Signed Roman domination in digraphs - Semantic Scholar
Signed Roman domination in digraphs S.M. Sheikholeslami∗ Department of Mathematics Azarbaijan Shahid Madani University Tabriz, I.R. Iran [email protected] L. Volkmann Lehrstuhl II f¨ ur Mathematik RWTH Aachen University 52056 Aachen, Germany [email protected]
Abstract Let D be a finite and simple digraph with vertex set V (D) and arc set A(D). A signed Roman dominating function (SRDF) on the digraph D is a function f : V (D) → {−1, 1, 2} P satisfying the conditions that (i) x∈N − [v] f (x) ≥ 1 for each v ∈ V (D), where N − [v] consists of v and all inner neighbors of v, and (ii) every vertex u for which Pf (u) = −1 has an inner neighbor v for which f (v) = 2. The weight of an SRDF f is w(f ) = v∈V (D) f (v). The signed Roman domination number γsR (D) of D is the minimum weight of an SRDF on D. In this paper we initiate the study of the signed Roman domination number of digraphs, and we present different bounds on γsR (D). In addition, we determine the signed Roman domination number of some classes of digraphs. Some of our results are extensions of well-known properties of the signed Roman domination number γsR (G) of graphs G. Keywords: Digraph, Signed Roman dominating function, Signed Roman domination number. MSC 2010: 05C20, 05C69
1
Terminology and introduction
In this paper we continue the study of Roman dominating functions in graphs and digraphs. Let G be a simple graph with vertex set V (G), and let N [v] = NG [v] be the closed neighborhood of the vertex v. A signed Roman dominating function on a graph G is defined in [1] as a function P f : V (G) −→ {−1, 1, 2} such that f (NG [v]) = x∈NG [v] f (x) ≥ 1 for every v ∈ V (G), and every vertex u for which f (u) = −1 Pis adjacent to a vertex v for which f (v) = 2. The weight of an SRDF f on a graph G is w(f ) = v∈V (G) f (v). The signed Roman domination number γsR (G) of G is the minimum weight of an SRDF on G. Following the ideas in [1], we initiate the study of signed Roman dominating functions on digraphs D. Let now D be a finite and simple digraph with vertex set V (D) and arc set A(D). The integers n(D) = |V (D)| and m(D) = |A(D)| are the order and the size of the digraph D, respectively. − + − We write d+ D (v) = d (v) for the out-degree of a vertex v and dD (v) = d (v) for its in-degree. The − − minimum and maximum in-degree are δ (D) and ∆ (D) and the minimum and maximum out-degree + − are δ + (D) and ∆+ (D). The sets ND (v) = N + (v) = {x | (v, x) ∈ A(D)} and ND (v) = N − (v) = ∗ Research
supported by the Research Office of Azarbaijan Shahid Madani University
1
{x | (x, v) ∈ A(D)} are called the out-neighborhood and in-neighborhood of the vertex v. Likewise, + − ND [v] = N + [v] = N + (v) ∪ {v} and ND [v] = N − [v] = N − (v) ∪ {v}. If X ⊆ V (D), then D[X] is the subdigraph induced by X. For an arc (x, y) ∈ A(D), the vertex y is an out-neighbor of x and x is an in-neighbor of y, and we also say that x dominatesP y or y is dominated by x. For a real-valued function f : V (D) −→ R, the weight of f is w(f ) = v∈V (D) f (v), and for S ⊆ V (D), we define P f (S) = v∈S f (v), so w(f ) = f (V (D)). Consult [3] and [4] for notation and terminology which are not defined here. We define a set S ⊆ V (D) to be a dominating set of D if for all v 6∈ S, v is dominated by a vertex s ∈ S. The minimum cardinality of a dominating set in D is the domination number γ(D). A signed dominating function (abbreviated SDF) on D is a function f : V (D) → {−1, 1} such P that f (N − [v]) = x∈N − [v] f (x) ≥ 1 for every v ∈ V (D). The signed domination number of a digraph D is γs (D) = min{w(f ) | f is a SDF of D}. A γs (D)-function is an SDF on D of weight γs (D). The signed domination number of a digraph was introduced by Zelinka [7] in 2005 and has been studied by several authors (see, for example, [2, 5, 6]). A signed Roman dominating function (abbreviated SDRF) on D is defined as a function f : P V (D) −→ {−1, 1, 2} such that f (N − [v]) = x∈N − [v] f (x) ≥ 1 for every v ∈ V (D) and every vertex u for which f (u) = −1 P has an in-neighbor v for which f (v) = 2. The weight of an SRDF f on a digraph D is w(f ) = v∈V (D) f (v). The signed Roman domination number γsR (D) of D is the minimum weight of an SRDF on D. A γsR (D)-function is a signed Roman dominating function on D of weight γsR (D). For an SRDF f on D, let Vi = Vi (f ) = {v ∈ V (D) : f (v) = i}. A signed Roman dominating function f : V (D) −→ {−1, 1, 2} can be represented by the ordered partition (V−1 , V1 , V2 ) of V (D). The associated digraph D(G) of a graph G is the digraph obtained from G when each edge e of − G is replaced by two oppositely oriented arcs with the same ends as e. Since ND(G) [v] = NG [v] for each vertex v ∈ V (G) = V (D(G)), the following useful observation is valid. Observation 1. If D(G) is the associated digraph of a graph G, then γsR (D(G)) = γsR (G). Let Kn and Kn∗ = D(Kn ) be the complete graph and complete digraph of order n, respectively. In [1], the authors determine the signed Roman domination number of complete graphs. Proposition 2. ([1]) If n 6= 3, then γsR (Kn ) = 1 and γsR (K3 ) = 2. Using Observation 1 and Proposition 2 , we obtain the signed Roman domination number of complete digraphs. Corollary 3. If n 6= 3, then γsR (Kn∗ ) = 1 and γsR (K3∗ ) = 2. Let G be a bipartite graph with partite sets L and R (standing for “left” and “right”). We define a left dominating set of G to be a set of vertices in R that dominate L. The minimum cardinality of a left dominating set in G is called the left domination number, denoted γL (G), of G. Let δL (G) and ∆L (G) denote the minimum and maximum degree, respectively, of a vertex of L in G and let δR (G) and ∆R (G) denote the minimum and maximum degree, respectively, of a vertex of R in G. In [1], the authors proved the following result. Proposition 4. ([1]) Let G be a bipartite graph of order n with partite sets L and R. If δL (G) ≥ 2, then γL (G) ≤ n/3, with equality if and only if every component of G is isomorphic to the path P3 or cycle C6 .
2
Preliminary results
In this section we present basic properties of the signed Roman dominating functions and the signed Roman domination numbers of digraphs. The definitions immediately lead to our first proposition. 2
Proposition 5. If f = (V−1 , V1 , V2 ) is an SRDF on a digraph D of order n, then (a) |V−1 | + |V1 | + |V2 | = n. (b) ω(f ) = |V1 | + 2|V2 | − |V−1 |. (c) Every vertex in V−1 is dominated by a vertex of V2 . (d) V1 ∪ V2 is a dominating set of D. Proposition 6. Assume that f = (V−1 , V1 , V2 ) is an SRDF on a digraph D of order n. If ∆+ (D) = ∆+ and δ + (D) = δ + , then (a) (2∆+ + 1)|V2 | + ∆+ |V1 | ≥ (δ + + 2)|V−1 |. (b) (2∆+ + δ + + 3)|V2 | + (∆+ + δ + + 2)|V1 | ≥ (δ + + 2)n. (c) (∆+ + δ + + 2)ω(f ) ≥ (δ + − ∆+ + 2)n + (δ + − ∆+ )|V2 |. (d) ω(f ) ≥ (δ + − 2∆+ + 1)n/(2∆+ + δ + + 3) + |V2 |. Proof. (a) It follows from Proposition 5 (a) that X X |V−1 | + |V1 | + |V2 | = n ≤
f (x) =
v∈V (D) x∈N − [v]
X
=
2(d+ D (v)
+ 1) +
v∈V2
X
(d+ D (v) + 1)f (v)
v∈V (D)
X
(d+ D (v)
+ 1) −
v∈V1
X
(d+ D (v) + 1)
v∈V−1
2(∆+ + 1)|V2 | + (∆+ + 1)|V1 | − (δ + + 1)|V−1 |.
≤
This inequality chain yields to the desired bound in (a). (b) Proposition 5 (a) implies that |V−1 | = n − |V1 | − |V2 |. Using this identity and Part (a), we arrive at (b). (c) According to Proposition 5 and Part (b) , we obtain Part (c) as follows (∆+ + δ + + 2)ω(f )
=
(∆+ + δ + + 2)(2(|V1 | + |V2 |) − n + |V2 |)
≥ 2(δ + + 2)n − 2(∆+ + 1)|V2 | + (∆+ + δ + + 2)(|V2 | − n) =
(δ + − ∆+ + 2)n + (δ + − ∆+ )|V2 |.
