Iteratioal Joural of Mathematics a Soft Computig Vol.5, No.1 (015), 65-71. ISSN Prit : 49-338 Prime labelig of geeralize Peterse graph ISSN Olie: 319-515 U. M. Prajapati 1, S. J. Gajjar 1 Departmet of Mathematics, St. Xavier College Ahmeaba - 380009, Iia. uaya64@yahoo.com Geeral Departmet, Govermet Polytechic Himmatagar - 383001, Iia. sachi@gmail.com Abstract I the preset work we have iscusse primality of geeralize Peterse graph P (, k). We have prove that geeralize Peterse graph P (, k) is prime the must be eve a k must be o. We have also fou some classes of geeralize Peterse graph P (, k) which amits prime labelig. Keywors: Graph labelig, prime labelig, geeralize Peterse graph. AMS Subject Classificatio(010): 05C78. 1 Itrouctio We cosier oly simple, fiite, uirecte a o-trivial graph G = (V, E) with the vertex set V a the ege set E. The umber of elemets of V, eote as V is calle the orer of the graph G while the umber of elemets of E, eote as E is calle the size of the graph G. P (, k) eotes the geeralize Peterse graph. For various graph theoretic otatios a termiology we follow Gross a Yelle [3 whereas for umber theory we follow D. M. Burto [1. We will give brief summary of efiitios a other iformatio which are useful for the preset ivestigatios. Defiitio 1.1. For a graph G = (V, E) a fuctio f havig omai V, E or V E is sai to be a graph laebilg of G. If the omai is V, E or V E the the correspoig labelig is sai to be a vertex labelig, a ege labelig or a total labelig. For latest survey o graph labelig we refer to J. A. Gallia [. Vast amout of literature is available o ifferet types of graph labelig a more tha 1000 research papers have bee publishe so far i last four ecaes. The preset work is aime to iscuss oe such labelig kow as prime labelig. Defiitio 1.. A prime labelig of a graph G of orer is a ijective fuctio f : V {1,, 3,..., } such that for every pair of ajacet vertices u a v, gc(f(u), f(v)) = 1. The graph which amits prime labelig is calle a prime graph. 65
66 U. M. Prajapati a S. J. Gajjar al [6. The otio of a prime labelig was origiate by Etriger a was iscusse i a paper by Tout et Defiitio 1.3. The Geeralize Peterse graphsp (, k) with 3 a 1 k [ are efie to be a graph with V (P (, k)) = {v i, u i : 1 i } a E(P (, k)) = {v i v i+1, v i u i, u i u i+k : 1 i, subscripts moulo }. P (, k) has vertices a 3 eges. These graphs were itrouce by Coxeter(1950) a ame by Watkis(1969). Kh. M. Momiul Haque, Li Xiaohui, Yag Yuasheg a Zhao Pigzhog [4, 5 have prove that geeralize Peterse graph P (, 1) is prime for eve 500 a P (, 3) is prime for eve 100. Prime Labelig of Geeralize Peterso graph Theorem.1. P (, k) is ot a prime graph for o. Proof: Here the orer of P (, k) is. So we have to use 1 to itegers to label these vertices. Thus we have eve itegers. Now we ca assig eve umbers to at most 1 (as is o) vertices out of the outer rim vertices. Now out of ier vertices, every ier vertex is ajacet to at least oe other ier vertex. So we ca assig eve umbers to at most 1 (as is o) vertices out of the ( ier vertices. Thus we ca assig at most 1 = 1 + 1 ) eve umbers to the vertices. But we have eve umbers, so it is ot possible. Thus P (, k) is ot a prime graph for o. Theorem.. If gc(, k) = the i P (, k) there are vertex isjoit ier cycles of orer, where. Proof: By the efiitio of P (, k), let v 1, v,..., v be the outer vertices, u 1, u,..., u be