On the signed total Roman domination and domatic numbers of graphs

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On the signed total Roman domination and domatic numbers of graphs Lutz Volkmann Lehrstuhl II für Mathematik, RWTH Aachen University, 52056 Aachen, Germany

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Article history: Received 19 August 2015 Received in revised form 11 March 2016 Accepted 7 June 2016 Available online xxxx Keywords: Signed total Roman dominating function Signed total Roman domination number Signed total Roman domatic number

abstract A signed total  Roman dominating function on a graph G is a function f : V (G) −→ {−1, 1, 2} such that u∈N (v) f (u) ≥ 1 for every vertex v ∈ V (G), where N (v) is the neighborhood of v , and every vertex u ∈ V (G) for which f (u) = −1 is adjacent to at least one vertex w for which f (w) = 2. The signed total Roman domination number of a graph G, denoted by γstR (G), equals the minimum weight of a signed total Roman dominating function. A set {f1 , f2 , . . . , fd } of distinct signed total Roman dominating functions on G with the property d that i=1 fi (v) ≤ 1 for each v ∈ V (G), is called a signed total Roman dominating family (of functions) on G. The maximum number of functions in a signed total Roman dominating family on G is the signed total Roman domatic number of G, denoted by dstR (G). In this paper we continue the investigation of the signed total Roman domination number, and we initiate the study of the signed total Roman domatic number in graphs. We present sharp bounds for γstR (G) and dstR (G). In addition, we determine the total signed Roman domatic number of some graphs. © 2016 Elsevier B.V. All rights reserved.

1. Terminology and introduction For notation and graph theory terminology, we in general follow Haynes, Hedetniemi and Slater [3]. Specifically, let G be a simple graph with vertex set V = V (G) and edge set E = E (G). The order |V | of G is denoted by n = n(G). For every vertex v ∈ V , the open neighborhood N (v) is the set {u ∈ V (G) | uv ∈ E (G)} and the closed neighborhood of v is the set N [v] = N (v) ∪ {v}. The degree of a vertex v ∈ V is d(v) = |N (v)|. The minimum and maximum degree of a graph G are denoted by δ = δ(G) and ∆ = ∆(G), respectively. A graph G is regular or r-regular if d(v) = r for each vertex v of G. Let S be a set of vertices, and let u ∈ S. We say that v is a private neighbor of u (with respect to S) if N [u] ∩ S = {u}. The complement of a graph G is denoted by G. We write Kn for the complete graph of order n, Kp,q for the complete bipartite graph with partite sets X and Y , where |X | = p and |Y | = q, Cn for the cycle of length n, and Pn for the path of order n. In this paper we continue the study of Roman dominating functions in graphs and digraphs. Following the ideas in [1], we defined in [7] the signed total Roman dominating function (STRDF) on a graph G as a function f : V (G) −→ {−1, 1, 2}  such that f (N (v)) = f (u) = −1 u∈N (v) f (u) ≥ 1 for each v ∈ V (G), and such that every vertex u ∈ V (G) for which  is adjacent to at least one vertex w for which f (w) = 2. The weight of an STRDF f is the value ω(f ) = v∈V (G) f (v). The signed total Roman domination number of a graph G, denoted by γstR (G), equals the minimum weight of an STRDF on G. We note that this parameter is only defined for graphs without isolated vertices. Thus we assume throughout this paper that δ(G) ≥ 1. A γstR (G)-function is a signed total Roman dominating function of G with weight γstR (G). A signed total Roman dominating function f : V (G) −→ {−1, 1, 2} can be represented by the ordered partition (V−1 , V1 , V2 ) of V (G), where Vi = {v ∈ V (G) | f (v) = i} for i = −1, 1, 2. In this representation, its weight is ω(f ) = |V1 | + 2|V2 | − |V−1 |. E-mail address: [email protected]. http://dx.doi.org/10.1016/j.dam.2016.06.006 0166-218X/© 2016 Elsevier B.V. All rights reserved.

