CC BY
3URAL MATHEMATICAL JOURNAL, Vol. 8, No. 2, 2022, pp. 81-93
DOI: 10.15826/umj.2022.2.007
RESTRAINED ROMAN REINFORCEMENT NUMBER
IN GRAPHS
Saeed Kosari
Institute of Computing Science and Technology, Guangzhou University,
Guangzhou 510006, China saeedkosari38@gzhu.edu.cn
Seyed Mahmoud Sheikholeslami
Department of Mathematics, Azarbaijan Shahid Madani University,
Tabriz, I.R. Iran s.m.sheikholeslami@azaruniv.ac.ir
Mustapha Chellali
LAMDA-RO Laboratory, Department of Mathematics, University of Blida,
B.P. 270, Blida, Algeria m_chellali ©Jyahoo. com
Maryam Hajjari
Department of Mathematics, Azarbaijan Shahid Madani University,
Tabriz, I.R. Iran m.hajjari260@gmail.com
Abstract: A restrained Roman dominating function (RRD-function) on a graph G = (V, E) is a function f from V into {0,1, 2} satisfying: (i) every vertex u with f (u) = 0 is adjacent to a vertex v with f (v) = 2; (ii) the subgraph induced by the vertices assigned 0 under f has no isolated vertices. The weight of an RRD-function is the sum of its function value over the whole set of vertices, and the restrained Roman domination number is the minimum weight of an RRD-function on G. In this paper, we begin the study of the restrained Roman reinforcement number ttr(G) of a graph G defined as the cardinality of a smallest set of edges that we must add to the graph to decrease its restrained Roman domination number. We first show that the decision problem associated with the restrained Roman reinforcement problem is NP-hard. Then several properties as well as some sharp bounds of the restrained Roman reinforcement number are presented. In particular it is established that rrR(T) = 1 for every tree T of order at least three.
Keywords: Restrained Roman domination, Restrained Roman reinforcement.
1. Introduction
For definitions and notations not given here we refer the reader to [8]. We consider simple graphs G with vertex set V = V(G) and edge set E = E(G). The order of G is n = n(G) = |V|. The open neighborhood of a vertex v, denoted by N(v) (or NG (v) to refer to G) is the set {u € V(G) | uv € E} and its closed neighborhood is the set N[v] = NG[v] = N(v) U {v}. The degree of vertex v € V is d(v) = dG(v) = |N(v)|. The maximum and minimum degree in G are denoted by A = A(G) and 5 = 5(G), respectively. A vertex of degree one is called a leaf and its neighbor is
called a support vertex. As usual, the path (cycle, complete p-partite graph, respectively) of order n is denoted by Pn (Cn, Kni)n2)...,np, respectively). A star of order n > 2 is the graph K1;n-1. For a subset S C V, the subgraph induced by S in G is denoted as G[S].
A subset S C V is a dominating set of G if every vertex in V \ S has a neighbor in S. The domination number y(G) is the minimum cardinality of a dominating set of G.
As an application, in the design of networks for example, it is essential to study the effect of some modifications of the graph parameters on its structure. These modifications can be deletion or addition of vertices, deletion or addition of edges. We refer the reader to chapter 7 of [8] when the graph parameter is the domination number. The reinforcement number r(G) of a graph G is the minimum number of edges that have to be added to the graph G in order to decrease the domination number. Of course for graphs G with domination number one it was assumed that r(G) = 0. The concept of the reinforcement number was introduced in 1990 by Kok and Mynhardt [10], and since then it has been defined and studied for several other domination parameters, such as Roman domination [9], total Roman domination [1], quasi-total Roman domination [5], Italian domination [7], double Roman domination [4] and rainbow domination [3, 13].
In 2015, Leely Pushpam and Padmapriea [11] introduced the concept of restrained Roman domination as a new variation of Roman domination. A restrained Roman dominating function (RRD-function, for short) on a graph G is a function f : V —> {0,1,2} having the properties that (i) every vertex u with f (u) = 0 is adjacent to a vertex v with f (v) = 2; and (ii) the subgraph induced by the vertices assigned 0 under f has no isolated vertices. The weight of an RRD-function f is the sum
w(f)= £ f(v)
vev (G)
and the restrained Roman domination number of G denoted by YrR(G), is the minimum weight of an RRD-function on G. Any RRD-function f on G can simply be referred as f = (V0, Vi, V2), where V = {v € V(G) : f (v) = i} for i € {0,1, 2}. For further studies on restrained Roman domination and its variants, see [2, 12, 14-16].
