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,∗ , 1 Department 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 γts (G). 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 t (G). A study of total strong dominating functions is carried out in this paper. denoted by γsf 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. 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)) = X f (v) ≥ 1, ∀ u ∈ v∈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. 71 *Corresponding author: M. Kavitha, E-mail: [email protected] 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 MTSDFs of a graph G with Bfs = Bgs and Pf = Pg . Let δ(v) = f (v)−g(v), for every v ∈ V . Then (i) If f (v) = 0 or f (v) = 1, then δ(v) = 0. X (ii) δ(v) = 0, ∀ v ∈ Bfs . u∈Ns (v) (iii) X δ(v) = 0, ∀ v ∈ Bgs . u∈Ns (v) 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 ∈ Bfs . Then v ∈ Bgs . X X δ(v) = (f (u) − g(u)) v∈Ns (v) u∈Ns (v) X = X f (u) − u∈Ns (v) g(u) u∈Ns (v) = 1 − 1 (since v ∈ Bfs and v ∈ Bgs ) = 0. Therefore, (ii) follows. (iii) is similar to (ii). Lemma 0.6: Let f and g be convex linear combination of MTSDFs g1 , g2 ,...,gn such that f is minimal. Then n n \ [ Bfs = Bgs = Bgsi , Pf = Pg = Pgi and g is minimal. i=1 i=1 Proof: Let v ∈ Pf . Then f (v) > 0. n n X X Let f = λi gi , 0 < λi < 1, λi = 1. i=1 i=1 Suppose gi (v) = 0, ∀ i. Then f (v) = 0, a contradiction. n [ Therefore, gi (v) > 0, for at least one i. Therefore, v ∈ Pgi . Therefore, Pf ⊆ n [ Pgi . Suppose v ∈ i=1 i=1 n [ Pgi . i=1 Then g(vi ) > 0, for some i. Therefore, f (v) > 0. n [ Therefore, v ∈ Pf . Therefore, Pgi ⊆ Pf . Therefore, Pf = n [ i=1 Pgi . Similarly, Pg = i=1 Therefore, Pf = Pg = n [ n [ Pgi . i=1 Pgi . Let v ∈ Bfs . i=1 Then f (Ns (v)) = 1. Therefore, n X λi gi (Ns (v)) = 1. i=1 Suppose v ∈ / Bgsi , for some i, 1 ≤ i ≤ n. Then gi (Ns (v)) > 1. Since gi (Ns (v)) ≥ 1, for all j , 1 ≤ j ≤ n, 72 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 X λi gi (Ns (v)) > i=1 λi . i=1 = 1, a contradiction. Therefore, v ∈ Bgsi , Therefore, Bfs ⊆ Let v ∈ n X ∀ i. Therefore, v ∈ n \ Bgsi . i=1 Bgsi . i=1 n \ Bgsi . Then gi (Ns (v)) = 1, for all i, 1 ≤ i ≤ n. i=1 Therefore, n \ n X λi gi (Ns (v)) = i=1 n X λi = 1. i=1 Therefore, f (Ns (v)) = 1. Therefore, v ∈ Bfs . n n \ \ Therefore, Bgsi ⊆ Bfs . Therefore, Bfs = Bgsi . i=1 Similarly, Bgs = n \ Bgsi . Hence Bfs = i=1 Bgs i=1 n \ = Bgsi . i=1 Since f is minimal, Bfs weakly dominates Pf . n n \ [ s That is, Bgi weakly dominates Pgi . i=1 i=1 That is, Bgs weakly dominates Pg . 