Chapter 3. Antimagic Gl'aphs

Chapter 3 Antimagic Gl'aphs

CHAPTER III ANTIMAGIC GRAPHS 3.1 Introduction In 1970 Kotzig and Rosa [81] defined the magic labeling of a graph G as a bijection f from VuE to {1,2,... IV u EI } such that for all edges xy, f(x) + f(y) + f(xy) is constant. In 1990 Hartsfield and Ringel [65] introduced antimagic labeling. Let G (V,E) be a (p,q) graph. Let f : E -+ {1,2,..., q} be a bijection. For each vertex v in V define f+(v) sum of the labels of the edges incident at v. If for any two distinct vertices u,v, f+(u) 1 f+ (v) then f is called an antimagic labeling. A graph is said to be antimagic if it admits an antimagic labeling. Let us investigate some antimagic graphs. 3.2 On Cycle Related Graphs First we prove In this section we give some results on cycle related graphs. Theorem 3.2.1: The crown C n 0 K 1 is antimagic for all n.. Proof: Let {v 1, v 2,...,v n ' w 1 ' w 2 '...,w n } be the vertex set of G Cn0 K1, Let {ei 11 ~ i ~ n} {vi Vi + 1 II:::; i ~ n where vn +1 v1} Label the edges of C n as follows.

f(ej) i, 1 ~ i ~ n Let us relabel the vertir.es Vj +1 by ui, 1 ~ i ~ n. Let wi be the pendant vertex adjacent with Uj, 1 ~ i ~ n, Label the edges Uj wi such that f(uj, wi) 2n - ( i - 1 ), 1 ~ i ~ n. Since each wi's are end vertices, f+ (Wj) f+ (Uj) 2n - ( i - 1 ), 1 ~ i ~ n. f( Ui, wi) + f (ej) + f(ei+1) 2n - (i - 1 ) + i + i + 1 2n + 2 + i, 1~i~n-1 f(u n, W n ) + f(e 1 ) + f(e n ) 2n - (n -1 ) + 1 + n 2n + 2 Clearly f+(u) *- f+(v) for all u, v E V(G) Hence C n 0 K 1 is antimagic for all n o.9l1ustration: C 11 0 K 1 is antimagic. W 6 6 we v 7 W g w 3 12 1 w 2 w 11 w 1 48

.9l1ustration : C 6 0 K 1 is antimagic. 10 ~,3 1 7 W 6 Remark: This result otherwise can be stated that C n 0 P1is antimagic. Then naturaly one may ask whether C n 0 Pm is antimagic. We have the following Theorem 3.2.2: C n 0 Pm is antimagic if either m nand m n +1. Proof: Case (i) Suppose m n Let v 1 v 2...,v n be the vertices of C n The set of edges are defined as e j {vjvj + l' 1 ~ i ~ n -1 and v n v1e n } Label the edges of C n by f( ej ) i, 1 ~ i ~ n Hence the edge labels are 1, 2,..., n. Next relabel the cycle vertices of C n as follows, ie., wi Vj+1' i 1,2,..., n -1 and w n v1 49

Let Pn be a path on n vertices. We can obtain C n 0 Pn by attaching Pn to each vertex of Cn' Note that C n 0 Pn is a graph with n(n +1) vertices and 2n 2 edges. Let uj j, i, j 1, 2,..., n be the path vertex adjacent with wi, i 1, 2,..., n. Label the remaining edges of C n 0 Pn as follows f (wj, u j j ) n + (2i - 2) n - (i - 2) + ( j - 1), 1 ~ j ~ n, 1 ~ i ~ n f (u jj,' U i j + 1) n + 2i n - i - ( j - 1), 1~ j ~ n -1, 1~ i ~ n Then the set of edge labels are {1, 2,..., n, n + 1,..., 2n 2 } Now f(w j, Uj 1 ) + f (wj, U j 2 ) +... + f (Wi' Uj n ) f+ (w 1 ) n [ n + (2i - 2) n - (i - 2)] + [n (n -1)] /2, n 2 (2i -1) - n (i - 2) + ( n (n - 1) ) /2 [n (n - 1)] / 2 + 2n - n 2 + i n (2n - 1) (3n - n 2 ) / 2 + i n (2n - 1), 3 + n 2 + n (n + 1)] / 2 (3n 2 + n + 6) / 2 2 ~ i ~ n-1 2 ~i~n-1. f+ (w n ) f+ (Wi) f+ (Ui1) f+ (Ujn) (n +1) + n [n + (2n-2) n - (n - 2)] + [n (n - 1)] / 2 n + n 2 (2n - 2) + 2n + (n 2 - n) /2 ( 4n 3-3n 2 + 5n ) / 2 (i + 1) + i + (3n - n 2 ) /2 + i n (2n - 1) (2 + 3n - n 2) / 2 + i ( 2 n 2 - n + 2), 2 ~ i ~ n - 1 2n + (4i - 2) n - (2i - 2), 2 + i (4n - 2), 1~ i ~ n 1~ i ~ n 2n + (4i - 2) n + (3-2i) 50