(d) The inequality chain in the proof of Part (a) and Proposition 5 (a) show that n
≤ 2(∆+ + 1)|V1 ∪ V2 | − (δ + + 1)|V−1 | =
2(∆+ + 1)|V1 ∪ V2 | − (δ + + 1)(n − |V1 ∪ V2 |)
=
(2∆+ + δ + + 3)|V1 ∪ V2 | − (δ + + 1)n
and thus
n(δ + + 2) . 2∆+ + δ + + 3 Using this inequality and Proposition 5, we obtain |V1 ∪ V2 | ≥
ω(f )
= ≥ =
2|V1 ∪ V2 | − n + |V2 | 2n(δ + + 2) − n(2∆+ + δ + + 3) + |V2 | 2∆+ + δ + + 3 n(δ + − 2∆+ + 1) + |V2 | 2∆+ + δ + + 3
This is the bound in Part (d), and the proof is complete.
3
3
Special families of digraphs
A tournament is a digraph in which for every pair u, v of different vertices, either (u, v) ∈ A(D) or (v, u) ∈ A(D), but not both. Next we determine the exact value of the signed Roman domination number for two particular types of tournaments. The acyclic tournament AT (n) with n vertices has the vertex set V (AT (n)) = {u1 , u2 , . . . , un }. An arc goes from ui into uj if and only if i < j. Let n be an odd positive integer such n = 2r + 1 with a positive integer r. We define the circulant tournament CT(n) with n vertices as follows. The vertex set of CT(n) is V (CT(n)) = {u0 , u1 , . . . , un−1 }. For each i, the arcs are going from ui to the vertices ui+1 , ui+2 , . . . , ui+r , where the indices are taken modulo n. Proposition 7. Let AT (n) for n ≥ 3 be an acyclic tournament. If n 6= 3, then γsR (AT (n)) = 1 and γsR (AT (3)) = 2. Proof. It is clear that γsR (AT (3)) = 2. Let n ≥ 4 and let f be a γsR (AT (n))-function. Since V (AT (n)) = N − [un ], we have γsR (AT (n)) = w(f ) = f (V (AT (n))) = f (N − [un ]) ≥ 1. If n is even, then define f : V (AT (n)) → {−1, 1, 2} by f (u1 ) = 2, f (ui ) = 1 for 2 ≤ i ≤
n n and f (ui ) = −1 when + 1 ≤ i ≤ n. 2 2
It is easy to see that f is an SRDF on AT (n) with ω(f ) = 1. Therefore γsR (AT (n)) ≤ 1 and so γsR (AT (n)) = 1. Assume that n is odd. If n = 5, then define f (u1 ) = f (u2 ) = 2 and f (u3 ) = f (u4 ) = f (u5 ) = −1. If n ≥ 7, then define the function f : V (AT (n)) → {−1, 1, 2} by f (u1 ) = f (u2 ) = 2, f (ui ) = 1 for 3 ≤ i ≤
n−1 n+1 and f (ui ) = −1 when ≤ i ≤ n. 2 2
Obviously, f is an SRDF on AT (n) with ω(f ) = 1. Hence γsR (AT (n)) ≤ 1 and thus γsR (AT (n)) = 1. This completes the proof. Proposition 8. Let n = 2r + 1 where r is a positive integer. If r 6= 2, then γsR (CT(n)) = 3 and γsR (CT(5)) = 4. Proof. It is clear that γsR (CT(3)) = 3 and γsR (CT(5)) = 4. Let r ≥ 3 and let f be a γsR (CT (n))function. If f (x) = 1 for each x ∈ V (CT (n)), then ω(f ) = n ≥ 3. We may assume, without loss of generality, that f (u0 ) = −1. Consider the sets N − [u0 ] = {ur+1 , ur+2 , . . . , un−1 , u0 } and N − [ur ] = {u0 , u1 , . . . , ur }. As f is an SRDF, we have f (N − [u0 ]) ≥ 1 and f (N − [ur ]) ≥ 1. Therefore ω(f ) = f (V (CT(n))) = f (N − [u0 ]) + f (N − [ur+1 ]) − f (u0 ) ≥ 3. This implies that γsR (CT(n)) ≥ 3. First let r be an odd integer. Define the function f : V (CT (n)) → {−1, 1, 2} by f (u r+1 ) = 2 f (ur+ r+1 ) = 2, f (ui ) = −1 for 0 ≤ i ≤ r−1 or r + 1 ≤ i ≤ r + r−1 and f (ui ) = 1 otherwise. 2 2 2 Obviously, f is an SRDF on CT (n) with ω(f ) = 3. Hence γsR (CT (n)) ≤ 3 and thus γsR (CT (n)) = 3. Assume now that r is even. If r = 4, then define f (u3 ) = f (u4 ) = f (u7 ) = f (u8 ) = 2 and f (x) = −1 otherwise. If r ≥ 6, then define the function f : V (CT (n)) → {−1, 1, 2} by f (u r+2 ) = 2 f (u r+4 ) = f (u 3r+2 ) = f (u 3r+4 ) = 2, f (ui ) = −1 for 0 ≤ i ≤ 2r or r + 1 ≤ i ≤ r + 2r and f (ui ) = 1 2 2 2 otherwise. Obviously, f is an SRDF on CT (n) with ω(f ) = 3. We deduce that γsR (CT (n)) ≤ 3 and thus γsR (CT (n)) = 3. This completes the proof. ∗ Let Kp,p be the complete bipartite digraph of order n = 2p.
4
∗ ∗ ∗ Proposition 9. If p ≥ 1 is an integer, then γsR (K1,1 ) = 1, γsR (K2,2 ) = 3 and γsR (Kp,p ) = 4 when p ≥ 3. ∗ ∗ Proof. It is a simple matter to verify that γsR (K1,1 ) = 1 and γsR (K2,2 ) = 3. Assume now that ∗ p ≥ 3. Let {u1 , u2 , . . . , up } and {v1 , v2 , . . . , vp } be the partite sets of D = Kp,p , and let f be a γ (D)-function. Assume, without loss of generality, that f (u ) = f (v ) = −1. Then the conditions sR 1 1 P P f (x) ≥ 1 and f (x) ≥ 1 imply that − − x∈N [v1 ] x∈N [u1 ] X X X ω(f ) = f (x) = f (x) + f (x) − f (v1 ) − f (u1 ) ≥ 4 x∈N − [v1 ]
x∈V (D)
x∈N − [u1 ]
and thus γsR (D) ≥ 4. Next we distinguish three cases. Case 1. Assume that p = 3t for an integer t ≥ 1. Define the number s = (2p − 3)/3 = 2t − 1 and the function g : V (D) −→ {−1, 1, 2} by g(u1 ) = g(u2 ) = . . . = g(us ) = g(v1 ) = g(v2 ) = . . . = g(vs ) = −1, g(us+1 ) = g(vs+1 ) = 1 and g(x) = 2 otherwise. If w is an arbitrary vertex of D, then X g(x) ≥ −s − 1 + 1 + 2(p − s − 1) = 1 x∈N − [w]
and therefore g is an SRDF on D such that X ω(g) = g(x) = −2s + 2 + 4(p − s − 1) = 4. x∈V (D)
This leads to γsR (D) ≤ 4 and thus γsR (D) = 4 in Case 1. Case 2. Assume that p = 3t + 1 for an integer t ≥ 1. Define s = (2p − 2)/3 = 2t and the function g : V (D) −→ {−1, 1, 2} by g(u1 ) = g(u2 ) = . . . = g(us ) = g(v1 ) = g(v2 ) = . . . = g(vs ) = −1 and g(x) = 2 otherwise. If w is an arbitrary vertex of D, then X g(x) ≥ −s − 1 + 2(p − s) = 1 x∈N − [w]
and therefore g is an SRDF on D such that X ω(g) = g(x) = −2s + 4(p − s) = 4. x∈V (D)