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 Bfs = Bgs and Pf = Pg . Proof: Suppose f is a BMTSDF. Suppose there exists a MTSDF g such that Bfs = Bgs and Pf = Pg . Let S = {a ∈ < : ha = (1 + a)g − af } be a TSDF and Bhs a = Bfs and Pha = Pf . Then S is a bounded open interval. Let S = (k1 , k2 ). Then k1 < −1 < 0 < k2 . Also hk1 and hk2 are MTSDFs . hk1 = (1 + k1 )g − k1 f . (1 + k1 )g hk1 − . Therefore, f = k1 k1 1 + k1 −1 and λ2 = . Let λ1 = k1 k1 Therefore, λ1 and λ2 are 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 Bfs = Bgs and Pf = Pg . Conversely, suppose f is not a BMTSDF. n X Then there exists MTSDFs g1 , g2 ,...,gn such that f = λi gi , where 0 < λi < 1 and n X i=1 λi = 1. i=1 73 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 Let g = n X µi gi , 0 < µi < 1 and i=1 Then by lemma 0.6, Bfs = Bgs = n X µi = 1. i=1 n \ n [ i=1 i=1 Bgsi and Pf = Pg = Pgi and since f is a MTSDF, g is a MTSDF. Thus there exists a MTSDF g such that Bfs = Bgs and Pf = Pg . Hence the theorem. Theorem 0.8: Let f be a MTSDF of a graph G = (V, E) with Bfs = {v1 , v2 , ..., vm } and Pf0 = {u ∈ V : 0 < f (u) < 1} = {u1 , u2 , ..., un }. Let A = [aij ] be a m × n matrix defined by 1, if v weakly dominates u i j aij = 0, otherwise. n X Consider the system of linear equations given by aij xj = 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 Bfs = Bgs and Pf = Pg . 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 xj 6= 0. Let f (uj1 ) − g(uj1 ) 6= 0. n n X X aij xj = aij (f (uj ) − g(uj )) j=1 j=1X = (f (u) − g(u)) u∈Ns (vi ) (since aij = 1 because vi weakly dominates u) = 0, by lemma0.5. Since xj 6= 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: f (v), if v ∈ / Pf0 g(v) = f (v) + xj , if v = uj , 1 ≤ j ≤ n, where M is to be suitably chosen. M Since {x1 , x2 , ..., xn } is a non-trivial solution, g 6= f . xj Since 0 < f (uj ) < 1, choose Mj > 0 such that 0 < (f (uj ) + M ) < 1, for each j, 1 ≤ j ≤ n. j 0 Let M = max{M1 , M2 , ..., Mn }. Choose M to be equal to M’. 74 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 g(u) + u∈Ns (v)∩P = X 0 f g(u) 0 f u∈Ns (v)−P X xi (f (vi ) + 0 ) + M u∈Ns (v)−Pf0 ui ∈Ns (v)∩Pf0 X f (u) 1 X xi M0 i u∈Ns (u) X = f (Ns (v)) + M1 0 xi = If v ∈ Bfs , then X xi = n X i X f (u) + i aij xi = 0. j=1 (since v weakly dominates Pf and hence aij = 1). Therefore, g(Ns (v)) = f (Ns (v)) = 1. Suppose v ∈ / Bfs . Then f (Ns (v)) > 1. Choose M 00 > 0 such that g(Ns (v)) > 1, ∀ v ∈ / Bfs . Let M = max{M 0 , M 00 }. For this choice of M , we have X 0 ≤ g(v) ≤ 1 and g(v) ≥ 1, ∀ v ∈ V u∈Ns (v) Therefore, g is a TSDF. From what we have seen above, Bfs = Bgs and Pf = Pg . Since f is a MTSDF, Bfs weakly dominates Pf . Therefore, Bgs weakly dominates Pg . Therefore, g is a MTSDF. 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 χs is a BMTSDF. Proof: Clearly, χs is a MTSDF. Let f = χs . Pf0 = φ. Therefore from the above theorem, χs is a BMTSDF. Example 0.10: Consider Hajo’s Graph H3 : vs 1 sv3 v2 s v4 s s v5 Define f1 and f2 as follows: f1 (v1 ) = f1 (v4 ) = f1 (v2 ) = f1 (v6 ) = 0. f1 (v3 ) = f1 (v5 ) = 1. 