k-equitable mean labeling

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1 Joural of Algorithms ad Comutatio joural homeage: htt://jac.ut.ac.ir k-euitable mea labelig P.Jeyathi 1 1 Deartmet of Mathematics, Govidammal Aditaar College for Wome, Tiruchedur ,Idia ABSTRACT Let G=(V, E) be a grah with vertices ad edges. Let f : V {0, 1, 2,, k} (1 k ) be a vertex labelig of G that iduces a edge labelig f : E {0, 1, 2,, k} be give by f (uv)= f (u)+ f (v) 2. A labelig f is called (k+1)- euitable mea labelig ( (k+1)-eml) if v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1, 2,, k where v f (x) ad e f (x), x=0, 1, 2,, k are the umber of vertices ad edges of G resectively with label x. A ew cocet amely k-euitable mea labelig of a grah is itroduced i this aer. ARTICLE INFO Article history: Received 0 Jue 2012 Received i revised form 20 February 201 Acceted 2 March 201 Available olie 01 July 201 Keyword: Mea labelig, euitable labelig, euitable mea labelig. AMS subject Classificatio: 05C78. 1 Itroductio Cahit [1] roosed the idea of distributig the vertex ad edge labels amog{0, 1, 2,, k 1} as evely as ossible to obtai a geeralizatio of graceful labelig as follows: For ay grah G(V, E) ad for ay ositive iteger k, assig vertex labels from{0, 1, 2,, k 1} so that whe the edge labels iduced by the absolute value of the differece of the vertex labels, the umber of vertices labeled with Corresodig author: P. Jeyathi. jeyajeyathi@rediffmail.com Joural of Algorithms ad Comutatio 44 (201) PP. 21-0

2 22 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP i ad the umber of vertices labeled with j differ by at most oe ad the umber of edges labeled with i ad the umber of edges labeled with j differ by at most oe. A grah with such a assigmet of labels is called k-euitable [2]. I [5] the otio of roduct cordial labelig was itroduced. A roduct cordial labelig of a grah G with the vertex set V is a fuctio f from V to{0, 1}such that if each edge uv is assiged the label f (u) f (v), the umber of vertices labeled with 0 ad the umber of vertices labeled with 1 differ by at most 1, ad the umber of edges labeled with 0 ad the umber of edges labeled with 1 differ by at most 1. A grah with a roduct cordial labelig is called roduct cordial grah. Somasudaram ad Poraj [4] itroduced the otio of mea labelig of grahs. A grah G with vertices ad edges is called a mea grah if there is a ijective fuctio f from the vertices of G to ( f (u)+ f (v) {0, 1, 2,, } such that whe each edge uv is labeled with, the the resultig edge labels 2 are distict. A ew cocet amely k-euitable mea labelig of a grah is itroduced i this aer. The grahs cosidered i this aer are fiite simle grahs. Let G=(V(G), E(G)) be a grah of order ad size. The vertex set ad the edge set of a grah G are deoted by V(G) ad E(G) resectively. The disjoit uio of two grahs G 1 ad G 2 is the grah G 1 G 2 with V(G 1 G 2 )=V(G 1 ) V(G 2 ) ad E(G 1 G 2 )=E(G 1 ) E(G 2 ). The disjoit uio of m coies of the grah G is deoted by mg.the grah G@P is obtaied by idetifyig a ed vertex of a ath P with ay vertex of G. Terms ad otatios ot defied here are used i the sese of Harary[]. For ay iteger, deotes the greatest iteger less tha or eual to ad deotes the least iteger greater tha or eual to. 2 k-euitable mea labelig Defiitio 2.1. Let G=(V, E) be a grah with vertices ad edges. Let f : V {0, 1, 2,, k}(1 k ) be a vertex labelig of G that iduces a edge labelig f : E {0, 1, 2,, k} be give by f (uv)= f (u)+ f (v) 2. A labelig f is called (k+1)-euitable mea labelig ((k+1)-eml) if v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1, 2,, k where v f (x) ad e f (x), x=0, 1, 2,, k are the umber of vertices ad edges of G resectively with label x. A grah G that admits (k )-euitable mea labelig is called a (k )-euitable mea grah ((k + 1) emg). Theorem 2.2. Let G be a (, )-coected grah. The G is a ( )-emg iff G is a mea grah. Proof: Suose G is a (+1)-emg. The there is a vertex labelig f : V {0, 1, 2,, } that iduces a edge labelig f : E {0, 1, 2,, } give by f (uv)= f (u)+ f (v) 2 ad satisfies v f (i) v f ( j) 1 ad e f (i) e f ( j) 1for i, j=0, 1, 2,,. Sice G has edges ad e f (i) e f ( j) 1, the edge labels are distict. Otherwise if e f (i) 2 for some i, the e f ( j)=0for at least oe label j i. Sice G is a coected grah, 1 ad hece +1. If =+1 the all the vertex labels must be distict ad v f (i)=1for all i=0, 1, 2,,.