In this paper, we are interested in starting the study of the restrained Roman reinforcement number rrR(G) of a graph G defined as the cardinality of a smallest set of edges F C E(G) such that jrn(G + F) < jrn(G), where G denotes the complement graph of G. If there is no subset of edges F satisfying YrR(G + F) < YrI(G), then we define rrR(G) = 0. Since for any nontrivial connected graph G, YrR(G) > 2, we deduce that rrR(G) = 0 for all nontrivial connected graphs with jrR(G) = 2. Moreover, a subset E' C E(G) is called an rrR(G)-set if \E'\ = rrR(G) and
YrR(G + E') <YrR(G).
Further, we will prove that the decision problem associated with the Restrained Roman reinforcement is NP-hard. Then various properties of the restrained Roman reinforcement number are investigated and some sharp bounds on it are presented.
We finish this section by observing that any rrR(G)-set of a connected graph G with YtR(G) > 3 can decrease the restrained Roman domination number of G by at most two.
Proposition 1. Let G be a connected graph with YrR(G) > 3. If F is an rrR(G)-set, then
YrR(G) - 2 < YrR(G + F) < YrR(G) - 1.
Both bounds are sharp.
Proof. By assumption, YrR(G + F) < YrR(G), whence the upper bound follows. To show the lower bound, let us assume that
YrR(G + F) < YrR(G) - 3.
Let / be a YrR(G + F)-function and let uv € F such that 0 € {/(u), / (v)}. If such an edge does not exist, then / is an RRD-function of G leading to the contradiction
YrR(G) < 7rR(G + F).
Hence we suppose that uv exists, and let F' = F — {uv}. Without loss of generality, suppose that /(u) = 0. If /(v) = 1, then / is an RRD-function of G leading to the contradiction
YrR(G) < YrR(G + F)
too. Hence assume that /(v) = 1.
First let /(v) = 2. If u has a neighbor w in G + F' with /(w) > 1, then the function g defined by g(w) = 2 and g(x) = /(x) otherwise, is an RRD-function of G + F' yielding as above to the contradiction YrR(G + F') < YrR(G). Hence we assume that each neighbor u in G + F' is assigned 0 under /. Let x1,..., xk be the neighbors of u in G + F'. If k = 1 and x1 has a neighbor assigned 0 other than u, then the function g(u) = 1 and g(x) = /(x) otherwise, is an RRD-function of G + F' yielding
YrR(G + F') < YrR(G + F) + 1 < YrR(G),
this is a contradiction. If k = 1 and x1 has no neighbor assigned 0 other than u, then the function g(u) = g(x1) = 1 and g(x) = /(x) otherwise, is an RRD-function of G + F' and thus
YrR(G + F') < YrR(G + F) + 2 < YrR(G),
it is a contradiction too. Hence assume that k > 2. If some xj has no neighbor assigned 0 other than u, then the function g(xj) = 2 and g(x) = /(x) otherwise, is an RRD-function of G + F' yielding again YrR(G + F') < YrR(G). Hence we assume that for each i, xj has at least two neighbors assigned 0 under /. In this case, we have g(u) = 1 and g(x) = /(x) otherwise, it is an RRD-function of G + F' and thus
YrR(G + F') <YrR(G).
Finally, assume that /(v) = 0. Since F is an rrR(G)-set, we can suppose, without loss of generality, that all neighbors of u in G + F' have positive labels under /. Now, if v has a neighbor with weight 0 in G+F', then the function g(u) = 1 and g(x) = /(x) otherwise, it is an RRD-function of G + F' while if v has no neighbor with weight 0 in G + F', then the function g(u) = g(v) = 1 and g(x) = /(x) otherwise, is an RRD-function of G + F'. Both situations yield the contradiction YrR(G + F') < YrR(G). Consequently,
YrR(G + F) > YrR(G) — 2.