75 sv6 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 0 vs 1 f1 : v2 0s 0 v4 s f2 (v1 ) = f2 (v4 ) = f2 (v6 ) = 0. 1 f2 (v2 ) = f2 (v3 ) = f2 (v5 ) = . 2 sv13 s sv6 0 1 v5 0vs 1 1 1 2v v22s f2 : 0 v4 s s3 s v5 1 2 s0v6 f1 is a TSDF. Bfs1 = {v1 , v3 , v4 , v5 }. Pf1 = {v3 , v5 }. Bfs1 weakly dominates Pf1 . Therefore, f1 is a MTSDF. Therefore, f2 is a TSDF. Bfs2 = V . Bfs2 weakly dominates Pf2 . f2 is a MTSDF. Pf01 = φ. Let A = [aij ]m×n be a m × n matrix defined by 1, if vi ∈ Bfs weakly dominates uj in Pf0 aij = . 0, otherwise. Consider the system of linear equations given by a11 x1 + a12 x2 + . . . + a1n xn = 0 . . . am1 x1 + am2 x2 + . . . + a1n xn = 0. Since Pf01 = φ, the system of equations does not occur and hence does not have a non-trivial solution. Therefore, f1 is a BMTSDF. f2 is a TSDF. Bfs2 = V, Pf2 = {v2 , v3 , v4 }. Pf02 = {v2 , v3 , v4 }. 76 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 A = [aij ]6×3 1 0 0 = 1 0 0 1 1 0 0 0 1 0 0 0 1 0 1 6×3 x1 + x2 = 0 x =0 2 imply x1 = 0 , x2 = 0 x = 0 1 x2 = 0 Therefore, f2 is a BMTSDF. Example 0.11: 0 1 u P5 : 1 u v1 v2 1 0 u u u v3 v4 v5 f1 is a TSDF. Bfs1 = {v1 , v2 , v4 , v5 }. Pf1 = {v2 , v3 , v4 }. Bfs1 weakly dominates Pf1 . v2 v3 v4 A = [aij ]4×3 = v1 v2 v4 v5 1 0 0 0 0 1 1 0 0 0 0 x1 x2 x3 1 x1 = 0, x2 = 0, x3 = 0. Therefore, f1 is a BMTSDF. Example 0.12: 0 t P9 : Consider v1 1 1 0 0 1 1 1 0 v2 v3 v4 v5 v6 v7 v8 v9 t t t t t t Define f1 on V (P6 ) as follows: f1 (v1 ) = f1 (v4 ) = f1 (v5 ) = f1 (v9 ) = 0. f1 (v2 ) = f1 (v3 ) = f1 (v6 ) = f1 (v7 ) = f1 (v8 ) = 1. f1 is a TSDF. Bfs1 = {v1 , v2 , v3 , v4 , v5 , v6 , v8 , v9 }. Pf1 = {v2 , v3 , v6 , v7 , v8 }. Bfs1 weakly dominates Pf1 . Therefore, f1 is a MTSDF. 77 t t 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 A = v2 v3 v6 v7 v8 v1 1 0 0 0 0 v2 v3 0 0 0 0 1 1 0 0 0 0 v4 0 1 0 0 0 v5 0 0 1 0 0 v6 0 0 0 1 0 v8 0 0 0 1 0 v9 0 0 0 0 1 xi = 0, ∀ i, 1 ≤ i ≤ 5. Therefore, f1 is a BMTSDF. Example 0.13: 0 t P9 : Consider v1 1 1 0 0 v2 v3 v4 v5 t t t t x1 x2 x3 x4 x5 1 1 1 0 v6 v7 v8 v9 t 0 0 0 0 0 0 0 0 x1 x2 x1 x2 x3 x4 x4 x5 t t t Define f1 on V (P9 ) as follows: f1 (v1 ) = f1 (v4 ) = f1 (v5 ) = f1 (v9 ) = 0. f1 (v2 ) = f1 (v3 ) = f1 (v6 ) = f1 (v7 ) = f1 (v8 ) = 1. f1 is a TSDF. Bfs1 = {v1 , v2 , v3 , v4 , v5 , v6 , v7 , v8 , v9 }. Pf1 = {v2 , v3 , v6 , v7 , v8 }. Bfs1 weakly dominates Pf1 . Therefore, f1 is a MTSDF. x1 x2 1 0 0 0 0 0 x1 0 1 0 0 0 0 x x 1 0 0 0 0 1 2 0 x2 x3 0 1 0 0 0 0 A= x3 = x4 = 0 0 1 0 0 0 0 0 0 1 0 x4 x5 0 x5 x6 0 0 0 0 1 0 x7 0 0 0 0 1 0 x 8 x9 xi = 0, ∀ i, 1 ≤ i ≤ 5. Therefore, f1 is a BMTSDF. Example 0.14: 1 1 0 0 t t t t Let P11 : v1 v2 v3 v4 0 1 1 0 1 1 0 v5 v6 v7 v8 v9 v10 v11 t t Define f1 on V (P11 ) as follows: f1 (v1 ) = f1 (v4 ) = f1 (v5 ) = f1 (v8 ) = f1 (v11 ) = 0. 