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A concept dual in a certain sense to the domination number is the domatic number, introduced by Cockayne and Hedetniemi [2]. They have defined the domatic number d(G) of a graph G by means of sets. A partition of V (G), all of whose classes are dominating sets in G, is called a domatic partition. The maximum number of classes of a domatic partition of G is the domatic number d(G) of G. But Rall has defined a variant of the domatic number of G, namely the fractional domatic number of G, using functions on V (G). (This was mentioned by Slater and Trees in [6].) Analogous to the fractional domatic number we may define the signed total Roman domatic number. d A set {f1 , f2 , . . . , fd } of distinct signed total Roman dominating functions on G with the property that i=1 fi (v) ≤ 1 for each v ∈ V (G), is called a signed total Roman dominating family (of functions) on G. The maximum number of functions in a signed total Roman dominating family (STRD family) on G is the signed total Roman domatic number of G, denoted by dstR (G). The signed total Roman domatic number is well-defined and dstR (G) ≥ 1 for all graphs G with δ(G) ≥ 1, since the set consisting of any STRDF forms an STRD family on G. In [5], Sheikholeslami and Volkmann investigated the signed Roman domatic number in graphs. Our purpose in this paper is to continue the investigations of the signed total Roman domination number and to initiate the study of the signed total Roman domatic number in graphs. First, we characterize the graphs G with γstR (G) = n(G). Second, we derive basic properties and bounds for the signed total Roman domatic number of a graph. In particular, we obtain different upper bounds on the sum γstR (G) + dstR (G). In addition, we determine the signed total Roman domatic number of some classes of graphs. Finally, we derive the Nordhaus–Gaddum type result dstR (G) + dstR (G) ≤ n − 1, with equality if and only if G = C4 and G = K2 ∪ K2 . We make use of the following known results in this paper. Proposition A ([7]). If G is a δ -regular graph of order n with δ ≥ 1, then

γstR (G) ≥

n δ

.

Proposition B ([7]). If Cn is a cycle of order n ≥ 3, then γstR (Cn ) = n/2 when n ≡ 0 (mod 4), γstR (Cn ) = (n + 3)/2 when n ≡ 1, 3 (mod 4) and γstR (Cn ) = (n + 6)/2 when n ≡ 2 (mod 4). Proposition C ([7]). If Pn is a path of order n ≥ 3, then γstR (Pn ) = n/2 when n ≡ 0 (mod 4), and γstR (Pn ) = ⌈(n + 3)/2⌉ otherwise. Proposition D ([7]). For p ≥ 1, γstR (Kp,p ) = 2, unless p = 3 in which case γstR (K3,3 ) = 4. Proposition E ([7]). If n ≥ 3 is an integer, then γstR (Kn ) = 3. Proposition F ([7]). Let G be a graph of order n. If δ(G) ≥ 1, then γstR (G) ≤ n and if δ(G) ≥ 3, then γstR (G) ≤ n − 1. 2. Signed total Roman domination number First we improve Proposition A for 3-regular graphs considerably. Theorem 1. If G is a 3-regular graph of order n, then

γstR (G) ≥

2n 3

,

with equality if and only if |V−1 | = |V1 | = |V2 | for every γstR -function f = (V−1 , V1 , V2 ) on G. Proof. Let f = (V−1 , V1 , V2 ) be a γstR (G)-function. If V−1 = ∅, then γstR (G) = n ≥ (2n)/3. Thus let now V−1 = {u1 , u2 , . . . , ut } ̸= ∅. As every vertex of V−1 is adjacent to at least one vertex in V2 , we choose to every vertex ui ∈ V−1 exactly one vertex u′i ∈ V2 for 1 ≤ i ≤ t. Since G is 3-regular, the condition f (N (v)) ≥ 1 for each vertex v , shows that no vertex of G has two neighbors in V−1 . If V2′ = {u′1 , u′2 , . . . , u′t }, then we thus deduce that |V2′ | = |V−1 |. Furthermore, every vertex of V−1 has a private neighbor in (V2 − V2′ ) ∪ V1 and therefore |(V2 − V2′ ) ∪ V1 | ≥ |V−1 |. Combining these observations, we obtain 3ω(f ) = 3(2|V2 | + |V1 | − |V−1 |) ≥ 3(|V2 | + |V1 |)

= 2|V2 | + 2|V1 | + |V2′ | + |(V2 − V2′ ) ∪ V1 | ≥ 2|V2 | + 2|V1 | + 2|V−1 | = 2n and so γstR (G) = ω(f ) ≥ (2n)/3. If γstR (G) = 2n , then the two inequalities occurring in the inequality chain above become 3 equalities, and this implies |V−1 | = |V1 | = |V2 |. Conversely, if |V−1 | = |V1 | = |V2 |, then γstR (G) = 2|V2 | + |V1 | − |V−1 | = |V2 | + |V1 | = 2n/3. 

