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Discrete Applied Mathematics

journal homepage:www.elsevier.com/locate/dam

On algorithmic complexity of double Roman domination

Abolfazl Poureidi

^a^,^∗

, Nader Jafari Rad

^b

aFaculty of Mathematical Sciences, Shahrood University of Technology, Shahrood, Iran

bDepartment of Mathematics, Shahed University, Tehran, Iran

a r t i c l e i n f o

Article history:

Received 16 March 2019

Received in revised form 20 June 2020 Accepted 28 June 2020

Available online xxxx Keywords:

Dominating set

Double Roman dominating function Algorithm

3-SAT

Combinatorial problems

a b s t r a c t

A double Roman dominating function (DRDF) on a graphG = (V,E) is a function f : V −→ {0,1,2,3}such that every vertexv ∈V withf(v) =0 is either adjacent to a vertexuwithf(u)=3 or two distinct verticesxandywithf(x)=f(y)=2, and every vertexv∈V withf(v⁾=1 is adjacent to a vertexuwithf(u)≥2. The weight off is the sumf(V)=∑

v∈Vf(v). The minimum weight of a DRDF onGis the double Roman domination number ofG, denoted byγdR(G). A graphGis a double Roman Graph ifγdR(G)=3γ^{(G), where}γ(G) is the domination number ofG.

In this paper, we first show that the decision problem associated to double Roman domination is NP-complete even when restricted to planar graphs. Then, we study the complexity issue of a problem posed in [R.A. Beeler, T.W. Haynesa and S.T. Hedetniemi, Double Roman domination, Discrete Appl. Math. 211 (2016), 23–29], and show that the problem of deciding whether a given graph is double Roman is NP-hard even when restricted to bipartite or chordal graphs. Then, we give linear algorithms that compute the domination number and the double Roman domination number of a given unicyclic graph. Finally, we give a linear algorithm that decides whether a given unicyclic graph is double Roman.

1. Introduction

For notation and terminology not given here we refer to [10]. LetG

=

(V

,

E) be a graph with vertex setVof ordern and edge setE. Theopen neighborhoodof a vertex

v ∈

V isN(

v

⁾

= {

u

∈

V

:

u

v ∈

E

}

and theclosed neighborhoodof

v

^is N

[ v ] =

N(

v

⁾

∪ { v }

. Thedegreeof

v

^{is deg(}

v

⁾

= |

N(

v

⁾

|

. A vertex of degree one is referred as aleaf and its unique neighbor is called asupport vertex. A treeT of ordern

≥

2 is called astarifn

=

2 orn

≥

3 andT contains exactly one vertex that is not leaf. Adouble staris a tree with precisely two vertices (as central vertices) that are not leaves. A path (respectively, cycle) of ordernis denoted byP_n(respectively,C_n). Aunicyclicgraph is a graph obtained from a treeT of order at least three by joining precisely two non-adjacent vertices ofT. Aplanar graph is a graph that can be drawn on the plane in such a way that its edges intersect only at their endpoints.

For a graph G

=

(V

,

^{E), a set}^S

⊆

V is called adominating set(DS) ofGif every

v ∈

V

−

S is adjacent to at least one vertexu

∈

S. Furthermore, ifS induces a connected subgraph ofG, thenS is aconnected dominating set (CDS) of G. Thedomination number(respectively,connected domination number) ofG, denoted by

γ

(G) (respectively,

γ

c(G)), is the minimum cardinality of a dominating set (respectively, connected dominating set) ofG. A DS ofGof minimum cardinality is referred as a

γ

(G)-set. A connected DS ofGof minimum cardinality is referred as a

γ

c(G)-set.

∗ Corresponding author.

E-mail addresses: [email protected](A. Poureidi),[email protected](N. Jafari Rad).

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A functionf

:

V

−→ {

0

,

¹

,

²

}

is aRoman dominating function(RDF) ofGif every vertexuwithf(u)

=

0 is adjacent to at least one vertex

v

^with^f⁽

v

⁾

=

2. Theweightof an RDFf, denoted by

w

^(f), is the sumf(V)

=

∑

v∈_Vf(

v

). The mathematical concept of Roman domination, defined and discussed by Stewart [16], and ReVelle and Rosing [15], and subsequently developed by Cockayne et al. [8]. The results on Roman domination and its variations up to now, have recently been collected in two book chapters and three surveys, (See [3–7]). One of the new variants of Roman domination is introduced by Beeler et al. [2]. Adouble Roman dominating function(DRDF)f

:

V

−→ {

0

,

¹

,

²

,

³

}

ofGhas the property that for every vertex

v ∈

V withf(

v

⁾

=

0 either there isu

∈

N(

v

^{) with} ^f^(u)

=

3 or there arex

,

^y

∈

N(

v

^{) with}^f^(x)

=

f(y)

=

2 and for every vertex

v ∈

V withf(

v

⁾

=

1 there isu

∈

N(

v

^{) with}^f^(u)

≥

2. The weight of a DRDF f onG is the sum f(V)

=

∑

v∈_Vf(

v

) and the minimum weight of a DRDFf is thedouble Roman domination numberofG, denoted by

γ

dR(G).

