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Extremal problems on weak Roman domination number
Information Processing Letters 138 (2018) 12–18
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Information Processing Letters www.elsevier.com/locate/ipl
Extremal problems on weak Roman domination number Enqiang Zhu ∗ , Zehui Shao Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China
a r t i c l e
i n f o
Article history: Received 6 April 2017 Received in revised form 5 May 2018 Accepted 29 May 2018 Available online 31 May 2018 Communicated by Jinhui Xu
a b s t r a c t We show that the weak Roman domination number of a connected n-vertex graph is at most 2n and characterize the graphs achieving equality. In addition, we provide a 3 constructive characterization of the trees for which the weak Roman domination number equals the Roman domination number and reveal several structural properties of these trees. This answers a problem posed in M. Chellali et al. (2014) [4]. © 2018 Elsevier B.V. All rights reserved.
Keywords: Weak Roman domination Roman domination Upper bound Combinatorial problems
1. Introduction All graphs considered in this paper are finite, simple and undirected. Let V (G ) and E (G ) be the vertex set and edge set of a graph G, respectively. For a vertex v ∈ V (G ), we use d G ( v ) to denote the degree of v in G and let N G ( v ) denote the neighborhood of v. The diameter of G is the maximum distance between vertices of G, denoted by diam(G ). An isolated vertex is a vertex with degree zero. A vertex of degree one is called a leaf, and its neighbor is called a support vertex. A vertex with at least two leaf neighbors is called a strong support vertex. We denote the star with one central vertex and k leaves by S k and the double star with exactly two adjacent central support vertices having p and q leaf neighbors by S p ,q . For two integers i , j such that i ≤ j, we use [i , j ] to denote the set {i , i + 1, i + 2, . . . , j }. For a graph G, let f be a function from V (G ) to {0, 1, 2}. Denote by V i , i = 0, 1, 2 the set of vertices assigned the label i under f . Thus, f can be viewed as a vertex partition of G such that V (G ) = { V 0 , V 1 , V 2 }, and we can equivalently write f = ( V 0 , V 1 , V 2 ). We call f an Ro-
*
Corresponding author. E-mail address: [email protected] (E. Zhu).
https://doi.org/10.1016/j.ipl.2018.05.009 0020-0190/© 2018 Elsevier B.V. All rights reserved.
man dominating function (RDF) of G if every vertex u ∈ V 0 is adjacent to at least one vertex v ∈ V 2 . A vertex u ∈ V 0 is said to be undefended with respect to f if it is not adjacent to a vertex v ∈ V 1 ∪ V 2 . We call f a weak Roman dominating function (WRDF) of G if each vertex u ∈ V 0 is adjacent to a vertex v ∈ V 1 ∪ V 2 , such that the function f defined by f (u ) = 1, f ( v ) = f ( v ) − 1, and f ( w ) = f ( w ) for all w ∈ V (G ) \ {u , v } has no undefended vertex. The weight of an RDF (resp. a WRDF) f of G, denoted by w ( f ), is the value v ∈ V (G ) f ( v ). The Roman domination number γ R ( G ) (resp. Weak Roman domination number γr (G )) is the minimum weight of an RDF (resp. a WRDF) of G. Obviously, γr (G ) ≤ γ R (G ). An RDF or a WRDF of G with weight ω is called a ω -RDF or ω -WRDF of G. For a subgraph G of G, we use f |G to denote the restriction of f to G . For a vertex v of a graph G, the open neighborhood of v is N G ( v ) = {u ∈ V (G )|uv ∈ E (G )} and the closed neighborhood of v is N G [ v ] = N G ( v ) ∪ { v }. For a set S ∈ V (G ), the open neighborhood of S is N G ( S ) = ∪ v ∈ S N G ( v ) and the closed neighborhood of S is N G [ S ] = N G ( S ) ∪ S. A vertex u is called a private neighbor of v with respect to S or simply an S-pn of v if N G [u ] ∩ S = { v }. Note that when v ∈ / S, v has no S-pn. The set pn( v , S ) = N G [ v ] − N G [ S \ { v }] of all S-pns of v is called the private neighbor set of v with respect to S. The external private neighbor set of v, denoted by epn( v , S ), is defined as epn( v , S ) = pn( v , S ) − { v }. Hence,
E. Zhu, Z. Shao / Information Processing Letters 138 (2018) 12–18
the set epn(v , S) consists of all S-pns of v that belong to V − S. For ease of presentation, we sometimes consider rooted trees. For a vertex v in a (rooted) tree T , we let C T ( v ) and D T ( v ) denote the sets of children and descendants of v, respectively. The maximal subtree at v is the subtree of T induced by D T ( v ) ∪ { v } and denoted by T v . Roman domination was first studied by Cockayne et al. [5] and has attracted the attention of many scholars; see [3,7,6,9]. Weak Roman domination is a less restrictive version of Roman domination, which was introduced by Henning and Hedetniemi [8]. Regarding the weak Roman domination number, Arumugam [2] and Chellali [4] proposed two problems to ask for a characterization of n-vertex trees T such that γr ( T ) = 2n and γr ( T ) = γ R ( T ), respec3 tively. Recently, José [1] provided a constructive characterization of the trees for which the Roman domination number strongly equals the weak Roman domination numfor a ber. In this paper, we first show that γr (G ) ≤ 2n 3 connected n-vertex graph G and characterize the graphs achieving equality, which answers Problem 3.6 proposed by Arumugam et al. [2]. Then, we provide a necessary and sufficient condition to characterize the trees T satisfying γ R ( T ) = γr ( T ) and therefore answer Problem 15 posed by Chellali et al. [4]. 2. Weak Roman domination number of connected graphs This section is devoted to showing that the weak Roman domination number of any n-vertex connected graph is at most 2n/3 and characterizing the n-vertex graph G satisfying γr (G ) = 2n/3. Since adding an edge cannot increase γr (G ), we sufficiently prove the bound for trees. Theorem 2.1. If T is an n-vertex tree, n ≥ 2, then γr ( T ) ≤
2n . 3
Proof. We proceed by induction on n. If n = 2 or 3, then T is a path on two vertices or three vertices, and γr ( T ) = 1 or 2. Let n ≥ 4. If T is a star, then γr ( T ) = 2 ≤ 23 n. If diamT = 3, then T is a double star S p ,q . In this case, when p ≥ 2 and q ≥ 2, γr ( T ) = 4. When p = 1 and q ≥ 2, γr ( T ) = 3. When p = 1 and q = 1, γr ( T ) = 2. Therefore, γr ( T ) ≤ 23 n. In the following, we assume that diamT ≥ 4 and every n -vertex tree T with n > n (≥ 2) has a WRDF f with weight at most 2n . Let P = u 0 u 1 . . . um be a longest path 3 in T , where m = diam(T). Clearly, d T (u i ) ≥ 2 for any i ∈ [1, m − 1], and d T (u 0 ) = d T (um ) = 1. Case 1. d T (um−1 ) = 3. Let T be the subtree of T by removing vertices um−1 and its leaf neighbors. Since diamT ≥ 4, we have n ≥ 3. Define
⎧ f ( v ), v ∈ V ( T ), ⎪ ⎪ ⎪ 2, v = um−1 and ⎪ ⎪ ⎨ d T ( u m −1 ) ≥ 4, f : V ( T ) → {0, 1, 2} : v → 1 , v = um−1 and ⎪ ⎪ ⎪ ⎪ d T ( u m −1 ) = 2, ⎪ ⎩ 0, v ∈ N T (um−1 ).
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Then, f is a WRDF of T , while w ( f ) = w ( f ) + 2 ≤ 23 (n −
4) + 2 < 23 n (when d T (um−1 ) ≥ 4) or w ( f ) = w ( f ) + 1 ≤ 2 (n 3
− 2) + 1 < 23 n (when d T (um−1 ) = 2). Case 2. d T (um−1 ) = 3. Let N T (um−2 ) \ {um−1 , um−3 } = { w 1 , w 2 , . . . , w }. Since P is the longest path, every vertex in N T ( w i ) \ {um−2 } is a leaf, and by Case 1, we may
assume w i has either zero or two leaf neighbors. Without loss of generality, let X 1 = { w 1 , w 2 , . . . , w } and X 2 = { w +1 , w +2 , . . . , w } be the sets of vertices with zero and two leaf neighbors, respectively, where 0 ≤ ≤ . Here, X 1 = ∅ when = 0. Consider the graph by removing edge um−2 um−3 from T ; let T 1 be the component containing um−3 , and T 2 be the component containing um−2 . We now define a function f from V ( T ) to {0, 1, 2} by letting f ( v ) = f ( v ) for any v ∈ T 1 , and when | X 1 | ≤ 1,
⎧ ⎨ 2, v = um−1 or v ∈ X 2 , f ( v ) = 1, v = w 1 and | X 1 | = 1, ⎩ 0, v ∈ V ( T 2 ) \ ( X 2 ∪ {um−1 , w 1 }),
and when | X 1 | ≥ 2,
f (v ) =
2, v ∈ ({um−1 , um−2 } ∪ X 2 ), 0, v ∈ V ( T 2 ) \ ( X 2 ∪ {um−1 , um−2 }).
