Parameterized Complexity of Globally Minimal Defensive Alliances

1 Introduction

During the last 20 years, the Defensive Alliance problem has been studied extensively. A defensive alliance in an undirected graph is a non-empty set of vertices with the property that each vertex has at least as many neighbours in the alliance (including itself) as outside the alliance. In 2000, Kristiansen, Hedetniemi, and Hedetniemi [kris] introduced defensive, offensive, and powerful alliances, and further studied by Shafique [HassanShafique2004PartitioningAG] and other authors [BAZGAN2019111, BLIEM2018334, small, Cami2006OnTC, Enciso2009AlliancesIG, Fernau, FERNAU2009177, ICDCIT2021, Lindsay, ROD, SIGARRETA20091687, SIGARRETA20061345, SIGA]. In this paper, we will focus on defensive alliances. A defensive alliance is strong if each vertex of the alliance has at least as many neighbours in the alliance (not counting itself) as outside the alliance. The theory of alliances in graphs have been studied intensively [Cami2006OnTC, 10.5614/ejgta.2014.2.1.7, frick] both from a combinatorial and from a computational perspective. As mentioned in [BAZGAN2019111], the focus has been mostly on finding small alliances, although studying large alliances do not only make a lot of sense from the original motivation of these notions, but was actually also delineated in the very first papers on alliances [kris].

Note that a defensive alliance is not a hereditary property, that is, a subset of defensive alliance is not necessarily a defensive alliance. Shafique [HassanShafique2004PartitioningAG] called an alliance a locally minimal alliance if the set obtained by removing any vertex of the alliance is not an alliance. Bazgan et al. [BAZGAN2019111] considered another notion of alliance that they called a globally minimal alliance which has the property that no proper subset is an alliance. In this paper we are interested in globally minimal alliances of maximum size. Clearly, the motivation is that big communities where every member still matters somehow are of more interest than really small communities. Also, there is a general mathematical interest in such type of problems, see [Manlove1998MinimaximalAM].

2 Basic Notations

Throughout this article, $G=(V,E)$ denotes a finite, simple and undirected graph of order $|V|=n$ . For a non-empty subset $S\subseteq V$ and a vertex $v\in V(G)$ , let $N_{S}(v)=\{u\in S~{}:~{}(u,v)\in E(G)\}$ , $N_{S}[v]=N_{S}(v)\cup\{v\}$ , and $d_{S}(v)$ denote its open neighborhood, closed neighborhood, and degree respectively in $S$ . The complement of the vertex set $S$ in $V$ is denoted by $S^{c}$ .

Definition 1.

A non-empty set $S\subseteq V$ is a defensive alliance in $G$ if for each $v\in S$ , $|N[v]\cap S|\geq|N(v)\setminus S|$ , or equivalently, $d_{S}(v)+1\geq d_{S^{c}}(v)$ .

A vertex $v\in S$ is said to be protected if $d_{S}(v)+1\geq d_{S^{c}}(v)$ . A non-empty set $S\subseteq V$ is a defensive alliance if every vertex in $S$ is protected.

Definition 2.

A vertex $v\in S$ is said to be marginally protected if it becomes unprotected when any of its neighbour in $S$ is moved from $S$ to $V\setminus S$ . A vertex $v\in S$ is said to be strongly protected if it remains protected when any of its neighbours is moved from $S$ to $V\setminus S$ .

Definition 3.

An alliance $S$ is called a locally minimal alliance if for any $v\in S$ , $S\setminus\{v\}$ is not an alliance.

Definition 4.

A defensive alliance $S$ is globally minimal defensive alliance or shorter minimal defensive alliance if no proper subset is an alliance.

In literature, a defensive alliance $S$ is called global defensive alliance if $S$ is a dominating set. It is to be noted that globally minimal defensive alliance is different from global defensive alliance.

Observation 1.

Let $S$ be a globally minimal defensive alliance of size at least two in graph $G$ . Then $S$ can never contain a vertex of degree one.

This can be proved by contradiction. Suppose $S$ contains a vertex $v$ of degree one. Note that $\{v\}$ is a proper subset of $S$ and it is a defensive alliance, a contradiction to the fact that $S$ is a globally minimal defensive alliance.

Figure 1: The set

S_{1}=\{7,2,9,3,11,4,13,5,15,6\}

is a locally minimal defensive alliance of size 10 and

S_{2}=\{1,2,3\}

is a globally minimal defensive alliance of size 3 in

G

.

A defensive alliance $S$ is connected if the subgraph induced by $S$ is connected. Notice that any globally minimal defensive alliance is also connected. In this paper, we consider Globally Minimal Defensive Alliance problem under structural parameters. We define the problem as follows:
Globally Minimal Defensive Alliance
Input: An undirected graph $G=(V,E)$ and an integer $k\geq 2$ . Question: Is there a globally minimal defensive alliance $S\subseteq V(G)$ such that $|S|\geq k$ ?

We now review the concept of a tree decomposition, introduced by Robertson and Seymour in [Neil]. Treewidth is a measure of how “tree-like” the graph is.

Definition 5.

[Downey] A tree decomposition of a graph $G=(V,E)$ is a tree $T$ together with a collection of subsets $X_{t}$ (called bags) of $V$ labeled by the vertices $t$ of $T$ such that $\bigcup_{t\in T}X_{t}=V$ and (1) and (2) below hold:

1.

For every edge $uv\in E(G)$ , there is some $t$ such that $\{u,v\}\in X_{t}$ .
2.

(Interpolation Property) If $t$ is a vertex on the unique path in $T$ from $t_{1}$ to $t_{2}$ , then $X_{t_{1}}\cap X_{t_{2}}\subseteq X_{t}$ .

Definition 6.

[Downey] The width of a tree decomposition is the maximum value of $|X_{t}|-1$ taken over all the vertices $t$ of the tree $T$ of the decomposition. The treewidth $tw(G)$ of a graph $G$ is the minimum width among all possible tree decomposition of $G$ .

For the standard concepts in parameterized complexity, see the recent textbook by Cygan et al. [marekcygan]. In this paper we prove that the Globally Minimal Defensive Alliance problem is polynomial time solvable on trees and it is W[1]-hard when parameterized by the treewidth of the graph.

