A Classification of Cyclic Nodes and Enumeration of Components of a Class of Discrete Graphs
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1 Appl. Math. Inf. Sc. 9, No. 1, ) 103 Appled Mathematcs & Informaton Scences An Internatonal Journal A Classfcaton of Cyclc Nodes and Enumeraton of Components of a Class of Dscrete Graphs M. Khald Mahmood 1, and Farooq Ahmad 1 Department of Mathematcs, Unversty of the Punjab, Lahore, Pakstan Faculty of Informaton Technology, Unversty of Central Punjab, Lahore, Pakstan Receved: 15 Mar. 01, Revsed: 15 Jun. 01, Accepted: 16 Jun. 01 Publshed onlne: 1 Jan. 015 Abstract: Let Z n be the rng of resdue classes modulo n. Defne f : Z n Z n by fx) = x. Acton of ths map s studed by means of dgraphs whch produce an edge from the resdue classes a to b f fa) b. For every nteger n, an explct formula s gven for the number of fxed ponts of f. It s shown that the graph Gp k ), k 1 has four fxed ponts f and only f 3 p 1 and has two fxed ponts f and only f 3 p 1. A classfcaton of cyclc vertces of the graph Gp k ) has been determned. A complete enumeraton of non-somorphc cycles and components of Gp k ) has been explored. Keywords: Congruences, Multplcatve Order, Cyclc Vertces, Components 1 Introducton The noton of congruence s of great nterest n number theory. A strong emphass on modular arthmetc leads n a natural way to jump over many of new destnatons especally n pure mathematcs. It has become a useful devce to solve most of the mathematcal problems whch are ntegral based. Use of modular arthmetc n studyng dscrete graphs and dgraphs s becomng an ncreasngly useful devce to explore a broad range of applcatons. Let f be any functon assumng ts values as the resdues after dvson by an nteger n. We can draw a graph that has the remanders as vertces when dvded by n and a drected edgea,b) f and only f fa) bmod n). For fx)=x k, the assocated dgraph s denoted by Gn, k). The dgraph of squarng modulo a prme p, has been studed n []. Earle L. Blanton [], L. Somer and M.Křížek [7], L. Szalay [9], T.D. Rogers [1], Troy Vasga [13] and Y. Meemark [1] have consdered and nvestgated propertes of a varety of dgraphs correspondng to the congruence a bmod n). The condtons for regularty, sem regularty and symmetrcally structured dgraphs have been dscussed n [1] and [8]. The structures of graphs of exponental congruences has been dscussed n [10]. Though many fascnatng features lke regularty, sem-regularty and symmetry of such dgraphs by means of sub-dagraphs have been explored, yet there are some topographes lke classfcatons of vertces, number of all possble cycles of all lengths, number of components etc for whch no explct formula s present. In ths pece of work, a complete characterzaton free from sub-dagraphs n terms of explct formulas for the number of fxed ponts, classfcatons of cyclc vertces, number of non-somorphc cycles and number of non-somorphc components of the graph G, p k ), k 1, where p s an odd prme, has been dscussed n detal. Prelmnares The vertces v 1,v,...,v t 1,v t form a cycle of length t f and only f v 1 v mod n) v v 3 mod n). v t v 1 mod n) The graph Gn) s sad to be connected f for each vertex par u and v, there exst some ntegral number m such that u m vmod n). A maxmal connected subgraph s termed as component, [5]. A vertex z s referred to be a fxed pont f z zmod n). Correspondng author e-mal: khald.math@pu.edu.pk c 015 NSP
2 10 M. K. Mahmood, F. Ahmad: A Classfcaton of Cyclc Nodes and Enumeraton of Components Fg. 1: G169) In Fg 1, the dgraph G169) has ten components. Among these sx are cyclc and four are the rooted trees. The graph has four fxed ponts. That s, the vertces whch has self loops. These are 0, 1, and 16. Precsely the graph G169) has three non-somorphc components, one of whch contans a non-somorphc cycle of length sx, one s a rooted tree wth root at 0 and one wth a non-zero fxed pont contanng somorphc cycles of length one). We frst recall a few defntons and some prevous results for use n the sequel. Defnton.1. [3] The functon φn) s defned as the number of dvsors d such that d n, where 1 d < n and d,n)=1. It s trval that φ1)=1, as gcd1,1)=1. Theorem.. [3] Euler) Let a be any nteger such that a,m)=1, where m 1, then a φm) 1mod m). The followng result can be derved usng Theorem.. Theorem.3. [3] Let k>0 and p be a prme, then φp k )= p k 1 1 p ) Defnton.. [3] Let n > 1 and gcda,n)=1. The order of a modulo n s the smallest postve nteger k such that a k 1 mod n). It s denoted as ord n a=k. Theorem.5. [6] Let fx) be a fxed polynomal wth ntegral coeffcents, and for any postve nteger m let Nm) denote the number of solutons of the congruence fx) 0mod m). If m=m 1 m wherem 1,m )=1, then Nm) = Nm 1 )Nm ). If m = p α s the canoncal factorzaton of m, then Nm)= Np α ). Theorem.6. [11] Consder an odd prme p, and let p a ±1, where ord p a = d. Let k 0 be the greatest nteger such that p k 0 a d 1. Then ord k p a=d for 1 k k 0 and d p k k 0 for k k 0. Theorem.7. [11] Let p, q, p q be odd prmes. Then, p q ) q p p 1 q 1 )= 1) Theorem.8. [7] There exsts a cycle of length t n G 1 n) f and only f t = ord d k for some dvsor d of λn). 3 Basc Results and Fxed Ponts of the Map A pont x s sad to be a fxed pont of the map f f fx) x mod p). To fnd the fxed pont of the map, we gve the followng elementary result whose proof s smple and straght forward. Lemma 3.1. Let x 0,1 be a fxed pont of the mappng f over Z p, then ) x+x p 1mod p) ) xx 1mod p) The followng result descrbes a relatonshp between nontrval fxed ponts of the map f. Theorem 3.. Let x 0,1. Then, x s a fxed pont of the mappng f over Z p f and only f x s a fxed pont of f over Z p. Proof. Let x 0,1. Suppose x s a fxed pont of f. That s, x x mod p). Then, x s a fxed pont of f as x ) =x ) x mod p). Conversely, Suppose x s a fxed pont of f. That s, x ) x mod p). Or x 8 x mod p). 1) Now x 0 mod p) mples that x 0mod p). Let x α mod p),α Z p,α 0. ) Then by 1), x α mod p). Ths further mples that x α mod p). 3) Thus p x + α or p x+α or p x α. We wll exhbt, the frst two are not possble. Ths wll complete the proof. Let p x + α. That s, x + α 0 mod p). Then by equaton ), x + x 8 0 mod p) or x 8 x mod p), a contradcton aganst 1). Otherwse, x x mod p) yelds that x 0 mod p). But p s an odd prme, so x 0mod p) reveals that x 0mod p), a contradcton as x 0mod p). Hence p x + α s not possble. Now, let p x+α. That s, x+α 0 mod p). Then by equaton ), x + x 0 mod p). Ths gves, x 1 mod p) or x x mod p). But x 1 mod p) snce 1) 1 mod p). Thus x x+1 0 mod p). Ths further can be wrtten as, x + x + 1 x mod p). Also by Lemma 3.1, x+x mod p). Hence, x 0 mod p). Snce p s an odd prme, so x 0 mod p) yelds a contradcton as x 0mod p). Hence p x+α. c 015 NSP
3 Appl. Math. Inf. Sc. 9, No. 1, ) / The followng corollares are the smple consequences of Theorem 3.. Corollary 3.3 Fxed ponts of Gp k ) are the quadratc resdues of p k. Corollary 3. The non zero fxed ponts of the graph Gp k ), k 1 form a cyclc subgroup of the group Z p. Before gvng the cardnalty of fxed ponts of Gn) for any nteger n, we need the followng mportant lemmas. Lemma 3.5. The numbers 0, 1, 3 k and.3 k are the fxed ponts of the graph G3 k ). Proof. For k=1, t s easy to see that 0 and 1 are the only solutons of the congruence x xmod 3 k ). Let k>1 and suppose x s a fxed pont of G3 k ), k > 1. Clearly 0 and 1 are the fxed ponts for k > 1 as well. For the numbers 3 k and.3 k 1 + 1, t s easy to see that 3 k 1 + 1) +3 k 1 + 1)+1 3mod 3 k ) as k>1 ) Usng ), we get, 3 k 1 + 1) 3 k 1 + 1) = 3 k 1 3 k 1 + 1){3 k 1 + 1) + 3 k 1 + 1)+1)} 3 k 1 3 k 1 + 1)3)mod 3 k ) 3 k 3 k 1 + 1)mod 3 k ) 0mod 3 k ), k>1 Ths shows that 3 k s a fxed pont of the graph G3 k ), k > 1. For the number.3 k 1 + 1, we note that.3 k 1 + 1) 3 k mod 3 k ). Hence by Theorem 3.,.3 k s a fxed pont of the graph G3 k ), k > 1. Lemma 3.6. Let p > 3 be any prme. Then the graph Gp k ), k 1 has four fxed ponts f and only f 3 p 1. proof. Suppose 3 p 1. Let x be a fxed pont of the graph Gp k ), k 1. Then, x x mod p k ) yelds that x x mod p). Ths gves, p x or p x 1 or p x + x+1. Thus the graph Gp k ), k 1 has four fxed ponts f and only f the congruence x + x mod p) s solvable. Now x + x mod p) s solvable f and only f y 3 mod p), where x y 1 mod p) s solvable for y. By Theorem.7, t s easy to establsh that -3 s a quadratc resdue modulo p f and only f p s a quadratc resdue modulo 3. But 3 p 1 mples that p 1 mod 3). Ths clearly shows that p s a quadratc resdue modulo 3 and hence -3 s a quadratc resdue modulo p. consequently, the congruence y 3 mod p) s solvable. Thus the graph Gp k ), k 1 has four fxed ponts f and only f 3 p 1. Lemma 3.7. Let p be a prme of the form 6k+1. Then 3k+1 3k+1 3) 1 3) +1 the