k-remainder Cordial Graphs

Transcription

1 Journal of Algorihms and Compuaion journal homepage: hp://jac.u.ac.ir k-remainder Cordial Graphs R. Ponraj 1, K. Annahurai and R. Kala 3 1 Deparmen of Mahemaics, Sri Paramakalyani College, Alwarkurichi 67 41, India. Deparmen of Mahemaics, Thiruvalluvar College, Papanasam 67 45, India. 3 Deparmen of Mahemaics, Manonmaniam Sundaranar Universiy, Tirunelveli 67 01, India. ABSTRACT In his paper we generalize he remainder cordial labeling, called k-remainder cordial labeling and invesigae he 4-remainder cordial labeling behavior of cerain graphs. ARTICLE INFO Aricle hisory: Received 30, June 017 Received in revised form 18, December 017 Acceped 0 December 017 Available online 5, December 017 Keyword: Pah; Cycle; Sar; Bisar; Crown; Comb; Complee graph. AMS subjec Classificaion: 05C78. 1 Inroducion Graphs considered here are finie and simple. Graph labeling is used in several areas of science and echnology like coding heory, asronomy, circui design ec. For more deails refer Gallian []. The origin of graph labeling is graceful labeling which was inroduced by Rosa (1967. Le G 1, G respecively be (p 1, q 1, (p, q graphs. The corona of G 1 wih G, G 1 G is he graph obained by aking one copy of G 1 and p 1 copies of G and joining he i h verex of G 1 wih an edge o every verex in he i h copy of G. The Corresponding auhor: R. Ponraj. ponrajmahs@gmail.com kannahuraivcmahs@gmail.com karhipyi91@yahoo.co.in Journal of Algorihms and Compuaion 49, issue, December 017, PP. 41-5

2 4 R. Ponraj / JAC 49, issue, December 017, PP bisar B m,n is he graph obained by making adjacen he wo cenral verices of K 1,m and K 1,n. A graph S(G derived from a graph G by a sequence of edge subdivisions is called a subdivision of a graph G. Cahi[1], inroduced he concep of cordial labeling of graphs. Recenly Ponraj e al. [4], inroduced he remainder labeling of graphs and invesigaed he remainder cordial labeling behavior of several graphs like pah, cycle, complee graph, sar, bisar ec. Moivaed by hese conceps, in his paper we generalize he remainder cordial labeling, called k-remainder cordial labeling and invesigae he 4- remainder cordial labeling behavior of cerain graphs. Terms are no defined here follows from Harary [3] and Gallian []. k-remainder cordial labeling Definiion 1. Le G be a (p, q graph. Le f be a map from V (G o he se {1,,..., k} where k is an ineger < k V (G. For each edge uv assign he label r where r is he remainder when f(u is divided by f(v (or f(v is divided by f(u according as f(u f(v or f(v f(u. f is called a k-remainder cordial labeling of G if v f (i v f (j 1, i, j {1,..., k} where v f (x denoe he number of verices labelled wih x and e f (0 e f (1 1 where e f (0 and e f (1 respecively denoe he number of edges labeled wih even inegers and number of edges labelled wih odd inegers. A graph wih a k-remainder cordial labeling is called a k-remainder cordial graph. Remark. When k =, number of edges wih label 0 is q. So here does no exiss a -remainder cordial labeling. Theorem 3. Every graph is a subgraph of a conneced k-remainder cordial graphs for k 4. Proof. Le G be a (p, q graph. Consider he k-copies of he complee graph K p. Le G i denoes he i h copy of K p and V (G i = {u i j : 1 j p}. Le s = ( p 1. Nex consider he s copies of he pah on k verices and denoes i h copy as Pk i : vi 1v i... vk i (1 i s. We now consruc he super graph G of he graph G as given below; Le V (G = k V (G i s V (Pk i and E(G = k E(G i s E(Pk i {ui 1v1 i1 : 1 i i=1 i=1 k 1} {u v3} 1 {vv i 3 i1 : 1 i s 1} {v3v i i1 : 1 i s 1} {v3v i 4 i1 : 1 i s 1} {u u 3, u 3u 3 3, u 4u 3 4} {u 3 u 4, u 3 3u 4 3}. Clearly G has kp k ( p k verices and (k 1 ( p edges. Le f be his verex labeling. We now check he verex and edge condiion of he remainder cordialiy. v f (1 = v f ( =... = v f (k = p s and e f (0 = k ( p s 1, ef (1 = k 1 5 (k s s 1 s 1 s 1 = k ( p s 1. Hence f is a k-remainder cordial labeling of G. i=1 i=1

