CHAPTER - 3 SOME DOMINATION PARAMETERS OF THE DIVISOR CAYLEY GRAPH

Transcription

1 CHAPTER - 3 SOME DOMINATION PARAMETERS OF THE DIVISOR CAYLEY GRAPH

2 Madhavi [231 has introduced the concept of divisor Cayley graph and studied some of its properties. She also gave methods of enumeration of disjoint Hamilton cycles. In this Chapter we determined certain domination sets. namely, vertex cover, edge cover, independent set. makh'ig and related domination parameters of the divisor Cayley graph. In section 3.1 we introduce the concept of divisor Cayley graph and briefly sketch certain basic properties needed for our discussion, which were studied in detail in [23]. 3.1 DIVISOR CAYLEY GRAPH AN11 ITS PROPERTIES : Let n 2 I be an integer and let S be the set of divisors of n. Then the set S* = f s, n - s I s E S } is a symmetric subset of the group ( Z,,, 8 ), the additive abelian group of integers modulo n. The Cayley graph of ( ) associated with the above symmetric subset S* is called the divisor of Cayiey graph and it is denoted by G( Z,, D). That is, the graph Ci(Z,, D) is the graph whose vertex set is V -. { 0,1,2, , n-1 } and the edge scl E is the set of all ordered pairs of vertices x, y such that either x - y E s*, or, y - x E s*. For n = 2,3,4,6,10,11 the divisor Cayley graphs are given at the end ofthe Chapter in fig: 3,1, 3.2,3.3, 3.4,3.5 and 3.6,

3 It is immediate from the Theorem in Chapter I that the graph nls'i 2 G( Z, D) is ls'l- regular, and the number of edges in G( Z, D) is -. The divisor Cayley graph G(Z,,D) has following properties. (i) The graph G( Z,, D) is Hamiltonian and hence connected. (Sce, Lemma of [23]). (ii) (a) Degree of each vertex in G( Z, D) is odd and if, and only if, n is even. (b) Degree of each vertex in G( Z, D) is even and if, and only if, n is odd. ( See, Lemma of [23] ). (iii) The graph G( Z, D) is not bipartite. (See, Lemma of P31). (iv) If n is a prime, then the graph G( Z,, D) is the outer Hamilton cycle. ( See, Lemma of [23] ). (v) The divisor Cayley graph G( Z, D) can be decomposed into edge disjoint Hamilton cycles if, and only if, n is odd. (See, Theorem of [23] ). Remark: For n=2,3,4,6, the symmetric set s*= {I, 2, , n - I). Since each of these graphs is IS*/ - regular. This shows that each vertex of G( Z, D) is adjacent to all other vertices of G( Z,,D). So G( Z, D) is a complete graph for n = 2, 3, 4,6. This can see in figures at the end of the Chapter.

4 3.2 WJU'EX DOMINATION IN DIVISOR CAYLEY GRAPH In this section the vertex cover and vertex covering number of the divisor Cayley graph G( Z,, D) are determined. Definition : Let G be a graph. A vertex \ md an edge e of Ci nre said to cover each other if they are incident. A set S of vertices which co\ers all the edges of a graph G is called vertex cover of 6, in the sense thsr every edge of G is incident with some vertex in S A minimum vertex cover is the one with minimull cardinality. Definition : : The cardinality of a minimum vertex cover of a graph G is called the covering number and is denoted by P(G). Remark : : The vertex covering number is n - 1, for n - 2, 3, 4, 6, since G( Z,, D) is a complete graph for n = 2,3,4,6 (See page 94 ot'l l HI 1. Theorem : 3.2.4: Let n > 5 be a prime. Then the minimum vertex covcr of G( Z,.D) is {1,3,5,... n}. - Proof : Let us consider the divisor Cayley graph of G( Z,,D).