numbers 0, 1, and are the fxed ponts of the of the graph Gp k ). Proof. Snce p 1 mod 6), -3 s a quadratc resdue of p. Then by Euler s Theorem, 3) p 1 1 mod p). Ths means that 3) 3k+1 3 mod p). Ths can also be wrtten as 3) 3k+1 ) 3 mod p). Ths shows that 3) 3k+1 s a soluton of the congruence y 3k+1 3) 1 3 mod p). Hence by Lemma 3.6, s a fxed pont of the graph Gp). Also by Theorem 3., 3) 3k+1 +1 s a fxed pont snce 3) 3k+1 1 ) = 3) 3k Lemma 3.8. Let p > 3 be any prme. Then the graph Gp k ), k 1 has two fxed ponts f and only f 3 p 1. Proof. Snce p 1mod 3), so p s not a quadratc resdue modulo 3. Hence the congruence x + x+1 0 mod p) s not solvable. Thus 0 and 1 are the only fxed ponts of Gp k ), k 1. For any nteger n. We defne the functons ξn) and ωn) as { 0, f 3 n or 3 n ξn)= 1, f 3 k n, k>1 and { 0, f 3 n or 3 ωn)= k n, k>1 1, f 3 n Theorem 3.9. Let p k 1 1 pk...pk r r be the canoncal representaton of any nteger n, where p 1, p,..., p r are dstnct odd prmes. Let Ln) denote the number of fxed ponts of the graph Gn), then, r+ξn), f 3 p 1, 1 r Ln)= r+ωn), f 3 p 1, 1 r r t+ξn), f 3 p 1, 1 t and 3 p 1, t+ 1 r Proof. To fnd the number Ln), we need to count the number of solutons of the congruence x x mod n). We note that the congruence x x mod p k 1 1 pk...pk r r ) s solvable f and only f x x mod p k ) s solvable for each = 1,,...,r. Let 3 p 1 for each = 1,,...,r, then by Lemma 3.8, Lp k ) = for each. Snce p 1 < p <...< p r are dstnct prmes, so by Theorem.5, Ln)= r. Now f 3 p 1 for each =1,,...,r, then by Lemma 3.6, Lp k )= for each. Hence by Theorem.5, Ln) = r. Fnally, wthout any loss, we assume that 3 p 1 for each = 1,,...,t, then 3 p 1 for each = t + 1,t +,...,r. Hence by Lemmas 3.6, 3.8, Lp k ) = for = 1,,...,t and Lpk ) = for = t + 1,t +,...,r. Thus agan by Theorem.5, Ln) = t r t = r t. Now we dscuss the followng three cases. 1) Let 3 n. Then by defnton, ξn) = ωn) = 0. Thus result s true n ths case. c 015 NSP
4 106 M. K. Mahmood, F. Ahmad: A Classfcaton of Cyclc Nodes and Enumeraton of Components... ) Let 3 n. Then by defnton ξn)=0 and ωn)= 1. Now snce 3 n, so one of the p k = 3 for some th factor of n. Wthout any loss, we let p k 1 1 = 3. Then the condton 3 p 1, 1 r must reduces to 3 p 1, r. Thus n ths case, we must get, Ln) = r 1 = r+ωn) when 3 p 1, r. Therefore, we set ξn)=0 and ωn)= 1 to get the desred result. 3) Let 3 k n, k > 1. Then by defnton ξn) = 1 and ωn) = 0. Now snce 3 k n, k > 1, so one of the p k = 3 k, k>1 for some th factor of n. Wthout any loss, we let p k 1 1 = 3k, k > 1. Then by Theorem.5 and by Lemma 3.5, we have, Ln)=L3 k )Lp k...pk r r )= Lp k...pk r r ). Now f 3 p 1, r, then Lp k...pk r r ) = r 1. Hence, Ln) = Lp k...pk r r ) = r+1 = r+ξn). Thus n ths case we set ξn)=1 and ωn)=0 to get the desred result. Remark The number Ln) s always even. Corollary Let p be an odd prme and α 0,1, a fxed pont of the graph Gp k ) such that α,α Z p = {0,1,,..., p 1}. Then α p 1 f and only f p=3+α+ 1). Classfcatons of Cyclc Vertces In ths secton we present explct formulas to enumerate cyclc vertces of the graph Gp k ), where p s prme. The followng nequaltes can easly be proved usng mathematcal nducton. Lemma.1. For k, k αk ), α =,3, Theorem.. The vertces 1+ l 3 k and 1+. l 3 k for l = 0,1, form cycles of length 3 n the graph G3 k ). Proof. The vertces a 0,a 1 and a form a cycle of length 3 n G3 k ) f and only f a 0 a 1 mod 3 k ), a 1 a mod 3 k ) and a a 0 mod 3 k ). Now, 1+ l 3 k ) = 1+ l+1 3 k + α= α= α) αl 3 αk ) 5) Snce k, by Lemma 3.1, k αk ), α =,3,. Then, 3 k 3 αk ). Hence, α) αl 3 αk ) 0mod 3 k ). Puttng n 5), we obtan, 1+ l 3 k ) 1+ l+1 3 k mod 3 k ), l = 0,1, 6) Fnally,we note that k = 1+1+3) 3 3 k = )3 k = 1+3 k + 3 k +.3 k k mod 3 k ) 7) Equatons 6) and 7) yelds that the vertces 1+ l 3 k for l = 0, 1, form a cycle of length 3 n the graph G3 k ), k. Smlarly, t s easy to see that the vertces 1+. l 3 k for l = 0,1, form a cycle of length 3 n the graph G3 k ), k 3. Corollary.3. If the vertces 1+ l 3 k for l = 0,1, are at a 3-cycle n G3 k ), k 3 then 1+ l 3 k 1 for l = 0,1, are at a 3-cycle n G3 k+1 ), k 3. The proof of above corollary s evdent f we take k=r+1 and apply Theorem.. However, the mportance of ths result s of great nterest as we are lftng the vertces of a 3-cycle of a graph to a 3-cycle n ts hgher modulo graph. Theorem.. ) The vertces 1 + l 3 k r 1 for l = 0,1,,...,3 r 1 form cycles of length 3 r, r=1,,...,q n the graph G3 k ), k, where, f k= q= k 1 k, f k s odd, k> 1, f k s even, k> ) For q < r k, the vertces 1+. l 3 k r 1 form cycles of length 3 r n the graph G3 k ), k 5. Proof. Frst, we prove that there exst cycles of length 3 r, 0 r k n the graph G3 k ). Let s denote the order of modulo 3 k by ord 3 r. Clearly ord 3 = 1 and ord 3 = 3. Then by Theorem.6, k 0 = 1. Hence ord 3 r = 3 r 1. But the only odd dvsors of φ3 r ) = 3 r 1 are 1,3,3,...,3 r 1, where, ord 3 r 1=3 r. Thus by Theorem.8, there exst cycles of length 3 r, 0 r k. Next for k > and r=1,,...,q, t s easy to see that Also for α =, k αk r 1), where, α = 3,. Thus, 3 k 3 αk r 1) for α = 3,. 3 k 3 ) αk r 1) Ths shows that α) 3 αk r 1) 0mod 3 k ), α =,3, Then the followng equaton 1+ l 3 k r 1 ) = 1+ l+1 3 k r 1 + reduces to α= α) αl 3 αk r 1) 1+ l 3 k r 1 ) 1+ l+1 3 k r 1 mod 3 k ) 8) For k =, proof s smple and straght forward. So we complete the proof n the followng two cases. Case a). Let k be even. Then by defnton of q, q= k 1. We dscuss the cycle of length 3 q. In ths case, r= q, then equaton 8) becomes 1+ l 3 k ) 1+ l+1 3 k mod 3 k ), l = 0,1,...,3 r 1.9) c 015 NSP
5 Appl. Math. Inf. Sc. 9, No. 1, ) / Fnally, f l = 3 r 1=3 q 1, where q= k 1, then 1+ l+1 3 k = 1+1+3) 3 k 1 3 k = k ) 3 k = 1+3 k + 3 k + terms nvolvng 3 k 1+3 k mod 3 k ) 10) Caseb). Let k be odd. Then q= k 1. For a cycle of length 3 q. Agan by equaton 8), we obtan, 1+ l 3 k 1 ) 1+ l+1 3 k 1 mod3 k ),l = 0,...,3 q 111) Take l = 3 q 1, then 1+ l+1 3 k 1 = 1+1+3) 3 k 1 3 k 1 = k ) 3 k 1 = 1+3 k k + terms nvolvng 3 k 1+3 k 1 mod 3 k ) 1) Let a 0 = k q 1, a 1 = k q 1,..., a 3 q 1 = 1+ 3q 1 3 k q 1. Then equatons 8), 10) and 1) mples that a 0 a 1 mod 3 k ) a 1 a mod 3 k ). a 3 q 1 a 0 mod 3 k ) Ths clearly shows that the vertces a 0, a 1,..., a 3 q 1 form a cycle of length 3 q. The proof of part ) s smlar. The below results can be proved usng Theorem.. Corollary.5. The vertces 1 + l 3 k r 1 and 1+. l 3 k r 1 for l = 0,1,,...,3 r 1 are always at a cycle of length 3 r where, 1 r k n the graph G3 k ), k 3. Corollary.6. There exst k non-somorphc cycles of length greater than one n G3 k ). Corollary.7. The maxmum length of any cycle n G3 k ) s 3 k. The followng theorem gves the classfcatons of noncyclc vertces n the graph G3 k ). Theorem.8 ) The vertces 3 k 1+ l 3 k r 1 ) and 3 k 1+. l 3 k r 1 ) for l= 0,1,,...,3 r 1 are the non-cycle vertces of the graph G3 k ) and are always mapped on the cyclc vertces. ) The vertces ±1, ±1 + 3 k ) and ±1 +.3 k ) are mapped on 1, 1+3 k and 1+.3 k respectvely. Proof.) Ths s smple snce 3 k 1+ l 3 k r 1 )) 1+ l+1 3 k r 1 mod 3 k ) and 1+ l+1 3 k r 1 ) 1+ l 3 k r 1 ) mod 3 k ) Fg. : shows the 3,3 and 3 3 Cycles of G3 5 ) where the vertces 1+ l 3 k r 1 for l= 0,1,,...,3 r 1 are always at a cycle of length 3 r. The proof for the vertces 3 k 1+. l 3 k r 1 ) s smlar. ) By Lemma 3.5, the vertces 1, 1+3 k and 1+.3 k are the fxed ponts of G3 k ), so they mapped on themselves. Snce 1) 1. Fnally we classfy the vertces n G3 k ) whch are not prme to 3. We see that these vertces always form a tree n G3 k ) wth root at 0. Frst we note that the vertces 3α, α = 1,,...,3 k 1 has 3 k 5 branch ponts except root. For β = 1,,,..., 3k 1, where gcd3,β) = 1, the vertces 3β) are the branch ponts of the graph G3 k ). That s the vertces 3α, α = 1,,...,3 k 1 are mapped on any of the branch ponts. It can easly be seen that f 3α are mapped on3β), then by defnton α) =3β) mod 3 k ) gves, α β)α+ β)α + β ) 0mod 3 k ) But α + β 0 mod 3 k ) s not solvable for any α as proved n the proof of Theorem 3.. Hence, the only possbltes are α = ±β + 3 k t. That s, α β mod 3 k ) where β = ±1,±,±,...,± 3k 1, k and gcd3,β) = 1. Note that there are.3 k 5 resdues namely ±1.3,±.3,,...,±3 k 5.3 modulo 3 k whch are dvsble by 3. Thus to count the branch ponts excludng 0, we count the number of β s such that β =±1,±,±,...,± 3k 1, k n CRS modulo 3 k. Thus the total number of such β s s 3 k.3 k 5 = 3 k 5. Ths shows that there are 3 k 5 branch ponts other than 0 n the tree component of the graph G3 k ). c 015 NSP
