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Research article

On the Italian reinforcement number of a digraph

  • Received: 16 January 2021 Accepted: 12 April 2021 Published: 15 April 2021
  • MSC : 05C69, 05C20

  • The Italian reinforcement number of a digraph is the minimum number of arcs that have to be added to the digraph in order to decrease the Italian domination number. In this paper, we present some new sharp upper bounds on the Italian reinforcement number of a digraph. We also determine the exact values of the Italian reinforcement number of the Cartesian products of directed paths and directed cycles: P2Pn, P3Pn, P3Cn, C3Pn and C3Cn.

    Citation: Zhihong Xie, Guoliang Hao, S. M. Sheikholeslami, Shuting Zeng. On the Italian reinforcement number of a digraph[J]. AIMS Mathematics, 2021, 6(6): 6490-6505. doi: 10.3934/math.2021382

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  • The Italian reinforcement number of a digraph is the minimum number of arcs that have to be added to the digraph in order to decrease the Italian domination number. In this paper, we present some new sharp upper bounds on the Italian reinforcement number of a digraph. We also determine the exact values of the Italian reinforcement number of the Cartesian products of directed paths and directed cycles: P2Pn, P3Pn, P3Cn, C3Pn and C3Cn.



    For notation and graph theory terminology, we in general follow [10] and [11]. Specifically, let D be a finite and simple digraph with vertex set V(D) and arc set A(D). For two vertices x,yV(D), we use (x,y) to denote the arc with direction from x to y, that is, x is incident to (x,y) and y is incident from (x,y), and we also say x is an in-neighbor of y and y is an out-neighbor of x. For vV(D), the out-neighborhood and in-neighborhood of v, denoted by N+D(v)=N+(v) and ND(v)=N(v), are the sets of out-neighbors and in-neighbors of v, respectively. Likewise, N+D[v]=N+[v]=N+(v){v} and ND[v]=N[v]=N(v){v}. In general, for a set XV(D), we denote N+D(X)=N+(X)=vXN+(v). We write d+D(v)=d+(v)=|N+(v)| for the out-degree of a vertex v and dD(v)=d(v)=|N(v)| for its in-degree. The maximum out-degree of D is denoted by Δ+(D)=Δ+. Let D1=(V1,A1) and D2=(V2,A2) be two digraphs. The Cartesian product of D1 and D2 is the digraph D1D2 with vertex set V1×V2 and for (x1,y1),(x2,y2)V(D1D2), ((x1,y1),(x2,y2))A(D1D2) if and only if either (x1,x2)A1 and y1=y2, or x1=x2 and (y1,y2)A2.

    A directed star is a digraph of order n2 with vertex set {u1,u2,,un} and arc set {(u1,ui):2in}. We call the center of a directed star to be a vertex of maximum out-degree. A leaf of a digraph D is a vertex of out-degree 0 and in-degree 1. For a vertex vXV(D), the private neighborhood pn(v,X) of v is defined by pn(v,X)=N+(v)N+(X{v}). The complement of a digraph D denoted by ¯D is the digraph with vertex set V(D) defined by (u,v)A(¯D) if and only if (u,v)A(D). An oriented tree is a digraph that can be obtained from a tree by assigning a direction to (that is, orienting) each edge of the tree. We write Cn for the directed cycle of length n, Pn for the directed path of order n. For a real-valued function f:V(D)R, the weight of f is ω(f)=xV(D)f(x), and for XV(D), we define f(X)=xXf(x), and hence ω(f)=f(V(D)).

    Let G be a finite, simple and undirected graph with vertex set V(G) and edge set E(G). A function f:V(G){0,1,2} is an Italian dominating function (IDF) on G if each vertex vV(G) assigned 0 under f satisfies xN(v)f(x)2, where N(v)={xV(G):vxE(G)}. The minimum value of xV(G)f(x) of an IDF f on G is called the Italian domination number of G. The Italian domination was introduced in [2] and has been studied by several authors [1,6,7,12,13,14,17]. For more details on Italian domination, we refer the reader to the recent book chapter [5] and survey paper [3].

    A function f:V(D){0,1,2} is an Italian dominating function (IDF) on a digraph D if each vertex vV(D) assigned 0 under f satisfies xN(v)f(x)2. The minimum weight of an IDF on D is called the Italian domination number of D and is denoted by γI(D). And we say that a function f is a γI(D)-function if it is an IDF with weight γI(D). Hao et al. [8] and Volkmann [18] independently introduced the concept of Italian domination in digraphs. This parameter has been studied in [16,19]. For more details on Roman domination parameters in digraphs, we refer the reader to [4].

    Let ¯G be the complement of an undirected graph G. A subset F of E(¯G) is an Italian reinforcement set (IR-set) of G if γI(G+F)γI(G)1. The Italian reinforcement number of a graph G is the minimum size of an IR-set of G. In [9], Hao et al. introduced this concept.

    Following the ideas in [9], Kim [15] initiated the study of Italian reinforcement number in digraphs. For a digraph D, a subset F of A(¯D) is an Italian reinforcement set (IR-set) of D if γI(D+F)γI(D)1. The Italian reinforcement number of a digraph D, denoted by rI(D), is the minimum size of an IR-set of D. An rI(D)-set is an IR-set F of D with |F|=rI(D). It is clear that if 1γI(D)2, then addition of arcs does not reduce the Italian domination number. Thus if 1γI(D)2, then the Italian reinforcement number is defined to be 0. Consequently we always assume that when we discuss rI(D), all digraphs involved satisfy γI(D)3.

    For an IDF f on D, let Vi={vV(D):f(v)=i} for i{0,1,2} and hence f can be represented by the ordered triple (V0,V1,V2) (or (Vf0,Vf1,Vf2) to refer f) of V(D) induced by f.