Evidently, f is a WRDF of T . When | X 1 | ≤ 1, w ( f ) = min{ w ( f ) + 2 + 1 + 2( − 1) (the case of | X 1 | = 1), w ( f ) + 2 + 2 (the case of | X 1 | = 0)} = min{ 23 (n − 5) + 3, 23 (n −
4) + 2} < 23 n. When | X 1 | ≥ 2, w ( f ) = f ( w ) + 2 + 2 + 2( − 1 ) ≤ 23 [n − (3 + 1 + 1 + 3( − 1 ))] + 2( − 1 ) + 4 = 2 n + 43 − 23 1 . Hence, w ( f ) ≤ 23 n. In particular, when 3 | X 1 | ≥ 3, we have w ( f ) < 23 n. This also implies that if w ( f ) = 23 n, it must be the case that d T (um−1 ) = 3 and | X 1 | = 2. 2
Let f be a γ R (G )-RDF (or a γr (G )-WRDF) of a graph G, such that the number of vertices labeled with 1 is the minimum. We call such function f a special γ R (G )-RDF (or a special γr (G )-WRDF) of G and use F R (G ) (or Fr (G )) to denote the set of all special γ R (G )-RDFs (or γr (G )-WRDFs) of G. One can readily check that under a special γ R (G )-RDF (or a γr (G )-WRDF), any strong support vertex is labeled with 2. Otherwise, we can assign to the support vertex a weight of 2 and its two leaf neighbors a weight of 0. Particularly, let g be a special γ R (G )-RDF of G. Then, G does not contain three vertices u , v , w, such that vu , v w ∈ E (G ) and f ( v ) = 0, f (u ) = f ( w ) = 1, or vu ∈ E (G ) and f ( v ) = 2, f (u ) = 1. Theorem 2.2. Let T be a tree on n vertices, n ≥ 3. Then, γr ( T ) = 2n/3 if and only if every vertex of degree at least two in T has exactly two leaf neighbors. Proof. Let V and V be the set of leaves and vertices with exactly two leaf neighbors in T , respectively. If V = V \ V , then n = 3| V |. Let f ∈ Fr ( T ). Then, every vertex in V is labeled with 2, and every vertex in V is labeled with 0 under f . Therefore, γr ( T ) = w ( f ) = 2| V | = 2n/3. Conversely, if γr ( T ) = 2n/3, then let g be a special 2n -WRDF of T . Let P = u 0 u 1 . . . um be the longest path 3
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in T . Recall the proof of Theorem 2.1, which is by induction on n. In the base cases and Case 1, we obtain a WRDF with weight less than 23 n except for the cases where T = S 2 or T is the double star S 2,2 . In Case 2 and diamT ≥ 4, equality requires that um−1 , um−2 ∈ V . Therefore, g (um−1 ) = g (um−2 ) = 2. Let T 1 be the subtree obtained by deleting um−1 and its two leaf neighbors. Obviously, γr ( T 1 ) = 23 n1 , where n1 = | V ( T 1 )| (otherwise, let g be a γr ( T 1 )-WRDF of T 1 , based on which we can obtain a new WRDF of T with weight less than 2n by labeling um−1 3 with 2 and its two leaf neighbors with 0 and a contraction). Let P 1 be the longest path of T 1 , P 1 = u 0 u 1 . . . u p . With the same analysis, both u p −1 and u p −2 have exactly two leaf neighbors in T 1 . We then delete u p −1 and its two leaf neighbors from T 1 . The resulting graph is denoted by T 2 , and then γr ( T 2 ) = 23 n2 , where n2 = | V ( T 2 )|. Repeatedly conduct this procedure until some T q satisfies diamT q < 4. Then, we can deduce that T q is either star S 2 or double star S 2,2 , in which each vertex of degree at least 2 has exactly two leaf neighbors. Therefore, by the procession of deleting the vertices above, we can see that T is the tree with the specified structure of the theorem. 2 We then extend this characterization to all connected graphs. Theorem 2.3. If G is a connected n-vertex graph, then γr (G ) ≤ 2n/3, with equality if and only if every vertex with degree at least 2 is adjacent to exactly two leaf neighbors. Proof. By Theorem 2.1, the weak Roman domination number of every spanning tree of G is at most 2n , which im3 plies that γr (G ) ≤ 2n/3. Let V be the set of strong support vertices of G. Let f ∈ Fr . Then, every vertex of V is labeled with 2 under f , and every leaf adjacent to a vertex of V is labeled with 0. If G has the specified structure, then n = 3| V |, and γr (G ) = w ( f ) = 2| V | = 2n/3. Now, suppose that γr (G ) = 2n . Since adding edges can3 not increase γr , every spanning tree T of G has the struc-WRDF of T . ture in Theorem 2.2. Let g ∈ Fr ( T ) be a 2n 3 Denote by v 1 , v 2 , . . . , v m the m strong support vertices of T . It is the case that n = 3m, and every v i is adjacent to exactly 2 leaf neighbors. Obviously, V ⊆ { v 1 , v 2 , . . . , v m }. Since g ∈ Fr ( T ), we have g ( v i ) = 2 and g ( w ) = 0 for any w∈ / { v 1 , v 2 , . . . , v m } (i.e. w is a leaf of T ). We claim that E (G ) \ E ( T ) does not contain an edge e = xy such that g (x) = 2 or g ( y ) = 2. Otherwise, suppose that x is a leaf of T , and let x be the support vertex of x in T . Consider the graph T obtained from T by deleting edge xx and adding edge xy; it is clear to see that T is also a spanning tree of G. Since x has exactly two leaf neighbors in T , it follows that either x has one leaf neighbor, which contradicts Theorem 2.2, or d T (x ) = 1, which implies that T = T = S 2 and G is a triangle. However, we can readily check that the weak Roman domination number of a triangle is 1, which contradicts the assumption of γr (G ) = 2n/3. Hence, each edge in E (G ) \ E ( T ) joins two strong support vertices of T , which indicates that v i is also adjacent to exactly two leaf neighbors in G, and V = { v i |i ∈ [1, m]}. This completes the proof. 2
Fig. 1. Operations
σ1 , σ2 , σ3 , σ4 , σ5 .
3. Trees with γ R = γr 3.1. Construction of the family T In this section, we describe a procedure to construct trees T satisfying γr ( T ) = γ R ( T ). For this purpose, five operations, denoted by σ1 , σ2 , σ3 , σ4 , σ5 , are proposed. For a tree T with γr ( T ) = γ R ( T ) = γ , let V t ( T ) denote the set of vertices v such that γr ( T − v ) = γr ( T ). Further, let f be a special γ -RDF of T and u ∈ V t ( T ) be a vertex labeled with 0 under f . We refer to u as an e-vertex of T if any γ -WRDF g = ( V 0 , V 1 , V 2 ) of T with g (u ) = 1 satisfying N T (u ) ∩ epn(u , V 1 ∪ V 2 ) = ∅. We now present the specific procedures to construct trees of T . Let T 1 , T 2 , . . . , T m (m ≥ 1) be a sequence of stars such that T i , i = 1, 2, . . . , m is either S 0 or S k with k ≥ 2 (i.e., T i is a star with either 0 leaves or at least 2 leaves), and when m ≥ 2 T 1 = S 0 . First, define an RDF for each T i , denoted by f ∗ , such that if T i = S 0 , assign 1 to the unique vertex; otherwise, assign 2 to the central vertex and 0 to its leaves. Then, trees in T can be obtained from T 1 , T 2 , . . . , T m by adding m − 1 edges e 1 , e 2 , . . . , em−1 among them, for which e i connect T i and T i +1 by one of the five operations σ1 , σ2 , σ3 , σ4 , σ5 , where T 1 = T 1 and T i , i ≥ 2 is the tree built by adding edges e 1 , e 2 , . . . , e i −1 to T 1 , T 2 , . . . , T i according to the corresponding rules. The illustration of the definitions of these operations are shown in Fig. 1.
• σ1 : T i +1 = S 0 , and e i = uv satisfying f ∗ (u ) = 0, f ∗ ( v ) = 1 and u ∈ V t ( T i ); • σ2 : T i +1 = S k with k ≥ 2, and e i = uv satisfying f ∗ (u ) = 0, f ∗ ( v ) = 2, and u ∈ V t ( T i ); • σ3 : T i +1 = S k with k ≥ 3, and e i = uv such that f ∗ ( v ) = 0; • σ4 : T i +1 = S k with k ≥ 2, and e i = uv satisfying f ∗ (u ) = f ∗ ( v ) = 2; • σ5 : T i +1 = S 2 , and e i = uv satisfying that f ∗ (u ) = f ∗ ( v ) = 0, and u is an e-vertex of T i . Lemma 3.1. [8] If T is a path on n ≥ 3 vertices, then
2n , and γr ( T ) = 3n . 3 7
γR (T ) =
Lemma 3.2. Let T be a tree satisfying γ R ( T ) = γr ( T ) = γ , and let g = ( V 0 , V 1 , V 2 ) be a special γ -RDF of T . For a vertex u ∈ V ( T ), if g (u ) = 0, then T does not contain any γ -WRDF, under which u is labeled with 2.
E. Zhu, Z. Shao / Information Processing Letters 138 (2018) 12–18
Proof. Suppose to the contrary that there exists a γ -WRDF f of T such that f (u ) = 2. Let T be a rooted tree at u. Let C T (u ) = {u 1 , u 2 , . . . , u }. Obviously, g | T u is an RDF of T u i i for any i ∈ [1, ], and w ( f | T u ) ≤ w ( g | T u ). Since g (u ) = 0 i i and f (u ) = 2, there exists either one i ∈ [1, ] that satisfies w ( f | T u ) = w ( g | T u ) − 2 and w ( f | T u ) = w ( g | T u ) i
i
i
i
for any i ∈ [1, ] \ {i } or exactly two i , i ∈ [1, ] such that w ( f | T u ) = w ( g | T u ) − 1, w ( f | T u ) = w ( g | T u ) − 1, i
i
i
i
and w ( f | T u ) = w ( g | T u ) for any i ∈ [1, ] \ {i , i }. For i
i
the former case, assume i = 1, and let f = ∪j =2 ( g | T u ) ∪ j
f | T u1 ∪ ( f (u ) = 1). Then, f is a WRDF of T with weight w ( f ) − 1 = γ − 1 and a contradiction. For the latter case, we assume i = 1, i = 2. Then, for j = 1, 2, when | f (u j ) − g (u j )| = 1, we can deduce that u j has exactly one child, denoted by u 2 j , satisfying that | w ( f | T u ) − w ( g | T u )| = 1 and w ( f | T u ) = w ( g | T u ) 2j
2j
2j
2j
for any u 2 j ∈ C T (u j ) \ {u 2 j }. Furthermore, if | f (u 2 j ) − g (u 2 j )| = 1, then u 2 j has exactly one child, denoted by u 3 j , satisfying that | w ( f 3 j ) − w ( g 3 j )| = 1 and w ( f | T u ) = w ( g |T u
3j
3 j
) for any u 3 j ∈ C T (u 2 j ) \ {u 3 j }. Implementing this
procedure repeatedly, since T is finite, we can finally find a sequence of vertices u 1 , u 21 , . . . , u p 1 1 , u 2 , u 22 , . . . , u p 2 2 , such that | f (ukj ) − g (ukj )| = 1 for k ∈ [1, p j − 1], | f (u p j j ) − g (u p j j )| = 1, and | w ( f | T u ) − w ( g | T u )| = 1 for k ∈ k j
k j
[1, p j ]. Here, j ∈ [1, 2] and u 1 j = u j .
Claim 1. f (ukj ) = 1 and g (ukj ) = 1 for k ∈ [1, p j − 1]. Proof of the Claim 1. Otherwise, we have f (ukj ) = g (ukj ) = 1 since | f (ukj ) − g (ukj )| = 1. Then, for any child x of ukj , it is the case that g | T x is an RDF of T x , which implies that w ( g |T u ) ≥ w ( f |T u ). In addition, if w ( g | T u )> (k+1) j
w ( f |T u
(k+1) j
(k+1) j
(k+1) j
), then g = g | T −T u(k+1) j ∪ f | T u(k+1) j is a WRDF of
T with weight less than γ , and a contradiction. Therefore, w ( f | T u ) = w ( g | T u ), which yields a contradiction. 2 kj
kj
Claim 2. For any k ∈ [1, p j ], if f (uk j ) = 2, then g (uk j ) = 2. Proof of the Claim 2. Otherwise, suppose that g (uk j ) = 2. Since | w ( f | T uk j ) − w ( g | T uk j )| = 1, we have either w ( f | T uk j ) − w ( g | T uk j ) = 1 or w ( f | T uk j ) − w ( g | T uk j ) = −1. Then, f | T −T u ∪ g | T uk j or g | T −T u ∪ f | T uk j is a WRDF k j
of T with weight f. 2
k j
γ − 1, contradicting the assumption of
Claim 3. There does not exist any k ∈ [1, p j ], satisfying f (uk j ) = g (uk j ) = 0. Proof of the Claim 3. Suppose to the contrary that f (uk j ) = g (uk j ) = 0. Then, g | T u and f | T u is an (k +1) j
(k +1) j
RDF and a WRDF of T u (k +1) j , respectively. Therefore, we have that w ( g | T u ) = w ( f |T u ), which implies that (k +1) j
w ( g | T u k j ) = w ( f | T u
k j
(k +1) j
), and a contradiction. 2
Now, consider any two integers k, k + 1 ∈ [1, p j − 1]; if g (uk ) = g (uk+1 ) = 0 (or g (uk ) = g (uk+1 ) = 2),
15
then f (uk ) = f (uk+1 ) = 2 (or f (uk ) = f (uk+1 ) = 0) by Claims 1, 2, 3, and g | T u and f | T u are WRDFs of (k+1) j
T u (k+1) j . Since w ( g | T u (k+1) j ) = w ( f | T u g |T −T u
(k+1) j
(k+1) j
), we have that ∪ f | T u(k+1) j or f | T −T u(k+1) j ∪ g | T u(k+1) j is a WRDF (k+1) j
of T with weight less than γ , and a contradiction. This shows that { g (uk ), g (uk+1 )} (or { f (uk ), f (uk+1 )})={0,2}. Furthermore, if g (u ( p j −1) j ) = g (u p j j ) = 0 (or f (u ( p j −1) j ) = f (u p j j ) = 0), then f (u ( p j −1) j ) = 2 and f (u p j j ) = 1 (or g (u ( p j −1) j ) = 2 and g (u p j j ) = 1). With the analogous discussion, we can also obtain a WRDF of T with weight less than γ and yield a contradiction. In addition, we can also deduce for j = 1, 2, that f (u j ) = 0, and g (u j ) = 2 (when p j ≥ 2), or 1 (when p j = 1). This shows that the restriction of g and f to the subtree induced by V ∗ = {u , uk j | j = 1, 2, k ∈ [1, p j ]}, denoted by T , is the RDF and WRDF of T , respectively. Let v = u (k+1) j be an arbitrary child of ukj , k ∈ [1, p j − 1]. By the above analysis, we obtain that w ( f | T v ) = w ( g | T v ), and for any child v of u p j j , j = 1, 2, w ( f | T v ) = w ( g | T v ). Denote by G 1 , G 2 , . . . , G q the q components of T − V ∗ , and let G = G 1 ∪ G 2 ∪ . . . ∪ G q . Clearly, for each G i , i ∈ [1, q], it is the case that G i = T z for some child z of vertices of V ∗ . Then, we define a WRDF g of G as follows, where j ∈ [1, 2], k ∈ [1, p j − 1] and k ∈ [1, p j ]. Note that { f (ukj ), g (ukj )} ∩ {1} = ∅ by Claim 1, and there does not exist the cases f (uk j ) = g (uk j ) = 2 and f (uk j ) = g (uk j ) = 0 by Claims 2 and 3, respectively. (1) When z ∈ C T (u ), let g | T z = g | T z ; (2) When z ∈ / C T (u ) but z ∈ C T (ukj ), k ∈ [1, p j − 1], let g | T z = f | T z if f (ukj ) = 0, and let g | T z = g | T z if f (ukj ) = 0 but g (ukj ) = 0. (3) When z ∈ C T (u p j j ), it has that f (u p j j ) = 1 or g (u p j j ) = 1. If g | T z or f | T z is a WRDF of T z , let g | T z = g | T z or g | T z = f | T z . If both g | T z and f | T z are not WRDFs of T z , then f (u p j j ) = 1, g (u p j j ) = 2, and f ( z) = g ( z) = 0. In this case, we add z to V ∗ and exchange the labels of u p j j and z based on f . Clearly, the resulting label, denoted still by f , is also a γ -WRDF of T , and g (u p j j ) − f (u p j j ) = 2, f ( z) − g ( z) = 1. So, we can assume that this case does not exist under f . Now, we can see that g is a WRDF of G with weight w ( f |G )(= w ( g |G )). Since γ R ( T ) = γr ( T ), it has that T satisfies γ R ( T ) = γr ( T ) = w ( g | T ) = w ( f | T ). Given that T is a path, we can deduce from Lemma 3.1 that T is a path with 3 vertices, which implies that g (u 1 ) = g (u 2 ) = 1. However, in this case, we can relabel u with 2 and u 1 , u 2 with 0 based on g to gain a γ -WRDF of T with fewer vertices labeled 1. This contradicts the assumption that g is a special γ -WRDF of T . This completes the proof. 2 By Lemma 3.2, we can easily deduce that f ∗ is a γ R ( T i )-RDF of T i for i = 1, 2, . . . , m; in particular, f ∗ is a special γ R ( T i )-RDF of T i . Lemma 3.3. Let T i be a tree with γ R ( T i ) = γr ( T i ) = γ , and g ∈ F R ( T i ) with g (u ) = 0. If T i+1 is obtained from T i by operation σ1 , then γ R ( T i+1 ) = γr ( T i+1 ).
