# Parameterized Complexity of Problems in Coalitional Resource Games

Coalition formation is a key topic in multi-agent systems. Coalitions enable agents to achieve goals that they may not have been able to achieve on their own. Previous work has shown problems in coalitional games to be computationally hard. Wooldridge and Dunne (Artificial Intelligence 2006) studied the classical computational complexity of several natural decision problems in Coalitional Resource Games (CRG) - games in which each agent is endowed with a set of resources and coalitions can bring about a set of goals if they are collectively endowed with the necessary amount of resources. The input of coalitional resource games bundles together several elements, e.g., the agent set Ag, the goal set G, the resource set R, etc. Shrot, Aumann and Kraus (AAMAS 2009) examine coalition formation problems in the CRG model using the theory of Parameterized Complexity. Their refined analysis shows that not all parts of input act equal - some instances of the problem are indeed tractable while others still remain intractable. We answer an important question left open by Shrot, Aumann and Kraus by showing that the SC Problem (checking whether a Coalition is Successful) is W[1]-hard when parameterized by the size of the coalition. Then via a single theme of reduction from SC, we are able to show that various problems related to resources, resource bounds and resource conflicts introduced by Wooldridge et al are 1. W[1]-hard or co-W[1]-hard when parameterized by the size of the coalition. 2. para-NP-hard or co-para-NP-hard when parameterized by |R|. 3. FPT when parameterized by either |G| or |Ag|+|R|.

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07/09/2020

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## 1 Introduction

### 1.1 Coalitions

In multi-agent systems (MAS), where each agent has limited resources, the formation of coalitions of agents is a very powerful tool [6]. Coalitions enable agents to accomplish goals they may not have been able to accomplish individually. As such, understanding and predicting the dynamics of coalitions formation, e.g., which coalitions are more beneficial and/or more likely to emerge, is a question of considerable interest in multi-agent settings. Unfortunately, a range of previous studies have shown that many of these problems are computationally complex [7, 8]. Nonetheless, as noted by Garey and Johnson [4]

, hardness results, such as NP-completeness, should merely constitute the beginning of the research. NP-hardness just indicates that a general solution for all instances of the problem most probably does not exist. Still, efficient solutions for important sub-classes may well exist.

### 1.2 Formal Model of Coalition Resource Games

The framework we use to model coalitions is the CRG model introduced in [8], defined as follows. The model contains a non-empty, finite set of agents. A coalition, typically denoted by , is simply a set of agents, i.e., a subset of . The grand coalition is the set of all agents, . There is also a finite set of goals . Each agent is associated with a subset  of the goals. Agent  is satisfied if at least one member of  is achieved, and unsatisfied otherwise. Achieving the goals requires the expenditure of resources, drawn from the total set of resource types . Achieving different goals may require different quantities of each resource type. The quantity denotes the amount of resource  required to achieve goal . It is assumed that is a non-negative integer. Each agent is endowed certain amounts of some or all of the resource types. The quantity denotes the amount of resource  endowed to agent . Again, it is assumed that is a non-negative integer. Formally, a Coalition Resource Game is a -tuple given by

 Γ=⟨Ag,G,R,G1,G2,…,Gn,en,req⟩

where:

• is the set of agents

• is the set of possible goals

• is the set of resources

• For each , is a subset of such that any of the goals in  would satisfy  but  is indifferent between the members of

• en : is the endowment function

• req : is the requirement function

The endowment function en extends to coalitions by summing up endowments of its members as

The requirement function req extends to sets of goals by summing up requirements of its members as

A set of goals  satisfies agent  if and satisfies a coalition  if it satisfies every member of . A set of goals  is feasible for coalition  if that coalition is endowed with sufficient resources to achieve all goals in , i.e., for all we have . Finally we say that a coalition is successful if there exists a non-empty set of goals  that satisfies  and is feasible for it. In general, we use the notation and is successful for . The CRG models many real-world situations like the virtual organizations problem [1] and voting domains.

## 2 Problem Definitions and Previous Work

### 2.1 Problems Related to Coalition Formation

Shrot et al. [5] considered the following four problems related to coalitions.

1. Successful Coalition (SC)
Instance: A CRG and a coalition
Question: Is successful?

2. Exists a Successful Coalition of Size (ESCK)
Instance: A CRG and an integer
Question: Does there exist a successful coalition of size exactly ?