G 3;)<;;60 3 + i (4n - 2), 1:::; i :::; n f+ (Uj j) 3n + (6i - 2) n - [3i + U- 1) - 3], 1:::; i :::; n, n + (6n -3) i - ( j - 4), 1:::; i :::; n, 2:::;j:::;n-1 2:::;j:::;n-1 Clearly f+ (u) of f+ (v) V U, V E V (C n 0 Pn) :. C n 0 Pn is antimagic..9llustration : C s 0 Psis antimagic 107 108 109t~~~~ 110 111 112 113 92 93 94 95 9697 98 77 78 79 80 81 82 83 Case (ii) m n + 1 If we replace m by n + 1 in the above setup we can get C n 0 Pn+1 is antimagic. 0 51

r!/);w11aifl' Theorem 3.2.3: The double crown C n 0 ~ is antimagic for all odd n, n ~ 3. Proof: Let C n be a cycle. Let V 1 'V 2 '...,v n be the vertices of Cn' Let ~ be the complete graph on two vertices. Now attach ~ to each vertex of the cycle Cn' The newly obtained graph is denoted by C n 0 K 2. Note that C n 0 K 2 is a graph with 3n vertices and 4n edges. Let aj, b i, i 1,2,,n be the vertices adjacent to the rim vertices of Cn' Let{ejl i 1,2,,n}{VjVj+11 i1,2,...,n-1}and e n v n v 1 be the cycle edge and {(aj, bj), (Vj, aj), (Vj, bj), i 1 to n be the other edges of C n 0 K 2 Define f on the edge set of C n 0 K 2 as follows f (a 1,b 1 ) (i+1)/2 ifiisodd n+1 i {--+- if i is even 2 2 2n f (aj,bj) f (vj,aj) f(v1,a1) f (vi, bj) f(v 1,b 1 ) n + (i -1), 2n + (i -1), 3n 3n + (i -1), 4n 25 is n 2 5 i 5 n 2 5 i 5 n Now f+ (Vj) (i +1) I 2 + (n + 2) 12+ (i - 1) 12 ie f+ (vi) f+ (V1) + 2n + (i -1) + 3n + (i -1) [6 i +11 n -3] 12. [6 i +11 n - 3] /2, 2 5 i 5 n f (en) + f (e1) + f(v 1 ' a 1 ) + f(v 1 ' b 1 ) (n +1) /2 + (1 + 1) I 2 + 3n + 4n 52

CR~3 r!!jgw'wfu/ ~ (15n + 3) /2 f+ (aj) f(vj, aj) +f(aj, bj) f+ (a 1 ) 5n, f+ (b 1 ) 6n 2n + (i - 1) + n + (i - 1) 3n + 2i - 2, 2 ~ i ~n f+ (b j) f (aj, bj) + f(vj, b j ) 4n + 2i - 2, 2 ~ i ~ n. Clearly f+ (u) 1:- f+ (v) Vu, V E V (C n 0 K 2 ). Hence C n 0 K 2 is antimagic for all odd n. 0..9Ilustration: C 11 0 K 2 is antimagic. by bg 6----411 22 53

Construction 2.1: Let C 4 be a cycle on 4 vertices V 1 'V 2 'V 3,V4. Attach mi, i 1, 2, 3, 4 pendant vertices with Vi, such that m 1 $; m 2 $; m 3 $; m4 Then the resulting graph is denoted!:)y G. Clearly G has m 1 + m 2 + m 3 + Theorem 3.2.4: The Graph G as per the above construction is antimagic. Proof: Let vl' v2' V3' V4 be the vertices of C4 Let v1 l' V1 2' ' V1m 1 be the vertices attached with v1 V 2 l' V 2 2' ' V2 m2 be the vertices attached with v 2 V 3 l' V 3 2' ' V 3m3 be the vertices attached with v 3 v4 l' V4 2'..., V4 m. be the vertices attached with v4 Label the edges of G as follows f (vi, Vi +1) i, 1$;i$; 3,. f (v4'v 1 ) 4 f (V 1,V 1 j) 4 + j, 1 ~ j $; m 1 f (V 2,V 2 j) 4 + m 1 +j, 1 $; j $; m 2 f (v 3,V 3 j) 4 + m 1 + m 2 + j, 1 $;j $; m 3 f (V4,V 4 j) 4 + m 1 + m 2 + m 3 + j, 1 $; j $; m 4 The resulting edge labels are { 1, 2,..., 4 + m 1 + m 2 + m 3 + m4} Now f+ (v 1 j) 4 + j, 1 $; j $; m 1 f+ (v 2 j) 4 + m 1 + j, 1 ~ j $; m 2 f+ (v 3 j) 4 + m 1 + m 2 + j, 1 $;j $; m 3 f+ (v4j) 4 + m 1 + m 2 + m 3 + j, 1 $;j $; m 3 54