A

γ

dR(G)-function is a DRDFf onGwith

w

^(f⁾

= γ

dR(G). A graphGis calleddouble Romanif

γ

dR(G)

=

3

γ

(G). The concept of double Roman domination was further studied, see for example [1,12,14,18].

Problem 1(Beeler et al. [2]).Characterize the double Roman graphs.

For a DRDFf onG, we denote byV_i (orV_i^f to refer tof) the set of all the vertices ofGwith labeliunderf. Thus a DRDFf can be represented by (V₀

,

^V1

,

^V2

,

^V3), and we can use the notationf

=

(V₀

,

^V1

,

^V2

,

^V3).

Corollary 1(Beeler et al. [2]).For any graph G, there is a

γ

dR(G)-function f

=

(V₀

,

^V1

,

^V2

,

^V3)with V₁

= ∅

.

Ahangar et al. [1] and Jafari Rad et al. [12] showed that the associated decision problem for double Roman domination is NP-complete, even for bipartite or chordal graphs. Zhang et al. [18] proposed a linear algorithm that computes the double Roman domination number of trees. Yue et al. [17] showed a linear algorithm that computes the double Roman domination number of cographs. Henning et al. [11] characterized the double Roman trees. In this paper we study the complexity issue of the double Roman domination and prove its NP-hardness in planar graphs. We also study the complexity issue ofProblem 1and study it in trees and unicyclic graphs.

The organization of the paper is as follows. We proceed as follows. In Section3we show that the associated decision problem for double Roman domination is NP-complete even when restricted to planar graphs. Then, we prove that the decision problem related toProblem 1is NP-hard even when restricted to bipartite and chordal graphs. In Section4, we give a linear algorithm that compute the double Roman domination number of a given unicyclic graph. In Section5, we give a linear algorithm that compute the domination number of a given unicyclic graph. Finally, we give a linear algorithm that decides whether a given unicyclic graph is double Roman.

2. Preliminary

Consider the following family of graphs related toProblem 1:

•

FamilyF_dR3: the family of all graphsGwith

γ

dR(G)

=

3

γ

^(G).

•

FamilyF_dR33c: the family of all graphsGwith

γ

dR(G)

=

3

γ

^(G)

=

3

γ

c(G).

Note thatF_dR33cis an infinite family even when restricted to chordal and bipartite graphs, since for any positive integer n, ifT_nis a tree obtained fromP_nby adding five new leaves to any vertex ofP_n, then it can be seen thatT_n

∈

F_dR33c. The following is obvious.

Corollary 2. F_dR33c

⊆

F_dR3.

Thus, to prove the NP-hardness of problem of whether a given graph belongs toF_dR3 we only need to prove the NP-hardness of problem of whether a given graph belongs toF_dR33c. To this end, we introduce a reduction from the 3-SAT problem. Recall that 3-SAT is the problem of deciding whether a given boolean formula in 3-conjunctive normal form is satisfiable. It is well-known that 3-SAT problem is NP-complete [9]. Let Φ

= {

C

,

^X

}

be an instance in 3-SAT Problem, that is,Φis a boolean formula in 3-conjunctive normal form. LetC

= {

c₁

,

^c2

, . . . ,

^cl

}

be a set ofl

≥

1 clauses over a setX

= {

x₁

, . . . ,

^xk

}

ofk

≥

3 variables. For each 1

≤

j

≤

l, the clausec_j(consisting of exactly three literals) is of the formc_j

= {

y_1j

,

^y2j

,

^y3j

}

, where each ofy_1j,y_2jandy_3jis either a variable or the negative of a variable inX.

3. NP-hardness results

Consider the following decision problems.

Double Roman Domination (DRD) Problem:

Instance:A graphGand a positive integerm.

Question:Does there exist a DRDFf onGwith

w

^(f⁾

≤

m?

Double Roman Graph (DRG) Problem:

Instance:A graphG.

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Fig. 1. Illustration of replacing clause-vertexcjbyzjfor eachj∈ {1,³,⁴}, variable-vertexxibyHi, clause-edgexic3by edgesu²_iz3, clause-edgexic4

by edgeu⁵_iz4and clause-edgexic1by edgesu⁷_iz1for whichl=4,NH(xi)= {c3,^c4,^c1},xi∈c3,xi∈c4and¬xi∈c1.

Question:IsG

∈

F_dR3?

Double Roman 3Connected-Domination (DR3CD) Problem:

Instance:A graphG.

Question:IsG

∈

F_dR33c?