3. Maximal Coalition (MAXC)
Instance: A CRG and a coalition
Question: Is every proper superset of not successful?

4. Maximal Successful Coalition (MAXS)
Instance: A CRG and a coalition
Question: Is successful and every proper superset of  not successful?

The results from Shrot et al. [5] are summarized in Figure 1.

In this work we consider the problems which were defined by Wooldridge et al. [8] but were not considered by Shrot et al. [5]. We define these problems in detail in the following sections.

1. Necessary Resource (NR)
Instance: A CRG , coalition and resource
Question: Is ?

2. Strictly Necessary Resource (SNR)
Instance: A CRG , coalition and resource
Question: Is and we have ?

3. -Optimality (CGRO)
Instance: A CRG , coalition , goal set and resource
Question: Is ?

4. R-Pareto Efficient Goal Set (RPEGS)
Instance: A CRG , coalition and a goal set
Question: Is R-Pareto Efficient for coalition ?

5. Successful Coalition With Resource Bound (SCRB)
Instance: A CRG , coalition and a resource bound b
Question: Does such that respects b?

6. Conflicting Coalitions (CC)
Instance: A CRG , coalitions and a resource bound b
Question: If and we have ?

## 3 Parameterized Complexity

We now provide a brief introduction to the key relevant concepts from the theory of parameterized complexity. The definitions in this section are taken from [3] and [2]. The core idea of parameterized complexity is to single out a specific part of the input as the parameter and ask whether the problem admits an algorithm that is efficient in all but the parameter. In most cases the parameter is simply one of the elements of the input (e.g., the size of the goal set), but it can actually be any computable function of the input:

###### Definition 3.1.

Let be a finite alphabet.

1. A parametrization of is a mapping that is polynomial time computable.

2. A parameterized problem (over ) is a pair consisting of a set of strings over  and a parameterization  of .

As stated, given a parameterized problem we seek an algorithm that is efficient in all but the parameter. This is captured by the notion of fixed parameter tractability, as follows:

###### Definition 3.2.

A parameterized problem is fixed parameter tractable (FPT) if there exist an algorithm , a constant , and a computable function , such that decides in time .

Thus, while the fixed-parameter notion allows inefficiency in the parameter , by means of the function , it requires polynomial complexity in all the rest of the input. In particular, a problem that is FPT is tractable for any bounded parameter value. While the core aim of parameterized complexity is to identify problems that are fixed-parameter tractable, it has also developed an extensive complexity theory, allowing to prove hardness results, e.g., that certain problems are (most probably) not FPT. To this end, several parameterized complexity classes have been defined. Two of these classes are the class W[1] and the class para-NP. We will formally define these classes shortly, but the important point to know is that there is strong evidence to believe that both classes are not contained in FPT (much like NP is probably not contained in P). Thus, W[1]-hard and para-NP-hard problems are most probably not fixed-parameter tractable. The class W[1] can be defined by its core complete problem, defined as follows:

Short Nondeterministic Turing Machine Computation

Instance: A single-tape, single-head non-deterministic Turing machine

, a word  and an integer
Question: Is there a computation of  on input  that reaches the accepting state in at most  steps?
Parameter:

Note that this definition is analogous to that of NP, with the addition of the parameter .

###### Definition 3.3.

The class W[1] contains all parameterized problems FPT-reducible (defined hereunder) to Short-Nondeterministic-Turing-Machine-Computation.

The class para-NP is defined as follows :

###### Definition 3.4.

A parameterized problem is in para-NP if there exists a non-deterministic Turing machine , constant  and an arbitrary computable function , such that for any input ,  decides if in time .

Establishing hardness results most frequently requires reductions. In parameterized complexity, we use FPT-reduction, defined as follows:

###### Definition 3.5.

Let and be parameterized problems over the alphabets and respectively. An FPT-reduction (FPT many-to-one reduction) from to is a mapping such that:

1. For all we have .

2. is computable in time for some constant and an arbitrary function .

3. There is a computable function such that for all .

Point (1) simply states that is indeed a reduction. Point (2) says that it can be computed in the right amount of time - efficient in all but the parameter. Point (3) states that the parameter of the image is bounded by (a function of) that of the source. This is necessary in order to guarantee that FPT-reductions preserve FPT-ness, i.e. with this definition we obtain that if reduces to and FPT then is also in FPT.