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Proof. Suppose to the contrary that γ R ( T i+1 ) = γr ( T i+1 ). Let f be a special γr ( T i+1 )-WRDF of T i+1 . Clearly, w ( f ) ≥ γ . Additionally, g together with g ( v ) = 1 is an RDF of T i+1 , which indicates that γ ≤ w ( f ) = γr ( T i+1 ) ≤ γ R ( T i+1 ) ≤ γ + 1. This implies that γr ( T i+1 ) = γ and γ R ( T i+1 ) = γ + 1. Since v is a leaf of T i+1 , we assume that f ( v ) = 0. (If f ( v ) = 2, v can be relabeled with 1. If f ( v ) = 1, then f (u ) = 0; otherwise, f | T is a WRDF of T i with weight i γ − 1 and a contradiction. Therefore, the restriction of f to T i − u is a WRDF of T i − u with weight γ − 1, which contradicts the assumption of u ∈ V t ( T i ).) Thus, f (u ) = 0. By Lemma 3.2, we have f (u ) = 2. Now, we assume f (u ) = 1. Then, v is the unique ( V 1 ∪ V 2 )-pn of u, i.e., epn(u , V 1 ∪ V 2 ) \ { v } = ∅. Therefore, f restrict to T i − u is a (γ − 1)-WRDF of T i − u, and a contradiction. 2 Lemma 3.4. Let T i be a tree with γ R ( T i ) = γr ( T i ) = γ and g ∈ F R ( T i ) such that g (u ) = 0. If T i+1 is obtained from T i by operation σ2 , then γ R ( T i+1 ) = γr ( T i+1 ). Proof. Suppose to the contrary, that γ R ( T i+1 ) = γr ( T i+1 ). Let f ∈ Fr ( T i+1 ). Then, f ( v ) = 2, and w ( f ) < γ + 2 since g together with g ( v ) = 2 is a (γ + 2)-RDF of T i+1 . Therefore, f (u ) = 0, and f | T −u is a WRDF of T i with weight i
w ( f ) − 2 < γ . This contradicts u ∈ V t ( T i ).
2
Lemma 3.5. If T i is a tree with γ R ( T i ) = γr ( T i ) = γ and g ∈ F R ( T i ). If T i+1 is obtained from T i by operation σ3 or σ4 , then γ R ( T i+1 ) = γr ( T i+1 ), where g (u ) ∈ {0, 1, 2} for σ3 and g (u ) = 2 for σ4 . Proof. Suppose to the contrary that γr ( T i+1 ) < γ R ( T i+1 ). Let f ∈ Fr ( T i+1 ). Since g together with g ( w ) = 2 or g ( v ) = 2 is a (γ + 2)-RDF of T i+1 , we can deduce that w ( f ) = γ + 1, where w is the support vertex of v in T i +1 in σ3 . As for σ3 , it has that f ( w ) = 2, and we can assume f ( v ) = 0 (since if f ( v ) = 0, then f (u ) = 0, and we can exchange the labels of u and v). Therefore, f | T is a WRDF i
of T i with weight w ( f ) − 2 < γ , and a contradiction. As for σ4 , it is clear that f ( v ) = 2. Since w ( f | T ) = i w ( f ) − 2 = γ − 1, we can see that f | T is not a WRDF i
of T i . Then, f (u ) = 0 and N T (u ) contains no vertices lai beled with 2 under f . Let N T (u ) = {u 1 , u 2 , . . . , u }. Root i T at u; then f | T u is a WRDF of T u i for i ∈ [1, ]. Therei fore, w ( f | T u ) ≥ w ( g | T u ). Since g (u ) = 2 and f (u ) = 0, i i we can see that there is exactly one u i , say u 1 , such that w ( f | T u1 ) = w ( g | T u1 )+1, and w ( f | T u ) = w ( g | T u ) for j ∈ j
j
[2, ]. We now define g as: g | T u j = f | T u j , g | T u1 = g | T u1 , and g (u ) = 1. Clearly, g is a (γ − 1)-WRDF of T i , and a contradiction. 2
Lemma 3.6. Let T i be a tree with γ R ( T i ) = γr ( T i ) = γ and u be a vertex labeled with 0 under a special γ -RDF g of T i . If T i+1 is obtained from T i by operation σ5 , then γ R ( T i+1 ) = γr ( T i+1 ). Proof. To the contrary, we assume γ R ( T i+1 ) = γ + 2 and γr ( T i+1 ) = γ + 1. Root T = T i at u. Let f = ( V 0 , V 1 , V 2 ) be a (γ + 1)-WRDF of T i+1 . Then, f (u ) = 0; otherwise, we have f ( v ) + f ( w ) + f (x) = 2, and f | T −u is a (γ − 1)-WRDF
of T − u, and a contradiction with u ∈ V t ( T ). By Lemma 3.2 f (u ) = 2. Suppose that f (u ) = 1. Since u is an e-vertex, we can see that epn(u , V 1 ∪ V 2 ) = ∅, say u ∈ epn(u , V 1 ∪ V 2 ). Then, f |u uv wx is not a WRDF, which implies that f is not a WRDF of T i+1 , and a contradiction. 2 Theorem 3.7. If T ∈ T , then γ R ( T ) = γr ( T ). Proof. Let T 1 , T 2 , . . . , T m be the m stars constructing T . Then, according to Lemmas 3.3, 3.4, 3.5 and 3.6, the conclusion can be proved easily by induction on m. 2 3.2. Constructive characterization of trees with γ R = γr In this section, we mainly characterize the structure of trees T such that γ R = γr . For this, we first present the following observation. Lemma 3.8. Let G be a graph with a cut edge e = uv, such that γ R (G ) = γr (G ) = γ . Let f = ( V 0 , V 1 , V 2 ) be a γ -RDF of G. If v ∈ V 1 ∪ V 2 and u ∈ / epn-(v, V 2 ); then γ R (G 1 ) = γr (G 1 ), γ R (G 2 ) = γr (G 2 ), and γ = γ R (G 1 ) + γ R (G 2 ), where G 1 and G 2 are the two components of G − e.
/ epn-( v , V 2 ), f |G i is an Proof. Since v ∈ V 1 ∪ V 2 and u ∈ RDF (also a WRDF) of G i , i = 1, 2. Then, w ( f ) = w ( f |G 1 ) + w ( f |G 2 ), and γ R (G 1 ) = γr (G 1 ) = w ( f |G 1 ) and γ R (G 2 ) = γr (G 2 ) = w ( f |G 2 ); otherwise, it will yield a WRDF of G with weight less than γ , and a contradiction. Therefore, γ R ( G ) = γ R ( G 1 ) + γ R ( G 2 ). 2 The following conclusion is a major structural characteristic for trees T satisfying γ R ( T ) = γr ( T ). Theorem 3.9. Let T be a tree on n vertices, n ≥ 2. If γr ( T ) = γ , then T contains a strong support vertex.