3 43 R. Ponraj / JAC 49, issue, December 017, PP We now invesigae he 4 remainder cordial labeling behaviors of some graphs. Theorem 4. The complee graph K n is 4 remainder cordial iff n 3. Proof. Suppose f is a 4 remainder cordial labeling of K n. The proof is divided ino four cases. Case(i. n > 3 Subcase(i. n 0 (mod 4 Le n = 4. Then v f (1 = v f ( = v f (3 = v f (4 = and we find also e f (0 = ( ( ( ( = 4 4 (. and e f (1 = =. Then e f (0 e f (1 = 4 4 ( = 4 ( = ( 1 = = 4 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Subcase(ii. n 1 (mod 4 Le n = 4 1. Then any one of he following four possibiliies are occurs. Type A : v f (1 = 1, v f ( =, v f (3 =, v f (4 =. Type B : v f (1 =, v f ( = 1, v f (3 =, v f (4 =. Type C : v f (1 =, v f ( =, v f (3 = 1, v f (4 =. Type D : v f (1 =, v f ( =, v f (3 =, v f (4 = 1. Type A : v f (1 = 1, v f ( =, v f (3 =, v f (4 =. Then e f (0 = ( 1 ( 1 ( 1 ( ( 1 ( ( = 4 4 (1(1 1 ( 1 ( 1 ( 1 = 5 4 ( 3 ( = 7 3. and e f (1 = =. Then we find e f (0 e f (1 = 7 3 = 5 3 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type B : v f (1 =, v f ( = 1, v f (3 =, v f (4 =. Then e f (0 = ( 1 ( 1 ( ( ( ( 1 = 4 3 ( 1 (1(1 1 = 6. and e f (1 = ( 1 =. Then we find e f (0 e f (1 = (6 ( = 4 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion.

4 44 R. Ponraj / JAC 49, issue, December 017, PP Type C : v f (1 =, v f ( =, v f (3 = 1, v f (4 =. Then e f (0 = ( 1 ( ( ( ( 1 = 4 3 ( 1 (1(1 1 = 6. and e f (1 = ( 1 ( 1 =. Then we find e f (0 e f (1 = 6 ( = 4 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type D : v f (1 =, v f ( =, v f (3 =, v f (4 = 1. Then e f (0 = ( 1 ( 1 ( ( ( ( 1 = 4 3 ( 1 (1(1 1 = 6. and e f (1 = ( 1 =. Then we find e f (0 e f (1 = (6 ( = 4 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Subcase(iii. n (mod 4 Le n = 4. In his case any one of he following arises. Type A : v f (1 = 1, v f ( = 1, v f (3 =, v f (4 =. Type B : v f (1 = 1, v f ( =, v f (3 = 1, v f (4 =. Type C : v f (1 = 1, v f ( =, v f (3 =, v f (4 = 1. Type D : v f (1 =, v f ( = 1, v f (3 = 1, v f (4 =. Type E : v f (1 =, v f ( = 1, v f (3 =, v f (4 = 1. Type F : v f (1 =, v f ( =, v f (3 = 1, v f (4 = 1. Type A : v f (1 = 1, v f ( = 1, v f (3 =, v f (4 =. Then we find e f (0 = ( 1 ( 1 ( 1 ( 1 ( ( 1 1 ( ( = 1 3( 1 (1(1 1 ( 1 = ( ( = and also e f (1 = ( 1 =. We ge e f (0 e f (1 = (6 5 1 ( = > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type B : v f (1 = 1, v f ( =, v f (3 = 1, v f (4 =. Now we find e f (0 = ( 1 ( 1 ( 1 ( ( 1 ( 1 ( = ( 1 ( 1 ( ( 1 = 1 (1(1 1 ( 1 = ( ( = and also e f (1 = ( 1 ( 1 =. We ge e f (0 e f (1 = (6 4 1 ( = 4 1 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion.