## 4 Our Results & Techniques

We consider problems regarding resources bounds and resource conflicts which were shown to be computationally hard in Wooldridge et al. ([8]) but were not considered in Shrot et al. [5]. We also solve three open questions posed in Shrot et al. by showing that

1. SC parameterized by is W[1]-hard

2. ESCK parameterized by is FPT

3. ESCK parameterized by is para-NP-hard

We study the complexity of NR, SNR, CGRO, RPEGS, SCRB and CC problems when parameterized by natural parameters and . We also give a general integer program which with slight modifications for each problem shows that these problems are FPT when parameterized by or (except CC parameterized by which is open). We note that Shrot et al. showed that SC parameterized by is para-NP-hard. We complete this hardness result by showing that SC parameterized by is W[1]-hard and thus answer their open question. Using these hardness results and via a single theme of parameter preserving reductions we show that hardness results for all of the above problems when parameterized by and . We also show that Theorem 3.2 of Shrot et al. [5] is false - which claims that ESCK is FPT when parameterized by . We give a counterexample to their proposed algorithm and show that the problem is indeed para-NP-hard.

These results help us to understand the role of various components of the input and identify which ones actually make the input hard. Since all the problems we considered remain intractable when parameterized by or , there is no point in trying to restrict these parameters. On the other hand, most of the problems are FPT when parameterized by or and thus we might enforce this restriction in real-life situations to ensure the tractability of these problems.

We summarize all the results in Figure 2. The results from [8] are in green, from [5] in black and our results are in red color. We use the abbreviations NPC for NP-complete, and pNP for para-NP.

## 5 Problems Left Open in Shrot et al. [5]

First we show that SC parameterized by is W[1]-hard.

###### Theorem 5.1.

SC is W[1]-hard when parameterized by .

###### Proof.

We prove this by reduction from Independent Set (parameterized by size of independent set) which is a well-known W[1]-complete problem. Let be a graph with and . Let be a given integer. We also assume that H has no isolated points as we can just add those points to the independent set and decrease the parameter appropriately. We build a CRG as follows:

 Γ=⟨Ag,G,R,G1,G2,…,Gk,en,req⟩

where

• for all

• For all ,

• For all and , we have if and are incident in and otherwise

We claim that has an independent set of size if and only if the grand coalition is successful in .

Suppose Independent Set answers YES, i.e., has an independent set of size say . Consider the goal set given by . Clearly satisfies as for all . Now consider any edge . Let be the number of vertices from incident on . Clearly but as is independent set we have . Now, for every we have . Thus is feasible for . Summing up, is successful for and hence SC answers YES for .

Suppose now that SC answers YES for . Let be successful for . Claim is that both and cannot be in if . To see this, let be any edge incident on (we had assumed earlier that graph has no isolated vertices). Then which contradicts the fact that is successful for . Since ’s are disjoint and is successful (hence also satisfiable) for , we know that contains at least one goal from each . Also we have seen before that and implies that . From each we pick any goal that is in . Let us call this as . We know that when . We claim that is an independent set in . Suppose not and let be an edge between and for some . Then which contradicts the fact that is successful for . Thus is an independent set of size in and so Independent Set also answers YES.

Note that and so this reduction shows that the SC problem is W[1]-hard. ∎

We note that the SC problem can be solved in time (since we only need to check the subsets of size at most of ) and thus SC parameterized by  is not para-NP-hard. Now we answer the only remaining open problem by Shrot et al. by showing that ESCK parameterized by  is para-NP-hard.

###### Theorem 5.2.

Checking whether there exists a successful coalition of size  (ESCK) is para-NP-hard when parameterized by .

###### Proof.

We prove this by reduction from SC which was shown to be para-NP-hard with respect to the parameter  in Theorem 3.8 of [5].
Let be a given instance of SC. We consider an instance of ESCK

• for all

We claim that SC answers YES if and only if ESCK answers YES.

Suppose SC answers YES, i.e., is a successful coalition in . In we just remove all agents not belonging to from . All the resources and the en and req functions carry over. So is a successful coalition for also. But we had chosen and so ESCK answers YES.

Suppose that ESCK answers YES. So there exists a successful coalition of size in . But and we had chosen and so the only coalition of size in is the grand coalition . As ESCK answered YES we know that is successful in . So it is also successful in and so SC also answers YES.

Note that and so this reduction shows that the ESCK problem is para-NP-hard. ∎

## 6 Problems Related to Resources

For a coalition , we recollect the notation we use: and both satisfies and is feasible for it. In this section we show hardness results for three different problems related to resources.