Known Results: The decision version for several types of alliances have been shown to be NP-complete. For an integer $r$ , a nonempty set $S\subseteq V(G)$ is a defensive $r$ -alliance if for each $v\in S$ , $|N(v)\cap S|\geq|N(v)\setminus S|+r$ . A set is a defensive alliance if it is a defensive $(-1)$ -alliance. A defensive $r$ -alliance $S$ is global if $S$ is a dominating set. The defensive $r$ -alliance problem is NP-complete for any $r$ [SIGARRETA20091687]. The defensive alliance problem is NP-complete even when restricted to split, chordal and bipartite graph [Lindsay]. For an integer $r$ , a nonempty set $S\subseteq V(G)$ is an offensive $r$ -alliance if for each $v\in N(S)$ , $|N(v)\cap S|\geq|N(v)\setminus S|+r$ . An offensive 1-alliance is called an offensive alliance. An offensive $r$ -alliance $S$ is global if $S$ is a dominating set. Fernau et al. showed that the offensive $r$ -alliance and global offensive $r$ -alliance problems are NP-complete for any fixed $r$ [FERNAU2009177]. They also proved that for $r>1$ , $r$ -offensive alliance is NP-hard, even when restricted to $r$ -regular planar graphs. There are polynomial time algorithms for finding minimum alliances in trees [CHANG2012479, Lindsay]. Bliem and Woltran [BLIEM2018334] proved that deciding if a graph contains a defensive alliance of size at most $k$ is W[1]-hard when parameterized by treewidth of the input graph. Bazgan et al. [BAZGAN2019111] proved that deciding if a graph contains a globally minimal strong defensive alliance of size at least $k$ is NP-complete, even for cubic graphs. Moreover, deciding if a graph contains a globally minimal defensive alliance of size at least $k$ is NP-complete, even for graphs of degree 3 or 4 [BAZGAN2019111].

3 FPT algorithm parameterized by neighbourhood diversity

In this section, we present an FPT algorithm for Globally Minimal Defensive Alliance problem parameterized by neighbourhood diversity. We say two vertices $u$ and $v$ have the same type if and only if $N(u)\setminus\{v\}=N(v)\setminus\{u\}$ . The relation of having the same type is an equivalence relation. The idea of neighbourhood diversity is based on this type structure.

Definition 7.

[Lampis] The neighbourhood diversity of a graph $G=(V,E)$ , denoted by ${\tt nd}(G)$ , is the least integer $k$ for which we can partition the set $V$ of vertices into $k$ classes, such that all vertices in each class have the same type.

If neighbourhood diversity of a graph is bounded by an integer $k$ , then there exists a partition $\{C_{1},C_{2},\ldots,C_{k}\}$ of $V(G)$ into $k$ type classes. It is known that such a minimum partition can be found in linear time using fast modular decomposition algorithms [Tedder]. Notice that each type class could either be a clique or an independent set by definition. For algorithmic purpose it is often useful to consider a type graph $H$ of graph $G$ , where each vertex of $H$ is a type class in $G$ , and two vertices $C_{i}$ and $C_{j}$ are adjacent iff there is complete bipartite clique between these type classes in $G$ . It is not difficult to see that there will be either a complete bipartite clique or no edges between any two type classes. The key property of graphs of bounded neighbourhood diversity is that their type graphs have bounded size. In this section, we prove the following theorem:

Theorem 1.

The Globally Minimal Defensive Alliance problem is fixed-parameter tractable when parameterized by the neighbourhood diversity.

Let $G$ be a connected graph such that ${\tt nd}(G)=k$ . Let $C_{1},\ldots,C_{k}$ be the partition of $V(G)$ into sets of type classes. We assume $k\geq 2$ since otherwise the problem becomes trivial. Let $x_{i}=|C_{i}\cap S|$ where $S$ is a globally minimal defensive alliance. We define $I_{0}=\{C_{i}~{}|~{}x_{i}=0\}$ , $I_{1}=\{C_{i}~{}|~{}x_{i}=1\}$ and $I_{2}=\{C_{i}~{}|~{}x_{i}\geq 2\}$ . We next guess if $C_{i}$ belongs to $I_{0}$ , $I_{1}$ , or $I_{2}$ . There are at most $3^{k}$ possibilities as each $C_{i}$ has three options: either in $I_{0}$ , $I_{1}$ , or $I_{2}$ . We reduce the problem of finding a globally minimal defensive alliance to an integer linear programming optimization with $k$ variables. Since integer linear programming is fixed parameter tractable when parameterized by the number of variables [lenstra], we conclude that our problem is FPT when parameterized by the neighbourhood diversity $k$ .

ILP formulation: Our goal here is to find a largest globally minimal defensive alliance $S$ of $G$ , with $C_{i}\cap S=\emptyset$ when $C_{i}\in I_{0}$ , $|C_{i}\cap S|=1$ when $C_{i}\in{I_{1}}$ , and $|C_{i}\cap S|\geq 2$ when $C_{i}\in{I_{2}}$ where $I_{0},I_{1},I_{2}$ are given. For each $C_{i}$ , we associate a variable $x_{i}$ that indicates $|S\cap C_{i}|=x_{i}$ . As the vertices in $C_{i}$ have the same neighbourhood, the variables $x_{i}$ determine $S$ uniquely, up to isomorphism. The objective here is to maximize $\sum\limits_{C_{i}\in I_{1}\cup I_{2}}{x_{i}}$ under the conditions given below. Let $\mathcal{C}$ be a subset of $I_{1}\cup I_{2}$ consisting of all type classes which are cliques; and $\mathcal{I}={I_{1}\cup I_{2}}\setminus\mathcal{C}$ . We consider two cases:

Case 1: Suppose $v\in C_{j}$ where $C_{j}\in\mathcal{I}$ . Then the degree of $v$ in $S$ , that is,

\displaystyle d_{S}(v)

\displaystyle=\sum\limits_{C_{i}\in N_{H}(C_{j})\cap{(I_{1}\cup I_{2})}}{x_{i}}

(1)

Thus, including itself, $v$ has $1+\sum\limits_{C_{i}\in N_{H}(C_{j})\cap{(I_{1}\cup I_{2})}}{x_{i}}$ defenders in $G$ . Note that if $C_{i}\in I_{1}\cup I_{2}$ , then only $x_{i}$ vertices of $C_{i}$ are in $S$ and the the remaining $n_{i}-x_{i}$ vertices of $C_{i}$ are outside $S$ . The number of neighbours of $v$ outside $S$ , that is,

\displaystyle d_{S^{c}}(v)

\displaystyle=\sum\limits_{C_{i}\in N_{H}(C_{j})\cap{(I_{1}\cup I_{2})}}{(n_{i}-x_{i})}+\sum\limits_{C_{i}\in N_{H}(C_{j})\cap I_{0}}{n_{i}}

(2)

Therefore, a vertex $v$ from an independent type class $C_{j}\in\mathcal{I}$ is protected if and only if $1+\sum\limits_{C_{i}\in N_{H}(C_{j})\cap{(I_{1}\cup I_{2})}}{x_{i}}\geq\sum\limits_{C_{i}\in N_{H}(C_{j})\cap{(I_{1}\cup I_{2})}}{(n_{i}-x_{i})}+\sum\limits_{C_{i}\in N_{H}(C_{j})\cap{I_{0}}}{n_{i}}$ .
Case 2: Suppose $v\in C_{j}$ where $C_{j}\in\mathcal{C}$ . The number of neighbours of $v$ in $S$ , that is,

