DeepAI

# Group Control for Procedural Rules: Parameterized Complexity and Consecutive Domains

We consider Group Control by Adding Individuals (GCAI) for two procedural rules – the consensus-start-respecting rule and the liberal-start-respecting rule. It is known that GCAI for both rules are NP-hard, but whether they are fixed-parameter tractable with respect to the number of distinguished candidates remained open. We resolve both open problems in the affirmative. In addition, we strengthen the NP-hardness of GCAI by showing that, with respect to the natural parameter the number of added individuals, GCAI for both rules is W[2]-hard. Notably, the W[2]-hardness for the liberal-start-respecting rule holds even when restricted to a very special case where the qualifications of individuals satisfy the so-called consecutive ones property. However, for the consensus-start-respecting rule, the problem becomes polynomial-time solvable in this special case. We also study a duality restriction where the disqualifications of individuals fulfill the consecutive ones property, and we show that under this restriction GCAI for both rules turn out to be polynomial-time solvable. Our reductions for showing W[2]-hardness also imply several other lower bounds concerning kernelization and exact algorithms.

• 8 publications
• 1 publication
10/03/2018

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## 1 Introduction

In the model of group identification, we have a group of individuals each of whom holds binary valuations on all individuals including herself, and the model aims to determine who among these individuals are socially qualified by utilizing a certain social aggregation rule (cf. KasherRwhoisj ). Since the initial works of Kasher Kasher1993 and Kasher and Rubinstein KasherRwhoisj , group identification has been extensively explored from the perspective of economics, with the main focus being on axiomatic characterizations of different social aggregation rules (cf. DBLP:series/sfsc/Dimitrov11 ; DBLP:journals/mss/DimitrovSX07 ; DBLP:journals/mss/Nicolas07 ; DBLP:journals/geb/Miller08 ; DBLP:journals/jet/SametS03 ).

As social aggregation rules can be seen as special voting systems where the individuals are both voters and candidates, inspired by the pioneering work of Bartholdi, Tovey, and Trick Bartholdi92howhard on voting control problems, Yang and Dimitrov DBLP:journals/aamas/YangD18 initiated the study of group identification from a computer science perspective by investigating the complexity of the

(

GCAI

/

###### Problem 9

GCDI

) problems. In particular,

GCAI

/

###### Problem 11

GCDI

consists in determining if a given (valuation) profile can be modified by adding/deleting a limited number of individuals to make a given subset of distinguished individuals all socially qualified. Yang and Dimitrov studied the consensus-start-respecting rule and the liberal-start-respecting rule, and showed that

###### Problem 12

GCAI

for both rules are NP-hard, while

###### Problem 13

GCDI

for both rules turned out to be polynomial-time solvable. However, it is left open whether

###### Problem 14

GCAI

for these two rules are fixed-parameter tractable (FPT) with respect to the number of distinguished candidates. We resolve the open questions in the affirmative by reducing

###### Problem 15

GCAI

to a variant of the

###### Problem 16

Directed Steiner Tree

problem (Theorems 2 and 3). In addition, we strengthen the NP-hardness by showing that

###### Problem 17

GCAI

for both rules are W[2]-hard with respect to the number of added individuals (Theorems 4 and 6). Particularly, the W[2]-hardness for the liberal-start-respecting rule holds even when restricted to a very special case where the qualifications of individuals satisfy the so-called consecutive ones property. However, for the consensus-start-respecting rule, the problem is polynomial-time solvable in this special case (Theorem 5). We also study a duality restriction where the disqualifications fulfill the consecutive ones property, and show that under this restriction

###### Problem 18

GCAI

for both rules turn out to be polynomial-time solvable (Theorem 7). Our hardness reductions also lead to numerous lower bounds concerning kernelizations and exact algorithms (Corollaries 16).

## 2 Related Works

Since the first work of Yang and Dimitrov DBLP:journals/aamas/YangD18 on the complexity of group control problems, several related problems have been proposed and studied very recently. In particular, Erdélyi, Reger, and Yang DBLP:journals/aamas/ErdelyiRY20 ; DBLP:conf/aldt/ErdelyiRY17 studied the destructive counterpart of the group control problems, group bribery problems, and the problems of determining socially qualified individuals with incomplete information. Additionally, Boehmer et al. DBLP:conf/ijcai/BoehmerBKL20 also considered numerous bribery problems in group identification through the lens of parameterized complexity.

