Iteratioal J.Math. Combi. Vol. (011), 40-51 Absolutely Harmoious Labelig of Graphs M.Seeivasa (Sri Paramakalyai College, Alwarkurichi-6741, Idia) A.Lourdusamy (St.Xavier s College (Autoomous), Palayamkottai, Idia) E-mail: msvasa @yahoo.com, lourdugaam@hotmail.com Abstract: Absolutely harmoious labelig f is a ijectio from the vertex set of a graph G with q edges to the set {0, 1,,..., q 1}, if each edge uv is assiged f(u) + f(v) the the resultig edge labels ca be arraged as a 0, a 1, a,..., a q 1 where a i = q i or q + i, 0 i q 1. However, whe G is a tree oe of the vertex labels may be assiged to exactly two vertices. A graph which admits absolutely harmoious labelig is called absolutely harmoious graph. I this paper, we obtai ecessary coditios for a graph to be absolutely harmoious ad study absolutely harmoious behavior of certai classes of graphs. Key Words: Graph labelig, Smaradachely k-labelig, harmoious labelig, absolutely harmoious labelig. AMS(010): O5C78 1. Itroductio A vertex labelig of a graph G is a assigmet f of labels to the vertices of G that iduces a label for each edge xy depedig o the vertex labels. For a iteger k 1, a Smaradachely k-labelig of a graph G is a bijective mappig f : V {1,,, k V (G) + E(G) } with a additioal coditio that f(u) f(v) k for uv E. particularly, if k = 1, i.e., such a Smaradachely 1-labelig is the usually labelig of graph. Amog them, labeligs such as those of graceful labelig, harmoious labelig ad mea labelig are some of the iterestig vertex labeligs foud i the dyamic survey of graph labelig by Gallia []. Harmoious labelig is oe of the fudametal labeligs itroduced by Graham ad Sloae [3] i 1980 i coectio with their study o error correctig code. Harmoious labelig f is a ijectio from the vertex set of a graph G with q edges to the set {0, 1,,..., q 1}, if each edge uv is assiged f(u) + f(v)(mod q) the the resultig edge labels are distict. However, whe G is a tree oe of the vertex labels may be assiged to exactly two vertices. Subsequetly a few variatios of harmoious labelig, amely, strogly c-harmoious labelig [1], sequetial labelig [5], elegat labelig [1] ad felicitous labelig [4] were itroduced. The later three labeligs were itroduced to avoid such exceptios for the trees give i harmoious labelig. A strogly 1 Received November 08, 010. Accepted May, 011.
Absolutely Harmoious Labelig of Graphs 41 1-harmoious graph is also kow as strogly harmoious graph. It is iterestig to ote that if a graph G with q edges is harmoious the the resultig edge labels ca be arraged as a 0, a 1, a,, a q 1 where a i = i or q + i, 0 i q 1. That is for each i, 0 i q 1 there is a distict edge with label either i or q + i. A aother iterestig ad atural variatio of edge label could be q i or q + i. This prompts to defie a ew variatio of harmoious labelig called absolutely harmoious labelig. Defiitio 1.1 A absolutely harmoious labelig f is a ijectio from the vertex set of a graph G with q edges to the set {0, 1,,..., q 1}, if each edge uv is assiged f(u)+f(v) the the resultig edge labels ca be arraged as a 0, a 1, a,..., a q 1 where a i = q i or q+i, 0 i q 1. However, whe G is a tree oe of the vertex labels may be assiged to exactly two vertices. A graph which admits absolutely harmoious labelig is called absolutely harmoious graph. The result of Graham ad Sloae [3] states that C, = 1(mod 4) is harmoious, but we show that C, = 1(mod 4) is ot a absolutely harmoious graph. O the other had, we show that C 4 is a absolutely harmoious graph, but it is ot harmoious. We observe that a strogly harmoious graph is a absolutely harmoious graph. To iitiate the ivestigatio o absolutely harmoious graphs, we obtai ecessary coditios for a graph to be a absolutely harmoious graph ad prove the followig results: 1. Path P, 3, a class of baaa trees, ad P Km c are absolutely harmoious graphs.. Ladders, C K c m, Triagular sakes, Quadrilateral sakes, ad mk 4 are absolutely harmoious graphs. 3. Complete graph K is absolutely harmoious if ad oly if = 3 or 4. 