Open Journal of Discrete Mathematics, 2012, 2, 24-34 http://dx.doi.org/10.4236/ojdm.2012.21006 Published Online January 2012 (http://www.SciRP.org/journal/ojdm)
On the Line Graph of the Complement Graph for the Ring of Gaussian Integers Modulo n
Manal Ghanem1, Khalida Nazzal2 1Department of Mathematics, Irbid National University, Irbid, Jordan 2Department of Mathematics, Palestine Technical University (Kadoorie), Tulkarm, Palestine Email: [email protected], [email protected]
Received October 13, 2011; revised November 2, 2011; accepted November 22, 2011
ABSTRACT The line graph for the complement of the zero divisor graph for the ring of Gaussian integers modulo n is studied. The diameter, the radius and degree of each vertex are determined. Complete characterization of Hamiltonian, Eulerian, planer, regular, locally H and locally connected is given. The chromatic number when n is a power Li n[] of a prime is computed. Further properties for and are also discussed. Li n[] n[]i
Keywords: Complement of a Graph; Chromatic Index; Diameter; Domination Number; Eulerian Graph; Gaussian Integers Modulo n; Hamiltonian Graph; Line Graph; Radius; Zero Divisor Graph
1. Introduction of Gaussian integers modulo n is investigated in [4]. In this paper it should be kept in mind that The line graph LG of a graph G is defined to be Vi [] ={1 i }, and hence, its line graph is K , the graph whose vertex set constitutes of the edges of G , 2 0 Where two vertices are adjacent if the corresponding q []i is an integral domain, so q []iK = 0 . Further, edges have a common vertex in G . The importance of []i is a complete graph whose complement is line graphs stems from the fact that the line graph q2 transforms the adjacency relations on edges to adjacency totally disconnected and thus its line graph is K . While relations on vertices. For example, the chromatic index 0 of a graph leads to the chromatic number of its line graph. p []iK = pp1, 1 , so its complement is disconnected The zero divisor graph of a commutative ring R , with two components each of which is isomorphic to denoted by R , is defined as the graph whose vertex K p1 . Finally, note that the graph 2q []i is bipartite, set is the set of all non-zero zero divisors of R and [1] and qq []iK = 22. 12 qq121, 1 edge set ER= xyxyR : , 0 and xy = 0 . This In this paper, we investigate properties of the graph type of graphs provides an example showing that algebraic Lin [] . We find the diameter, the radius of methods could be applied to problems about graphs. The Li [] . We determine which Li [] is Eu- set of Gaussian integers, denoted by []i , is defined as n n the set of complex numbers abi , where ab, . If lerian, Hamiltonian, regular, locally H , locally connec- x is a prime Gaussian integer, then x is either ted or planer. Furthermore, the chromatic index and the 1) 1 i or 1 i , or edge domination number of n[]i where n is a 2) q where q is a prime integer and q 3mod4, or power of a prime are computed. While the domination 3) abi , abi where ab22 = p, p is a prime number of n[]i is given. On the other hand, a for- integer and p 1mod4 . mula which gives the degree of each vertex in []i Throughout this paper, p and p denote prime n i is derived, thus the degree of its complement as well as integers which are congruent to 1 modulo 4, while q its line graph could easily be found. and and qi denote prime integers which are congruent to 3 modulo 4. All rings in this paper are assumed to be commutative with unity. The zero divisor graph for the 2. When Is L n[]i Eulerian or ring of Gaussian integers modulo n is studied in [1] Planner and [2], the complement of this graph is discussed in [3]. While the line graph of the zero divisor graph for the ring If G is a connected graph. Then G is Eulerian if and