    The aim of this paper is to continue the study of Italian reinforcement number in digraphs. We establish some new sharp upper bounds on the Italian reinforcement number. Also, we determine the exact values of rI(P2Pn), rI(P3Pn), rI(P3Cn), rI(C3Pn) and rI(C3Cn).

    In this section, some fundamental results that are used in this paper are herein recalled. The reader is referred to [15] for more information and details.

    Proposition A. Let D be a digraph with γI(D)3, F be an rI(D)-set and let f be a γI(D+F)-function. Then

    (a) For each arc (u,v)F, f(u){1,2} and f(v)=0.

    (b) γI(D+F)=γI(D)1.

    Proposition B. Let D be a digraph with γI(D)3. Then rI(D)=1 if and only if there exist a γI(D)-function f=(V0,V1,V2) and a vertex vV1 such that one of the following conditions holds:

    (a) f(ND(v))=0, f(ND(x){v})2 for each xN+D(v)V0 and V2.

    (b) f(ND(v))=1 and f(ND(x){v})2 for each xN+D(v)V0.

    Proposition C. For any digraph D of order n with γI(D)3,

    rI(D)nΔ+γI(D)+2.

    Proposition D. If D is a digraph of order n3 with Δ+(D)1 and γI(D)=n, then rI(D)=1.

    Hao et al. [8] showed that for any integer n3, γI(Pn)=γI(Cn)=n, which, together with Proposition D, would yield the following result directly.

    Corollary 2.1. For any integer n3, rI(Pn)=rI(Cn)=1.

    Our aim in this section is to derive some sharp upper bounds for Italian reinforcement number in digraphs.

    Proposition 3.1. Let D be a digraph and let H be a subdigraph of D with γI(H)=γI(D)3. Then rI(D)rI(H).

    Proof. Let F be an rI(H)-set. Since H+F is a subdigraph of D+(FA(D)), we conclude from Proposition A(b) that

    γI(D+(FA(D)))γI(H+F)<γI(H)=γI(D).

    This implies that FA(D) is an IR-set of D and so rI(D)|FA(D)||F|=rI(H), as desired.

    Note that Pn is a subdigraph of Cn and γI(Pn)=γI(Cn)=n for n3. Moreover, it follows from Corollary 2.1 that rI(Pn)=rI(Cn)=1. This demonstrates the sharpness of Proposition 3.1.

    Theorem 3.2. For any digraph D with γI(D)3, let f=(V0,V1,V2) be a γI(D)-function. Then

    (a) If vV1, then rI(D)|(N+(v)pn(v,V1))V0|+2.

    (b) If vV2, then rI(D)|pn(v,V2)V0|.

    Proof. (a) Let vV1. First, assume that V2. Let wV2 and let F=({(w,x):x(N+(v)pn(v,V1))V0}{(w,v)})A(D). Note that every vertex in pn(v,V1)V0 has at least one in-neighbor in V2. Thus the function h1 defined by h1(v)=0 and h1(x)=f(x) for each xV(D){v} is an IDF on D+F with ω(h1)=ω(f)1 and hence F is an IR-set of D, implying that

    rI(D)|F||(N+(v)pn(v,V1))V0|+1.

    Second, assume that V2=. This forces pn(v,V1)V0=. Note that there must exist two distinct vertices v1,v2V1{v} since γI(D)3. Let U1=N+(v)N+(v1)V0, U2=(N+(v)N+(v1))V0 and let F=({(v1,v),(v2,v)}{(v2,x):xU1}{(v1,x):xU2})A(D). One can check that the function h1 defined earlier is an IDF on D+F with ω(h1)=ω(f)1 and hence F is an IR-set of D, implying that

    rI(D)|F||{(v2,x):xU1}{(v1,x):xU2}|+2=|(N+(v)V0)|+2=|(N+(v)pn(v,V1))V0|+2,

    where the last '=' holds since pn(v,V1)V0=.

    As a result, (a) is true.

    (b) Let vV2. Then there must exist a vertex wV1V2 since γI(D)3. Let F={(w,x):xpn(v,V2)V0}A(D). It is obvious that the function h2 defined by h2(v)=1 and h2(x)=f(x) for each xV(D){v} is an IDF on D+F with ω(h2)=ω(f)1 and hence F is an IR-set of D, implying that

    rI(D)|F||pn(v,V2)V0|,

    as desired.

    This completes the proof.

    Remark 1. The upper bounds in Theorem 3.2 are sharp. To see this, consider the following two examples.

    (a) Let D be the digraph with vertex set XY, where X={xi:1i4} and Y={yi:1i4}, and arc set {(xi,xj),(yi,yj):1i2,3j4}. One can check that the function f=({x3,x4,y3,y4},{x1,x2,y1,y2},) is a γI(D)-function and hence γI(D)=4.

    We next prove that rI(D)4. Let F be an rI(D)-set and let g be a γI(D+F)-function. We deduce from Proposition A(b) that ω(g)=γI(D+F)=γI(D)1=3. If g(X)=0 (resp., g(Y)=0), then every vertex of X (resp., Y) is incident from one arc in F and so |F||X|=4 (resp., |F||Y|=4). So in the following we may assume that g(X)1 and g(Y)1. Recall that g(X)+g(Y)=ω(g)=3. By symmetry, assume that g(X)=2 and g(Y)=1.

    First, assume that exactly one vertex of X is assigned 2 under g and the others are assigned 0 under g. Without loss of generality, assume that g(x2)=0. Since dD(x2)=0, x2 is incident from one arc in F. Moreover, since g(Y)=1, each of exactly three vertices of Y assigned 0 under g is incident from at least one arc in F. As a result, we have |F|4.