5 45 R. Ponraj / JAC 49, issue, December 017, PP Type C : v f (1 = 1, v f ( =, v f (3 =, v f (4 = 1. Now we find e f (0 = ( 1 ( 1 ( 1 ( 1 ( ( 1 1 ( ( = 3( 1 ( 1 ( ( 1 = (1(1 1 ( 1 = ( = and also e f (1 = ( 1 =. We ge e f (0 e f (1 = (6 5 1 ( = > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type D : v f (1 =, v f ( = 1, v f (3 = 1, v f (4 =. Now we find e f (0 = ( 1 ( 1 ( 1 ( ( ( 1 = 3( 1 ( ( 1 = 3 3 ( 1 (1(1 1 = 4 3 ( ( ( 1 = 6 3. and also e f (1 = ( 1 ( 1 = 1 = 3 1. We ge e f (0 e f (1 = (6 3 ( 3 1 = 4 1 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type E : v f (1 =, v f ( = 1, v f (3 =, v f (4 = 1. Now we find e f (0 = ( 1 ( 1 ( 1 ( ( ( 1 = ( 1 ( 1 ( ( 1 = 1 ( 1 (1(1 1 = ( ( ( 1 = and also e f (1 = ( 1 ( 1 = ( 1 =. We ge e f (0 e f (1 = (6 4 1 ( = 4 1 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type F : v f (1 =, v f ( =, v f (3 = 1, v f (4 = 1. Now we find e f (0 = ( 1 ( 1 ( 1 ( ( ( 1 = 3( 1 ( ( 1 = 4 3 ( 1 (1(1 1 = 4 3 ( ( ( 1 = 6 3. and also e f (1 = ( 1 ( 1 = 1 = 3 1. We ge e f (0 e f (1 = (6 3 ( 3 1 = 4 1 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Subcase(iv. n 3 (mod 4

6 46 R. Ponraj / JAC 49, issue, December 017, PP Le n = 4 3. In his case any one of he following arises. Type A : v f (1 = 1, v f ( = 1, v f (3 = 1, v f (4 =. Type B : v f (1 = 1, v f ( = 1, v f (3 =, v f (4 = 1. Type C : v f (1 = 1, v f ( =, v f (3 = 1, v f (4 = 1. Type D : v f (1 =, v f ( = 1, v f (3 = 1, v f (4 = 1. Type A : v f (1 = 1, v f ( = 1, v f (3 = 1, v f (4 =. Now we find e f (0 = ( 1 ( 1 ( 1 ( 1 ( ( 1 1 = ( 1 ( 1 3 ( ( 1 = 4 6 ( 1 3 (1(1 1 = 4 6 ( 3 ( ( 1 ( = 6 7. and also e f (1 = ( 1 ( 1 = 1 = 3 1. We ge e f (0 e f (1 = (6 7 ( 3 1 = > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type B : v f (1 = 1, v f ( = 1, v f (3 =, v f (4 = 1. Now we find e f (0 = ( 1 ( 1 ( 1 ( ( 1 1 = ( 1 ( 1 3 ( ( 1 = 4 6 ( 1 3 (1(1 1 = 4 6 ( 3 ( ( 1 ( ( 1 = 6 7. and also e f (1 = ( 1 ( 1 = 1 = 3 1. We ge e f (0 e f (1 = (6 7 ( 3 1 = > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type C : v f (1 = 1, v f ( =, v f (3 = 1, v f (4 = 1. Now we find e f (0 = ( 1 ( 1 ( 1 ( 1 ( ( 1 1 = 3( 1 ( 1 3 ( ( 1 = ( 1 3 (1(1 1 = ( 3 ( ( 1 ( = and also e f (1 = ( 1 ( 1 =. We ge e f (0 e f (1 = (6 8 3 ( = > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Type D : v f (1 =, v f ( = 1, v f (3 = 1, v f (4 = 1. Now we find e f (0 = ( 1 ( 1 ( 1 ( 1 ( ( 1 1 = 3( 1 ( 1 3 ( ( 1 = ( 1 3 (1(1 1 = ( 3 ( ( 1 (