\displaystyle d_{S}(v)=(x_{j}-1)+\sum\limits_{C_{i}\in N_{H}(C_{j})\cap(I_{1}\cup I_{2})}{x_{i}}

(3)

This is to ensure that when $v$ is picked in the solution it contributes to the $x_{j}$ value and hence it itself cannot be accounted as its own neighbour. The number of neighbours of $v$ outside $S$ , that is,

\displaystyle d_{S^{c}}(v)=\sum\limits_{C_{i}\in N_{H}[C_{j}]\cap(I_{1}\cup I_{2})}{(n_{i}-x_{i})}+\sum\limits_{C_{i}\in N_{H}[C_{j}]\cap{I_{0}}}{n_{i}}

(4)

Thus a vertex $v$ from clique type class $C_{j}\in\mathcal{C}$ is protected if and only if $d_{S}(v)+1\geq d_{S^{c}}(v)$ , that is, $\sum\limits_{C_{i}\in N_{H}[C_{j}]\cap(I_{1}\cup I_{2})}{x_{i}}\geq\sum\limits_{C_{i}\in N_{H}[C_{j}]\cap(I_{1}\cup I_{2})}{(n_{i}-x_{i})}+\sum\limits_{C_{i}\in N_{H}[C_{j}]\cap I_{0}}{n_{i}}$ .

Let ${\bf x}=(x_{1},\ldots,x_{k})$ be the vector corresponds to the set $S\subseteq V(G)$ . We want to make sure that the vector ${\bf x}=(x_{1},\ldots,x_{k})$ or the set $S$ forms a defensive alliance, but no proper subset of $S$ forms defensive alliance. We now characterize all proper subsets of $S$ in terms of $k$ -length vectors. We define a new variable $y_{i}$ as follows: $0<y_{i}<x_{i}$ for all $i$ . Let $L({\bf x})$ be the set of all length $k$ vectors where the $i$ th entry be either $0$ , $y_{i}$ or $x_{i}$ . Note that each vector in $L({\bf x})$ represents a proper subset of $S$ unless the $i$ th entry is $x_{i}$ for all $i$ . The number of vectors in $L({\bf x})$ is $\prod\limits_{i=1}^{k}{(x_{i}+1)}$ . We define another set $L^{\prime}({\bf x})$ as follows: let $L^{\prime}({\bf x})$ be the set of all length $k$ vectors where the $i$ th entry is either $0$ , $x_{i}-1$ or $x_{i}$ ; note that $x_{i}-1$ is possible only if $x_{i}\geq 2$ , that is, $C_{i}\in I_{2}$ . Clearly, $L^{\prime}({\bf x})\subseteq L({\bf x})$ and $L^{\prime}({\bf x})$ has at most $3^{k}$ vectors.

Lemma 2.

Let ${\bf x}=(x_{1},\ldots,x_{k})$ be the vector that represent $S\subseteq V(G)$ . If no vector in $L^{\prime}({\bf x})$ forms a defensive alliance then no vector in $L({\bf x})$ forms a defensive alliance.

Proof.

Assume, for the sake of contradiction, that the vector ${\bf x}_{1}\in L({\bf x})$ forms a defensive alliance. Without loss of generality, let ${\bf x}_{1}=(y_{1},x_{2}-1,y_{3},\ldots,x_{k})$ , then we obtain the vector ${\bf x}^{\prime}_{1}=(x_{1}-1,x_{2}-1,x_{3}-1,\ldots,x_{k})\in L^{\prime}({\bf x})$ from ${\bf x}_{1}$ by replacing $y_{i}$ by $x_{i}-1$ for all $i$ . As ${\bf x}^{\prime}_{1}\in L^{\prime}({\bf x})$ , we know ${\bf x}^{\prime}_{1}$ does not form a defensive alliance. This means, there is a vertex $u\in C_{i}$ which is not protected in ${\bf x}^{\prime}_{1}$ (assume that the $i$ th entry of ${\bf x}^{\prime}_{1}$ is non-zero). We observe that the number of neighbours of $u$ in ${\bf x}_{1}$ , is less than or equal to the number of neighbours in ${\bf x}^{\prime}_{1}$ . In other words, $u$ is not protected in ${\bf x}_{1}$ either, a contradiction to that assumption that ${\bf x}_{1}\in S$ forms a defensive alliance. This proves the lemma.

In order to ensure that $S$ is a globally minimal defensive alliance, we check ${\bf x}=(x_{1},x_{1},\ldots,x_{k})$ forms a defensive alliance but none of the vectors in $L^{\prime}({\bf x})$ forms a defensive alliance. Let ${\bf x}^{\prime}_{1},{\bf x}^{\prime}_{2},\ldots,{\bf x}^{\prime}_{3^{k}}$ be the vectors in $L^{\prime}({\bf x})$ . We make guesses in two phases. In the first phase, we guess if $C_{i}$ belongs to $I_{0},I_{1}$ or $I_{2}$ . There are at most $3^{k}$ possibilities as each $C_{i}$ has three options: either $I_{0},I_{1}$ or $I_{2}$ . In the second phase, we guess if an unprotected vertex of ${\bf x}^{\prime}_{i}$ belongs to type class either $C_{1},C_{2},\ldots,C_{k-1}$ or $C_{k}$ . We define

R_{j}=\Big{\{}{\bf x}^{\prime}_{i}\in L^{\prime}({\bf x})~{}|~{}\text{ an unprotected vertex of }{\bf x}^{\prime}_{i}\text{ is in type class }C_{j}\Big{\}}.

There are at most $k^{3^{k}}$ possibilities as each ${\bf x}^{\prime}_{i}$ has at most $k$ options: $R_{1},R_{2},\ldots,R_{k}$ . If $C_{j}$ is an independent type class, then it contains an unprotected vertex if,

1+\sum\limits_{C_{i}\in N_{H}(C_{j})\cap(I_{1}\cup I_{2})}{2x_{i}^{\prime}}<\sum\limits_{C_{i}\in N_{H}(C_{j})}{n_{i}}.

If $C_{j}$ is a clique type class, then it contains an unprotected vertex if,

\sum\limits_{C_{i}\in N_{H}[C_{j}]\cap(I_{1}\cup I_{2})}{2x_{i}^{\prime}}<\sum\limits_{C_{i}\in N_{H}[C_{j}]}{n_{i}}.