We first introduce a polynomial time reduction from PLANAR 3-SAT Problem to DRD Problem to show that DRD Problem is NP-complete even when restricted to planar graphs. A natural graph associated to 3-SAT Problem is the bipartite graphG{C,^X}that hasC

∪

X as its vertex set and has an edge between the verticesx_i andc_jifc_jcontains either x_i or

¬

x_i, where

¬

x_i refers to thenegationofx_i. PLANAR 3-SAT is 3-SAT restricted to those instances

{

C

,

^X

}

for which G{C,^X}is planar. It is well-known that PLANAR 3-SAT Problem is NP-complete [9,13]. We introduce two polynomial time reductions from 3-SAT Problem to DR3CD Problem to show that the problem of deciding whether a given graph belongs to F_dR33cis NP-hard even when restricted to bipartite and chordal graphs. We construct graphsH_Φ,G_ΦandI_Φcorresponding toΦas follows.

3.1. The first reduction: H_Φ

LetΦ

= {

C

,

^X

}

be an instance of 3-SAT Problem such that the associated graphG{C,^X}toΦis planar. Assume thatHis a planar embedding ofG{C,^X}. We replace each variable-vertexx_iofH, where 1

≤

i

≤

k, by a graphH_i as variable gadget, whereH_i is obtained from a cycle of order 3lwith verticesu¹_i

, . . . ,

^u^3l_i such that each of verticesu^3j_i ⁻²

,

^u^3j_i ⁻¹is adjacent to new vertices

v

3j−₂,ⁱ

, v

3j^′−₂,ⁱ

, v

3j^′′−₂,ⁱfor each 1

≤

j

≤

l. We replace each clause-vertexc_jofH, where 1

≤

j

≤

l, by a new vertexz_j. In the rest we fix indicesiandj, where 1

≤

i

≤

kand 1

≤

j

≤

l. Assume thatc_i₁

,

^ci2

, . . . ,

^cim, where 1

≤

m

≤

l, is the sequence of all clause-vertices adjacent tox_i inHin clockwise direction starting from an arbitrary clause-vertex inN_H(x_i), that is,x_ic_j^′ is an edge ofHfor eachj^′

∈ {

i₁

,

ⁱ2

, . . . ,

ⁱm

}

. Ifx_i

∈

c_i_r (respectively,

¬

x_i

∈

c_i_r), where 1

≤

r

≤

m, then we replacex_ic_i_r by new edgeu^3r_i ⁻¹z_i_r (respectively,u^3r_i ⁻²z_i_r). LetH_Φ be the resulting graph. SeeFig. 1. It is easy to see thatH_Φ is a planar graph.

Lemma 1. The boolean formulaΦis satisfiable if and only if there is a DRDF f on H_Φ with

w

^(f⁾

≤

3kl.

Proof. Assume thatΦis satisfiable. Lett be an assignment of truth values for the variables ofX for whichΦ^evaluates totrue. We construct a setV₃on the vertex set ofH_Φ as follows. Ift assigns the valuetrue(respectively, the valuefalse) tox_i, then we add all verticesu²_i

,

^u⁵_i

, . . . ,

^u^3l_i⁻¹ (respectively, all vertices u¹_i

,

^u⁴_i

, . . . ,

^u^3l_i ⁻²^{) to}^V3. It is easy to see that f

=

(V(H_Φ)

−

V₃

, ∅ , ∅ ,

^V3) is a DRDF onH_Φ with

w

^(f⁾

=

3kl.

Assume that there is a DRDF f

=

(V₀

,

^V1

,

^V2

,

^V3) on H_Φ with

w

^(f⁾

≤

3kl. Consider values f(

v

3j−₂,i)

,

^f⁽

v

^′_3j−₂,i)

,

f(

v

_3j^′′−₂,i)

,

^f^(u^3j_i ⁻²⁾

,

^f^(u^3j_i ⁻¹) for each 1

≤

i

≤

k and 1

≤

j

≤

l. Since all vertices

v

3j−₂,i

, v

^′_3j−₂,i

, v

_3j^′′−₂,i are only adjacent to verticesu^3j_i ⁻²andu^3j_i ⁻¹, we find thatS_ij

=

f(

v

3j−₂,i)

+

f(

v

_3j^′−₂,i)

+

f(

v

_3j^′′−₂,i)

+

f(u^3j_i⁻²)

+

f(u^3j_i ⁻¹)

≥

3 for each 1

≤

i

≤

k and 1

≤

j

≤

l. So,

w

^(f⁾

≥

∑

i∈{₁,...,k},j∈{₁,...,l}S_ij

≥

3kl. Since

w

^(f⁾

≤

3kl, we haveS_ij

=

3 andf(z_j)

=

f(u^3j_i )

=

0 for each 1

≤

i

≤

kand 1

≤

j

≤

l. Sincef(z_j)

=

0 for each 1

≤

j

≤

l, iff(

v

3j−₂,i)

+

f(

v

^′_3j−₂,i)

+

f(

v

_3j^′′−₂,i)

̸=

0 for some 1

≤

i

≤

kand 1

≤

j

≤

l, thenS_ij

>

3, a contradiction. So,f(

v

3j−₂,i)

+

f(

v

_3j^′−₂,i)

+

f(

v

^′′_3j−₂,i)

=

0 for each 1

≤

i

≤

kand 1

≤

j

≤

l. Thus, either bothf(u^3j_i ⁻²)

=

0 andf(u^3j_i ⁻¹)

=

3 or bothf(u^3j_i ⁻²)

=

3 andf(u^3j_i ⁻¹)

=

0 for each 1

≤

i

≤

kand 1

≤

j

≤

l.