7 47 R. Ponraj / JAC 49, issue, December 017, PP = and also e f (1 = ( 1 ( 1 = 4. We ge e f (0 e f (1 = (6 6 1 ( 4 = 4 1 > 1 for any posiive ineger. Therefore e f (0 e f (1 > 1. which is a conradicion. Hence he complee graph K n is no 4 remainder cordial for n > 3. Nex is he Pah. Theorem 5. Any pah P n is 4 remainder cordial. Proof. Le P n be a pah u 1 u... u n. We now divide he proof ino he following four cases. Case(i. n 0 (mod 4 Assign he labels 1,, 3, 4 respecively o he verices u 1, u, u 3, and u 4. Now we consider he nex four verices u 5, u 6, u 7, and u 8. Assign he labels 1,, 3, 4 o he verices u 5, u 6, u 7, u 8. The same paern is coninued for he nex four verices. Proceeding like his assign he labels, unil we reach he las verex u n. Noe ha in his process he las four verices namely u n 3, u n, u n 1, and u n received he labels 1,, 3, and 4. Case(ii. n 1 (mod 4 As in he case(i, assign he labels o he verices u 1, u,... u n 1. Nex assign he label 1 o he verex u n. Case(iii. n (mod 4 Assign he labels o he verices u i, (1 i n 1,as in he case(ii. Finally assign he label o he verex u n. Case(iv. n 3 (mod 4 In his case assign he labels o he verices u i, (1 i n 1,as in he case(iii. Finally assign he label 3 o he verex u n. The Table 1, esablish ha his labeling f is a 4 remainder cordial labeling. Table 1: Edge condiion of 4-remainder cordial labeling of a pah Naure of n r (mod 4 e f (0 e f (1 n n n 0, (mod 4 n 1 n 1 n 1 (mod 4 n n n (mod 4 n 3 (mod 4 n 1 n 1 Nex invesigaion is he cycle graph. Theorem 6. All cycles C n is 4 remainder cordial. Proof. Le C n = u 1 u... u n be a cycle. Case(i. n 0 (mod 4