We now formulate ILP formulation of globally minimal defensive alliance problem, for given $I_{0},I_{1},I_{2}$ and $R_{1},R_{2},\ldots,R_{k}$ . There are at most $3^{k}k^{3^{k}}$ ILPs:

$\begin{split}&\text{Maximize }\sum\limits_{C_{i}\in I_{1}\cup I_{2}}{x_{i}}\\ &\text{Subject to~{}~{}~{}}\\ &x_{i}=1\text{ for all }i~{}:~{}C_{i}\in I_{1};\\ &x_{i}\in\{2,\ldots,|C_{i}|\}\text{ for all }i~{}:~{}C_{i}\in I_{2}\\ &1+\sum\limits_{C_{i}\in N_{H}(C_{j})\cap(I_{1}\cup I_{2})}{2x_{i}}\geq\sum\limits_{C_{i}\in N_{H}(C_{j})}{n_{i}},~{}~{}\text{for all }C_{j}\in\mathcal{I},\\ &\sum\limits_{C_{i}\in N_{H}[C_{j}]\cap(I_{1}\cup I_{2})}{2x_{i}}\geq\sum\limits_{C_{i}\in N_{H}[C_{j}]}{n_{i}},~{}~{}\text{for all }C_{j}\in\mathcal{C},\\ &\text{ for }j=1\text{ to }k;\\ &1+\sum\limits_{C_{i}\in N_{H}(C_{j})\cap(I_{1}\cup I_{2})}{2x_{i}^{\prime}}<\sum\limits_{C_{i}\in N_{H}(C_{j})}{n_{i}},\forall~{}{\bf x}^{\prime}_{i}\in R_{j};C_{j}\text{ is an independent class}\\ &\sum\limits_{C_{i}\in N_{H}[C_{j}]\cap(I_{1}\cup I_{2})}{2x_{i}^{\prime}}<\sum\limits_{C_{i}\in N_{H}[C_{j}]}{n_{i}},\forall~{}{\bf x}^{\prime}_{i}\in R_{j};C_{j}\text{ is a clique class}\\ \end{split}$

Solving the ILP: Lenstra [lenstra] showed that the feasibility version of $p$ -ILP is FPT with running time doubly exponential in $p$ , where $p$ is the number of variables. Later, Kannan [kannan] proved an algorithm for $p$ -ILP running in time $p^{O(p)}$ . In our algorithm, we need the optimization version of $p$ -ILP rather than the feasibility version. We state the minimization version of $p$ -ILP as presented by Fellows et. al. [fellows].

$p$ -Variable Integer Linear Programming Optimization ( $p$ -Opt-ILP): Let matrices $A\in\ Z^{m\times p}$ , $b\in\ Z^{p\times 1}$ and $c\in\ Z^{1\times p}$ be given. We want to find a vector $x\in\ Z^{p\times 1}$ that minimizes the objective function $c\cdot x$ and satisfies the $m$ inequalities, that is, $A\cdot x\geq b$ . The number of variables $p$ is the parameter. Then they showed the following:

Lemma 3.

[fellows] $p$ -Opt-ILP can be solved using $O(p^{2.5p+o(p)}\cdot L\cdot log(MN))$ arithmetic operations and space polynomial in $L$ . Here $L$ is the number of bits in the input, $N$ is the maximum absolute value any variable can take, and $M$ is an upper bound on the absolute value of the minimum taken by the objective function.

In the formulation for Globally Minimal Defensive Alliance problem, we have at most $k$ variables. The value of objective function is bounded by $n$ and the value of any variable in the integer linear programming is also bounded by $n$ . The constraints can be represented using $O(k^{2}\log{n})$ bits. Lemma 3 implies that we can solve the problem with one guess in FPT time. There are at most $3^{k}k^{3^{k}}$ guesses, and the ILP formula for each guess can be solved in FPT time. Thus Theorem 1 holds.

4 W[1]-hardnes parameterized by treewidth

Theorem 4.

The Globally Minimal Defensive Alliance problem is W[1]-hard when parameterized by the treewidth, pathwidth, treedepth and feedback vertex set of the graph.

Proof.

The approach for using Multi-Colored Clique in reductions is described in [?], and has been proven to be very useful in showing hardness results in the parameterized complexity setting. Before giving details of our construction, we will need to introduce some new terminology. We use G to denote a graph colored with $k$ colors given in an instance of Multi-Colored Clique, and $G^{\prime}$ to denote the graph in the reduced instance of Globally minimal defensive alliance. For a color $c\in[k]$ , we let $V_{c}$ denote the subset of vertices in $G$ colored with color $c$ and for a pair of distinct colors $c_{1},c_{2}\in[k]$ , we let $E_{c_{1},c_{2}}$ denote the subset of edges in $G$ with endpoints colored $c_{1}$ and $c_{2}$ . In general, we use $u$ and $v$ for denoting arbitrary vertices in $G$ , and $x$ to denote an arbitrary vertex in $G^{\prime}$ .

We construct $G^{\prime}$ using two types of gadgets. Our goal is to guarantee that any globally minimal defensive alliance in $G^{\prime}$ with a specific size encodes a multi-colored clique in $G$ . These gadgets are the selection and validation gadgets. The selection gadgets encode the selection of $k$ vertices and $k\choose 2$ edges that together encode a vertex and edge set of some multi-colored clique in $G$ . The selection gadgets also ensure that in fact $k$ distinct vertices are chosen from $k$ distinct color classes, and that distinct $k\choose 2$ edges are chosen from $k\choose 2$ distinct edge color classes. The validation gadgets validate the selection done in the selection gadgets in the sense that they make sure that the edges chosen are in fact incident to the selected vertices. In the following we sketch the construction of these gadgets:

Selection: For each color-class $c\in[k]$ , and each pair of distinct colors $c_{1},c_{2}\in[k]$ , we construct a $c$ -selection gadget and a $\{c_{1},c_{2}\}$ -selection gadget which respectively encode the selection of a vertex colored $c$ and an edge colored $\{c_{1},c_{2}\}$ in $G$ . The $c$ -selection gadget consists of a vertex $x_{v}$ for every vertex $v\in V_{c}$ , and likewise, the $\{c_{1},c_{2}\}$ -selection gadget consists of a vertex $x_{\{u,v\}}$ for every edge $\{u,v\}\in E_{\{c_{1},c_{2}\}}$ . There are no edges between the vertices of the selection gadgets, that is, the union of all vertices in these gadgets is an independent set in $G^{\prime}$ .

