consider n rbitrry lbeling of K(2, 3), s illustrted in Figure 3. b c d e f For K(2, 3) to be h-mgic, we require tht Figure 3: An rbitrry lbeling of K(2, 3). + c + e + b (mod h); b + d + f c + d (mod h). If we dd these equtions, we get e + f = 0. Hence, the induced sum cnnot be nonzero. 43
Definition 1.6. The null set of grph G, denoted by N(G), is the set of ll nturl numbers h N such tht G is h-mgic nd dmits zero-sum lbeling in Z h. One cn introduce number of opertions mong zero-sum grphs which produce mgic grphs. Frucht nd Hrry [2] introduced the coron of two G nd H, denoted by G c H, to be the grph with bse G such tht ech verte v V (G) is joined to ll vertices of seprte copy of H. Observtion 1.7. If G hs zero-sum in Z h, then G c K 1 is h-mgic. A grph G with fied verte u V (G) will be denoted by the order pir (G, u). Given two ordered pir (G, u) nd (H, v), one cn construct nother grph by linking these two grphs through identifying the vertices u nd v. We will use the nottion (G, u) (H, v) for this construction or simply G H if there is no mbiguity bout the choices of u nd v. Definition 1.8. Given n grphs G i (i = 1, 2, n), the chin G 1 G 2 G n is the grph in which one of the vertices of G i is identified with one of the vertices of G i+1. If G i = G, we use the nottion G n for the n-link chin ll of whose links re G. Observtion 1.9. If grphs G i hve zero sum, so does the chin G 1 G 2 G n, hence it is mgic grph. Moreover, if G i = G, then the null set of the chin G n is the sme s N(G). Figure 4: A 7-link chin whose links re K 4 With the nottion in 1.8, if we further identify one of the vertices of G n by nother verte of G 1, the resulting grph is necklce. Similrly, ll the bids of this necklce cn be the sme grph G, for which we hve the following observtion: Observtion 1.10. If the grphs G i hve zero sum, so does the necklce formed by these grphs. Moreover, if G i = G, then the null set of this necklce is the sme s N(G). These re just few opertions mong the grphs tht preserve the mgic property, when the grphs re zero-sum. In mgic lbeling of grphs, knowing the components of grph nd the null sets of the components will be etremely helpful. For emple, consider the grph G illustrted in Figure 5. This grph is constructed by five copies of K 4. In the net section, (theorem 2.1), it 44
Figure 5: Find the integer-mgic spectrum of this grph!!! is shown tht the null set of K 4 is N {2}. With this informtion nd the fct tht the pplied construction preserves the zero-sum property, one cn esily see tht N(G) = IM (G) = N {2}. In the following sections the null sets of few well known clsses of grphs will be chrcterized. 2 Null Sets of Complete Grphs Complete grphs being regulr re fully mgic, hence their integer-mgic spectrum is N. In this section we will determine the null sets of these grphs. Note tht K 3 C 3 nd N(K 3 ) = 2N. In wht follows we will ssume tht n 4. { N if n is odd; Theorem 2.1. If n 4, then N(K n ) = N {2} if n is even. Proof. Let u 1, u 2,, u n be the vertices of K n nd ssume tht they re rrnged counterclockwise round circle. If n is even, then deg(u i ) is odd nd K n cnnot hve zero sum in Z 2. Also, with the { following convention, we will use u j s one of the vertices even if j 1, 2,, n : Let uj n if j > n ; u j = To prove the theorem, we will consider the following five cses, in ech u j+n if j 0. cse we will introduce n pproprite lbeling l : E(K n ) Z 3 with sum 0. Cse 1. n is odd nd n = 4p + 1. In this cse, lbeling of the edges re done by { 1 if j = i ± r (1 r p); l(u i u j ) = 1 otherwise. Since the deg(u i ) = n 1 = 4p, there re 4p edges tht re incident with verte u i, hlf of which re lbeled 1 nd the other hlf 1. Therefore, l + (u i ) = 0 for ll i = 1, 2,,, n. Cse 2. n is odd nd n = 4p + 3. In this cse, we lbel the edges by 2 if j = i + 1; l(u i u j ) = 1 if j = i ± r (2 r p); 1 otherwise. 45