This yields to γsR (D) ≤ 4 and so γsR (D) = 4 in Case 2. Case 3. Assume that p = 3t + 2 for an integer t ≥ 1. Define the integer s = (2p − 4)/3 = 2t and g : V (D) −→ {−1, 1, 2} by g(u1 ) = g(u2 ) = . . . = g(us ) = g(v1 ) = g(v2 ) = . . . = g(vs ) = −1, g(us+1 ) = g(us+2 ) = g(vs+1 ) = g(vs+2 ) = 1 and g(x) = 2 otherwise. If w is an arbitrary vertex of D, then X g(x) ≥ −s − 1 + 2 + 2(p − s − 2) = 1 x∈N − [w]
and therefore g is an SRDF on D such that X ω(g) = g(x) = −2s + 4 + 4(p − s − 2) = 4. x∈V (D)
This leads to γsR (D) ≤ 4 and hence γsR (D) = 4 in Case 3. Since we have discussed all possible cases, the proof is complete. 5
Observation 1 and Proposition 9 immediately imply the signed Roman domination number for complete bipartite graphs Kp,p . Corollary 10. If p ≥ 1 is an integer, then γsR (K1,1 ) = 1, γsR (K2,2 ) = 3 and γsR (Kp,p ) = 4 when p ≥ 3. Proposition 11. Let Cn be an oriented cycle of order n ≥ 2. Then γsR (Cn ) = n/2 when n is even and γsR (Cn ) = (n + 3)/2 when n is odd. Proof. Let Cn = v1 v2 . . . vn v1 . Assume first that n is even. Since ∆+ (Cn ) = δ + (Cn ) = 1, Proposition 6 (c) implies that γsR (Cn ) ≥ n/2. Define the function g : V (Cn ) −→ {−1, 1, 2} by g(v2i−1 ) = −1 and g(v2i ) = 2 for 1 ≤ i ≤ n/2. Then g is an SRDF on Cn of weight n/2. Therefore γsR (Cn ) ≤ n/2 and thus γsR (Cn ) = n/2 when n is even. Next assume that n is odd. Let f be a γsR (Cn )-function. First we observe that f (vi )+f (vi+1 ) ≥ 1 and if f (vi ) = −1, then f (vi−1 ) = f (vi+1 ) = 2, where the indices are taken modulo n. If we assume, without loss of generality, that f (v2 ) = −1, then we deduce that ω(f ) ≥ 3 + (n − 3)/2 = (n + 3)/2. Now define g : V (Cn ) −→ {−1, 1, 2} by g(v2i−1 ) = −1 and g(v2i ) = 2 for 1 ≤ i ≤ (n − 1)/2 and g(vn ) = 2. Then g is an SRDF on Cn of weight (n + 3)/2. Therefore γsR (Cn ) ≤ (n + 3)/2 and thus γsR (Cn ) = (n + 3)/2 when n is odd. Proposition 12. Let Pn be an oriented path of order n. Then γsR (Pn ) = n/2 when n is even and γsR (Pn ) = (n + 1)/2 when n is odd. Proof. Let Pn = v1 v2 . . . vn be a directed path where (vi , vi+1 ) ∈ A(Pn ) for 1 ≤ i ≤ n − 1. If n = 1, then γsR (P1 ) = 1 and we are done. Let now n ≥ 2, and let f be a γsR (Pn )-function. First we observe that f (vi ) + f (vi+1 ) ≥ 1 and hence γsR (Pn ) ≥ n/2 when n is even. Define the function g : V (Pn ) −→ {−1, 1, 2} by g(v2i ) = −1 and g(v2i−1 ) = 2 for 1 ≤ i ≤ n/2. Then g is an SRDF on Pn of weight n/2. Therefore γsR (Pn ) ≤ n/2 and thus γsR (Pn ) = n/2 when n is even. Next assume that n is odd. We observe that f (v1 ) + f (v2 ) + f (v3 ) ≥ 2. Since f (vi ) + f (vi+1 ) ≥ 1, we deduce that ω(f ) ≥ (n + 1)/2. Now define g : V (Pn ) −→ {−1, 1, 2} by g(v2i+1 ) = −1 and g(v2i ) = 2 for 1 ≤ i ≤ (n − 1)/2 and g(v1 ) = 1. Then g is an SRDF on Pn of weight (n + 1)/2. Therefore γsR (Pn ) ≤ (n + 1)/2 and thus γsR (Pn ) = (n + 1)/2 when n is odd.
4
Bounds on the signed Roman domination number
We start with a simple but sharp lower bound on the signed Roman domination number of a digraph. Theorem 13. If D is a digraph of order n, then γsR (D) ≥ 2 + ∆− (D) − n. Proof. Let w ∈ V (D) be a vertex of maximum in-degree, and let f be a γsR (D)-function. Then the definitions imply X X X γsR (D) = f (v) = f (x) + f (x) x∈N − [w]
x∈V (D)
≥ 1+
X
x∈V (D)−N − [w]
f (x) ≥ 1 − (n − (∆− (D) + 1)) = 2 + ∆− (D) − n,
x∈V (D)−N − [w]
and the proof of the desired lower bound is complete. Example 14. Let n ≥ 3 be an integer, and let K1,n−1 be a star with the center u and the leaves v1 , v2 , . . . , vn−1 . Now let H1 be an orientation of K1,n−1 such ∆+ (H1 ) = n − 1. Define the function f : V (H1 ) −→ {−1, 1, 2} by f (u) = 2 and f (vi ) = −1 for 1 ≤ i ≤ n − 1. Then f is an SRDF on H1 of weight 3 − n. Therefore γsR (H1 ) ≤ 3 − n and so γsR (H1 ) = 3 − n by Theorem 13. Thus Theorem 13 is sharp for ∆− (D) = 1. 6
Now let t ≥ 1 be an integer such that 2t + 1 ≤ n − 1. Let H2t+1 be the digraph consisting of H1 with the additional arcs (vi , v1 ) for 2 ≤ i ≤ 2t+1. Define g : V (H2t+1 ) −→ {−1, 1, 2} by g(u) = 2 and g(v1 ) = g(v2 ) = . . . = g(vt ) = 1 and g(vt+1 ) = g(vt+2 ) = . . . g(vn−1 ) = −1. Then g is an SRDF on H2t+1 of weight 2+(2t+1)−n = 2+∆− (H2t+1 )−n. Therefore γsR (H2t+1 ) ≤ 2+∆− (H2t+1 )−n and so γsR (H2t+1 ) = 2 + ∆− (H2t+1 ) − n by Theorem 13. Thus Theorem 13 is sharp for ∆− (D) = 2t + 1 odd. Next let t ≥ 2 be an integer such that 2t ≤ n − 1. Let H2t be the digraph consisting of H1 with the additional arcs (vi , v1 ) for 2 ≤ i ≤ 2t. Define h : V (H2t ) −→ {−1, 1, 2} by h(u) = h(v1 ) = 2 and h(v2 ) = h(v3 ) = . . . = h(vt−1 ) = 1 and h(vt ) = h(vt+1 ) = . . . = h(vn−1 ) = −1. Then h is an SRDF on H2t of weight 2 + (2t) − n = 2 + ∆− (H2t ) − n. Therefore γsR (H2t ) ≤ 2 + ∆− (H2t ) − n and so γsR (H2t ) = 2 + ∆− (H2t ) − n by Theorem 13. Thus Theorem 13 is sharp for ∆− (D) = 2t ≥ 4 even. In the case that ∆− (D) = 2, we will improve Theorem 13 a little bit. Theorem 15. If D is a digraph of order n with ∆− (D) = 2, then γsR (D) ≥ 3 + ∆− (D) − n = 5 − n. Proof. Let w ∈ V (D) be a vertex of maximum in-degree such that N − (w) = {u, v}, and let f be a γsR (D)-function. We distinguish three cases. Assume first that f (w) = −1. It follows that f (u) = 2 or P f (v) = 2. Assume, without loss of generality, that f (u) = 2. We deduce that f (v) ≥ 1 and thus x∈N − [w] f (x) ≥ 2 and therefore γsR (D) =
X
f (x) =
x∈V (D)
X x∈N − [w]
X
f (x) +
f (x) ≥ 2 − (n − 3) = 5 − n.
