Iteratioal J.Math. Combi. Vol.2 2009), 53-57 The Forcig Domiatio Number of Hamiltoia Cubic Graphs H.Abdollahzadeh Ahagar Departmet of Mathematics, Uiversity of Mysore, Maasagagotri, Mysore- 570006 Pushpalatha L. Departmat of Mathematics, Yuvaraja s College, Mysore-570005) E-mail: ha.ahagar@yahoo.com, pushpakrisha@yahoo.com Abstract: A set of vertices S i a graph G is called to be a Smaradachely domiatig k-set, if each vertex of G is domiated by at least k vertices of S. Particularly, if k = 1, such a set is called a domiatig set of G. The Smaradachely domiatio umber γ k G) of G is the miimum cardiality of a Smaradachely domiatig set of G. For abbreviatio, we deote γ 1G) by γg). I 1996, Reed proved that the domiatio umber γg) of every -vertex graph G with miimum degree at least 3 is at most 3/8. Also, he cojectured that γh) /3 for every coected 3-regular -vertex graph H. I [?], the authors preseted a sequece of Hamiltoia cubic graphs whose domiatio umbers are sharp ad i this paper we study forcig domiatio umber for those graphs. Key Words: Smaradachely domiatig k-set, domiatig set, forcig domiatio umber, Hamiltoia cubic graph. AMS2000): 05C69 1. Itroductio Throughout this paper, all graphs cosidered are fiite, udirected, loopless ad without multiple edges. We refer the reader to [12] for termiology i graph theory. Let G be a graph, with vertices ad e edges. Let N v) be the set of eighbors of a vertex v ad N[v] = N v) {v}. Let d v) = N v) be the degree of v. A graph G is r regular if d v) = r for all v. Particularly, if r = 3 the G is called a cubic graph. A vertex i a graph G domiates itself ad its eighbors. A set of vertices S i a graph G is called to be a Smaradachely domiatig k-set, if each vertex of G is domiated by at least k vertices of S. Particularly, if k = 1, such a set is called a domiatig set of G. The Smaradachely domiatio umber γ k G) of G is the miimum cardiality of a Smaradachely domiatig set of G. For abbreviatio, we deote γ 1 G) by γg). A subset F of a miimum domiatig set S is a forcig subset for S if S is the uique miimum domiatig set cotaiig F. The forcig domiatio umber f G, γ) of S is the miimum cardiality amog the forcig subsets of S, ad the forcig domiatio umber f G, γ) of G is the miimum forcig domiatio umber amog 1 Received April 3, 2009. Accepted Jue 2, 2009.
54 H.Abdollahzadeh Ahagar ad Pushpalatha L. the miimum domiatig sets of G [1], [2], [5]-[7]). For every graph G, f G, γ) γ G). Also The forcig domiatio umber of several classes of graphs are determied, icludig complete multipartite graphs, paths, cycles, ladders ad prisms. The forcig domiatio umber of the cartesia product G of k copies of the cycle C 2k+1 is studied. The problem of fidig the domiatio umber of a graph is NP-hard, eve whe restricted to cubic graphs. Oe simple heuristic is the greedy algorithm, [11]). Let d g be the size of the domiatig set retured by the greedy algorithm. I 1991 Parekh [9] showed that d g + 1 2e + 1. Also, some bouds have bee discovered o γ G) for cubic graphs. Reed [10] proved that γ G) 3 8. He cojectured that γh) 3 for every coected 3- regular cubic) -vertex graph H. Reed s cojecture is obviously true for Hamiltoia cubic graphs. Fisher et al. [3]-[4] repeated this result ad showed that if G has girth at least 5 the γ G) 5 14. I the light of these bouds o γ, i 2004 Seager cosidered bouds o d g for cubic graphs ad showed that [11]): For ay graph of order, 1+ G γ G) see [4]) ad for a cubic graph G, d g 4 9. I this paper, we would like to study the forcig domiatio umber for Hamiltoia cubic graphs. I [8], the authors showed that: Lemma A. If r 2 or 3 mod 4), the γ G ) = γ G). Lemma B. If r 0 or 1 mod 4), the γ G ) = γ G) 1. Theorem C. If r 1 mod 4), the γ G 0 ) = m 4 m 3. 2. Forcig domiatio umber Remark 2.1 Let G = V, E) be the graph with V = {v 1, v 2,..., v } for = 2r ad E = {v i v j i j = 1 or r}. So G has two vertices v 1 ad v of degree two ad 2 vertices of degree three. By the graph G is the graph described i Fig.1. Fig.1. The graph G. For the followig we put N p [x] = {z z is oly domiated by x} {x}. Remark 2.2 Suppose that the graphs G ad G are two iduced subgraphs of G such that V G ) = V G) {v 1, v } ad V G ) = V G) {v 1 } or V G ) = V G) {v 2r }). Remark 2.3 Let G 0 be a graph of order m that = 2r, V G 0 ) = {v 11, v 12,..., v 1, v 21, v 22,..., v 2,..., v m1, v m2..., v m } ad E = m i=1 {v ijv il j l = 1 or r} {v i v i+1)1 i = 1, 2,..., m 1} {v 11 v m }. By the graph G 0 is 3-regular graph. Suppose that the graph G i