    Second, assume that exactly two vertices of X are assigned 1 under g. This forces Vg2=. Recall that g(Y)=1. If g(y1)=1 (the case g(y2)=1 is similar), then g(y2)=g(y3)=g(y4)=0 and hence y2 is incident from two arcs in F and each of y3 and y4 is incident from one arc in F, implying that |F|4. If g(y3)=1 (the case g(y4)=1 is similar), then g(y1)=g(y2)=0 and hence each of y1 and y2 is incident from two arcs in F, implying that |F|4.

    Therefore, the above arguments mean that rI(D)=|F|4. Now let F={(x1,y2),(x2,y2), (x1,y3),(x1,y4)}. It is not difficult to verify that the function h=({x3,x4,y2,y3,y4},{x1,x2,y1}, ) is an IDF on D+F with ω(h)=3<γI(D) and hence F is an IR-set of D. Thus rI(D)|F|=4. As a result, we obtain rI(D)=4.

    Note that f=({x3,x4,y3,y4},{x1,x2,y1,y2},) is a γI(D)-function. We have that for 1i2,

    rI(D)=|(N+(xi)pn(xi,Vf1))Vf0|+2=|(N+(yi)pn(yi,Vf1))Vf0|+2=4.

    (b) Let ts3 and p1 be arbitrary integers and let D be the digraph with vertex set {ui:1is}{vi:1it}{wi:1ip} and arc set {(u1,ui):2is}{(v1,vi):2it}{(u1,wi),(v1,wi):1ip}.

    Let g1 be a γI(D)-function. If there exists some vertex in {u2,u3,,us} assigned 0 under g1, then g1(u1)=2 and if each vertex in {u2,u3,,us} is assigned 1 or 2 under g1, then si=2g1(ui)s1. In either case, we have si=1g1(ui)2. Similarly, we have ti=1g1(vi)2. Therefore, γI(D)4. On the other hand, the function f=(V(D){u1,v1},,{u1,v1}) is an IDF on D and so γI(D)ω(f)=4. Consequently, we have γI(D)=4. This also implies that the function f is a γI(D)-function.

    We next prove that rI(D)=s1. Let F={(v1,ui):2is}. Then the function g2 defined by g2(u1)=1, g2(v1)=2 and g2(x)=0 otherwise, is an IDF on D+F with ω(g2)=3<γI(D), and hence F is an IR-set of D. Thus rI(D)|F|=s1. Therefore it is enough to prove that rI(D)s1. Let F be an rI(D)-set and let h be a γI(D+F)-function. We conclude from Proposition A(b) that ω(h)=γI(D+F)=γI(D)1=3. If ti=1h(vi){0,1}, then at least t1 vertices in {v1,v2,,vt} is incident from an arc in F and hence |F|t1s1. If ti=1h(vi){2,3}, then this forces si=1h(ui){0,1} and hence at least s1 vertices in {u1,u2,,us} is incident from an arc in F, implying that |F|s1. As a result, we obtain rI(D)=s1.

    Note that the function f=(V(D){u1,v1},,{u1,v1}) is a γI(D)-function, |pn(u1,Vf2)Vf0|=s1 and |pn(v1,Vf2)Vf0|=t1. Thus if s=t, then

    rI(D)=s1=|pn(u1,Vf2)Vf0|=|pn(v1,Vf2)Vf0|.

    As an immediate consequence of Theorem 3.2, we have the following result on the Italian reinforcement number of a digraph in terms of its maximum out-degree.

    Corollary 3.3. For any digraph D with γI(D)3, rI(D)Δ++2.

    The example of (a) in Remark 1 demonstrates that Corollary 3.3 is sharp. Using Corollary 3.3 and Proposition C, we obtain the following result.

    Corollary 3.4. Let D be a digraph of order n with γI(D)3. Then rI(D)n/2.

    Proof. If Δ+n/22, then it follows from Corollary 3.3 that rI(D)Δ++2n/2. If Δ+n/21, then by Proposition C, we get

    rI(D)nΔ+γI(D)+2n(n/21)3+2=n/2,

    which completes our proof.

    The upper bound of Corollary 3.4 is sharp. For example, let D be the digraph with vertex set {ui,vi:0i3} and arc set {(ui,vi),(ui,vi+1):0i3}, where the indices are taken modulo 4. It is not difficult to check that the function f=({vi:0i3},{ui:0i3},) is the unique γI(D)-function, the set F={(u0,u3),(u1,u3),(u0,v3),(u1,v0)} is an rI(D)-set and g=({u3}{vi:0i3},{ui:0i2},) is a γI(D+F)-function. Therefore rI(D)=|F|=4.

    The upper bound of Corollary 3.3 can be improved if we restrict our attention to any digraph D with at least one leaf and γI(D)3.

    Theorem 3.5. For any digraph D with at least one leaf and γI(D)3, rI(D)max{2,Δ+}.

    Proof. Let u be a leaf of D, v be the unique in-neighbor of u and let f be a γI(D)-function. Clearly f(u)1. If f(u)=0, then this forces f(v)=2 and it follows from Theorem 3.2(b) that

    rI(D)|pn(v,Vf2)Vf0|Δ+max{2,Δ+}.

    So in the following we may assume that f(u)=1. It is not different to verify that f(v)1. Define the function g by g(u)=0 and g(x)=f(x) for each xV(D){u}.

    Assume first that f(v)=1. Since γI(D)3, there exists some vertex wV(D){u,v} such that f(w)1 and hence the function g defined earlier is an IDF on D+{(w,u)} with ω(g)=ω(f)1<γI(D), and so {(w,u)} is an IR-set of D, implying that rI(D)=1<max{2,Δ+}.