3 2 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP If <+1the there is at least oe label say i with v f (i)=0. If ay label j i occurs more tha oce, we get a cotradictio to v f (i) v f ( j) 1. Hece, the vertex labels are distict. Thus f is a mea labelig. The coverse art follows from the defiitio of the mea labelig of a grah. Theorem 2.. G is a 2-emg iff G is a roduct cordial grah. Proof: Let f be a 2-eml of G. The f : V(G) {0, 1} is a vertex labelig of G that iduces a edge labelig f give by f (uv)= f (u)+ f (v) 2 ad it satisfies v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1. Defie g : V(G) {0, 1} by g(v)=1 f (v) ad g (uv)=1 f (uv). The, g is a vertex labelig of G with v g (0)=v f (1) ad v g (1)=v f (0). Hece, v g (0) v g (1) 1. Now g(u)g(v)=(1 f (u))(1 f (v))= 1 ( f (u)+ f (v))+ f (u) f (v). If f (u)=0= f (v) the f (uv)=0 ad f (u)+ f (v) f (u) f (v)=0. If both f (u)=1= f (v) the f (uv)=1 ad f (u)+ f (v) f (u) f (v)=2 1=1. If oe of f (u) ad f (v) is zero, say f (u)=0 ad f (v)=1 the f (uv)=1 ad f (u)+ f (v) f (u) f (v)=1+0=1. Hece, 1 (f(u)+ f (v))+ f (u) f (v)= 1 f (uv). Thus, we have g (uv)=1 f (uv)=g(u)g(v). Therefore, e f (0)=e g (1) ad e f (1)=e g (0) which imlies that e g (0) e g (1) 1. Hece, g is a roduct cordial labelig of G. The roof of the coverse is similar to the revious argumet. -euitable mea labelig of some stadard grahs. I this sectio, -euitable mea labelig of some families of grahs are exhibited. Lemma.1. If a (, )-grah G admits a -eml f the v f (i) Theorem.2. ad e f (i), i=0, 1, 2. (i) For ay (, )-grah G, the grah mg is a -euitable mea grah. (ii) For ay (, )--euitable mea grah G, the grah (m+1)g is a - euitable mea grah. Proof: (i) Assig 0 to all the vertices of the first m coies of G, assig 1 to all the vertices of ext m coies of G ad assig 2 to all vertices of the remaiig m coies of G. Thus, we have v(0) = v(1) = v(2)=m ad e(0)=e(1)=e(2)=m. (ii) Assig 0 to all the vertices of first m coies of G, assig 1 to all the vertices of ext m coies of G ad assig 2 to all the vertices of last m coies of G. The remaiig oe coy of G has the give -eml. Hece, (m )G is a -emg. Theorem.. Let H be a (, ) grah ad cosider m coies of H as H i, 1 i m. Let G be a grah obtaied by idetifyig a vertex of H i with a vertex of H i+1 for 1 i m 1. The G is a -emg.