f+ (v 1 ) f (V 4J V 1 ) + f (v 1, V 2 ) + 2)(V j,v i ;) (m 12 + 9m 1 + 10) /2 m, j1 m2 f+ (V 2 ) f (V 1, V 2 ) + f (V 2, V 3 ) + 2)(V 2, V 2j ) (ml + 9m 2 + 2m 1 m 2 + 6) /2 f+ (V 3 ) f (V 21 V 3 ) + f (V 3, V 4 ) + "L f (V 31 V 3j ) (m 2 3 + 9m 3 + 2m 1 m 3 + 2m 2 m 3 + 10) / 2 f+ (v4) f(v 31 V 4 ) +f(v 4, V 1 ) + "L f (V 4,V 4j ) (m 4 2 + 9m 4 + 2m 1 m 4 +2m 2 m 4 + 2m 3 m 4 +14) /2 Clearly f+ (u) -:j:: f+ (v) \fu,v E V (G). Hence G is antimagic. j1 m3 j1 m. j1 Sllustration: C 4 with m 1 5, m 2 6, m 3 7, m 4 8. 55

Next we consider the graph < Cn,t > consisting of t copies of C n sharing a vertex in common. Then we!lave the following theorem. Theorem 3. 2.5: < C n, t> is antimagic V t ~ 1 and V n ~ 3 Proof: <Cn, t > has tn - (t - 1) vertices and (tn) edges. Let vi j, i 1,2,...,n, j 1,2,...t be the vertices of < Cn, t >. Let w V1j 1 ~ j ~ t be the common vertex. Let {ejj, 11 ~j~t, 1 ~i~n} {Vjjvj+111 ~j~t, 1~i~n-1} and en j vn j w Define fan E «C n, t» as follows f( ej j) f( en j) Now f+ (w) i + U- 1) n, n J, f(eij) +f(enj), 1~i~n-1, 1~j~t. j1,2,...,t. I I 2)(e jj ) +2)(e nj ) j1 j1 I I L[1+(j-1)n]+L(nj) nt2 + t j1 j1 f (ei-1j) + f (ejj) (i-1) + ( j-1) n + i + U-1) n 2i - 1 + 2n U-1), 2 ~ i ~ n, 1~j~t. Clearly f+ (u) "* f+ (v) V u, v in V( < Cn, t> ) Hence < C n, t > is antimagic. 56

3llustration < C a, 6 > is antimagic. 10 15 20 11 14 Theorem 3.2.6: The grid C n x C n is antimagic for all even n. Proof: C n x C n has 2n 2 edges and n 2 vertices. Let the vertices of C n x C n be as follows, a 1 1 a 1 2 a 21 a 22 a 1 n a 2 n 57

Here each row and each column are repectively adjacent and a 1 j, an j, 1 ~ i ~ n are adjacent, aj l' aj n, 1~ i ~ n are adjacent. Define the labeling f on E (G) as follows, From * it is clear that 3i -2, (3n - 2) + (2i - 1), 1 ~ i ~ n. (3n - 2) + (r - 3) (2n -1) + 2i-1, 1 ~ i ~ n, 4 ~ r~ n.. * f (an n - l' an n ) 3n - 2 + (n - 3) (2n - 1) + (2n - 1) To label the remaining edges Set K3n - 2 + (n - 2) (2n - 1) 3n - 2 + (n - 2) (2n - 1) (i i) f(aj l' aj+11) 3i -1, 1~i~n-1 f (aj 2, aj +1 2) I f (aj 3 ' a i + 1 3) I 3i, (3n - 3) + 3 + (i - 1) 2, 1~i~n-1 1~i~n-1 f (aj r ' aj + 1r ), (3n - 3) + (r - 3) (2n -4) + (r - 2)3+(i -1) 2, 1~ i ~ n - 1, 4 ~ r ~ n. K+ i, (iv) f(ai1,ajn) K + n + [n- (i -1)], 1 ~ i ~ n. The set of edge labels are {1,2,...,2n 2 } 2 (K + n + 2) 58

Cff--.~3 ~~~wpf~ f+ (a 1 2) 3n + K + 5 f+ (a 1 j) K + ( 6 U- 2 ) + 5) n - (2j - 6), 3~ j ~ n - 1. Also f+ (aj 1) 2n + K + 8 (i - 1) +2, 2 ~ i ~ n-1 Also f+ (an j) (6 U- 1) + 5) n + K - (2j + 2), 2~ j ~ n-1 Thus f+ (a 1 n ) 4n 2 + 2k - 5n + 5 f+ (a 2 n ) K + 6n 2-10n + 11) f+ (a 3 n) K + 6n 2-10n + 16 f+ (a 4 n) K + 6n 2-10n + 21 Hence f+ (aj n) K + 6n 2-10n + 5i +1, 2 ~i~n-1 Also f+ (aj 2) 3n + 11 (i - 1) + 3, 2 ~i~n-1 Similarly f+ (aj 3) 14n + 8 (i - 1) -5, 2 ~i~n-1 f+ (aj 4) 22n + 8 (i-1) - 9, 2 ~i~n-1 f+ (aj 5) 30n + 8 (i-1) -13, 2 ~i~n-1 f+ (aj 6) 38n + 8 (i-1) -17, 2 ~i~n-1 f+ (aj 7) 46n + 8 (i-1) -21, 2 ~ i ~ n -1 In general f+ (aij) (8j -10) n + 8 (i -1) - (4U -1)-3), 2~i~n-1, 3~j~n-1. Thus f+ (u) 1:- f+ (v) V U,V E V(Cn x Cn ) :. C n x C n is antimagic 0 59