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In view of Proposition D, we have γstR (K3,3 ) = 4. If G = tK3,3 for an integer t ≥ 1, then it follows that γstR (G) = 4t = (2n(G))/3, and hence Theorem 1 is sharp. Next we improve Proposition F. Theorem 2. Let G be a graph of order n with δ(G) ≥ 1. If ∆(G) ≥ 3, then γstR (G) ≤ n − 1. Proof. Let u be a vertex of maximum degree. If there is a vertex v ∈ N (u) with d(v) = 1, then the function f : V (G) −→ {−1, 1, 2} with f (v) = −1, f (u) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1 and thus γstR (G) ≤ n − 1. If d(x) ≥ 3 for all x ∈ N (u), then the function f : V (G) −→ {−1, 1, 2} with f (u) = −1, f (v) = 2 for one vertex v ∈ N (u) and f (x) = 1 otherwise is an STRDF on G of weight n − 1 and thus γstR (G) ≤ n − 1. So assume that there exists at least one vertex v ∈ N (u) with d(v) = 2. If v is adjacent to a vertex w ∈ N (u), then the function f : V (G) −→ {−1, 1, 2} with f (v) = −1, f (u) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1 and thus γstR (G) ≤ n − 1. Now we assume that v is adjacent to a vertex w ̸∈ N (u). If d(w) ≥ 3, then the function f : V (G) −→ {−1, 1, 2} with f (v) = −1, f (u) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1. Next let d(w) = 2, and let z ̸= v be a neighbor of w . If d(z ) ≥ 3, then f (v) = f (w) = −1, f (u) = f (z ) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 2. If d(z ) = 2, then let y ̸= w a further neighbor of z. Then the f (v) = f (w) = −1, f (u) = f (z ) = f (y) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1. Therefore there remain the cases that d(w) = 1 or d(w) = 2 and d(z ) = 1. Now rename v = v1 , w = w1 and z = z1 . Assume that v2 , v3 , . . . , vt with t ≥ 2 are further neighbors of u with d(vi ) = 2 for 2 ≤ i ≤ t, and assume that all other neighbors of u have degree at least three. Let wi ̸= u adjacent to vi . As above, we observe that 1 ≤ d(wi ) ≤ 2. Assume first that d(wi ) = 2 for 1 ≤ i ≤ t, and let zi ̸= vi be adjacent to wi for 1 ≤ i ≤ t. As above, we see that d(zi ) = 1. Then the function f : V (G) −→ {−1, 1, 2} with f (u) = f (z1 ) = f (z2 ) = · · · = f (zt ) = −1, f (vi ) = f (wi ) = 2 for 1 ≤ i ≤ t and f (x) = 1 otherwise is an STRDF on G of weight n − 2. Assume second that, without loss of generality, d(w1 ) = d(w2 ) = 1. Then f (w1 ) = f (w2 ) = −1, f (u) = f (v1 ) = f (v2 ) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1. Assume third that, without loss of generality, d(w1 ) = 1 and d(wi ) = 2 for 2 ≤ i ≤ t. Then the function f : V (G) −→ {−1, 1, 2} with f (w1 ) = 2, f (v1 ) = 1, f (u) = f (z2 ) = f (z3 ) = · · · = f (zt ) = −1, f (vi ) = f (wi ) = 2 for 2 ≤ i ≤ t and f (x) = 1 otherwise is an STRDF on G of weight n − 1. Consequently, we only have to discuss the case that d(x) ≥ 3 for all neighbors x ∈ N (u) − {v}. In the case that d(w) = 2 and d(z ) = 1, the function f : V (G) −→ {−1, 1, 2} with f (u) = f (z ) = −1, f (v) = f (w) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 2. Thus we now investigate the case that d(v) = 2, d(w) = 1 and d(x) ≥ 3 for x ∈ N (u) − {v}. Let y ̸= v be a neighbor of u. If y has a neighbor a of degree 1, then f (a) = f (w) = −1, f (u) = f (v) = f (y) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1. If all neighbors of y have degree at least three, then f (y) = −1, f (u) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1. So assume that y has a neighbor a of degree two, and let b ̸= y be a neighbor of a. If d(b) = 1, then f (b) = f (u) = −1, f (w) = f (y) = f (a) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1. If d(b) ≥ 3, then f (a) = −1, f (y) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1. Now let d(b) = 2 and c ̸= a a further neighbor of b. If d(c ) ≥ 3, then f (a) = f (b) = −1, f (c ) = f (y) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 2. Let next d(c ) = 2, and let z ̸= b be adjacent to c. Then f (a) = f (b) = −1, f (y) = f (c ) = f (z ) = 2 and f (x) = 1 otherwise is an STRDF on G of weight n − 1. It remains the case that d(c ) = 1. Now rename a = a1 , b = b1 and c = c1 . Assume that a2 , a3 , . . . , as with s ≥ 1 are further neighbors of y with d(ai ) = 2 for 2 ≤ i ≤ s, and assume that all other neighbors of y have degree at least three. Let bi ̸= y adjacent to ai . As above, we observe that d(bi ) = 2. Let ci ̸= ai adjacent to bi . As above, we see that d(ci ) = 1. Now the function f : V (G) −→ {−1, 1, 2} with f (y) = −1, f (ci ) = −1, f (ai ) = f (bi ) = 2 for 1 ≤ i ≤ s and f (x) = 1 otherwise is an STRDF on G of weight n − 2, and the proof is complete.  Theorem 3. Let G be a graph of order n with δ(G) ≥ 1. Then γstR (G) = n if and only if the components of G are K2 , K3 , P3 or C6 . Proof. If the components of G are K2 , K3 , P3 or C6 , then γstR (G) = n (see Propositions B and C). Conversely, assume that γstR (G) = n. If ∆(G) ≥ 3, then it follows from Theorem 2 that γstR (G) ≤ n − 1. Let now 1 ≤ ∆(G) ≤ 2. This implies that the components of G are paths or cycles. Thus Propositions B and C show that the components of G are K2 , K3 , P3 or C6 .  The next result is an extension of Proposition F Theorem 4. If G is a graph of order n with δ(G) ≥ 3, then

γstR (G) ≤ n − 2



δ(G) − 1 2



+ 1.

