Rainbow domination in the Cartesian product of directed paths

Transcription

1 AUSTRALASIAN JOURNAL OF COMBINATORICS Volume 70() (018), Pages 9 61 Rainbow domination in the Cartesian product of directed paths Guoliang Hao College of Science, East China University of Technology Nanchang 001, Jiangxi, PR China guoliang-hao@16com Doost Ali Mojdeh Department of Mathematics University of Mazandaran Babolsar, Iran damojdeh@umzacir Shouliu Wei Department of Mathematics, Minjiang University Fuzhou 5011, Fujian, PR China wslwillow@16com Zhihong Xie College of Science, East China University of Technology Nanchang 001, Jiangxi, PR China xiezh168@16com Abstract For a positive integer k, ak-rainbow dominating function (krdf) of a digraph D is a function f from the vertex set V (D) tothesetofall subsets of the set {1,,,k} such that for any vertex v with f(v), u N (v) f(u) {1,,,k}, wheren (v) is the set of in-neighbors of v The weight of a krdf f of D is the value v V (D) f(v) Thek-rainbow domination number of a digraph D, denoted by γ rk (D), is the minimum weight of a krdf of D LetP m P n denote the Cartesian product of P m and P n,wherep m and P n denote the directed paths of order m and n, respectively In this paper, we determine the exact values of γ rk (P m P n ) for any positive integers k, m and n Corresponding author

2 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), Introduction and notation The concept of domination in graphs, with its many variations, has been extensively studied (see, for example, [,, 6, 7, 8, 9]) One of the variations on the domination theme is rainbow domination There are many results on rainbow domination in undirected graphs; for example, see [, 5, 10, 11] However, there exists a smaller number of results on rainbow domination in digraphs Our aim in this paper is to study the rainbow domination in digraphs Throughout this paper, D (V (D),A(D)) denotes a digraph with vertex set V (D) andarcseta(d) For two vertices u, v V (D), we use (u, v) todenotethe arc with direction from u to v, and we say that u is an in-neighbor of v Forv V (D) we denote the set of in-neighbors of v by N (v), and we define the in-degree of v by d (v) N (v) For a positive integer k, weusep({1,,,k}) to denote the set of all subsets of the set {1,,,k} Ak-rainbow dominating function (krdf) of a digraph D is a function f : V (D) P({1,,,k}) such that for any vertex v with f(v), u N (v) f(u) {1,,,k} The weight of a krdf f of D is the value ω(f) v V (D) f(v) The k-rainbow domination number of a digraph D, denoted by γ rk (D), is the minimum weight of a krdf of D AkRDF f of D with ω(f) γ rk (D) is called a γ rk (D)-function Let D 1 (V 1,A 1 )andd (V,A ) be two digraphs with disjoint vertex sets V 1 and V and disjoint arc sets A 1 and A, respectively The Cartesian product D 1 D is the digraph with vertex set V 1 V and for (x 1,y 1 ), (x,y ) V (D 1 D ), ((x 1,y 1 ), (x,y )) A(D 1 D ) if and only if either (x 1,x ) A 1 and y 1 y,or x 1 x and (y 1,y ) A For any y V,wedenotebyD y 1 the subdigraph of D 1 D induced by the vertex set {(x, y) :x V 1 } Then it is easy to see that D y 1 is isomorphic to D 1 Let P n denote the directed path of order n We emphasize that V (P n ) {0, 1,n 1} and A(P n ){(i, i +1):i 0, 1,,n }, throughout this paper As defined earlier, for each j {0, 1,,n 1}, wedenotebypm j the subdigraph of P m P n induced by the vertex set {(i, j) :i 0, 1,,m 1} In 01, rainbow domination in digraphs was introduced by Amjadi et al [1] However, to date no research has been done for the Cartesian product of two directed paths In this paper, we give the exact values of γ rk (P m P n ) for any positive integers k, m and n The -rainbow domination number In this section, we will determine the exact values of the -rainbow domination number in Cartesian products of directed paths Theorem 1 For any positive integer n, γ r (P 1 P n )γ r (P n )n