6 108 M. K. Mahmood, F. Ahmad: A Classfcaton of Cyclc Nodes and Enumeraton of Components Fg. 3: shows the three branch ponts of the tree component of G3 6 ) wth root at 0. The above dscusson leads to the followng result. Theorem.9. The undrected graph of the vertces 3α, α = 1,,...,3 k 1 form a tree wth root at 0 and have 3 k 5 branch ponts. Espousng the steps elucdated above, we can cheer to extant the formula for fndng the cycle vertces even to hgher powers of any prme p. Ths can be entertaned n Theorem.11. The proof of the followng proposton s analogous to Theorem.6. Proposton.10. Let p > 3 be any prme. The possble cycle length of the graph Gp k ) s at most d.p r,r= 0,1,,...,k 1, where ord p =d. Theorem.11. ) Let p > 3 be any prme and d > 0 dvdes φp). If d 1mod p k ). Then the vertces 1 + l p k r 1, k > 1 form cycles of length d.p r,r = 0,1,,..., k 1 n the graph Gpk ) where, l = 0,1,,...,d.p r 1. ) For k r k, the vertces 1+.l 3 k r 1 form cycles of length d.p r n the graph Gp k ), k 5. Proof. We prove ). The proof of ) can be derved n a smlar fashon. Note that k { k 3 1=, f k s odd k 1, f k s even Now for any r, we have, { k k 3 k r 1 = 1, f k s odd k k 1) 1, f k s even = { k+1 k, f k s odd, f k s even Then, k αk r 1),α =,3,. So, p k p αk r 1) for any prme p. Ths means that, followng equaton 1+ l p k r 1 ) = 1+ l+1 p k r 1 + reduces to p αk r 1) 0 mod p k ). Then the α= α) αl p αk r 1) 1+ l p k r 1 ) 1+ l+1 p k r 1 mod p k ) 13) Also f d 1mod p k ) where, d > 0 dvdes φp). Then d pr = d ) pr 1) pr 1mod p k ) 1) Fnally we dscuss a cycle of maxmum length d p r. By equaton 13) and 1), we obtan 1+ d pr 1 p k r 1 ) 1+ d pr p k r 1 mod p k ) 1+ p k r 1 mod p k ) That s, 1+ d pr 1 p k r 1 ) 1+ p k r 1 mod p k )15) Let a 0 = 1+ 0 p k r 1, a 1 = 1+ 1 p k r 1,..., and a d p r 1= 1+ d pr 1 p k r 1. Then equatons 13) and 15) yelds that a 0 a 1 mod p k ) a 1 a mod p k ). a d p r 1 a 0 mod p k ) Ths clearly shows that the vertces a 0, a 1,..., a d p r 1 form a cycle of length d p r. The below results can be proved usng Theorem.11. Corollary.1. Let p be any prme and d > 0 dvdes φp). If d 1mod p k ). Then for k > 1, the sets of vertces{1+ l p : 0 l d.p k 1},{1+ l p : 0 l d.p k 3 1},...,{1+ l p : 0 l d 1} form cycles of length d.p k,d.p k 3,..., and d respectvely n the graph Gp k ). Corollary.13. If v 1,v,...,v r form a cycle of length r n Gp k ) then for any nteger s, the vertces v s 1,vs,...,vs r also form a cycle of length r n Gp k ). Proof. Snce v s 1 ) = v 1 )s v s,vs ) = v )s v s 3,...,vs r) =v r) s v s 1. Hence vs 1,vs,...,vs r form a cycle of length r n Gp k ). Corollary.1. If v 1,v,...,v r form a cycle of length r n Gp) then v 1,v,...,v r are the quadratc resdues of p. Proof. Snce v n) v 1,v 1 ) v,...,v n 1 ) v n. Hence v 1,v,...,v r are the quadratc resdues of p. 5 Enumeraton of Cycles and Components The vertces v 1,v,...,v s. consttutes a component Gp k ) f for each, 1 s, there exst some j, 1 j s such c 015 NSP
7 Appl. Math. Inf. Sc. 9, No. 1, ) / that v v j mod p k ), for all j. It s well known that each component contans exactly one cycle n the graph Gp k ), k 1 but yet t s desrable to know that how many components does Gp k ), k 1 have? In ths secton we address ths problem and enumerate number of cycles of all lengths, and hence number of non-somorphc components of the graph Gp k ), k 1. In the followng theorems, we dscuss the possble cases and enumerate the number of non-somorphc components n detal. Before gvng the detal of non-somorphc components, we frst gve the followng smple Lemma. Lemma ) There are two non-somorphc possble components correspondng to all fxed ponts of the graph Gp k ). ) The graph G3 k ),k > 1 has k non-somorphc components. Proof. Let 3 p, then by Lemma 3.8, the vertces 0 and 1 are the only fxed ponts of the graph Gp k ). Snce 0 1mod p k ), so 0 and 1 are not adjacent to each other. Let f possble) x and y be the vertces of the components contanng the fxed ponts 0 and 1 respectvely such that x ymod p k ). That s, x and y are the adjacent vertces. By defnton, there must exst ntegers t 1,t,...,t r such that x t 1 pmod p k ), t 1 p) t pmod p k ),..., t r p) 0mod p k ). Then clearly x t 0mod p k ) for some nteger t. Smlarly there exst ntegers s 1,s,...,s t such that y s 1 mod p k ) and s 1 s mod p k ) and so on s r 1mod pk ). Then for some nteger s, we obtan, y s 1mod p k ). Let l be the least common multple of the ntegers t and s. Now f x ymod p k ), then x ) l y l mod p k ). Ths means that x l ) y l mod p k ) or 0 1mod p k ) whch s not possble. Thus the fxed ponts 0 and 