The upper bound of Proposition 1 is attained for the cycle while the lower bound is attained for the cycle C6. □
2. NP-hardness result
The aim of this section, is to show that the decision problem associated with the Restrained Roman reinforcement is NP-hard. Consider the following decision problem.
x,
Li
Hi
Ho
Figure 1. The graphs Li and H = Hi U H2.
Restrained Roman reinforcement problem (RR-reinforcement)
Instance: A nonempty graph G and a positive integer k. Question: Is rrR(G) < k?
We show that the NP-hardness of the RR-reinforcement problem by transforming the well-known 3-SAT problem to it in polynomial time. Recall that the 3-SAT problem specified below was proven to be NP-complete in [6].
z
3-SAT problem
Instance: A collection C = {Ci, C2,..., Cm} of clauses over a finite set X of variables such that |Cj| = 3 for every j € {1, 2,... , m}.
Question: Is there a truth assignment for X that satisfies all the clauses in C?
Theorem 1. Problem RR-reinforcement is NP-hard for an arbitrary graph.
Proof. Let X = {xi, x2,..., xn} and C = {Ci, C2,..., Cm} be an arbitrary instance of 3-SAT problem. We will build a graph G and a positive integer k such that rrR(G) < k if and only if C is satisfiable.
For each i € {1,2,..., n}, we associate to the variable x € X a copy of the graph Lj as depicted in Figure 1, and for each j € {1,2,...,m}, we associate to the clause Cj = {uj, Vj, Wj} € C a vertex cj by adding the edge-set Ej = {cjUj, CjVj, CjWj}. Finally, we enclose the graph H illustrated in Figure 1 by connecting vertices si, si to every vertex Cj. Clearly, the resulting graph G is of order 8n+m + 19 and size 11n+5m + 27 and hence G can be built in polynomial time. Set k = 1. Figure 2 provides an example of the resulting graph when X = {xi, x2, x4} and C = {Ci,C2,C3}, where Ci = {xi,x2,x3}, C2 = {xi,x2,x4} and C3 = {x2,x3,x4}.
It is easy to verify that for any YrR(G)-function g we must have
E g(v) > 4
vev (Lj)
for each j € {1,2,..., n}. Moreover, to restrained Roman dominate all vertices of V(H), we need that
m
Eg(Cj)+ g(V(H)) > 6. i=i
Therefore
YrR(G) = w(g) > 4n + 6.
Figure 2. An instance of the restrained Roman reinforcement number problem resulting from an instance of 3-SAT. Here k = 1 and YrR(G) = 22, where the black vertex p means there is a RRDF f with f (p) = 2.
Basing on the assignment given to the graph in Figure 2, one can easily define an RRD-function of G with weight 4n + 6, which consequently leads to YrR(G) = 4n + 6.
In the following, we show that C is satisfiable if and only if rrR(G) = 1. Let C be satisfiable and t : X ^ {T, F} a satisfying function for C. We build a subset S of vertices of G as follows. If t(xi) = T, then put the vertices Xi and iji in S; while if t(xi) = F, then put the vertices xl and Z in S. So |S| = 2n. Define the function h on V(G) by h(x) = 2 for every x € S, h(si) = 1, h(s3) = h(s'3) = 2 and h(y) = 0 for the remaining vertices. It is easy to verify that h is an RRD-function of G + s4s3 of weight
4n + 5 < YrR(G) = 4n + 6,
and hence rrR(G) = 1.
Conversely, let rrR(G) = 1. Then, there is an edge e = uv € E(G) for which
YrR(G + e) < 4n + 6.
Let g = (V0,V1, V2) be a YrR(G + e)-function. Since whatever the added edge e, we have g(V(Lj)) > 4, and thus vertices u and v cannot both belong to V(L) (for otherwise YrR(G + e) > 4n + 6). On the other hand, since rrR(G) = 1 and g(V(Lj)) > 4, we must have
m
Eg(cj)+ g(V(H)) < 6. j=i
Since also whatever the added edge e, we have g(V(H)) > 5, we conclude that g(V(H)) = 5. In particular, this is only possible if g(s1) = 0, g(s1) < 1 and
m
Eg(cj ) = o. j=i
In addition, we note that if {x», x»} C V2 or {x», x»} n Vi = 0 for some i, then g(V(Lj)) > 5 which results in the contradiction
YrR(G + e) > 4n + 6.