    Assume next that f(v)=0. Then f(N(v))2. If there exists some vertex wN(v) such that f(w)=2, then the function g defined earlier is an IDF on D+{(w,u)} with ω(g)=ω(f)1<γI(D), and so {(w,u)} is an IR-set of D, implying that rI(D)=1<max{2,Δ+}. If there exist two different vertices w1,w2N(v) such that f(w1)=f(w2)=1, then the function g defined earlier is an IDF on D+{(w1,u),(w2,u)} with ω(g)=ω(f)1<γI(D), and so {(w1,u),(w2,u)} is an IR-set of D, implying that rI(D)2max{2,Δ+}.

    This completes the proof.

    It is worth pointing out that the upper bound of Theorem 3.5 is sharp. We shall construct infinitely many digraphs T with rI(T)=max{2,Δ+(T)}. Let Δ2 be an arbitrary integer and let T denote an oriented tree with 1Δ+(T)Δ. For any vertex uV(T), let Su denote a directed star of order Δ with center C(Su). Define T(Δ,T) to be the digraph obtained from T(uV(T)Su) by adding a new arc (C(Su),u) for each uV(T).

    Proposition 3.6. For an arbitrary integer Δ2, let T be an oriented tree with 1Δ+(T)Δ and let T=T(Δ,T). Then rI(T)=max{2,Δ+(T)}.

    Proof. We now claim that γI(T)=2|V(T)|. It is easy to see that the function h1 defined by h1(C(Su))=2 for each uV(T) and h1(x)=0 otherwise, is an IDF on T and hence γI(T)ω(h1)=2|V(T)|. Thus it is enough for us to prove that γI(T)2|V(T)|. Let h2 be a γI(T)-function. Noting that dT(C(Su))=0 for each uV(T), we have h2(C(Su))1 and hence if there exists a leaf x of Su such that h2(x)1, then h2(V(Su))h2(C(Su))+h2(x)2 and if there exists a leaf x of Su such that h2(x)=0, then h2(V(Su))h2(C(Su))=2. Consequently, we get

    γI(T)=ω(h2)uV(T)h2(V(Su))2|V(T)|.

    We next prove that rI(T)=max{2,Δ+(T)}. It follows from Theorem 3.5 that rI(T)max{2,Δ+(T)}. Note that max{2,Δ+(T)}=Δ. Thus it is enough for us to prove that rI(T)Δ. Let F be an rI(T)-set and f be a γI(T+F)-function.

    Claim 3.7. For any vertex uV(T) with f(V(Su))1, there must exist a vertex in V(Su) incident from an arc in F.

    Proof of Claim 3.7. Let u be any vertex of V(T) with f(V(Su))1. Note that f(C(Su))f(V(Su))1. If f(C(Su))=0, then C(Su) is incident from an arc in F since dT(C(Su))=0, and if f(C(Su))=1, then f(V(Su){C(Su)})=0 and hence each vertex of V(Su){C(Su)} is incident from an arc in F. Thus Claim 3.7 is true.

    Claim 3.8. For any vertex uV(T) with f(V(Su){u})1,

    (a) There exist Δ1 vertices in V(Su){u} incident from an arc in F.

    (b) If the number of vertices in V(Su){u} incident from an arc in F is Δ1, then f(V(Su))=1 and u is not adjacent from an arc in F.

    Proof of Claim 3.8. Let u be any vertex of V(T) with f(V(Su){u})1. It is obvious that f(V(Su))1. If f(V(Su))=0, then each vertex of Su is adjacent from an arc in F, and if f(V(Su))=1, then each of exactly Δ1 vertices of Su assigned 0 under f is adjacent from an arc in F. Thus Claim 3.8 is also true.

    We deduce from Proposition A(b) that γI(T+F)=γI(T)1=2|V(T)|1 and hence there exists a vertex uV(T) satisfying f(V(Su){u})1. Moreover, if there exists a vertex vV(T){u} satisfying f(V(Sv){v})1, then we conclude from Claim 3.8(a) that |F|2(Δ1)Δ. So in the following we may assume that u is the unique vertex satisfying f(V(Su){u})1. This implies that f(V(Sw){w})2 for each wV(T){u}.

    If f(V(Su))=0 or u is incident from an arc in F, then we deduce from Claim 3.8 that rI(T)=|F|Δ. Suppose, next, that f(V(Su))=1 and u is not incident from an arc in F. This forces f(V(Su){u})=1 and f(u)=0. Furthermore, since ω(f)=γI(T+F)=2|V(T)|1 and f(V(Sw){w})2 for each wV(T){u}, one can verify that f(V(Sw){w})=2. Since f(V(Su))=1, f(u)=0 and u is not incident from an arc in F, there must exist an in-neighbor, say v, of u in T with f(v)1. Moreover, it is clear that f(V(Sv){v})=2 since vV(T){u} and hence f(V(Sv))=f(V(Sv){v})f(v)1. Thus it follows from Claim 3.7 that there must exist a vertex of V(Sv) incident from an arc in F. Furthermore, since f(V(Su){u})=1, Claim 3.8(a) yields that there exist Δ1 vertices in V(Su){u} incident from an arc in F. Therefore, we get rI(T)=|F|Δ.

    This completes the proof.

    Let r be a positive integer and let n=2r+1. We define the circulant tournament CT(n) of order n with vertex set {u0,u1,,un1} as follows. For each i{0,1,,n1}, the arcs are going from ui to the vertices ui+1,ui+2,,ui+r, where the indices are taken modulo n. For Italian domination number, Hao et al. [8] showed the following result.

    Proposition E. ([8]) Let r be a positive integer and let n=2r+1. Then

    γI(CT(n))={3,if  r=1,4,if  r2.

    The following result, which can be deduced from Propositions C and E, indicates that the upper bound in Proposition C is sharp.

    Proposition 3.9. Let r be a positive integer and let n=2r+1. Then

    rI(CT(n))={1, if r=1,r1, if r2.