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Proof. Let ⌊(δ(G) − 1)/2⌋ = k, and let v ∈ V (G) be an arbitrary vertex. Now let A = {u1 , u2 , . . . , uk } be a set of k neighbors of v . Define the function f : V (G) −→ {−1, 1, 2} such that f (v) = 2, f (ui ) = −1 for 1 ≤ i ≤ k and f (x) = 1 for x ∈ V (G) − (A ∪ {v}). Then f (N (x)) ≥ −k + (δ(G) − k) = δ(G) − 2



δ(G) − 1



2

≥1

for each vertex x ∈ V (G). Therefore f is an STRDF on G of weight n − 2k + 1 and thus γstR (G) ≤ n − 2k + 1.



Proposition E shows that Theorem 4 is sharp for Kn when n ≥ 4 is even. 3. Bounds on the signed total Roman domatic number In this section we present basic properties of dstR (G) and sharp bounds on the signed total Roman domatic number of a graph. Theorem 5. For every graph G with δ(G) ≥ 1, dstR (G) ≤ δ(G). Moreover, if dstR (G) = δ(G), then for each STRD family {f1 , f2 , . . . , fd } on G with d = dstR (G) and each vertex v of minimum  d degree, x∈N (v) fi (x) = 1 for each function fi and i=1 fi (x) = 1 for all x ∈ N (v). Proof. Let {f1 , f2 , . . . , fd } be an STRD family on G such that d = dstR (G). If v is a vertex of minimum degree δ(G), then we deduce that dstR (G) = d ≤

d  

fi (x) =

i=1 x∈N (v)

d  

fi (x) ≤

x∈N (v) i=1



1 = δ(G).

x∈N (v)

If dstR (G) = δ(G), then the two inequalities occurring in the proof become equalities. Hence for the STRD family   {f1 , f2 , . . . , fd } on G and for each vertex v of minimum degree, x∈N (v) fi (x) = 1 for each function fi and di=1 fi (x) = 1 for all x ∈ N (v).  For the special case δ(G) = 3 we will improve Theorem 5 slightly. Corollary 6. If G is a graph with minimum degree δ(G) = 3, then dstR (G) ≤ 2. Proof. Suppose to the contrary that {f1 , f2 , f3 } is an STRD family on G. Let v be a vertex of minimum degree. By Theorem 5,  3 we deduce that x∈N (v) fi (x) = 1 for each function fi and i=1 fi (x) = 1 for all x ∈ N (v). It follows that fi (x) ≤ 1 for all x ∈ N (v) and 1 ≤ i ≤ 3. This implies fi (v) ≥ 1 for 1 ≤ i ≤ 3, and we obtain the contradiction dstR (G) ≤ 2. 

3

i=1 fi

(v) ≥ 3. Thus

Example 7. If C4t is a cycle of length 4t with an integer t ≥ 1, then dstR (C4t ) = 2. Proof. According to Theorem 5, dstR (C4t ) ≤ 2. Let C4t = v1 v2 . . . v4t v1 . Define the functions f1 and f2 by f1 (v4i+1 ) = f1 (v4i+4 ) = −1,

f1 (v4i+2 ) = f1 (V4i+3 ) = 2

and f2 (v4i+1 ) = f2 (v4i+4 ) = 2,

f2 (v4i+2 ) = f2 (V4i+3 ) = −1

for 0 ≤ i ≤ t − 1. It is easy to see that fi is a signed total Roman dominating function on C4t for 1 ≤ i ≤ 2 and {f1 , f2 } is a signed total Roman dominating family on C4t . Therefore dstR (C4t ) ≥ 2 and so dstR (C4t ) = 2.  Example 8. If p ≥ 4 is an integer, then dstR (Kp,p ) = p. Proof. We deduce from Theorem 5 that dstR (Kp,p ) ≤ p. Now let A = {u1 , u2 , . . . , up } and B = {v1 , v2 , . . . , vp } be the bipartition of Kp,p . First let p = 2t be even. Define f1 (ui ) = f1 (vi ) = −1 for 1 ≤ i ≤ t, f1 (ui ) = f1 (vi ) = 1 for t + 1 ≤ i ≤ 2t − 1, and f1 (u2t ) = f1 (v2t ) = 2. In addition define fj (ui ) = fj (vi ) = f1 (ui+j−1 ) for 2 ≤ j ≤ 2t and 1 ≤ i ≤ 2t, where the indices are taken modulo 2t. It is straightforward to verify that fi is a signed total Roman dominating function on K2t ,2t for 1 ≤ i ≤ 2t and {f1 , f2 , . . . , f2t } is a signed total Roman dominating family on K2t ,2t . Therefore dstR (K2t ,2t ) ≥ 2t and hence dstR (K2t ,2t ) = 2t.