Thus, |{xj, x»} n V2| < 1 and {x», x»} n Vi = 0 for every i € {1,... ,n}. Therefore each vertex Cj must have a neighbor in {x», Xj} for some i which is assigned a 2. In this case, define the mapping t : X ^{T,F} by
t(x ) = {T if f (xj) = 2, (2.1)
1 F otherwise
for i € {1 , . . . , n} .
We show that t satisfies the truth assignment for C. It is enough to show that every clause in C is satisfied by t. Consider an arbitrary clause Cj € C for some j € {1, . . . , m}. If Cj is dominated by x», then g(x») = 2 and so t(x») = T. If Cj is dominated by x, then g(xj) = 2 and hence t(x») = F and t(afj) = T. Therefore, in either case the clause Cj is satisfied. The arbitrariness of j shows that all clauses in C are satisfied by t, that is, C is satisfiable. This completes the proof of the theorem. □
3. Exact values
In this section, we determine the restrained Roman reinforcement number of some classes of graphs including paths, cycles and complete p-partite graphs for any integer p > 2. As observed in [11], for every connected graph G of order n > 2, we have 2 < 7rR(G) < n. A characterization of all connected graphs of order n with YrR(G) € {2,3,n} was provided in [11, 14] as follows.
The graph B4. The graph £3,2.
Figure 3. Graphs B4 and B3,2.
Let C := (uiu2u3u4u5) be a cycle of length 5 and let Bp be the graph obtained from C by adding p > 1 new vertices attached by edges at ui and let Bp,q be the graph obtained from C by adding p > 1 new vertices attached by edges at ui and q > 1 other new vertices attached by edges at u3 (see Fig. 3). Recall that the diameter, diam(G), of a graph G is the maximum distance between the pair of vertices.
Proposition 2 [11]. Let G be a connected graph of order n > 2. Then
(a) YrR(G) =2 if and only if n = 2 or A(G) = n — 1 and ¿(G) > 2;
(b) YrR(G) = n if and only if G ~ C4, C5, Bp, or G is a tree with diam(G) < 5.
Proposition 3 [14]. Let G be a connected graph of order n > 4. Then YrR(G) = 3 if and only if G satisfies one of the following conditions :
(i) A(G) = n — 1 and G has exactly one leaf;
(ii) A(G) = n — 2 and G has a vertex u of degree n — 2 such that the induced subgraph G[N (u)] has no isolated vertex.
On the other hand, the exact values of the restrained Roman domination number have been established in [11] for paths, cycles and complete p-partite graphs.
Proposition 4 [11]. The following conditions holds:
(a) YrR(Pn) = n for 1 < n < 6 and YrR(Pn) = T(2n + 1)/3] +1 for n > 7;
(b) YrR(Cn) = 2 [n/3] when n ^ 2 (mod 3) and YrR(Cn) = 2 [n/3] + 1 otherwise;
(c) YrR(Km,n) = 4 for m, n > 2;
(d) if Kni)n2)...)np is the complete p-partite graph such that p > 3 and n1 < n2 < ... < np, then YrR(Ki)n2,...,np) = 2, YrR(K2,„2,...,np) = 3 and YrR(Kni,n2,...,np) = 4 for ni > 3.
Now we are ready to find the restrained Roman reinforcement number for paths, cycles and complete p-partite graphs, p > 2.
Proposition 5. For n > 3, rrR(Pn) = 1.
Proof. Let Pn := wiw2 ... wn. If n = 0 (mod 3), then the function g defined by
g(w3i+i) = 2
for 0 < i < (n — 3)/3 and g(w) = 0 otherwise, is an RRD-function of Pn + wiwn of weight 2n/3. If n = 2 (mod 3), then the function g defined by g(wn) = 2, g(w3i+i) = 2 for 0 < i < (n — 5)/3 and g(x) = 0 otherwise, is an RRD-function of Pn + wiwn-2 of weight (2n + 2)/3. Finally, if n = 1 (mod3), then the function g defined by g(wn) = 1, g(w3i+i) = 2 for 0 < i < (n — 4)/3 and g(w) = 0 otherwise, is an RRD-function of Pn + wiwn-i of weight (2n + 1)/3. All considered cases show that rrR(Pn) = 1. □
Proposition 6. For n > 4,
2 if n = 0 (mod 3), 1 otherwise.