4 24 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP Proof: For the give grah G, we have V(G) =m (m 1)=m( 1)+1 ad E(G) =m. Let u i be a vertex of H i ad u i+1 be a vertex of H i+1 such that u i is idetified with u i+1 for 1 i m 1. Now, assig 0 to all the vertices of Hi for 1 i m, assig 1 to all the vertices of H i for m+1 i 2m excet for u m+1 ad assig 2 to all the vertices of H i for 2m+1 i m excet for the vertex u 2m+1. The we have v(0)=m( 1)+1, v(1)=m( 1), v(2)=m( 1) ad e(0)=m, e(1)=m, e(2)=m. Hece, G is a -emg. Theorem.4. If G(, ) is a -emg the (G) 2r+ t where =r+ t, t {0, 1, 2}. Proof: Let f be a -eml of G ad let S 0, S 1 ad S 2 be the subgrahs of G iduced by the edges of G that have labels 0,1 ad 2 resectively. The E(S 0 ) + E(S 1 ) + E(S 2 ) =. Let v V(G). If f (v)=0 the v V(S 0 ) V(S 1 ), if f (v)=1 the v V(S 1 ) V(S 2 ) ad if f (v)=2 the v V(S 1 ) V(S 2 ). Hece, deg(v) E(S 0 ) + E(S 1 ) or E(S 1 ) + E(S 2 ). If = r the E(S 0 ) = E(S 1 ) = E(S 2 ) = r ad hece deg(u) 2r. If = r+1 the { E(S 0 ), E(S 1 ), E(S 2 ) }={r, r, r+1} ad hece deg(u) 2r+1. If =r+2 the{ E(S 0 ), E(S 1 ), E(S 2 ) }= {r, r+1, r+1} ad hece deg(u) 2r+2. Thus (G) 2r+ t. Theorem.5. The cycle C is a -emg iff 0(mod). Proof: Suose C is a -emg ad 0(mod). The =r ad hece e(0)=e(1)=e(2)=rad v(0)=v(1)=v(2)=r. If v(0)=rthe e(0) r 1 which is a cotradictio. Therefore 0(mod). Coversely assume 0(mod). Let C be the cycle v 1 v 2 v v v 1. We cosider the followig two cases. Case (i): 1(mod). Hece, =r+1. Defie a vertex labelig f : V {0, 1, 2} by 0 if 1 i r f (v i )= 1 if r+ 2 i 2r+1 2 if 2r+ 2 i r+1 Now, v f (0)=r+1, v f (1)=v f (2)=r, e f (0)=e f (2)=rad e f (1)=r+1. Thus v f (i) v f ( j) 1 ad e f (i) e f ( j) 1for i, j=0, 1, 2. Therefore f is a -eml. Case (ii): 2(mod). Hece, =r+ 2. Defie f by 0 if 1 i r f (v i )= 1 if r+ 2 i 2r+1 2 if 2r+ 2 i r+2 The, v f (0)=r+1=v f (2), v f (1)=r, ad e f (0)=r, e f (1)=e f (2)=r+1. Hece, C is -emg. Theorem.6. The ath P is a -emg for all 2. Proof: Let a ath P be v 1 v 2 v v. Defie a vertex labelig f : V {0, 1, 2} as follows. Case (i): 0(mod). Take =r. 0 if 1 i r f (v i )= 1 if r i 2r 2 if 2r i r

5 25 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP Now, v f (0)=v f (1)=v f (2)=r, e f (0)=r 1 ad e f (1)=r=e f (2). Case (ii): 1(mod). Take =r. 0 if 1 i r f (v i )= 1 if r+ 2 i 2r+1 2 if 2r+ 2 i r+1 Thus, v f (0)=r,= v f (1)=v f (2)=rad e f (0)=e f (1)=e f (2)=r. Case (iii): 2(mod). Take =r if 1 i r f (v i )= 1 if r+ 2 i 2r+2 2 if 2r+ i r+2 Now, v f (0)=v f (1)=r, v f (2)=rad e f (0)=e f (2)=r, e f (1)=r. I the above three cases f satisfies v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1, 2. Hece, P is a -emg G P Figure 1 Theorem.7. If G is a -emg the G@P, where 1(mod) is a - emg. Proof: Let P be a ath u 1 u 2 u u ad f be a -eml of P as Theorem.6. By the Case (ii) of Theorem.6 v f (0)=r,=v f (1)=v f (2)=rad e f (0)=e f (1)=e f (2)=r. Let g be a -eml of G ad u V(G) with g(u)=0. Now, idetify the vertex u with a ed vertex of P whose label is 0. Defie a labelig h : V(G@P) {0, 1, 2} by g(v) if v V(G) h(v)= f (v) if v P

6 26 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP Now v h (0)=v g (0)+v f (0) 1=v g (0)+r, v h (1)=v g (1)+v f (1)=v g (0)+r, v h (2)=v g (2)+v f (2)=v g (2)+ r, e h (0)=e g (0)+r, e h (1)=e g (1)+r ad e h (2)=e g (2)+r. Thus v h (i) v h ( j) = v g (i) v g ( j) 1 ad e h (i) e h ( j) = e g (i) e g ( j) 1. Hece, h is a -eml of G@P. A examle for a - emg with G= C 7 ad =4is give i Figure 1 ad Figure G@P 4 Figure 2 Theorem.8. The bistar B(m, ) with m is -emg iff. Proof: Let V(B(m, ))={u, u i : 1 i } {v, v i : 1 i m} ad E(B(m, ))={uv} {uu i : 1 i } {vv i : 1 i m}. Thus =m++2 ad =m++1. Defie a vertex labelig f as follows: Case (i): Suose =r. 0 if 1 i r 1 if 1 i 2r 1 f (u)=0; f (u i )= ; f (v)=1; f (v i )= 1 if r+1 i 2 if 2r i m Hece, v f (0)=r+1, v f (1)=v f (2)=rad e f (0)=e f (1)=e f (2)=r. Case (ii): Suose =r. 0 if 1 i r 1 if 1 i 2r f (u)=0; f (u i )= ; f (v)=1; f (v i )= 1 if r+1 i 2 if 2r +1 i m Hece, v f (0)=r+1, v f (1)=r, v f (2)=rad e f (0)=r, e f (1)=r+1ad e f (2)=r. Case (iii): Suose =r+2. 0 if 1 i r 1 if 1 i 2r f (u)=0; f (u i )= ; f (v)=1; f (v i )= 1 if r+1 i 2 if 2r +1 i m Hece, v f (0)=r+1, v f (1)=r, v f (2)=r+1 ad e f (0)=r, e f (1)=e f (2)=r+1. I all the above three cases f satisfies v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1, 2. Thus f is a -eml of B(m, ). Coversely, suose that m ad <. We have = + t where t {0, 1, 2}. The 2 + t= < =m++1 =m+1= (G). Hece, (G)>2 + t. By Theorem.4, B(m, ) is ot -emg.