### 6.1 Necessary Resource (NR)

The idea of a necessary resource is similar to that of a veto player in the context of conventional coalition games. A resource is said to be necessary if the accomplishment of any set of goals which is successful for the coalition would need a non-zero consumption of this resource. Thus if a necessary resource is scarce then the agents possessing the resource become important. We consider the Necessary Resource problem: Given a coalition and a resource answer YES if and only if for all . NR was shown to be co-NP-complete in Wooldridge et al. [8]. We note that if is not successful, then NR vacuously answers YES. We give a reduction from SC to .

###### Lemma 6.1.

Given an instance of SC we can construct an instance of NR such that SC answers YES iff NR answers NO.

###### Proof.

We keep everything the same except . We extend the en and req functions to by for all and for all . Now claim is that SC answers YES iff NR answers NO.

Suppose SC answers YES. So such that . Now and so and thus . But and so NR answers NO.

Suppose NR answers NO. So as such that and . Now is obtained from by only adding a new resource and so clearly . Thus SC will answer YES. ∎

###### Theorem 6.2.

The parameterized complexity status of Necessary Resource is as follows :

• FPT when parameterized by

• co-W[1]-hard when parameterized by

• co-para-NP-hard when parameterized by

###### Proof.

When parameterized by , we consider all subsets of . For each subset, we can check in polynomial time if it is a member of and if it requires non-zero quantity of the resource given in the input.

The other two claims follow from Lemma 6.1, Theorem 3.8 in Shrot et al., and Theorem 5.1. ∎

### 6.2 Strictly Necessary Resource (SNR)

The fact that a resource is necessary does not mean that it will be used. Because the coalition in question can be unsuccessful and hence the resource is trivially necessary. So we have the Strictly Necessary Resource problem: Given a coalition and a resource answer YES if and only if and we have . SNR was shown to be strongly -complete in Wooldridge et al. [8]. To prove the parameterized hardness results, we give a reduction from SC to SNR.

###### Lemma 6.3.

Given an instance of SC we can construct an instance of SNR such that SC answers YES iff SNR answers YES.

###### Proof.

We keep everything the same except . We extend the en and req functions to by for all and for all . Now claim is that SC answers YES iff SNR answers YES.

We first show that . As is obtained from by just adding one resource and keeping everything else the same, we have . Now let . Any coalition has at least one member and hence at least one endowment of resource . But and so . Summing up we have .

Suppose SC answers YES. This implies . So . For every as . Therefore SNR answers YES

Suppose SNR answers YES. So as otherwise SNR would have said NO. Hence and SC so answers YES. ∎

###### Theorem 6.4.

The parameterized complexity status of Strictly Necessary Resource is as follows :

• FPT when parameterized by

• W[1]-hard when parameterized by

• para-NP-hard when parameterized by

###### Proof.

When parameterized by , we consider all subsets of . For each subset, we can check in polynomial time if it is a member of and if it requires non-zero quantity of the resource given in the input.

The other two claims follow from Lemma 6.3, Theorem 3.8 in Shrot et al., and Theorem 5.1. ∎

### 6.3 (C,G0,r)-Optimality (CGRO)

We may want to consider the issue of minimizing usage of a particular resource. If satisfaction is the only issue, then a coalition will be equally happy between any of the goal sets in . However in practical situations we may want to choose a goal set among which minimizes the usage of some particular costly resource. Thus we have the -Optimality problem: Given a coalition , resource and a goal set answer YES if and only if for all . CGRO was shown to be strongly co-NP-complete in Wooldridge et al. [8]. To prove the parameterized hardness results, we give a reduction from SC to .

###### Lemma 6.5.

Given an instance of SC we can construct an instance of CGRO such that SC answers YES iff CGRO answers NO.

###### Proof.

Define , and . We extend the en to as follows: for all and if . We extend req to and as follows: , for all and for all . Let . Now claim is that SC answers YES iff CGRO answers NO.

Suppose SC answers YES. So, . Claim is that because as . Note also that as and for every , . Therefore and hence CGRO answers NO.

Suppose CGRO answers NO. So such that . Claim is otherwise . So and we already had . Therefore and so SC answers YES. ∎

###### Theorem 6.6.

The parameterized complexity status of -Optimality is as follows :

• FPT when parameterized by

• co-W[1]-hard when parameterized by

• co-para-NP-hard when parameterized by

###### Proof.

When parameterized by , we consider all subsets of . For each subset, we can check in polynomial time if it is a member of and if it requires atleast quantity of resource where and are given in the input.