    Proof. If r=1, that is, if CT(n) is a directed cycle of length 3, then Corollary 2.1 yields rI(CT(n))=1. So in the following we may assume that r2. By Propositions C and E, we have

    rI(CT(n))nΔ+(CT(n))γI(CT(n))+2=r1.

    Hence it suffices to prove that rI(CT(n))r1. Let F be an rI(CT(n))-set and let f=(V0,V1,V2) be a γI(CT(n)+F)-function. Then by Propositions A(b) and E, we have

    |V1|+2|V2|=ω(f)=γI(CT(n)+F)=γI(CT(n))1=3.

    This implies that |V1|=|V2|=1, or |V1|=3 and |V2|=0.

    Suppose now that |V1|=|V2|=1. Without loss of generality, assume that u0V2 and ui0V1 for some i0{1,2,,2r}. Then for each j{r+1,r+2,,2r}{i0}, ujN+CT(n)(u0) and hence by the definition of γI(CT(n)+F)-function, we have uj is adjacent from one arc in F. Thus rI(CT(n))=|F|r1. Suppose next that |V1|=3 and |V2|=0. Without loss of generality, assume that u0,uk,ulV1 for 0<k<l2r.

    First, assume that lk<r, where k,l{1,2,,r}. Then NCT(n)(uj){u0,uk,ul}={u0} for each j{1,2,,k1}, NCT(n)(uj){u0,uk,ul}={ul} for each j{k+r+1,k+r+2,,l+r} and NCT(n)(uj){u0,uk,ul}= for each j{l+r+1,l+r+2,,2r}. This implies that rI(CT(n))=|F|r1.

    Second, assume that lk<r, where k{2,3,,r} and l{r+1,r+2,,2r1}. Clearly NCT(n)(uj){u0,uk,ul}={u0} for each j{lr,lr+1,,k1}, NCT(n)(uj){u0,uk,ul}={uk} for each j{r+1,r+2,,l1} and NCT(n)(uj){u0,uk,ul}={ul} for each j{k+r+1,k+r+2,,2r}. This implies that rI(CT(n))=|F|r1.

    Third, assume that lk<r, where k,l{r+1,r+2,,2r}. Clearly NCT(n)(uj){u0,uk,ul}={u0} for each j{lr,lr+1,,r}, NCT(n)(uj){u0,uk,ul}= for each j{r+1,r+2,,k1} and NCT(n)(uj){u0,uk,ul}={uk} for each j{k+1,k+2,,l1}. This implies that rI(CT(n))=|F|r1.

    Finally assume that lkr. Then 1kr and r+1l2r. It is clear that NCT(n)(uj){u0,uk,ul}={uk} for each j{r+1,r+2,,k+r}{l}, NCT(n)(uj){u0,uk,ul}= for each j{k+r+1,k+r+2,,l1} and NCT(n)(uj){u0,uk,ul}={ul} for each j{l+1,l+2,,2r}. This implies that rI(CT(n))=|F|r1.

    This completes the proof.

    In this section, we first determine the exact values of γI(P2Pn), γI(P3Pn), γI(P3Cn) and γI(C3Pn), based on which we further determine the exact values of Italian reinforcement number of some classes of cartesian products.

    Let m{2,3} and nm be an integer. We emphasize that V(PmPn)={(i,j):1im,1jn} and A(PmPn)={((i,j),(i+1,j)):1im1,1jn}{((i,j),(i,j+1)):1im,1jn1}. For notational convenience, if f is an IDF on PmPn, then we denote aj=mi=1f((i,j)) for each 1jn.

    Now we consider the exact value of Italian domination number of P2Pn. We begin with the following lemma.

    Lemma 4.1. Let n2 be an integer and let f be a γI(P2Pn)-function. Then a12 and aj1 for each j{2,3,,n}.

    Proof. Since d((1,1))=0, we have f((1,1))1. If f((1,1))=2, then a1f((1,1))=2. If f((1,1))=1, then clearly f((2,1))1 since (1,1) is the unique in-neighbor of (2,1) and so a1=f((1,1))+f((2,1))2. We next show that aj1 for each j{2,3,,n}. Suppose, to the contrary, that there exists some j{2,3,,n} such that aj=f((1,j))+f((2,j))=0. Then by the definition of γI(P2Pn)-function, we have f((1,j1))=f((2,j1))=2. One can check that the function g defined by g((2,j1))=0, g((2,j))=1 and g(x)=f(x) otherwise, is an IDF on P2Pn with ω(g)=ω(f)1<γI(P2Pn), a contradiction. As a result, we obtain aj1 for each j{2,3,,n}.

    Theorem 4.2. For any integer n2, γI(P2Pn)=4n3.

    Proof. Let f be a γI(P2Pn)-function, aj=f((1,j))+f((2,j)) for each j{1,2,,n} and let bj=aj2+aj1+aj for each j{3,4,,n}.

    Claim 4.3. For each j{3,4,,n}, bj4.

    Proof of Claim 4.3. Suppose, to the contrary, that there exists some j0{3,4,, n} such that bj03. Note that for each j{1,2,,n}, aj1 by Lemma 4.1. Therefore, we have bj0=aj02+aj01+aj03. This forces aj02=aj01=aj0=1.

    Assume first that f((2,j0))=1. This implies that f((1,j0))=aj0f((2,j0))=0. Since (1,j01) is the unique in-neighbor of (1,j0), we have f((1,j01))=2, a contradiction to the fact that f((1,j01))aj01=1.

    Assume second that f((2,j0))=0. This implies that f((1,j0))=aj0f((2,j0))=1. Since f is a γI(P2Pn)-function, f((2,j01))1. Moreover, since f((2,j01))aj01=1, we have f((2,j01))=1 and so f((1,j01))=0. Noting that (1,j02) is the unique in-neighbor of (1,j01), we obtain f((1,j02))=2, a contradiction to the fact that f((1,j02))aj02=1.