3 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), Proof Let f be a γ r (P n )-function Since d (0) 0, it follows from the definition of γ r (P n )-function that f(0) Moreover, if there exists some i {1,,,n 1} such that f(i), then clearly f(i 1) {1, } Thisimpliesthatγ r (P n )ω(f) n 1 i0 f(i) n On the other hand, it is easy to see that γ r(p n ) n Therefore, γ r (P 1 P n )γ r (P n )n Lemma Let n be any integer and let f be a γ r (P P n )-function such that {(i, j) V (P P n ):f((i, j)) } is minimum Then for each j {0, 1,,n 1}, f((0,j)) + f((1,j)) 1 Proof If f((1, 0)), then clearly f((0, 0)) + f((1, 0)) 1 Otherwise, f((1, 0)) Then by the definition of γ r (P P n )-function, we have f((0, 0)) {1, } and hence f((0, 0)) + f((1, 0)) 1 We now claim that for each j {1,,,n 1}, f((0,j)) + f((1,j)) 1 Suppose, to the contrary, that there exists some j {1,,, n 1} such that f((0,j)) + f((1,j)) 0Thisimpliesthatf((0,j)) f((1,j)) Then by the definition of γ r (P P n )-function, we have f((1,j 1)) {1, } Define a function g : V (P P n ) P({1, }) by { {1}, if v (1,j 1), (1,j), g(v) f(v), otherwise Then it is easy to see that g is a RDF of P P n with weight ω(g) ω(f) γ r (P P n ), implying that g is also a γ r (P P n )-function Moreover, it is easy to see that {(i, j) V (P P n ):f((i, j)) } {(i, j) V (P P n ):g((i, j)) } 1, contradicting the choice of f This completes the proof Theorem For any integer n, γ r (P P n ) n Proof Let f be a γ r (P P n )-function such that {(i, j) V (P m P n ):f((i, j)) } is minimum For each j {0, 1,,n 1}, leta j f((0,j)) + f((1,j)) and for each j {,,,n 1}, letb j a j + a j 1 + a j Then by Lemma, we have that for each j {1,,,n 1}, a j 1 Moreover, by the definition of γ r (P P n )-function, it is easy to verify that a 0 Claim 1 For each j {,,,n 1}, b j Proof of Claim 1 Suppose, to the contrary, that there exists some j 0 {,,, n 1} such that b j0 Note that for each j {0, 1,,n 1}, a j 1 Therefore, we have b j0 and hence a j0 a j0 1 a j0 1 Assume that f((1,j 0 )) 1, implying that f((0,j 0 )) Sincef is a γ r (P P n )-function, f((0,j 0 1)) {1, } Then we have a j0 1 f((0,j 0 1)), which is a contradiction Assume next that f((0,j 0 )) 1, implying that f((1,j 0 )) Sincef is a γ r (P P n )-function, {1, }\f((0,j 0 )) f((1,j 0 1)) and hence f((1,j 0 1)) 1 This implies that

4 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), f((0,j 0 1)) and hence f((0,j 0 )) {1, } Thus,a j0 f((0,j 0 )), which is also a contradiction So, this claim is true Therefore, if n 0(mod), then we have γ r (P P n ) and if n 1(mod), then we have γ r (P P n )a 0 + n 1 b k+ n n n k0 b k + k1 (n 1), n + n In both cases, we provide a RDF f : V (P P n ) P({1, }) defined by {1, }, if v (0, k) for0 k n 1, f {1}, if v (1, k +1)for0 k n (v), {}, if v (0, k +)for0 k n,, otherwise, and hence γ r (P P n ) ω(f )( n 1 If n (mod), then we have γ r (P P n )a 0 + a 1 + n +1 n + n +) b k+1 + k1 n (n ) n In this case, we also provide a RDF f : V (P P n ) P({1, }) defined by {1, }, if v (0, k) for0 k n, f {1}, if v (1, k +1)for0 k n (v), {}, if v (0, k +)for0 k n 5,, otherwise, and hence γ r (P P n ) ω(f ) ( n n +1 which completes our proof +1)+( n n, + n 5 +)