the ier vertices a E(P (, k)) = {v i v i+1, v i u i, u i u i+k : 1 i, subscripts moulo }. Thus every ier vertex u i is ajacet with ier vertex u i+k (subscript moulo ). Now costruct a sequece of vertices as follows: Let us start with u 1. By efiitio of P (, k), u 1 is ajacet with u 1+k, u 1+k is ajacet with u 1+k a so o. Thus we have a sequece u 1, u (1+k), u (1+k),..., u (1+ k) of ier vertices. This sequece repeats after steps as 1+ k 1(mo ). Also if 1+ik 1+jk ( mo ) for some i a j less tha, the (i j)k 0 ( mo ). Therefore, (i j)k 1 0 (mo 1 ). [As gc(, k) =, let = 1, k = k 1 a gc( 1, k 1 ) = 1. This implies that i j 0 ( mo 1 ) a hece i j ( mo 1 ), which is ot possible as i a j are istict a less tha = 1. Thus we have a cycle with orer whose vertex set is C 1 = {u 1, u (1+k),..., u (1+( 1)k) }. Now choose the smallest t such that u t / C 1 a costruct aother set C = {u t, u (t+k), u (t+k),..., u (t+( 1)k) } of ier vertices. Clearly o two vertices of C are same. Also C 1 a C are isjoit, as if possible let u (t+ik) = u (1+jk) for some i a j less tha. The t + ik 1 + jk ( mo ) which
Prime labelig of geeralize Peterse graph 67 implies that t 1 + (j i)k ( mo ). Therefore u t is a elemet of C 1 as j i is less tha, which cotraicts the choice of u t. Thus C 1 a C are isjoit sets. Now choose aother ier vertex which is ot i C 1 a C both a costruct a ew set as above. By repeatig this process times, we ca fi such sets of orer such that all the sets are pair wise isjoit. Thus we have isjoit ier cycles each with orer. Theorem.3. P (, k) is ot a prime graph if both a k are eve umbers. Proof: Let gc(, k) = the i P (, k) there is a outer cycle C a isjoit ier cycles C Case 1: Let [ be o. The we ca label at most vertices of each ier cycles C by eve umbers a we ca label at most vertices of outer cycle C by eve umbers. Thus the umber of vertices of P (, k) that ca be labele by eve umbers is [ +. Now [ + < + =. But we have eve umbers to label the vertices. So it is ot possible.. Case : Let be eve. Here we must have to label the vertices of outer cycle C alteratively by o a eve umbers. Also we must have to label half of the ier vertices by eve umbers a half by o umbers. Now choose a outer vertex say u which is labele by eve umber a fi the correspoig ier vertex say v which is ajacet to this outer vertex u. Let C be the ier cycle cotaiig the vertex v. The all the outer vertices which are ajacet with the vertices of this ier cycle C are labele by eve umbers as k is eve. So o vertices of this ier cycle C ca be label by eve umber. Thus prime labelig is ot possible i this case. Thus i each case P (, k) is ot a prime graph if both a k are eve umbers. Theorem.4. If geeralize Peterse graph P (, k) is prime the must be eve a k must be o. Theorem.4 follows from Theorem.1 a Theorem.3. Theorem.5. If for t, t + 1 is prime the geeralize Peterse graph P (, k) is a prime graph, where = t + a k = t 1 + 1. Proof: Let v 1, v,..., v be the outer vertices a u 1, u,..., u be the ier vertices of P (, k). Now efie a fuctio f : V (P (, k)) {1,,..., } as follows: i for u = v i, 1 i t + ; t+1 + 4 for u = u 1 ; t + 4i for u = u f(u) = (i 1), i =, 3,..., t + 1; t + 4i for u = u ( t 1 +i+1), i = 1,, 3,..., t ; t + 4i + 1 for u = u i, i = 1,, 3,..., t ; t + 4i 1 for u = u ( t 1 +i), i = 1,, 3,..., t + 1. Clearly f is a ijective fuctio. Let e be a arbitrary ege of P (, k). To prove f is a prime labelig of P (, k) we have the followig cases:
68 U. M. Prajapati a S. J. Gajjar Case 1: If e = vi vi+1, the gc(f (vi ), f (vi+1 )) = gc(i, i + 1) = 1, for i = 1,,..., t + 1. Case : If e = v1 vt +, the gc(f (v1 ), f (vt + )) = gc(1, t + ) = 1. Case 3: If e = v1 u1, the gc(f (v1 ), f (u1 )) = gc(1, t+1 + 4) = 1. Case 4: If e = v(i 1) u(i 1), the gc(f (vi 1 ), f (ui 1 )) = gc(i 1, t +4i ) = gc(i 1, t ) = 1 for i =, 3,..., t + 1. Case 5: If e = v(t 1 +i+1) u(t 1 +i+1), the gc(f (v(t 1 +i+1) ), f (u(t 1 +i+1) )) = gc(t 1 + i + 1, t + 4i) = gc(t 1 + i + 1, 1) = 1, for i =, 3,..., t. Case 6: If e = vi ui, the gc(f (vi ), f (ui )) = gc(i, t + 4i + 1) = gc(i, t + 1) = 1, as t + 1 is prime, for i = 1,,..., t. Case 7: If e = v(t 1 +i) u(t 1 +i), the gc(f (v(t 1 +i) ), f (u(t 1 +i) )) = gc(t 1 + i, t + 4i 1) = gc(t 1 + i, 1) = 1, for i = 1,,..., t + 1. Case 8: If e = u1 u(t 1 +), the gc(f (u1 ), f (u(t 1 +) )) = gc(t+1 +4, t +3) = gc(, t +3) = 1. Case 9: If e = u(i 1) u(t 1 +i), the gc(f (u(i 1) ), f (u(t 1 +i) )) = gc(t +4i, t +4i 1) = 1, for i =,..., t + 1. Case 10: If e = ui u(t 1 +i+1), the gc(f (ui ), f (u(t 1 +i+1) )) = gc(t + 4i + 1, t + 4i) = 1, for i = 1,,..., t. Thus P (, k) amits a prime labelig. So P (, k) is a prime graph for = t + a k = t 1 + 1, where t + 1 is prime. Illustratio.6. Prime labelig of the geeralize Peterse graph P (18, 9) is show i Figure 1. Figure 1: Prime labelig of geeralize Peterse graph P (18, 9). Theorem.7. If 8 + 5 a 8 + 9 are prime the geeralize Peterse graph P (4 +, + 1) is a prime graph. Proof: Let v1, v,..., v4+ be the outer vertices a u1, u,..., u4+ be the ier vertices of P (4+, + 1). Now efie a fuctio f : V (P (4 +, + 1)) {1,,..., 8 + 4} as follows:
Prime labelig of geeralize Peterse graph 69 4i + 1 for u = v (i+1), 0 i ; 8 4i + 4 for u = v (i+), 0 i 1; 4i + 4 for u = v (+i+), 0 i ; 8 4i + 1 for u = v f(u) = (+i+3), 0 i 1; 4i + for u = u (i+1), 0 i ; 8 4i + 3 for u = u (i+), 0 i 1; 4i + 3 for u = u (+i+), 0 i ; 8 4i + for u = u (+i+3), 0 i 1. Clearly f is a ijective fuctio. Let e be a arbitrary ege of P (4 +, + 1). To prove f is a prime labelig of P (4 +, + 1) we have the followig cases: Case 1: If e = v (i+1) v (i+), the gc(f(v (i+1) ), f(v (i+) )) = gc(4i + 1, 8 4i + 4) = gc(4i + 1, 8 + 5) = 1, for 0 i 1 as 8 + 5 is prime. Case : If e = v (i+) v (i+3), the gc(f(v (i+) ), f(v (i+3) )) = gc(f(v (i+) ), f(v ((i+1)+1) )) = gc(8 4i + 4, 4i + 5) = gc(8 + 9, 4i + 5) = 1, for 0 i 1 as 8 + 9 is prime. Case 3: If e = v (+1) v (+), the gc(f(v (+1) ), f(v (+) )) = gc(4 + 1, 4) = 1. Case 4: If e = v (+i+) v (+i+3), the gc(f(v (+i+) ), f(v (+i+3) )) = gc(4i + 4, 8 4i + 1) = gc(4i + 4, 8 + 5) = 1, for 0 i 1, as 8 + 5 is prime. Case 5: If e = v (+i+3) v (+i+4), the for 0 i 1, gc(f(v (+i+3) ), f(v (+i+4) )) = gc(f(v (+i+3) ), f(v (+(i+1)+) )) = gc(8 4i + 1, 4i + 8) = gc(8 + 9, 4i + 8) = 1. ( as 8 + 9 is prime). Case 6: If e = v (4+) v 1, the gc(f(v (4+) ), f(v 1 )) = gc(4 + 4, 1) = 1. Case 7: If e = v (i+1) u (i+1), the gc(f(v (i+1) ), f(u (i+1) )) = gc(4i + 1, 4i + ) = 1, for 0 i. Case 8: If e = v (i+) u (i+), the gc(f(v (i+) ), f(u (i+) )) = gc(8 4i + 4, 8 4i + 3) = 1, for 0 i 1. Case 9: If e = v (+i+) u (+i+), the gc(f(v (+i+) ), f(u (+i+) )) = gc(4i + 4, 4i + 3) = 1, for 0 i. Case 10: If e = v (+i+3) u (+i+3), the for 0 i 1, gc(f(v (+i+3) ), f(u (+i+3) )) = gc(8 4i + 1, 8 4i + ) = 1. Case 11: If e = u (i+1) u (+i+), the gc(f(u (i+1) ), f(u (+i+) )) = gc(4i +, 4i + 3) = 1, for 0 i. Case 1: If e = u (i+) u (+i+3), the gc(f(u (i+) ), f(u (+i+3) )) = gc(8 4i + 3, 8 4i + ) = 1, for 0 i 1. Thus, P (4 +, + 1) amits a prime labelig. So P (4 +, + 1) is a prime graph, where