Validation: We assign to every vertex $v$ in $G$ two unique identification numbers, $\text{low}(v)$ and $\text{high}(v)$ , with $\text{low}(v)\in[n]$ and $\text{high}(v)=2n-\text{low}(v)$ . For every pair of distinct colors $c_{1},c_{2}\in[k]$ , we construct validation gadgets between the $\{c_{1},c_{2}\}$ -selection gadget and the $c_{1}$ - and $c_{2}$ -selection gadget. Let $c_{1}$ and $c_{2}$ be any pair of distinct colors. We describe the validation gadget between the $c_{1}$ - and ${c_{1},c_{2}}$ -selection gadgets. It consists of two vertices $\{\lambda^{1}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{2}_{\{c_{1},(c_{1},c_{2})\}}\}$ , the validation-pair of this gadget. The first vertex $\lambda^{1}_{\{c_{1},(c_{1},c_{2})\}}$ of this pair is connected to each vertex $x_{v},v\in V_{c_{1}}$ , by $\text{low}(v)$ parallel edges, and to each edge-selection vertex $x_{\{u,v\}},\{u,v\}\in E_{\{c_{1},c_{2}\}}$ and $v\in V_{c_{1}}$ , by $\text{high}(v)$ parallel edges. The other $\lambda^{2}_{\{c_{1},(c_{1},c_{2})\}}$ vertex is connected to each $x_{v},v\in V_{c_{1}}$ , by $\text{high}(v)$ parallel edges, and to each $x_{\{u,v\}},\{u,v\}\in E_{\{c_{1},c_{2}\}}$ and $v\in V_{c_{1}}$ , by $\text{low}(v)$ parallel edges. We next subdivide the edges between the selection and validation gadgets to obtain a simple graph, where all new vertices introduced by the subdivision are referred to as the connection vertices. The connection vertices adjacent to $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ and $x_{v}$ are denoted by $Y_{\{x_{v},\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}\}}$ and similarly the connection vertices adjacent to $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ and $x_{\{u,v\}}$ are denoted by $Y_{\{x_{\{u,v\}},\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}\}}$ .

We next add two one degree vertices adjacent to all connection vertices. For every vertex $x_{v}$ in $c$ -selection gadget and $x_{\{u,v\}}$ in $\{c_{1},c_{2}\}$ -selection gadget, we add equal number of one degree vertices adjacent to $x_{v}$ and $x_{\{u,v\}}$ as much as the degree of $x_{v}$ and $x_{\{u,v\}}$ until this point in construction respectively. Next, we add the following gadget corresponding to every vertex $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ in validation pair gadget for all distinct colors $c_{1},c_{2}\in[k]$ and $1\leq j\leq 2$ . Let us assume that the degree of vertex $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ until this point in construction is $d$ . First, we add a set $V_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}^{\triangle}$ of $N$ vertices adjacent to $\lambda^{j}_{\{c_{1},c_{2}\}}$ and to a new vertex $\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$ . We make the vertex $\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$ adjacent to a set $V_{\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}}}^{\square}$ of $N$ one degree vertices. We also make every vertex in $V_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}^{\triangle}$ adjacent to two one degree vertices. Next, we introduce two vertices $\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}$ and $\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$ . Next add an edge between $\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}$ and $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ . Again, we add a new set $V_{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}^{\triangle}$ of $N$ vertices adjacent to $\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}$ and $\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$ . Similar to before, We make every vertex in $V_{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}^{\triangle}$ adjacent to two one degree vertices. We also make the vertex $\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$ and $\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}$ adjacent to a set $V_{\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}}}^{\square}$ and $V_{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}^{\square}$ of $N$ and $N+1$ one degree vertices respectively. Finally, for the vertex $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ , we add a set $V_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}^{\square}$ of $N+4n-d+1$ many one degree vertices adjacent to $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ . We define $N=100n^{2}$ . This completes the construction of graph $G^{\prime}$ . We set $k^{\prime}=(4N+8).2.{k\choose 2}+{k\choose 2}+k$ . We observe that removing the vertices

	$\displaystyle R=$	$\displaystyle\{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}},$
		$\displaystyle\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{2},(c_{1},c_{2})\}}$
		$\displaystyle\|1\leq j\leq 2,\ \text{distinct colors}\ \ c_{1},c_{2}\in[k]\}$

of $8.{k\choose 2}$ from graph $G^{\prime}$ , we are left with trees of height at most $3$ . Now, we will prove that both instances are equivalent. Let us assume that there exists a multi-colored clique $C$ of size $k$ . We claim that the following set

	$\displaystyle S$	$\displaystyle=\{x_{v},x_{\{v_{i},v_{j}\}}~{}\|~{}v,v_{i},v_{j}\in C\}$
		$\displaystyle\bigcup\limits_{1\leq j\leq 2}\{Y_{\{x_{v},\lambda^{j}_{\{c_{1},c_{2}\}}\}},Y_{\{x_{\{v_{i},v_{j}\}},\lambda^{j}_{\{c_{1},c_{2}\}}\}}~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2},\{v,v_{i},v_{j}\}\in C\}$
		$\displaystyle\bigcup\limits_{1\leq j\leq 2}\{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$
		$\displaystyle V^{\triangle}_{{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}}~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2}\}$
		$\displaystyle\bigcup\limits_{1\leq j\leq 2}\{\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{2},(c_{1},c_{2})\}}$
		$\displaystyle V^{\triangle}_{{\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}}}}~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2}\}$

is a globally minimal defensive alliance of size at exactly $k^{\prime}$ . We observe that $|S|=k^{\prime}$ . To prove that $S$ is a globally minimal defensive alliance, we will prove that $S$ is a connected defensive alliance such that every vertex is marginally protected. It is easy to see every vertex in the set $\{x_{v},x_{\{v_{i},v_{j}\}}~{}|~{}v,v_{i},v_{j}\in C\}$ is marginally protected as it has all the connection vertices adjacent to it inside the solution and equal number of one degree neighbours outside the solution. It is also easy to see that all the vertices in the set

	$\displaystyle\bigcup\limits_{1\leq j\leq 2}$	$\displaystyle\{Y_{\{x_{v},\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}\}},Y_{\{x_{\{v_{i},v_{j}\}},\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}\}},Y_{\{x_{v},\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}\}},Y_{\{x_{\{v_{i},v_{j}\}},\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}\}}$
		$\displaystyle~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2},\{v,v_{i},v_{j}\}\in C\}$

are marginally protected as they have two neighbours inside the solution and two neighbours outside the solution. Similarly, we observe that all the vertices in the set

	$\displaystyle\bigcup\limits_{1\leq j\leq 2}$	$\displaystyle\{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}},V^{\triangle}_{{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}},$
		$\displaystyle\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{2},(c_{1},c_{2})\}},V^{\triangle}_{{\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}}}}$
		$\displaystyle~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2}\}$

are also marginally protected. Lastly, we prove that the vertices in the set $\bigcup\limits_{1\leq j\leq 2}\{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}~{}|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2}\}$ are marginally protected. Let us take any vertex $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ for some fixed $c_{1}$ and $c_{2}$ . We denote degree of $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ inside the $c_{1}$ -selection gadget, $c_{2}$ -selection gadget and $\{c_{1},c_{2}\}$ -selection gadget together by $d$ . Without loss of generality, we assume that the neighbours $x_{u}$ and $x_{\{u,v\}}$ of $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ are inside the solution where $u\in V_{c_{1}}$ and $v\in V_{c_{2}}$ . It implies that the vertices in the set $Y_{\{x_{u},\lambda^{j}_{\{c_{1},c_{2}\}}\}}$ and $Y_{\{x_{\{u,v\}},\lambda^{j}_{\{c_{1},c_{2}\}}\}}$ are inside the solution. Therefore, $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ has total $2n$ connection vertices neighbours inside the solution as $\text{high}(u)+\text{low}(n)=2n$ . Since $V^{\triangle}_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}\subseteq S$ and $V^{\square}_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}\cap S=\emptyset$ , we have $d_{S}(\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}})=2n+1+N$ and $d_{S^{c}}(\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}})=(d-2n)+(N+4n-d+2)=N+2n+1$ , therefore the vertex $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ is marginally protected. It is easy to observe that $S$ is connected. This shows that $S$ is a globally minimal defensive alliance.