γR (T ) =
Proof. When n = 2, it is the case that γ R ( T ) = 2 and γr ( T ) = 1. When n = 3, T is a path on three vertices, which has a strong support vertex, and γ R ( T ) = γr ( T ) = 2. When
n ≥ 4, T is not a path by Lemma 3.1. In this case, if T is a star, then γ R ( T ) = γr ( T ) = 2, and T contains a strong support vertex. If T is a double star S p ,q , then when p or q is equal to 1, γ R ( T ) = γr ( T ) = 3; when p ≥ 2 and q ≥ 2, γ R ( T ) = γr ( T ) = 4. Since T is not a path when n ≥ 4, T contains a strong support vertex. In what follows, we assume diamT ≥ 4 and n > 4. We proceed by induction on n. Let T be a tree with diamT = m ≥ 4 and n > 4, such that γ R ( T ) = γr ( T ). And any subtree T of T on n (≥ 3) vertices, such that n < n and γ R ( T ) = γr ( T ) contains a strong support vertex. Let P = u 0 u 1 . . . um be a longest path of T , where m = diam(T). If d T (um−1 ) > 2, then um−1 is a strong support vertex. Therefore, we assume d T (um−1 ) = 2. Let N T (um−2 ) \ {um−1 , um−3 } = { w 1 , w 2 , . . . , w }. Evidently, N T ( w i ) \ {um−2 } contains only leaves, and d T ( w i ) ≤ 2; otherwise, w i is a strong support vertex. Let f be a γ -RDF of T , T 1 and T 2 be the two components of T − um−1 um−2 , which contain um−1 and um−2 , respectively. Then, T 1 is a path on two vertices, and γ R ( T 1 ) = γr ( T 1 ). By Lemma 3.8,
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it follows that either f (um−1 ) = 0 and um−1 is a V 2 -sp of um−2 , which implies f (um−2 ) = 2, or f (um−2 ) = 0 and um−2 is a V 2 -sp of um−1 , which implies f (um−1 ) = 2. Case 1. f (um−1 ) = 0, f (um−2 ) = 2, and um−1 is a V 2 -sp of um−2 . Then f (um ) = 1, and f | T 2 is an RDF of T 2 with weight γ − 1. Therefore, γ R ( T 2 ) ≤ γ − 1. Indeed, γr ( T 2 ) = γ R ( T 2 ) = γ − 1; otherwise, suppose that γr ( T 2 ) < γ − 1. Let f be a γr ( T 2 )-RDF of T 2 . Then, w ( f ) < γ − 1, and f can be extended to a WRDF of T with weight less than γ by letting f (um−1 ) = 0 and f (um ) = 1. This contradicts the assumption γr ( T ) = γ . Applying the inductive hypothesis to T 2 , we obtain a strong support vertex, denoted by x, of T 2 . If x is a strong support vertex of T , the conclusion holds; otherwise, x = um−3 , d T (um−2 ) = 2, and um−3 has exactly one leaf-neighbor in T , say y. If f (um−3 ) ∈ {0, 1}, then f ( y ) = 1. We relabel um−3 with 2, and y, um−2 with 0; if f (um−3 ) = 2, then f ( y ) = 0. We relabel um−2 with 0. In this way, we gain a WRDF of T with weight less than γ , which yields a contraction. Case 2. f (um−1 ) = 2 and f (um−2 ) = 0. Then, f (um ) = 0, and um−2 is a V 2 -sp of um−1 . Let T 1 and T 2 be the two subtrees of T − um−2 um−3 , which contain um−2 and um−3 , respectively. By Lemma 3.8, γ R ( T i ) = γr ( T i ) for i = 1, 2, which indicates | V ( T i )| ≥ 3. By the inductive hypothesis, T i , i = 1, 2 contains a strong support vertex. We assume T 1 = S 2 , i.e., d T (um−2 ) = 2; otherwise, the strong support vertex of T 1 is also a strong support vertex of T . In addition, one can readily check that any strong support vertex u of T 2 is a strong support vertex of T if u = um−4 . Therefore, we assume that T 2 contains only one strong support vertex um−4 , and um−4 is not a strong support vertex of T . Thus, d T (um−3 ) = 2 and in T um−4 has exactly one leaf neighbor, say z. We claim that f (um−3 ) = 0; otherwise, we can gain a WRDF of T with weight γ − 1 by relabeling um−1 with 1, a contradiction. This shows that f (um−4 ) = 2 and f ( z) = 0. Let T 3 and T 4 be the two components of T − um−3 um−4 , where T 3 contains um−3 and T 4 contains um−4 . Evidently, f | T is an RDF of T 4 . Since w ( f | T ) = 4
4
w ( f ) − 2 = γ − 2, it has that γ R ( T 4 ) ≤ γ − 2 and γr ( T 4 ) ≤ γ − 2. If γr ( T 4 ) < γr ( T ) − 2, then let f be a γr ( T 4 )-WRDF of T 4 . After relabeling um−3 and um−1 with 1, f can be extended to a WRDF of T with weight w ( f ) + 2 < γ , and a contradiction. Therefore, γ R ( T 4 ) = γr ( T 4 ) = γ − 2. Now, by the inductive hypothesis, T 4 contains a strong support vertex, which is also a strong support vertex of T since um−4 is adjacent to a leaf vertex. This completes the theorem. 2
Theorem 3.10. A tree T with γ R ( T ) = γr ( T ) = γ if and only if T ∈T . Proof. By Theorem 3.7, we sufficiently prove the necessity. If T is a star or a double star, the conclusion can be readily checked. Therefore, we assume diamT ≥ 4. We proceed by induction on | V ( T )|. Since γ R ( T ) = γr ( T ), it follows from Theorem 3.9 that T contains a strong support vertex v. Let f = ( V 0 , V 1 , V 2 ) be a special γ -RDF of T . Then, f ( v ) = 2, and f ( y ) = 1 for any y ∈ N T ( v ). Suppose that d T ( v ) = m. Without loss of generality, let { v 1 , v 2 , . . . , v } ⊆ V 0 , and { v +1 , v +2 , . . . , v m } ⊆
17
V 2 . Let N T ( v i ) \ { v } = { v i1 , v i2 , . . . , v ip i }, i = 1, 2, . . . , m. Note that when v i is a leaf of T , N T ( v i ) \ { v } = ∅. Now, (R1) delete v i v i j for i ∈ [1, ] and j ∈ [1, p i ]; (R2) delete v v i for i ∈ [ + 1, m]. By the above two rules, T is decomposed into a star S k , k ≥ 2, with central vertex v, and some components G i j and G k , which are components containing v i j and v k , respectively, where i ∈ [1, ], j ∈ [1, p i ] and k ∈ [ + 1, m]. Claim. f (u ) = 2 for every strong support vertex u of G ∈ { G i j , G k }. Proof of the Claim. Let u 1 , u 2 be two leaf neighbors of u in G . When u is a strong support vertex of T , f (u ) = 2. We assume u is not a support vertex of T . Then, u 1 or u 2 is a vertex of T with degree 2, say u 1 , and u 1 is adjacent to a leaf u of S k . Since T is a tree, there is only one edge u 1 u between G and S k . Therefore, u 2 is a leaf of T . Clearly, f (u 1 ) = 2 and f (u 2 ) = 2. If f (u ) = 1, then f (u 2 ) = 1. We relabel u with 2 and u 2 with 0; if f (u ) = 0, then f (u 1 ) = f (u 2 ) = 1. We relabel u with 2 and u i with 0 for i = 1, 2. Thus, we obtain a γ -RDF of T with less vertices labeled with 1, which contradicts the assumption of f . 2 Consider the deleted edges v i v i j and v v k ; we can see that v i ∈ / epn-(v i j , V 2 ) and v k ∈ / epn-(v, V 2 ). Therefore, by applying Lemma 3.8, we deduce that γ R (G i j ) = γr (G i j ) =
γi j , γ R (G k ) = γr (G k ) = γk , and γ = i=1 pj=i 1 γi j + m k=+1 γk . Thus, for each G i j and G k that is not a star, we
repeatedly conduct rules (R1) and (R2) until all of the resulting components are stars. In this way, T can be decomposed into a sequence of stars, denoted by T 1 , T 2 , . . . , T q such that either | V ( T i )| = 1 or | V ( T i )| ≥ 3, by deleting q − 1 edges, say e 1 , e 2 , . . . , eq−1 . Among these edges, we can always select an edge e such that one of the two components of T − e is isomorphic to some T i . Without loss of generality, we assume that e = e 1 , and T − e = T ∪ T 1 . By Lemma 3.8, it is easy to see that γ R ( T ) = γr ( T ), and by the inductive hypothesis, T ∈ T . Then, to prove the theorem, we sufficiently show that the operation of adding e 1 between T and T 1 belongs to one of the five operations. Let e 1 = xy such that x ∈ T , y ∈ T 1 . Case 1. f ( y ) = 1. It follows that f (x) = 0 and T 1 is an isolated vertex. Obviously, γ ( T ) = γ − 1, and f | T is a (γ − 1) RDF of T . Suppose that x ∈ / V t ( T ). Then, T − x has a WRDF f such that w ( f ) < w ( f | T ) = γ − 1. However, based on f , we label x and y with 1 and 0, respectively. The resulting labeling is a WRDF of T with weight w ( f ) + 1 < γ , which contradicts the selection of f . Therefore, x ∈ V t ( T ), and T ∈ T by σ1 . Case 2. f ( y ) = 2. Since f ∈ F R ( T ), f (x) = 1. If f (x) = 2, then T 1 is a star with at least two leaves, and T ∈ T by σ4 . If f (x) = 0, then T 1 is a star with central vertex y and at least two leaves. By (R1) and (R2), we can see that x is adjacent to a vertex labeled with 2 in T . Therefore, f | T is a (γ − 2)-RDF of T , which indicates that γr ( T ) ≤ γ − 2. Suppose that x ∈/ V t ( T ); let f be a WRDF of T − x such that w ( f ) < γr ( T ) ≤ γ − 2. Based on f , we label x and y with 0 and 2, respectively and label leaves of T 1 with 0. The resulting labeling is a WRDF of T with weight
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w ( f ) + 2 < γ and a contradiction. Therefore, x ∈ V t ( T ), and T ∈ T by σ2 . Case 3. f ( y ) = 0. Then, f | T is an RDF of T , y is a leaf of T 1 , and T 1 is a star with at least two leaves. When T 1 has at least three leaves, by σ3 T ∈ T . When T 1 = S 2 , i.e., T 1 is a star with a central vertex, say y 1 , and two leaves, say y , y 2 . By the claim, we have f ( y 1 ) = 2 and f ( y ) = f ( y 2 ) = 0. By ( R1) and ( R2), we have f (x) = 2. If f (x) = 1, then f | T together with labeling y 1 with 1 and labeling y , y 2 with 0 is a WRDF of T with weight γ − 1 and a contradiction. Hence, we assume f (x) = 0 in the following. In this case, suppose that γr ( T − x) = γr ( T ). Let f be a WRDF of T − x with weight γr ( T − x). Obviously, w ( f ) < γ − 2. Based on f , we label x and y 1 with 1 and label y and y 2 with 0. The resulting labeling is a WRDF of T with weight w ( f ) + 2 < γ , which contradicts the assumption γr ( T ) = γ . Therefore, γr ( T − x) = γr ( T ), i.e., x ∈ V t ( T ). In addition, let g = ( V 0 , V 1 , V 2 ) be a (γ − 2)-WRDF of T such that g (x) = 1. If epn(x, V 1 ∪ V 2 )∩ N T (x) = ∅, then g together with g ( y ) = g ( y 2 ) = 0 and g ( y 1 ) = 1, is a (γ − 1)-WRDF of T and a contradiction. Therefore, epn(x, V 1 ∪ V 2 )∩ N T (x) = ∅. This shows that x is an e-vertex of T , and T ∈ T by σ5 . This completes the proof. 2 4. Conclusion The approach in Section 3.1 can be adapted to construct trees with the same Roman and weak Roman domination number. In the process, we need to determine whether a vertex belongs to V t ( T ) or is a e-vertex, which can be addressed by simple linear time algorithms based on standard approaches [8]. Moreover, many constructive characteristics are present in section 3.2, by which we can easily check whether a tree possesses the desired property. On the basis of these results, we finally give necessary and sufficient conditions to characterize the trees T satisfying γ R ( T ) = γr ( T ).
In [4], the authors also put forward another problem to determine a characterization of the trees T with γr ( T ) = γr2 ( T ), where γr2 ( T ) is the 2-rainbow domination number of T . Based on a simple derivation from the result that γr (G ) ≤ γr2 (G ) ≤ γ R (G ) [4], we can readily see that any tree G with γ R (G ) = γr (G ) satisfies γr2 (G ) = γr (G ). However, how to determine these trees T such that γr (G ) = γr2 (G ) but γr2 (G ) = γ R (G ) remains an interesting and challenging work. Acknowledgements The project is supported by National Science Foundation of China under Grant (61672051, 61702075); China Postdoctoral Science Foundation under grant 2017M611223. References [1] J. Alvarado, S. Dantas, D. Rautenbach, Strong equality of Roman and weak roman domination in trees, Discrete Appl. Math. 208 (2016) 19–26. [2] S. Arumugam, K. Ebadi, M. Manrique, Co-Roman domination in graphs, Proc. Math. Sci. 125 (1) (2015) 1–10. [3] E.W. Chambers, B.P. Kinnersley, D.B. West, Extremal problems for Roman domination, SIAM J. Discrete Math. 23 (3) (2009) 1575–1586. [4] M. Chellali, T.W. Haynes, S.T. Hedetniemi, Bounds on weak Roman and 2-rainbow domination numbers, Discrete Appl. Math. 178 (2014) 27–32. [5] E.J. Cockayne, P.A.D. Jr, S.M. Hedetniemi, S.T. Hedetniemi, On Roman domination in graphs, Discrete Math. 278 (2004) 11–22. [6] S.T. Hedetniemi, R.R. Rubalcaba, P.J. Slater, M. Walsh, Few compare to the great Roman Empire, Congr. Numer. 217 (2013) 129–136. [7] M.A. Henning, A characterization of Roman trees, Discuss. Math., Graph Theory 22 (2002) 325–334. [8] M.A. Henning, S.T. Hedetniemi, Defending the Roman Empire—A new strategy, Discrete Math. 266 (2003) 239–251. [9] Y. Wu, H. Xing, Note on 2-rainbow domination and Roman domination in graphs, Appl. Math. Lett. 23 (2010) 706–709.