4. Cycle C, = 1 or (mod 4), C m C where m ad are odd, mk 3, m are ot absolutely harmoious graphs.. Necessary Coditios Theorem.1 If G is a absolutely harmoious graph, the there exists a partitio (V 1, V ) of the vertex set V (G), such that the umber of edges coectig the vertices of V 1 to the vertices q of V is exactly. Proof If G is a absolutely harmoious graph,the the vertices ca be partitioed ito two sets V 1 ad V havig respectively eve ad odd vertex labels. Observe that amog the q edges q q edges or edges are labeled with odd umbers, accordig as q is eve or q is odd. For a edge to have odd label, oe ed vertex must be odd labeled ad the other ed vertex must be eve labeled. Thus, the umber of edges coectig the vertices of V 1 to the vertices q of V is exactly. Remark. A simple ad straight forward applicatio of Theorem.1 idetifies the o absolutely harmoious graphs. For example, complete graph K has ( 1) edges. If we assig
4 M.Seeivasa ad A.Lourdusamy m vertices to the part V 1, there will be m( m) edges coectig the vertices of V 1 to the vertices of V. If K has a absolutely harmoious labelig, the there is a choice of m for which m( m) =. Such a choice of m does ot exist for = 5, 7, 8.10,... 4 A graph is called eve graph if degree of each vertex is eve. Theorem.3 If a eve graph G is absolutely harmoious the q = 0 or 3 (mod 4). Proof Let G be a eve graph with q = 1 or (mod 4) ad d(v) deotes the degree of the vertex v i G. Suppose f be a absolutely harmoious labelig of G. The the resultig edge labels ca be arraged as a 0, a 1, a,..., a q 1 where a i = q i or q + i, 0 i q 1. I other words, for each i, the edge label a i is (q i) + ib i, 0 i q 1 where b i {0, 1}. Evidetly v V (G) q 1 d(v)f(v) kb k = k=0 As d(v) is eve for each v ad q = 1 or (mod 4), q + 1. v V (G) q 1 d(v)f(v) kb k = 0 (mod ) but q + 1 = 1 (mod ). This cotradictio proves the theorem. k=0 Corollary.4 A cycle C is ot a absolutely harmoious graph if = 1 or (mod 4). Corollary.5 A grid C m C is ot a absolutely harmoious graph if m ad are odd. Theorem.6 If f is a absolutely harmoious labelig of the cycle C, the edges of C ca be partitioed ito two sub sets E 1, E such that uv E 1 f(u) + f(v) = ( + 1) 4 ad uv E f(u) + f(v) = ( 3). 4 Proof Let v 1 v v 3...v v 1 be the cycle C, where e i = v i 1 v i, i ad e 1 = v v 1. Defie E 1 = {uv E/f(u) + f(v) is o egative} ad E = {uv E/f(u) + f(v) is egative}. Sice f is a absolutely harmoious labelig of the cycle C, I other words, uv E uv E 1 f(u) + f(v) + f(u) + f(v) = ( 1). uv E f(u) + f(v) = ( 1). (1) Sice uv E (f(u) + f(v) ) =, we have f(u) + f(v) f(u) + f(v) =. () uv E 1 uv E
Absolutely Harmoious Labelig of Graphs 43 Solvig equatios (1) ad (), we get the desired result. Remark.7 If ( + 1) ( 3) = 1 or (mod 4) the both ad caot be itegers. Thus 4 4 the cycle C is ot a absolutely harmoious graph if = 1 or (mod 4). Remark.8 Observe that the coditios stated i Theorem.1, Theorem.3, ad Theorem.6 are ecessary but ot sufficiet. Note that C 8 satisfies all the coditios stated i Theorems.1,.3, ad.6 but it is ot a absolutely harmoious graph. For, checkig each of the 8! possibilities reveals the desired result about C 8. 3. Absolutely Harmoious Graphs Theorem 3.1 The path P +1,where is a absolutely harmoious graph. r + 1 if = 0 (mod 4) Proof Let P +1 : v 1 v...v +1 be a path, r =, s =, r otherwise s 1 if = 0 or 1 (mod 4) t = s if = or 3 (mod 4), T t + if = 0 or 1 (mod 4) 1 =, T = t + 1 if = or 3 (mod 4) ad T 3 = 1 if = 0 or 1 (mod 4) if = or 3 (mod 4). The r + s + t = + 1. Defie f : V (P +1 ) {0, 1,, 3,, 1} by: f(v i ) = T 1 i if 1 i r, f(v r+i ) = T i if 1 i s ad f(v r+s+i ) = T 3 + i if 1 i t. Evidetly f is a absolutely harmoious labelig of P +1. For example, a absolutely harmoious labelig of P 1 is show i Fig.3.1. 9 7 5 3 0 4 a 8 a 6 a 4 a a 0 a 1 a 3 a 7 a 10 a 9 a 5 10 8 6 5 1 Fig.3.1 The tree obtaied by joiig a ew vertex v to oe pedat vertex of each of the k disjoit stars K 1,1, K 1,, K 1,3,..., K 1,k is called a baaa tree. The class of all such trees is deoted by BT( 1,, 3,..., k ). Theorem 3. The baaa tree BT(,,,..., ) is absolutely harmoious.