5 Suppose that n 2 5 is a prime. Then 1 is the only divisor of n other than n so that S' = { 1, n - I }. Let V, = 11, 3, 5,..., n }. Let e be an edge of G( Z, e = (r, s), where 1 I; r, s 5 n and either r - s, or, s - r is in s*. D). Then We claim that one of r and s is in VI. For, if r e VI and s e VI, then r = 2a and s = 2P, for some positive integers a and p so that r - s = 2 (a- p) < n - 1 +! s*. Similarly s - r = 2 (P-a) < n -1 e s*. (since n - 1 is the largest even number < n). This is a contradiction. So atleast one of r and s must belong to VI, which implies that V, is a vertex cover of G( Z,, D). Let us now show that Vl is the minimal vertex cover of G( Z,, D). For this, consider the set V, - {i}, for any i ~v,. Then i = 2t-1 for some positive integer t. Now the edge (2t-1, 2t), ( this is an edge, since 2t - (2t-1) = 1 E S' ), is not covered by the set V, - {i), since 2t is even so that 2t e Vl - (ij and also 2t - 1 = i +! VI - {ij. So VI is the minimum vertex cover of G( Z,, D). The following Corollary is immediate from the Theorem. Corollarv : 3.2.5: If n 2 5 is a prime, then the vertex covering number of G( Z,, D) is ntl given by P( G( Z, D)) = 7;-.

6 Theorem : 3.2.6: Let n be not a prime and n 2 8. Let do be the smallest positive integer that does not divide n. (i) The set VO = V - Uo, where IJ~={kd~ll<kSr sucht~~atr&f~*andfort<r.tde s*] is a vertex cover of G( Z,,. D). (ii) For any positive integer d -. 6 that does not divide n, let Ud={~Iu=dtkdo,ifusnnndu=d+kd~-n. if u>n. 0 5 k 5 r - 1 } then the set Vd = V - Ud is also a vertex: cover of G( Z". D). (iii) I If, I = /Vdj - Proof : (i) Let us consider the divisor Cayley graph G( Z,,,D). Suppose [hat n is not a prime and n 2 8. Let do he the s~~lallest positive integer that does not divide n. Let Vo = V - UD, where Uo={kdo/ ~~k<rsuchthatrd~~~*andfort~r,td,e~*}. Let e be an edge of G( Z, D). Then e = (i, i), where I r; i, j 5 n and either i - j, or, j -i is in s'. We claim that one of i and j is in Vo. For, if iccvo and j e Vo, then i = & and j = tdo, 1 5 s, t 5 r. Suppose i > j, then i - j = & - tdo = (3-t)& < rdo. So (S - t) do e s*. Hence i-j (E. s'. Similarly, if i < j, then j - i e S*. This shows e = ( i, j ) is not an edge of G( Z, D),

7 which is a contradiction. So, atleast one of i and j must belong to Vo which implies that Vo is a vertex cover of G( Z, D). (ii) Let d be the positive integer that does not divide n. LetVduV-Ud. where Ud= (u/u=dtkdoifu5nandu=d+kdo-n ifu>n,o<ksr-1). Let e be an edge of G( Z,,D). Then e = (i, j) where 1 5 i, j 5 n and either i - j, or, j - i is in s*. We claim that one of i and j is in Vd. Suppose ist Vd and j e Vd Here three cases will arise. Case i : Let i, j 5 n. Then i=d + s&, j -= d + tdo, for some s and t, Oss, t~r-1,nowi-j=(d+sdo)-(d+t&)=sdo-t&=(s-t)& e s*, since s - t r. Similarly i < j, j - i = ( d + tdo) - (d + sdo) = tdo - sdo = (t - s)do e s*, since t - s < r. This shows e = (i, j) is not an edge of G( Z, D), which is a contradiction. So, atleast one of i and j must belong to Vd, which implies that Vd is a vertex cover of G( Z,,,D). - Case ii : Let i, j>nandi>j.theni=(d+sdo)-n,j=(dttdo)-n,for somesandt,ost<si r- 1.Nowi-j=[(d+s&)-n]-[(d+tdo)-n] =s&-tdo=(s-t)do E s'sinces-t<r. Thisshowse=(i,j)isnotan edge of G( Z,, D), which is a contradiction. So, atleast one of i and j must belong to Vd, which implies that Vd is a vertex cover of G( Z,, D).

8 Case iii : Let isnandjsn. Then i=d+sx&,andj=d+t&-n,tas.now i - j =(d + sdo) - ( d + tq - n ) = n - (t - s )Q. We claim b t n-(t -s )do+? s'. For, if n - ( t - s ) d, e s'. nlerl n - [ n - ( t - s ) do] E s*, or, ( t - s ) do E s', a contradiction since t - s < r. So, i - j +? s*. This shows e = (i, j) is not an edge of G( &, I)), which is a contradiction. So, atleast one of i and j must belong to Vd which implies that Vd is a vettex cover of G( Z", D). (iii) Since / li,,, I = ludl, we have / Ifd<, 1 = IVdl. The following corollary is immediate from the Theorem. Corollary : 3.2.7: Let n be [lot a prime and n ;r: 8.Let do he the smallesc positive integer that does not divide n.then the vertex covering number of Ci( Z,, D) is given by P( G( Z, D)) = n - 1, where I = /U, ( and - Proof : 11, = ( kdo k 5 r such that rdo E S* and for t.: r, tdo e S' }. From part (ii) of Theorem 3.2.6, we conclude that for every integer d z 0, which is not a divisor of n, we gel a vertex cover Vd, namely, Vd = V Ud where Ud= { ~/u=d+kd,ifu~nandu=dtk~-nifu>n,o4k5r-1 } and &, is the smallest positive integer that does not divide n.