Contents lists available at ScienceDirect
Information Processing Letters www.elsevier.com/locate/ipl
Extremal problems on weak Roman domination number Enqiang Zhu ∗ , Zehui Shao Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China
a r t i c l e
i n f o
Article history: Received 6 April 2017 Received in revised form 5 May 2018 Accepted 29 May 2018 Available online 31 May 2018 Communicated by Jinhui Xu
a b s t r a c t We show that the weak Roman domination number of a connected n-vertex graph is at most 2n and characterize the graphs achieving equality. In addition, we provide a 3 constructive characterization of the trees for which the weak Roman domination number equals the Roman domination number and reveal several structural properties of these trees. This answers a problem posed in M. Chellali et al. (2014) [4]. © 2018 Elsevier B.V. All rights reserved.
Keywords: Weak Roman domination Roman domination Upper bound Combinatorial problems
1. Introduction All graphs considered in this paper are finite, simple and undirected. Let V (G ) and E (G ) be the vertex set and edge set of a graph G, respectively. For a vertex v ∈ V (G ), we use d G ( v ) to denote the degree of v in G and let N G ( v ) denote the neighborhood of v. The diameter of G is the maximum distance between vertices of G, denoted by diam(G ). An isolated vertex is a vertex with degree zero. A vertex of degree one is called a leaf, and its neighbor is called a support vertex. A vertex with at least two leaf neighbors is called a strong support vertex. We denote the star with one central vertex and k leaves by S k and the double star with exactly two adjacent central support vertices having p and q leaf neighbors by S p ,q . For two integers i , j such that i ≤ j, we use [i , j ] to denote the set {i , i + 1, i + 2, . . . , j }. For a graph G, let f be a function from V (G ) to {0, 1, 2}. Denote by V i , i = 0, 1, 2 the set of vertices assigned the label i under f . Thus, f can be viewed as a vertex partition of G such that V (G ) = { V 0 , V 1 , V 2 }, and we can equivalently write f = ( V 0 , V 1 , V 2 ). We call f an Ro-
*
Corresponding author. E-mail address: [email protected] (E. Zhu).
https://doi.org/10.1016/j.ipl.2018.05.009 0020-0190/© 2018 Elsevier B.V. All rights reserved.
man dominating function (RDF) of G if every vertex u ∈ V 0 is adjacent to at least one vertex v ∈ V 2 . A vertex u ∈ V 0 is said to be undefended with respect to f if it is not adjacent to a vertex v ∈ V 1 ∪ V 2 . We call f a weak Roman dominating function (WRDF) of G if each vertex u ∈ V 0 is adjacent to a vertex v ∈ V 1 ∪ V 2 , such that the function f defined by f (u ) = 1, f ( v ) = f ( v ) − 1, and f ( w ) = f ( w ) for all w ∈ V (G ) \ {u , v } has no undefended vertex. The weight of an RDF (resp. a WRDF) f of G, denoted by w ( f ), is the value v ∈ V (G ) f ( v ). The Roman domination number γ R ( G ) (resp. Weak Roman domination number γr (G )) is the minimum weight of an RDF (resp. a WRDF) of G. Obviously, γr (G ) ≤ γ R (G ). An RDF or a WRDF of G with weight ω is called a ω -RDF or ω -WRDF of G. For a subgraph G of G, we use f |G to denote the restriction of f to G . For a vertex v of a graph G, the open neighborhood of v is N G ( v ) = {u ∈ V (G )|uv ∈ E (G )} and the closed neighborhood of v is N G [ v ] = N G ( v ) ∪ { v }. For a set S ∈ V (G ), the open neighborhood of S is N G ( S ) = ∪ v ∈ S N G ( v ) and the closed neighborhood of S is N G [ S ] = N G ( S ) ∪ S. A vertex u is called a private neighbor of v with respect to S or simply an S-pn of v if N G [u ] ∩ S = { v }. Note that when v ∈ / S, v has no S-pn. The set pn( v , S ) = N G [ v ] − N G [ S \ { v }] of all S-pns of v is called the private neighbor set of v with respect to S. The external private neighbor set of v, denoted by epn( v , S ), is defined as epn( v , S ) = pn( v , S ) − { v }. Hence,
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the set epn(v , S) consists of all S-pns of v that belong to V − S. For ease of presentation, we sometimes consider rooted trees. For a vertex v in a (rooted) tree T , we let C T ( v ) and D T ( v ) denote the sets of children and descendants of v, respectively. The maximal subtree at v is the subtree of T induced by D T ( v ) ∪ { v } and denoted by T v . Roman domination was first studied by Cockayne et al. [5] and has attracted the attention of many scholars; see [3,7,6,9]. Weak Roman domination is a less restrictive version of Roman domination, which was introduced by Henning and Hedetniemi [8]. Regarding the weak Roman domination number, Arumugam [2] and Chellali [4] proposed two problems to ask for a characterization of n-vertex trees T such that γr ( T ) = 2n and γr ( T ) = γ R ( T ), respec3 tively. Recently, José [1] provided a constructive characterization of the trees for which the Roman domination number strongly equals the weak Roman domination numfor a ber. In this paper, we first show that γr (G ) ≤ 2n 3 connected n-vertex graph G and characterize the graphs achieving equality, which answers Problem 3.6 proposed by Arumugam et al. [2]. Then, we provide a necessary and sufficient condition to characterize the trees T satisfying γ R ( T ) = γr ( T ) and therefore answer Problem 15 posed by Chellali et al. [4]. 2. Weak Roman domination number of connected graphs This section is devoted to showing that the weak Roman domination number of any n-vertex connected graph is at most 2n/3 and characterizing the n-vertex graph G satisfying γr (G ) = 2n/3. Since adding an edge cannot increase γr (G ), we sufficiently prove the bound for trees. Theorem 2.1. If T is an n-vertex tree, n ≥ 2, then γr ( T ) ≤
2n . 3
Proof. We proceed by induction on n. If n = 2 or 3, then T is a path on two vertices or three vertices, and γr ( T ) = 1 or 2. Let n ≥ 4. If T is a star, then γr ( T ) = 2 ≤ 23 n. If diamT = 3, then T is a double star S p ,q . In this case, when p ≥ 2 and q ≥ 2, γr ( T ) = 4. When p = 1 and q ≥ 2, γr ( T ) = 3. When p = 1 and q = 1, γr ( T ) = 2. Therefore, γr ( T ) ≤ 23 n. In the following, we assume that diamT ≥ 4 and every n -vertex tree T with n > n (≥ 2) has a WRDF f with weight at most 2n . Let P = u 0 u 1 . . . um be a longest path 3 in T , where m = diam(T). Clearly, d T (u i ) ≥ 2 for any i ∈ [1, m − 1], and d T (u 0 ) = d T (um ) = 1. Case 1. d T (um−1 ) = 3. Let T be the subtree of T by removing vertices um−1 and its leaf neighbors. Since diamT ≥ 4, we have n ≥ 3. Define
⎧ f ( v ), v ∈ V ( T ), ⎪ ⎪ ⎪ 2, v = um−1 and ⎪ ⎪ ⎨ d T ( u m −1 ) ≥ 4, f : V ( T ) → {0, 1, 2} : v → 1 , v = um−1 and ⎪ ⎪ ⎪ ⎪ d T ( u m −1 ) = 2, ⎪ ⎩ 0, v ∈ N T (um−1 ).
13
Then, f is a WRDF of T , while w ( f ) = w ( f ) + 2 ≤ 23 (n −
4) + 2 < 23 n (when d T (um−1 ) ≥ 4) or w ( f ) = w ( f ) + 1 ≤ 2 (n 3
− 2) + 1 < 23 n (when d T (um−1 ) = 2). Case 2. d T (um−1 ) = 3. Let N T (um−2 ) \ {um−1 , um−3 } = { w 1 , w 2 , . . . , w }. Since P is the longest path, every vertex in N T ( w i ) \ {um−2 } is a leaf, and by Case 1, we may
assume w i has either zero or two leaf neighbors. Without loss of generality, let X 1 = { w 1 , w 2 , . . . , w } and X 2 = { w +1 , w +2 , . . . , w } be the sets of vertices with zero and two leaf neighbors, respectively, where 0 ≤ ≤ . Here, X 1 = ∅ when = 0. Consider the graph by removing edge um−2 um−3 from T ; let T 1 be the component containing um−3 , and T 2 be the component containing um−2 . We now define a function f from V ( T ) to {0, 1, 2} by letting f ( v ) = f ( v ) for any v ∈ T 1 , and when | X 1 | ≤ 1,
⎧ ⎨ 2, v = um−1 or v ∈ X 2 , f ( v ) = 1, v = w 1 and | X 1 | = 1, ⎩ 0, v ∈ V ( T 2 ) \ ( X 2 ∪ {um−1 , w 1 }),
and when | X 1 | ≥ 2,
f (v ) =
2, v ∈ ({um−1 , um−2 } ∪ X 2 ), 0, v ∈ V ( T 2 ) \ ( X 2 ∪ {um−1 , um−2 }).
Evidently, f is a WRDF of T . When | X 1 | ≤ 1, w ( f ) = min{ w ( f ) + 2 + 1 + 2( − 1) (the case of | X 1 | = 1), w ( f ) + 2 + 2 (the case of | X 1 | = 0)} = min{ 23 (n − 5) + 3, 23 (n −
4) + 2} < 23 n. When | X 1 | ≥ 2, w ( f ) = f ( w ) + 2 + 2 + 2( − 1 ) ≤ 23 [n − (3 + 1 + 1 + 3( − 1 ))] + 2( − 1 ) + 4 = 2 n + 43 − 23 1 . Hence, w ( f ) ≤ 23 n. In particular, when 3 | X 1 | ≥ 3, we have w ( f ) < 23 n. This also implies that if w ( f ) = 23 n, it must be the case that d T (um−1 ) = 3 and | X 1 | = 2. 2
Let f be a γ R (G )-RDF (or a γr (G )-WRDF) of a graph G, such that the number of vertices labeled with 1 is the minimum. We call such function f a special γ R (G )-RDF (or a special γr (G )-WRDF) of G and use F R (G ) (or Fr (G )) to denote the set of all special γ R (G )-RDFs (or γr (G )-WRDFs) of G. One can readily check that under a special γ R (G )-RDF (or a γr (G )-WRDF), any strong support vertex is labeled with 2. Otherwise, we can assign to the support vertex a weight of 2 and its two leaf neighbors a weight of 0. Particularly, let g be a special γ R (G )-RDF of G. Then, G does not contain three vertices u , v , w, such that vu , v w ∈ E (G ) and f ( v ) = 0, f (u ) = f ( w ) = 1, or vu ∈ E (G ) and f ( v ) = 2, f (u ) = 1. Theorem 2.2. Let T be a tree on n vertices, n ≥ 3. Then, γr ( T ) = 2n/3 if and only if every vertex of degree at least two in T has exactly two leaf neighbors. Proof. Let V and V be the set of leaves and vertices with exactly two leaf neighbors in T , respectively. If V = V \ V , then n = 3| V |. Let f ∈ Fr ( T ). Then, every vertex in V is labeled with 2, and every vertex in V is labeled with 0 under f . Therefore, γr ( T ) = w ( f ) = 2| V | = 2n/3. Conversely, if γr ( T ) = 2n/3, then let g be a special 2n -WRDF of T . Let P = u 0 u 1 . . . um be the longest path 3
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in T . Recall the proof of Theorem 2.1, which is by induction on n. In the base cases and Case 1, we obtain a WRDF with weight less than 23 n except for the cases where T = S 2 or T is the double star S 2,2 . In Case 2 and diamT ≥ 4, equality requires that um−1 , um−2 ∈ V . Therefore, g (um−1 ) = g (um−2 ) = 2. Let T 1 be the subtree obtained by deleting um−1 and its two leaf neighbors. Obviously, γr ( T 1 ) = 23 n1 , where n1 = | V ( T 1 )| (otherwise, let g be a γr ( T 1 )-WRDF of T 1 , based on which we can obtain a new WRDF of T with weight less than 2n by labeling um−1 3 with 2 and its two leaf neighbors with 0 and a contraction). Let P 1 be the longest path of T 1 , P 1 = u 0 u 1 . . . u p . With the same analysis, both u p −1 and u p −2 have exactly two leaf neighbors in T 1 . We then delete u p −1 and its two leaf neighbors from T 1 . The resulting graph is denoted by T 2 , and then γr ( T 2 ) = 23 n2 , where n2 = | V ( T 2 )|. Repeatedly conduct this procedure until some T q satisfies diamT q < 4. Then, we can deduce that T q is either star S 2 or double star S 2,2 , in which each vertex of degree at least 2 has exactly two leaf neighbors. Therefore, by the procession of deleting the vertices above, we can see that T is the tree with the specified structure of the theorem. 2 We then extend this characterization to all connected graphs. Theorem 2.3. If G is a connected n-vertex graph, then γr (G ) ≤ 2n/3, with equality if and only if every vertex with degree at least 2 is adjacent to exactly two leaf neighbors. Proof. By Theorem 2.1, the weak Roman domination number of every spanning tree of G is at most 2n , which im3 plies that γr (G ) ≤ 2n/3. Let V be the set of strong support vertices of G. Let f ∈ Fr . Then, every vertex of V is labeled with 2 under f , and every leaf adjacent to a vertex of V is labeled with 0. If G has the specified structure, then n = 3| V |, and γr (G ) = w ( f ) = 2| V | = 2n/3. Now, suppose that γr (G ) = 2n . Since adding edges can3 not increase γr , every spanning tree T of G has the struc-WRDF of T . ture in Theorem 2.2. Let g ∈ Fr ( T ) be a 2n 3 Denote by v 1 , v 2 , . . . , v m the m strong support vertices of T . It is the case that n = 3m, and every v i is adjacent to exactly 2 leaf neighbors. Obviously, V ⊆ { v 1 , v 2 , . . . , v m }. Since g ∈ Fr ( T ), we have g ( v i ) = 2 and g ( w ) = 0 for any w∈ / { v 1 , v 2 , . . . , v m } (i.e. w is a leaf of T ). We claim that E (G ) \ E ( T ) does not contain an edge e = xy such that g (x) = 2 or g ( y ) = 2. Otherwise, suppose that x is a leaf of T , and let x be the support vertex of x in T . Consider the graph T obtained from T by deleting edge xx and adding edge xy; it is clear to see that T is also a spanning tree of G. Since x has exactly two leaf neighbors in T , it follows that either x has one leaf neighbor, which contradicts Theorem 2.2, or d T (x ) = 1, which implies that T = T = S 2 and G is a triangle. However, we can readily check that the weak Roman domination number of a triangle is 1, which contradicts the assumption of γr (G ) = 2n/3. Hence, each edge in E (G ) \ E ( T ) joins two strong support vertices of T , which indicates that v i is also adjacent to exactly two leaf neighbors in G, and V = { v i |i ∈ [1, m]}. This completes the proof. 2
Fig. 1. Operations
σ1 , σ2 , σ3 , σ4 , σ5 .