Proof. Assume that Cn := (wiw2 ... wn) be a cycle on n vertices. If n = r (mod 3) with r € {1,2}, then by a similar argument to that used in the proof of Proposition 5, we can see that rrR(Cn) = 1. Hence we assume that n = 0 (mod 3). First, since the function g defined by g(wn-2) = 1, g(w3i+i) = 2 for 0 < i < (n — 6)/3 and g(x) = 0 otherwise, is an RRD-function of Cn + {wiwn-i,wiwn-3} of weight (2n — 3)/3 = YrR(Cn) — 1 (Proposition 4-(b)), we deduce that
rrR(Cra) < 2.
Now we prove the inverse inequality. For this purpose, we need only to show that adding an arbitrary edge e cannot decrease jrR(Cn)- Observe that for any edge e € Cn,
YrR(Cn + e) < YrR(Cn).
Let e be an arbitrary edge in Cn and let / be a jrR(Cn + e)-function. Suppose first that there are three consecutive vertices Wi,wi+i,wi+2 such that f(wi) = f(wi+i) = f(wi+2) = 0, say for i = 1.
Then the edge e must join w2 to some vertex assigned 2, say wk, with k / {1,3}. Also, to restrained Roman dominate w1 and w3, we must also have f (w4) = f (wn) = 2.
Consider the cycles C' := (w2w3 ...wk) of order k — 1 and C'' := (w2wk ...wnw1) of order n — k + 3. Let k — 1 = s1 (mod 3) and n — k + 3 = s2 (mod 3). Notice that s1 = 0 and s2 = 2; s1 = 2 and s2 = 0 or s1 = s2 = 1. Assume that k — 1 = 0 (mod 3) (the case n — k + 3 = 0 (mod 3) is similar). Then n — k + 3 = 2 (mod 3), and since the restrictions of f on V(C) and V(C'') are RRD-functions, we deduce from Proposition 4-(b), that
YrR(Cn + e) = f (V (C')) + f (V (C'')) — 2
^ , ^ O 2(k — 1) , 2(n — k + 3) + 3 + 2 2n + 3 .
> 1tr{C) + 7rfl(C") - 2 = —+ -i-^--2 = —> YrR(Cn).
Assume now that s1 = s2 = 1. Then, as above, it follows from Proposition 4-(b) that
7rR(C„ + e) = f (V (C')) + f (V (C')) — 2
> WO) + WO") - 2 = 2№-1)3+3 + 1 + 2("-fc+33) + 3 + 1 - 2 = 22 + 8 > lrR{C,t).
Thus in either case we obtain a contradiction. Next suppose there are three consecutive vertices Wi, Wi+1, Wi+2 such that f (wi) + f (wi+1) + f (wi+2) = 1, say for i = 1.
If f (w2) = 1, then f (w1) = f (w3) = 0 and each of w1 and w3 must be adjacent a vertex assigned 2 as well as to a vertex assigned 0. This possible only if e = w1w3 and so
H = (Cn + e) — W2
is a cycle on n — 1 vertices, where the restriction of f to H is an RRD-function. It follows that
YrR(Cn + e) = f (V(H)) + 1 > YrR(H) + 1, and by Proposition 4-(b), we obtain
tn , ^ ^ 2(n — 1) + 3 + 2
7rfl(C„ + e) > -i-L-+ 1 > yrR(Cn)
which is a contradiction. Hence we can assume that f (w2) = 0. Without loss of generality, let f(w1) = 1 and f(w3) = 0. To restrained Roman dominate w2, the edge e must join w2 to a vertex with label 2, say wk. Likewise for w3 we must have f (w4) = 2. Now, consider the cycles C' := (w2w3 ... wk) of order k — 1 and the path P' := wk ... wnw1 of order n — k + 2.
Let k — 1 = s1 (mod 3) and n — k + 2 = s2 (mod 3). Notice that s1 = 0 and s2 = 1; s1 = 1 and s2 = 0 or s1 = s2 = 2. Notice also that the restrictions of f on V(C') and V(P') are RRD-functions, and thus
YrR(Cn + e) = f (V(C')) + f (V(P')) — 2 > YrR(C') + YrR(P') — 2.