x∈V (D)−N − [w]
Assume second that f (w) = 2. It follows that f (u) + f (v) ≥ 0 and so X X X γsR (D) = f (x) = f (x) + f (x) ≥ 2 − (n − 3) = 5 − n. x∈V (D)
x∈N − [w]
x∈V (D)−N − [w]
Finally, assume that f (w) = 1. It follows that f (u) + f (v) ≥ 0. If f (u) + f (v) ≥ 1, then X X X γsR (D) = f (x) = f (x) + f (x) ≥ 2 − (n − 3) = 5 − n. x∈V (D)
x∈N − [w]
x∈V (D)−N − [w]
In the remaining case that f (u) + f (v) = 0, we assume, without loss of generality, that f (u) = −1 and f (v) = 1. Then there exists an in-neighbor y of u such that f (y) = 2, and we obtain X X γsR (D) = f (x) = f (w) + f (y) + f (x) ≥ 3 − (n − 2) = 5 − n. x∈V (D)
x∈V (D)−{w,y}
Since we have discussed all possible cases, the proof is complete. Example 16. Now let H2 be the digraph H1 defined in Example 14 together with the arc (v2 , v1 ). Then ∆− (H2 ) = 2. Define f : V (H2 ) −→ {−1, 1, 2} by f (u) = 2 and f (v1 ) = 1 and f (vi ) = −1 for 2 ≤ i ≤ n − 1. Then f is an SRDF on H2 of weight 5 − n. Therefore γsR (H1 ) ≤ 5 − n and so γsR (H1 ) = 3 + ∆− (H2 ) − n = 5 − n by Theorem 15. Thus Theorem 15 is sharp. A digraph D is out-regular or r-out-regular if δ + (D) = ∆+ (D) = r. As an application of Proposition 6 (c), we obtain a lower bound on the signed Roman domination number for r-outregular digraphs, which improves the lower bound given in Theorem 13 considerably. Corollary 17. If D is an r-out-regular digraph of order n with r ≥ 1, then γsR (D) ≥ n/(r + 1). Using Corollary 17 and Observation 1, we obtain the next known result. Corollary 18. ([1]) If G is an r-regular graph of order n with r ≥ 1, then γsR (G) ≥ n/(r + 1). 7
If D is not out-regular, then the next lower bound on the signed Roman domination number is valid. Corollary 19. Let D be a digraph of order n, minimum out-degree δ + and maximum out-degree ∆+ . If δ + < ∆+ , then −2(∆+ )2 + 2∆+ δ + + ∆+ + 2δ + + 3 γsR (D) ≥ n. (∆+ + 1)(2∆+ + δ + + 3) Proof. Multiplying both sides of the inequality in Proposition 6 (d) by ∆+ − δ + and adding the resulting inequality to the inequality in Proposition 6 (c), we obtain the desired lower bound. Since ∆+ (D(G)) = ∆(G) and δ + (D(G)) = δ(G), Corollary 19 and Observation 1 lead to the next known corollary. Corollary 20. ([1]) Let G be a graph of order n, minimum degree δ and maximum degree ∆. If δ < ∆, then −2∆2 + 2∆δ + ∆ + 2δ + 3 n. γsR (G) ≥ (∆ + 1)(2∆ + δ + 3) Now we present a sharp upper bound on the signed Roman domination number. Theorem 21. Let D be a digraph of order n. Then γsR (D) ≤ n, with equality if and only if D is the disjoint union of isolated vertices and oriented triangles C3 . Proof. If we define f : V (D) −→ {−1, 1, 2} by f (x) = 1 for each x ∈ V (D), then f is an SRDF on D and so γsR (D) ≤ n. By Proposition 11, γsR (C3 ) = 3 and consequently γsR (D) = n when D is the disjoint union of isolated vertices and oriented triangles. Conversely, assume that γsR (D) = n, and suppose to the contrary that D is not the disjoint union of isolated vertices and oriented triangles. Let P = v1 v2 . . . vt be a longest oriented path in D. If t = 2, then define g : V (D) −→ {−1, 1, 2} by g(v2 ) = −1, g(v1 ) = 2 and g(x) = 1 for x ∈ V (D) − {v1 , v2 }. Then g is an SRDF on D of weight n − 1, a contradiction. Assume next that t ≥ 3, and that P is contained in a component different from a C3 . Define g : V (D) −→ {−1, 1, 2} by g(vt ) = −1, g(vt−1 ) = 2 and g(x) = 1 for x ∈ V (D) − {vt−1 , vt }. Then g is an SRDF on D of weight n − 1 with exception of the case that vt dominates v1 . Next we assume that vt dominates v1 . Since P is a longest oriented path, we observe that the vertices of P induce a component H of D. First assume that t = 3. Since H is not an oriented triangle, there exists, without loss of generality, an arc v1 v3 . Then define g : V (D) −→ {−1, 1, 2} by g(v2 ) = −1, g(v1 ) = 2 and g(x) = 1 for x ∈ V (D) − {v1 , v2 }. Then g is an SRDF on D of weight n − 1, a contradiction. Next assume that t ≥ 4. If H is isomorphic to an oriented cycle Ct , then define g : V (H) −→ {−1, 1, 2} as in the proof of Proposition 11 and g(x) = 1 for x ∈ V (D) − V (H). Then g is an SRDF on D of weight at most n − 1, a contradiction. Finally, assume that H is not isomorphic to Ct . Then there exists an arc (vi+s , vi ) for some 1 ≤ s ≤ t − 2, where the indices are taken modulo t. Assume first that s = 1 and, without loss of generality, i = 1. Then define g : V (D) −→ {−1, 1, 2} by g(v1 ) = −1, g(v2 ) = 2 and g(x) = 1 for x ∈ V (D) − {v1 , v2 }. Then g is an SRDF on D of weight n − 1, a contradiction. Assume second that t − 2 ≥ s ≥ 2 and, without loss of generality, i = 1. Then define g : V (D) −→ {−1, 1, 2} by g(vt ) = −1, g(vt−1 ) = 2 and g(x) = 1 for x ∈ V (D) − {vt−1 , vt }. Then g is an SRDF on D of weight n − 1, a contradiction. This completes the proof. Theorem 22. Let D be a digraph of order n. Then γsR (D) ≥ 2γ(D) − n, with equality if and only if D is the disjoint union of isolated vertices. Proof. Let f = (V−1 , V1 , V2 ) be a γsR (D)-function. If V2 = ∅, then V (D) = V1 and thus γsR (D) = n. But then, by Theorem 21, D is the disjoint union of isolated vertices and oriented triangles. Assume that D contains exactly p oriented triangles and r isolated vertices. Then γ(D) = 2p + r and n = 3p + r. We deduce that n = γsR (D) = 3p + r ≥ 4p + 2r − n = p + r = 2γ(D) − n 8
with equality if and only if p = 0. This is the desired result for the case that V2 = ∅. Second assume that |V2 | ≥ 1. Since |V−1 | = n − |V1 | − |V2 |, Proposition 5 implies that γsR (D) = |V1 | + 2|V2 | − |V−1 | = 2|V1 | + 3|V2 | − n > 2|V1 ∪ V2 | − n ≥ 2γ(D) − n. Therefore γsR (D) > 2γ(D) − n in that case, and the proof is complete. Finally, we derive an upper bound on the signed Roman domination number of a digraph in terms of the signed domination number and the order. Theorem 23. If D is a digraph of order n, then γsR (D) ≤ γs (D) +
n . 3
Furthermore, this bound is sharp. Proof. Let f be a γs (D)-function and let L = {v ∈ V (D) | f (v) = −1} and R = {v ∈ V (D) | f (v) = 1}. Then, n = |L| + |R| and γs (D) = w(f ) = |R| − |L|. Let D1 be the bipartite spanning subdigraph of D (of order n) with partite sets L and R, where A(D1 ) consists of all arcs in D that go from R into L. Since f is an SDF on D, each vertex in L has − at least two in-neighbors in R in D and so δL (D1 ) ≥ 2. Assume that H is the graph obtained from D1 by replacing any arc with an edge. Then δL (H) ≥ 2 and by Proposition 4, γL (H) ≤ n/3. Let R2 be a γL (H)-set. Thus R2 is a subset of vertices in R that dominate L in H and |R2 | = γL (H) ≤ n/3. It follows that for any vertex v ∈ L, there is at least one arc in D1 which goes from R2 to v. Hence R2 dominates all vertices of L in D1 and so in D. Let R1 = R \ R2 . Then g = (L, R1 , R2 ) is an SRDF on D, and therefore we obtain γsR (D) ≤ w(g) = 2|R2 | + |R1 | − |L| = 2|R2 | + (|R| − |R2 |) − |L| = γs (D) + |R2 | ≤ γs (D) + n/3 as desired. To prove sharpness, let D be a digraph with vertex set V (D) = {ui , vi | 1 ≤ i ≤ 3} and arc set A(D) = {(v1 , u1 ), (v1 , u2 ), (v2 , u1 ), (v2 , u3 ), (v3 , u2 ), (v3 , u3 )} ∪ {(vi , vj ) | 1 ≤ i 6= j ≤ 3}. It is easy to see that γs (D) = 0, γsR (D) = 2, and the proof is complete. We conclude this paper with an open problem. Problem. Characterize all digraphs that attain the bound in Theorem 23.