7 27 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP Theorem.9. K 1, is -emg iff 2. Proof: Suose that 2. Whe =1, K 1, P 2 ad whe =2, K 1, P. Hece, by Theorem.6, K 1, is -emg. Suose K 1, is -emg. The (K 1, ) 2 + t where t {0, 1, 2}. Here (K1, )==. By Theorem.4, 2 + t + t 2 + t 0 = 0. Hece, =t euitable mea labelig of T (k) (>1) We defie the grah T (k) to be the grah with the vertex set V(T (k) )={v 1, v 2, v,, v k ; v k+1 v k+2,, v 2k 1 ; v 2k, v 2k+1, v k 2 ; ; v (k 1)( 1)+2 ; v (k 1)+1 } ad with the edge set E(T (k) )={v i v i+1 : 1 i (k 1)} {v 1 v k, v k v 2k 1, v 2k 1 v k 2,, v (k 1)( 1)+1 v (k 1)+1 }. Hece, we have =(k 1)+1 ad =k. Lemma 4.1. If k 1(mod) the T (k) is a -emg. Proof: Sice k 1(mod), 1(mod). Defie a labelig f : V(T (k) ) {0, 1, 2} as follows: 0 if 1 i f (v i )= 1 if + 2 i 2 2 if i Thus v f (0)=, v f (1)=, v f (2)=. To fid the values of e f (0), e f (1) ad e f (2) we cosider the followig three cases. Case (i): Suose 0(mod). Take =r. The =r(k 1)+1 ad = r(k 1). Hece, e f (0)=rk ad v f (1)=v f (2)=r(k 1) which imlies that e f (1)=e f (2)=rk. Case (ii): Suose 1(mod). Take =r+1. The = (r+1)(k 1)+1 = r(k 1)+k 1+1 = r(k 1)+ k = r(k 1)+ k 1. Hece, v f (0)=r(k 1)+ k 1 which imlies e f (0)=rk+ k 1. Agai v f (1)=v f (2)=r(k 1)+ k 1 imlies that e f (1)=(r)k rk k 1 rk k 2 = rk+ k 1 ad e f (2)=rk 1+ k 1 + 2=rk+ k+2. Case (iii): Suose 2(mod). Take =r+2. The = (r+2)(k 1)+1 = r(k 1)+ 2(k 1). imlies Hece, v f (0)=r(k 1)+ 2(k 1) +1 imlies that e f (0)=rk+ 2(k 1), v f (1)=v f (2)=r(k 1)+ 2(k 1) that e f (2)=rk 1+ 2(k 1) + 2=rk+ 2(k 1) ad e f (1)=rk+ 2k 2rk 1 4(k 1) = rk+ 2(k 1). I the above three cases f satisfies v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1, 2. Hece, f is a -eml of T (k). Lemma 4.2. If k 0(mod) ad 0(mod) the T (k) is a -emg. Proof: If k 0(mod) the 0(mod). Sice 0(mod), take =r. Defie a labelig f as follows:

8 28 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP if 1 i f (v i )= 1 if + 2 i 2 2 if i Thus v f (0)= +1, v f (1)=v f (2)=. Here = (k 1)r+1 = r(k 1). Hece, v f (0)=r(k 1)+1, v f (1)=v f (2)=r(k 1). Thus e f (1)=rk 2rk=rk, e f (2)=rk ad e f ()=rk. Hece, f satisfies v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1, 2. Therefore, f is a -eml of T (k). Lemma 4.. If k 0(mod) ad 0(mod) the T (k) is a -emg. Proof: If k 0(mod) the 0(mod). Sice 0(mod), take =r. Defie a labelig f as follows: 0 if 1 i f (v i )= 1 if + 2 i 2 2 if i Thus v f (0)= +1, v f (1)=v f (2)=. Here = (k 1)r+1 = r(k 1). Hece, v f (0)=r(k 1)+1, v f (1)=v f (2)=r(k 1). Thus e f (1)=rk 2rk=rk, e f (2)=rk ad e f ()=rk. Hece, f satisfies v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1, 2. Therefore, f is a -eml of T (k). Lemma 4.4. If k 0(mod) ad 1(mod) the T (k) is ot a -emg. Proof: Let f be a -eml of T (k). The v f (0) is either or. If we take =r the v f (0) is either r(k 1)+ k or r(k 1)+ k. Hece, e f (0) is either rk+ k 1 or rk+ k. Sice k 0(mod), we must have e f (0)= k. Lemma 4.5. If k 0(mod) ad 2(mod) the T (k) Proof: Let f be a -eml of T (k) is ot a -emg.. The v f (0) is either or. If we take =r+ 2 the v f (0) 2k or rk 1+. Sice is ot a -emg. is either r(k 1)+ 2k or r(k 1)+ 2k. Hece, e f (0) is either rk 2+ 2k k 0(mod), we must have e f (0)= k(r+2). which gives a cotradictio. Thus T (k) Lemma 4.6. If k 2(mod) ad 0(mod) the T (k) is a -emg. Proof: Defie a labelig f as follows: 0 if 1 i f (v i )= 1 if + 2 i 2 2 if i Thus v f (0)=, v f (1)=v f (2)= ad e f (0)=e f (1)=e f (2)= = k Hece, f satisfies v f (i) v f ( j) 1ad e f (i) e f ( j) 1for i, j=0, 1, 2. Therefore, f is a -eml of T (k). Lemma 4.7. If k 2(mod) ad 1(mod) the T (k) is a -emg.

9 29 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP Proof: Defie a labelig f as follows: 0 if 1 i f (v i )= 1 if + 2 i if 2 + i The v f (0) = v f (1) =, v f (2) =. If we take = r+1 the e f (0) = rk+ k 2 ad e f (1)=e f (2)=rk+ k 2. Hece, f satisfies v f (i) v f ( j) 1 ad e f (i) e f ( j) 1 for i, j=0, 1, 2. Therefore, f is a -eml of T (k). Lemma 4.8. If k 2(mod) ad 2(mod) the T (k) is ot a -emg. Proof: Sice k 2(mod) ad 2(mod), k 1 1(mod) ad (k 1) (mod). Hece, v f (0)=v f (1)=v f (2)= = (k 1)(r+2)+1 = r(k 1)+ 2k 1 = r(k 1)+ 2(k 2). So, e f (0)=rk+ 2(k 2) ad e f (2)=rk+2+ 2(k 2) which imlies that e f (0) e f (2) =2. Thus, T (k) is ot a -emg. From the above lemmas we have the followig theorem. Theorem 4.9. T (k) (i) k 1(mod) is a -emg if (ii) k 0(mod) ad 0(mod) (iii) k 2(mod) ad 1(mod) (iv) k 2(mod) ad 0(mod). Theorem K 1,2 K 1, is a -emg. Proof: Let u ad v be the cetral vertices of the star grahs K 1,2 ad K 1, resectively, u 1, u 2,, u 2 be the vertices icidet with u ad v 1, v 2,, v be the vertices icidet with v. Hece, =+2 ad =. Now assig 0 to all the vertices of K 1,, 1 to the vertices u, u 1, u 2,, u ad 2 to the vertices u +1, u +2,, u 2. Hece, v f (0)=+1, v f (1)=v f (2)=, e f (0)=e f (1)=e f (2)=. Thus, K 1,2 K 1, is a -emg. Refereces [1] I. Cahit, Cordial grahs: a weaker versio of graceful ad harmoious grahs, Ars Combi., 2 (1987), [2] I. Cahit, O cordial ad -euitable labelligs of grahs, Util. Math., 7 (1990), [] F. Harary, Grah Theory, Addiso Wesley, Massachusetts, (1972).

10 0 P. Jeyathi/Joural of Algorithms ad Comutatio 44 (201) PP [4] S. Somasudaram ad R. Poraj, Mea labeligs of grahs, Natl Acad, Sci. Let., 26 (200), [5] M. Sudaram, R. Poraj ad S. Somasudaram, Product cordial labelig of grahs, Bull. Pure ad Alied Scieces (Mathematics & Statistics), 2E (2004),