Now using Propositions 4-(a,b), we get a contradiction as before. Finally, let
f (Wi) + f (Wi+1) + f (Wi+2) > 2 for each 1 < i < n, where the sum in indices is taken modulo n. Then we have
1 " 2n
7rfl(C„ + e) = - J2if{wi) + /(wi+i) + /(«'i+2)) > y = JrR{C„), i=1
and therefore, YrR(Cn + e) = YrR(G). Consequently, rrR(Cn) = 2 as desired. □
Proposition 7. For integers 1 < r < s with r + s > 3,
1 if r = 1, 2, 3,
rrR(Kr,s) = , r - 2 if r > 4.
Proof. Let X = (xi, ..., xr} and Y = {yi, • • •, ys} be the partite sets of If r = 1, then the function g defined by g(x1 ) = 2, g(y1) = g(y2) = 0 and g(x) = 1 otherwise, is an RRD-function of K1>s + y1y2 of weight n — 1 and it follows from Proposition 2-(b) that rrR (KM) = 1.
If r = 2, then the function g defined by g(x1 ) = 2 and g(x) = 0 otherwise, is an RRD-function of K2>s + x1x2 of weight 2 and we get from Proposition 2-(a) that rrR(K2;S) = 1.
If r = 3, then the function g defined by g(x1 ) = 2,g(x2) = 1 and g(x) = 0 otherwise, is an RRD-function of K3>s + x1 x3 of weight 3 and by Proposition 4-(c), we have rrR(K3;S) = 1.
Let r > 4. First we observe that the function g defined by g(x1) = 2, g(x2) = 1 and g(x) = 0 otherwise, is an RRD-function of + (x1xi | 3 < i < r} of weight 3 and thus by Proposition 4-(c), rrR(Kr,s) < r — 2.
To show that rrR(Kr,s) > r — 2, let F be an rrR(Kr,s)-set. Then
2 < 7rR(Kr,s + F) < 3.
By Propositions 2-(a) and 3 we must have A(Kr;S + F) > r + s — 2 and this implies that |F| > r — 2. Therefore rrR(Kr,s) = r — 2 and the proof is complete. □
Proposition 8. Let Kni)n2)...>np be the complete p-partite graph such that p > 3 and 3 < m < n2 < ... < np. Then rrR(Krai;ra2,...,np) = n1 — 2.
Proof. Let G = Kniin2,...,nP and X1 = (x1, ..., xm}, X2 = (y1,... ,y„2},... ,XP be the partite sets of G. Let F be an rrR(G)-set. By Proposition 4-(d) we deduce that YrR(G+F) e (2,3}, and by Propositions 2-(a) and 3 we must have
A(G + F) > m + ■ ■ ■ + np — 2
implying that |F| > n1 — 2. On the other hand, the function g defined by f (x1) = 2, f (x2) = 1 and f (x) = 0 otherwise, is an RRD-function of G + (xix1 | 3 < i < n1} yielding rrR(G) < |F| = n1 — 2. Consequently, rrR(G) = n1 — 2. □
4. Graphs with small restrained Roman reinforcement number
In this section, we study graphs with small restrained Roman reinforcement number. We begin with the following lemma.
Lemma 1. If G is a connected graph of order n > 3 with YrR(G) = n, then rrR(G) = 1.
Proof. By Proposition 2, G ~ C4,C5,Bp,BP;q or G is a tree with diam(G) < 5. If G € {C4,C5}, then the desired result follows from Proposition 6. If G € {Bp,BP;q}, then the function g defined by g(ui) = 2, g(u2) = g(u5) = 0 and g(x) = 1 otherwise, is an RRD-function of G + u2u5 and hence rrR(G) = 1. Hence, we assume that G is a tree with diameter at most 5.
Let viv2 ... vk (k > 3) be a diametral path in G. Define the function f by f (vi) = f (v3) = 0, f (v2) = 2 and f (x) = 1 for the remaining vertices. Clearly, f is an RRD-function of G + viv3 and hence rrR(G) = 1. □
Proposition 9. Let G be a connected graph of order n > 4 with Yr/(G) > 3. If f = (Vo, Vi, V2) is a YrR(G)-function with Vi = 0, then
rrR(G) = 1.
Proof. Let f = (Vo, Vi, V2) be a YrR(G)-function such that Vi = 0. If YrR(G) = n, then the desired result comes from Lemma 1.