Copyright © 2012 SciRes. OJDM M. GHANEM ET AL. 25 only if every vertex of G has even degree. For a finite [],im 3 are the union of a nullgraph and a ring R , the line graph LR of a connected graph qm R is Eulerian if and only if all vertices of R connected graph [3], we have the following. have the same parity ( see the proof of Lemma 3.10, [5]). Theorem 3.1 Li n[] is connected if and only if On the other hand, if G has both even and odd vertices, 2 npqqq 2, , , 12. then so is its complement. So, for a connected graph Theorem 3.2 If nm=2m , 2 or nqm=,m 3, then []i , the graph Li [] is Eulerian if and n n diam L [] i =2. n only if all vertices in n []i are either even or all Proof. 1) Assume that nm=2m , 2 and vertices in n []i are all odd. But n[]i is con- mm nected if np ,2 , qqq , 12 [3] and n []i is x =,=,=,=xxiyyyizzziwwwi12 12 12 1 2 Eulerian if np=2, or n is a product of distinct odd primes [1]. It is easy to show that all vertices of are two nonadjacent vertices in VL [] i . Since 2 n n []i are odd if and only if nq= . This proves the following theorem. for every abi [] i, a and b are both even or 2m Theorem 2.1 Li [] is Eulerian if and only if n odd [1], we have three cases: n is a product of distinct odd primes. Case I: for i =1,2, xiii,,yz and wi are odd. Then A planar graph is a graph that can be embedded in the we have the path [,x yxzzw ] [,] [, ]. plane, i.e., it can be drawn on the plane in such a way Case II: for i =1,2, x or y is odd(even) and z that its edges intersect only at their endpoints. i i i or wi is even (odd). Assume that x12, x are even and Next we determine when the graph Li n [] is zz12, are odd. Then we have the path planar. [,x yxzzw ] [,] [, ]. In a graph G the maximum vertex degree and the Case III: for i =1,2, xiii,,yz and wi are even. minimum vertex degree will be denoted by G and tst s Then [,x yii ]= 2112 2 , 2 2 2 and G , respectively. 112 2 [,zw ]= 2ts33 2i , 2ts44 2 i where , The following theorem characterizes graphs G whose 3344 ii line graph LG is planer. are odd and 1, tsii m for 14i . If tst112,,, or Theorem 2.2 [6] m A nonempty graph G has a planer line graph LG s2 < , say t1 , then tst334,, or s41< mt , say t3 . if and only if 2 1) G is planer. So, we have the path [,x yxzzw ] [,] [, ]. Now 2) G 4 , and suppose that m is odd. Then 3) if deg v =4, then v is a cut vertex. m 1 G a) If ts== , , for i =1 or 2, say for ii2 i i The graph n[]i is planer if and only if n =2,5 i =1, then tst,, or s < mt , say t . Hence, we or q2 [3]. For n =2, q2 , LiK [] = . While 334 41 3 n 0 have the path [,x yxzzw ] [,] [, ]. for n =5, n[]iKK = 44, this graph is regular of degree 3. m 1 b) If ti or si = and tsii , for i =1 or 2, say Thus we obtain the following. 2 for i =1, then we have a path Theorem 2.3 The graph Lin [] is planer if and only is n =5. [,x yxzzw ] [,] [, ] or [,x yxwzw ] [, ] [, ]. 3. The Diameter of m 1 L []i c) If tsii== ,= i i, for i =1 or 2, say for n 2 For a connected graph G , the distance, duv, , m 1 i =1, then ts== implies that . Other- between two vertices u and v is the minimum of the 22 2 22 lengths of all uv paths of G . The eccentricity of a m 1 wise t or s . Then we have a path vertex v in G is the maximum distance from v to 2 2 2 any vertex in G . The diameter of G , diam G , is the [,x yyzzw ] [,] [, ] or maximum eccentricity among the vertices of G . Since [,x yywzw ] [, ] [, ]. mm m n[]i is connected if np ,2 , qqq , 12 and each 2) Assume that nqm=, 3 and of []i and []i is the union of two qqiqqitsts1122,, p qq12 112 2
ts33ts44 complete graphs, while []i and 3344qqiqqiVLi,[] n 2m
Copyright © 2012 SciRes. OJDM 26 M. GHANEM ET AL.