9 Also from part (iii) of the Theorem we get I V, / = IVdl. So I V, I is the vertex covering number of G( &, D), which is n - I, where I = IU INDEPENDENT SET OF DIVISOR CAYLEY GRAPH Definition :3,3.1 A subset S of V is called an independent set of a graph G if no two vertices of S are adjacent in G. Definition : The number of vertices in the largest independent set of a graph G is called the independence number and it is denoted by a (G). A single vertex of any graph G constitutes an independent set Theorem: Let n = 5, or, n r 7 and let do be the smallest positive integer that does not divide n. For any integer t 2 1, the set of vertices Vt in G(Z,, D), which is of the form V, = { t + rd, 10 I r I k - 1, where k is the least positive integer such that rd,& S* and t + rdo < n 1, is an independent set of ~(z,,d). &Q& Let n = 5, or, n 2 7 and let do be the smallest positive integer that does not divide n. For any integer t 2 1, consider the set V, = { t + rdo / 0 2 r 5 k - 1, where k is the least positive integer such that rdocc S* and t+rdo<n}.

10 Let t + Ido. t + mdo be any two distinct vertices in V,. Then 1 t m. For definiteness let us take m < I. Then I, m I k - 1, so that I - m < k - 1. Sobythedefinitionof~,,(i-m)& e s*. Now(t+/&)-(t+m&)= ( I - m ) do +z s'. So, t + I do and t t indo, If m are not adjacent and thus V, is an independent set. Next, let s be the smallest integer such that t < s and s es V,. Consider the set V, = { s + rdo 1 0 r. r s k - 1, where k is the least positive integer such that r4 +z S* and s + rdo < n ). As in the case of V,, one can show that V, is also an independen1 set. Proceeding in this way we obtain a finite number of independent sets. say, V,, V2,..., V,. Now we claim that for t.c: s (i) V, n V, = 4 (ii) I V, I 2 IVd. (i) Let t i s. If possible let V, n V, r 41. Then there is u vertex x E VI n V,, so that x E V, and x E V,. Then x = I t pdo, x = s I- qdo for some positive integers p and q. Since t ~t s, we have p r q. Let p > q. 'Then t t pdo = s.t q4 implies that t + ( p q ) do = s, where p - q is an integer. So s = t + ( p q ) do E VI and p -- q ' P. This shows that s E V,, which is a contradiction to the fact that s e VI.

11 Let p < q. Now s = s + Odo E V, shows that s is the smallest number inv, and s>t. Againt+p&=s+qdoimpliesthatt=s+(q-p)do E V,, where q - p.: q is an integer. This shows that t E V,. This is a contradiction to the fact that s is the smallest number in V, and s > t. So, t r V,. These show that, there is no common elements between V, and V, so that V, n V, = 4 + (ii) Let I V, I = kl and ( V, I = k2. Then V,=(t,t+do,tt2do,..., t+(kl-l)do),whereklistheleast positive integer such that t + ( kl - I )do < n. Similarly V, = { s s + &, s + 2do,..., s t (k2-1 )do ), where k2 is the least positive integer such that s + ( k2 - I )do < n. Since t < s and t + rdo < s + rdo. This together with t + ( kl - 1 ) do < n and s + (k2-l)do n implies that k, > k2. Remark : : The following is the procedure for finding the independent sets and the independence number of G(Z,D). Consider the graph G(Z,D). Let do be the smallest positive integer that does not divide n. We start with vertex 1 and find the corresponding independent set, VI = ( I + r$ 1 0 O r r k - 1, where k is the least positive integer such that r& e S* and 1 + rdo < n 1.