The Forcig Domiatio Number of Hamiltoia Cubic Graphs 55 is a iduced subgraph of G 0 with the vertices v i1, v i1,..., v i. By the graph G 0 is the graph described i Fig. 2. Fig. 2. The graph G 0. Propositio 2.4 If r 0 mod 4), the fg, γ) 2, otherwise fg, γ) = 1. proof First we suppose that r 1 mod 4). It is easy to see that fg, γ) > 0, because G has at least two miimum domiatig set. Suppose F = {v 1 } S where S is a miimum domiatig set. Sice γg) = 2 r/4 + 1, for two vertices v x ad v y i S, N[v x ] N[v y ] 6. This implies that {v 2, v r+1 } S =, the v r+3 S. A same argumet shows that v 5 S. Thus S must be cotais {v r+7, v 9,..., v 2r 2, v r }, therefore fg, γ) = 1. If r 2 mod 4), we cosider S = {v 2, v 6, v 10,..., v r, v r+4, v r+8,..., v 2r 6, v 2r 2 }. Assig the set F = {v 2 } the it follows fg, γ) 1, because N p [x] = 4 to each vertex x S. O the other had sice G has at least two miimum domiatig set. Hece fg, γ) = 1. If r 3 mod 4), for S = {v 1, v 5, v 9,..., v r 2, v r+3, v r+7,..., v 2r 4, v 2r }, the set F = {v 1 } shows that f G, γ) 1. Further, sice G has at least two miimum domiatig set, the it follows f G, γ) = 1. Fially let r 0 mod 4), we cosider S = {v 1, v 5, v 9,..., v r 3, v r+1, v r+3, v r+7,..., v 2r 5, v 2r 1 }. If F = {v 1, v r+1 }, a simple verificatio shows that f G, γ) 2. Propositio 2.5 If r 1 mod 4) the f G, γ = 0. Proof By Lemma B, we have γ G ) = 2 r/4. Now, we suppose that S is a arbitrary miimum domiatig set for G. Obviously for each vertex v x S, N p [v x ] = 4, so {v r 1, v r+2 } S. But {v 2r 2, v r 2 } S = therefore v 2r 3 S. Thus S must be cotais ) {v r 5, v r 9,..., v r+10, v r+6 }, the S is uiquely determied ad it follows that f G, γ = 0. Propositio 2.6 If r 0 mod 4) the f G, γ = 0. Proof Let r 0 mod 4) ad S be a arbitrary miimum domiatig set for G with V G ) = V G) {v 1 }. If {v 2r, v 2r 1 } S. Without loss of geerality, we assume that v 2r S the S must be cotais {v r+2, v r 2, v r 6,..., v 10, v 6, v 2r 4, v 2r 8,..., v r+8 }. O the other had by Lemma B, γ G ) = 2 r/4 Note that by Proof of Lemma B oe ca see
56 H.Abdollahzadeh Ahagar ad Pushpalatha L. γg ) = γg ) where r 0 mod 4)). So the vertices v 3, v 4, v r+4 ad v r+5 must be domiated by oe vertex ad this is impossible. Thus ecessarily v r S, but {v r 1, v 2r 1 } S = which implies v 2r 2 S. Fially the remaiig o-domiated vertices {v r+1, v r+2, v 2 } is just domiated by v r+2. Therefore the ) set S = {v 4, v 8,..., v r 4, v r, v r+2, v r+6,..., v 2r 2 } is uiquely determied which implies f G, γ = 0. 3. Mai Results Theorem 3.1 If r 2 or 3 mod 4), the f G 0, γ) = m. Proof Let r 2 mod 4) ad S be a miimum domiatig set for G 0. If there exists i {1, 2,..., m} such that S {v i1, v i } the it implies S G i > 2 r/4 + 1. Moreover γ G 0 ) = m 2 r/4 + 1). From this it immediately follows that there exists j {1, 2,..., m} {i} such that S G j < 2 r/4 + 1 ad this is cotrary to Lemma A. Hece S {v i1, v i } = for 1 i m. O the other had f G i, γ) = 1 for 1 i m which implies f G 0, γ) = m. Now we suppose that r 3 mod 4) ad S is miimum domiatig set for G 0, such that F = {v i1 1 i m} S. Sice v i1 S ad γ G 0 ) = 2 r/4 +2 the {v i2, v i3 } S = ad this implies v ir+3) S. With similar descriptio, we have {v i5, v i9,..., v ir 2), v ir+6), v ir+11),..., v i2r 4) } S. But for the remaiig o-domiated vertices v ir, v i2r) ad v i2r 1) ecessarily implies that v i2r) S. Hece S is the uique miimum domiatig set cotaiig F. Thus f G 0, γ) m. A trivial verificatio shows that f G, γ, f G, γ 1 for i {1, 2,..., m}, therefore f G 0, γ) = m. 1 if m 0 mod 3) Theorem 3.2 f G 0, γ) = 2 otherwise for r 1 mod 4). Proof If m 0 mod 3), we suppose that F = {v 1 } S ad S is a miimum domiatig set for G 0. By Theorem C, we have γ G 0 ) = m /4 m/3, the v 3,1 S. Here, we use the proof of Propositios 4 ad 5. From this the sets S V G 1 ), S V G 2 ), S V G 3 ) uiquely characterize. By cotiuig this process the set S uiquely obtai, the f G 0, γ) = 1. If m 1 or 2 mod 3), the the set F = {v 1, v m } uiquely characterize the miimum domiatig set for G 0, therefore f G 0, γ) = 2. m 3 + 1 if m 0 mod 3) Theorem 3.3 f G 0, γ) = for r 0 mod 4). + 3 otherwise m 3 Proof If m 0 mod 3) the set F = {v 21, v 2r+4), v 5r+4), v 8r+4),..., v m 1r+4) } determie the uique miimum domiatig set for G 0 the f G 0, γ) m/3 + 1. But γ G i ) = 2 r/4 for m/3 of G i s. Hece f G 0, γ) = m/3 + 1. The proof of the case m 1 or 2 mod 3) is similar to the previous case.
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