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Second let p = 2t + 1 be odd. Define f1 (ui ) = f1 (vi ) = −1 for 1 ≤ i ≤ t + 1, f1 (ui ) = f1 (vi ) = 1 for t + 2 ≤ i ≤ 2t − 1, and f1 (u2t ) = f1 (v2t ) = f1 (u2t +1 ) = f1 (v2t +1 ) = 2. In addition, define fj (ui ) = fj (vi ) = f1 (ui+j−1 ) for 2 ≤ j ≤ 2t + 1 and 1 ≤ i ≤ 2t + 1, where the indices are taken modulo 2t + 1. It is easy to see that fi is a signed total Roman dominating function on K2t +1,2t +1 for 1 ≤ i ≤ 2t + 1 and {f1 , f2 , . . . , f2t +1 } is a signed total Roman dominating family on K2t +1,2t +1 . Therefore dstR (K2t +1,2t +1 ) ≥ 2t + 1 and hence dstR (K2t +1,2t +1 ) = 2t + 1.  Examples 7 and 8 show that Theorem 5 is sharp. Theorem 9. If G is a graph of order n with δ(G) ≥ 1, then

γstR (G) · dstR (G) ≤ n. Moreover, if γstR (G) · dstR (G) = n, then for each STRD family {f1 , f2 , . . . , fd } on G with d = dstR (G), each function fi is a γstR (G)d function and i=1 fi (v) = 1 for all v ∈ V (G). Proof. Let {f1 , f2 , . . . , fd } be an STRD family on G such that d = dstR (G) and let v ∈ V (G). Then d · γstR (G) =

d 

γstR (G) ≤

fi (v)

i=1 v∈V (G)

i =1

=

d  

d  

fi (v) ≤

v∈V (G) i=1



1 = n.

v∈V (G)

If γstR (G) · dstR (G) = n, then the two inequalities occurring in the proof become equalities. Hence for the STRD family   {f1 , f2 , . . . , fd } on G and for each i, v∈V (G) fi (v) = γstR (G). Thus each function fi is a γstR (G)-function, and di=1 fi (v) = 1 for all v ∈ V (G).  Using Proposition D and Example 8 for p ≥ 4, we deduce that γstR (Kp,p ) · dstR (Kp,p ) = 2p and therefore equality in Theorem 9. Consequently, Theorem 9 is sharp. Example 10. If k ≥ 0 is an integer, then dstR (K9k+6 ) = 3k + 2. Proof. According to Theorem 9 and Proposition E, we have dstR (K9k+6 ) ≤

9k + 6 9k + 6 = = 3k + 2. γstR (K9k+6 ) 3

Now let v1 , v2 , . . . , v9k+6 be the vertices of K9k+6 . For 1 ≤ i ≤ 3k + 2 define the function fi : V (K9k+6 ) −→ {−1, 1, 2} by f1 (v1 ) = f1 (v2 ) = · · · = f1 (v6k+3 ) = −1, f1 (v6k+4 ) = f1 (v6k+5 ) = · · · = f1 (v9k+6 ) = 2 and fp (vi ) = fp−1 (vi+3 ) for 2 ≤ p ≤ 3k + 2, where the indices are taken modulo 9k + 6. It is a simple matter to verify that fi is a signed total Roman domination function for 1 ≤ i ≤ 3k + 2 and {f1 , f2 , . . . , f3k+2 } is a signed total Roman dominating family on K9k+6 . Therefore dstR (K9k+6 ) ≥ 3k + 2 and thus dstR (K9k+6 ) = 3k + 2.  This is a further example that demonstrates the sharpness of Theorem 9. The next result is a supplement to Example 7. Corollary 11. If Cn is a cycle of length n such that n ̸≡ 0 (mod 4), then dstR (Cn ) = 1. Proof. Since n ̸≡ 0 (mod 4), Proposition B implies γstR (Cn ) ≥ (n + 3)/2. Applying this bound and Theorem 9, we obtain dstR (Cn ) ≤

n

γstR (Cn )

≤

2n n+3

< 2.

Thus dstR (Cn ) = 1, and the proof is complete.