12 Next let t be least positive integer such that t 6 VI. Consider the corresponding independent set, V, = { t + rdo I 0 < r I; k - 1. wlrcrc k is the least positive integer such that rdo~~*andt+rd,,<n). Proceeding in this way, we pet a finite number of independent sets, say. VI, V2,..., Vs, which are such that (i) (ii) V, nv,=+ iftics IVl(>IV~12... ZIV,/. So, Vl is one of the largest independent set of G(Z,, D) so that IVII is the independence number of G(Z,,, D) This discussion leads to the following corollaries, which give the independence nuinher and the chromatic number of G(Z,D). Corollarv : 3.3.5: Let n = 5, or, n 2 7 and 4 be the smallest positive integer that does not divide n. Then the independence numkr a (G(Z,,D) ) is given by a (G(Zn,D)) = k, where k is the least positive integer such that 1 1 kdu c: n and either kd,, divides n, or, ( n - k 6) divides n. - Proof : Let n = 5. or. n L 7 and do he the smallest positive integer that does not divide n.

13 VI = { 1 + rdo 10 I r S k - 1, where k is the least positive integer such thatrdo e S' and 1 +r&<n). From this, we see that k is the least positive integer such that k& E S*. That is, k is the least positive integer such that either k 4 divides n, or, ( n - k4) divides n and this k is the independence number of G(Z,D). Remark :3.3.6 For n = 2, 3, 4, 6, the independence number of G(Z,D) is 1 since G(Z,,D) is a complete graph. Definition : 3.3.7: 1. A k - vertex colouring of a graph G is an assignment of k - colours, 1, 2,.., k, to the vertices of G, So, a k - vertex colouring of G partitions V into the partition ( VI, Vz,..., Vk ), where V, is the subset of vertices of V which are coloured by i. 2. The colouring of G is proper if no two adjacent vertices have the same colour. 3. Thus a proper k - vertex colouring -of a loopless graph G is a partition (V,, V2,..., Vk ) of the vertex set V of G into k independent sets. 4. G is called k-colourable jf G has a proper k-vertex colouring. 5. The chromatic number, x(g), of G is the minimum k for which G is k-colourable. If x(g) = k, G is said to be k-chromatic.

14 Corollarv : 3.3.8: For n = 5, or, n 2 7 the Chromatic number of divisor Cayley graph of G(Z,,,D) is s, where s is the number of disjoint independent subsets of 0). && Proof is immediate from the Remark Remark : 3.3.9: For n = 2, 3, 4, 6, the chronlatic number of G(Z,,,D) is n. since G(Z,D) is a complete graph. 3.4 : EDGE DOMINATION IN DIVISOR CAYLEY GRAPH This section is devoted for the determination of edge cover, edge domination and the related domination parameters of the divisor Chiyley graph. Definition : An edge cover of a graph G is a set of edges covering all the vertices of G. A minimum edge cover is one with minimum cardinality. The number of edges in a minimum edge cover of G is called the edge covering number of G and it is denoted by P1(G). Theorem : 3.4.2: If n > 1, the minimum edge covering of the divisor Cayley graph G(Z,, D) is given by

15 (i) ((0,1),(2,3),...,(n-4,n-3),(n-2,n-l)),ifniseven. (ii) {(0,1),(2,3),,..., (n-3,n-2),(n-1,o)},ifnis odd. - Proof : (i) Suppose n is an even number. Consider the set of ordered pairs of vertices given by FI = {(0,1),(2,3),..., (n-4,n-3),(n-2,n-l)}. For each ordered pair ( 2i, 2itl), 0 5 i 5 "-2, (2i+l) - 2i = 1 E S' so that 2 (2i, 2i+l) is an edge of G(Zn, D). So F1 is a set of edges in G. Further the edges in FI cover all the vertices of G(Zn, D), So F1 forms an edge covering of G(Zn, D). Furthermore, the end vertices of the edges in FI are distinct. To show that F1 is the minimum edge covering of G(Z,, D). Let us consider the edge set FI - {el), where e, E F,, then e, = ( 2i, 2itl). Clearly, the vertices 2i, 2itl are not covered by the remaining edges of the edges set FI - {e,}, so that F, - {e,} cannot form an edge covering of G(Zn, D). Hence, F, is the minimum edge covering of G(Z,, D). Since n is even, the number of distinct pairs of distinct vertices of the form (2i, 2i+l), n i 5 -, so that the cardinality of the set FI is n. 2 2 (ii) Suppose n is an odd number. Consider the set ordered pairs of vertices given by

