Basic Maximal Total Strong Dominating Functions

1. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 Basic Maximal Total Strong Dominating Functions Dr. M. Kavitha1,∗, 1Department of Mathematics, KPR Institute of Engineering and Technology, Coimbatore - 641407, India. Let G = (V,E) be a simple graph. A subset D of V (G) is called a total strong dominating set of G, if for every u ∈ V (G), there exists a v ∈ D such that u and v are adjacent and deg(v) ≥ deg(u). The minimum cardinality of a total strong dominating set of G is called total strong domination number of G and is denoted by γs Corresponding to total strong dominating set of G, total strong dominating function can be defined. The minimum weight of a total strong dominating function is called the fractional total strong domination number of G and is denoted by γt Keywords: Total Strong Dominating Function, Maximal Total Strong Dominating Function, Basic Maximal Total Strong Dominating Function Introduction: Corresponding to total strong dominating sets in a graph, total strong dominating functions may be defined. The minimality of a total strong dominating function can be characterised. Convex combination of minimal total strong dominating function is defined and studied. A total strong dominating function is basic, if it cannot be expressed as a covex combination of minimal total strong dominating functions. Basic total strong dominating functions are characterised. t(G). sf(G). A study of total strong dominating functions is carried out in this paper. Definition 0.1: Let G = (V,E) be a simple graph without strong isolates. A function f : V (G) → [0,1] is called a Total Strong Dominating Function (TSDF), if f(Ns(u)) = V (G), where Ns(u) = {x ∈ N(u) : deg (x) ≥ deg (u)}. Definition 0.2: A TSDF is called a Minimal Total Strong Dominating Function (MTSDF), if whenever g : V (G) → [0,1] and g < f, g is not a TSDF. Definition 0.3: Let G be a graph without isolated vertex. A TSDF (MTSDF) is called a basic TSDF (basic MTSDF) denoted by BTSDF (BMTSDF) if it cannot be expressed as a proper convex combination of two distinct TSDFs(MTSDFs). Remark 0.4: A BTSDF need not be a BMTSDF. X f(v) ≥ 1, ∀ u ∈ v∈Ns(u) 71 *Corresponding author: M. Kavitha, E-mail: kavinandhu7204@gmail.com

2. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 Lemma 0.5: Let f and g be two distinct MTSDFsof a graph G with Bs for every v ∈ V . Then (i) If f(v) = 0 or f(v) = 1, then δ(v) = 0. (ii) δ(v) = 0, ∀ v ∈ Bs X Proof: (i) We have, f(v) = 1 if and only if g(v) = 1 and f(v) = 0 if and only if g(v) = 0. Therefore, (i) follows. (ii) Let v ∈ Bs Then v ∈ Bs X = f(u) − = 1 − 1 (since v ∈ Bs = 0. Therefore, (ii) follows. (iii) is similar to (ii). Lemma 0.6: Let f and g be convex linear combination of MTSDFsg1, g2,...,gn such that f is minimal. Then n\ Proof: Let v ∈ Pf. Then f(v) > 0. n X Suppose gi(v) = 0, ∀ i. Then f(v) = 0, a contradiction. Therefore, gi(v) > 0, for at least one i. Therefore, v ∈ i=1 n[ Then g(vi) > 0, for some i. Therefore, f(v) > 0. n[ Therefore, Pf= i=1 i=1 n[ Then f(Ns(v)) = 1. Therefore, i=1 Suppose v / ∈ Bs Since gi(Ns(v)) ≥ 1, for all j, 1 ≤ j ≤ n, gand Pf= Pg. Let δ(v) = f(v)−g(v), f= Bs X f. u∈Ns(v) δ(v) = 0, ∀ v ∈ Bs (iii) g. u∈Ns(v) f. g. X X (f(u) − g(u)) δ(v) = v∈Ns(v) u∈Ns(v) X g(u) u∈Ns(v) u∈Ns(v) fand v ∈ Bs g) n[ Bs gi, Pf= Pg= Pgiand g is minimal. Bs f= Bs g= i=1 i=1 n X Let f = λigi, 0 < λi< 1, λi= 1. i=1 i=1 n[ Pgi. n[ Therefore, Pf⊆ Pgi. Suppose v ∈ Pgi. i=1 i=1 Therefore, v ∈ Pf. Therefore, Pgi⊆ Pf. i=1 n[ n[ f. Pgi. Similarly, Pg= Pgi. Therefore, Pf= Pg= Pgi. Let v ∈ Bs n X i=1 λigi(Ns(v)) = 1. gi, for some i, 1 ≤ i ≤ n. Then gi(Ns(v)) > 1. 72

3. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 n n X X = 1, a contradiction. λi. λigi(Ns(v)) > i=1 i=1 n\ Bs gi. Therefore, v ∈ Bs gi, ∀ i. Therefore, v ∈ n\ gi. Then gi(Ns(v)) = 1, for all i, 1 ≤ i ≤ n. X Therefore, f(Ns(v)) = 1. Therefore, v ∈ Bs Therefore, Bs i=1 Bs Therefore, Bs gi. f⊆ i=1 n\ Bs Let v ∈ i=1 n n X Therefore, λi= 1. λigi(Ns(v)) = i=1 i=1 f. n\ n\ g= gi⊆ Bs n\ Bs f. Therefore, Bs gi. f= i=1 i=1 n\ Bs Bs Similarly, Bs gi. Hence Bs gi. f= Bs g= i=1 i=1 Since f is minimal, Bs n\ That is, Bs Hence g is a MTSDF. Theorem 0.7: Let f be a MTSDF. Then f is a BMTSDF if and only if there does not exist an MTSDF g such that Bs Proof: Suppose f is a BMTSDF. Suppose there exists a MTSDF g such that Bs Let S = {a ∈ < : ha= (1 + a)g − af} be a TSDF and Bs Then S is a bounded open interval. Let S = (k1,k2). Then k1< −1 < 0 < k2. Also hk1and hk2are MTSDFs. hk1= (1 + k1)g − k1f. Therefore, f =(1 + k1)g k1 k1 Let λ1=1 + k1 k1 Therefore, λ1and λ2are positive and λ1+ λ2= 1. Therefore, f is a convex combination of g and hk1. Therefore, f is not a BMTSDF, a contradiction. Therefore, there does not exist a MTSDF g such that Bs Conversely, suppose f is not a BMTSDF. n X where 0 < λi< 1 and i=1 fweakly dominates Pf. n[ Bs That is, giweakly dominates Pgi. i=1 i=1 gweakly dominates Pg. gand Pf= Pg. f= Bs gand Pf= Pg. f= Bs fand Pha= Pf. ha= Bs −hk1 k1. . and λ2=−1 gand Pf= Pg. f= Bs Then there exists MTSDFsg1, g2,...,gnsuch that f = λigi, i=1 n X λi= 1. 73

4. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 n n X X n\ Let g = µigi, 0 < µi< 1 and µi= 1. i=1 i=1 n[ Bs Then by lemma 0.6, Bs giand Pf= Pg= Pgiand since f is a MTSDF, g is a MTSDF. f= Bs g= i=1 i=1 Thus there exists a MTSDF g such that Bs Hence the theorem. Theorem 0.8: Let f be a MTSDF of a graph G = (V,E) with Bs f(u) < 1} = {u1,u2,...,un}. Let A = [aij] be a m × n matrix defined by   Consider the system of linear equations given by gand Pf= Pg. f= Bs f= {v1,v2,...,vm} and P0 f= {u ∈ V : 0 <  1, if viweakly dominates uj 0, otherwise. aij= n X aijxj= 0, 1 ≤ i ≤ m. j=1 Then f is a BMTSDF if and only if the above system does not have a non-trivial solution. Proof: Suppose f is not a BMTSDF. Then there exists a MTSDF g such that Bs Let xj= f(uj) − g(uj), 1 ≤ j ≤ n. Suppose xj= 0, ∀ j, 1 ≤ j ≤ n. If f(v) = 0, then v / ∈ Pf= Pg. Therefore, g(v) = 0. Therefore, f(v) − g(v) = 0, ∀ v / ∈ Pf. That is, f(v) = g(v), ∀ v / ∈ Pf. If f(v) = 1, then by Theorem ??, g(v) = 1 and hence f(v) − g(v) = 0. Therefore, f = g, a contradiction. Therefore, there exists, some j, 1 ≤ j ≤ n such that xj6= 0. Let f(uj1) − g(uj1) 6= 0. n X = (f(u) − g(u)) (since aij= 1 because viweakly dominates u) = 0, by lemma0.5. Since xj6= 0, the left hand side has a non-trivial solution. Conversely, let {x1,x2,...,xn} be a non-trivial solution for the system of linear equations. Define g : V (G) → [0,1] as follows:   Since 0 < f(uj) < 1, choose Mj> 0 such that 0 < (f(uj) + Let M0= max{M1,M2,...,Mn}. Choose M to be equal to M’. gand Pf= Pg. f= Bs n X aij(f(uj) − g(uj)) X aijxj= j=1 j=1 u∈Ns(vi)  if v / ∈ P0 f(v), f(v) +xj f g(v) = M, if v = uj, 1 ≤ j ≤ n, where M is to be suitably chosen. Since {x1,x2,...,xn} is a non-trivial solution, g 6= f. xj Mj) < 1, for each j, 1 ≤ j ≤ n. 74

5. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 X = For any v ∈ V , g(Ns(v)) = g(u) u∈Ns(v) X X X X M0) + X xi g(u) + g(u) u∈Ns(v)∩P0 u∈Ns(v)−P0 X f f (f(vi) +xi = f(u) ui∈Ns(v)∩P0 u∈Ns(v)−P0 f f 1 = f(u) + xi M0 X i u∈Ns(u) = f(Ns(v)) + 1 M0 i n X X If v ∈ Bs (since v weakly dominates Pfand hence aij= 1). Therefore, g(Ns(v)) = f(Ns(v)) = 1. Suppose v / ∈ Bs Then f(Ns(v)) > 1. Choose M00> 0 such that g(Ns(v)) > 1, ∀ v / ∈ Bs Let M = max{M0,M00}. For this choice of M, we have 0 ≤ g(v) ≤ 1 and u∈Ns(v) Therefore, g is a TSDF. From what we have seen above, Bs Since f is a MTSDF, Bs Therefore, Bs Hence f is not a BMTSDF. Corollary 0.9: Let G = (V,E) be a graph without isolated vertices. Let S be a minimal total strong dominating set of G. Then χsis a BMTSDF. Proof: Clearly, χsis a MTSDF. Let f = χs. P0 Therefore from the above theorem, χsis a BMTSDF. Example 0.10: Consider Hajo’s Graph H3: s s s Define f1and f2as follows: f1(v1) = f1(v4) = f1(v2) = f1(v6) = 0. f1(v3) = f1(v5) = 1. f, then aijxi= 0. xi= i j=1 f. f. X g(v) ≥ 1, ∀ v ∈ V gand Pf= Pg. f= Bs fweakly dominates Pf. gweakly dominates Pg. Therefore, g is a MTSDF. f= φ. v1 s v3 v2 s s s s s v6 v4 v5 75

6. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 s 0 v1 s s f1 : 0 v3 1 v2 s s s s s s 0 v6 v4 1 v5 0 f2(v1) = f2(v4) = f2(v6) = 0. f2(v2) = f2(v3) = f2(v5) =1 2. s 0 v1 1 2 1 2 s s f2 : v3 v2 s s s s s s 0 0 v6 v4 v5 1 2 f1is a TSDF. Bs f1= {v1,v3,v4,v5}. Pf1= {v3,v5}. Bs Therefore, f1is a MTSDF. Therefore, f2is a TSDF. Bs Bs f2is a MTSDF. P0 Let A = [aij]m×nbe a m × n matrix defined by   Consider the system of linear equations given by a11x1+ a12x2+ ... + a1nxn= 0 . . . am1x1+ am2x2+ ... + a1nxn= 0. Since P0 Therefore, f1is a BMTSDF. f2is a TSDF. Bs f2= V, Pf2= {v2,v3,v4}. P0 f1weakly dominates Pf1. f2= V . f2weakly dominates Pf2. f1= φ.   fweakly dominates ujin P0 1, if vi∈ Bs      f . aij= 0, otherwise. f1= φ, the system of equations does not occur and hence does not have a non-trivial solution. f2= {v2,v3,v4}. 76

7. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470   1 0 0 1 0 0 1 1 0 0 0 1 0 0 0 1 0 1   A = [aij]6×3=   6×3    Therefore, f2is a BMTSDF. Example 0.11:   x1+ x2= 0         x2= 0 imply x1= 0 , x2= 0 x1= 0 x2= 0 1 1 1 0 u u u u u v5 0 P5 : v1 v4 v2 v3 f1is a TSDF. Bs Pf1= {v2,v3,v4}. Bs f1= {v1,v2,v4,v5}. f1weakly dominates Pf1. v2 v3 v4 v1 v2 v4 v5 0 1 1 0 0 0 1 0 x1 x2 x3 A = [aij]4×3 = 0 0 0 1 x1= 0, x2= 0, x3= 0. Therefore, f1is a BMTSDF. Example 0.12: t t t t t t t t t 1 1 1 0 1 0 1 0 0 Consider P9 : v5 v7 v3 v4 v9 v1 v2 v8 v6 Define f1on V (P6) as follows: f1(v1) = f1(v4) = f1(v5) = f1(v9) = 0. f1(v2) = f1(v3) = f1(v6) = f1(v7) = f1(v8) = 1. f1is a TSDF. Bs Pf1= {v2,v3,v6,v7,v8}. Bs Therefore, f1is a MTSDF. f1= {v1,v2,v3,v4,v5,v6,v8,v9}. f1weakly dominates Pf1. 77

8. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 v3 v3 v6 v7 v8 v2 v1 v2 v3 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 x1 x2 x3 x4 x5 x1 x2 1 0 0 0 v4 1 0 0 0 x1 x2 x3 x4 0 A = v5 v6 0 0 0 1 0 0 0 0 1 0 x4 x5 v8 0 0 0 0 1 v9 1 0 0 0 0 xi= 0, ∀ i, 1 ≤ i ≤ 5. Therefore, f1is a BMTSDF. Example 0.13: t t t t t t t t t 1 1 1 0 1 0 1 0 0 Consider P9 : v5 v7 v3 v4 v9 v1 v2 v8 v6 Define f1on V (P9) as follows: f1(v1) = f1(v4) = f1(v5) = f1(v9) = 0. f1(v2) = f1(v3) = f1(v6) = f1(v7) = f1(v8) = 1. f1is a TSDF. Bs Pf1= {v2,v3,v6,v7,v8}. Bs Therefore, f1is a MTSDF. f1= {v1,v2,v3,v4,v5,v6,v7,v8,v9}. f1weakly dominates Pf1.   x1 x2 x1 x2 x3 x4 x5 x6 x7 x8 x9       1 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0       x1 x2 x3 x4 x5   A = = =         xi= 0, ∀ i, 1 ≤ i ≤ 5. Therefore, f1is a BMTSDF. Example 0.14: t t t t t t t t t t t 1 1 0 1 0 1 0 1 0 0 1 Let P11 : v5 v7 v3 v4 v9 v1 v2 v8 v10 v11 v6 Define f1on V (P11) as follows: f1(v1) = f1(v4) = f1(v5) = f1(v8) = f1(v11) = 0. 78

9. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 f1(v2) = f1(v3) = f1(v6) = f1(v7) = f1(v9) = f1(v10) = 1. f1is a TSDF. Bs f1= {v1,v2,v3,v4,v5,v6,v7,v9,v10,v11}. Pf1= {v2,v3,v6,v7,v9,v10}. Bs Therefore, f1is a MTSDF. f1weakly dominates Pf1. v2 v3 v6 v7 v8 v2 v1 v2 v3 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 x1 x2 x3 x4 x5 x1 x2 1 0 0 0 v4 1 0 0 0 x1 x2 x3 x4 0 A = v5 v6 0 0 0 1 0 0 0 0 1 0 x4 x5 v8 0 0 0 0 1 v9 1 0 0 0 0 xi= 0, ∀ i, 1 ≤ i ≤ 6. Therefore, f1is a BMTSDF. Again consider t t t t t t t t t t t 1 1 0 1 1 0 0 1 0 1 1 P11 : v5 v7 v3 v4 v9 v1 v2 v8 v10 v11 v6 Define f2on V (P11) as follows: f2(v1) = f2(v5) = f2(v6) = f2(v11) = 0. f2(v2) = f2(v3) = f2(v4) = f2(v7) = f2(v8) = f2(v9) = f2(v10) = 1. f2is a TSDF. Bs f2= {v1,v2,v4,v5,v6,v7,v10,v11}. Pf2= {v2,v3,v4,v7,v8,v9,v10}. Bs f2weakly dominates Pf2. 79

10. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 v2 v3 v4 v7 v9 v10 v8 v1 1 0 0 0 0 0 0 v2 0 0 0 0 1 0 0 x1 x2 x3 x4 x5 x6 x7 v4 v5 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 v6 B = 0 0 0 0 1 0 0 0 v7 1 0 0 0 0 0 0 0 0 1 v10 0 0 0 0 0 v11 0 0 0 0 0 1 0 xi= 0, ∀ i, 1 ≤ i ≤ 7. Therefore, f2is a BMTSDF. Example 0.15: Let G = C2n+1. Let V (G) = {v1,v2,...,v2n+1}. Case (i) Let n ≡ 0 (mod 2). Let n = 2k. Then 2n + 1 = 4k + 1.   Bs f= {v2,v3,v4,v5,...,v4k+1} Pf= {v1,v5,...,v4k+1,v2,v6,...,v4k−2} Clearly, Bs Therefore, f is a MTSDF.  1, if i ≡ 1,2 (mod 4) 0, if i ≡ 3,0 (mod 4). Let f(vi) = Then f is a TSDF. fweakly dominates Pf. v1 v2 v5 v6... v4k−1v4k−2 v4k+1 ... v2 1 x1 0 0 0 0 0 0 0 v3 ... x2 0 0 0 0 0 1 0 0 v4 v5 ... x3 0 0 0 0 0 1 0 0 0 0 . 0 1 ... x4 0 0 0 0 A = . . . . . . . . . . . v4k+1 x2k+1 0 0 0 ... 0 0 0 0 1 80

11. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 Therefore, xi= 0, ∀ i, 1 ≤ i ≤ 2k + 1. Therefore, f is a BTMSDF. Case (ii) Let n ≡ 1 (mod 2). Let n = 2k + 1. Then 2n + 1 = 4k + 3.   Bs Pf= {v1,v2,v5,v6,...,v4k+1,v4k+2}. Clearly, Bs Therefore, f is a MTSDF.  1, if i ≡ 1,2 (mod 4) 0, if i ≡ 3,0 (mod 4). Let f(vi) = Then f is a TSDF. f= {v1,v2,v5,v6,...,v4k+2}. fweakly dominates Pf. v1 v2 0 v4k+2 v5 v6... v4k+1 1 0 ... 0 0 v1 0 0 x1 ... 1 0 0 0 v2 0 0 x2 0 ... 0 0 v5 1 0 0 0 0 x3 v6 0 0 . 1 ... 0 x4 0 0 0 A = . . . . . . . . . . . x2k+1 1 0 0 v4k+2 ... 0 0 0 0 Therefore, xi= 0, ∀ i, 1 ≤ i ≤ 2k + 2. Therefore, f is a BTMSDF. Case (iii) Let G = C2n. Let V (G) = {v1,v2,...,v2n}. Let f(vi) =  Bs f=  Pf=    1, if i ≡ 1,2 (mod 4) 0, otherwise. Clearly, f is a TSDF.  V − {v1,v2n}, if n ≡ 1 (mod 4).  {v1,v5,...,v2n−1,v2,v6,...,v2n},   if n ≡ 2 (mod 4) V, {v1,v5,...,v2n−3,v2,v6,...,v2n−2}, if n ≡ 2 (mod 4) if n ≡ 1 (mod 4). 81

12. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 Clearly, Bs Therefore, f is a MTSDF. Let n ≡ 2 (mod 4). fweakly dominates Pf. v1 v2 v5 v6... v2n−3 v2n−2 ... 1 v1 0 0 0 0 0 ... v2 1 0 0 0 0 0 . A = . . . . . v2n 1 0 0 ... 0 0 0 AX = 0 implies X = 0. Therefore, f is a BMTSDF. Let n ≡ 1 (mod 4). v1 v2 v5 v6... v2n−1 v2n v2 ... 1 0 0 0 0 0 v3 ... 1 0 0 0 0 0 . A = . . . . . v2n−1 0 1 0 ... 0 0 0 AX = 0 implies X = 0. Therefore, f is a BMTSDF. REFERENCES [1] Acharya, B.D., The Strong Domination Number of a Graph and Related Concepts, J.Math. Phys. Sci., 14,471-475(1980). [2] Allan R.B., and Laskar.R.C., On Domination and Independent Domination Number of a Graph, Dicrete Math., 23, 73-76(1978). [3] Arumugam.S. and Regikumar.K., Basic Minimal Dominating Functions, Utilitas Mathematica, 77, pp. 235-247(2008). [4] Arumugam.S. and Regikumar.K., Fractional Independence and Fractional Domination Chain in Graphs, AKCE J.Graphs Combin., 4, No.2, pp.161-169(2007). [5] Arumugam.S. and Sithara Jerry.A., A Note On Independent Domination in Graphs, Bulletin of the Allahabad Mathematical Society, Volume 23, Part 1.(2008). [6] Arumugam.S. and Sithara Jerry.A., Fractional Edge Domination in Graphs, Appl. Anal. Discrete Math.,(2009). 82