Sllustration: Cs X C s is antimagic. 113./ 128 ------------------ --... 53 3_--- 24 ~9 127 ( ~ ~~-t 23 I- 38 I 5 "- j ~_6~j 83 '- I ~-- ~---'- 4 25 40 55 70 85 100 ~-- - 26 41 e- 56 171 _- I- 1--- - 86 ---- 126 ---I-- I-- _._.-- I... 7 27 42 57 72 87 102 8 9 28 r!3 ~8 88 ~-- -- 1--- - - 1-- -- 125 - -.---- 29 --"'-. '- 10 44 59 74 89 104 30 ~5 60 I- 75 90 I-- 1--- - - -- - 124 --- - - 11 1~_...- 31 - '- 13 46 61 76 91 106 f-- -- I-- -123 l- I- ---I- -- - ---. 14 114 15 115 U 116 47...--HI 62 118 77 119 92 120 - -- 16 33 48 63 78 93 108 \17 _ 18..-- -.--- 1--- - 1---_ I-...-- 34 122 49 64 ~-- 94 ~ /.-- 19 35 50 65 80 95 110 20.-- -- \ ~--\ 3L_-\-, L- 1 ~L_-\ 121 r!1 ~ 96 ~ 22 37 52 67 82 97 112 99 101 103 105 107 109 111 3.3. On Path Related Graphs related graphs. In this section we study the role of antimagic labeling on path Theorem 3. 3.1: The graph G Pn + K t is antimagic. Proof: Let v 1 ' v 2,..., v n be the n vertices of P n. Let w 1 ' w 2,.,Wt be the t isolated vertices of K t Join v 1 ' v 2,...,v n to each wi, 1~ i ~ t. The newly obtained graph is G Pn + ~. Clearly G has (n + t) vertices and (n - 1 + t n) edges. 60

Define f on E (G) by f (Vj, Vj+1) f (v w) J, I j, 1 ~ j ~ n - 1 i n + U- 1), 1 ~ j ~ n, 1 ~ i ~ t. Now I f (V1,V 2 ) + ~)(V1Wi) i1 [nt 2 + nt + 2] / 2 [n t 2 + 3nt + 2n - 2t - 2] / 2 t f(vj_1,vj) +f(vj,vj + 1) + 'Lf(vj,w i ) i1 I (j -1 ) + j + 'L(in + j -1) i2 [nt 2 + (n - 2)t + 2 + 2j (t + 2)] / 2, 2 ~ j ~ n, n 'Lf(vj,w j ) j1 n 'L(in+j-1) j1 (n 2 (2i + 1) - n) /2, 1~ i ~ t. Clearly f+ (u) I; f+ (v) V U, V E V (G). Hence G is antimagic. 0 $llustration: P3 + K s is antimagic. Then immediately we get the following corollary. 61

Corollary 3.3.1: Fans are antimagic. Proof: By theorem 3.1, Pn + K t is antimagic. When t 1, P n + K 1 P n + K 1 is antimagic. o Construction 3. 3.1: Let F n be a fan. Take m isomorphic copies of F n Let w be an isolated vertex. Then the one point union of the centre vertex of each F i with w. is denoted as F~m). Note that F~m) has 2mn edges and m (n +1) + 1 vertices Theorem 3.3.3: The graph F~m) is antimagic. Proof: Let V [F~m)] {Uj} U {Vji} U {w}, 1 $ j $ m, 1 $ i $ n. Label the edges of F~m) f (Vj, Vj j) f ( Vj i' Vj i +1 ) f (Vj, w) as follows. 2n (j - 1) - ( j - 1) + i, 1 $ i $ n, 1$ j $ m. (2j-1)n-(j-1)+i, 1 $i$n-1,1 $j$m. 2mn - (j - 1), 1 $ j $ m. Now f+ (w) n Lf(Vj,w) j1 [m 2 (4n - 1) + m]./2 f (Vj, Vj 1 ) + f (Vj 1,' Vj 2 ) 2n (j - 1) - (j - 1) + 1 + (2j - 1) n - (j-1) + 1 4-3n + (4n - 2) j, 1 $ j $ m f (Vj, Vj n) + f (Vj n -1,Vj n) 1-n+j(4n-2), 1 $j$m t f( V j ' vjj +f( w, Vj) j 1 (j -1) (2n2 - n + 1) + 2mn +(1/2) [n (n + 1)], 1 $j $ m 62