3. Trees with γ R = γr 3.1. Construction of the family T In this section, we describe a procedure to construct trees T satisfying γr ( T ) = γ R ( T ). For this purpose, five operations, denoted by σ1 , σ2 , σ3 , σ4 , σ5 , are proposed. For a tree T with γr ( T ) = γ R ( T ) = γ , let V t ( T ) denote the set of vertices v such that γr ( T − v ) = γr ( T ). Further, let f be a special γ -RDF of T and u ∈ V t ( T ) be a vertex labeled with 0 under f . We refer to u as an e-vertex of T if any γ -WRDF g = ( V 0 , V 1 , V 2 ) of T with g (u ) = 1 satisfying N T (u ) ∩ epn(u , V 1 ∪ V 2 ) = ∅. We now present the specific procedures to construct trees of T . Let T 1 , T 2 , . . . , T m (m ≥ 1) be a sequence of stars such that T i , i = 1, 2, . . . , m is either S 0 or S k with k ≥ 2 (i.e., T i is a star with either 0 leaves or at least 2 leaves), and when m ≥ 2 T 1 = S 0 . First, define an RDF for each T i , denoted by f ∗ , such that if T i = S 0 , assign 1 to the unique vertex; otherwise, assign 2 to the central vertex and 0 to its leaves. Then, trees in T can be obtained from T 1 , T 2 , . . . , T m by adding m − 1 edges e 1 , e 2 , . . . , em−1 among them, for which e i connect T i and T i +1 by one of the five operations σ1 , σ2 , σ3 , σ4 , σ5 , where T 1 = T 1 and T i , i ≥ 2 is the tree built by adding edges e 1 , e 2 , . . . , e i −1 to T 1 , T 2 , . . . , T i according to the corresponding rules. The illustration of the definitions of these operations are shown in Fig. 1.
• σ1 : T i +1 = S 0 , and e i = uv satisfying f ∗ (u ) = 0, f ∗ ( v ) = 1 and u ∈ V t ( T i ); • σ2 : T i +1 = S k with k ≥ 2, and e i = uv satisfying f ∗ (u ) = 0, f ∗ ( v ) = 2, and u ∈ V t ( T i ); • σ3 : T i +1 = S k with k ≥ 3, and e i = uv such that f ∗ ( v ) = 0; • σ4 : T i +1 = S k with k ≥ 2, and e i = uv satisfying f ∗ (u ) = f ∗ ( v ) = 2; • σ5 : T i +1 = S 2 , and e i = uv satisfying that f ∗ (u ) = f ∗ ( v ) = 0, and u is an e-vertex of T i . Lemma 3.1. [8] If T is a path on n ≥ 3 vertices, then
2n , and γr ( T ) = 3n . 3 7
γR (T ) =
Lemma 3.2. Let T be a tree satisfying γ R ( T ) = γr ( T ) = γ , and let g = ( V 0 , V 1 , V 2 ) be a special γ -RDF of T . For a vertex u ∈ V ( T ), if g (u ) = 0, then T does not contain any γ -WRDF, under which u is labeled with 2.
E. Zhu, Z. Shao / Information Processing Letters 138 (2018) 12–18
Proof. Suppose to the contrary that there exists a γ -WRDF f of T such that f (u ) = 2. Let T be a rooted tree at u. Let C T (u ) = {u 1 , u 2 , . . . , u }. Obviously, g | T u is an RDF of T u i i for any i ∈ [1, ], and w ( f | T u ) ≤ w ( g | T u ). Since g (u ) = 0 i i and f (u ) = 2, there exists either one i ∈ [1, ] that satisfies w ( f | T u ) = w ( g | T u ) − 2 and w ( f | T u ) = w ( g | T u ) i
i
i
i
for any i ∈ [1, ] \ {i } or exactly two i , i ∈ [1, ] such that w ( f | T u ) = w ( g | T u ) − 1, w ( f | T u ) = w ( g | T u ) − 1, i
i
i
i
and w ( f | T u ) = w ( g | T u ) for any i ∈ [1, ] \ {i , i }. For i
i
the former case, assume i = 1, and let f = ∪j =2 ( g | T u ) ∪ j
f | T u1 ∪ ( f (u ) = 1). Then, f is a WRDF of T with weight w ( f ) − 1 = γ − 1 and a contradiction. For the latter case, we assume i = 1, i = 2. Then, for j = 1, 2, when | f (u j ) − g (u j )| = 1, we can deduce that u j has exactly one child, denoted by u 2 j , satisfying that | w ( f | T u ) − w ( g | T u )| = 1 and w ( f | T u ) = w ( g | T u ) 2j
2j
2j
2j
for any u 2 j ∈ C T (u j ) \ {u 2 j }. Furthermore, if | f (u 2 j ) − g (u 2 j )| = 1, then u 2 j has exactly one child, denoted by u 3 j , satisfying that | w ( f 3 j ) − w ( g 3 j )| = 1 and w ( f | T u ) = w ( g |T u
3j
3 j
) for any u 3 j ∈ C T (u 2 j ) \ {u 3 j }. Implementing this
procedure repeatedly, since T is finite, we can finally find a sequence of vertices u 1 , u 21 , . . . , u p 1 1 , u 2 , u 22 , . . . , u p 2 2 , such that | f (ukj ) − g (ukj )| = 1 for k ∈ [1, p j − 1], | f (u p j j ) − g (u p j j )| = 1, and | w ( f | T u ) − w ( g | T u )| = 1 for k ∈ k j
k j
[1, p j ]. Here, j ∈ [1, 2] and u 1 j = u j .
Claim 1. f (ukj ) = 1 and g (ukj ) = 1 for k ∈ [1, p j − 1]. Proof of the Claim 1. Otherwise, we have f (ukj ) = g (ukj ) = 1 since | f (ukj ) − g (ukj )| = 1. Then, for any child x of ukj , it is the case that g | T x is an RDF of T x , which implies that w ( g |T u ) ≥ w ( f |T u ). In addition, if w ( g | T u )> (k+1) j
w ( f |T u
(k+1) j
(k+1) j
(k+1) j
), then g = g | T −T u(k+1) j ∪ f | T u(k+1) j is a WRDF of
T with weight less than γ , and a contradiction. Therefore, w ( f | T u ) = w ( g | T u ), which yields a contradiction. 2 kj
kj
Claim 2. For any k ∈ [1, p j ], if f (uk j ) = 2, then g (uk j ) = 2. Proof of the Claim 2. Otherwise, suppose that g (uk j ) = 2. Since | w ( f | T uk j ) − w ( g | T uk j )| = 1, we have either w ( f | T uk j ) − w ( g | T uk j ) = 1 or w ( f | T uk j ) − w ( g | T uk j ) = −1. Then, f | T −T u ∪ g | T uk j or g | T −T u ∪ f | T uk j is a WRDF k j
of T with weight f. 2
k j
γ − 1, contradicting the assumption of
Claim 3. There does not exist any k ∈ [1, p j ], satisfying f (uk j ) = g (uk j ) = 0. Proof of the Claim 3. Suppose to the contrary that f (uk j ) = g (uk j ) = 0. Then, g | T u and f | T u is an (k +1) j
(k +1) j
RDF and a WRDF of T u (k +1) j , respectively. Therefore, we have that w ( g | T u ) = w ( f |T u ), which implies that (k +1) j
w ( g | T u k j ) = w ( f | T u
k j
(k +1) j
), and a contradiction. 2
Now, consider any two integers k, k + 1 ∈ [1, p j − 1]; if g (uk ) = g (uk+1 ) = 0 (or g (uk ) = g (uk+1 ) = 2),
15
then f (uk ) = f (uk+1 ) = 2 (or f (uk ) = f (uk+1 ) = 0) by Claims 1, 2, 3, and g | T u and f | T u are WRDFs of (k+1) j
T u (k+1) j . Since w ( g | T u (k+1) j ) = w ( f | T u g |T −T u
(k+1) j
(k+1) j
), we have that ∪ f | T u(k+1) j or f | T −T u(k+1) j ∪ g | T u(k+1) j is a WRDF (k+1) j
of T with weight less than γ , and a contradiction. This shows that { g (uk ), g (uk+1 )} (or { f (uk ), f (uk+1 )})={0,2}. Furthermore, if g (u ( p j −1) j ) = g (u p j j ) = 0 (or f (u ( p j −1) j ) = f (u p j j ) = 0), then f (u ( p j −1) j ) = 2 and f (u p j j ) = 1 (or g (u ( p j −1) j ) = 2 and g (u p j j ) = 1). With the analogous discussion, we can also obtain a WRDF of T with weight less than γ and yield a contradiction. In addition, we can also deduce for j = 1, 2, that f (u j ) = 0, and g (u j ) = 2 (when p j ≥ 2), or 1 (when p j = 1). This shows that the restriction of g and f to the subtree induced by V ∗ = {u , uk j | j = 1, 2, k ∈ [1, p j ]}, denoted by T , is the RDF and WRDF of T , respectively. Let v = u (k+1) j be an arbitrary child of ukj , k ∈ [1, p j − 1]. By the above analysis, we obtain that w ( f | T v ) = w ( g | T v ), and for any child v of u p j j , j = 1, 2, w ( f | T v ) = w ( g | T v ). Denote by G 1 , G 2 , . . . , G q the q components of T − V ∗ , and let G = G 1 ∪ G 2 ∪ . . . ∪ G q . Clearly, for each G i , i ∈ [1, q], it is the case that G i = T z for some child z of vertices of V ∗ . Then, we define a WRDF g of G as follows, where j ∈ [1, 2], k ∈ [1, p j − 1] and k ∈ [1, p j ]. Note that { f (ukj ), g (ukj )} ∩ {1} = ∅ by Claim 1, and there does not exist the cases f (uk j ) = g (uk j ) = 2 and f (uk j ) = g (uk j ) = 0 by Claims 2 and 3, respectively. (1) When z ∈ C T (u ), let g | T z = g | T z ; (2) When z ∈ / C T (u ) but z ∈ C T (ukj ), k ∈ [1, p j − 1], let g | T z = f | T z if f (ukj ) = 0, and let g | T z = g | T z if f (ukj ) = 0 but g (ukj ) = 0. (3) When z ∈ C T (u p j j ), it has that f (u p j j ) = 1 or g (u p j j ) = 1. If g | T z or f | T z is a WRDF of T z , let g | T z = g | T z or g | T z = f | T z . If both g | T z and f | T z are not WRDFs of T z , then f (u p j j ) = 1, g (u p j j ) = 2, and f ( z) = g ( z) = 0. In this case, we add z to V ∗ and exchange the labels of u p j j and z based on f . Clearly, the resulting label, denoted still by f , is also a γ -WRDF of T , and g (u p j j ) − f (u p j j ) = 2, f ( z) − g ( z) = 1. So, we can assume that this case does not exist under f . Now, we can see that g is a WRDF of G with weight w ( f |G )(= w ( g |G )). Since γ R ( T ) = γr ( T ), it has that T satisfies γ R ( T ) = γr ( T ) = w ( g | T ) = w ( f | T ). Given that T is a path, we can deduce from Lemma 3.1 that T is a path with 3 vertices, which implies that g (u 1 ) = g (u 2 ) = 1. However, in this case, we can relabel u with 2 and u 1 , u 2 with 0 based on g to gain a γ -WRDF of T with fewer vertices labeled 1. This contradicts the assumption that g is a special γ -WRDF of T . This completes the proof. 2 By Lemma 3.2, we can easily deduce that f ∗ is a γ R ( T i )-RDF of T i for i = 1, 2, . . . , m; in particular, f ∗ is a special γ R ( T i )-RDF of T i . Lemma 3.3. Let T i be a tree with γ R ( T i ) = γr ( T i ) = γ , and g ∈ F R ( T i ) with g (u ) = 0. If T i+1 is obtained from T i by operation σ1 , then γ R ( T i+1 ) = γr ( T i+1 ).