13. International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available Online www.ijtsrd.com e-ISSN 2456 - 6470 [7] Berge.C., Theory of Graphs and its Applications, Dunod, Paris(1958). [8] Berge.C., Graphs and Hypergraphs, North-Holland, Amsterdam(1973). [9] Cockayne E.J., Dawes R.M. and Hedetniemi.S.T., Total domination in graphs, Networks, 10, 211-219(1980). [10] Cockayne E.J., Favaron O., Payan C. and Thomasaon A.G., Contributions to the theory of domination, independence and irredundance in graphs, Discrete Math., 33, 249-258(1981). [11] Cockayne E.J., MacGillivray G. and Mynhardt C.M., Convexity of minimal dominating functions and universal in graphs, Bull. Inst. Combin. Appl., 5, 37-38(1992). [12] Cockayne E.J. and Mynhardt C.M., Convexity of minimal dominating functions of trees: A survey, Quaestiones Math., 16, 301-317(1993). [13] Cockayne E.J., MacGillivray G. and Mynhardt C.M., Convexity of minimal dominating functions of trees-II, Discrete Math., 125, 137-146(1994). [14] Cockayne E.J. and Mynhardt C.M., A characterization of universal minimal total dominating functions in trees, Discrete Math., 141, 75-84(1995). [15] Cockayne.E.J., Fricke.G., Hedetniemi.S.T. and Mynhardt.C.M., Properties of Minimal Dominating Functions of Graphs, Ars Combin., 41,107-115(1995). [16] Cockayne E.J., MacGillivray G. and Mynhardt C.M., Convexity of minimal dominating functions of trees, Utilitas Mathematica, 48, 129-144(1995). [17] Cockayne.E.J. and Hedetniemi.S.T., Towards a Theory of Domination in Graphs, Networks, 7,247-261(1977). [18] Cockayne.E.J. and Mynhardt.C.M., Minimality and Convexity of dominating and related functions in Graphs, A unifying theory, Utilitas Mathematica, 51,145-163(1997). [19] Cockayne.E.J. and Mynhardt.C.M. and Yu.B., Universal Minimal Total Dominating Functions in Graphs, Networks, 24, 83-90(1994). [20] Grinstead D. and Slater P.J., Fractional Domination and Fractional Packing in Graphs, Congr., Numer., 71, 153-172(1990). [21] Harary.F., Graph Theory, Addison-Wesley, Reading, Mass(1969). [22] Hedetniemi.S.T. and Laskar.R.L., Bibliography on Domination in Graphs and Some Basic Definitions of Domination Parameters, Discrete Math., 86,pp 257-277(1990). [23] Ore.O, Theory of Graphs, AMS(1962). [24] Regikumar K., Dominating Functions - Ph.D thesis, Manonmaniam Sundaranar University(2004). [25] Sampathkumar.E and Pushpa Latha.L., Strong Weak Domination and Domination Balance in a Graph, Discrete Math., 161,235- 242(1996). [26] Scheinerman E.R. and Ullman D.H., Fractional Graph Theory: A rational approach to the theory of graphs, John Wiley and Sons Inc.,(1997). [27] Terasa W. Haynes, Stephen T. Hedetneimi, Peter J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker Inc.,(1998). [28] Terasa W. Haynes, Stephen T. Hedetneimi, Peter J. Slater, Domination in Graphs: Advanced Topics Marcel Dekker Inc., New York(1998). [29] Yu.B., Convexity of Minimal Total Dominating Functions in Graphs, J.Graph Theory, 24,313-321(1997). 83