m diam L [] i =3. Then tst112,, or s2 < , say t1 . Hence tst334,, or n 2 s41< mt , say t3 . Then we have the path tsts 4. The Radius and the Girth of the Graph qqiqqi1122, 112 2 L []i ts11ts33 n [,1133qqiqqi ]. qqiqqits33, ts44 For a connected graph G , the radius of G , rad G , is 3344 the minimum eccentricity among the vertices of G . So, Theorem 3.3 Let R be a ring that is a product of rad G diam G . Since for any two rings R1 and R2 with at least one of them is not [,]ab V L [] i , [,]ab and [,]ai bi are non ad- ID with more than one regular element and the other has n more than two regular elements. Then jacent, rad L [] i >1. Using Theorem 3.2 gives diam L R =3. n Proof. Suppose that RRR= and R is not ID, 12 1 for nm=2m , 2 or nqm=,m 3, reg R1 2 and reg R2 3 . Let zV11 R rad L [] i =2. and uregR221 . Clearly, n dz1,0 , ,0 , 0,1 , 0, u = 3 in LR. So, m m 12 Theorem 4.1 If npm=, 2 or nts= where diam L R 3 . Now, let m 1 , t is prime integer, gcd.. t , s =1 and nqq 12,
aa,,, bb , cc ,,, dd VL R , then rad L n [] i =2. 1 2 12 12 1 2 then ab 0 or ab 0 and cd 0 or cd 0 . 11 22 11 22 Proof. Since rad L n [] i >1 to show that So, we have three cases: Case I: ab11 0 and cd11 0 . Then rad L [] i =2 it is enough to find a vertex n ac11, regR 1 implies that zaazccELR,0 , , ,0 , , . vVL [] i with eccentricity 2. If 112112 n m 22 And a1 or cZR11 , say a1 implies that nppabm=,= , 2, then uaauccELR,0 , , ,0 , , dabiabixy[,],[,]2 for every 112112 where uregRc. [,x yVL ] [] i . So rad L [] i =2. 111 n pm Case II: ab22 0 and cd22 0 . Then there exists m vregRac2222, and hence Now, assume that ntsm=, 1 and
aa12, , 0, v 2 cc 12 , , 0, v 2 EL R . (,x ywzVL ),(,) [] i . n Case III: ab11 0 and cd22 0 or ab22 0 and Then we have four cases: cd11 0 . Let ab11 0 and cd22 0 . Then Case I: t =2. Then aregR11 implies that aa,,, zc VL R and 12 12 di1,1,1,0, xywz ,,, 2. zc,=, dd or dd,,, zc VL R . 12 1 2 12 12 Case II: tp = . Then And if aZR , then ac,=, bb or 11 12 12 dabiabixywz ,1 , ,1 , , , , 2 . ac,,, bb VL R and 12 12 Case III: tq= 1 and m =1. Then s q2 and hence ac12,=, dd 1 2 or ac,,, dd VL R . 12 1 2 there exists aV s [] i . So,
daxywz0,1 , 1, , , , , 2 . For np= m , []i [7] and for nmmp p nnn= with gcd.. n , n =1, []iii [] []. Case IV: tqm=, 2. Then 12 12 nnn12 Moreover reg [] i =2 and reg [] i 3 for 2 m dq,1 , 1,0 , xywz , , , 2 . m 2 . An immediate consequence of Theorem 3.3 is the following. Theorem 4.2 rad L [] i =2 if and only if m n Theorem 3.4 Let npm=, 2 or n is a composite npqq 2, , , 2 or qq . such that nqq 12. Then 12
Copyright © 2012 SciRes. OJDM M. GHANEM ET AL. 27
Vising [8], proved that for a connected simple graph i =1,2 and either R1 or R2 is not ID, then G with n-vertices and radius 2, the upper bound of the LR is locally connected. nn 2 Proof. Suppose that R1 is not ID, zV11 R number of edges of G is . Then Golberg [9] 2 and xy,,, zw V L R , then we have three proved that the lower bound of numbers of edges of a cases: 31n simple connected graph G with radius 2 is . Case I: xz==0. Then 4 zxyzzwELR,1 , , ,1 , , . So we can conclude the following. 11 Theorem 4.3 For n 2, pqq , , 2 or qq , 12 Case II: yw==0. If x, zregR 1 , then Lit [] = implies that zxyzzwELR,1 , , ,1 , , . n 11 Otherwise there exists uregRxz{,}. So, 31ttt2 11 EL [] i . n uxyuzwELR,0 , , ,0 , , . 42 11 The girth of a graph G , g G is the length of a Case III: xy,0 or zw,0 . Assume that xy 0 , shortest cycle contained in the graph. If the graph does then z 0 implies that there exists uregRx11 {} not contain any cycles (i.e.. it’s an acyclic graph), its satisfies girth is defined to be infinity. If abca,,, is a cycle of uxyuzwELR,0 , , ,0 , , . 11 length three in G . Then [ab , ],[ bc , ],[ ca , ],[ ab , ] is a While w 0 implies that that there exists cycle of length 3 in LG. So, gLG =3 when- vregRw {} satisfies ever gG=3. In [3] it is proved that the girth of 22 0,vxyvzwELR , , 0, , , . 2 22 n[]i equals 3 for nqq 2, , . So, we have the fol- lowing. From Theorem 5.1 and Theorem 5.2 we conclude the 2 following. Theorem 4.4 For nqq 2, , , gL n [] i =3. Theorem 5.3 If npm=,m 1 or n is a composite integer such that nqq , then both []i and 5. The Locally Connected Property of the 12 n are locally connected. Li n [] Graphs n[]i and L n[]i