Combining Theorems 1 and 9, we obtain the following corollary. Corollary 12. If G is a 3-regular graph, then dstR (G) = 1. For some regular graphs we will improve the upper bound given in Theorem 5. Corollary 13. Let G be a δ -regular graph of order n such that n is not a multiple of δ . Then dstR (G) ≤ δ − 1.

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Proof. Let n = pδ + r with integers p ≥ 1 and 1 ≤ r ≤ δ − 1. According to Proposition A, we have

γstR (G) ≥

n δ

 =

pδ + r



δ

= p + 1.

Now Theorem 9 yields dstR (G) ≤

n

γstR (G)

≤

n

<δ

p+1

and therefore dstR (G) ≤ δ − 1.



Example 8 demonstrates that Corollary 13 is not valid in general. Theorem 14. Let G be a graph of order n ≥ 3 with δ(G) ≥ 1. Then dstR (G) ≤ n − 2, with equality if and only if G is isomorphic to K3 , P3 or C4 . Proof. Let δ = δ(G). If δ ≤ n − 2, then Theorem 5 implies that dstR (G) ≤ δ ≤ n − 2. If δ = n − 1, then G is isomorphic to the complete graph, and we deduce from Proposition E that γstR (G) = 3. Hence it follows from Theorem 9 that dstR (G) ≤ n/γstR (G) = n/3 ≤ n − 2, as desired. If G is isomorphic to K3 , P3 or C4 , then it is easy to verify that dstR (G) = n − 2. Conversely, assume that dstR (G) = n − 2. If δ ≤ n − 3, then Theorem 5 leads to the contradiction n − 2 = dstR (G) ≤ δ ≤ n − 3. Thus there remain the cases that δ = n − 1 or δ = n − 2. If δ = n − 1, then we observe as above that n − 2 = dstR (G) ≤ n/3 and so n = 3. This yields to G = K3 . In the case δ = n − 2 we distinguish two cases. Case 1. Assume that G is δ -regular. If δ = 1, then n = 3, a contradiction. If δ = 2, then n = 4 and so G = C4 . Now let δ ≥ 3. Applying Corollary 13, we obtain the contradiction δ = n − 2 = dstR (G) ≤ δ − 1. Case 2. Assume that δ = n − 2 and ∆(G) = n − 1. If n = 3, then G = P3 . If n = 4, then G = K4 − e, where e is an arbitrary edge of K4 . It is easy to verify that γstR (K4 − e) = 3. Using Theorem 9, we arrive at the contradiction n − 2 = 2 = dstR (K4 − e) ≤ n/3 = 4/3. Let now n ≥ 5, and let f be a γstR (G)-function. If f (x) = 1 for all x ∈ V (G), then γstR (G) = n > 2. If f (v) = −1 for one vertex v ∈ V (G), then there exists a vertex w ∈ V (G) with f (w) = 2. If d(w) = n − 1, then it follows that γstR (G) = f (w) + f (N (w)) ≥ 2 + 1 = 3 > 2. If d(w) = n − 2, then let u be the vertex not adjacent to w . This leads to γstR (G) = f (w) + f (N (w)) + f (u) ≥ 2 + 1 − 1 = 2. Altogether, we have shown that γstR (G) ≥ 2. Using again Theorem 9, we obtain n n ≤ , n − 2 = dstR (G) ≤ γstR (G) 2 a contradiction to n ≥ 5. This completes the proof.



For regular graphs we can improve Theorem 14. Theorem 15. If G is a δ -regular graph of order n with δ ≥ 1, then dstR (G) ≤ n/2. Proof. It follows from Proposition A that γstR (G) ≥ ⌈n/δ⌉ ≥ 2. Thus Theorem 9 implies that dstR (G) ≤

n

γstR (G)

≤

and the proof is complete.

n 2

, 

Example 8 demonstrates that Theorem 15 is sharp. 4. Upper bounds on γstR (G ) + dstR (G ) The upper bound on the product γstR (G) · dstR (G) leads to an upper bound on the sum of these two parameters. Theorem 16. Let G be a graph of order n with δ(G) ≥ 1. Then

γstR (G) + dstR (G) ≤ n + 1, with equality if and only if the components of G are K2 , K3 , P3 or C6 . Proof. It follows from Theorem 9 that n γstR (G) + dstR (G) ≤ + dstR (G). dstR (G) According to Theorem 5,√we have 1 ≤ dstR (G) ≤ √ n − 1. Using these bounds and the fact that the function g (x) = x + n/x is decreasing for 1 ≤ x ≤ n and increasing for n ≤ x ≤ n − 1, we obtain

γstR (G) + dstR (G) ≤

n dstR (G)

and the desired bound is proved.