f (Vj _ 1,Vj j) + f (Vj j. Vj j + 1 ) + f (Vj, Vj j ) (6n - 3) j - 4n + 2 + 3i, 1 ~ j ~ m, Clearly f+ (u) 7:- f+ (v) \j U, v in V (F~m»). 2 ~ i ~ n-1. Hence F~m) is antimagic. o 3Ilustration: F~5) is antimagic. V45 V 21 v 1 v 44 40 v22 v 43 v 42 v 4 W 39 v23 37 v 2 38 v 3 v 24 v 41 Theorem 3 3.4: Pn x C 3 is antimagic for even n. Proof: The graph G Pn x C 3 has 3n vertices and 6n - 3 edges. Let the vertices of G be { aj j, 11 ~ i ~ 3, 1 ~ j ~ n}. Let the edges of G be {aj j aj j+1 I 1 ~ i ~ 3, 1 ~ j ~ n - 1} u 63

{ a1 j a2 j I 1:::; i :::; n} u {a 1 j a 3 j I 1:::; i :::; n} u {a 2 j a 3 j I 1:::; i :::; n}. Suppose n. 2r. Define fan E (G) as follows. f (a12i-l' a12i) 6i -5, 1 :::; i :::; r f(a12j a12j+1) 6i, 1:::;i :::; r - 1, f(a2 2i - l'a2 2i ) 6i -4, 1 :::; i :::; r f (a2 2i, a2 2i + 1) 6i -1, 1 :::; i :::; r - 1 f (a3 2i - 1,a3 2i) 6i -3, 1 :::; i :::; r f (a3 2i, a3 2i + 1) 6i -2, 1 :::; i :::; r - 1 f (a1 2i a. ) 3 (n - 1) + i, 1 :::; i :::; r - l' 321-1 3 (2r - 1) + i, 1:::;i :::; r f (a12i,a3 2i ) 3 (2r - 1) + r + i, 1 :::; i :::; r f (a1j. a2 i ) (4n - 2) + 2(i - 1), 1 :::; i... n f (a2j, a3i) (4n - 1) + 2 (i - 1), 1 :::; i :::; n. Clearly the set of edge values are {1, 2,..., 6n - 3}. 1 + (4n - 2) + 3(n - 1) + 1 7n - 3 (4n - 2) + (4n - 1) + 6-4 8n-1 (4n-1)+(6-3)+3(n-1)+1 7n 64

f (a 1 n - 1 ) + f (a 1 n, a 2 n) + f ( a 1 n, a 3 n) f (a 1 2r _ 1 a 1 2r )+(4n-2) + 2(n-2)+3(2r - 1) + 2r,, 14r + 6n -12 26 r-12 4n - 2 + 2 (n - 1) + 4n - 1 + 2 (n - 1) + 6r - 4 12n + 6r - 11 30 r - 11 [3 (2r - 1) + r + r] + [(4n - 1) + 2 (n -1 )]+ 6r - 3 (6r - 3 + 2r) + 6n - 3 + 6r - 3 14r + 6n - 9 26 r - 9 + f(a 1 2i' a 32 i) 6i - 5 + 6i + 4n - 2 + 2 (i - 1) + 3 (2r - 1) +r + i 17i + 4n + 7r -12 15r + 17i -12, 1 ~ i ~ r - 1 f (a 1 2i' a 1 2i + 1) + f (a1 2i + l' a1 2i + 2 ) + f (a 1 2i + l' a 22 i + 1 ) + f (a1 2i+1' a32i+1) [6i + 6 (i + 1) - 5] + 4n - 2 + 2 (2i + 1-1) + 3 (n - 1) + (i +1) 17i + 7n 14r+17i, 1~i~ r-1 f (a 3 2i _ l' a 3 2i ) + f (a3 2i' a3 2i + 1) + f (a 22 i. a 32 i) + f (a1 2i> a3 2i ) 65

6i-3 + 6i-2 + 4n -1 + 2(2i -1) + 3(2n-1) + r + i 15r + 17i -11 f (a 3 2i, a 3 2i+1) + f (a 3 2i+1' a 3 2i+2) + f (a 2 2i+1' a3 2i+1) + f (a 1 2i+1' a3 2i+1) [6i - 2 + 6 (i + 1) - 3] + 4n - 1 + 2 (2i + 1-1) + 3 (2r - 1) + i + 1 ie.. 12i + 4-3 + 4n - 1 + 4i + 6r - 3 + i +1 17i-2+4n+6r 14r+17i-2. 1~i~ r-1 f (a 2 2i _ 1. a 2 2i ) + f (a 2 2i. a 2 2i + 1) + f (a 1 2i. a 2 2i) + f (a 2 2i. a 32 i) 6i - 4 + 6i - 1 + 4n - 2 + 2 (2i - 1) + 4n - 1 + 2 (2i - 1) 12i - 5 + 4n - 2 + 4i - 2 + 4n - 1 + 4i 2 20i + 8n -12 16r + 20i -12, 1 ~ i ~ r - 1 f (a 2 i> a 2 2i + 1) + f (a 2 2i + l' a3 2i + 2) + f (a 1 2i + 1. a 2 2i + 1) + f (a2 2i + 1. a3 2i +1) 6i -1 + 6 (i +1) - 4 + 4n - 2 + 2 (2i +1-1) + (4n -1) + 2 (2i + 1-1) 12i + 1 + 4n - 2 + 4i + 4n - 1 + 4i 20i + 8n - 2 16r + 20i - 2. 1 ~ i ~ r - 1 Clearly f+ (u) :f- f+ (v) V U V E V (G). Hence G is antimagic. 0 66