8 48 R. Ponraj / JAC 49, issue, December 017, PP Fix he labels 1,, 3, 4 respecively o he four consecuive verices u 1, u, u 3, and u 4. Nex assign he labels 4, 3,, 1 respecively o he verices u 5, u 6, u 7, and u 8. Nex assign he labels 4, 3,, 1 o he verices u 9, u 10, u 11, u 1. In his manner assign he labels, unil we reach he las verex u n. I is easy o verify ha he las four verices u n 3, u n, u n 1, and u n received he labels 4, 3,, 1. Case(ii. n 1 (mod 4 As in he case(i, assign he labels o he verices u 1, u,... u n 1. Nex assign he label 4 o he verex u n. Case(iii. n (mod 4 Assign he labels o he verices u 1, u,... u n 1,as in he case(ii. Finally assign he label 3 o he verex u n. Case(iv. n 3 (mod 4 In his case assign he labels o he verices u 1, u,... u n 1,as in he case(iii. Finally assign he label o he verex u n. The Table, esablish ha his labeling f is a 4 remainder cordial labeling. Table : Edge condiion for 4 remainder cordial labeling of a cycle Naure of n r (mod 4 e f (0 e f (1 n n n 0, (mod 4 n1 n 1 n 1 (mod 4 n 3 (mod 4 n 1 n1 Nex we invesigae any comb is 4 remainder cordial. Theorem 7. Any comb P n K 1 is 4 remainder cordial. Proof. Le P n = u 1 u... u n be a Pah. Le v i be he pendan verices aached o u i, 1 i n. Assign he labels o he verices u 1, u,... u n as in heorem 6. Case(i. n 0 (mod 4 We now consider he pendan verices, fix he labels 4, 3,, 1 respecively o he verices v 1, v, v 3, and v 4. Nex assign he labels 1,, 3, 4 o he four verices v 5, v 6, v 7, and v 8. In similar fashion assign he labels 1,, 3, 4 respecively o he nex four consecuive verices v 9, v 10, v 11, v 1. Proceed as above and labels he nex four verices and so on. In his he las four verices v n 3, v n, v n 1, v n received he labels 1,, 3, 4. Case(ii. n 1 (mod 4 As in he case(i, assign he labels o he pendan verices v 1, v,... v n 1. Nex assign he label 1 o he verex v n. Case(iii. n (mod 4 Assign he labels o he verices v 1, v,... v n 1,as in he case(ii. Finally assign he label o he verex v n. Case(iv. n 3 (mod 4

9 49 R. Ponraj / JAC 49, issue, December 017, PP In his case assign he labels o he verices v 1, v,... v n 1,as in he case(iii. Finally assign he label 3 o he verex v n. The Table 3, esablish ha his labeling f is a 4 remainder cordial labeling. Table 3: Edge condiion for 4 remainder cordial labeling of a comb Naure of n r (mod 4 e f (0 e f (1 n n n 0, (mod 4 n1 n 3 n 1, 3 (mod 4 4-remainder cordial labeling of P 5 K 1 is given in Figure 1. Nex is he Crown C n K 1. Figure 1: Theorem 8. All crowns are 4 remainder cordial. Proof. The crown C n K 1 is obained from he comb P n K 1, and by adding he edge u n u 1. Case(i. n 0, (mod 4 The verex labeling as in heorem 7, is also a 4 remainder cordial labeling of crown. Case(ii. n 1, 3 (mod 4 Assign he labels, 3 o he verices u 1, u respecively and assign he labels, 3 o he nex wo verices u 3, u 4. Coninuing in his way unil we reach he verex u n 1. Tha is assign he labels, 3,, 3,..., 3 o he verices u 1, u,..., u n 1. Now assign he label o he las verex u n. Nex we consider he pendan verices, assign he labels o he verices v 1, v,... v n 1 in he paern 1, 4, 1, 4,... 1, 4. Finally assign he label 4 o he verex v n. The following able 4, shows ha his labeling f is a 4 remainder cordial labeling.