 + dstR (G) ≤ max n + 1,

n n−1

 + n − 1 = n + 1,

L. Volkmann / Discrete Applied Mathematics (

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If the components of G are K2 , K3 , P3 or C6 , then Theorem 3 implies that γstR (G) = n. Hence it follows by Theorem 9 that dstR (G) = 1 and so γstR (G) + dstR (G) = n + 1. Conversely, assume that γstR (G) + dstR (G) = n + 1. The inequality above leads to n + 1 = γstR (G) + dstR (G) ≤

n dstR (G)

+ dstR (G) ≤ n + 1.

This implies dstR (G) = 1 and thus γstR (G) = n. Now Theorem 3 leads to the desired result.



Theorem 17. Let G be a graph of order n ≥ 4 with δ(G) ≥ 1. If G has a component different from K2 , K3 , P3 and C6 , then

γstR (G) + dstR (G) ≤ n, with equality if and only if G = C4 or γstR (G) = n − 1 and dstR (G) = 1. Proof. It follows from Theorem 16 that γstR (G) + dstR (G) ≤ n. Clearly, if G = C4 (see Proposition B and Example 7) or γstR (G) = n − 1 and dstR (G) = 1, then γstR (G) + dstR (G) = n. Conversely, assume that γstR (G) + dstR (G) = n. If γstR (G) > n/2, then Theorem 9 implies that dstR (G) = 1. This yields γstR (G) = n − 1. Let now γstR (G) ≤ n/2. If γstR (G) ≤ 1, then it follows from Theorem 14 that γstR (G) + dstR (G) ≤ n − 1, a contradiction. So assume that γstR (G) ≥ 2. We deduce from Theorem 9 that

γstR (G) + dstR (G) ≤ γstR (G) +

n

γstR (G)

.

Using the bounds √ 2 ≤ γstR (G) ≤ n/2, and the fact that the function g (x) = x + n/x is decreasing for 2 ≤ x ≤ increasing for n ≤ x ≤ n/2, we obtain n = γstR (G) + dstR (G) ≤ γstR (G) +

n

γstR (G)

√

n and

  n n n ≤ max 2 + , + 2 = 2 + . 2 2

2

This implies n = 4 and therefore γstR (G) = 2. Now it is easy to verify that G = C4 .



Theorem 18. If G is a graph of order n with δ(G) ≥ 5, then

γstR (G) + dstR (G) ≤ n − 2, unless n = 6, in which case G = K6 with γstR (G) + dstR (G) = 5 = n − 1. Proof. If γstR (G) > n/2, then Theorem 9 implies that dstR (G) = 1. According to Theorem 4 and the hypothesis δ(G) ≥ 5, we deduce that γstR (G) + dstR (G) ≤ (n − 3) + 1 = n − 2. Let now γstR (G) ≤ n/2. If γstR (G) ≤ 1, then it follows from Theorem 14 that γstR (G) + dstR (G) ≤ 1 + (n − 3) = n − 2. So assume that γstR (G) ≥ 2. As in the proof of Theorem 17, we obtain γstR (G) + dstR (G) ≤ 2 + n/2. If n ≥ 8, then we deduce that γstR (G) + dstR (G) ≤ 2 + n/2 ≤ n − 2. Assume now that n = 7. It follows that 2 ≤ γstR (G) ≤ 3. If γstR (G) = 2, then by Theorem 9, dstR (G) ≤ 3 and so γstR (G)+ dstR (G) ≤ 5 = n − 2. If γstR (G) = 3, then Theorem 9 implies that dstR (G) ≤ 2 and hence γstR (G)+ dstR (G) ≤ 5 = n − 2. Finally assume that n = 6. Clearly, then G = K6 . In view of Proposition E, we have γstR (G) = 3. Therefore Theorem 9 leads to dstR (G) ≤ 2. Now let v1 , v2 , . . . , v6 be the vertex set of G. For 1 ≤ i ≤ 2 define the function fi : V (G) −→ {−1, 1, 2} by f1 (v1 ) = f1 (v2 ) = f1 (v3 ) = −1, f1 (v4 ) = f1 (v5 ) = f1 (v6 ) = 2 and f2 (v1 ) = f2 (v2 ) = f2 (v3 ) = 2, f2 (v4 ) = f2 (v5 ) = f2 (v6 ) = −1. Then f1 and f2 are STRD functions on G, and {f1 , f2 } is a signed total Roman domination family on G. Therefore dstR (G) ≥ 2 and thus dstR (G) = 2. Altogether we obtain γstR (G) + dstR (G) = 5 = n − 1, and the proof is complete.  5. Nordhaus–Gaddum type results Results of Nordhaus–Gaddum type study the extreme values of the sum or product of a parameter on a graph and its complement. In their current classical paper [4], Nordhaus and Gaddum discussed this problem for the chromatic number. We present such inequalities for the signed total Roman domatic number. Theorem 19. If G is a graph of order n such that δ(G), δ(G) ≥ 1, then dstR (G) + dstR (G) ≤ n − 1, with equality if and only if G = C4 . Proof. It follows from Theorem 5 that dstR (G) + dstR (G) ≤ δ(G) + δ(G)

= δ(G) + (n − ∆(G) − 1) ≤ n − 1, and this is the desired bound.