1 are the vertces of dsjont components. Fnally, as deg1) s the number of ncongruent solutons of the congruence x 1mod p k ). Thus deg1) whereas deg0) s at least p k 1 as t p) 0mod p k ). Thus the components contanng the fxed ponts 0 and 1 are the non-somorphc components. Agan by Lemma 3.6, there are four fxed ponts of the graph Gp k ) f and only f 3 p. Then by Lemma 3.7, f α s a non zero fxed pont of the graph Gp k ) then 1,α and α are the possble non zero fxed ponts of the graph Gp k ). As α s a fxed pont so,p k α) α) α α mod p k ). Ths shows that α and p k α are adjacent to each other. The rest of the proof can easly be completed adoptng the steps explaned above. ) By Theorem., there exst k possble cycles of length greater then one. Also by Theorem.8, each component contans a cycle, so there exst k non-somorphc components n G3 k ). Moreover by Theorem 5.1 ), there exst two non-somorphc components correspondng to all fxed ponts of the graph G3 k ). Thus, G3 k ) has k + = k non-somorphc components. Theorem 5.. Let p > 3 be any prme such that φp) = l q r, where q s an odd prme and ord q = α. If 3 φp) then Gp) contans r + 1 non-somorphc components. If 3 φp) then Gp) contans r k non-somorphc components, where k 0 s the largest nteger such that α 1mod p k 0). Proof. If 3 φp) then q = 3. As ord 3 r=3 r 1 φp)= l 3 r, so by Theorem.8, we nfer that there exst cycles of lengths 3,3,...,3 r 1 n Gp). That s, there exst r 1 possble cycles of length > 1. Also by Lemma 5.1 ), there exst two non-somorphc components correspondng to cycles of lengths one. Snce each component contans a cycle, we conclude that there exst r 1+= r+1 non-somorphc components when 3 φp). Next we suppose that 3 φp), then q > 3. Clearly ord q >1. Snce k 0 s the largest nteger such that α 1 mod p k 0), so by Theorem.6, order of modulo q r s α for r = 1,,...,k 0 and α p r k 0 for r k 0. Ths shows that there exst cycles of lengths α,αq,...,αq r k 0. Thus there exst r k non-somorphc cycles of length > 1. By Lemma 5.1, there exst two non-somorphc components correspondng to cycles of lengths one. Hence, we fnd that there exst r k non-somorphc cycles of dfferent lengths. Consequently, f 3 p, then the graph Gp) has r k non-somorphc components. Example 5.3 1) Take p = Then φp)= 3511). Take q=3511. Here, r= and k 0 = as ord q = ord q = 1755, ord q 3 = = 1755q and ord q = = 1755q. Thus there exst three non-somorphc components contanng cycles of lengths 1755, and Snce 3 φp), so by Lemma 3.8, 0 and 1 are the only fxed ponts of the graph Gp). By Lemma 5.1, there are two non-somorphc components contanng the fxed ponts 0 and 1. Hence, there are 5 = = r k possble non-somorphc components. ) Take p=1373. Then φp)= 7 3. Here, q=7, r = 3 and k 0 = 1 as ord 7 = 3, ord 7 = 1 and ord 7 3 = 17. Thus there exst three non-somorphc components contanng cycles of lengths 3, 1 and 17 plus two non-somorphc components contanng correspondng to fxed ponts 0 and 1. Hence, there are 5 = = r k possble non-somorphc components n Gp). Theorem 5.. Let p > 3 be any prme such that φp)= l q k 1 1 qk...qk r r, where l > 0,k 1 and q 1,q,...,q r are dstnct odd prmes. Let ord q =α > 1. Then, a) If α for each, are parwse relatvely prme ntegers. Then the graph Gp) has r + 1 non-somorphc components. b) If α α j for each < j. Then there does not exst any cycle of length α j, j > 1. In ths case Gp) has r+ non-somorphc components. c) Let α α j for all and j. If lcmα, j 1 α j )=l α k for all k=1,,...,r. Then there exst a cycle of length l. In ths case Gp) has r + 1 non-somorphc components as well. c 015 NSP
8 110 M. K. Mahmood, F. Ahmad: A Classfcaton of Cyclc Nodes and Enumeraton of Components... Proof. a) Let ord q = α and ord q j = α j where α,α j ) = 1 for j. That s α and α j are the least postve ntegers such that α 1mod q ) and α j 1 mod q j ) 16) Let t be a least postve nteger such that t 1mod q q j ) 17) Snce q,q j ) = 1, so by equaton 16), t s easy to fnd that α α j 1 mod q q j ) 18) But t s the least postve nteger such that t 1mod q q j ), we must get, By equaton 17), we get t α α j 19) t 1mod q ) and t 1mod q j ) 0) But by equaton 16), α t and α j t as α,α j are least postve ntegers. Snce α,α j )=1, so α α j t 1) By equatons 19) and 1), we get α α j = t. Ths shows that there exst a cycle of length α α j n Gp). Now snce each α > 1 and are parwse prme to each