Claim 1.

|

V₃

∪ {

u^3j_i ⁻²

,

^u^3j_i ⁺²

}| ≤

1, for each1

≤

i

≤

k and1

≤

j

≤

l, where the addition is in modulo3l.

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Fig. 2. IllustratingG_Φcorresponding toΦ= {{c1,^c2},{x1,^x2,^x3}}, wherec1= {¬x1,¬x2,^x3}andc2= {x1,¬x2,^x3}.

Proof. Suppose for a contradiction that

|

V₃

∪ {

u^3j_i ⁻²

,

^u^3j_i⁺²

}| =

2. So, both u^3j_i⁻¹

,

^u^3j_i ⁺¹

∈

V₀ and so f(u^3j_i )

̸=

0, a contradiction. □

Claim 2.

|

V₃

∪ {

u^3j_i ⁻¹

,

^u^3j_i⁺¹

}| ≤

1, for each1

≤

i

≤

k and1

≤

j

≤

l, where the addition is in modulo3l.

Proof. Suppose for a contradiction that

|

V₃

∪ {

u^3j_i ⁻¹

,

^u^3j_i ⁺¹

}| =

2. Sincef(u^3j_i ⁺¹)

=

3, byClaim 1we havef(u^3j_i⁺⁵)

=

0 and sof(u^3j_i ⁺⁴)

=

3. Similarly, we find thatu^3j_i⁺⁷

,

^u^3j_i⁺¹⁰

, . . . ,

^u^3j_i⁺^3l⁻²⁽

=

u^3j_i ⁻²)

∈

V₃, that is, bothu^3j_i⁻¹

,

^u^3j_i⁻²

∈

V₃, a contradiction. _□

ByClaims 1and2eitheru¹_i

,

^u⁴_i

, . . . ,

^u^3l_i⁻²

∈

V₀andu²_i

,

^u⁵_i

, . . . ,

^u^3l_i⁻¹

∈

V₃oru¹_i

,

^u⁴_i

, . . . ,

^u^3l_i ⁻²

∈

V₃andu²_i

,

^u⁵_i

, . . . ,

^u^3l_i ⁻¹

∈

V₀ for each 1

≤

i

≤

k. We fix indicesiand j, where 1

≤

i

≤

k and 1

≤

j

≤

l. Ifu¹_i

,

^u⁴_i

, . . . ,

^u^3l_i ⁻²

∈

V₀ and u²_i

,

^u⁵_i

, . . . ,

^u^3l_i ⁻¹

∈

V₃ (respectively, u¹_i

,

^u⁴_i

, . . . ,

^u^3l_i ⁻²

∈

V₃ and u²_i

,

^u⁵_i

, . . . ,

^u^3l_i⁻¹

∈

V₀), then we assign the value true (respectively, the valuefalse) to the variablex_i. We claim thatΦis satisfiable for this assignment.

Assume without loss of generality thatc_j

= {

x₁

, ¬

x₂

,

^x6

}

. Since f(z_j)

=

0, we havef(u^3j

′−₁

1 )

=

3, f(u^3j

′′−₂

2 )

=

3 or f(u^4j

′′′−₁

6 )

=

1 for somej^′

,

^j^′′

,

^j^′′′

∈ {

1

,

²

, . . . ,

^l

}

. Assume without loss of generality thatf(u^3j

′−₁

1 )

=

3. So,x₁has the value true. It causes to satisfy the clausec_j, that is, the boolean formulaΦis satisfiable. This completes the proof. □

Clearly, we can compute H_Φ in polynomial time and the DRD Problem belongs to NP. By Lemma 1 we have the following.

Theorem 1. The DRD Problem is NP-complete even when restricted to planar graphs.

3.2. The second reduction: G_Φ

LetΦ

= {

C

,

^X

}

be an instance of 3-SAT Problem. For each variablex_i, where 1

≤

i

≤

k, we construct a graphG_i as variable gadget, whereG_iis obtained from a path graph of order 2 with verticesu¹_i

,

^u²_i such that each of verticesu¹_i

,

^u²_i ^is adjacent to new vertex

v

i^sfor eachs

∈ {

1

,

²

,

³

}

. For each clausec_j

= {

y_1j

,

^y2j

,

^y3j

}

, where 1

≤

j

≤

l, we add a new vertex zjsuch thatzjis adjacent to five new leaves. Fors

=

1

,

²

,

^{3, if}^ysj

=

xi, for some 1

≤

i

≤

k, then we add edgeu²_izjand if y_sj

= ¬

x_i, for some 1

≤

i

≤

k, then we add edgesu¹_iz_j. We add a new vertexosuch that is adjacent to five new leaves and add edgesou¹_i andou²_i for each 1

≤

i

≤

k. Finally add all edgesabfor eacha

∈ {

u¹_i

,

^u²_i

}

andb

∈ {

u¹_j

,

^u²_j

}

and for all 1

≤

i

<

^j

≤

k. LetG_Φ be the resulting graph. SeeFig. 2. It is easy to see thatG_Φ is a chordal graph.