We say that a vertex v is locally connected if the 6. When Is L n[]i Hamiltonian? neighborhood of v , Nv , is connected; and G is locally connected if every vertex of G is locally con- A Hamiltonian cycle is a cycle that visits each vertex nected. exactly once (except the vertex which is both the start Theorem 5.1 If RRR=,12 reg Ri 2 for and end, and so is visited twice). A graph that contains a i=1,2 and either R1 or R2 is not ID, then R is Hamiltonian cycle is called a Hamiltonian graph. The locally connected. line graph of a graph G with more than 4 vertices and Proof. Suppose that R is not ID and 1 diameter 2 is Hamiltonian [10]. But [],im 2 is x, yV R. Then we have two cases: 2m Case I: x =0 or y =0. If x =0, then there exists disconnected with one isolated vertex 22mm11 i zV11 R. So zabER1,1 , for all and the other component, call this component H , with ab,, N xy . And if y =0, then there exists diameter 2 [3]. So, LiLH [] . Similarly, 2m uregRx11{} such that uNxy1,0 , . There- fore, uabER,0 , for every 1 []i has a connected subgraph H with diameter ab,, N xy . So Nxy, is connected. qm Case II: x 0 and y 0 . Then there exist 2 and LiLH [] . Hence, the following vregRx{}, vregRy{} and qm 11 22 vzv,0 , , zV11 R such that 112 and result is obtained. m m 0,vNxy2 , . Moreover, Theorem 6.1 If nm=2 , 2 or nqm=, 3, then vzvvzvER,0 , , 0, , . And for every 112212 is Hamiltonian. Lin [] ts, Z R , vts1,0 , or 0,vtsE2 , R . So Nxy, is connected. Oberly and Sumner [11] proved that every connected, Theorem 5.2 If RRR=,12 reg Ri 2 for locally connected claw free graph (i.e. it does not contain
Copyright © 2012 SciRes. OJDM 28 M. GHANEM ET AL.
a complete bipartite graph K1,3 ) is hamiltonian. Since 22m 2 the line graph is claw free, using Theorem 5.3, we get the m []iq = q 1. q following. m Finally we find the chromatic index of Theorem 6.2 If npm=, 2 or n is a composite integer such that nqq , then Li [] is hamil- m [],im 2. 12 n p tonian. A subset D of the vertex set VG is said to be independent if no two vertices in this set are adjacent. A 7. The Chromatic Number of the Graph clique of a graph is a maximal complete subgraph. A graph G is said to be split if it’s vertex set can be L n[]i partitioned into two subsets A and B such that A induces a clique and B is independent in G . The edge coloring of a graph G is an assignment of Lemma 7.4 [13] Let G be a split graph. If G is colors to the edges of the graph so that no two adjacent odd, then GG = . edges have the same color. The minimum required num- Theorem 7.5 If np= 2 , then ber of colors for the edges of a given graph is called the []ippp =2 321. chromatic index of the graph denoted by G . n Lemma 7.1 [12] Proof. Since []i , it is enough to find p222pp If G has order 2s and Gs =2 1, then . First, we’ll show that is GG= . pp22 pp22 m Theorem 7.2 If nm=2 , 2, then a split graph. Let []i =221m 3. n AupuU=,: 2 and p } p
Proof. Note that in m []i , the induced subgraph, 2 pv , : v U 2 and p , p mm11 H , with VH=[]22 V m i i is con- Bpp= , : and 0,0 . 2 p 21m nected, VH=2 2, [1] and Clearly, VA = B, A induces a clique pp22 H =[] m i . Since the vertex 1 i is adja- 2 and B is independent. Therefore, 22 is a cent to all other vertices in H , we have pp Hdegi=1=23 21m . Using Lemma 6.1, split graph. Moreover, 21m m []i =2 3. 2 22 =1,deg p pp Since q []i is empty graph and =[]0,:2 ip p 01, p []iqK =2 1 is edgeless with q2 1 verti- p q21 =2ppp32 1 ces, we consider the case [],iq 3. qm is odd. From Lemma 7.4, m Theorem 7.3 If nqm=, 3, then []ippp =2 321. pm 22m 2 n []iq = q 1 A graph G is said to be critical if G is connected mm11 and GG =1 and for every edge e of G , Proof. Let Aq=: qi,0 q . we have Ge\{ } < G . The well-known Vizing’s Then A is the set of all isolated vertices in []i . qm theorem states that for a simple graph G , GG= or G 1 . So the induced subgraph, H , with the vertices Lemma 7.6 [14] VH=[] V i A is a connected graph, If G is a critical graph, then G has at least qm GG 2 of vertices of maximum degree. VH= q22m q 2. Clearly the vertex q is adjacent to Therefore, if G is a simple graph such that for every all other vertices in H and hence, vertex v of maximum degree there exists an edge vu deg q=1 q22m q 2 . Using Lemma 7.1, such that Gdegu 2 is more than the number