Since the deg(u i ) = n 1 = 4p + 2, there re 4p + 2 edges incident with this verte, two edges re lbeled 2, 2p 2 edges re lbeled 1, nd the remining 2p + 2 edges re lbeled 1. Therefore, l + (u i ) = 0 for ll i = 1, 2,,, n. This lbeling is illustrted in tble (2.1). u 1 u 2 u 3 u 4 u 5 u 6 u 7 u 1 2 1 1 1 1 2 u 2 2 2 1 1 1 1 u 3 1 2 2 1 1 1 u 4 1 1 2 2 1 1 u 5 1 1 1 2 2 1 u 6 1 1 1 1 2 2 u 7 2 1 1 1 1 2 Cse 3. n is even nd n = 6p + 4. In this cse, we lbel the edges by 2 if j i = 3p + 2; l(u i u j ) = 2 if j = i ± r (1 r p); 1 otherwise. Note tht u i u j (j i = 3p + 2) re the opposite vertices, u i u j (j = i + r) re on the left of u i, nd u i u j (j = i r) re on the right of u i. Since the deg(u i ) = n 1 = 6p + 3, there re 6p + 3 edges incident with this verte. We lbel 2p + 1 of them by 2 (opposite, p on the left, nd p on the right) nd the remining 4p + 2 by 1. Therefore, l + (u i ) = 0 for ll i = 1, 2,,, n. Cse 4. n is even nd n = 6p + 2. In this cse, we lbel the edges by l(u i u j ) = 2 if j i = 3p + 2; 2 if j = i ± r (2 r p); 1 if j = i ± 1; 1 otherwise. Since the deg(u i ) = n 1 = 6p + 1, there re 6p + 1 edges incident with this verte. We lbel 2p 1 of them by 2 (opposite, p 1 on the left, nd p 1 on the right), two edges by 1 (immedite left nd right), nd the remining 4p by 1. Therefore, l + (u i ) = 0 for ll i = 1, 2,,, n. Cse 5. n is even nd n = 6p. In this cse, we lbel the edges by l(u i u j ) = 2 if j i = 3p + 2; 2 if j = i ± r (2 r p); 1 if j = i + 1 (i = 2r 1); 1 otherwise. Since the deg(u i ) = n 1 = 6p 1, there re 6p 1 edges incident with this verte. We lbel 2p 1 of them by 2 (opposite, p 1 on the left, nd p 1 on the right), one edges by 1, nd the remining 4p by 1. Therefore, l + (u i ) = 0 for ll i = 1, 2,,, n. This lbeling is illustrted in tble (2.2). (2.1) 46
u 1 u 2 u 3 u 4 u 5 u 6 u 1 1 1 2 1 1 u 2 1 1 1 2 1 u 3 1 1 1 1 2 u 4 2 1 1 1 2 u 5 1 2 1 1 1 u 6 1 1 2 1 1 (2.2) 3 Null Sets of Complete Biprtite Grphs Theorem 3.1. Let m, n 2. Then { N if m + n is even; N(K(m, n)) = N {2} if m + n is odd. Proof. Let S = {u 1, u 2,, u m } nd T = {v 1, v 2,, v n } be the two prtite sets. In lbeling of edges u i v j, with elements of Z h (h 3), we will consider three cses: Cse I. m, n re both even. We lbel the edges by l(u i v j ) = ( 1) i+j. This will result in l + 0. Cse II. m is even nd n is odd. We lbel the edges by 2 ( 1) i 1 if j = 1 l(u i v j ) = ( 1) i if j = 2, 3 ( 1) i+j otherwise This lbeling is illustrted in tble (3.1). v 1 v 2 v 3 v 4... v n u 1 2 1 1 1... 1 u 2 2 1 1 1... 1 u 3 2 1 1 1... 1 u 4 2 1 1 1... 1......... u m 1 2 1 1 1... 1 u m 2 1 1 1... 1 Cse III. m, n re both odd. We lbel the edges by using the following tble (8): (3.1) 47
v 1 v 2 v 3 v 4 v 5... v n u 1 2 1 1 2 2... 2 u 2 1 2 1 1 1... 1 u 3 1 1 2 1 1... 1 u 4 2 1 1... u 5 2 1 1.. u m 2 1 1.. ( 1) i+j Finlly, we observe tht if m, n hve different prity, the grph would not be 2-mgic.... (3.2) 4 Null Sets of Cycle Relted Grphs There re different clsses of cycle relted grphs tht hve been studied for vriety of lbeling purposes. J. Gllin [3] hs nice collection of such grphs. In this section, the null sets of some of the cycle relted grphs re investigted. First, one useful observtion: Observtion 4.1. In ny mgic lbeling of cycle the edges should lterntively be lbeled the sme elements of the group. Proof. Let u 1, u 2, u 3, nd u 4 be the four consecutive vertices of cycle. l(u 1 u 2 ) + l(u 2 u 3 ) = l(u 2 u 3 ) + l(u 3 u 4 ) implies tht l(u 1 u 2 ) = l(u 3 u 4 ). The requirement of Since cycle is 2-regulr grph, it is fully mgic. Therefore, its integer-mgic spectrum is N. For the null-set of C n we hve the following theorem: { N if n is even; Theorem 4.2. N(C n ) = 2N if n is odd. Proof. If n is even, then there re even number of edges nd we lbel every other edge by 1 nd 1. If