10 50 R. Ponraj / JAC 49, issue, December 017, PP Table 4: Edge condiion for 4 remainder cordial labeling of crown Naure of n e f (0 e f (1 n n n is even n1 n 1 n is odd Theorem 9. All sars are 4 remainder cordial. Proof. Le K 1,n be he sar wih V (K 1,n = {u, u i : 1 i n} and E(K 1,n = {uu i : 1 i n}. we now give a 4 remainder cordial labeling o he sar K 1,n. Assign he label 3 o he cener verex u. Case(i. n 0 (mod 4 le n = 4 Assign he label 1 o he pendan verices u 1, u,..., u. Nex assign he label o he pendan verices u 1, u,..., u. We now assign he label 3 o he nex pendan verices u 1, u,..., u 3. Finally assign he label 4 o he remaining pendan verices. Case(ii. n 1 (mod 4 As in case(i, assign he label o he verices u, u i (1 i n 1. Nex assign he label 1 o he las verex u n. Case(iii. n (mod 4 Assign he label o he verices u, u i (1 i n 1 as in case(ii. Nex assign he label o he verex u n. Case(iv. n 3 (mod 4 As in he case(iii, assign he label o he verices u, u i (1 i n 1. Nex assign he label 4 o he verex u n. Obviously his verex labeling f is 4 remainder cordial labeling. Theorem 10. The bisar B n,n are 4 remainder cordial for all n. Proof. Le B n,n be he bisar wih V (B n,n = {u, v, u i, v i : 1 i n} and E(B n,n = {uv, uu i, vv i : 1 i n}. Clearly B n,n has n verices and n 1 edges. Assign he label 1, 3 respecively o he cenral verices u and v. Consider he pendan verices u i. Case(i. n o (mod 4 Le n = 4. Assign he label 1 o he pendan verices u 1, u,..., u and assign he label 3 o he verices u 1, u,..., u 4. Nex we move o he oher side pendan verices v i. Assign he label o he verices v 1, v,..., v and assign he label 4 o he remaining pendan verices v 1, v,..., v 4. Case(ii. n 1 (mod 4 Le n = 4 1. Assign he labels o he verices u, v, u i, v i (1 i n, as in he case(i. Nex assign he label 4, respecively o he verices u i and v i. Case(iii. n (mod 4 As in he case(ii, assign he label o he verices u, v, u i, v i (1 i n 1. Nex assign labels 1, 4 o he verices u n and v n respecively.

11 51 R. Ponraj / JAC 49, issue, December 017, PP Case(iv. n 3 (mod 4 Assign he labels o he verices u, v, u i, v i (1 i n 1 in case(iii. Finally assign he labels 3, o he remaining verices. This verex labeling is a 4 remainder cordial labeling follows from able 5. Table 5: Edge condiion of 4 remainder cordial labeling of bisar Naure of n e f (0 e f (1 n n n 0, (mod 4 n 1, 3 (mod 4 For illusraion, a 4 remainder cordial labeling of B 5,5 is shown in Figure. n n Figure : Theorem 11. The subdivision of he sar S(K 1,n are 4 remainder cordial. Proof. Le V (S(K 1,n = {u, u i, v i : 1 i n} and E(S(K 1,n = {uu i, u i v i : 1 i n}. The proof is divided in o four cases given below. Case(i. n 0 (mod 4 le n = 4. Assign he label 3 o he verex u. Nex we consider he verices of degree. Assign he label 3 o he verices u 1, u,..., u and assign he label o he verices u 1, u,..., u 4. Nex we move o he pendan verices. Assign he label 4 o he verices v 1, v,..., v and assign he label 1 o he verices v 1, v,..., v 4. Case(ii. n 1 (mod 4 Assign he labels o he verices u, u i, v i (1 i n 1 as in case(i. Nex assign he labels, 1 respecively o he verex u n and v n. Case(iii. n (mod 4 As in case(ii, assign o labels o he verices u, u i, v i (1 i n 1. Finally assign he labels 4, 3 o he verices u n and v n respecively. Case(iv. n 3 (mod 4 Assign he labels o he verices u, u i, v i (1 i n 1 as in case(iii. Nex assign he labels, 1 respecively o he remaining verices u n and v n. The able 6, esablish ha his verex labeling f is a 4 remainder cordial labeling.

12 5 R. Ponraj / JAC 49, issue, December 017, PP Table 6: Edge condiion of 4 remainder cordial labeling of subdivision of sar Naure of n e f (0 e f (1 n n n 0, (mod 4 n 1, 3 (mod 4 References [1] Cahi, I., Cordial Graphs : A weaker version of Graceful and Harmonious graphs, Ars combin., 3 (1987, [] Gallian, J.A., A Dynamic survey of graph labeling, The Elecronic Journal of Combinaorics., 19, (017. [3] Harary, F., Graph heory, Addision wesley, New Delhi, [4] Ponraj, R. and Annahurai, K., and Kala, R., Remainder cordial labeling of graphs, Journal of algorihm and Compuaion, 49 (017, n 1 n 1