References [1] H.A. Ahangar, M.A. Henning, Y. Zhao, C. L¨owenstein and V. Samodivkin, Signed Roman domination in graphs, J. Comb. Optim. DOI 10.1007/s10878-012-9500-0 [2] M. Atapour, R. Hajypory, S. M. Sheikholeslami, and L. Volkmann, The signed k-domination number of directed graphs, Cent. Eur. J. Math. 8 (2010), 1048-1057. [3] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, Inc., New York (1998). [4] T. W. Haynes, S. T. Hedetniemi, and P. J. Slater, editors, Domination in Graphs, Advanced Topics, Marcel Dekker, Inc., New York (1998). [5] H. Karami, S. M. Sheikholeslami, and A. Khodkar, Lower bounds on the signed domination numbers of directed graphs, Discrete Math. 309 (2009), 2567-2570. [6] L. Volkmann, Signed domination and signed domatic numbers of digraphs Discuss. Math. Graph Theory 31 (2011), 415-427. [7] B. Zelinka, Signed domination numbers of directed graphs, Czechoslovak Math. J. 55 (2005), 479–482.
9
Abstract Let D be a finite and simple digraph with vertex set V (D) and arc set A(D). A signed Roman dominating function (SRDF) on the digraph D is a function f : V (D) → {−1, 1, 2} P satisfying the conditions that (i) x∈N − [v] f (x) ≥ 1 for each v ∈ V (D), where N − [v] consists of v and all inner neighbors of v, and (ii) every vertex u for which Pf (u) = −1 has an inner neighbor v for which f (v) = 2. The weight of an SRDF f is w(f ) = v∈V (D) f (v). The signed Roman domination number γsR (D) of D is the minimum weight of an SRDF on D. In this paper we initiate the study of the signed Roman domination number of digraphs, and we present different bounds on γsR (D). In addition, we determine the signed Roman domination number of some classes of digraphs. Some of our results are extensions of well-known properties of the signed Roman domination number γsR (G) of graphs G. Keywords: Digraph, Signed Roman dominating function, Signed Roman domination number. MSC 2010: 05C20, 05C69
1
Terminology and introduction
In this paper we continue the study of Roman dominating functions in graphs and digraphs. Let G be a simple graph with vertex set V (G), and let N [v] = NG [v] be the closed neighborhood of the vertex v. A signed Roman dominating function on a graph G is defined in [1] as a function P f : V (G) −→ {−1, 1, 2} such that f (NG [v]) = x∈NG [v] f (x) ≥ 1 for every v ∈ V (G), and every vertex u for which f (u) = −1 Pis adjacent to a vertex v for which f (v) = 2. The weight of an SRDF f on a graph G is w(f ) = v∈V (G) f (v). The signed Roman domination number γsR (G) of G is the minimum weight of an SRDF on G. Following the ideas in [1], we initiate the study of signed Roman dominating functions on digraphs D. Let now D be a finite and simple digraph with vertex set V (D) and arc set A(D). The integers n(D) = |V (D)| and m(D) = |A(D)| are the order and the size of the digraph D, respectively. − + − We write d+ D (v) = d (v) for the out-degree of a vertex v and dD (v) = d (v) for its in-degree. The − − minimum and maximum in-degree are δ (D) and ∆ (D) and the minimum and maximum out-degree + − are δ + (D) and ∆+ (D). The sets ND (v) = N + (v) = {x | (v, x) ∈ A(D)} and ND (v) = N − (v) = ∗ Research
supported by the Research Office of Azarbaijan Shahid Madani University
1
{x | (x, v) ∈ A(D)} are called the out-neighborhood and in-neighborhood of the vertex v. Likewise, + − ND [v] = N + [v] = N + (v) ∪ {v} and ND [v] = N − [v] = N − (v) ∪ {v}. If X ⊆ V (D), then D[X] is the subdigraph induced by X. For an arc (x, y) ∈ A(D), the vertex y is an out-neighbor of x and x is an in-neighbor of y, and we also say that x dominatesP y or y is dominated by x. For a real-valued function f : V (D) −→ R, the weight of f is w(f ) = v∈V (D) f (v), and for S ⊆ V (D), we define P f (S) = v∈S f (v), so w(f ) = f (V (D)). Consult [3] and [4] for notation and terminology which are not defined here. We define a set S ⊆ V (D) to be a dominating set of D if for all v 6∈ S, v is dominated by a vertex s ∈ S. The minimum cardinality of a dominating set in D is the domination number γ(D). A signed dominating function (abbreviated SDF) on D is a function f : V (D) → {−1, 1} such P that f (N − [v]) = x∈N − [v] f (x) ≥ 1 for every v ∈ V (D). The signed domination number of a digraph D is γs (D) = min{w(f ) | f is a SDF of D}. A γs (D)-function is an SDF on D of weight γs (D). The signed domination number of a digraph was introduced by Zelinka [7] in 2005 and has been studied by several authors (see, for example, [2, 5, 6]). A signed Roman dominating function (abbreviated SDRF) on D is defined as a function f : P V (D) −→ {−1, 1, 2} such that f (N − [v]) = x∈N − [v] f (x) ≥ 1 for every v ∈ V (D) and every vertex u for which f (u) = −1 P has an in-neighbor v for which f (v) = 2. The weight of an SRDF f on a digraph D is w(f ) = v∈V (D) f (v). The signed Roman domination number γsR (D) of D is the minimum weight of an SRDF on D. A γsR (D)-function is a signed Roman dominating function on D of weight γsR (D). For an SRDF f on D, let Vi = Vi (f ) = {v ∈ V (D) : f (v) = i}. A signed Roman dominating function f : V (D) −→ {−1, 1, 2} can be represented by the ordered partition (V−1 , V1 , V2 ) of V (D). The associated digraph D(G) of a graph G is the digraph obtained from G when each edge e of − G is replaced by two oppositely oriented arcs with the same ends as e. Since ND(G) [v] = NG [v] for each vertex v ∈ V (G) = V (D(G)), the following useful observation is valid. Observation 1. If D(G) is the associated digraph of a graph G, then γsR (D(G)) = γsR (G). Let Kn and Kn∗ = D(Kn ) be the complete graph and complete digraph of order n, respectively. In [1], the authors determine the signed Roman domination number of complete graphs. Proposition 2. ([1]) If n 6= 3, then γsR (Kn ) = 1 and γsR (K3 ) = 2. Using Observation 1 and Proposition 2 , we obtain the signed Roman domination number of complete digraphs. Corollary 3. If n 6= 3, then γsR (Kn∗ ) = 1 and γsR (K3∗ ) = 2. Let G be a bipartite graph with partite sets L and R (standing for “left” and “right”). We define a left dominating set of G to be a set of vertices in R that dominate L. The minimum cardinality of a left dominating set in G is called the left domination number, denoted γL (G), of G. Let δL (G) and ∆L (G) denote the minimum and maximum degree, respectively, of a vertex of L in G and let δR (G) and ∆R (G) denote the minimum and maximum degree, respectively, of a vertex of R in G. In [1], the authors proved the following result. Proposition 4. ([1]) Let G be a bipartite graph of order n with partite sets L and R. If δL (G) ≥ 2, then γL (G) ≤ n/3, with equality if and only if every component of G is isomorphic to the path P3 or cycle C6 .
2
Preliminary results
In this section we present basic properties of the signed Roman dominating functions and the signed Roman domination numbers of digraphs. The definitions immediately lead to our first proposition. 2
Proposition 5. If f = (V−1 , V1 , V2 ) is an SRDF on a digraph D of order n, then (a) |V−1 | + |V1 | + |V2 | = n. (b) ω(f ) = |V1 | + 2|V2 | − |V−1 |. (c) Every vertex in V−1 is dominated by a vertex of V2 . (d) V1 ∪ V2 is a dominating set of D. Proposition 6. Assume that f = (V−1 , V1 , V2 ) is an SRDF on a digraph D of order n. If ∆+ (D) = ∆+ and δ + (D) = δ + , then (a) (2∆+ + 1)|V2 | + ∆+ |V1 | ≥ (δ + + 2)|V−1 |. (b) (2∆+ + δ + + 3)|V2 | + (∆+ + δ + + 2)|V1 | ≥ (δ + + 2)n. (c) (∆+ + δ + + 2)ω(f ) ≥ (δ + − ∆+ + 2)n + (δ + − ∆+ )|V2 |. (d) ω(f ) ≥ (δ + − 2∆+ + 1)n/(2∆+ + δ + + 3) + |V2 |. Proof. (a) It follows from Proposition 5 (a) that X X |V−1 | + |V1 | + |V2 | = n ≤
f (x) =
v∈V (D) x∈N − [v]
X
=
2(d+ D (v)
+ 1) +
v∈V2
X
(d+ D (v) + 1)f (v)
v∈V (D)
X
(d+ D (v)
+ 1) −
v∈V1
X
(d+ D (v) + 1)
v∈V−1
2(∆+ + 1)|V2 | + (∆+ + 1)|V1 | − (δ + + 1)|V−1 |.