44 M.Seeivasa ad A.Lourdusamy 1 a 15 a 10 a 5 a 0 4 9 14 19 a 16 a 6 a 4 a 14 0 5 10 15 a 19 a 17 a 9 a 18 a 8 a 7 a 1 a a 3 a 11 a 13 a 1 1 3 6 7 8 11 1 13 16 17 18 Fig.3. Proof Let V (BT(,,,, ) = {v} {v j, v jr : 1 j k ad 1 r } where d(v j ) = ad E(BT(,,,..., ) = {vv j : 1 j k} {v j v jr : 1 j k, 1 r }. Clearly BT(,,, ) has order ( + 1)k + 1 ad size ( + 1)k. Defie as follows: f : V (BT(,,, ) {1,, 3,..., ( + 1)k 1} f(v) = 1, f(v j ) = ( + 1)(j 1) : 1 j k, f(v jr ) = ( + 1)(j 1) + r : 1 r. It ca be easily verified that f is a absolutely harmoious labelig of BT(,,,..., ). For example a absolutely harmoious labelig of BT(4, 4, 4, 4)is show i Fig.3.. The coroa G 1 G of two graphs G 1 (p 1, q 1 ) ad G (p, q ) is defied as the graph obtaied by takig oe copy of G 1 ad p 1 copies of G ad the joiig the i th vertex of G 1 to all the vertices i the i th copy of G. Theorem 3.3 The coroa P Km C is absolutely harmoious. Proof Let V (P Km) C = {u i : 1 i } {u ij : 1 i, 1 j m} ad E(P Km) C = {u i u i+1 : 1 i 1} {u i u ij : 1 i, 1 j m}. We observe that P Km C has order (m + 1) ad size (m + 1) 1. Defie f : V (P Km C ) {0, 1,,..., m + } as follows: 0 if i = 1, f(u i ) = (m + 1)(i 1) if i = (m + 1)(i 1) 1 otherwise, (m + 1)i if 1 i, f(u im ) = (m + 1)i 1 if i = 1, (m + 1)i i, ad for 1 j m 1, (m + 1)(i 1) + j if 1 i 1, f(u ij ) = (m + 1)(i 1) + j 1 if i. It ca be easily verified that f is a absolutely harmoious labelig of P Km C. For example a absolutely harmoious labelig of P 5 K3 C is show i Fig. 3.3.