Copyright © 2012 SciRes. OJDM M. GHANEM ET AL. 29
of vertices with maximum degree in G , we have 22m 2 Liqq [] = 1. GG= [13]. n m Theorem 7.7 If npm=, 3, then 3) If npm=,m 2, then []ippp =2 21mm 22 1. n Lippp [] =2 21mm 22 1. n Proof. Let , V and uv, U . p pm 8. The Domination Number of []i Then the vertices of with maximum de- n ppmm A subset D of the vertex set VG of a graph G is gree have the form pv, or up, where 0 a dominating set in G if each vertex of G , not in D , and 0 and is adjacent to at least one vertex of D . The minimum NpvV ,= cardinality of all dominating sets in G , G , is called ppmm the domination number of G . ppvm1 ,0 : 0 , In [],im 2, the vertex 22mm11 i is an 2m and isolated vertex while the vertex 1 i dominates all vertices in the second component. Therefore, Nu,= p V ppmm 2 m []i =2. The graph 21[]iqK = 1 , 2 q 0,pupm1 : 0 , . thus []iq =2 1. In [],i m 3 the q2 qm 21mm 22 So, mm =2ppp 1 . And pp vertices qqimm11 are isolated while the vertex q is adjacent to all other vertices in the vertices of with minimum degree ppmm Viqqi[] mm11:, , m1 m1 qmq have the form p ,0 or 0, p where mi1 Np,0 = upi , : 1 and 2 so m []iq = . Since q Np0,mi1 = pvi , : 1 . So qq []iK = 22 K 12 qq1211 []ippmpmp = mm1 2 . pm and p []iK = pp11 K, Therefore, []ii = [] =2. qq12 p []i 2 ppmm pm The set D =1,0,0,1 is a minimum dominating set =2ppmm11 p m mpmp p21m p1. for . And if nnn= , where >2ppmm1 p1 ppmm 12 gcd.. n , n =1, then []iii [] []. This 12 nnn12 But the graph has only ppmm graph is connected and the set D =1,0,0,1 is a 21ppmm1 p vertices of maximum degree. So, minimum dominating set for []ii [] . nn12 21mm 22 mm=2ppp 1. m pp Theorem 8.1 1) If nq 2, , then Since []i , the result holds. pmmmpp []i =2. Since the edge coloring of any graph leads to a vertex n coloring of its line graph, we obtain the following. 2 m Corollary 7.8 1) If nm=2 , 2, then 2) 2 []iq = 1 and q 21m Li n [] =2 3. 2 m []iqm = , 3. m q 2) If nqm=, 3, then
Copyright © 2012 SciRes. OJDM 30 M. GHANEM ET AL.
9. The Domination Number of L []i and not both jk,= m. The set n mm TA= mm kj A mm is the maximum inde- The independence number of G , G , is the maxi- kj== 22 mum cardinality of all independent sets in G . A subset mm D of the edge set VG of a graph G is an edge 11 pendent set, while SA= 22 induces a dominating set in G if each edge of G , not in D , is jk=1 =1 kj adjacent to at least one edge of D . The minimum cardinality of all edge dominating sets in G , G , is maximum complete subgraph in []i . There are called the edge domination number of G . The minimum qm cardinality of all independent edge dominating sets, some edges joining S to T , and []i has no iG , is called the independence edge domination qm number of G . The study of the domination number of other adjacency. Easy calculations give the line graph of G leads to the study of edge or line 2 Aqq=1 22mk j when 1,kj m 1, kj domination number of G , i.e. LG= G . On mj mj1 mk mk1 Aqmj = q and Aqkm = q when the other hand, for any graph G , iGG= [15]. If S is an independent set in G , then S induces a m 2 complete graph in G . While if S induces a complete kj, m. While Tq=12 and graph in G , then it is independent in G . Recall that 2 mm []i [2]. Then the sets, 2 22mm2 Sq=122 q . AU=2:j , jm=1,2, ,2 1 form a jm22 j partition for the set V . Clearly, the set Thus we obtain the following theorem. 