    So, this claim is true.

    Therefore, by Lemma 4.1 and Claim 4.3, we have that if n0(mod3), then

    γI(P2Pn)=n3k=1b3k4n3=4n3,

    if n1(mod3), then

    γI(P2Pn)=a1+n13k=1b3k+12+4(n1)3=4n3,

    and if n2(mod3), then

    γI(P2Pn)=a1+a2+n23k=1b3k+23+4(n2)3=4n3.

    On the other hand, the function g defined by

    g((i,j))={1,if i=1 and j0(mod3),or i=2 and j2(mod3),2,if i=1 and j1(mod3),0,otherwise,

    is an IDF on P2Pn and hence

    γI(P2Pn)ω(g)=n3+n13+2×n3=4n3,

    which completes our proof.

    We shall give the exact value of the Italian reinforcement number of P2Pn. To our aim, the following lemmas are essential.

    Lemma 4.4. Let n2 be any positive integer and let f be an IDF on P2Pn. If there exists some k{1,2,,n1} such that f((1,k))=1 and f((2,k))=0, then the restriction f of f on {(i,j):1i2,k+1jn} is an IDF on P2Pnk.

    Proof. Let f be the restriction of f on {(i,j):1i2,k+1jn}. Since f((1,k))=1 and (1,k) is the unique in-neighbor of (1,k+1), we have f((1,k+1))1. If f((1,k+1))=2, then clearly f is an IDF on P2Pnk. And if f((1,k+1))=1, then we conclude from f((2,k))=0 and the fact (2,k+1) has exactly two in-neighbors (2,k) and (1,k+1), that f((2,k+1))1 and hence f is an IDF on P2Pnk.

    Lemma 4.5. Let n0(mod3) be any positive integer and let f be an IDF on P2Pn. If an2, then ω(f)4n3+1.

    Proof. It is easy to check that the restriction f of f on {(i,j):1i2,1jn1} is an IDF on P2Pn1 and hence by Theorem 4.2, ω(f)4(n1)3. Therefore, we obtain

    ω(f)=ω(f)+an4(n1)3+2=4n3+1,

    as desired.

    Lemma 4.6. Let n0(mod3) be any positive integer and let f be an IDF on P2Pn. If there exists some k{1,2,,n1} such that f((1,k))=f((2,k))=1 and ak+12, then ω(f)4n3+1.

    Proof. Observe that the restriction f1 of f on {(i,j):1i2,1jk1} is an IDF on P2Pk1 and hence by Theorem 4.2, ω(f1)4(k1)3. Let f2 be the restriction of f on {(i,j):1i2,k+1jn}. Since (1,k) is the unique in-neighbor of (1,k+1) and f((1,k))=1, we have f((1,k+1))1. If f((1,k+1))=1, then f((2,k+1))1 since ak+12 and hence f2 is an IDF on P2Pnk and if f((1,k+1))=2, then clearly f2 is also an IDF on P2Pnk. In either case, we conclude from Theorem 4.2 that ω(f2)4(nk)3. Therefore, we obtain

    ω(f)=ω(f1)+ω(f2)+ak4(k1)3+4(nk)3+24n3+1,

    as desired.

    Lemma 4.7. Let n0(mod3) be any positive integer, f be an IDF on P2Pn and let k{2,3,,n} such that ak1+ak3. If n=k, or n>k and the restriction f of f on {(i,j):1i2,k+1jn} is an IDF on P2Pnk, then ω(f)4n3+1.

    Proof. Observe that the restriction f1 of f on {(i,j):1i2,1jk2} is an IDF on P2Pk2 and hence by Theorem 4.2, ω(f1)4(k2)3. Consequently, if n=k, then

    ω(f)=ω(f1)+an1+an4(k2)3+3=4n3+1.

    So in the following we may assume that n>k. Since the restriction f of f on {(i,j):1i2,k+1jn} is an IDF on P2Pnk by our assumption, Theorem 4.2 yields ω(f)4(nk)3. Therefore, we obtain

    ω(f)=ω(f1)+ω(f)+ak1+ak4(k2)3+4(nk)3+3=4n3+1,

    as desired.

    In the next, we shall determine the Italian reinforcement number of P2Pn.

    Theorem 4.8. For any integer n2,

    rI(P2Pn)={2, if  n0(mod3),1, otherwise.

    Proof. If n1(mod3), then the function h1 defined by

    h1((i,j))={1, if  i=1  and  j0(mod3), or  i=1  and  j=n, or  i=2  and  j2(mod3),2, if  i=1, j1(mod3)  and  j<n,0, otherwise,

    is an IDF on P2Pn+{((1,1),(2,n))} and so ω(h1)=4n31<γI(P2Pn) by Theorem 4.2, implying that the set {((1,1),(2,n))} is an IR-set of P2Pn and hence rI(P2Pn)=1. If n2(mod3), then the function h2 defined by

    h2((i,j))={1, if  i=1  and  j0(mod3), or  i=2, j2(mod3)  and  j<n,2, if  i=1  and  j1(mod3),0, otherwise,

    is an IDF on P2Pn+{((1,1),(2,n))} and so ω(h2)=4n31<γI(P2Pn) by Theorem 4.2, implying that the set {((1,1),(2,n))} is an IR-set of P2Pn and hence rI(P2Pn)=1. If n0(mod3), then the function h3 defined by

    h3((i,j))={1, if  i=1, j0(mod3)  and  j<n, or  i=2  and  j2(mod3),2, if  i=1  and  j1(mod3),0, otherwise,

    is an IDF on P2Pn+{((1,1),(1,n)),((1,1),(2,n))} and so ω(h3)=4n31<γI(P2Pn) by Theorem 4.2, implying that the set {((1,1),(1,n)),((1,1),(2,n))} is an IR-set of P2Pn and hence rI(P2Pn)2.