16 s"- For each ordered pair ( 2i, 2i+l), 0, 5 l i (2i+l) - 2i = 1 G S, so that 2 (2i, 2i+l) is an edge of G(Z, D). Further the edges in F1 cover all the vertices of G(Z,. D). So F2 forms an edge covering of G(Z,,, D)., To show that Fz is the minimum edge covering of G(Z,,,D). let. us consider the edge set Fz - { e, ). Thcn e, = ( 2i, 2itl). Clearly, the vertices 2i. 2i+l are not covered by the edge set Fz - ( e, ), so that F2 is the minin~um edge covering of G(5, D). Since the n + 1 vertices 0, I. 2,..., n - 1, 0 can be paired into ntl n-l 2 2 ' - distinct pairs of vertices ( 2i, 2i+l), 0 5 i 5 - the cardinality of The following Corollary is immediate from the 'Theorem. Corollarv : 3.4.3: Jf n :. 1. the edge covering numher of divisor Caylcy graph p'( G(Z,, D) is given by n (i), if n is even. n*l ( ) - if n is odd..2

17 Definition : 3,4.4: ' A subset F of the edge set E in a graph G is an edge dominating set if each edge E not in F ( that is in E - F ) is adjacent to atleast one edge in F. The minimum cardinality among all edge dominating sets of G is called an edge domination number of G and is denoted by yl(~). Theorem : 3.4.5: The edge dominating set of the divisor Cayley graph G(Z,,D), n > 2 is the set of edges. (i) {(0,1),(2,3),...,(n-2,n-l)},ifniseven. (ii) { (1,2), (3, 4),....., ( n - 2, n - 1) J, if n is odd. Proof : (i) Lct n be even. Consider the set of ordered pairs of' vertices given by E,={(O,1),(2,3),...,(n-2,n-I)}. n-2 Foreachorderedpair(2i,2itl), O < i -, (2i-t-1)-2i= 1 E sb,so 2 that, ( 2i, 2i+l) is an edge of G(Z,, D). So El is a set of edges in G(Z,, D). Clearly no two edges in El are adjacent. Let ( r, s) E E - El. Then r 1 0 and s # r t 1 Here two cases will arise, namely either r is even, or, r is odd.

18 Cpsei: Suppose r is even. Then r = 2t, for some integer and ( r, s) = (2t, s) Then the edge (24 2t*1) is in El clearly it is adjacent with the edge (21, s). - Case ii: Suppose r is odd. Then I 2tt1, for some integer t? 0 and (r, s) = (2t +I. s). Then the edge ( ,2t) which is same as (2t, 214 1) is in El and this is adjacent with ( 2tt1, s). So, El is an edge dominating set of G(Zn, D). Let us now show that El IS the minimum edge dominating set of G(Zn, D). To see this, let us delete the edge (i, it]) from El and form the edge set E ~' = El - ( i, 1t1 ). Now ( i, itl) is not adjacent to any edge of the edge set El1, since any edge ( r, s) E El1 is such that r # i, i I and s # i, it]. So El1 is not an edge dominating set of G(Z,,D). Ilence, El is tile minimum edge dominating set of G(Zn,D). (ii) Let n be odd. Consider the set of ordered pairs of vertices given by E2= {(1,2),(3,4),..., (n-2,n-1 )] n - l For each ordered pair ( 2i, Zitl), 0 5 ii -, 2 ( 2itl) - 2i = 1 E S', so that ( 2i, 2it 1) is an edge of G(Z,, D), SO E2 is a set of edges in Ci(Z,,, D).

19 Let ( r, s) E E -'E2. As in (i), we may assume r? 1 and s # r + 1 Here two cases will arise. Case i : Suppose r is odd. Then r 2tt1, for some integer t > 0 and ( r, S) = (2t +1, s ). Then the edge ( Zttl, 2t+2) is in E2 and this is adjacent with the edge ( 2tt1, s). Case ii : 7 Suppose r is even. Let r = 2t, for some integer t > 0. Then the edge (2t, s). Consider the edge (2t, 2t-1) which is same as ( 2t - 1, 2t) in E2. This is adjacent with the edge (2t, s). If r = 0, then the edge ( 0, s) is adjacent with ( s - 1, s) in El, if s is even and it is adjacent with ( s, stl) in E2, if s is odd. Thus E2 is an edge dominating set. As in (i), we can see that E2 is the minimum edge dominating set. The following Corollary is immediate from the above Theorem. CoroUarv : 3.4,6: Proof : If n > 2, the edge domination number y'( G(Z,,D) ) is given by n i) - if n is even. 2 ' (ii)!a, if n is odd. 2 If n is even, the minimum edge dominating set of G(Z,,,D) is