22m Theorem 9.2 If nqm=,m 3, then 21m 2 TA= is the maximum independent set in mm jm= j 2 1) []iqq = 22 1 . n , while the set SA= m1 induces a maxi- 2m j=1 j 2 m m1 2) n[]iq = 1 if m is even and q if mum complete subgraph in . There are some m is odd. 22m Li[] = [] i edges joining S to T , no other adjacency exists in nin 2 2m . Any edge dominating set for 2m must 3) mm 2 2 1 2 =[]=1. iqq22 contain at least S 2 element in order to dominate n 2 S . On the other hand, this dominating set for S do- Now, we move to the case np= m . Let minates all other edges in . Since 22m kj 21mj AppUkj=,: m k and U m j . Aj =2 , then S and T , could easily be com- p p puted to get the following theorem. m Theorem 9.1 For nm=2 , 2. Clearly, the sets Akj where 0,kj m and not both kj,= m or 0, partition the vertices of mm1 1) n[]i =2 2 1 . 22mk j 2 mm and Apkj =1 p . Let m pp 2) n[]i =2 1. mm1 SA10= kk A 0 Li[] = [] i kk=1 =1 nin 3) mm 11 mm11 22 mm =[]=221. n i SASA23=,=,kj kj kj=1 =1 mm kj== To study the graph [],im 3, consider the 22 qm mm 11 partition of []i given by 22mm qm SASA45=and=.kj kj jk=1 mm=1 kj kj== AqqiU=: and U , 22 kj q m k q m j
1,kj m . Note that S1 induces a complete graph in
Copyright © 2012 SciRes. OJDM M. GHANEM ET AL. 31
m1 annabi =[]:0mod x ixabi n . . Vertices in A are adjacent to all ppmm k =1 k 0 abi n m1 So,xa . bi 0 mod n x . 0 mod . vertices except some vertices in Akm . Similarly, cdi cdi m1 k =1 vertices in A0k are adjacent to all vertices except k =1 m1 abi n some vertices in A , and vertices in A are But Ui [] . So, x 0mod k =1 mk m0 n cdi cdi cdi adjacent to all vertices except vertices in A0m . On the other hand A induces a complete subgraph and n 0m and hence there exists mi [] such that x = m . vertices in this set are adjacent to all other vertices except cdi those of A . Clearly S induces a complete subgraph. m0 2 Since mtcdir= where tr,[] i and the norm Vertices in S3 form an independent set, and are of r is less than the norm of cdi , adjacent to some vertices in SSSS A. 1245 0m ann a bi =: r r []= i [] i . By Theorem Each of S and S induces a complete subgraph and cdicdi 4 5 22 2 of [7], cdi []=icd22 = , so the result are adjacent to some vertices in SSS123 A 0m . cd Besides, there are some edges between S4 and S5 . On holds. the other hand, Theorem 10.2 Let vV n[] i and mm g..cd v , n = c di. Then SAA3 =. kj mm mm 22 2 kj== cd 1, if v 0 22 deg v = 22 2 . cd2, if v = 0 The above argument shows that The order of []i can be easily computed using n Limim[] = [] i formulas given in [1]. Thus we can find the degree of pp each vertex in the complement of n []i , here we 1 give the degree of each vertex in the line graph of =[]=[] iiS ppmm 3 []i , an analogous formula for the degree of 2 n vertices in Li [] could be obtained. m n 2m 1 21mm 22 =2Ppp2 2. Corollary 10.3 Let uv,[ V L n i] , 2 g..cd u , n = a bi and g..cd v , n = c di. Then deg[,] u v 10. The Degree of the Vertices in []i n abcd222 2 4, if u 2 0 and v 2 0 . 222 2 2 2 and L n[]i =5,if=0and0abcd u v abcd222 26, if u 2 = 0 and v 2 = 0 Now, we determine the cardinality of the annihilator of the element abi , ann a bi in []i . This helps n Proof. Note that, for any graph G and uv E G , find the degree of each vertex in []i , its n deg[,]= u v deg u deg v 2. complement, as well as the degree of each vertex in their LG() G G corresponding line graphs. In the following we determine the degree of every
Theorem 10.1 If abi n[] i, then vertex in the graphs n []i when ann a bi= c22 d where g..cd a bin , = c di. nmnqm=2mm , 2, = , 3 and npm=,m 1. m Proof. Let abi n[] i and Theorem 10.4 Let nm=2 , 3 and , are odd. g..cd a bin , = c di. Then Then in n []i ,