    It remains to prove that if n0(mod3), then rI(P2Pn)2. For the sake of contradiction, we may assume that rI(P2Pn)=1. Using Proposition B, there exists a γI(P2Pn)-function f=(V0,V1,V2) and a vertex vV1 such that one of the conditions (a) and (b) in Proposition B is true.

    Suppose that (a) is true. Then V2 and there exists some integer k{1,2,,n} such that one of the following holds:

    (I) f((1,k))=1, f((1,k1))=f((2,k))=0, f((2,k1))=2 and if n>k, then f((1,k+1))1, where v=(1,k).

    (II) f((1,k))=1, f((1,k1))=0, f((2,k))1 and if n>k, then f((1,k+1))1, where v=(1,k).

    (III) f((2,k))=1, f((2,k1))=f((1,k))=0 and if n>k, then f((1,k+1))=2 and f((2,k+1))=0, where v=(2,k).

    (IV) f((2,k))=1, f((2,k1))=f((1,k))=0 and if n>k, then f((2,k+1))1, where v=(2,k).

    First, assume that (I) holds. Notice that ak1+ak=3. Thus if n=k, then by Lemma 4.7, ω(f)4n3+1, a contradiction to Theorem 4.2. And if n>k, then we deduce from Lemma 4.4 that the restriction f of f on {(i,j):1i2,k+1jn} is an IDF on P2Pnk and hence Lemma 4.7 yields ω(f)4n3+1, a contradiction to Theorem 4.2.

    Second, assume that (II) holds. Since (1,k2) is the unique in-neighbor of (1,k1) and f((1,k1))=0, we have ak2f((1,k2))=2. And by Lemma 4.1, ak11 and so ak2+ak13. Moreover, since f((1,k))=1 and f((2,k))1, the restriction f of f on {(i,j):1i2,kjn} is an IDF on P2Pnk+1 and hence Lemma 4.7 yields ω(f)4n3+1, a contradiction to Theorem 4.2.

    Finally, assume that (III) or (IV) holds. Since (1,k1) is the unique in-neighbor of (1,k) and f((1,k))=0, we have f((1,k1))=2 and hence ak1+ak=3. Therefore, if n=k, then by Lemma 4.7, ω(f)4n3+1, a contradiction to Theorem 4.2. Suppose next that n>k. Let f be the restriction of f on {(i,j):1i2,k+1jn}. If (III) is true, then clearly f is an IDF on P2Pnk and hence by Lemma 4.7, ω(f)4n3+1, a contradiction to Theorem 4.2. Assume now that (IV) is true. Since (1,k) is the unique in-neighbor of (1,k+1) and f((1,k))=0, we have f((1,k+1))1. Moreover, since f((2,k+1))1, f is an IDF on P2Pnk. Thus again by Lemma 4.7, ω(f)4n3+1, a contradiction to Theorem 4.2.

    Suppose, next, that (b) is true. Then there exists some integer k{1,2,,n} such that one of the following holds:

    (V) f((1,k1))=f((1,k))=1, f((2,k1))=2, f((2,k))=0 and if n>k, then f((1,k+1))1, where v=(1,k).

    (VI) f((1,k1))=f((1,k))=1, f((2,k))1 and if n>k, then f((1,k+1))1, where v=(1,k).

    (VII) f((1,k))=0, f((2,k1))=f((2,k))=1 and if n>k, then f((1,k+1))=2 and f((2,k+1))=0, where v=(2,k).

    (VIII) f((1,k))=0, f((2,k1))=f((2,k))=1 and if n>k, then f((2,k+1))1, where v=(2,k).

    (IX) f((1,k))=f((2,k))=1, f((2,k1))=0 and if n>k, then f((1,k+1))=2 and f((2,k+1))=0, where v=(2,k).

    (X) f((1,k))=f((2,k))=1, f((2,k1))=0 and if n>k, then f((2,k+1))1, where v=(2,k).

    First, assume that (V) holds. One can check that the function g1 defined by g1((2,k1))=1 and g1(x)=f(x) otherwise, is an IDF on P2Pn and hence ω(g1)=ω(f)1<γI(P2Pn), a contradiction.

    Second, assume that (VI) holds. If n=k, then an2 and hence by Lemma 4.5, ω(f)4n3+1, a contradiction to Theorem 4.2. Hence we may assume that n>k. Since f((1,k+1))1, we have that if f((2,k))=2, then the function g2 defined by g2((2,k))=1 and g2(x)=f(x) otherwise, is an IDF on P2Pn and hence ω(g2)=ω(f)1<γI(P2Pn), a contradiction. As a result, we have f((2,k))1. Moreover, since f((2,k))1, f((2,k))=1. If ak+12, then by Lemma 4.6, ω(f)4n3+1, a contradiction to Theorem 4.2. Thus ak+11. Moreover, since ak+1f((1,k+1))1, we have ak+1=1. This implies that f((1,k+1))=1 and f((2,k+1))=0. If n=k+1, then an1+an=3 and so by Lemma 4.7, ω(f)4n3+1, a contradiction to Theorem 4.2. Suppose, next, that n>k+1. Then Lemma 4.4 yields that the restriction f of f on {(i,j):1i2,k+2jn} is an IDF on P2Pnk1. Recall that ak+ak+1=3. Then by Lemma 4.7, ω(f)4n3+1, a contradiction to Theorem 4.2.