.9l1ustration: P12 X C 3 is antimagic 45 39 32 66 44 29 38 26 62 43 23 37 20 58 42 36 54 12 41 7 35 8 50 6 40 5 34 2 117 Definition 3.3.1: Triangular snake is a connected graph in which all blocks are triangles and the block-cut point graph is a path. If it has n blocks then it is said to be of length n. 67

Theorem 3.3.5: Triangular snakes are antimagic. Proof: Let R n be a triangular snake with length n. Let V(R n ) {aj I0 ~ i ~ n} u {bj1' 1 ~ i ~ n} E(R n ) {ajai+1' 10 ~i~n-1} u {aj bj + 11 1 0 ~i~n-1} u {aj b i1 1~ i ~ n}. Define f on E (R n ) as follows, f(aj,ai+1) i+1, 0 ~i~n-1 f(aj,b i + 11 ) (n+1)+2i, 0 ~i~n-1 f(aj,b j1 ) n+2+2(i-1), 1~i~n. With the above labeling the set of edge values are {1.2,...,3n}. Now f+ (a o ) f (a o b 11 ) + (a o a1) (n+1)+1 n + 2 f (a n -1. an) + f (an. bn1 ) (n -1) + n + 2 + 2 (n - 1) 4n f(aj, aj1) +f(aj, aj + 1) +f(ai1' bi1 ) +f(aj, bj + 11) (i-1 +1)+(i+1)+(n+2)+2(i-1 )+n+1 +2i 2n + 6i +2, 1 ~ i ~ n - 1 f(aj, b i1 ) +f(aj_1' bi1 ) n + 2 + 2 (i-1) n + 1 + 2 (i - 1) 2n - 1 + 4i, 1 ~ i ~ n. Clearly f+ (u) ~ f+ (v) V U,V E V (R n ). Hence Rn is antimagic. 0 311ustration: R s is antimagic. b b 21 b31 b 41 bs1 11 6 7 8 68

Theorem 3.3.6: Quadrilateral snakes are antimagic. Proof: Let Q n stand for the quadrilateral snake whose length is n. {aj I 0 :s; i :s; n} u {bj l' b j 2 I 1:s; i :s; n } { aj,aj+1 I 0 :s; i :s; n-1 } u {bj l' bj 2 I 1 :s; i :s; n } u { aj, bj + 11 I 0 :s; i :s; n-1 } u {ai, bj 2I 1:s; i :s; n } Clearly Q n has (3n + 1) vertices and 4n edges. Label the 'edges of Q n as follows, f(aj,aj+1) i +1, o :s;i:s;n-1 f (b j l' b j2) (n + 2) + (i -1)3, f(aj,b i + 11 ) (n + 1) + 3i, f (aj, bj 2) (n +3) + (i -1)3, o :s;i:s;n-1 1 :s; i :s; n. Then the set of edge labels are {1,2,3,...,4n}. Now f+ (aj) f (aj, aj + 1) + f (aj _ l' aj) + f (ai, bj + 1,1) + f (aj, b j2) (i +1) + (n + 1) + 3i + (n + 3) + (i - 1)3 + (i -1) +1 2n + 8i + 2, f (a o, a 1 ) + f (a o, b11 ) 1 + (n + 1) n + 2 f (an _ 1 ' an) + f (an, bn2) (n - 1 +1) + (n + 3) + (n -1) 3 1 :s; i :s; n - 1. n + n + 3 + 3n - 3 5n f (aj _ l' b i1 ) + f (b j1, bj2) (n + 1) + 3 (i - 1) + (n + 2) + 3 (i - 1) 2n + 3 + 6i - 6 2n + 6i - 3, 1:s; i :s; n 69