n is odd, then in ny mgic lbeling of C n, ll the edges re lbeled the sme element of Z h. As result, for C 2k+1 to be zero-sum, one needs 2 0 (mod h) or 2 h. On the other hnd, if h = 2r, then the choice of = r will result to the zero-sum mgic lbeling of C 2k+1. A cycles with P k chord is cycle with the pth P k joining two nonconsecutive vertices of the cycle. Since the degree set of these grphs is {2, 3}, they re not 2-mgic. Bsed on Observtion 4.1, it is enough to consider the cses when k = 2, 3. The chord P k splits C n into two subcycles. Depending on the number of edges of these subcycles, we will hve different results for the null set. The net lemm is bout cycles with P 2 chord: 48
Lemm 4.3. { Let G n,2 be the cycle C n with P 2 chord. Then N {2} both subcycles re even; N(G n,2 ) = 2N {2} otherwise. Proof. Since the degree set of G n,2 is {2, 3}, the grph is not 2-mgic. Now bsed on the observtion 4.1, it is enough to consider C 3 nd C 4 s the two subcycles. 1 1-2 1 1 z+ b b z z+b Figure 6: G n,2 consists of two even subcycles. Cse I. Both subcycles re even. The lbeling illustrted in Figure 6, proves tht the integer-mgic spectrum of G n,2 is the sme s its null set; tht is, N(G n,2 ) = IM(G n,2 ) = N {2}. c d z b z+ z+b z b r+1 r r 1 Figure 7: G n,2 consists of two odd subcycles. Cse II. Both subcycles re odd. The typicl lbeling of G n,2 in Z h is illustrted in Figure 7. The requirement + z + d = c + d nd b + z + c = c + d imply tht c = + z nd d = b + z. Also, + b = c + d will result to 2z 0 (mod h) or 2 h. On the other hnd, if h = 2r, then the choice of z = r, = 1, nd b = 1 provides zero sum result. Therefore, IM(G n,2 ) = N(G n,2 ) = 2N {2}. Cse III. Subcycles hve different prities. The typicl lbeling of G n,2 in Z h is illustrted in Figure 8. The condition + + z = + y + z implies = y. Also, the requirements + z + = 2 will result to z = nd b = 2. Therefore, given Z h {0}, we need nother nonzero element, 2, hence h 4. Therefore, the integer-mgic spectrum of such grphs would be N {2, 3}, while the null set is 2N {2}. b z y 2- - Figure 8: G n,2 consists of one odd nd one even subcycles. Corollry 4.4. C n with P 2 chord is not uniformly null. 49
Lemm 4.5. { Let G n,3 be the cycle C n with P 3 chord. Then N {2} both subcycles re even; N(G n,3 ) = 2N {2} otherwise. Proof. Bsed on the observtion 4.1, it is enough to consider C 4 nd C 5 s the two subcycles. 1-2 1 2 Figure 9: G n,3 consists of two even subcycles. Cse I. Both subcycles re C 4. The lbeling illustrted in Figure 9, shows tht the integer-mgic spectrum of G n,3 is the sme s its null set; tht is, N(G n,3 ) = IM(G n,3 ) = N {2}. 2- - + - Figure 10: G n,3 consists of two odd subcycles. Cse II. Both subcycles re C 5. The typicl mgic lbeling of G n,3 in Z h is illustrted in Figure 10, which hs sum 2. Here, given Z h, one needs nother nonzero element,. Hence, the grph cnnot be 3-mgic, nd its integer-mgic spectrum is N {2, 3}. However, for the grph to hve zero sum, we need 2 0 (mod h); tht is, h hs to be even. Therefore, its null set in contined in 2N {2}. On the other hnd, if h = 2r, then the choices of = r nd = 1 provide zero sum result. Therefore, N(G n,3 ) = 2N {2}. Cse III. Subcycles hve different prities. The typicl mgic lbeling of G n,3 in Z h is illustrted in Figure 8. For the grph to be mgic, we need 3 + c + = + c + or 2 0 (mod h); tht is, h is even nd the integer-mgic spectrum of the grph would be 2N {2}. For the grph to hve zero sum, we need the dditionl condition + c + 0 (mod h), tht is lwys possible. One such lbeling hs been provided in Figure 11. Thus IM(G n,3 ) = N(G n,3 ) = 2N {2}. Corollry 4.6. C n with P 3 chord is not uniformly null. We summrize the bove 4.3 nd 4.5 in the following theorem: 50