    Third, assume that (VII) or (VIII) holds. Since (1,k1) is the unique in-neighbor of (1,k) and f((1,k))=0, this forces f((1,k1))=2. One can check that the function g3 defined by g3((2,k1))=0 and g3(x)=f(x) otherwise, is an IDF on P2Pn and hence ω(g3)=ω(f)1<γI(P2Pn), a contradiction.

    Finally, assume that (IX) or (X) holds. If n=k, then Lemma 4.5 yields ω(f)4n3+1 since an=2, a contradiction to Theorem 4.2. Hence we may assume that n>k. Note that f((1,k))=f((2,k))=1. If (IX) holds, then ak+1=2 and so we conclude from Lemma 4.6 that ω(f)4n3+1, a contradiction to Theorem 4.2. Assume, next, that (X) holds. Since (1,k) is the unique in-neighbor of (1,k+1) and f((1,k))=1, we have f((1,k+1))1 and hence ak+1=f((1,k+1))+f((2,k+1))2. Moreover, since f((1,k))=f((2,k))=1, we have ω(f)4n3+1 by Lemma 4.6, a contradiction to Theorem 4.2.

    The proof is completed.

    Next we determine the exact value of γI(P3Pn). We begin with the following lemmas. Analogous to Lemma 4.1, we first obtain the following result.

    Lemma 4.9. Let n3 be an integer and let f be a γI(P3Pn)-function. Then a13 and aj1 for each j{2,3,,n}.

    Lemma 4.10. Let n3 be an integer and let f be a γI(P3Pn)-function. If aj=1 for some j{2,3,,n}, then aj13.

    Proof. Suppose that there exists some j{2,3,,n} such that aj=1. Then f((1,j))+f((2,j))+f((3,j))=1. If f((1,j))=1, then f((2,j))=f((3,j))=0. This forces f((2,j1))1 and f((3,j1))=2, implying that aj13. If f((2,j))=1, then f((1,j))=f((3,j))=0. This forces f((1,j1))=2 and f((3,j1))1, implying that aj13. If f((3,j))=1, then f((1,j))=f((2,j))=0. This forces f((1,j1))=f((2,j1))=2, implying that aj14, which completes our proof.

    Theorem 4.11. For any integer n3, γI(P3Pn)=2n.

    Proof. Let f be a γI(P3Pn)-function. By Lemmas 4.9 and 4.10, we obtain γI(P3Pn)=ω(f)=nj=1aj2n. Hence it suffices for us to prove that γI(P3Pn)2n. If n1(mod3), then the function g1 defined by

    g1((i,j))={2,if i=1 and j=3k2 for 1k(n1)/3,1,if i{1,2} and j=n,or i{1,3} and j=3k for 1k(n1)/3,or i=2 and j=3k1 for 1k(n1)/3,or i=3 and j=3k2 for 1k(n1)/3,0,otherwise,

    is an IDF on P3Pn and hence

    γI(P3Pn)ω(g1)=2×(n1)/3+2+4×(n1)/3=2n.

    If n1(mod3), then the function g2 defined by

    g2((i,j))={2,if i=1 and j=3k2 for 1kn/3,1,or i{1,3} and j=3k for 1kn/3,or i=2 and j=3k1 for 1kn/3,or i=3 and j=3k2 for 1kn/3,0,otherwise,

    is an IDF on P2Pn and hence

    γI(P3Pn)ω(g2)=4×n/3+2×n/3=2n,

    which completes our proof.

    Kim [16] determined the exact value of the cartesian product of two directed cycles C3 and Cn as follows.

    Proposition F. For any integer n3, γI(C3Cn)=2n.

    As a consequence of Theorem 4.11 and Proposition F, we have the following result.

    Corollary 4.12. For any integer n3, γI(P3Cn)=γI(C3Pn)=2n.

    Proof. Since P3Pn is a subdigraph of P3Cn and P3Cn is a subdigraph of C3Cn, we deduce from Theorem 4.11 and Proposition F that

    2n=γI(P3Pn)γI(P3Cn)γI(C3Cn)=2n.

    This implies that γI(P3Cn)=2n. Similarly, we have γI(C3Pn)=2n.

    Theorem 4.13. For any integer n3,

    rI(P3Pn)=rI(P3Cn)=rI(C3Pn)=rI(C3Cn)=1.

    Proof. By Theorem 4.11, Proposition F and Corollary 4.12, we have γI(P3Pn)=γI(P3Cn)=γI(C3Pn)=γI(C3Cn)=2n. Moreover, we note that P3Pn is a subdigraph of P3Cn, C3Pn and C3Cn. Thus by Proposition 3.1, it is enough to prove rI(P3Pn)=1. One can check that if n1(mod3) (resp., n1(mod3)), then the function g1 (resp., g2) defined as in Theorem 4.11 is a γI(P3Pn)-function, and g1 (resp., g2) and the vertex (3,3) satisfy the condition (a) of Proposition B. Thus by Proposition B, rI(P3Pn)=1, as desired.

    The main objective of this paper is to study the Italian reinforcement number of a digraph D defined to be the minimum number of arcs which must be added to D in order to decrease the Italian domination number of D. We first establish some new sharp upper bounds on the Italian reinforcement number and then we determine the exact values of rI(P2Pn), rI(P3Pn), rI(P3Cn), rI(C3Pn) and rI(C3Cn). Analogous work can be carried out for other digraph parameters such as double Italian domination.

    This study was supported by the National Natural Science Foundation of China (Nos. 12061007, 11861011) and the Open Project Program of Research Center of Data Science, Technology and Applications, Minjiang University, China.

    The authors declare no conflict of interest.



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  • This article has been cited by:

    1. Ahlam Almulhim, Bana Al Subaiei, Saiful Rahman Mondal, Survey on Roman {2}-Domination, 2024, 12, 2227-7390, 2771, 10.3390/math12172771
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