Lemma 2.

γ

^(GΦ)

=

k

+

l

+

1.

Proof. LetSbe a

γ

^(G_Φ)-set. Since each of verticesoandz_j, where 1

≤

j

≤

l, is adjacent to five leaves, botho

,

^zj

∈

S.

Since all vertices

v

_i¹

, v

_i²

, v

³_i^{, where 1}

≤

i

≤

k, are only adjacent to verticesu¹_i

,

^u²_i, at least one of verticesu¹_i

,

^u²_i ^{belongs to} S. So,

γ

^(G_Φ⁾

= |

S

| ≥

k

+

l

+

1.

LetS

= {

o

,

^zj

,

^u¹_i

|

1

≤

i

≤

k

,

¹

≤

j

≤

l

}

. Clearly,S is a DS onG_Φ with

|

S

| =

k

+

l

+

1. So,

γ

^(G_Φ⁾

≤

k

+

l

+

1. This completes the proof. □

Lemma 3.

γ

dR(G_Φ)

=

3(k

+

l

+

1).

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Proof. Let f be a

γ

dR(G_Φ)-function. Since each of vertices oand z_j, where 1

≤

j

≤

l, is adjacent to five leaves, we havef(o)

=

f(z_j)

=

3. Since all vertices

v

¹_i

, v

²_i

, v

³_i^{, where 1}

≤

i

≤

,

^u²_i, we find that

∑2

s=1f(u^s_i)

+

∑3

s=1f(

v

^si)

≥

3. So,

γ

dR(G_Φ)

= w

^(f⁾

≥

3(k

+

l

+

1).

LetV₃

= {

o

,

^zj

,

^u¹_i

|

1

≤

i

≤

k

,

¹

≤

j

≤

l

}

. Clearly,f

=

(V(G_Φ)

−

V₃

, ∅ , ∅ ,

^V3) is a DRDF onG_Φ with

w

^(f⁾

=

3(k

+

l

+

1).

So,

γ

dR(G_Φ)

≤

3(k

+

l

+

1). This completes the proof. □

Lemma 4. The boolean formulaΦis satisfiable if and only if G_Φ

∈

F_dR33c.

Proof. Assume thatΦis satisfiable. Lettbe an assignment of truth values for the variables ofXfor whichΦevaluates to true. We construct a setSon the vertex set ofG_Φ as follows. InitializeSto be

{

o

,

^zj

:

1

≤

j

≤

l

}

. Ift assigns the valuetrue (respectively, the valuefalse) tox_i, then we add vertexu²_i (respectively, vertexu¹_i) toS. It is easy to see thatSis a CDS on G_Φ with

|

S

| =

k

+

l

+

1. So,

γ

c(G_Φ)

≤

k

+

l

+

1. ByLemma 2we have

γ

^(GΦ)

=

k

+

l

+

1. By the fact

γ

^(G)

≤ γ

c(G) for any graphG, it obtains that

γ

c(G_Φ)

=

k

+

l

+

1. ByLemma 3we have

γ

dR(G_Φ)

=

3(k

+

l

+

1). So,

γ

dR(G_Φ)

=

3

γ

c(G_Φ)

=

3

γ

^(GΦ), that is,G_Φ

∈

F_dR33c.

LetG_Φ

∈

F_dR33c. ByLemma 2we have

γ

^(GΦ)

=

k

+

l

+

1. LetSbe a CDS onG_Φ. So,

|

S

| =

k

+

l

+

1. Clearly, bothoand z_j, where 1

≤

j

≤

l, belong toS. SinceSis a connected dominating set andobelongs toS, at least one of verticesu¹_i and u²_i belongs toSfor each 1

≤

i

≤

k. If bothu¹_i

,

^u²_i

∈

Sfor some 1

≤

i

≤

k, then

|

S

| >

^k

+

l

+

1, a contradiction. So, either bothu¹_i

∈

S andu²_i

∈ /

Sor bothu¹_i

∈ /

Sandu²_i

∈

S for each 1

≤

i

≤

k.

We fix indicesiandj, where 1

≤

i

≤

kand 1

≤

j

≤

l. Recall that either bothu¹_i

∈

Sandu²_i

∈ /

S or bothu¹_i

∈ /

Sand u²_i

∈

S. Ifu¹_i

∈ /

Sandu²_i

∈

S(respectively,u¹_i

∈

Sandu²_i

∈ /

S), then we assign the valuetrue(respectively, the valuefalse) to the variablex_i. We claim thatΦis satisfiable for this assignment.