Hence assume that YrR(G) < n. Then Vo = 0 and so V2 = 0. Since G is connected and Vi = 0, there exists a vertex w € Vi such that w is dominated by Vo U V2. Note that if w has a neighbor in Vo and another one in V2, then reassigning w provides an RRD-function with weight YrR(G) — 1, a contradiction.
Now, if w is adjacent to a vertex in V2, then the function g defined by g(w) = 0 and g(x) = f (x) otherwise, is an RRD-function of G + wz where z € Vo, of weight less than YrR(G), and thus rrR(G) = 1. If w is adjacent to a vertex in Vo, then the function g defined by g(w) = 0 and g(x) = f (x) otherwise, is an RRD-function of G + wu where u € V2, of weight less than YrR(G) and so rrR(G) = 1. This completes the proof. □
Proposition 10. Let G be a connected graph of order n with YrR(G) > 3. Then rrR(G) = 1 if and only if YrR(G) = n or G has a function f = (Vo, Vi, V2) of weight less than YrR(G) such that one of the following conditions holds:
(i) G[Vo] has at most two isolated vertices and V2 dominates Vo;
(ii) G[Vo] has no isolated vertices and there is exactly one vertex v € Vo which is not dominated by V2.
Proof. If YrR(G) = n, then by Lemma 1 we have rrR(G) = 1. Hence suppose that YrR(G) < n, and let f = (Vo, Vi, V2) be a function on G with weight less than YrR(G) satisfying (i) or (ii). Since w(f) < YrR(G) < n — 2, we have |Vo| > 2. In the case V2 is non-empty, let u € V2. Now, if (ii) holds, then V2 = 0 and f is an RRD-function of G + uv.
Assume now that (i) holds. If G[Vo] has two isolated vertices w,v, then f is an RRD-function of G + {wv} and if G[Vo] has exactly one isolated vertex, say w, then f is an RRD-function of G + {wz}, where z is any vertex in Vo — {w}. Hence in either case rrR(G) = 1.
Conversely, let rrR(G) = 1 and suppose that {uv} is an rrR(G)-set. If YrR(G) = n, then we are done. Hence suppose that YrR(G) < n — 1 and let f be a YrR(G + {uv})-function. Notice that vertices u and v cannot be assigned both positive values under f (otherwise f is an RRD-function of G). Without loss of generality, assume that f (u) = 0. If f (v) = 0, then f is a function satisfying (i). Hence assume that f (v) > 1. If u is adjacent to a vertex with label 2 other than v, then f is an RRD-function of G. Hence u is not dominated by V2 in G and so f is a function satisfying (ii). This completes the proof. □
Proposition 11. Let G be a connected graph of order n with YrR(G) > 3. If ¿(G) = 1, then
rrR(G) = 1.
Proof. First note that n > 3, since YrR(G) > 3. If YrR(G) = n, then the result comes from Lemma 1. Hence we assume that YrR(G) < n, and let f = (Vo, Vi, V2) be a YrR(G)-function. We have Vo = 0 (because YrR(G) < n) and thus V2 = 0.
Let u be a support vertex of G and ui a leaf neighbor of u. By definition we have f(ui) > 1. If f (ui) = 1 or f (u) = 1, then the desired result comes from Proposition 9.
Hence we assume that f (ui) = 2 and f (u) = 1. The minimality of f implies that f (u) = 0. Note that ui is the only neighbor of u which assigned a 2, for otherwise ui can be reassigned the
value 1 instead of 2. Let w be a neighbor of u with label 0. To Roman dominate w, there is a vertex v such that f (v) = 2. Then the function g defined on G + uv by g(u1) = 1 and g(x) = f (x) otherwise, is an RRD-function of G + {uv} with weight w(f) — 1. Consequently, rrR(G) = 1. □
Corollary 1. For any tree T of order n > 3, rrR(T) = 1.
5. Bounds on rrR(G)
In this section, we present some sharp upper bounds on the restrained Roman reinforcement number of a graph. Given a set S C V of vertices in a graph G and a vertex v € S, the external private neighborhood of v with respect to S in the set
epn(v, S) = {u € V — S | N(u) n S = {v}}.