78 t t t t t 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. f1 is a TSDF. Bfs1 = {v1 , v2 , v3 , v4 , v5 , v6 , v7 , v9 , v10 , v11 }. Pf1 = {v2 , v3 , v6 , v7 , v9 , v10 }. Bfs1 weakly dominates Pf1 . Therefore, f1 is a MTSDF. A = v2 v23 v6 v7 v8 v1 1 0 0 0 0 v2 v3 0 0 0 0 1 1 0 0 0 0 v4 0 1 0 0 0 v5 0 0 1 0 0 v6 0 0 0 1 0 v8 0 0 0 1 0 v9 0 0 0 0 1 1 1 1 0 0 1 1 1 1 0 v2 v3 v4 v5 v6 v7 v8 v9 v10 v11 xi = 0, ∀ i, 1 ≤ i ≤ 6. Therefore, f1 is a BMTSDF. Again consider 0 t P11 : v1 t t t t x1 x2 x3 x4 x5 t t t Define f2 on 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. f2 is a TSDF. Bfs2 = {v1 , v2 , v4 , v5 , v6 , v7 , v10 , v11 }. Pf2 = {v2 , v3 , v4 , v7 , v8 , v9 , v10 }. Bfs2 weakly dominates Pf2 . 79 0 0 0 0 0 0 0 0 x1 x2 x1 x2 x3 x4 x4 x5 t t t 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 v8 v9 v10 v1 1 0 0 0 0 0 0 v2 0 1 0 0 0 0 0 v4 v5 0 0 1 0 0 0 1 0 0 0 0 0 0 v6 0 0 0 1 0 v7 0 0 0 v10 0 0 0 0 0 1 0 0 0 1 0 v11 0 0 0 0 0 0 1 B = x1 x2 x3 x4 x5 x6 x7 0 0 0 0 0 0 0 0 0 0 xi = 0, ∀ i, 1 ≤ i ≤ 7. Therefore, f2 is 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. 1, if i ≡ 1, 2 (mod 4) Let f (vi ) = 0, if i ≡ 3, 0 (mod 4). Then f is a TSDF. Bfs = {v2 , v3 , v4 , v5 , ..., v4k+1 } Pf = {v1 , v5 , ..., v4k+1 , v2 , v6 , ..., v4k−2 } Clearly, Bfs weakly dominates Pf . Therefore, f is a MTSDF. v1 v2 v5 A = v2 1 0 0 0 v3 0 1 0 0 v4 0 0 0 0 . 1 0 0 1 v5 . v6 . . . ... ... ... ... v4k−1v4k−2 v4k+1 0 0 0 x1 0 0 0 0 x2 0 0 0 0 x3 0 0 0 0 x4 . 0 . . . . . . . . . v4k+1 1 x2k+1 0 0 0 0 ... 0 0 0 80 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. 1, if i ≡ 1, 2 (mod 4) Let f (vi ) = 0, if i ≡ 3, 0 (mod 4). Then f is a TSDF. Bfs = {v1 , v2 , v5 , v6 , ..., v4k+2 }. Pf = {v1 , v2 , v5 , v6 , ..., v4k+1 , v4k+2 }. Clearly, Bfs weakly dominates Pf . Therefore, f is a MTSDF. v2 v1 v2 v5 v6 . . . 0 1 0 0 ... 1 0 0 0 ... v5 0 0 v6 0. 0 v1 A = v4k+1 v4k+2 0 0 x1 0 x2 0 0 x3 0 0 x4 . 0 . . . . x2k+1 0 0 0 0 1 ... 0 1 0 ... 0 . . . . . v4k+2 0 0 1 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 }. 1, if i ≡ 1, 2 (mod 4) Let f (vi ) = 0, otherwise. Clearly, f is a TSDF. V, if n ≡ 2 (mod 4) Bfs = V − {v1 , v2n }, if n ≡ 1 (mod 4). {v , v , ..., v 1 5 2n−3 , v2 , v6 , ..., v2n−2 }, if n ≡ 2 (mod 4) Pf = {v1 , v5 , ..., v2n−1 , v2 , v6 , ..., v2n }, if n ≡ 1 (mod 4). 81 0 . 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, Bfs weakly dominates Pf . Therefore, f is a MTSDF. Let n ≡ 2 (mod 4). v1 v2 v5 v6 . . . v2n−3 v 2n−2 v1 0 1 0 0 ... 0 0 v2 1 0 0 ... 0 0 0 0 0 ... 0 0 0 . A = . . . . . v2n 1 AX = 0 implies X = 0. Therefore, f is a BMTSDF. 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