Assume without loss of generality that c_j

= {

x₁

, ¬

x₂

,

^x6

}

. Sincez_j

∈

S, we have u²₁

∈

S,u¹₂

∈

S oru²₆

∈

S. Assume without loss of generality that u²₁

∈

S. So,x₁has the value true. It causes to satisfy the clausec_j, that is, the boolean formulaΦis satisfiable. This completes the proof. □

Note that we can computeG_Φ in polynomial time. Thus byLemma 4and the fact thatG_Φ is a chordal graph we have the following.

Theorem 2. The DR3CD Problem is NP-hard even when restricted to chordal graphs.

ByCorollary 2andTheorem 2we have the following.

Corollary 3. The DRG Problem is NP-hard even when restricted to chordal graphs.

3.3. The third reduction: I_Φ

Let Φ

= {

C

,

^X

}

be an instance of 3-SAT Problem. For each variable x_i, where 1

≤

i

≤

k, we construct a graph I_i as variable gadget, whereI_i is obtained from a path graph of order 3 with verticesu¹_i

,

^u²_i

,

^u³_i ^{such that}^u²_i is not a leaf, u²_i is adjacent to new five leaves and both vertices u¹_i and u³_i are adjacent to

v

^s_i ^{for each}^s

∈ {

1

,

²

,

³

}

. For each clause c_j

= {

y_1j

,

^y2j

,

^y3j

}

, where 1

≤

j

≤

l, we add a new vertexz_jsuch thatz_jis adjacent to five new leaves. Fors

=

1

,

²

,

^{3, if} y_sj

=

x_i, for some 1

≤

i

≤

k, then we add edgeu³_iz_jand ify_sj

= ¬

x_i, for some 1

≤

i

≤

k, then we add edgesu¹_iz_j. We add a new vertexosuch that is adjacent to five new leaves and add edgesou¹_i andou²_i for each 1

≤

i

≤

k. LetI_Φ be the resulting graph. SeeFig. 3. It is easy to see thatI_Φ is a bipartite graph.

Lemma 5.

γ

^(IΦ)

=

2k

+

l

+

1.

Proof. LetSbe a

γ

^(IΦ)-set. Since each of verticeso,u²_i andz_j, where 1

≤

i

≤

kand 1

≤

j

≤

l, is adjacent to five leaves, we haveo

,

^u²_i

,

^zj

∈

S. Since all vertices

v

_i¹

, v

_i²

, v

_i³^{, where 1}

≤

i

≤

,

^u³_i, at least one of verticesu¹_i

,

^u³_i ^{belongs to}^{S. So,}

γ

^(I_Φ⁾

= |

S

| ≥

2k

+

l

+

1.

LetS

= {

o

,

^zj

,

^u¹_i

,

^u²_i

|

1

≤

i

≤

k

,

¹

≤

j

≤

l

}

. Clearly,Sis a DS onI_Φ with

|

S

| =

2k

+

l

+

1. So,

γ

^(I_Φ⁾

≤

2k

+

l

+

1. This completes the proof. _□

Lemma 6.

γ

dR(I_Φ)

=

6k

+

3l

+

3.

Proof. Letf be a

γ

dR(I_Φ)-function. Since each of verticeso,u²_i andz_j, where 1

≤

i

≤

kand 1

≤

j

≤

l, is adjacent to five leaves, we havef(o)

=

f(u²_i)

=

f(z_j)

=

3. Since all vertices

v

_i¹

, v

²_i

, v

³_i^{, where 1}

≤

i

≤

,

^u²_i^, we find thatf(u¹_i)

+

f(u³_i)

+

∑3

s=₁f(

v

_i^s⁾

≥

3. So,

γ

dR(I_Φ)

= w

^(f⁾

≥

6k

+

3l

+

3.

LetV₃

= {

o

,

^zj

,

^u¹_i

,

^u²_i

|

1

≤

i

≤

k

,

¹

≤

j

≤

l

}

. Clearly,f

=

(V(I_Φ)

−

V₃

, ∅ , ∅ ,

^V3) is a DRDF onI_Φwith

w

^(f⁾

=

6k

+

3l

+

3.

So,

γ

dR(I_Φ)

≤

6k

+

3l

+

3. This completes the proof. _□

(6)

Fig. 3. IllustratingI_Φ corresponding toΦ= {{c1,^c2},{x1,^x2,^x3}}, wherec1= {¬x1,¬x2,^x3}andc2= {x1,¬x2,^x3}.

Lemma 7. The boolean formulaΦis satisfiable if and only if I_Φ

∈

F_dR33c.