other, so there must exst a cycle of length d = α j for all j factors. As there are r dfferent odd prmes n φp). Therefore the total number of cycles of length greater then one must be r 1) + r ) r r) = r 1 ) Moreover, by Lemma 5.1 ), there exst two non-somorphc components contanng cycles of length one. Consequently, there exst r 1 + = r + 1 components. b) On contrary we suppose that there exst a cycle of length α α j correspondng to some odd dvsor q q j. That s α α j s the least postve nteger such that α α j 1 mod q q j ) 3) Snce α α j, so there exst some nteger α such that α j = αα. Then by equaton16), α j 1 mod q q j ). Now by equaton 18), α α j α j. Hence α α j = α j. Ths shows that α = 1, whch s a contradcton as α > 1 for each. Ths clearly shows that there does not exst any cycle of length α j, j > 1. Thus there exst only cycles of lengths α, =1,,...,r. Usng Lemma 5.1), the number of nonsomorphc components s r 1) + =r+ ) c) Note that f α t and α j t and α,α j ) 1, then ths s false n general that α α j t. However, lcmα,α j ) t s true n ether case. That s, l t. The rest of the proof s smlar to part a). Remark 5.5. In vew of Theorem 5., we note that the number of non-somorphc cycles of dfferent lengths>1 are actually the number of dstnct orders modulo prme powers appearng n the canoncal form of φp). Thus to count the non-somorphc cycles of length > 1, our task s to count the number of dstnct orders modulo all possble dvsors of φp). For nstance, f φp)= l q r 1 1 q r, r 1,r > 1. Suppose for each, there does not exst any nteger k 0 > 1 such that α 1 mod q k 0 ). Then there must exst cycles of lengths α,αq 1,αq 1,...,αqr and β,β q,β q,...,β qr 1 n Gp) f ord q1 = α and ord q = β provded α β. Fnally, we need to know whether there exst some more orders modulo q 1 q j, for all, j where, 1 r 1 1 and 1 j r 1. Thus by Theorem 5. c), we need to know all possble dstnct lcm s of all possble products of ntegers α,αq 1,αq 1,...,αqr and β,β q,β q,...,β qr 1 dfferent from αq 1 and β q j for all, j. To generalze the concept, we defne the followng notaton. Notaton. Let α q j, where, 1 r and 1 j k 1 be dstnct ntegers. Let l α q j for all, j. We denote Nl) as the number of dstnct lcm s of all possble products of the ntegers α q j for all, j. Theorem 5.6. Let q 1,q,...,q r and p > 3 be dstnct odd prmes such that φp) = l q k 1 1 qk...qk r r, l > 0, where ord q = α > 1. Suppose for each, there does not exst any nteger k 0 > 1 such that α 1 mod q k 0 ). Then the graph Gp) has r =1 k + Nl) + non-somorphc components, where Nl) s defned above. Proof. Let ord q = α. Snce for each, there does not exst any nteger k 0 > 1 such that α 1 mod p k 0 ). Then by Theorem.6, ord q k =α q k 1. Thus there exst cycles of length α q j, 1 j k 1, where =1,,...,r. That s, there exst r =1 k non-somorphc components contanng cycles of length α q j, 1 j k 1. Also f lcmord q u m, n 1 ord q v n ) = l α q j for all, j, where 1 u k m 1 and 1 v k n 1, then by Theorem 5. c), there exst a cycle of length l n Gp). Now f Nl) s the number of dstnct lcm s of all possble products of the orders α q j for all, j. Then there exst r =1 k + Nl) non-somorphc cycles of length > 1. Usng Lemma 5.1, we conclude that the graph Gp) contans r =1 k + Nl)+ non-somorphc components. Theorem 5.7. Let p > 3 be any prme such that φp k ) = l q r p k 1, where q s an odd prme. Let ord q = α and ord p = β. Suppose there does not exst any ntegers s 0,t 0 > 1 such that α 1 mod q s 0) and β 1mod p t 0) a) If αq 1 β for all =1,,...,r. Then the graph Gp k ) has r+k+1 non-somorphc components. b) If αq 1 and β p j are parwse relatvely prme ntegers for all = 1,,...,r and j =,3,...,k. Then the c 015 NSP
9 Appl. Math. Inf. Sc. 9, No. 1, ) / graph Gp k ) has r + 1)k + 1 non-somorphc components. Proof. Let ord q = α and ord p = β. Snce there does not exst any ntegers s 0,t 0 > 1 such that α 1 mod q s 0) and β 1 mod p t 0). Then by Theorem.6, we must obtan, ord q = αq 1 and ord p j 1=β p j for =1,,...,r and j =,3,...,k. Ths shows that there exst r + k 1 cycles of lengths α,αq,αq,...,αq r 1,β,β p,β p,...,β p k 3 and β p k. Next we show that there does not exst any other cycle of length> 1. Let t αβ q 1 p j for all, j. We suppose that there exst ntegers u and v such that ord q u pv= t, 1 u r 1, 1 v k. Then t s the least postve nteger such that t 1mod q u p v ) 5) Snce p