Theorem 4.7. { Let G n,k be the cycle C n with P k chord. Then N {2} both subcycles re even; N(G n,k ) = 2N {2} otherwise. Moreover, G n,k is not uniformly null grph. c + 1 +c +1 +c Figure 11: G n,3 consists of one odd nd one even subcycles. When k copies of C n shre common edge, it will form n n-gon book of k pges nd is denoted by B(n, k). N N {2} Theorem 4.8. N(B(n, k)) = 2N {2} 2N n is even, k is odd; n nd k re both even; n is odd, k is even; n nd k re both odd. Proof. Depending on whether n is even or odd it will be enough to consider C 4 nd C 3, respectively. If n is even nd k is odd, we will lbel the common edge by 1 nd top edges 1, 1 lterntively. This is zero sum mgic lbeling. If n nd k re both even, we will lbel the common edge by 1 nd one top edge by 2 the remining top edges 1, 1 lterntively. This provides zero sum mgic lbeling. Note tht, in this cse, the degrees of vertices do not hve the sme prity nd the book is not 2-mgic. Suppose n is odd (C 3 ). We lbel the common edge by z nd the i th cycle edges by i, i s illustrted in Figure 12. u 1 z - 1 v Figure 12: A typicl zero sum mgic lbeling of B(3, k). The requirements l + (u) = l + (v) = 0 will led us to the equtions z + i = z i = 0 or 2z 0 (mod h), which implies tht h is even (z 0). On the other hnd if h = 2r is even, then we consider two cses: 51
cse I. If k is odd, we lbel ll the edges by r which results in zero sum mgic lbeling. r u r+1 r r 1 v Figure 13: A zero sum mgic lbeling of B(3, 2k). cse II. If k is even, we choose 1 = r 1, 2 = 1, nd z = i = r (i 3), s illustrted in Figure 13. Finlly, we observe tht when n is odd nd k is even, the book cnnot be 2-mgic. Therefore, the null spce would be 2N {2}. There re mny other clsses of cycle relted grphs. Wheels W n = C n + K 1 nd Fns (lso known s Shells) re mong them. When n 3 chords in cycle C n shre common verte, the resulting grph is clled Fn (or Shell) nd is denoted by F n, which is isomorphic to P n 1 +K 1. We conclude this pper by the following problems: Problem 4.9. Find the null sets of W n nd F n. Problem 4.10. In 1.5 it ws shown tht K(2, 3) is uniformly null grph. Identify clss of grphs whose elements re uniformly null. References [1] G. Bchmn nd E. Slehi, Non-Mgic nd K-Nonmgic Grphs, Congressus Numerntium 160 (2003), 9708. [2] R. Frucht nd F. Hrry, On the Coron of Two Grphs, Aequtiones Mthemtice 4 (1970), 322-325. [3] J. Gllin, A Dynmic Survey in Grphs Lbeling (ninth edition), Electronic Journl of Combintorics (2005). [4] S. Jezny nd M. Trenkler, Chrcteriztion of Mgic Grphs, Czechoslovk Mthemticl Journl 33 (l08), (1983), 435-438. 52
[5] S-M Lee, E. Slehi, Integer-Mgic Spectr of Amlgmtions of Strs nd Cycles, Ars Combintori 67 (2003), 199-212. [6] S-M Lee, E. Slehi, nd H. Sun, Integer-Mgic Spectr of Trees with Dimeter t most Four Journl of Combintoril Mthemtics nd Combintoril Computing 50 (2004), 35. 199-212. [7] S-M Lee nd H. Wong, On Integer-Mgic Spectr of Power of Pths, Journl of Combintoril Mthemtics nd Combintoril Computing 42 (2002), 18794. [8] R.M. Low nd S-M Lee, On the Integer-Mgic Spectr of Tesselltion Grphs, Austrlsin Journl of Combintorics 34 (2006), 195-210. [9] E. Slehi, Integer-Mgic Spectr of Cycle Relted Grphs, Irnin Journl of Mthemticl Sciences nd Informtics 2 (2006), 53-63. [10] E. Slehi nd S-M Lee, Integer-Mgic Spectr of Functionl Etension of Grphs, to pper in the Journl of Combintoril Mthemtics nd Combintoril Computing. [11] J. Sedlcek, On Mgic Grphs, Mth. Slov. 26 (1976), 329-335. [12] J. Sedlcek, Some Properties of Mgic Grphs, in Grphs, Hypergrph, Bloc Syst. 1976, Proc. Symp. Comb. Anl. Zielon Gor (1976), 247-253. [13] R.P. Stnley, Liner Homogeneous Diophntine Equtions nd Mgic Lbelings of Grphs, Duke Mthemtics Journl 40 (1973), 607-632. [14] R.P. Stnley, Mgic Lbeling of Grphs, Symmetric Mgic Squres, Systems of Prmeters nd Cohen-Mculy Rings, Duke Mthemtics Journl 40 (1976), 511-531. [15] W.D. Wllis, Mgic Grphs, Birkhäuser Boston 2001. 53