5 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), We shall determine the exact value of γ r (P m P n ) for any integers m, n For this purpose, we need some observations Observation Let m, n be any integers and let f be a γ r (P m P n )-function Then n 1 f((0, 0)) + f((i, 0)) + f((0,j)) m + n i1 and the equality holds if f((0, 0)) {1, }, f((1, 0)) f((0, 1)) and f((i, 0)) f((0,j)) 1for i, j Proof If f((i, 0)) for each i, then i0 f((i, 0)) m If there exists some vertex, say (i, 0), such that f((i, 0)), then by the definition of γ r (P m P n )- function, we have f((i 1, 0)) {1, }, which implies that i0 f((i, 0)) m Similarly, we get n 1 j0 f((0,j)) n Therefore, it is easy to verify that f((0, 0)) + i1 j1 n 1 f((i, 0)) + f((0,j)) m + n, establishing the desired lower bound If f((0, 0)) {1, }, f((1, 0)) f((0, 1)) and f((0,j)) 1 f((i, 0)) for i, j, then clearly which completes our proof j1 n 1 f((0, 0)) + f((i, 0)) + f((0,j)) m + n, i1 Observation 5 Let m, n be any integers and let f be a γ r (P m P n )-function such that f((0, 0)) {1, }, f((1, 0)) f((0, 1)) and f((i, 0)) f((0,j)) 1 for i, j Then (i) f((i, 1)) f((i +1, 1)) for i 0 (ii) i1 f((i, 1)) m j1 (iii) The equality in (ii) holds if f((k 1, 1)) 1 for 1 k m and f((k, 1)) 0for 1 k Proof (i) If f((i, 1)) f((i +1, 1)) for some i 0, then f((i +1, 0)) {1, }, a contradiction to our assumption Consequently, f((i, 1)) f((i +1, 1)) for i 0 (ii) According to (i), (ii) is trivial (iii) By our assumption, we have f((i, 0)) {1} for i Let f((k 1, 1)) {} for 1 k m and let f((k, 1)) for 1 k Then clearly i1 f((i, 1)) m

6 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), The following observation can be deduced by a similar discussion to that for Observation 5 Observation 6 Let m, n be any integers and let f be a γ r (P m P n )-function such that f satisfies the conditions of Observation 5 and (iii) of Observation 5 Then (i) f((i, )) f((i +1, )) 0for i 0 (ii) i1 f((i, )) (iii) The equality of (ii) holds if f((k, )) 1for 1 k and f((k 1, )) 0for 1 k m Now by using induction we have the following result Observation 7 Let m, n be any integers and let f be a γ r (P m P n )-function such that f satisfies the conditions of Observations 5 Then (i) f((i, j)) f((i +1,j)) 0for i 0 and j 1 (ii) i1 f((i, j)) m for odd j 1, and i1 j 1 f((i, j)) for even (iii) The first equality of (ii) holds if f((k 1,j)) 1for 1 k m and f((k, j)) 0for 1 k when j is odd, and the second equality of (ii) holds if f((k, j)) 1for 1 k and f((k 1,j)) 0for 1 k m when j 1 is even Now we may derive the following theorem Theorem 8 For any integers m, n, m +1 n +1 m n γ r (P m P n ) + Proof Let f be a γ r (P m P n )-function such that f satisfies the conditions of Observation 5 Therefore, by Observations and 7, we obtain γ r (P m P n ) f((0, 0)) + m + n + m + n + i1 n 1 n 1 f((i, 0)) + f((0,j)) + n/ l1 n/ l1 i1 m + j1 f((i, l 1)) + (n 1)/ l1 (n + m ) + n 1 (m 1) m +1 n +1 m n + j1 (n 1)/ l1 m 1 i1 f((i, j)) f((i, l)) i1

7 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), To show the upper bound, we now provide a RDF g : V (P m P n ) P({1, }) defined by {1, }, if i j 0, {1}, if i k 1for1 k m and j l 1for1 l n, {}, if i 0and j n 1, or g((i, j)) if i m 1andj 0, or if i k for 1 k m and j l for 1 l n,, otherwise Therefore, we have γ r (P m P n ) ω(g) m n m n + +(n ) + (m ) + m +1 n +1 m n +, which completes our proof The -rainbow domination number In this section, we will derive the exact value of the -rainbow domination number in Cartesian products of directed paths Lemma 1 For any integers m, n, { mn n 1 γ r (P m P n ), if m is odd, mn+m, if both m and n are even with m n Proof Let f be a γ r (P m P n )-function such that {(i, j) V (P m P n ):f((i, j)) } is minimum and let a j i0 f((i, j)) for each j {0, 1,,n 1} Claim For each i {0, 1,,m 1} and j {0, 1,,n 1}, f((i, 0)) and f((0,j)) ProofofClaim If there exists some i {1,,,m 1} such that f((i, 0)), then by the definition of γ r (P m P n )-function, we have f((i 1, 0)) {1,, } Define a function g : V (P m P n ) P({1,, }) by {1}, if v (i 1, 0), (i, 0), g(v) f(v) {1}, if v (i 1, 1), f(v), otherwise