f(aj, b i2 )+f(b i1, bj2) n + 3 + (i - 1) 3 + (n + 2) + (i - 1) 3 2n + 5 + 6i - 6 (2n - 1) + 6i 1 ~ i ~ n Then f+ (u) I: f+ (v) V U, V EV (Q n ). Hence Q n is antimagic Vn. 0.9Ilustration: Q 6 is antimagic. b 11 8 b 12 b21 11 ~2 b 31 14 b b! 32 \/ \} \f 41 S/::' 23\_ 17;;:" 20 a o 1 a 1 2 a 3 a 3 4 a 4 5 a 6 96 2 5 Theorem 3. 3.7: K 4 snakes are antimagi~. Proof: Let R n stand for the K 4 snake whose length is n. R n has 6n edges and 3n + 1 vertices. {aj I 0 ~ i ~ n} u {bi1, bi2, I 1 ~ i ~ n }. { aj, ai+1 I0 ~ i ~ n -1} u {bj1, bj2 I 1 ~ i ~ n } u { aj, b i + 1 1I0 ~ i ~ n -1} u {aj, bi2 11 ~ i ~ n } u { aj, bj+ 1 2 I0 ~ i ~ n - 1} u {ai, bj, 111 ~ i ~ n }. Label the edges of R n as follows, f(aj,ai1) 6i+1, f (bj l' bj 2) 6i -2, f(aj, bj + 11) 6i + 2, o ~i~n-1 o sisn-1 f (aj, bj 2) 6i, 70

r.ql~... ~/I~ f(aj,b i + 12 ) 6i+5, 0 ~i~n-1 f(aj,bj1 ) 6i-3, 1~i~n. The set of edge values are {1,2,...,6n}. f+ (aj) f (aj, aj +1) + f (aj, bj +11) + f (aj, bj 2) + f (ai, bj +1 2 ) + f (ai, bj 1) + f (aj -1' aj) 6i +1 + 6i + 2 + 6i + 6i + 5 + 6i -3 + 6 (i - 1) + 1 36i, 1 ~ i ~ n - 1 f+ (a o ) f (a o, a 1 ) + f (a o, b 1 2) + f (ao, b1 1) 1+5+2 8 f+ (b j 2) f (b j l' bj 2) + f (aj, b i 2) + f (aj _ l' bj 2) 6i - 2 + 6i + 6 (i - 1) + 5 18 i - 3, 1 ~ i ~ n f+(b i1 ) f(bj1,bj2)+f(aj,bj1)+f(aj_1,bj1) 6i - 2 + 6i - 3 + 6 (i -1) + 2 18 i-9, 1~i~n f+ (an) f (an _1' an) +f( an, bn2 ) +f(an. bn 1) 6 (n-1) + 1 + 6n + 6n-3 18n-8 Hence f+ (u) ::j:. f+ (v) 'tj U,V E V (R n ). Thus Rn is antimagic. 0..9llustration: R 4 is antimagic. 71

Theorem 3.3.8: nk 2 is not antimagic. Proof: nk 2 has n edges and 2n vertices. If we label the edges as 1,2,...,n, then vertices incident with each edge has the same sum. Hence it is not antimagic. o 3.4. On Star Related Graphs In this section we give some results related to antimagic labeling on stars and the graphs which contain stars. Theorem 3.4.1: The graph G K 1 n + K t is antimagic., be the vertices of the star K 1 nand w 1 ', Define fan E (G) by f (va' Vj) j, 1 ~j ~n f (va' wk) (n +1) k, 1 ~k ~t f (Vj, wk) k (n +1) +j, 1 ~j ~ n, 1 ~k ~t. Now f+ (va) L f( v0' v j) +L f( v0' Wk ) n j1 k1 (n+1) (t2 + t + n) /2 f+ (Vj) f (va' Vj) +L f( v!' w k ) k1 t t [(n + 1) t 2 + t (n + 2j + 1) + 2j] /2 For 1 ~k ~ t, f+ (wk) f(v o, wk) +if(vi'w k ) 11 (n+1) [2k (k + 1) + n] /2 72

Clearly f+ (u) 7:- Hence G is antimagic. f+ (v) V u,v in V (G). o $llustration: K 1 5 +K 2 is antimagic., Theorem 3. 4.2: Book graph B n is antimagic for all n. the common vertices of Bn B n has 2 (n + 1) vertices and 3n + 1 edges. Label the edges of B n as follows, f (Xc,Yo) 1, f (aj, bj) 3i, 1 ~i ~ n f (x o, aj) 3i-1 I 1 ~i ~ n f (Yo, bj) 3i + 1 I 1 ~i ~ n. Then the set of edge labels are {1, 2,...,3n+1} 73

0:~3 ~.!/.,/UlfjM;/ wp/,v Now f+ (Xc) m :L f(x O,a 1 ) + f (Xc, Yo) jl 1 + [3 n (n + 1)] /2 - n [2 + 3n 2 + 3n - 2n] / 2 [3n 2 + n + 2] / 2 f+ (aj) f (ai, b j ) + f (Xo, aj) 3i + 3i-1 6i -1, 1 ~i ~ n f+ (b i ) f (ai. b j) + f (Yo, bj) 3i + 3i +1 6i + 1, 1 ~i ~ n. f+ (Yo) f (xo, Yo) +:L f(yo,b 1 ) il m 1 + n + [3 n (n +1)] /2 [2 + 2n + 3n 2 + 3n] / 2 [3n 2 + 5n + 2] / 2 Clearly f+ (u) t:- f+ (v) \;j u, V E V (B n ) Hence B n is antimagic. 0..9l1ustration: B 5 is antimagic. 1 2 a1 5 a2 ---::::: 14 8 a3 11 Cl4... as 3 6 9 12 15 4 b1 7 ~ b3 ~ 10 13 b 4 16 bs 74