2k mm 21,if1< Copyright © 2012 SciRes. OJDM 32 M. GHANEM ET AL. kk npq=2, , 2 . 2) deg2= deg 2 i r m Theorem 11.1 If n = π j where π s are j=1 j j 2k m 21,if1<k distinct Gaussian primes and m j 1 and mmm 2 npqm 2, , , 2, then []i is not regular. = . n Proof. Choose two vertices π and π such that 2k m t s 22,ifkm < ππ , then gcdn.. ,π = π gcdn.. ,π = π . So, 2 ts ttss the result follows. 3) deg i =1. Next, we discuss regularity of the graph Proof. 1) Note that, gcd.. n , 2ks 2 i =2 minks{,} if Li n[] and Li n [] . Clearly, if G is re- ks k ks or and g..cd n , 2 2 i =2 1 i 2 gular, then LG is also regular, so if npq=, , then if and only if ks= and = . Moreover the graph Li n[] is regular. On the other hand, if 2 kk m G is the complete bipartite graph K , then 22=0 i if and only if k . rs, 2 deg[,]= u v r s 2 for all vertices in LKrs, . Thus 2) Obvious. Li qq [] is regular. While 2[]iiq [] is a 3) Note that if , are odd, then 12 g..cd ( i , n )=1 i. bipartite graph with partite sets m Theorem 10.5 Let nqm=, 3, , are rela- Ai= 1 ,0 , 1,0 , i ,0 and tively prime with q . Then in n []i , BixxVi1,: q [] 2k m qksk1, if 1 and < ks2 0,xxV : [ i ] . deg q q i = . q 2k m qks2, if Moreover, Ni1,0= B, Ni1,1=1,0 i and 2 NA0,1 = . Thus, Theorem 10.6 Let npm=,m 1, pa= 22 b and deg1,0,1,1 i i deg 1,0,0,1 i , gcd.. , p =1. Then in n []i , ks and hence, L []i is not regular. deg a bi a bi 2q Theorem 11.2 If ntm=,m 2, t is a prime and 2 22ks m nq , then the graph Li [] is not regular. ab1, if kors < and ks , 1 n Proof. If nm=2m , 2, then 2 mm11 m1 ks 22 m deg1,22 i i deg 2,2 i . If ab2, if ks , =.2 m mm121 nqm=, 3, then degqqi ,, degqqi . 22s ab1, if k = 0 And if np= m , pa= 22 b, m 2 , then 22k ab1, if s = 0 mm deg a bi, a bi deg a bi,. a bimm a bi 1 11. When Is L n[]i , L n[]i Regular? Theorem 11.3 Let RRR= 12 where R1 and R2 are commutative rings with unity with at least one of A graph G in which all vertices have the same degree them is not ID. Then LR is not regular. is called regular graph. Proof. Suppose that R1 is not ID and Rrii= , for Regularity of []i was studied in [1]. However, 2 n i =1,2. Let x11VR . If x1 =0, then we provide our own proof, since it comes as an im- mediate consequence of Theorem 10.2. Clearly, if Nx 12,0 = 0, aaR : 0 2 npq=2, , , then n []i is regular. If m m y , a : y ann x11 0, x nm=2 , 2 or nqm=, 3, then the graph n []i has a vertex which is adjacent to all other and aR 2 if vertices and it is not complete graph, thus n []i is ann x11 0, x x 1, a : a R 2 0 , hence not regular. 2 deg x12,0 , 0,1 2 r r1 4 . And if x1 0 , Now, we show that n []i is regular if and only if Copyright © 2012 SciRes. OJDM M. GHANEM ET AL. 33 Nx12,0 = 0, a : a R 0 12. When is L n[]i , L n[]i y , a : y ann x0, x 11 Locally H? and aR 2 if ann x110, x , hence A simple graph G is said to be locally H if the neighborhood of each vertex in VG induces the same deg x ,0 , 0,1 r r 3 . But 121 graph H . The cartesian product GH of two graphs G and H is the graph with vertex set deg[(1,0),(0,1)] = r12 r 4 . So LR is not re- gular. VG H = VG VH and two vertices in So as a consequence of Theorem 11.2 and Theorem VG H are adjacent if and only if they are equal in 11.3, we conclude the following. one coordinate and adjacent in the other. Before we proceed, we give the following lemma. Theorem 11.4 The graph Li n[] is regular if 2 and only if npqqq=, ,12. Lemma 12.1 1) If GKn=,n 3, then LG is 2 locally Kn22K . Observe that, for nq=2, , n []i is the empty 2) If GK=,,2mn, mn , then LG is locally graph. []iN K , so the line graph Kmn11 K . qqqq3242 1 Proof. 1) Let [,]uv V L Kn , then Nuv[,]=[, uaaVK ]: {,} uv Li 3 [] is regular. While n q [av , ]: a V Kn { uv , } npp[]iK11 K each of the sets [,ua ]: a V Kn {,} uv and which is regular, so is Li [] . n [,]:av a V Kn {,} uv induces a copy of Kn2 and since we deal with an undirected graphs, then for a fixed a , [,ua ] and [,va ] are adjacent. Thus the result InLidegiidegi m [ ] , 1 ,1 3 1 , 2 . 2 holds. 