Absolutely Harmoious Labelig of Graphs 45 0 a 16 3 a 8 8 a 0 11 a 7 15 a 18 a 17 a 15 a a 10 a a 3 a a 1 a 4 a 5 a 6a1 9 11 a a 14 13 1 4 5 6 7 8 9 10 1 13 14 16 17 18 Fig.3.3 Theorem 3.4 The coroa C K C m is absolutely harmoious. Proof Let V (C K C m) = {u i : 1 i } {u ij : 1 i, 1 j m} ad E(C K C m) = {u i u i+1 : 1 i 1} {u u 1 } {u i u ij : 1 i, 1 j m}. We observe that C Km C has order (m + 1) ad size (m + 1). Defie f : V (C Km) C {0, 1,,..., m + 1} as follows: 0 if i = 1, f(u i ) = (m + 1)(i 1) 1 if i 1, (m + 1)i if 1 i 3, f(u im ) = (m + 1)i 1 otherwise (m + 1)(i 1) otherwise, ad for 1 j m 1 (m + 1)(i 1) + j if 1 i 1, f(u ij ) = (m + 1)(i 1) + j 1 if i. It ca be easily verified that f is a absolutely harmoious labelig of C Km C. For example a absolutely harmoious labelig of C 5 K3 C is show i Figure 3.4. 15 17 14 18 13 19 a 15 a 14 a 13 a 7 a 6 a5 1 4 a a 18 19 a 16 16 1 0 a 0 a 17 3 a 1 a11 8 a 10 5 a 9 a 3 a 1 a 6 11 7 10 9, Fig.3.4 Theorem 3.5 The ladder P P, where is a absolutely harmoious graph. Proof Let V (P P ) = {u 1, u, u 3,..., u } {v 1, v, v 3,..., v } ad E(P P ) = {u i u i+1 : 1 i 1} {v i v i+1 : 1 i 1} {u i v i : 1 i }. We ote that P P has order
46 M.Seeivasa ad A.Lourdusamy ad size 3. Case 1. 0(mod 4). Defie f : V (P P ) {0, 1,,..., 3 3} by f(v 1 ) = 0, f(v + Case. 1(mod 4). 3i if i is odd, 3i if i is eve ad i 4 f(u i ) =, 3i 1 if i is eve ad i =, 3i 3 if i is eve ad +4 i, ) = 3 6, f(v i+1 ) = f(u i ) + 1 if 1 i 1 ad i. Defie f : V (P P ) {0, 1,,..., 3 3} by 3i if i is odd ad 1 i 3, 3i 1 if i = +1 f(u i ) =, 3i 3 if i is odd ad +5 i, 3i if i is eve, f(v 1 ) = 0, f(v +3) = 3 3, f(v i+1 ) = f(u i ) + 1 if 1 i 1 ad i +1. Case 3. (mod 4). Defie f : V (P P ) {0, 1,,..., 3 3} by 3i if i is odd, f(u i ) = 3i if i is eve ad i, 3i 3 if i is eve ad + i, f(v 1 ) = 0, f(v i+1 ) = f(u i ) + 1 if 1 i 1. Case 4. 3(mod 4). Defie f : V (P P ) {0, 1,,..., 3 3} by 3i if i is odd ad 1 i 1, f(u i ) = 3i 3 if i is odd ad +3 i, 3i if i is eve. f(v 1 ) = 0, f(v i+1 ) = f(u i ) + 1 if 1 i 1.
Absolutely Harmoious Labelig of Graphs 47 I all four cases, it ca be easily verified that f is a absolutely harmoious labelig of P P. For example, a absolutely harmoious labelig of P 9 P is show i Fig.3.5. 1 4 7 10 14 16 18 4 a 0 a 14 a 8 a 1 a 5 a 9 a 15 a 1 a 4 a 19 a 13 a 7 a 0 a 3 a 10 a 16 a a 3 a 18 a 1 a 6 a a 4 a 11 a 17 0 5 8 11 1 17 19 3 Fig.3.5 A K -sake has bee defied as a coected graph i which all blocks are isomorphic to K ad the block-cut poit graph is a path. A K 3 -sake is called triagular sake. Theorem 3.6 A triagular sake with blocks is absolutely harmoious if ad oly if = 0 or 1 (mod 4). Proof The ecessity follows from Theorem.3.Let G be a triagular sake with blocks o p vertices ad q edges. The p = 1 ad q = 3. Let V (G ) = {u i : 1 i + 1 } {v i : 1 i } ad E(G ) = {u i u i+1, u i v i, u i+1 v i : 1 i }. Case 1. 0 (mod 4). Let m = 4. Defie f : V (G ) {0, 1,,..., 3 1} as follows: 0 if i = 1, i if i 3m ad i 0 or (mod 3), f(u i ) = i 1 if i 3m ad i 1 (mod 3), 6i 3 7 otherwise, 1 if i = 1, i 1 if i 3m 1 ad i or (mod 3), f(v i ) = i if i 3m 1 ad i 1 (mod 3), 6m + 1 if i = 3m, 6i 3 3 otherwise. Case. 1 (mod 4). Let m = 1. Defie f : V (G ) {0, 1,,..., 3 1} as follows: 4