≤
This inequality chain yields to the desired bound in (a). (b) Proposition 5 (a) implies that |V−1 | = n − |V1 | − |V2 |. Using this identity and Part (a), we arrive at (b). (c) According to Proposition 5 and Part (b) , we obtain Part (c) as follows (∆+ + δ + + 2)ω(f )
=
(∆+ + δ + + 2)(2(|V1 | + |V2 |) − n + |V2 |)
≥ 2(δ + + 2)n − 2(∆+ + 1)|V2 | + (∆+ + δ + + 2)(|V2 | − n) =
(δ + − ∆+ + 2)n + (δ + − ∆+ )|V2 |.
(d) The inequality chain in the proof of Part (a) and Proposition 5 (a) show that n
≤ 2(∆+ + 1)|V1 ∪ V2 | − (δ + + 1)|V−1 | =
2(∆+ + 1)|V1 ∪ V2 | − (δ + + 1)(n − |V1 ∪ V2 |)
=
(2∆+ + δ + + 3)|V1 ∪ V2 | − (δ + + 1)n
and thus
n(δ + + 2) . 2∆+ + δ + + 3 Using this inequality and Proposition 5, we obtain |V1 ∪ V2 | ≥
ω(f )
= ≥ =
2|V1 ∪ V2 | − n + |V2 | 2n(δ + + 2) − n(2∆+ + δ + + 3) + |V2 | 2∆+ + δ + + 3 n(δ + − 2∆+ + 1) + |V2 | 2∆+ + δ + + 3
This is the bound in Part (d), and the proof is complete.
3
3
Special families of digraphs
A tournament is a digraph in which for every pair u, v of different vertices, either (u, v) ∈ A(D) or (v, u) ∈ A(D), but not both. Next we determine the exact value of the signed Roman domination number for two particular types of tournaments. The acyclic tournament AT (n) with n vertices has the vertex set V (AT (n)) = {u1 , u2 , . . . , un }. An arc goes from ui into uj if and only if i < j. Let n be an odd positive integer such n = 2r + 1 with a positive integer r. We define the circulant tournament CT(n) with n vertices as follows. The vertex set of CT(n) is V (CT(n)) = {u0 , u1 , . . . , un−1 }. For each i, the arcs are going from ui to the vertices ui+1 , ui+2 , . . . , ui+r , where the indices are taken modulo n. Proposition 7. Let AT (n) for n ≥ 3 be an acyclic tournament. If n 6= 3, then γsR (AT (n)) = 1 and γsR (AT (3)) = 2. Proof. It is clear that γsR (AT (3)) = 2. Let n ≥ 4 and let f be a γsR (AT (n))-function. Since V (AT (n)) = N − [un ], we have γsR (AT (n)) = w(f ) = f (V (AT (n))) = f (N − [un ]) ≥ 1. If n is even, then define f : V (AT (n)) → {−1, 1, 2} by f (u1 ) = 2, f (ui ) = 1 for 2 ≤ i ≤
n n and f (ui ) = −1 when + 1 ≤ i ≤ n. 2 2
It is easy to see that f is an SRDF on AT (n) with ω(f ) = 1. Therefore γsR (AT (n)) ≤ 1 and so γsR (AT (n)) = 1. Assume that n is odd. If n = 5, then define f (u1 ) = f (u2 ) = 2 and f (u3 ) = f (u4 ) = f (u5 ) = −1. If n ≥ 7, then define the function f : V (AT (n)) → {−1, 1, 2} by f (u1 ) = f (u2 ) = 2, f (ui ) = 1 for 3 ≤ i ≤
n−1 n+1 and f (ui ) = −1 when ≤ i ≤ n. 2 2
Obviously, f is an SRDF on AT (n) with ω(f ) = 1. Hence γsR (AT (n)) ≤ 1 and thus γsR (AT (n)) = 1. This completes the proof. Proposition 8. Let n = 2r + 1 where r is a positive integer. If r 6= 2, then γsR (CT(n)) = 3 and γsR (CT(5)) = 4. Proof. It is clear that γsR (CT(3)) = 3 and γsR (CT(5)) = 4. Let r ≥ 3 and let f be a γsR (CT (n))function. If f (x) = 1 for each x ∈ V (CT (n)), then ω(f ) = n ≥ 3. We may assume, without loss of generality, that f (u0 ) = −1. Consider the sets N − [u0 ] = {ur+1 , ur+2 , . . . , un−1 , u0 } and N − [ur ] = {u0 , u1 , . . . , ur }. As f is an SRDF, we have f (N − [u0 ]) ≥ 1 and f (N − [ur ]) ≥ 1. Therefore ω(f ) = f (V (CT(n))) = f (N − [u0 ]) + f (N − [ur+1 ]) − f (u0 ) ≥ 3. This implies that γsR (CT(n)) ≥ 3. First let r be an odd integer. Define the function f : V (CT (n)) → {−1, 1, 2} by f (u r+1 ) = 2 f (ur+ r+1 ) = 2, f (ui ) = −1 for 0 ≤ i ≤ r−1 or r + 1 ≤ i ≤ r + r−1 and f (ui ) = 1 otherwise. 2 2 2 Obviously, f is an SRDF on CT (n) with ω(f ) = 3. Hence γsR (CT (n)) ≤ 3 and thus γsR (CT (n)) = 3. Assume now that r is even. If r = 4, then define f (u3 ) = f (u4 ) = f (u7 ) = f (u8 ) = 2 and f (x) = −1 otherwise. If r ≥ 6, then define the function f : V (CT (n)) → {−1, 1, 2} by f (u r+2 ) = 2 f (u r+4 ) = f (u 3r+2 ) = f (u 3r+4 ) = 2, f (ui ) = −1 for 0 ≤ i ≤ 2r or r + 1 ≤ i ≤ r + 2r and f (ui ) = 1 2 2 2 otherwise. Obviously, f is an SRDF on CT (n) with ω(f ) = 3. We deduce that γsR (CT (n)) ≤ 3 and thus γsR (CT (n)) = 3. This completes the proof. ∗ Let Kp,p be the complete bipartite digraph of order n = 2p.
4
∗ ∗ ∗ Proposition 9. If p ≥ 1 is an integer, then γsR (K1,1 ) = 1, γsR (K2,2 ) = 3 and γsR (Kp,p ) = 4 when p ≥ 3. ∗ ∗ Proof. It is a simple matter to verify that γsR (K1,1 ) = 1 and γsR (K2,2 ) = 3. Assume now that ∗ p ≥ 3. Let {u1 , u2 , . . . , up } and {v1 , v2 , . . . , vp } be the partite sets of D = Kp,p , and let f be a γ (D)-function. Assume, without loss of generality, that f (u ) = f (v ) = −1. Then the conditions sR 1 1 P P f (x) ≥ 1 and f (x) ≥ 1 imply that − − x∈N [v1 ] x∈N [u1 ] X X X ω(f ) = f (x) = f (x) + f (x) − f (v1 ) − f (u1 ) ≥ 4 x∈N − [v1 ]
x∈V (D)
x∈N − [u1 ]
and thus γsR (D) ≥ 4. Next we distinguish three cases. Case 1. Assume that p = 3t for an integer t ≥ 1. Define the number s = (2p − 3)/3 = 2t − 1 and the function g : V (D) −→ {−1, 1, 2} by g(u1 ) = g(u2 ) = . . . = g(us ) = g(v1 ) = g(v2 ) = . . . = g(vs ) = −1, g(us+1 ) = g(vs+1 ) = 1 and g(x) = 2 otherwise. If w is an arbitrary vertex of D, then X g(x) ≥ −s − 1 + 1 + 2(p − s − 1) = 1 x∈N − [w]
and therefore g is an SRDF on D such that X ω(g) = g(x) = −2s + 2 + 4(p − s − 1) = 4. x∈V (D)
This leads to γsR (D) ≤ 4 and thus γsR (D) = 4 in Case 1. Case 2. Assume that p = 3t + 1 for an integer t ≥ 1. Define s = (2p − 2)/3 = 2t and the function g : V (D) −→ {−1, 1, 2} by g(u1 ) = g(u2 ) = . . . = g(us ) = g(v1 ) = g(v2 ) = . . . = g(vs ) = −1 and g(x) = 2 otherwise. If w is an arbitrary vertex of D, then X g(x) ≥ −s − 1 + 2(p − s) = 1 x∈N − [w]