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Proof. Suppose to the contrary that γ R ( T i+1 ) = γr ( T i+1 ). Let f be a special γr ( T i+1 )-WRDF of T i+1 . Clearly, w ( f ) ≥ γ . Additionally, g together with g ( v ) = 1 is an RDF of T i+1 , which indicates that γ ≤ w ( f ) = γr ( T i+1 ) ≤ γ R ( T i+1 ) ≤ γ + 1. This implies that γr ( T i+1 ) = γ and γ R ( T i+1 ) = γ + 1. Since v is a leaf of T i+1 , we assume that f ( v ) = 0. (If f ( v ) = 2, v can be relabeled with 1. If f ( v ) = 1, then f (u ) = 0; otherwise, f | T is a WRDF of T i with weight i γ − 1 and a contradiction. Therefore, the restriction of f to T i − u is a WRDF of T i − u with weight γ − 1, which contradicts the assumption of u ∈ V t ( T i ).) Thus, f (u ) = 0. By Lemma 3.2, we have f (u ) = 2. Now, we assume f (u ) = 1. Then, v is the unique ( V 1 ∪ V 2 )-pn of u, i.e., epn(u , V 1 ∪ V 2 ) \ { v } = ∅. Therefore, f restrict to T i − u is a (γ − 1)-WRDF of T i − u, and a contradiction. 2 Lemma 3.4. Let T i be a tree with γ R ( T i ) = γr ( T i ) = γ and g ∈ F R ( T i ) such that g (u ) = 0. If T i+1 is obtained from T i by operation σ2 , then γ R ( T i+1 ) = γr ( T i+1 ). Proof. Suppose to the contrary, that γ R ( T i+1 ) = γr ( T i+1 ). Let f ∈ Fr ( T i+1 ). Then, f ( v ) = 2, and w ( f ) < γ + 2 since g together with g ( v ) = 2 is a (γ + 2)-RDF of T i+1 . Therefore, f (u ) = 0, and f | T −u is a WRDF of T i with weight i
w ( f ) − 2 < γ . This contradicts u ∈ V t ( T i ).
2
Lemma 3.5. If T i is a tree with γ R ( T i ) = γr ( T i ) = γ and g ∈ F R ( T i ). If T i+1 is obtained from T i by operation σ3 or σ4 , then γ R ( T i+1 ) = γr ( T i+1 ), where g (u ) ∈ {0, 1, 2} for σ3 and g (u ) = 2 for σ4 . Proof. Suppose to the contrary that γr ( T i+1 ) < γ R ( T i+1 ). Let f ∈ Fr ( T i+1 ). Since g together with g ( w ) = 2 or g ( v ) = 2 is a (γ + 2)-RDF of T i+1 , we can deduce that w ( f ) = γ + 1, where w is the support vertex of v in T i +1 in σ3 . As for σ3 , it has that f ( w ) = 2, and we can assume f ( v ) = 0 (since if f ( v ) = 0, then f (u ) = 0, and we can exchange the labels of u and v). Therefore, f | T is a WRDF i
of T i with weight w ( f ) − 2 < γ , and a contradiction. As for σ4 , it is clear that f ( v ) = 2. Since w ( f | T ) = i w ( f ) − 2 = γ − 1, we can see that f | T is not a WRDF i
of T i . Then, f (u ) = 0 and N T (u ) contains no vertices lai beled with 2 under f . Let N T (u ) = {u 1 , u 2 , . . . , u }. Root i T at u; then f | T u is a WRDF of T u i for i ∈ [1, ]. Therei fore, w ( f | T u ) ≥ w ( g | T u ). Since g (u ) = 2 and f (u ) = 0, i i we can see that there is exactly one u i , say u 1 , such that w ( f | T u1 ) = w ( g | T u1 )+1, and w ( f | T u ) = w ( g | T u ) for j ∈ j
j
[2, ]. We now define g as: g | T u j = f | T u j , g | T u1 = g | T u1 , and g (u ) = 1. Clearly, g is a (γ − 1)-WRDF of T i , and a contradiction. 2
Lemma 3.6. Let T i be a tree with γ R ( T i ) = γr ( T i ) = γ and u be a vertex labeled with 0 under a special γ -RDF g of T i . If T i+1 is obtained from T i by operation σ5 , then γ R ( T i+1 ) = γr ( T i+1 ). Proof. To the contrary, we assume γ R ( T i+1 ) = γ + 2 and γr ( T i+1 ) = γ + 1. Root T = T i at u. Let f = ( V 0 , V 1 , V 2 ) be a (γ + 1)-WRDF of T i+1 . Then, f (u ) = 0; otherwise, we have f ( v ) + f ( w ) + f (x) = 2, and f | T −u is a (γ − 1)-WRDF
of T − u, and a contradiction with u ∈ V t ( T ). By Lemma 3.2 f (u ) = 2. Suppose that f (u ) = 1. Since u is an e-vertex, we can see that epn(u , V 1 ∪ V 2 ) = ∅, say u ∈ epn(u , V 1 ∪ V 2 ). Then, f |u uv wx is not a WRDF, which implies that f is not a WRDF of T i+1 , and a contradiction. 2 Theorem 3.7. If T ∈ T , then γ R ( T ) = γr ( T ). Proof. Let T 1 , T 2 , . . . , T m be the m stars constructing T . Then, according to Lemmas 3.3, 3.4, 3.5 and 3.6, the conclusion can be proved easily by induction on m. 2 3.2. Constructive characterization of trees with γ R = γr In this section, we mainly characterize the structure of trees T such that γ R = γr . For this, we first present the following observation. Lemma 3.8. Let G be a graph with a cut edge e = uv, such that γ R (G ) = γr (G ) = γ . Let f = ( V 0 , V 1 , V 2 ) be a γ -RDF of G. If v ∈ V 1 ∪ V 2 and u ∈ / epn-(v, V 2 ); then γ R (G 1 ) = γr (G 1 ), γ R (G 2 ) = γr (G 2 ), and γ = γ R (G 1 ) + γ R (G 2 ), where G 1 and G 2 are the two components of G − e.
/ epn-( v , V 2 ), f |G i is an Proof. Since v ∈ V 1 ∪ V 2 and u ∈ RDF (also a WRDF) of G i , i = 1, 2. Then, w ( f ) = w ( f |G 1 ) + w ( f |G 2 ), and γ R (G 1 ) = γr (G 1 ) = w ( f |G 1 ) and γ R (G 2 ) = γr (G 2 ) = w ( f |G 2 ); otherwise, it will yield a WRDF of G with weight less than γ , and a contradiction. Therefore, γ R ( G ) = γ R ( G 1 ) + γ R ( G 2 ). 2 The following conclusion is a major structural characteristic for trees T satisfying γ R ( T ) = γr ( T ). Theorem 3.9. Let T be a tree on n vertices, n ≥ 2. If γr ( T ) = γ , then T contains a strong support vertex.