48 M.Seeivasa ad A.Lourdusamy 0 if i = 1, i if i 3m + ad i 0 or (mod 3), f(u i ) = i 1 if i 3m + ad i 1 (mod 3), 6i 3 7 otherwise, 1 if i = 1, i 1 if i 3m + 1 ad i 0 or (mod 3) f(v i ) = i if i 3m + 1 ad i 1 (mod 3) 6i 3 3 otherwise. I both cases, it ca be easily verified that f is a absolutely harmoious labelig of the triagular sake G. For example, a absolutely harmoious labelig of a triagular sake with five blocks is show i Fig.3.6. 1 3 5 6 1 a 14 a 1 a 10 a 8 a 6 a 3 a a 1 a 5 a 11 0 a 13 a 9 4 a 4 7 a 0 8 a 7 14 Fig.3.6 Theorem 3.7 K 4 -sakes are absolutely harmoious. Proof Let G be a K 4 -sake with blocks o p vertices ad q edges. The p = 3+1 ad q = 6. Let V (G ) = {u i, v i, w i : 1 i } {v +1 } ad E(G ) = {u i v i, u i w i, v i w i : 1 i } {u i v i+1, v i v i+1, w i v i+1 : 1 i } Defie f : V (G ) {0, 1,,..., 6 1} as follows: f(u i ) = 3i 3, f(v i ) = 3i, f(w i ) = 3i 1 where 1 i, ad f(v +1 ) = 3 + 1. It ca be easily verified that f is a absolutely harmoious labelig of G ad hece K 4 -sakes are absolutely harmoious. For example, a absolutely harmoious labelig of a K 4 -sake with five blocks is show i Fig.3.7.
Absolutely Harmoious Labelig of Graphs 49 a 8 a 16 0 6 8 1 a 4 14 a 3 a 6 a 9 a 4 a 19 a 7 a 17 a 15 a 1 a 5 a 0 a 7 1 a 5 4 a 1 7 a 13 10 a 8 13 a 1 16 a 3 a 0 a a 18 a 14 a 11 a 9 a 10 3 5 9 10 a 6 a Fig.3.7 A quadrilateral sake is obtaied from a path u 1 u...u +1 by joiig u i, u i+1 to ew vertices v i, w i respectively ad joiig v i ad w i. Theorem 3.8 All quadrilateral sakes are absolutely harmoious. Proof Let G be a quadrilateral sake with V (G ) = {u i : 1 i + 1 } {v i, w i : 1 i } ad E(G ) = {u i u i+1, u i v i, u i+1 w i, v i w i : 1 i }. The p = 3 + 1 ad q = 4. Let if 0 (mod ) m = 1 if 1 (mod ). Defie f : V (G ) {0, 1,,...4 1} as follows: 0 if i = 1, 4i 3 if 1 i m, f(u i ) = 4i 6 if i m + 1,, f(v i ) = 4i if m + 1 i, 4i 7 if m + i + 1 4i if 1 i m, f(w i ) = 4i 1 if m + 1 i. It ca be easily verified that f is a absolutely harmoious labelig of the quadrilateral sake G ad hece quadrilateral sakes are absolutely harmoious. For example, a absolutely harmoious labelig of a quadrilateral sake with six blocks is show i Fig.3.8. a 19 a 3 a 13 1 4 9 1 18 19 a 3 a 18 a 9 a a 1 a 7 a 1 a 16 a 14 0 a 6 a 8 10 13 a 6 17 1 a 17 a 10 a 0 a 4 a 15 a 0 a 11 a 5 a 1 5 8 14 15 3 Fig.3.8
50 M.Seeivasa ad A.Lourdusamy Theorem 3.9 The disjoit uio of m copies of the complete graph o four vertices, mk 4 is absolutely harmoious. Proof Let u j i where 1 i 4 ad 1 j m deotes the ith vertex of the j th copy of mk 4. We ote that that mk 4 has order 4m ad size 6m. Defie f : V (mk 4 ) {0, 1,,...6m 1} as follows:f(u 1 1 ) = 0, f(u1 ) = 1, f(u1 3 ) =, f(u1 4 ) = 4, f(u 1 ) = q 3, f(u ) = q 4, f(u 3 ) = q 5, f(u 4 ) = q 7, f(uj+ i ) = f(u j i ) + 6 if j is odd, ad f(uj+ i ) = f(u j i ) 6 if j is eve, where 1 i 4 ad 1 j m. Clearly f is a absolutely harmoious labelig. For example, a absolutely harmoious labelig of 5K 4 is show i Figure 11. Box Observatio 3.10 If f is a absolutely harmoious labelig of a graph G,which is ot a tree, the 1. Each x i the set {0, 1, } has iverse image.. Iverse images of 0 ad 1 are adjacet i G. 3. Iverse images of 0 ad are adjacet i G. Theorem 3.11 The disjoit uio of m copies of the complete graph o three vertices, mk 3 is absolutely harmoious if ad oly if m = 1. Proof Let u j i,where1 i 3 ad 1 j m deote the ith vertex of the j th copy of mk 