and therefore g is an SRDF on D such that X ω(g) = g(x) = −2s + 4(p − s) = 4. x∈V (D)
This yields to γsR (D) ≤ 4 and so γsR (D) = 4 in Case 2. Case 3. Assume that p = 3t + 2 for an integer t ≥ 1. Define the integer s = (2p − 4)/3 = 2t and g : V (D) −→ {−1, 1, 2} by g(u1 ) = g(u2 ) = . . . = g(us ) = g(v1 ) = g(v2 ) = . . . = g(vs ) = −1, g(us+1 ) = g(us+2 ) = g(vs+1 ) = g(vs+2 ) = 1 and g(x) = 2 otherwise. If w is an arbitrary vertex of D, then X g(x) ≥ −s − 1 + 2 + 2(p − s − 2) = 1 x∈N − [w]
and therefore g is an SRDF on D such that X ω(g) = g(x) = −2s + 4 + 4(p − s − 2) = 4. x∈V (D)
This leads to γsR (D) ≤ 4 and hence γsR (D) = 4 in Case 3. Since we have discussed all possible cases, the proof is complete. 5
Observation 1 and Proposition 9 immediately imply the signed Roman domination number for complete bipartite graphs Kp,p . Corollary 10. If p ≥ 1 is an integer, then γsR (K1,1 ) = 1, γsR (K2,2 ) = 3 and γsR (Kp,p ) = 4 when p ≥ 3. Proposition 11. Let Cn be an oriented cycle of order n ≥ 2. Then γsR (Cn ) = n/2 when n is even and γsR (Cn ) = (n + 3)/2 when n is odd. Proof. Let Cn = v1 v2 . . . vn v1 . Assume first that n is even. Since ∆+ (Cn ) = δ + (Cn ) = 1, Proposition 6 (c) implies that γsR (Cn ) ≥ n/2. Define the function g : V (Cn ) −→ {−1, 1, 2} by g(v2i−1 ) = −1 and g(v2i ) = 2 for 1 ≤ i ≤ n/2. Then g is an SRDF on Cn of weight n/2. Therefore γsR (Cn ) ≤ n/2 and thus γsR (Cn ) = n/2 when n is even. Next assume that n is odd. Let f be a γsR (Cn )-function. First we observe that f (vi )+f (vi+1 ) ≥ 1 and if f (vi ) = −1, then f (vi−1 ) = f (vi+1 ) = 2, where the indices are taken modulo n. If we assume, without loss of generality, that f (v2 ) = −1, then we deduce that ω(f ) ≥ 3 + (n − 3)/2 = (n + 3)/2. Now define g : V (Cn ) −→ {−1, 1, 2} by g(v2i−1 ) = −1 and g(v2i ) = 2 for 1 ≤ i ≤ (n − 1)/2 and g(vn ) = 2. Then g is an SRDF on Cn of weight (n + 3)/2. Therefore γsR (Cn ) ≤ (n + 3)/2 and thus γsR (Cn ) = (n + 3)/2 when n is odd. Proposition 12. Let Pn be an oriented path of order n. Then γsR (Pn ) = n/2 when n is even and γsR (Pn ) = (n + 1)/2 when n is odd. Proof. Let Pn = v1 v2 . . . vn be a directed path where (vi , vi+1 ) ∈ A(Pn ) for 1 ≤ i ≤ n − 1. If n = 1, then γsR (P1 ) = 1 and we are done. Let now n ≥ 2, and let f be a γsR (Pn )-function. First we observe that f (vi ) + f (vi+1 ) ≥ 1 and hence γsR (Pn ) ≥ n/2 when n is even. Define the function g : V (Pn ) −→ {−1, 1, 2} by g(v2i ) = −1 and g(v2i−1 ) = 2 for 1 ≤ i ≤ n/2. Then g is an SRDF on Pn of weight n/2. Therefore γsR (Pn ) ≤ n/2 and thus γsR (Pn ) = n/2 when n is even. Next assume that n is odd. We observe that f (v1 ) + f (v2 ) + f (v3 ) ≥ 2. Since f (vi ) + f (vi+1 ) ≥ 1, we deduce that ω(f ) ≥ (n + 1)/2. Now define g : V (Pn ) −→ {−1, 1, 2} by g(v2i+1 ) = −1 and g(v2i ) = 2 for 1 ≤ i ≤ (n − 1)/2 and g(v1 ) = 1. Then g is an SRDF on Pn of weight (n + 1)/2. Therefore γsR (Pn ) ≤ (n + 1)/2 and thus γsR (Pn ) = (n + 1)/2 when n is odd.
4
Bounds on the signed Roman domination number
We start with a simple but sharp lower bound on the signed Roman domination number of a digraph. Theorem 13. If D is a digraph of order n, then γsR (D) ≥ 2 + ∆− (D) − n. Proof. Let w ∈ V (D) be a vertex of maximum in-degree, and let f be a γsR (D)-function. Then the definitions imply X X X γsR (D) = f (v) = f (x) + f (x) x∈N − [w]
x∈V (D)
≥ 1+
X
x∈V (D)−N − [w]
f (x) ≥ 1 − (n − (∆− (D) + 1)) = 2 + ∆− (D) − n,
x∈V (D)−N − [w]
and the proof of the desired lower bound is complete. Example 14. Let n ≥ 3 be an integer, and let K1,n−1 be a star with the center u and the leaves v1 , v2 , . . . , vn−1 . Now let H1 be an orientation of K1,n−1 such ∆+ (H1 ) = n − 1. Define the function f : V (H1 ) −→ {−1, 1, 2} by f (u) = 2 and f (vi ) = −1 for 1 ≤ i ≤ n − 1. Then f is an SRDF on H1 of weight 3 − n. Therefore γsR (H1 ) ≤ 3 − n and so γsR (H1 ) = 3 − n by Theorem 13. Thus Theorem 13 is sharp for ∆− (D) = 1. 6
Now let t ≥ 1 be an integer such that 2t + 1 ≤ n − 1. Let H2t+1 be the digraph consisting of H1 with the additional arcs (vi , v1 ) for 2 ≤ i ≤ 2t+1. Define g : V (H2t+1 ) −→ {−1, 1, 2} by g(u) = 2 and g(v1 ) = g(v2 ) = . . . = g(vt ) = 1 and g(vt+1 ) = g(vt+2 ) = . . . g(vn−1 ) = −1. Then g is an SRDF on H2t+1 of weight 2+(2t+1)−n = 2+∆− (H2t+1 )−n. Therefore γsR (H2t+1 ) ≤ 2+∆− (H2t+1 )−n and so γsR (H2t+1 ) = 2 + ∆− (H2t+1 ) − n by Theorem 13. Thus Theorem 13 is sharp for ∆− (D) = 2t + 1 odd. Next let t ≥ 2 be an integer such that 2t ≤ n − 1. Let H2t be the digraph consisting of H1 with the additional arcs (vi , v1 ) for 2 ≤ i ≤ 2t. Define h : V (H2t ) −→ {−1, 1, 2} by h(u) = h(v1 ) = 2 and h(v2 ) = h(v3 ) = . . . = h(vt−1 ) = 1 and h(vt ) = h(vt+1 ) = . . . = h(vn−1 ) = −1. Then h is an SRDF on H2t of weight 2 + (2t) − n = 2 + ∆− (H2t ) − n. Therefore γsR (H2t ) ≤ 2 + ∆− (H2t ) − n and so γsR (H2t ) = 2 + ∆− (H2t ) − n by Theorem 13. Thus Theorem 13 is sharp for ∆− (D) = 2t ≥ 4 even. In the case that ∆− (D) = 2, we will improve Theorem 13 a little bit. Theorem 15. If D is a digraph of order n with ∆− (D) = 2, then γsR (D) ≥ 3 + ∆− (D) − n = 5 − n. Proof. Let w ∈ V (D) be a vertex of maximum in-degree such that N − (w) = {u, v}, and let f be a γsR (D)-function. We distinguish three cases. Assume first that f (w) = −1. It follows that f (u) = 2 or P f (v) = 2. Assume, without loss of generality, that f (u) = 2. We deduce that f (v) ≥ 1 and thus x∈N − [w] f (x) ≥ 2 and therefore γsR (D) =
X
f (x) =
x∈V (D)
X x∈N − [w]
X
f (x) +
f (x) ≥ 2 − (n − 3) = 5 − n.