Theorem 3.4.3: A book B n with n-pentegonal pages is antimagic Proof: Let (Xc, bj, ai, Cj, Yo) be the ith page of B n, 1 :::; i :::; n. B n with n-pentegonal pages has 4n + 1 edges and 3n + 2 vertices. Define f on E (B n ) as follows, f (xo. Yo) 4n + 1 f (xo. bj) 4i - 3, 1:::; i :::; n f (bj, aj) 4i - 2, 1:::; i :::; n f (aj, Cj) 4i - 1, 1:::; i :::; n f (Cj, Yo) 4i, 1 :::; i :::; n. Then the set of edge labels are {1,2,... An + 1} f (x o, Yo) +i)(x o,b 1 ) n i-1 (4n 2 + 6n + 2) /2 f (x o b j ) + f (b j, aj) 4i - 3 + 4i - 2 8i - 5, 1:::; i :::; n. f (b j aj) + f (aj Cj) 4i - 2 + 4i-1 8i - 3, 1:::; i :::; n f (ai, Cj) + f (Ci' Yo) 4i - 1 + 4i 8i -1, 1 :::; i :::; n. n f (Xc, Yo) +L)(c i, Yo) i 1 i1 4n+1 + 1/2 [4n (n + 1)] [8n + 2 + 4n2 + 4n] /2 4n 2 +12n + 2] /2 Clearly f+ (u) :f. f+ (v) V U,V E V (B n ). Thus Bn is antimagic V n. 0 75

$llustration: 21 Theorem 3.4.4: Let B n be a book graph with n square pages. Let Hn be a graph obtained by subdividing the free edges of the pages exactly once. Then H n is antimagic. Proof: Let xo,yo be the common edge. Let (X o, Ii, ai, mj, bj, ni, yo) be the jth page, 1 si s n. H n contain 6n + 1 edges and 5n + 2 vertices. Label the edges of H n as follows, f (xo, Yo) 6n + 1 f (Xc, Ij) 6i -5, 1 si sn f (Ii, aj) 6i - 4, 1 si sn f (aj, mj) 6i -3, 1 si sn f (mi, bj) 6i - 2, 1 si s n f (b i, ni) 6i -1, 1 si s n f (Yo, nj) 6i, 1 si s n. 76

(ff.,~/3 The set of edge labels are {1,2,...,6n +1} n Now f+ (x o ) f (Xc, Yo) + :Lf(xo,l,) ;1 6n + 1 + 112 [6 n (n + 1)] - 5n [12n + 2 + 6n 2 + 6n - 10 n] 12 (6n 2 + 8n + 2) 12 f+ (I j) f (x o, Ij) + f (Ii, aj) 6i - 5 + 6i - 4 12i - 9, 1 ~i ~ n f+ (aj) f (Ii, aj) + f (aj, mj) 6i - 4 + 6i - 3 12i - 7, 1 ~i ~ n f+ (mj) f (aj, mj) + f (mj, bj) 6i - 3 + 6i - 2 12i - 5, 1 ~i ~n f+ (b j) f (mi, bj) + f (b j, nj) 6i - 2 + 6i - 1 12i - 3, 1 ~i ~ n f+ (nj) f (bj, nj) + f (nj, Yo) 6i -1 + 6i 12i - 1, 1 ~i ~n f+ (Yo) f (Xc, Yo) + :Lf(yo,n i ) n i1 6n + 1 + [6n (n + 1) ] 12 12n + 2 + 6n 2 + 6n/2 [6n 2 + 18n + 2] 12, 1 ~i ~ n Clearly f+ (u) :1 f+ (v) V U,V E V (H n ) Hence H n is antimagic for all n. 77 0

..9'llustration: H 4 is antimagic. Theorem 4.5: C n u K 1 m is antimagic for all m and for all odd n., Proof: Let v 1 v 2,..., v n be the vertices of C n. Assume that n is odd. Let ej {Vj vi + 1 I 1 sis n} where v n + 1 v 1. be the edges of the cycle C n Label edges of K 1 m by 1,2,...,m., Label the cycle edges by f( ej) m + i, 1 sis n. Then the set of edge labels are {1,2,..., m + n}. Then all vertex labels are distinct. The central vertex has the sum 1 + 2 +... + m m( m + 1) / 2. Each end vertex has the sum as edge labels. Also f+(u) * f+(v) \if u,v in V(C n U K 1, m ). Thus Cn U K1, m is antimagic. o 78

.JIlustration: C s u K 1 6 is antimagic. J 1 79