70 U. M. Prajapati a S. J. Gajjar 8 + 5 a 8 + 9 are prime. Illustratio.8. Prime labelig of the geeralize Peterse graph P (6, 3) is show i Figure. Figure : Prime labelig of geeralize Peterse graph P (6, 3). Theorem.9. If + 1, 4 + 3 a 6 + 5 are prime the geeralize Peterse graph P (4 + 4, + 1) is a prime graph. Proof: Let v1, v,..., v4+4 be the outer vertices a u1, u,..., u4+4 be the ier vertices of P (4+ 4, + 1). Now efie a fuctio f : V (P (4 + 4, + 1)) {1,,..., 8 + 8} as follows: i for u = vi, 1 6 i 6 4 + 4; 4 + 3 + i for u = u, 1 6 i 6 + ; i f (u) = 4 + 4 + i for u = u (+3+i), 1 6 i 6 ; 4 + 4 + i for u = u (i 1), + 1 6 i 6 +. Clearly f is a ijective fuctio. Let e be a arbitrary ege of P (4 +, + 1). To prove f is a prime labelig of P (4 +, + 1) we have the followig cases: Case 1: If e = vi vi+1, the for 1 6 i 6 4 + 3, gc(f (vi ), f (vi+1 )) = gc(i, i + 1) = 1. Case : If e = v(4+4) v1, the gc(f (v(4+4) ), f (v1 )) = gc(4 + 4, 1) = 1. Case 3: If e = ui u(+3+i), the gc(f (ui ), f (u(+3+i) )) = gc(4 + 3 + i, 4 + 4 + i) = 1, for 1 6 i 6. Case 4: If e = ui u(i 1), the gc(f (ui ), f (u(i 1) ) = gc(4 + 3 + i, 4 + 4 + i) = 1 for + 1 6 i 6 +. Case 5: If e = u(+3+i) u(i+), the gc(f (u(+3+i) ), f (u(i+) ) = gc(4+4+i, 4+5+i) = 1, for 1 6 i 6. Case 6: If e = u(i 1) u(i+), the gc(f (u(i 1) ), f (u(i+) ) = gc(4+4+i, 4+5+i) = 1, for + 1 6 i 6 + 1. Case 7: If e = u(+3) u, the gc(f (u(+3) ), f (u ) = gc(8 + 8, 4 + 5) = gc(, 4 + 5) = 1. Case 8: If e = vi ui the gc(f (vi ), f (ui ) = gc(i, 4 + 3 + i) = gc(i, 4 + 3) = 1, for 1 6 i 6 + as 4 + 3 is prime.
Prime labelig of geeralize Peterse graph 71 Case 9: If e = v(+3+i) u(+3+i), the gc(f (v(+3+i) ), f (u(+3+i) ) = gc( + 3 + i, 4 + 4 + i) = gc( + 3 + i, + 1) = gc( + i, + 1) = 1, for 1 6 i 6, as + 1 is prime. Case 10: If e = v(i 1) u(i 1), the gc(f (v(i 1) ), f (u(i 1) ) = gc(i 1, 4 + 4 + i) = gc(i 1, 6 + 5) = 1, for + 1 6 i 6 +, as 6 + 5 is prime. Thus, P (4 + 4, + 1) amits a prime labelig. So P (4 + 4, + 1) is a prime graph, where + 1, 4 + 3 a 6 + 5 are prime. Illustratio.10. Prime labelig of geeralize Peterse graph P (8, 3) is show i Figure 3. Figure 3: Prime labelig of geeralize Peterse graph P (8, 3). Now we have the followig cojecture. Cojecture.11. If is eve, k is o a 1 6 k 6 is prime. hi the the geeralize Peterse graph P (, k) Ackowlegemet: The authors are highly thakful to the aoymous referee for valuable commets a costructive suggestios. Refereces [1 D. M. Burto, Elemetary Number Theory, Seco Eitio, 1990. [ J. A. Gallia, A Dyamic Survey of Graph Labelig, The Electroic Joural of Combiatorics, Vol. 19, 01. [3 J. Gross a J. Yelle, Habook of Graph Theory, CRC Press, 004. [4 Kh. M. Momiul Haque, Li Xiaohui, Yag Yuasheg a Zhao Pigzhog, O the Prime Labelig of Geeralize Peterse Graphs P (, 1), Utilitas Mathematica, I press. [5 Kh. M. Momiul Haque, Li Xiaohui, Yag Yuasheg a Zhao Pigzhog, O the Prime Labelig of Geeralize Peterse Graphs P (, 3), It. J. Cotemp. Math. Scieces, Vol. 6, No. 36(011), 1783-1800. [6 A. Tout, A. N. Dabboucy a K. Howalla, Prime labelig of Graphs, Nat. Aca. Sci. Letters, Vol. 11(198), 365-368.