The other two claims follow from Lemma 6.5, Theorem 3.8 in Shrot et al., and Theorem 5.1. ∎

## 7 Problems Related to Resource Bounds

### 7.1 R-Pareto Efficient Goal Set (RPEGS)

We use the idea of Pareto Efficiency to measure the optimality of a goal set w.r.t the set of all resources. In our model we say that a goal set  is R-Pareto Efficient w.r.t a coalition  if no goal set in requires at most as much of every resource and strictly less of some resource. More formally we say that a goal set is R-Pareto Efficient w.r.t a coalition  if and only if ,

We note that is not necessarily in . Thus we have the R-Pareto Efficient Goal Set problem: Given a coalition and a goal set answer YES if and only if is R-Pareto Efficient w.r.t . Wooldridge et al. [8] show that RPEGS is strongly co-NP-complete. To prove the parameterized hardness results, we give a reduction from SC to .

###### Lemma 7.1.

Given an instance of SC we can construct an instance of RPEGS such that SC answers YES iff RPEGS answers NO.

###### Proof.

Define and . We extend the en to as follows: for all and if . We extend req to as follows: for all ; and for all . Let . Now claim is that SC answers YES iff RPEGS answers NO.

We first show that . Let . Then claim is that because otherwise for all we have . Also claim is that any goal set in also is in . All other things carry over from and we have additionally that as . Hence we have .

Suppose SC answers YES, i.e., . As we have . Now for every , . Also . Therefore requires strictly more of every resource in than and hence RPEGS answers NO.

Suppose RPEGS answers NO. Claim is that otherwise it would have answered YES vacuously. As we have and hence SC answers YES. ∎

###### Theorem 7.2.

The parameterized complexity status of R-Pareto Efficient Goal Set is as follows :

• FPT when parameterized by

• co-W[1]-hard when parameterized by

• co-para-NP-hard when parameterized by

###### Proof.

When parameterized by , we consider all subsets of . For each subset, we can check in polynomial time if it is a member of and if it shows that is not R-Pareto Efficient.

The other two claims follow from Lemma 7.1, Theorem 3.8 in Shrot et al., and Theorem 5.1. ∎

### 7.2 Successful Coalition with Resource Bound (SCRB)

In real-life situations we typically have a bound on the amount of each resource. A resource bound is a function with the interpretation that each coalition has at most quantity of resource  for every . We say that a goal set respects a resource bound b w.r.t. a given CRG iff we have . Thus we have the Successful Coalition With Resource Bound problem: Given a coalition and a resource bound  b answer YES if and only if such that respects b. Wooldridge et al. [8] show that SCRB is strongly NP-complete. To prove the parameterized hardness results, we give a reduction from SC to .

###### Lemma 7.3.

Given an instance of SC we can construct an instance of SCRB such that SC answers YES if and only if SCRB answers NO.

###### Proof.

Define and . Let b

be a vector with

components whose first entries are 1 and the last entry is , i.e., . We extend the en to as follows: for all and if . We extend req to as follows: for all . Now the claim is that SC answers YES if and only if SCRB answers NO.

Suppose SC answers YES. So, there exists such that . In we have as and for all . Thus and so SCRB cannot vacuously answer YES. Now, for any such that we have . This means that no goal set in the non-empty set respects b which implies that SCRB answers NO.

Suppose SCRB answers NO. So such that and respects b. As was obtained from by adding a resource and keeping everything else same, we have and hence SC answers YES. ∎

###### Theorem 7.4.

The parameterized complexity status of Successful Coalition With Resource Bound (SCRB) is as follows:

• FPT when parameterized by

• co-W[1]-hard when parameterized by

• co-para-NP-hard when parameterized by

###### Proof.

When parameterized by , we consider all subsets of . For each subset,we can check in polynomial time if it is a member of and if it requires non-zero quantity of the resource given in the input.

The other two claims follow from Lemma 7.3, Theorem 3.8 in Shrot et al., and Theorem 5.1. ∎

## 8 Problems Related to Resource Conflicts

### 8.1 Conflicting Coalitions (CC)

When two or more coalitions desire to use some scarce resource, it leads to a conflict in the system. This issue is a classic problem in distributed and concurrent systems. In our framework we say that two goal sets are in conflict w.r.t a resource bound if they are individually achievable within the resource bound but their union is not. Formally a resource bound is a function with the interpretation that each coalition has at most quantity of resource  for every . We say that a goal set respects a resource bound b w.r.t. a given CRG if and only if we have . We denote by the fact that and are in conflict w.r.t b. Formally, is defined as . Thus we have the Conflicting Coalitions problem: Given coalitions and a resource bound b answer YES if and only if and we have . Wooldridge et al. [8] show that CC is strongly co-NP-complete. To prove the parameterized hardness results, we give a reduction from SC to .