Proof. Assume thatΦis satisfiable. Lett be an assignment of truth values for the variables ofX for whichΦ^evaluates totrue. We construct setSon the vertex set ofI_Φ as follows. InitializeSto be

{

o

,

^zj

,

^u²_i

:

1

≤

i

≤

k

,

¹

≤

j

≤

l

}

. Iftassigns the valuetrue(respectively, the valuefalse) tox_i, then we add vertexu³_i (respectively, vertexu¹_i) toS. It is easy to see thatSis a CDS onI_Φ with

|

S

| =

2k

+

l

+

1. So,

γ

c(I_Φ)

≤

2k

+

l

+

1. ByLemma 5we have

γ

^(I_Φ⁾

=

2k

+

l

+

1. By the fact

γ

^(G)

≤ γ

c(G) for any graphG, it obtains that

γ

c(I_Φ)

=

2k

+

l

+

1. ByLemma 6we have

γ

dR(I_Φ)

=

3(k

+

l

+

1). So,

γ

dR(I_Φ)

=

3

γ

c(I_Φ)

=

3

γ

^(IΦ), that is,I_Φ

∈

F_dR33c.

LetI_Φ

∈

F_dR33c. ByLemma 5we have

γ

^(IΦ)

=

2k

+

l

+

1. LetSbe a CDS onI_Φ. So,

|

S

| =

2k

+

l

+

1. Clearly, all vertices o,u²_i andz_j, where 1

≤

i

≤

kand 1

≤

j

≤

l, belong toS. SinceSis a connected dominating set andobelongs toS, at least one of verticesu¹_i andu³_i belongs toSfor each 1

≤

i

≤

k. If bothu¹_i

,

^u³_i

∈

S for some 1

≤

i

≤

k, then

|

S

| >

^2k

+

l

+

1, a contradiction. So, either bothu¹_i

∈

Sandu³_i

∈ /

Sor bothu¹_i

∈ /

Sandu³_i

∈

Sfor each 1

≤

i

≤

k.

We fix indicesiandj, where 1

≤

i

≤

kand 1

≤

j

≤

l. Recall that either bothu¹_i

∈

S andu³_i

∈ /

Sor bothu¹_i

∈ /

S and u³_i

∈

S. Ifu¹_i

∈ /

Sandu³_i

∈

S(respectively,u¹_i

∈

Sandu³_i

∈ /

S), then we assign the valuetrue(respectively, the valuefalse) to the variablex_i. We claim thatΦis satisfiable for this assignment.

Assume without loss of generality thatc_j

= {

x₁

, ¬

x₂

,

^x6

}

. Sincez_j

∈

S, we haveu³₁

∈

S,u¹₂

∈

S oru³₆

∈

S. Assume without loss of generality thatu³₁

∈

S. So,x₁ has the value true. It causes to satisfy the clausec_j, that is, the boolean formulaΦis satisfiable. This completes the proof. □

We can computeI_Φin polynomial time. Thus byLemma 7and the fact thatI_Φis a bipartite graph we have the following.

Theorem 3. The DR3CD Problem is NP-hard even when restricted to bipartite graphs.

ByCorollary 2andTheorem 3we have the following.

Corollary 4. The DRG Problem is NP-hard even when restricted to bipartite graphs.

4. Computing double Roman domination number of unicyclic graphs

In this section, we give a linear algorithm that computes the double Roman domination number of unicyclic graphs.

LetG

=

(V

,

E) be a graph, letu

∈

V, let

v

be a vertex not inVand leta

∈ {

0

,

¹

,

²

,

³

}

. We define the following.

• γ

dR(G

,

^u

=

a)

=

min

{ w

^(f⁾

|

f is a DRDF onGwithf(u)

=

a

}

,

• γ

dR(G

,

^u

, v

⁾

=

min

{ w

^(f⁾

|

f is a DRDF onG

+

u

v

^with^f^(u)

=

0 andf(

v

⁾

=

2

}

.

A

γ

dR(G

,

^u

=

a)-function (respectively,

γ

dR(G

,

^u

, v

)-function) is a DRDFf onGwith

w

^(f⁾

= γ

dR(G

,

^u

=

a) andf(u)

=

a (respectively,

w

^(f⁾

= γ

dR(G

,

^u

, v

^),^f^(u)

=

0 andf(

v

⁾

=

2).

Lemma 8. Let H₁

=

(V₁

,

^E1)and H₂

=

(V₂

,

^E2)be two graphs with V₁

∩

V₂

= ∅

such that u

∈

V₁,

v ∈

V₂and a vertex

w / ∈

V₁

∪

V₂. Let G

=

(V₁

∪

V₂

,

^E1

∪

E₂

∪ {

u

v }

). Then, we have the following.