8 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), Then it is easy to see that g is a RDF of P m P n with weight ω(g) ω(f) γ r (P m P n ), implying that g is also a γ r (P m P n )-function Moreover, clearly {(i, j) V (P m P n ):f((i, j)) } {(i, j) V (P m P n ):g((i, j)) } 1or, a contradiction to the choice of f Note that f((0, 0)) since d ((0, 0)) 0 Therefore, we have f((i, 0)) for each i {0, 1,,m 1} Similarly, we have f((0,j)) for each j {0, 1,,n 1} So, this claim is true Therefore, by Claim, we have a 0 i0 f((i, 0)) m Claim There do not exist two vertices (i, j) and(i +1,j), where i, j 1, of P m P n such that f((i, j)) f((i +1,j)) Proof of Claim Suppose, to the contrary, that there exist two vertices (i, j) and (i +1,j), where i, j 1, of P m P n such that f((i, j)) f((i +1,j)) Notethat f is a γ r (P m P n )-function Therefore, f((i+1,j 1)) {1,, } Define a function g : V (P m P n ) P({1,, }) by {1}, if v (i +1,j 1), (i +1,j), g(v) f(v) {1}, if v (i +,j 1), f(v), otherwise Then it is easy to see that g is a RDF of P m P n and ω(g) ω(f) γ r (P m P n ), implying that g is also a γ r (P m P n )-function Moreover, clearly {(i, j) V (P m P n ): f((i, j)) } {(i, j) V (P m P n ):g((i, j)) } 1 or, a contradiction to the choice of f So, this claim is true For each j {0, 1,,n 1}, letb j {(i, j) V (Pm j ):f((i, j)) } Thenby Claims and, we have that b 0 0andb j m for each j {1,,,n 1} Claim a 0 + a 1 m and a j 1 + a j m for each j {,,,n 1} Proof of Claim Let j {1,,,n 1} If f((i, j)) for each i, then clearly a j i0 f((i, j)) m and hence if j 1,thena 0 + a 1 a 0 + a j m + m m; if j {,,,n 1}, then a j 1 + a j (m b j 1 )+a j (m m )+m m Hencewemayassumethatthereexistssomei such that f((i, j)), implying that b j 1 For each k {1,,,b j },letf((i k,j)) Then it is easy to see that b j k1 f((i k,j)) 0 and by Claim, we have that for each k {1,,,b j }, f((i k 1,j)) Note that f is a γ r (P m P n )-function Therefore, for each k {1,,,b j }, f((i k,j 1)) f((i k 1,j)) {1,, } and hence f((i k,j 1)) +

9 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), f((i k 1,j)) Thus, a j 1 + a j f((i, j 1)) + i0 b j k1 i0 f((i, j)) [ f((i k,j 1)) + f((i k 1,j)) ]+ i X j f((i, j 1)) + i Y j f((i, j)) + b j k1 f((i k,j)) b j +(m b j 1 b j )+(m b j ) m b j 1, where X j {0, 1,,m 1}\{i 1,i,,i bj }, Y j X j \{i 1 1,i 1,,i bj 1} and i X j f((i, j 1)) m b j 1 b j Note that b 0 0andb j m for each j {1,,,n 1} Therefore, we have a 0 + a 1 m b 0 mand for each j {,,,n 1}, a j 1 + a j m b j 1 m m m So, this claim is true Therefore, it follows from Claim that if n is odd, then γ r (P m P n )ω(f) a 0 + m + n 1 implying that if both m and n are odd, then if n is even, then (a k 1 + a k ) n 1 k1 m, (m +1)(n 1) γ r (P m P n ) m + mn m 1 n 1 ; γ r (P m P n )ω(f) (a 0 + a 1 )+ m + n implying that if m is odd and n is even, then m, (a k + a k+1 ) n k1 (m +1)(n ) γ r (P m P n ) m + mn m 1 n 1