and q are dstnct prmes, so we have t 1mod q u ) and t 1 mod p v ) 6) But ord p v = β p v 1, so by 5), β p v 1 t. Also ord q u=αq u 1 and αq u 1 β, hence, ord q u p v=β pv 1, snce lcmαq u 1,β p v 1 )=β p v 1. As ord q u pv=t. Ths clearly shows that t β p v 1. Consequently t = β p v 1, whch s a contradcton as t αβ q 1 p j for all, j. Thus the only cycles of lengths > 1 are of lengths α,αq,αq,...,αq r 1,β,β p,β p,...,β p k 3 and β p k. Moreover there exst two non-somorphc components contanng cycles of length one. Thus n ths case, graph Gp k ) contans r+k+1 non-somorphc components. For the proof of part b), t s enough to show that there exst a cycle of length αβ q u 1 p v, 1 u r, v k. Snce ord q u = αq u 1 and ord p v = β p v 1. That s αq u 1 and β p v 1 are the least postve ntegers such that αqu 1 1 mod q u ) and β pv 1 1mod p v ) 7) Snceαq 1,β p j )=1, for all, j so we deduce that then by equaton 5), αβ q 1 p j 1mod q u p v ) 8) t αβ q 1 p j 9) Aa ord q u = αq u 1 and ord p v = β p v 1, so 7) yelds that, αq 1 t and β p j t. Snceαq 1,β p j )=1, so αβ q 1 p j t 30) Hence, αβ q 1 p j = t. Ths shows that there exst a cycle of length αβ q 1 p j for all, j. Snce there exst rk 1) such products, so we must get rk 1) more cycles of lengths > 1. Fnally, the total number of non-somorphc components s r+k+1+rk 1)=r+1)k+1. The followng corollary s a smple consequence of Theorem 5.7. Corollary 5.8. If αq 1 β p j and β p j αq 1 for all =1,,...,r and j =,3,...,k. Then the graph Gp k ) has r + 1)k + 1 non-somorphc components. The below result s the more general case of Theorem 5.6, and can be establshed easly by adoptng the technque explaned n the proof of Theorems 5.6 and 5.7. Theorem 5.9 Let q 0 > 3 be any prme and q 1,q,...,q r be dstnct odd prmes such that φq k 0 0 ) = l q k 1 1 qk...qk r r qk 0 1 0, l > 0. If ord q = α > 1. Then the graph Gp k ) has at most r =0 k + Nl) + 1 non-somorphc components, where Nl) denote the number of dstnct lcm s of all possble products of the ntegers α q j, where, 0 r and 1 j k 1. References [1] B. Wlson, Power Dgraphs Modulo n, Fbonacc Quart, 36, ). [] C. Lucheta, E. Mller and C. Reter, Dgraphs from Powers Modulo p, Fbonacc Quart, 3, ). [3] D.M. Burton, Elementary Number Theory, McGraw-Hll, 007. [] Earle L. Blanton, Jr., Spencer P. Hurd, and Judson S. McCrane, On a Dgraph Defned by Squarng Modulo n, The Fbonacc Quarterly, 30, ). [5] G. Chartrand and L. Lesndk, Graphs and Dgraphs, thrd edton, Chapman Hall, London, 1996). [6] Ivan Nvan, Herbert S. Zuckerman, An ntroducton to the Theory of Numbers, John Wley, Inc., 005). [7] L. Somer and M.Křížek, On a connecton of number theory wth graph theory, Czechoslovak Math. J., 5, ). [8] L. Somer and M.Křížek, On symmetrc dgraphs of the congruence x k ymod n). Dscrete Math., ). [9] L. Szalay, A dscrete teraton n number theory. BDTF Tud. Közl., 8, ). [10] M. Aslam Malk and M. Khald Mahmood, On Smple Graphs Arsng From Exponental Congruences, Journal of Appled Mathematcs, 01, Artcle ID 9895, 10 pages. do: /01/9895. [11] Melvyn B. Nathanson, Methods n Numbers Theory, Sprnger, 005. [1] T. D. Rogers, The graph of the square mappng on the prme felds. Dscrete Math., 18, ). [13] Troy Vasga, Jeffrey Shallt, On the teraton of certan quadratc maps over GFp), Dscrete Mathematcs 77, ). [1] Y. Meemark, N. Wroonsr, The quadratc dgraph on polynomal rngs over fnte felds. Fnte Felds Appl., 16, ). c 015 NSP
10 11 M. K. Mahmood, F. Ahmad: A Classfcaton of Cyclc Nodes and Enumeraton of Components... M. Khald Mahmood s Assstant Professor of Mathematcs at Unversty of the Punjab, Pakstan. He s pursung hs PhD degree n Mathematcs at Unversty of the Punjab, Pakstan. Hs research nterests are n the areas of Pure Mathematcs and Combnatorcs. Specfcally, hs man nterests are Number Theory, Graph Theory, Analyss and Algebra. Farooq Ahmad s Assocate Professor of Computer scence at Unversty of Central Punjab, Lahore, Pakstan. He receved hs PhD degree n Computer Scence at HIT, Chna. He s referee and Edtor of several nternatonal journals n the frame of Dscrete Algorthms, Graph Theory and Combnatorcs. He has publshed research artcles n reputed nternatonal journals of Computer and engneerng scences. Hs man research nterests are: Petrnets, Graph Theory, Dscrete Mathematcs and Computer Algorthms. c 015 NSP
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