20 and this contains "edges. 2 If n is odd, the minimum edge dominating set of G(Z,,,D) is E2=((1,2),(3,4),..., (n-2,n-i)} and this contains 2 edges. 3.5 : THE MATCHING NUMBER OF DIVlSOR CAYLEY GRAPH : Definition : 3.5.1: A matchingf of a graph G is a subset of E such that no two cdyes of F are adjacent. A matching F is called a perfect matchin~if it covers all the \tcnices of the graph G. Definition : 3.5.2: A maximal matching is a matching with maximum number of edges and the cardinality of a maximum lvatching is known as lhc matching number. Theorem 3.5.3: The matching number of the divisor Cayley graph G(Z,,D) is n (i) --,, ifnisevon. (ii) [", if n is odd. 2

21 - Proof : (i) Let n be even. Consider the set of ordered pairs of vertices given by FI=((0,1),(2,3),..., (n-2,n-1)). For each ordered pair ( 2i, Zit I), 0 5 is "-2, ( 2i+ I) - 2i = 1 2 S*, so that ( 2i, 2itl) is an edge of G(Z,, D). So FI is a set of edges in G(Z,, D). It is easy to see that no two edges in FI are adjacent. So, F, is the matching of the graph G(Z,,D). Let us now show that FI is a maximal matching of G(Z,, D). To see this let us consider the edge set { (0,1), (2,3),...,(n-2, n -1 ) } u { r, s), Suppose r is even. Then r = 2t, for some integer t 3 0 and ( r, s) = (2t, s). Then the edge (2t, 2ti-1) is in FI and is adjacent with (2t, s). Suppose r is odd, then r = 2tt1, for some integer t 1: 0 and (r,s) = (2t+l, s). Then the edge (2tt1, 2t), which is same as (2t, 2t+l), is in FI and is adjacent with (Zttl, s). So, the edge set { (0,1), (2,3),...,(n-2, n-i)} u {r,s} where r 2 0 and s f r +I is not matching of the graph G(Z,,D). Hence, the edge set ( (0,1), (2,3),..., (n-2, n-1)) is a maximal matching of the graph G(Z,,D). In this case the cardinality of a maximal matching of the graph

22 (ii) Let n be odd. Consider thc set of ordered pairs of vertices given by Fz= ((0,1),(2,3),..., (n-3,n-2 )}. For each ordered pair ( 2i, 2i+l), 0 5 is * ( 2i+l) - 2i = 1 G s*, 2 ' so that ( 2it?i+l) is an edge of G(&, I)). So F2 is a set of edges in G(Z,,, D). By inspection, it is easy to see drat no two edges in F2 are adjacent. So, F1 is a matching of the graph G(Z,,.I>). To show that F2 is a maximal matching of the graph G(Z,,D), let us consider the edge set (2,3),.,(n-3, n-2 ) ) u { r. s) where r? 0 ands#r+ 1. Suppose r is even Then r = 2t, for some integer t 2 0 and ( r, s) = (2t, s). Then the edge (2t.2tt I) is in F2 and is ad~acerit with (3, s). Suppose r is odd, then r = 2t+1, for some integer t 2 0 and (r, s) = (2t+ 1, s). Then the edge (3tt I, 2t), which is same as (2t, 2t+ 1) is in FI and is adjacent with (2t+I, s). If r = n - I. then the edge ( n - 1, s ) is adjacent with ( s - I, s) in F1, if s is even and it is adjacent with ( s, s+l) in FL. ii's is odd. So, the cdgu set { (0,1), (2,3),...., (n - 3, n - 2 )} v {r, s}, where r :. 0 and s 1- r +.I is no1 matching of the graph G(Z,,D). Hence, the edge set ( (0,1), (2.3),......,..., (n - 3, n - 2 )} is a maximal matching of the graph G(Z,,D). n The cardinality of a maximal matching of the graph G(Z,,D) is 7'

23 Fig: 3. I : The divisor Cayley Graph - G (2,,D) Fig: 3.2 : The divisor Caylav Graph - G (2;.D) Fig: 3.3 :?'tie divisor Caylay Graph - G (Z,,I)) Fig: 3.4 : Tllc divisor Caylay Grapl~ - G (2,,I))

24 Fig: 3.5 : The divisor Cayla>> Graph - G (L,,),I>) Fig: 3 6 I'hc divisor C'aylay Graph - CJ (7,,,Dl