In the reverse direction, we assume that there exists a globally minimal defensive alliance of size at least $k^{\prime}$ . First, We observe that no vertex of degree 1 can be part of globally minimal defensive alliance of size greater than or equal to two as the vertex itself forms defensive alliance. Next, we will prove that

	$\displaystyle\bigcup\limits_{1\leq j\leq 2}$	$\displaystyle\{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}},V^{\triangle}_{{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}}$
		$\displaystyle\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{2},(c_{1},c_{2})\}},V^{\triangle}_{{\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}}}}$
		$\displaystyle~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2}\}\subseteq S.$

If $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}\not\in S$ or $\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}\not\in S$ then we cannot add any vertex from the set $V^{\triangle}_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}$ inside the solution. If we cannot add the set $V^{\triangle}_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}$ then $|S|\leq k^{\prime}$ . This is true because every globally minimal defensive alliance is always connected and if we add any vertex from $V^{\triangle}_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}$ inside the solution then $|S|\leq N+1$ which is a contradiction. Next, we observe that protection of $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ require at least one vertex from the set $V^{\triangle}_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}$ inside the solution. As every vertex in $V^{\triangle}_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}$ has two one degree neighbours, it implies that the protection of that vertex requires $\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$ inside the solution. Now, the protection of $\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$ forces the full set $V^{\triangle}_{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}$ inside the solution as the one degree neighbours are always outside $S$ . Similarly, we observe that $V^{\triangle}_{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}$ and $\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}$ will be inside the solution. Symmetrically, we argue for $c_{2}$ color as well. This proves that

	$\displaystyle\bigcup\limits_{1\leq j\leq 2}$	$\displaystyle\{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}},V^{\triangle}_{{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}}$
		$\displaystyle\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{2},(c_{1},c_{2})\}},V^{\triangle}_{{\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}}}}$
		$\displaystyle~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2}\}\subseteq S.$

Observe that the above set has size exactly equal to $(4N+8).2.{k\choose 2}$ . We need to add at least $k+{k\choose 2}$ vertices from selection gadgets. Next, we observe that $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ must be marginally protected inside the solution as otherwise $S\setminus\{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}},V^{\triangle}_{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}\}$ will form defensive alliance. This is equivalent to say that the protection of $\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}$ requires exactly $2n$ neighbours from connection vertices inside the solution corresponding to the set $c_{1}$ -selection gadget and $\{c_{1},c_{2}\}$ -selection gadget. Now, we will prove that every vertex or edge selection gadget contributes at most one vertex inside the solution. Let us assume that there exists a vertex or edge selection gadget which contributes at least two vertices inside the solution. Without loss of generality, if $x_{u_{1}}$ and $x_{u_{2}}$ are inside the solution where $u_{1},u_{2}\in c_{1}$ . It implies that the protection of $x_{u_{1}}$ and $x_{u_{2}}$ requires $Y_{\{x_{u_{1}},\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}\}}$ and $Y_{\{x_{u_{2}},\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}\}}$ inside the solution. Note that $\lambda^{2}_{\{c_{2},(c_{1},c_{2})\}}$ will get more than $2n$ connection vertex neighbours inside the solution as $\text{high}(u_{1})+\text{high}(u_{2})>2n$ . This is a contradiction as $\lambda^{2}_{\{c_{2},(c_{1},c_{2})\}}$ is not marginally protected inside $S$ . We can argue similarly for edge selection gadget and other color classes as well. Since the contribution from selection gadgets is at least $k+{k\choose 2}$ , it implies that each selection gadget contributes exactly one vertex inside the solution. Next, we will show that if $c_{1}$ -selection gadget contributes $x_{u}$ and $c_{2}$ -selection gadget contributes $x_{v}$ then $c_{1},c_{2}$ -selection gadget must contributes $x_{\{u,v\}}$ . For the sake of contradiction assume that $c_{1},c_{2}$ -selection gadget contributes $x_{\{u^{\prime},v^{\prime}\}}$ such that $u\neq u^{\prime}$ . In this case, we get

	$\displaystyle\{$	$\displaystyle Y_{\{x_{u},\lambda^{j}_{\{c_{1},(c_{1},c_{2})}\}},Y_{\{x_{v},\lambda^{j}_{\{c_{2},(c_{1},c_{2})}\}},Y_{\{x_{\{u^{\prime},v^{\prime}\}},\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}\}},Y_{\{x_{\{u^{\prime},v^{\prime}\}},\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}\}}$
		$\displaystyle~{}\|~{}1\leq j\leq 2\}\subseteq S.$

We observe that $|Y_{\{x_{u},\lambda^{1}_{\{c_{1},(c_{1},c_{2})\}}\}}|=\text{low}(u),|Y_{\{x_{u},\lambda^{2}_{\{c_{1},(c_{1},c_{2})\}}\}}|=\text{high}(u),\\ |Y_{\{x_{v},\lambda^{1}_{\{c_{2},(c_{1},c_{2})\}}\}}|=\text{low}(v),|Y_{\{x_{v},\lambda^{2}_{\{c_{2},(c_{1},c_{2})\}}\}}|=\text{high}(v),\\ |Y_{\{x_{\{u,v\}},\lambda^{1}_{\{c_{1},(c_{1},c_{2})\}}\}}|=\text{high}(u),|Y_{\{x_{\{u,v\}},\lambda^{2}_{\{c_{1},(c_{1},c_{2})\}}\}}|=\text{low}(u),|Y_{\{x_{\{u,v\}},\lambda^{1}_{\{c_{2},(c_{1},c_{2})\}}\}}|=\text{high}(v),|Y_{\{x_{\{u,v\}},\lambda^{2}_{\{c_{2},(c_{1},c_{2})\}}\}}|=\text{low}(v)$ . We see that one of the vertex from the set $\{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}~{}|~{}1\leq j\leq 2\}$ is not protected because when $u\neq u^{\prime}$ either high $(u)$ +low $(u^{\prime})<2n$ or low(u)+high $(u^{\prime})<2n$ . This is a contradiction. It implies that we get a multicolored clique in $G$ .

Theorem 5.