3) Let [,]uv V L Kmn, , with partite sets A and In [i ] , m > 3, deg q , qi deg q2 ,. q B and with uA , vB . Then qm Nuv[,]=[,]: ubb B {}} v . [av , ]: a A { u } And in Lim [] , 2, pm Each set induces a complete graph Knm11, K , res- deg a bi,, a bi deg a bi2 a bi . So, the graph pectively. And Nuv[,] has no other edges. Thus Nuv[,] induces KKnm11 . m t Lin [] is not regular for ntm=, 2, is a In order for a graph to be locally H , it should be prime and nqq 23, . regular graph. Thus for the graph Li n[] , it Theorem 11.5 Let RRR= where R and R 2 12 1 2 suffices to check the cases npqqq=, ,12, and for are commutative rings with unity such that 3 Lin [] , we consider only the cases npq=, . VR= t, Rrii= for i =1,2. If reg Ri 2 Since []iK = and []iK = K, and rr12 , then LR is not regular. P pp1, 1 P pp11 Proof. Since reg R 2 , for i =1,2, there exist i Li P [] is locally KKpp22 and uregR1 and uregR1 . Therefore 11 22 LiP [] is locally K p22K . In the same manner 1,0 ,uuVLR ,0 , 0,1 , 0, . Since 12 we can show that Li [] is locally qq12 rr12 , KK , Li [] is locally K K and qq2222 q2 q222 12 deg1,0 , u121 ,0 = 2 t 2 r 4 2 t 2 r 4 . Li [] is locally K K . =deg 0,1 , 0, u q3 qq422 2 2 Theorem 12.2 The following statements are equi- So, LR is not regular. valent. Theorem 11.6 The graph Li n[] is regular if 1) The graph LiLinn[] [] is regular, and only if np= or q3 . Copyright © 2012 SciRes. OJDM 34 M. GHANEM ET AL. Mathematical Monthly, Vol. 112, No. 7, 2005, pp. 602- 2) The graph is locally LiLinn[] [] 611. doi:10.2307/30037545 H . [8] V. G. Vising, “The Number of Edges in a Graph of a Given Radius,” Soviet Mathematics—Doklady, Vol. 8, 1967, pp. 535-536. REFERENCES [9] C. Berg, “Graphs and Hypergraphs,” American Elsevier [1] E. Abu Osba, S. Al-Addasi and N. Abu Jaradeh, “Zero Publishing Co, Inc., New York, 1976. Divisor Graph for the Ring of Gaussian Integers Modulo [10] H. J. Veldman, “A Result on Hamiltonian Line Graphs n,” Communication in Algebra, Vol. 36, No. 10, 2008, pp. Involving Restrictions on Induced Subgraphs,” Journal of 3865-3877. doi:10.1080/00927870802160859 Graph Theory, Vol. 12, No. 3, 1988, pp. 413-420. [2] E. Abu Osba, S. Al-Addasi and B. Al-Khamaiseh, “Some doi:10.1002/jgt.3190120312 Properties of the Zero Divisor Graph for the Ring of [11] D. J. Oberly and D. P. Sumner, “Every Connected, Lo- Gaussian Integers Modulo n,” Glasgow Journal of Math, cally Connected Nontrivial Graph with No Induced Claw Vol. 53, No. 1, 2011, pp. 391-399. Is Hamiltonian,” Journal Graph Theory, Vol. 3, No. 4, doi:10.1017/S0017089511000024 1979, pp. 351-356. doi:10.1002/jgt.3190030405 [3] E. Abu Osba, “The Complement Graph for Gaussian [12] M. J. Plantholt, “The Chromatic Index of Graphs with Integers Modulo n,” Communication in Algebra, accepted. Large Maximum Degree,” Discrete Mathematics, Vol. 47, [4] K. Nazzal and M. Ghanem, “On the Line Graph of the 1983, pp.91-96. doi:10.1016/0012-365X(83)90074-2 Zero Divisor Graph for the Ring of Gaussian Integers [13] B.-L. Chen and H.-L. Fu, “Total Chromatic Number and Modulo n,” International Journal of Combinatorics, Vol. Chromatic Index of Split Graphs,” The Journal of Com- 2012, Article ID 957284. doi:10.1155/2012/957284 binatorial Mathematics and Combinatorial Computing, [5] P. F. Lee, “Line Graph of Zero Divisor Graph in Com- Vol. 17, 1995, pp. 137-146. mutative Rings,” Master’s Thesis, Colorado Christian [14] S. Akbari and A. Mohamamadaian, “On the Zero Divisor University, 2007. Graph of a Commutative Ring,” Journal of Algebra, Vol. [6] J. Sedlàĉek, “Some Properties of Interchange Graphs, 274, No. 2, 2004, pp. 847-855. Theory of Graphs and Its Applications,” Academic Press, doi:10.1016/S0021-8693(03)00435-6 New York, 1962, pp. 145-150. [15] S. Arumugam and S. Velammal, “Edge Domination in [7] G. Dersden and W. M. Dymcek, “Finding Factors of Graphs,” Taiwanese Journal of Mathematics, Vol. 2, No. Factor Rings over the Gaussian Integers,” American 2, 1998, pp. 173-179. Copyright © 2012 SciRes. OJDM