x∈V (D)−N − [w]
Assume second that f (w) = 2. It follows that f (u) + f (v) ≥ 0 and so X X X γsR (D) = f (x) = f (x) + f (x) ≥ 2 − (n − 3) = 5 − n. x∈V (D)
x∈N − [w]
x∈V (D)−N − [w]
Finally, assume that f (w) = 1. It follows that f (u) + f (v) ≥ 0. If f (u) + f (v) ≥ 1, then X X X γsR (D) = f (x) = f (x) + f (x) ≥ 2 − (n − 3) = 5 − n. x∈V (D)
x∈N − [w]
x∈V (D)−N − [w]
In the remaining case that f (u) + f (v) = 0, we assume, without loss of generality, that f (u) = −1 and f (v) = 1. Then there exists an in-neighbor y of u such that f (y) = 2, and we obtain X X γsR (D) = f (x) = f (w) + f (y) + f (x) ≥ 3 − (n − 2) = 5 − n. x∈V (D)
x∈V (D)−{w,y}
Since we have discussed all possible cases, the proof is complete. Example 16. Now let H2 be the digraph H1 defined in Example 14 together with the arc (v2 , v1 ). Then ∆− (H2 ) = 2. Define f : V (H2 ) −→ {−1, 1, 2} by f (u) = 2 and f (v1 ) = 1 and f (vi ) = −1 for 2 ≤ i ≤ n − 1. Then f is an SRDF on H2 of weight 5 − n. Therefore γsR (H1 ) ≤ 5 − n and so γsR (H1 ) = 3 + ∆− (H2 ) − n = 5 − n by Theorem 15. Thus Theorem 15 is sharp. A digraph D is out-regular or r-out-regular if δ + (D) = ∆+ (D) = r. As an application of Proposition 6 (c), we obtain a lower bound on the signed Roman domination number for r-outregular digraphs, which improves the lower bound given in Theorem 13 considerably. Corollary 17. If D is an r-out-regular digraph of order n with r ≥ 1, then γsR (D) ≥ n/(r + 1). Using Corollary 17 and Observation 1, we obtain the next known result. Corollary 18. ([1]) If G is an r-regular graph of order n with r ≥ 1, then γsR (G) ≥ n/(r + 1). 7
If D is not out-regular, then the next lower bound on the signed Roman domination number is valid. Corollary 19. Let D be a digraph of order n, minimum out-degree δ + and maximum out-degree ∆+ . If δ + < ∆+ , then −2(∆+ )2 + 2∆+ δ + + ∆+ + 2δ + + 3 γsR (D) ≥ n. (∆+ + 1)(2∆+ + δ + + 3) Proof. Multiplying both sides of the inequality in Proposition 6 (d) by ∆+ − δ + and adding the resulting inequality to the inequality in Proposition 6 (c), we obtain the desired lower bound. Since ∆+ (D(G)) = ∆(G) and δ + (D(G)) = δ(G), Corollary 19 and Observation 1 lead to the next known corollary. Corollary 20. ([1]) Let G be a graph of order n, minimum degree δ and maximum degree ∆. If δ < ∆, then −2∆2 + 2∆δ + ∆ + 2δ + 3 n. γsR (G) ≥ (∆ + 1)(2∆ + δ + 3) Now we present a sharp upper bound on the signed Roman domination number. Theorem 21. Let D be a digraph of order n. Then γsR (D) ≤ n, with equality if and only if D is the disjoint union of isolated vertices and oriented triangles C3 . Proof. If we define f : V (D) −→ {−1, 1, 2} by f (x) = 1 for each x ∈ V (D), then f is an SRDF on D and so γsR (D) ≤ n. By Proposition 11, γsR (C3 ) = 3 and consequently γsR (D) = n when D is the disjoint union of isolated vertices and oriented triangles. Conversely, assume that γsR (D) = n, and suppose to the contrary that D is not the disjoint union of isolated vertices and oriented triangles. Let P = v1 v2 . . . vt be a longest oriented path in D. If t = 2, then define g : V (D) −→ {−1, 1, 2} by g(v2 ) = −1, g(v1 ) = 2 and g(x) = 1 for x ∈ V (D) − {v1 , v2 }. Then g is an SRDF on D of weight n − 1, a contradiction. Assume next that t ≥ 3, and that P is contained in a component different from a C3 . Define g : V (D) −→ {−1, 1, 2} by g(vt ) = −1, g(vt−1 ) = 2 and g(x) = 1 for x ∈ V (D) − {vt−1 , vt }. Then g is an SRDF on D of weight n − 1 with exception of the case that vt dominates v1 . Next we assume that vt dominates v1 . Since P is a longest oriented path, we observe that the vertices of P induce a component H of D. First assume that t = 3. Since H is not an oriented triangle, there exists, without loss of generality, an arc v1 v3 . Then define g : V (D) −→ {−1, 1, 2} by g(v2 ) = −1, g(v1 ) = 2 and g(x) = 1 for x ∈ V (D) − {v1 , v2 }. Then g is an SRDF on D of weight n − 1, a contradiction. Next assume that t ≥ 4. If H is isomorphic to an oriented cycle Ct , then define g : V (H) −→ {−1, 1, 2} as in the proof of Proposition 11 and g(x) = 1 for x ∈ V (D) − V (H). Then g is an SRDF on D of weight at most n − 1, a contradiction. Finally, assume that H is not isomorphic to Ct . Then there exists an arc (vi+s , vi ) for some 1 ≤ s ≤ t − 2, where the indices are taken modulo t. Assume first that s = 1 and, without loss of generality, i = 1. Then define g : V (D) −→ {−1, 1, 2} by g(v1 ) = −1, g(v2 ) = 2 and g(x) = 1 for x ∈ V (D) − {v1 , v2 }. Then g is an SRDF on D of weight n − 1, a contradiction. Assume second that t − 2 ≥ s ≥ 2 and, without loss of generality, i = 1. Then define g : V (D) −→ {−1, 1, 2} by g(vt ) = −1, g(vt−1 ) = 2 and g(x) = 1 for x ∈ V (D) − {vt−1 , vt }. Then g is an SRDF on D of weight n − 1, a contradiction. This completes the proof. Theorem 22. Let D be a digraph of order n. Then γsR (D) ≥ 2γ(D) − n, with equality if and only if D is the disjoint union of isolated vertices. Proof. Let f = (V−1 , V1 , V2 ) be a γsR (D)-function. If V2 = ∅, then V (D) = V1 and thus γsR (D) = n. But then, by Theorem 21, D is the disjoint union of isolated vertices and oriented triangles. Assume that D contains exactly p oriented triangles and r isolated vertices. Then γ(D) = 2p + r and n = 3p + r. We deduce that n = γsR (D) = 3p + r ≥ 4p + 2r − n = p + r = 2γ(D) − n 8
with equality if and only if p = 0. This is the desired result for the case that V2 = ∅. Second assume that |V2 | ≥ 1. Since |V−1 | = n − |V1 | − |V2 |, Proposition 5 implies that γsR (D) = |V1 | + 2|V2 | − |V−1 | = 2|V1 | + 3|V2 | − n > 2|V1 ∪ V2 | − n ≥ 2γ(D) − n. Therefore γsR (D) > 2γ(D) − n in that case, and the proof is complete. Finally, we derive an upper bound on the signed Roman domination number of a digraph in terms of the signed domination number and the order. Theorem 23. If D is a digraph of order n, then γsR (D) ≤ γs (D) +
n . 3
Furthermore, this bound is sharp. Proof. Let f be a γs (D)-function and let L = {v ∈ V (D) | f (v) = −1} and R = {v ∈ V (D) | f (v) = 1}. Then, n = |L| + |R| and γs (D) = w(f ) = |R| − |L|. Let D1 be the bipartite spanning subdigraph of D (of order n) with partite sets L and R, where A(D1 ) consists of all arcs in D that go from R into L. Since f is an SDF on D, each vertex in L has − at least two in-neighbors in R in D and so δL (D1 ) ≥ 2. Assume that H is the graph obtained from D1 by replacing any arc with an edge. Then δL (H) ≥ 2 and by Proposition 4, γL (H) ≤ n/3. Let R2 be a γL (H)-set. Thus R2 is a subset of vertices in R that dominate L in H and |R2 | = γL (H) ≤ n/3. It follows that for any vertex v ∈ L, there is at least one arc in D1 which goes from R2 to v. Hence R2 dominates all vertices of L in D1 and so in D. Let R1 = R \ R2 . Then g = (L, R1 , R2 ) is an SRDF on D, and therefore we obtain γsR (D) ≤ w(g) = 2|R2 | + |R1 | − |L| = 2|R2 | + (|R| − |R2 |) − |L| = γs (D) + |R2 | ≤ γs (D) + n/3 as desired. To prove sharpness, let D be a digraph with vertex set V (D) = {ui , vi | 1 ≤ i ≤ 3} and arc set A(D) = {(v1 , u1 ), (v1 , u2 ), (v2 , u1 ), (v2 , u3 ), (v3 , u2 ), (v3 , u3 )} ∪ {(vi , vj ) | 1 ≤ i 6= j ≤ 3}. It is easy to see that γs (D) = 0, γsR (D) = 2, and the proof is complete. We conclude this paper with an open problem. Problem. Characterize all digraphs that attain the bound in Theorem 23.
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