The Exact Globally Minimal Defensive Alliance problem is W[1]-hard when parameterized by the treewidth, pathwidth, treedepth and feedback vertex set of the graph.

5 No polynomial kernel parameterized by vertex cover

Theorem 6.

No polynomial kernel parameterized by vertex cover number of input graph.

To prove Theorem LABEL:ppt, we give a polynomial parameter transformation from the well-known Red Blue Dominating Set problem (RBDS) to Globally Minimal Defensive Alliance problem parameterized by vertex cover number. Recall that in RBDS we are given a bipartite graph $G=(T\cup S,E)$ and an integer $k$ , and we are asked whether there exists a vertex set $X\subseteq S$ of size at most $k$ such that every vertex in $T$ has at least one neighbour in $X$ . We also refer to the vertices of $T$ as terminals and to the vertices of $S$ as sources or nonterminals. The following theorem is known:

Theorem 7.

[fomin_lokshtanov_saurabh_zehavi_2019] RBDS parameterized by $|T|$ does not admit a polynomial compression unless coNP $\subseteq$ NP/poly.

5.1 Proof of Theorem LABEL:ppt

By Theorem 7, RBDS parameterized by $|T|$ does not admit a polynomial compression unless coNP $\subseteq$ NP/poly. To prove Theorem LABEL:ppt, we give a PPT from RBDS parameterized by $|T|$ to Globally Minimal Defensive Alliance parameterized by the vertex cover number. Given an instance $I=(G=(T\cup S,E),k)$ of RBDS, we construct an instance $I^{\prime}=(G^{\prime},k^{\prime})$ of Globally Minimal Defensive Alliance as follows. First, we take the graph $G$ . We add two vertices $x$ and $x^{\prime}$ such that $x$ is adjacent to all the vertices in $S$ . We also add a set $V_{x}^{\triangle}$ of $4n$ vertices adjacent to both $x$ and $x^{\prime}$ . We add two more sets $V_{x}^{\square}$ and $V_{x^{\prime}}^{\square}$ of $(|S|-k)+4n$ and $4n+1$ vertices adjacent to $x$ and $x^{\prime}$ respectively. Next, for every vertex $u\in T$ , we add two vertices $u_{1},u_{2}$ and a set $V_{u}^{\triangle}$ of $4n$ vertices. We make both $u_{1}$ and $u_{2}$ adjacent to all the vertices in $V_{u}^{\triangle}$ . Next, we add two sets $V_{u_{1}}^{\square}$ and $V_{u_{1}}^{\square}$ of size $4n+1$ and $4n$ adjacent to $u_{1}$ and $u_{2}$ respectively. Finally, we add an edge $(u,u_{1})$ . Next, we add two vertices $a$ and $b$ . We make $b$ adjacent to all the vertices in $T$ and the vertex $x^{\prime}$ . We add another set $V_{b}^{\square}$ of $|T|+1$ many vertices adjacent to $b$ . We also make $a$ adjacent to all the vertices in $T$ . Let $d_{S}(u)$ denote the neighbours of $u$ in set $S$ for all $u\in T$ . For every $u\in T$ , we add a set $V_{u}^{a}=\{u_{1}^{a},u_{2}^{a},\ldots,u_{d_{S}(u)-1}^{a}\}$ of $d_{S}(u)-1$ vertices adjacent to $u$ and $a$ . We also add a set of one degree vertices $V_{a}^{\square}$ of size $|T|+\sum\limits_{u\in T}(d_{S}(u)-1)+2$ . We make $a$ adjacent to all the vertices in $V_{a}^{\square}$ . Finally, we add a set $T^{\square}=T_{1}^{\square}\cup T_{2}^{\square}$ where $|T_{1}^{\square}|=|T|+1$ and $|T_{2}^{\square}|=2$ . We make every vertex in the set $\bigcup V_{x}^{\triangle}$ adjacent to two arbitrary vertices in the set $T_{1}^{\square}$ . Similarly, we make every vertex in the set $\bigcup V_{u}^{\triangle}$ adjacent to both the vertices in the set $T_{2}^{\square}$ . For every vertex in $s\in S$ , we make $s$ adjacent to $d_{T}(s)+1$ many arbitrary vertices in $T^{\square}$ . Now for every vertex $t\in T^{\square}$ , we add $d+2$ many one degree vertices adjacent to $t$ where $d$ is the degree of $t$ until this point in the construction. Finally, we set $k^{\prime}=(|S|-k)+4n+2+|T|.(4n+3)+1$ . Now, we will show that $I$ is a yes instance if and only if $I^{\prime}$ is a yes instance. Let us assume that $I$ is a yes instance, that is, there exists a set $X$ of size at most $k$ in $S$ such that every vertex $u\in T$ have at least one neighbour in $X$ . Without loss of generality, we can assume that $|X|=k$ . Let $d_{X}(u)$ denotes the number of neighbours of $u$ in $X$ . We claim that the set $H=S\setminus X\bigcup\limits_{u\in T}(\{u,u_{1},u_{2}\}\cup V_{u}^{\triangle})\bigcup V_{x}^{\square}\bigcup\limits_{u\in T}\{u_{1}^{a},\ldots,u_{d_{X}(u)-1}^{a}\}\bigcup\{b,x,x^{\prime}\}$ is a globally minimal defensive alliance. It is clear that $|H|\geq k^{\prime}$ . First, we observe that every vertex in $H$ is marginally protected. We also see that $G^{\prime}[H]$ is connected. This implies that $H$ is a globally minimal defensive alliance.

In the reverse direction, we assume that there exists a globally minimal defensive alliance $H$ of size at least $k^{\prime}$ . First, we observe that no vertex of degree one can be part of $H$ as one degree vertex forms a defensive alliance. As the one degree vertices are not part of the solution, we observe that no vertex from $T^{\square}$ is part of the solution. This is true because we will not be able to protect any vertex from $T^{\square}$ as more than half of the neighbours are one degree vertices. Next, we see that $T\subseteq H$ . For the sake of contradiction assume that there exists some $u\in T$ such that $u\not\in H$ . We observe that, we cannot use $V_{u}^{\triangle}\cup\{u_{1},u_{2}\}$ inside $H$ as the solution cannot be connected and if we do not add set $V_{u}^{\triangle}$ inside $H$ then $H$ cannot achieve size $k^{\prime}$ . Therefore, we must include $u$ in $H$ . From above, we also see that we must add $u_{1}$ and $u_{2}$ inside $H$ as otherwise vertices in $V_{u}^{\triangle}$ cannot be protected. Therefore, we have $\bigcup\limits_{u\in T}(V_{u}^{\triangle}\cup\{u,u_{1},u_{2}\})\subseteq H$ . Similarly, we can argue that $V_{x}^{\triangle}\cup\{x,x^{\prime},b\}\subseteq H$ . Next, we see that $u\in T$ must be marginally protected in $H$ as otherwise $H\setminus(V_{u}^{\triangle}\cup\{u_{1},u_{2}\})$ forms a defensive alliance, which is not possible. For a vertex $u\in T$ , we have $d(u)=2d_{S}(u)+2$ and $N(u)=N_{S}(u)\cup V_{u}^{a}\cup\{a,b,u_{1}\}$ . Since $a\not\in H$ and $b,u_{1}\in H$ , we must have added at most $d_{S}(u)-1$ from set $N_{S}(u)$ for all $u\in T$ . Let us denote the set $H^{c}\cap S=X$ . Clearly, every vertex $u\in T$ have at least one neighbour in $X$ . Next, we see that $|X|\leq k$ . As otherwise, if $|S\cap H|\leq|S|-k$ then the vertex $x$ cannot be protected. This implies that $I$ is a yes instance.