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If G is not regular, then ∆(G)−δ(G) ≥ 1, and hence the above inequality chain implies the better bound dstR (G)+ dstR (G) ≤ n − 2. Let δ(G) = δ and δ(G) = δ . Assume now that G is δ -regular. Then G is δ -regular such that δ + δ + 1 = n. Assume, without loss of generality, that δ ≤ δ . If δ = 1, then δ = 1 and we obtain the contradiction 3 = δ + δ + 1 = n. Let now δ ≥ 2. If n = pδ + r with integers p ≥ 1 and 1 ≤ r ≤ δ − 1, then Theorem 5 and Corollary 13 lead to the better bound dstR (G) + dstR (G) ≤ (δ − 1) + δ = n − 2. Thus assume that n = pδ with an integer p ≥ 2. As δ ≤ δ , we observe that pδ = n = δ + δ + 1 ≤ 2δ + 1 and so p = 2. Therefore n = 2δ and hence δ = n − δ − 1 = 2δ − δ − 1 = δ − 1. If δ = 1, then δ = 2 and n = 4. Therefore G = C4 and G = K2 ∪ K2 . Theorem 5 and Example 7 imply that dstR (G) + dstR (G) = 2 + 1 = 3 = n − 1 and hence equality in the proved Nordhaus–Gaddum bound. Now assume that δ ≥ 2 and thus δ = δ + 1 ≥ 3. If n = qδ + r with integers q ≥ 1 and 1 ≤ r ≤ δ − 1, then Theorem 5 and Corollary 13 lead to the better bound dstR (G) + dstR (G) ≤ δ + (δ − 1) = n − 2. Thus assume that n = qδ with an integer q ≥ 2. Altogether, we have n = 2δ = q(δ − 1) with δ ≥ 3. It is straightforward to verify that this identity is only possible for q = 3 and δ = 3 and thus δ = 2 and n = 6. However, in this case we conclude from Corollary 12 and Theorem 5 that dstR (G) + dstR (G) ≤ 1 + 2 = 3 = n − 3. Conversely, if G = C4 , then G = K2 ∪ K2 , and we deduce from Theorem 5 and Example 7 that dstR (G) + dstR (G) = 2 + 1 = 3 = n − 1.  Theorem 20. Let G be a graph of order n such that δ(G), δ(G) ≥ 1. If δ(G) = 3 or δ(G) = 3, then dstR (G) + dstR (G) ≤ n − 3. Proof. Assume, without loss of generality, that δ(G) = 3. It follows from Theorem 5 and Corollary 6 that dstR (G) + dstR (G) ≤ 2 + δ(G) = (δ(G) − 1) + δ(G) = (δ(G) − 1) + (n − ∆(G) − 1). If G is not regular, then ∆(G) − δ(G) ≥ 1, and hence the above inequality chain implies dstR (G) + dstR (G) ≤ n − 3. Assume now that G is 3-regular. Then G is (n − 4)-regular. We deduce from Theorem 5 and Corollary 12 that dstR (G) + dstR (G) ≤ 1 + (n − 4) = n − 3. This completes the proof.



Let G = K4,4 . Then G is 3-regular. It follows from Example 8 and Corollary 12 that dstR (G)+ dstR (G) = 4 + 1 = 5 = n(G)− 3, and therefore equality in Theorem 20. References [1] [2] [3] [4] [5] [6] [7]

H.A. Ahangar, M.A. Henning, Y. Zhao, C. Löwenstein, V. Samodivkin, Signed Roman domination in graphs, J. Comb. Optim. 27 (2014) 241–255. E.J. Cockayne, S.T. Hedetniemi, Towards a theory of domination in graphs, Networks 7 (1977) 247–261. T.W. Haynes, S.T. Hedetniemi, P.J. Slater, Fundamentals of Domination in Graphs, Marcel Dekker, Inc., New York, 1998. E.A. Nordhaus, J.W. Gaddum, On complementary graphs, Amer. Math. Monthly 63 (1956) 175–177. S.M. Sheikholeslami, L. Volkmann, The signed Roman domatic number of a graph, Ann. Math. Inform. 40 (2012) 105–112. P.J. Slater, E.L. Trees, Multi-fractional domination, J. Combin. Math. Combin. Comput. 40 (2002) 171–181. L. Volkmann, Signed total Roman domination in graphs, J. Comb. Optim. http://dx.doi.org/10.1007/s10878-015-9906-6.