10 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), As a consequence, we have that if m is odd, then γ r (P m P n ) mn m 1 and if both m and n are even, then which completes our proof γ r (P m P n ) m + n m n 1 ; mn +m, Note that γ r (P m P n )γ r (P n P m ) Therefore, we have the following corollary Corollary For any integers m, n, { mn γ r (P m P n ) n 1, if n is odd, mn+n, if both m and n are even with n m As an immediate consequence of Lemma 1 and Corollary, we have the following result Corollary For integers m, n, { mn n 1, if m is odd, or if n is odd, γ r (P m P n ) mn+ max{m,n}, if both m and n are even Lemma For integers m, n, { mn n 1, if mornis odd, γ r (P m P n ) mn+ max{m,n}, if both m and n are even Proof If m or n is odd, then we provide a RDF f : V (P m P n ) P({1,, }) defined by {1, }, if i k 1for1 k m and j l 1for1 l n,, if i k 1for1 k m and f((i, j)) j l for 1 l n 1, or if i k for 1 k and j l 1for1 l n, {}, otherwise, and hence γ r (P m P n ) ω(f) m n ( m n m n 1 m 1 n ) +mn + + m 1 n 1 mn

11 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), If both m and n are even, then we also provide a RDF g : V (P m P n ) P({1,, }) defined by {1, }, if i {1,,,min{n,m 1}} and j k 1 for 1 k n i 1, or if i {,,,min{n,m 1}} and j k for n i k n, or if m n +1,il for n l m g((i, j)) and j k for 0 k n ; {}, if i 0and0 j n 1, or if j 0and1 i min{n 1,m 1}, or if i {,,,min{n,m }} and j k for 1 k n i, or if i {1,,,min{n 1,m 1}} and j k 1for n i+1 k n, or if m n +1,il +1for n l m and j k +1for0 k n ;, otherwise and hence { n +( m ω(g) )( n mn+n ), if n m n +( n )( n )+(m n )( n+ ) mn+m, if m n Therefore, if both m and n are even, then γ r (P m P n ) ω(g) which completes our proof mn+ max{m,n}, Using Corollary and Lemma, we can derive the following result Theorem 5 For any integers m, n, (1) If m is odd, or n is odd, then m 1 n 1 γ r (P m P n )mn () If both m and n are even, then γ r (P m P n ) mn +max{m, n} The k-rainbow domination number (k ) We now determine the exact value of γ rk (P m P n )fork Theorem 1 For any integers m, n 1 and k, γ rk (P m P n )mn

12 GUOLIANG HAO ET AL / AUSTRALAS J COMBIN 70 () (018), Proof Let f be a γ rk (P m P n )-function such that {(i, j) V (P m P n ):f((i, j)) } is minimum and let a j i0 f((i, j)) for each j {0, 1,,n 1} Claim 5 If f((i, j)), wherei, j 1, then f((i 1,j)) Proof of Claim 5 Suppose, to the contrary, that f((i 1,j)) 1 Without loss of generality, we may assume that f((i 1,j)) or {1} Notethatf is a γ rk (P m P n )- function Therefore, f((i, j 1)) {1,,, k} or {,,, k} Define a function g : V (P m P n ) P({1,,,k}) by {1}, if v (i, j 1), (i, j), g(v) f(v) {1}, if v (i +1,j 1), f(v), otherwise We observe that g is a krdf of P m P n with weight ω(g) ω(f) (k ) ω(f) γ rk (P m P n ), implying that g is also a γ rk (P m P n )-function Moreover, clearly {(i, j) V (P m P n ):f((i, j)) } {(i, j) V (P m P n ):g((i, j)) } 1 or, a contradiction to the choice of f So, this claim is true Using the similar method to Claim of Lemma 1, we have that for each i and j, f((i, 0)) and f((0,j)) This implies that a 0 m Moreover, it follows from Claim 5 that for each j {1,,,n 1}, a j m Therefore, n 1 γ rk (P m P n )ω(f) a j mn On the other hand, it is easy to see that γ rk (P n P m ) mn As a result, we have γ rk (P m P n )mn, which completes our proof j0 Acknowledgments This work was supported by Research Foundation of Education Bureau of Jiangxi Province of China (No GJJ150561), Doctor Fund of East China University of Technology (No DHBK01519), Natural Science Foundation of Fujian Province (No 016J0105), Science Foundation for the Education Department of Fujian Province (No JA1595) and Science Foundation of Minjiang University (No MYK1500) References [1] J Amjadi, A Bahremandpour, S M Sheikholeslami and L Volkmann, The rainbow domination number of a digraph, Kragujevac J Math 7 (01), [] B Brešar, M A Henning and D F Rall, Rainbow domination in graphs, Taiwanese J Math 1 (008), 1 5

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