3. Assigmets of the values 0, 1, to the vertices of K 3 gives the desired absolutely harmoious labelig of K 3. For m, mk 3 has 3m vertices ad 3m edges. If mk 3 is a absolutely harmoious graph, we ca assig the umbers {0, 1,, 3m 1} to the vertices of mk 3 i such a way that its edges receive each of the umbers a 0,a 1,...,a q 1 where a i = q i or q+i, 0 i q 1. By Observatio 3.10, we ca assume, without loss of geerality that f(u 1 1 ) = 0, f(u1 ) = 1, f(u1 3 ) =. Thus we get the edge labels a q 1, a q ad a q 3. I order to have a edge labeled a q 4, we must have two adjacet vertices labeled q 1 ad q 3. we ca assume without loss of geerality that f(u 1) = q 1 ad f(u ) = q 3. I order to have a edge labeled a q 5, we must have f(u 3 ) = q 4. There is ow o way to obtai a edge labeled a q 6. This cotradictio proves the theorem. Theorem 3.1 A complete graph K is absolutely harmoious graph if ad oly if = 3 or 4. Proof From the defiitio of absolutely harmoious labelig, it ca be easily verified that K 1 ad K are ot absolutely harmoious graphs. Assigmets of the values 0, 1, ad 0, 1,, 4 respectively to the vertices of K 3 ad K 4 give the desired absolutely harmoious labelig of them. For > 4, the graph K has q 10 edges. If K is a absolutely harmoious graph, we ca assig a subset of the umbers {0, 1,, q 1} to the vertices of K i such a way that the edges receive each of the umbers a 0,a 1,...,a q 1 where a i = q i or q + i, 0 i q 1. By Observatio 3.10, 0, 1, ad must be vertex labels. With vertices labeled 0, 1, ad, we have edges labeled a q 1, a q ad a q 3. To have a edge labeled a q 4 we must adjoi the vertex label 4. Had we adjoied the vertex label 3 to iduce a q 4, we would have two edges labeled a q 3, amely, betwee 0 ad 3, ad betwee 1 ad. Had we adjoied the vertex labels q 1
Absolutely Harmoious Labelig of Graphs 51 ad q 3 to iduce a q 4, we would have three edges labeled a 1, amely, betwee q 1 ad 0, betwee q 1 ad, ad betwee q 3 ad. With vertices labeled 0, 1,,ad 4, we have edges labeled a q 1, a q, a q 3, a q 4, a q 5, ad a q 6. Note that for K 4 with q = 6, this gives the absolutely harmoious labelig. To have a edge labeled a q 7, we must adjoi the vertex label 7; all the other choices are ruled out. With vertices labeled 0, 1,, 4 ad 7, we have edges labeled a q 1, a q, a q 3, a q 4, a q 5, a q 6, a q 7, a q 8, a q 9, ad a q 11. There is ow o way to obtai a edge labeled a q 10, because each of the ways to iduce a q 10 usig two umbers cotais at least oe umber that ca ot be assiged as vertex label. We may easily verify that the followig boxed umbers are ot possible choices as vertex labels: 0 10 1 9 8 3 7 4 6 q 1 q 9 q q 8 q 3 q 7 q 4 q 6 This cotradictio proves the theorem. Refereces [1] G. J. Chag, D. F. Hsu, ad D. G. Rogerss, Additive variatios o a graceful theme: Some results o harmoious ad other related graphs, Cogressus Numeratium, 3 (1981) 181-197. [] J. A. Gallia, A dyamic survey of graph labelig, The Electroic Joural of Combiatorics, 16 (009), #DS6. [3] R. L. Graham ad N. J. A. Solae, O additive bases ad Harmoious Graphs, SIAM, J. Alg. Discrete Methods, 1(1980) 38-404. [4] S. M. Lee, E. Schmeichel, ad S. C. Shee, O felicitous graphs, Discrete Mathematics, 93 (1991) 01-09. [5] Thom Grace, O sequetial labeligs of graphs, Joural of Graph Theory, 7 (1983), 195-01.