Proposition 12. Let G be a connected graph with Yr/(G) > 3. If f = (Vo,V1,V2) is a YrR(G)-function with V2 = 0, then
rrR(G) < min {|epn(v,V>) n V0| : v € V2}.
Proof. Let f = (Vo, V1, V2) be a YrR(G)-function with V2 = 0. If |epn(v, V2) n Vo| = 0 for some vertex v € V2, then reassigning v the value 1 instead of 2 provides an RRD-function of weight less than YrR(G) leading to a contradiction. Hence |epn(v, V2) n V0| > 1 for every v € V2. Let u be a vertex in V2 such that
|epn(u, V2) n Vo| = min{|epn(v, V2) n Vo| : v € V2}
and let epn(u, V2) n V0 = {u1,..., ue}. If |V2| > 2 and w € V2 — {u}, then the function g defined by g(u) = 1 and g(x) = f (x) otherwise, is an RRD-function of G + {wx | x € epn(v, V2) n V0} of weight less than YrR(G) and so
rrR(G) < min{|epn(v, V2) n Vo| : v € V2}.
Hence assume that V2 = {u}. Then u dominates all vertices in Vo. Since YrR(G) > 3, we have V1 = 0 and the desired result follows from Proposition 9. □
We observe that for any YrR(G)-function f = (Vo, V1, V2), every vertex u of V2 can have at most dG(u) neighbors in Vo. Whence we have the following corollary.
Corollary 2. Let G be a connected graph with YrR(G) > 3 and f = (Vo,V1,V2) a YrR(G)-function with |V2| > 1. Then rrR(G) < A.
Corollary 3. Let G be a connected graph with YrR(G) > 3 containing a path v1v2v3v4v5 in which dG(vi) = 2 for i € {2,3,4}. Then rrR(G) < 2.
Proof. If YrR(G) = n, then the result is immediate from Lemma 1. Hence we assume that YrR(G) < n, and let f = (Vo, V1, V2) be a YrR(G)-function. By Proposition 9, we may assume that V1 = 0. Then we must have 2 € {f (v2), f (v3), f (v4)} and the result follows from Proposition 12. □
Using Propositions 9 and 12 we obtain the next result.
x
Figure 4. A graph G of order 18 and rrR(G) = 4.
Theorem 2. For any graph G of order n > 3, we have
rrR(G) < max{1, (2n - 7rR(G))/7rR(G)}. Moreover, the bound is sharp.
Proof. If YrR(G) = 2, then rrR(G) = 0 and the result is true. If YrR(G) = n, then by Lemma 1, rrR(G) = 1 and the desired result follows. Hence we assume that 3 < YrR(G) < n, and let f = (V0, V, V2) be a YrR(G)-function. If V = 0, then the result follows from Proposition 9. Thus suppose that V = 0. Then YrR(G)/2 = |V2| > 2 and clearly
|epn(u, V2) n Vo| < (2n - YrR(G))/YrR(G)
for some u € V2. Now, the result is immediate by Proposition 9.
To show the sharpness, consider the graph G illustrated in Figure 4. It is easy to see that YrR(G) = 4 and the function f on G defined by f (x) = f (y) = 2 and f (z) = 0 otherwise, is the unique YrR(G)-function. Then
rrR(G) < (2n - YrR(G))/YrR(G) = 8.
Now let F be an rrR(G)-set. Then YrR(G + F) < 3 and so A(G + F) > n - 2 (see Propositions 2-(a) and 3). This implies that |F| > 8, and consequently,
rrR(G) = 8 = (2n - YrR(G))/YrR(G).
□
y
6. Conclusion
The main objective of this paper was to start the study of the restrained Roman reinforcement number rrR(G) of a graph G. We first showed that the decision problem associated with the restrained Roman reinforcement problem is NP-hard, and then various properties as well as some sharp bounds of the restrained Roman reinforcement number have been established. In particular we showed that rrR(T) = 1 for every tree T of order at least three and that rrR(G) < A(G) for any connected graph G with YrR(G) > 3. As a future work, one can focus on the problem of characterizing all connected graphs G such that rrR(G) = A(G).
Acknowledgements
This research was financially supported by fund from Hubei Province Key Laboratory of Intelligent Information Processing and Real-time Industrial System (Wuhan University of Science and Technology).
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