###### Lemma 8.1.

Given an instance of SC we can construct an instance of CC such that SC answers YES if and only if CC answers NO.

###### Proof.

Define and . Let b be a vector with components whose first entries are and the last entry is , i.e., . We extend the en to as follows: for all and if . We extend req to as follows: for all . Now the claim is that SC answers YES if and only if CC answers NO.

First we claim that . We built from by just adding one resource and so clearly . Now let . Then and . Summarizing we have our claim.

Suppose SC answers YES. So, there exists such that . As we have . As the cgs condition fails for and so CC answers NO.

Suppose CC answers NO. Claim is that . If not then and in fact CC would have vacuously answered YES. But and so . Thus SC answers YES. ∎

###### Theorem 8.2.

The parameterized complexity status of Conflicting Coalitions (CC) is as follows :

• FPT when parameterized by

• co-W[1]-hard when parameterized by

• co-para-NP-hard when parameterized by

###### Proof.

When parameterized by , we consider all choices for and . Given a choice we can check in polynomial time if and are members of and respectively. Also we can check the condition in polynomial time.

The other two claims follow from Lemma 8.1, Theorem 3.8 in Shrot et al., and Theorem 5.1. ∎

## 9 The Parameter |Ag|+|R| : Case of Bounded Agents plus Resources

Considering the results in previous sections, we can see that even in the case that size of coalition or number of resources is bounded the problem still remains computationally hard. So a natural question is what happens if we have a bound on ? Can we do better if total number of agents plus resources is bounded? Shrot et.al [5] show that by this parameterization the problems SC, MAXC and MAXSC have FPT algorithms and they left the corresponding question for the ESCK open. We will generalize the integer program given in Theorem 3.1 of [5], to give a FPT algorithm for the open problem of Existence of Successful Coalition of size (ESCK). Then by using a similar approach we will design FPT algorithms for the four other problems (NR, SNR, CGRO, SCRB) considered in this paper.

The integer program we define is a satisfiability problem (rather than an optimization problem). It consists of a set of constraints, and the question is whether there exists an integral solution to this set. Consider the following integer program (which we will name as IP):

 ∀i∈Ag: ∑g∈Gixg≥yi (1) ∀r∈R: ∑g∈Gxg×req(g,r)≤∑i∈Agyi×en(i,r) (2) ∀g∈G: xg∈{0,1} ∀i∈Ag: yi∈{0,1}

In this setting, , for each , represents the situation that the agent is participating in the coalition and , for each , represents the situation that goal is achieved. The first constraint guarantees that any participating agent has at least one of his goals achieved. The second constraint ensures that the participating agents have enough endowment to achieve all of the chosen goals. It is clear that any solution for this integer program is a coalition of agents and a successful set of goals for that coalition.

The above integer program has constraints and in Flum and Grohe [3] it is shown that checking feasibility of

Integer Linear Programming

is FPT in the number of constraints or in the number of variables. Now for each of our problems we will add some constraints to get new integer programs which solve those problems.

###### Theorem 9.1.

Checking whether there is a Successful Coalition of size (ESCK) is FPT when parameterized by .

###### Proof.

For ESCK, the general integer program given above needs only one additional constraint: We have to ensure that exactly number of agents will be selected. Therefore adding the constraint gives us the integer program for the problem ESCK. The number of constraints, i.e., for this integer program is and as Integer Linear Programming is FPT w.r.t number of variables or constraints we have that ESCK parameterized by is FPT.

In the problems NR, SNR and CGRO the coalition is always given. So we will change the variables ’s to constants where if and 0 otherwise. We call this new integer program a Fixed Coalition Integer Program (FCIP). The coalition is successful if and only if FCIP is satisfiable.

###### Theorem 9.2.

Checking whether the Resource is Needed for a Coalition to be Successful (NR) is FPT when parameterized by .

###### Proof.

We start with the integer program FCIP. The answer to NR is YES, if and only if in any successful subset of goals, there is at least one goal with . So we just need to check and see if the coalition is successful by only using the goals which do not need the resource . Therefore in FCIP, for all goals where we will set the variable to zero. Now the answer to NR is YES iff the resulting integer program is not satisfiable. Note that the number of constraints is still same as previously -