γR (T ) =
Proof. When n = 2, it is the case that γ R ( T ) = 2 and γr ( T ) = 1. When n = 3, T is a path on three vertices, which has a strong support vertex, and γ R ( T ) = γr ( T ) = 2. When
n ≥ 4, T is not a path by Lemma 3.1. In this case, if T is a star, then γ R ( T ) = γr ( T ) = 2, and T contains a strong support vertex. If T is a double star S p ,q , then when p or q is equal to 1, γ R ( T ) = γr ( T ) = 3; when p ≥ 2 and q ≥ 2, γ R ( T ) = γr ( T ) = 4. Since T is not a path when n ≥ 4, T contains a strong support vertex. In what follows, we assume diamT ≥ 4 and n > 4. We proceed by induction on n. Let T be a tree with diamT = m ≥ 4 and n > 4, such that γ R ( T ) = γr ( T ). And any subtree T of T on n (≥ 3) vertices, such that n < n and γ R ( T ) = γr ( T ) contains a strong support vertex. Let P = u 0 u 1 . . . um be a longest path of T , where m = diam(T). If d T (um−1 ) > 2, then um−1 is a strong support vertex. Therefore, we assume d T (um−1 ) = 2. Let N T (um−2 ) \ {um−1 , um−3 } = { w 1 , w 2 , . . . , w }. Evidently, N T ( w i ) \ {um−2 } contains only leaves, and d T ( w i ) ≤ 2; otherwise, w i is a strong support vertex. Let f be a γ -RDF of T , T 1 and T 2 be the two components of T − um−1 um−2 , which contain um−1 and um−2 , respectively. Then, T 1 is a path on two vertices, and γ R ( T 1 ) = γr ( T 1 ). By Lemma 3.8,
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it follows that either f (um−1 ) = 0 and um−1 is a V 2 -sp of um−2 , which implies f (um−2 ) = 2, or f (um−2 ) = 0 and um−2 is a V 2 -sp of um−1 , which implies f (um−1 ) = 2. Case 1. f (um−1 ) = 0, f (um−2 ) = 2, and um−1 is a V 2 -sp of um−2 . Then f (um ) = 1, and f | T 2 is an RDF of T 2 with weight γ − 1. Therefore, γ R ( T 2 ) ≤ γ − 1. Indeed, γr ( T 2 ) = γ R ( T 2 ) = γ − 1; otherwise, suppose that γr ( T 2 ) < γ − 1. Let f be a γr ( T 2 )-RDF of T 2 . Then, w ( f ) < γ − 1, and f can be extended to a WRDF of T with weight less than γ by letting f (um−1 ) = 0 and f (um ) = 1. This contradicts the assumption γr ( T ) = γ . Applying the inductive hypothesis to T 2 , we obtain a strong support vertex, denoted by x, of T 2 . If x is a strong support vertex of T , the conclusion holds; otherwise, x = um−3 , d T (um−2 ) = 2, and um−3 has exactly one leaf-neighbor in T , say y. If f (um−3 ) ∈ {0, 1}, then f ( y ) = 1. We relabel um−3 with 2, and y, um−2 with 0; if f (um−3 ) = 2, then f ( y ) = 0. We relabel um−2 with 0. In this way, we gain a WRDF of T with weight less than γ , which yields a contraction. Case 2. f (um−1 ) = 2 and f (um−2 ) = 0. Then, f (um ) = 0, and um−2 is a V 2 -sp of um−1 . Let T 1 and T 2 be the two subtrees of T − um−2 um−3 , which contain um−2 and um−3 , respectively. By Lemma 3.8, γ R ( T i ) = γr ( T i ) for i = 1, 2, which indicates | V ( T i )| ≥ 3. By the inductive hypothesis, T i , i = 1, 2 contains a strong support vertex. We assume T 1 = S 2 , i.e., d T (um−2 ) = 2; otherwise, the strong support vertex of T 1 is also a strong support vertex of T . In addition, one can readily check that any strong support vertex u of T 2 is a strong support vertex of T if u = um−4 . Therefore, we assume that T 2 contains only one strong support vertex um−4 , and um−4 is not a strong support vertex of T . Thus, d T (um−3 ) = 2 and in T um−4 has exactly one leaf neighbor, say z. We claim that f (um−3 ) = 0; otherwise, we can gain a WRDF of T with weight γ − 1 by relabeling um−1 with 1, a contradiction. This shows that f (um−4 ) = 2 and f ( z) = 0. Let T 3 and T 4 be the two components of T − um−3 um−4 , where T 3 contains um−3 and T 4 contains um−4 . Evidently, f | T is an RDF of T 4 . Since w ( f | T ) = 4
4
w ( f ) − 2 = γ − 2, it has that γ R ( T 4 ) ≤ γ − 2 and γr ( T 4 ) ≤ γ − 2. If γr ( T 4 ) < γr ( T ) − 2, then let f be a γr ( T 4 )-WRDF of T 4 . After relabeling um−3 and um−1 with 1, f can be extended to a WRDF of T with weight w ( f ) + 2 < γ , and a contradiction. Therefore, γ R ( T 4 ) = γr ( T 4 ) = γ − 2. Now, by the inductive hypothesis, T 4 contains a strong support vertex, which is also a strong support vertex of T since um−4 is adjacent to a leaf vertex. This completes the theorem. 2
Theorem 3.10. A tree T with γ R ( T ) = γr ( T ) = γ if and only if T ∈T . Proof. By Theorem 3.7, we sufficiently prove the necessity. If T is a star or a double star, the conclusion can be readily checked. Therefore, we assume diamT ≥ 4. We proceed by induction on | V ( T )|. Since γ R ( T ) = γr ( T ), it follows from Theorem 3.9 that T contains a strong support vertex v. Let f = ( V 0 , V 1 , V 2 ) be a special γ -RDF of T . Then, f ( v ) = 2, and f ( y ) = 1 for any y ∈ N T ( v ). Suppose that d T ( v ) = m. Without loss of generality, let { v 1 , v 2 , . . . , v } ⊆ V 0 , and { v +1 , v +2 , . . . , v m } ⊆
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V 2 . Let N T ( v i ) \ { v } = { v i1 , v i2 , . . . , v ip i }, i = 1, 2, . . . , m. Note that when v i is a leaf of T , N T ( v i ) \ { v } = ∅. Now, (R1) delete v i v i j for i ∈ [1, ] and j ∈ [1, p i ]; (R2) delete v v i for i ∈ [ + 1, m]. By the above two rules, T is decomposed into a star S k , k ≥ 2, with central vertex v, and some components G i j and G k , which are components containing v i j and v k , respectively, where i ∈ [1, ], j ∈ [1, p i ] and k ∈ [ + 1, m]. Claim. f (u ) = 2 for every strong support vertex u of G ∈ { G i j , G k }. Proof of the Claim. Let u 1 , u 2 be two leaf neighbors of u in G . When u is a strong support vertex of T , f (u ) = 2. We assume u is not a support vertex of T . Then, u 1 or u 2 is a vertex of T with degree 2, say u 1 , and u 1 is adjacent to a leaf u of S k . Since T is a tree, there is only one edge u 1 u between G and S k . Therefore, u 2 is a leaf of T . Clearly, f (u 1 ) = 2 and f (u 2 ) = 2. If f (u ) = 1, then f (u 2 ) = 1. We relabel u with 2 and u 2 with 0; if f (u ) = 0, then f (u 1 ) = f (u 2 ) = 1. We relabel u with 2 and u i with 0 for i = 1, 2. Thus, we obtain a γ -RDF of T with less vertices labeled with 1, which contradicts the assumption of f . 2 Consider the deleted edges v i v i j and v v k ; we can see that v i ∈ / epn-(v i j , V 2 ) and v k ∈ / epn-(v, V 2 ). Therefore, by applying Lemma 3.8, we deduce that γ R (G i j ) = γr (G i j ) =
γi j , γ R (G k ) = γr (G k ) = γk , and γ = i=1 pj=i 1 γi j + m k=+1 γk . Thus, for each G i j and G k that is not a star, we
repeatedly conduct rules (R1) and (R2) until all of the resulting components are stars. In this way, T can be decomposed into a sequence of stars, denoted by T 1 , T 2 , . . . , T q such that either | V ( T i )| = 1 or | V ( T i )| ≥ 3, by deleting q − 1 edges, say e 1 , e 2 , . . . , eq−1 . Among these edges, we can always select an edge e such that one of the two components of T − e is isomorphic to some T i . Without loss of generality, we assume that e = e 1 , and T − e = T ∪ T 1 . By Lemma 3.8, it is easy to see that γ R ( T ) = γr ( T ), and by the inductive hypothesis, T ∈ T . Then, to prove the theorem, we sufficiently show that the operation of adding e 1 between T and T 1 belongs to one of the five operations. Let e 1 = xy such that x ∈ T , y ∈ T 1 . Case 1. f ( y ) = 1. It follows that f (x) = 0 and T 1 is an isolated vertex. Obviously, γ ( T ) = γ − 1, and f | T is a (γ − 1) RDF of T . Suppose that x ∈ / V t ( T ). Then, T − x has a WRDF f such that w ( f ) < w ( f | T ) = γ − 1. However, based on f , we label x and y with 1 and 0, respectively. The resulting labeling is a WRDF of T with weight w ( f ) + 1 < γ , which contradicts the selection of f . Therefore, x ∈ V t ( T ), and T ∈ T by σ1 . Case 2. f ( y ) = 2. Since f ∈ F R ( T ), f (x) = 1. If f (x) = 2, then T 1 is a star with at least two leaves, and T ∈ T by σ4 . If f (x) = 0, then T 1 is a star with central vertex y and at least two leaves. By (R1) and (R2), we can see that x is adjacent to a vertex labeled with 2 in T . Therefore, f | T is a (γ − 2)-RDF of T , which indicates that γr ( T ) ≤ γ − 2. Suppose that x ∈/ V t ( T ); let f be a WRDF of T − x such that w ( f ) < γr ( T ) ≤ γ − 2. Based on f , we label x and y with 0 and 2, respectively and label leaves of T 1 with 0. The resulting labeling is a WRDF of T with weight
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w ( f ) + 2 < γ and a contradiction. Therefore, x ∈ V t ( T ), and T ∈ T by σ2 . Case 3. f ( y ) = 0. Then, f | T is an RDF of T , y is a leaf of T 1 , and T 1 is a star with at least two leaves. When T 1 has at least three leaves, by σ3 T ∈ T . When T 1 = S 2 , i.e., T 1 is a star with a central vertex, say y 1 , and two leaves, say y , y 2 . By the claim, we have f ( y 1 ) = 2 and f ( y ) = f ( y 2 ) = 0. By ( R1) and ( R2), we have f (x) = 2. If f (x) = 1, then f | T together with labeling y 1 with 1 and labeling y , y 2 with 0 is a WRDF of T with weight γ − 1 and a contradiction. Hence, we assume f (x) = 0 in the following. In this case, suppose that γr ( T − x) = γr ( T ). Let f be a WRDF of T − x with weight γr ( T − x). Obviously, w ( f ) < γ − 2. Based on f , we label x and y 1 with 1 and label y and y 2 with 0. The resulting labeling is a WRDF of T with weight w ( f ) + 2 < γ , which contradicts the assumption γr ( T ) = γ . Therefore, γr ( T − x) = γr ( T ), i.e., x ∈ V t ( T ). In addition, let g = ( V 0 , V 1 , V 2 ) be a (γ − 2)-WRDF of T such that g (x) = 1. If epn(x, V 1 ∪ V 2 )∩ N T (x) = ∅, then g together with g ( y ) = g ( y 2 ) = 0 and g ( y 1 ) = 1, is a (γ − 1)-WRDF of T and a contradiction. Therefore, epn(x, V 1 ∪ V 2 )∩ N T (x) = ∅. This shows that x is an e-vertex of T , and T ∈ T by σ5 . This completes the proof. 2 4. Conclusion The approach in Section 3.1 can be adapted to construct trees with the same Roman and weak Roman domination number. In the process, we need to determine whether a vertex belongs to V t ( T ) or is a e-vertex, which can be addressed by simple linear time algorithms based on standard approaches [8]. Moreover, many constructive characteristics are present in section 3.2, by which we can easily check whether a tree possesses the desired property. On the basis of these results, we finally give necessary and sufficient conditions to characterize the trees T satisfying γ R ( T ) = γr ( T ).
In [4], the authors also put forward another problem to determine a characterization of the trees T with γr ( T ) = γr2 ( T ), where γr2 ( T ) is the 2-rainbow domination number of T . Based on a simple derivation from the result that γr (G ) ≤ γr2 (G ) ≤ γ R (G ) [4], we can readily see that any tree G with γ R (G ) = γr (G ) satisfies γr2 (G ) = γr (G ). However, how to determine these trees T such that γr (G ) = γr2 (G ) but γr2 (G ) = γ R (G ) remains an interesting and challenging work. Acknowledgements The project is supported by National Science Foundation of China under Grant (61672051, 61702075); China Postdoctoral Science Foundation under grant 2017M611223. References [1] J. Alvarado, S. Dantas, D. Rautenbach, Strong equality of Roman and weak roman domination in trees, Discrete Appl. Math. 208 (2016) 19–26. [2] S. Arumugam, K. Ebadi, M. Manrique, Co-Roman domination in graphs, Proc. Math. Sci. 125 (1) (2015) 1–10. [3] E.W. Chambers, B.P. Kinnersley, D.B. West, Extremal problems for Roman domination, SIAM J. Discrete Math. 23 (3) (2009) 1575–1586. [4] M. Chellali, T.W. Haynes, S.T. Hedetniemi, Bounds on weak Roman and 2-rainbow domination numbers, Discrete Appl. Math. 178 (2014) 27–32. [5] E.J. Cockayne, P.A.D. Jr, S.M. Hedetniemi, S.T. Hedetniemi, On Roman domination in graphs, Discrete Math. 278 (2004) 11–22. [6] S.T. Hedetniemi, R.R. Rubalcaba, P.J. Slater, M. Walsh, Few compare to the great Roman Empire, Congr. Numer. 217 (2013) 129–136. [7] M.A. Henning, A characterization of Roman trees, Discuss. Math., Graph Theory 22 (2002) 325–334. [8] M.A. Henning, S.T. Hedetniemi, Defending the Roman Empire—A new strategy, Discrete Math. 266 (2003) 239–251. [9] Y. Wu, H. Xing, Note on 2-rainbow domination and Roman domination in graphs, Appl. Math. Lett. 23 (2010) 706–709.