(i)

γ

dR(G

,

^u

=

0)

=

min

{ γ

dR(H₁

,

^u

=

0)

+ γ

dR(H₂

, v =

0)

, γ

dR(H₁

,

^u

, w

⁾

+ γ

dR(H₂

, v =

2)

−

2

, γ

dR(H₁

−

u)

+ γ

dR(H₂

, v =

3)

}

, (ii)

γ

dR(G

,

^u

=

2)

= γ

dR(H₁

,

^u

=

2)

+

min

{ γ

dR(H₂

, v, w

⁾

−

2

, γ

dR(H₂

, v =

2)

, γ

dR(H₂

, v =

3)

}

,

(7)

(iii)

γ

dR(G

,

^u

=

3)

= γ

dR(H₁

,

^u

=

3)

+

min

{ γ

dR(H₂

− v

⁾

, γ

dR(H₂

, v =

2)

, γ

dR(H₂

, v =

3)

}

, (iv)

γ

dR(G

−

u)

= γ

dR(H₁

−

u)

+

min

{ γ

dR(H₂

, v =

0)

, γ

dR(H₂

, v =

2)

, γ

dR(H₂

, v =

3)

}

,

(v)

γ

dR(G

,

^u

, w

⁾

=

min

{ γ

dR(H₁

,

^u

, w

⁾

+ γ

dR(H₂

, v =

0)

, γ

dR(H₁

−

u)

+ γ

dR(H₂

, v =

2)

+

2

, γ

dR(H₁

−

u)

+ γ

dR(H₂

, v =

3)

+

2

}

. Proof. Let f be a

γ

dR(G)-function. ByCorollary 1, we can assume thatf(u)

,

^f⁽

v

⁾

∈ {

0

,

²

,

³

}

. Clearly,f(u)

=

a, where a

∈ {

0

,

²

,

³

}

if and only if bothf(u)

=

aandf(

v

⁾

=

0, bothf(u)

=

aandf(

v

⁾

=

2 or bothf(u)

=

aandf(

v

⁾

=

3.

Let f₁,f₂,f₁^u and f₂^v be restrictions off toH₁,H₂, H₁

−

u and H₂

− v

, respectively. Let g₁â,g₂â,g₁^′, g₂^′,g₁û and g₂^v be a

γ

dR(H₁

,

^u

=

a)-function,

γ

dR(H₂

, v =

a)-function,

γ

dR(H₁

,

^u

, w

)-function,

γ

dR(H₂

, v, w

)-function,

γ

dR(H₁

−

u)-function and

γ

dR(H₂

− v

)-function, respectively, wherea

∈ {

0

,

²

,

³

}

and letg_u(u)

=

g_v(

v

⁾

=

0 andg_w(

w

⁾

=

2.

Letf(u)

=

0 and

γ

dR

=

min

{ γ

dR(H₁

,

^u

=

0)

+ γ

dR(H₂

, v =

0)

, γ

dR(H₁

,

^u

, w

⁾

+ γ

dR(H₂

, v =

2)

−

2

, γ

dR(H₁

−

u)

+ γ

dR(H₂

, v =

3)

}

. So,f1is a DRDF onH1withf1(u)

=

0 andf2is a DRDF onH2withf2(

v

⁾

=

0, functionh

=

f1

∪

g_wis a DRDF onH1

+

u

w

with bothh(u)

=

0 andh(

w

⁾

=

2 andf₂is a DRDF onH₂ withf₂(

v

⁾

=

2 orf₁^u is a DRDF onH₁

−

uandf₂is a DRDF onH₂ withf₂(

v

⁾

=

3. Hence,

γ

dR

≤ γ

dR(G

,

^u

=

0). Functiong₁

=

g₁⁰

∪

g₂⁰is a DRDF onGwithg₁(u)

=

0, the restriction of g₂

=

g₁^′

∪

g₂² toG is a DRDF onG withg₂(u)

=

0 andg₃

=

g₁^u

∪

g₂³

∪

g_u is a DRDF onGwithg₃(u)

=

0. Hence,

γ

dR(G

,

^u

=

0)

≤ γ

dR. This completes the proof of part (i).

Let f(u)

=

2 and

γ

dR

=

min

{ γ

dR(H₁

,

^u

=

2)

+ γ

dR(H₂

, v, w

⁾

−

2

, γ

dR(H₁

,

^u

=

2)

+ γ

dR(H₂

, v =

2)

, γ

dR(H₁

,

^u

=

2)

+ γ

dR(H₂

, v =

3)

}

. So,f₁ is a DRDF on H₁ withf₁(u)

=

2 andh

=

f₂

∪

g_w is a DRDF on H₂

+ vw

^with^h(

v

⁾

=

0 andh(

w

⁾

=

2, functionf₁ is a DRDF onH₁ withf₁(u)

=

2 andf₂ is a DRDF onH₂withf₂(

v

⁾

=

2 orf₁is a DRDF onH₁ withf₁(u)

=

2 andf₂is a DRDF onH₂withf₂(

v

⁾

=

3. Hence,

γ

dR

≤ γ

dR(G

,

^u

=

2). The restriction ofg₁

=

g₁²

∪

g₂^′ toGis a DRDF onGwithg₁(u)

=

2, functiong₂

=

g₁²

∪

g₂²a DRDF onGwithg₂(u)

=

2 andg₃

=

g₁<