Theorem 8.

Given a vertex $v$ , decide whether there exists a globally minimal defensive alliance containing vertex $v$ is a NP-complete problem.

Proof.

We will prove the above result by giving a polynomial reduction from r-regular clique problem. Let $I=(G,k)$ be any instance of clique problem. We construct an instance $I^{\prime}=(G^{\prime},x)$ of above problem.

Figure 2: Reduction from r-regular clique to globally minimal defensive alliance containing vertex

x

.

First, we add the vertex $x$ and a clique $K_{n-2k}$ of size $n-2k$ . We make every vertex in $K_{n-2k}$ adjacent to all the vertices in $G$ and the vertex $x$ . Next, we add a set $V_{x}^{\square}$ of $n-2k$ one degree vertices adjacent to $x$ . We also add a set $V_{u}^{\square}$ of $n-r-2k$ one degree vertices adjacent to $u$ for each $u\in V(G)$ . Finally, we add a set $V_{z}^{\square}$ of $2(n-k)$ one degree vertices adjacent to $z$ for each $z\in K_{n-2k}$ . This completes the construction of $G^{\prime}$ . Now, we will show that $I$ is yes instance if and only if $I^{\prime}$ is a yes instance. Let us assume that $I$ is a yes instance, that is, there exists a clique $C$ of size $k$ in graph $G$ . We claim that $S=C\cup K_{n-2k}\cup\{x\}$ is a globally minimal defensive alliance containing vertex $x$ in graph $G^{\prime}$ . We observe that all the vertices in $S$ are marginally protected in $S$ and $G[S]$ is connected. This implied that $S$ is a globally minimal defensive alliance in $G^{\prime}$ . In the reverse direction, let us assume that there exists a globally minimal defensive alliance $S$ containing vertex $x$ . First, we observe that the one degree vertices cannot part $S$ as any one degree vertex forms a defensive alliance. This implies that the protection of $x$ requires all the vertices of $K_{n-2k}$ in the solution. Therefore, we can assume that $K_{n-2k}\subseteq S$ . We observe that since one degree vertices cannot be part of solution, every vertex in $K_{n-2k}$ need at least $k$ vertices from $V(G)$ . We also see that, if we take more than $k$ vertices from $V(G)$ inside $S$ then all the vertices in $K_{n-2k}$ are overprotected. Then, we see that $S\setminus{x}$ is also defensive alliance. This is a contradiction as $S$ is globally minimal defensive alliance. This proves that $V(G)$ contributes exactly $k$ vertices in the solution. Lets denote this set by $C$ . Since $G$ is $r$ -regular, we see that every vertex $u\in G$ that is also part of the solution requires exactly $k-1$ neighbours from the $V(G)$ . This implies that $C$ is a clique of size exactly $k$ . This implies that $I$ is a yes instance.

Theorem 9.

Globally minimal defensive alliance problem is NP-complete even on planar graphs.

6 Conclusion

The main contributions in this paper are that the Globally Minimal Defensive Alliance problem is polynomial time solvable on trees, and the problem parameterized by treewidth is W[1]-hard. It would be interesting to consider the parameterized complexity with respect to parameters neighbourhood diversity, vertex cover and twin cover. The modular width parameter also appears to be a natural parameter to consider here; and since there are graphs with bounded modular-width and unbounded neighborhood diversity, we believe this is also an interesting open problem. The parameterized complexity of the Globally Minimal Defensive Alliance problem remains unsettle when parameterized by solution size and other important structural graph parameters like clique-width and modular width.

Acknowledgement

The first author gratefully acknowledges support from the Ministry of Human Resource Development, Government of India, under Prime Minister’s Research Fellowship Scheme (No. MRF-192002-211). The second author’s research is supported in part by the Science and Engineering Research Board (SERB), Govt. of India, under Sanction Order No. MTR/2018/001025.

	$\displaystyle S$	$\displaystyle=\{x_{v},x_{\{v_{i},v_{j}\}}~{}\|~{}v,v_{i},v_{j}\in C\}$
		$\displaystyle\bigcup\limits_{1\leq j\leq 2}\{Y_{\{x_{v},\lambda^{j}_{\{c_{1},c_{2}\}}\}},Y_{\{x_{\{v_{i},v_{j}\}},\lambda^{j}_{\{c_{1},c_{2}\}}\}}~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2},\{v,v_{i},v_{j}\}\in C\}$
		$\displaystyle\bigcup\limits_{1\leq j\leq 2}\{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{1},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{1},(c_{1},c_{2})\}}$
		$\displaystyle V^{\triangle}_{{\lambda^{j}_{\{c_{1},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{1},(c_{1},c_{2})\}}}}~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2}\}$
		$\displaystyle\bigcup\limits_{1\leq j\leq 2}\{\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{j\triangle}_{\{c_{2},(c_{1},c_{2})\}},\lambda^{\prime j\triangle}_{\{c_{2},(c_{1},c_{2})\}}$
		$\displaystyle V^{\triangle}_{{\lambda^{j}_{\{c_{2},(c_{1},c_{2})\}}}},V^{\triangle}_{{\lambda^{\prime j}_{\{c_{2},(c_{1},c_{2})\}}}}~{}\|~{}c_{1},c_{2}\in[k],c_{1}\neq c_{2}\}$

Parameterized Complexity of Globally Minimal Defensive Alliances

Abstract

keywords:

1 Introduction

2 Basic Notations

Definition 1.

Definition 2.

Definition 3.

Definition 4.

Observation 1.

Definition 5.

Definition 6.

3 FPT algorithm parameterized by neighbourhood diversity

Definition 7.

Theorem 1.

Lemma 2.

Proof.

Lemma 3.

4 W[1]-hardnes parameterized by treewidth

Theorem 4.

Proof.

Theorem 5.

5 No polynomial kernel parameterized by vertex cover

Theorem 6.

Theorem 7.

5.1 Proof of Theorem LABEL:ppt

Theorem 8.

Proof.

Theorem 9.

6 Conclusion

Acknowledgement