Voting problems restricted to special domains have been widely studied in the literature. Particularly, the consecutive domain has been studied as an analog of single-peaked domain for dichotomous preferences (cf. DBLP:journals/jair/BrandtBHH15 ; DBLP:journals/iandc/FaliszewskiHHR11 ). This domain has been also studied under the name candidate interval (cf. DBLP:conf/aaai/Peters18 ; DBLP:conf/ijcai/ElkindL15 ; DBLP:conf/atal/LiuG16 ; DBLP:conf/ijcai/Yang19a ). It is known that many voting problems which are NP-hard in general become polynomial-time solvable when restricted to this domain, with only a few exceptions (cf. DBLP:journals/jair/BetzlerSU13 ; DBLP:journals/tcs/SkowronYFE15 ). For further discussions on the complexity of voting problems in restricted domains, we refer to structuredpreferencesElkindLP ; Hemaspaandra2016 for comprehensive surveys.

It should be also noted that the consecutive domain is equivalent to the so-called consecutive ones property which has a long research history and has found significant applications in a broad range of areas (cf. the survey DBLP:journals/eatcs/Dom09 and references therein).

## 3 Preliminaries

Throughout this paper we will need the following basic ingredients. For an integer  is the set of positive integers no greater than .

### 3.1 Social Aggregation Rules

Let  be a set of  individuals. Each individual has an opinion who from the set  possess a certain qualification and who do not. For , we write for ’s qualification of , while will stand for ’s disqualification of . The mapping is called a profile over . A social aggregation rule is a function  assigning to each pair of a profile  over  and a subset a subset . The members of are called  socially qualified individuals in  at the profile .

In what follows we focus in our analysis on two procedural rules: the consensus-start-respecting rule and the liberal-start-respecting rule. The reader is referred to DBLP:journals/mss/DimitrovSX07 for axiomatic characterizations of these rules.

Consensus-start-respecting rule

This rule determines the socially qualified individuals iteratively. First, all individuals qualified by everyone are considered socially qualified. Then, in each iteration, all individuals who are qualified by at least one of the currently socially qualified individuals are added to the set of socially qualified individuals. The iterations terminate when no new individuals can be added this way. Formally, for every , let

 KC0(φ,T)={a∈T∣∀(a′∈T)[φ(a′,a)=1]}.

For each , …, let

 KCℓ−1(φ,T)∪{a∈T∣∃(a′∈KCℓ−1(φ,T))[φ(a′,a)=1]}.

Then, for some  such that .

Liberal-start-respecting rule

This rule is similar to  with only the difference that the initial socially qualified individuals are those who qualify themselves. In particular, for every , let

 KL0(φ,T)={a∈T∣φ(a,a)=1}.

For each integer , …, let

 KLℓ−1(φ,T)∪{a∈T∣∃(a′∈KLℓ−1(φ,T))[φ(a′,a)=1]}.

Then, for some  such that .

It should be noted that when (resp. ) we have that (resp. ).

### 3.2 Consecutive Domains

Let be a linear order over . For each individual , let

 φ⊳(a)=(φ(a,a1),φ(a,a2),…,φ(a,an)).

We say that  is qualifying consecutive (QC) with respect to the order  if s are consecutive in , i.e., there are such that , for all  such that , and for all other possible values of . We say that is disqualifying consecutive (DQC) with respect to   if s are consecutive in . We say that  is QC (resp. DQC) if there is at least one linear order  over  with respect to which every where is QC (resp. DQC).

It is immediately clear from the above definition that checking if a profile is QC or DQC is equivalent to checking if a - matrix satisfies the consecutive ones property, which can be done in polynomial-time (cf.  DBLP:journals/jair/PetersL20 ; DBLP:journals/jcss/BoothL76 ; DBLP:journals/jal/Hsu02 ).

### 3.3 Group Control

Let us now formally state the group control problem we study. Let  be a social aggregation rule.

(GCAI)
Input:

A -tuple of a set  of individuals, a profile  over , two nonempty subsets  such that , and an integer .

Question:

Is there a subset such that  and ?

In what follows, we call members of  distinguished individuals.

### 3.4 Parameterized Complexity

A parameterized problem is subset where  is a fixed alphabet. A parameterized problem is FPT if there is an algorithm so that for each instance of the problem the algorithm determines correctly if is a YES-instance in time , where  is a computable function in the parameter

. The following hierarchy has been developed to classify parameterized problems:

 FPT⊆W[1]⊆W[2]⊆⋯⊆XP.

A parameterized problem is W[2]-hard if all problems in W[2] are parameterized reducible to the problem. W[2]-hard problems do not admit any FPT-algorithms unless the above hierarchy collapses at some level.

A kernelization of an FPT problem  is an algorithm which takes an instance of  as input and outputs an instance of  such that (1) the algorithm runs in polynomial in the size of , (2) is a YES-instance if and only if is a YES-instance, and (3) for some computable function  in . If  has a kernelization where  is a polynomial, we say that  admits a polynomial kernel.

For further discussions on parameterized complexity, we refer to Cygan2015 ; DBLP:series/txcs/DowneyF13 .

### 3.5 Useful Graph Problems

In the following, we introduce some useful graph problems for our study. We assume the reader is familiar with the basics in graph theory. For a consult, we refer the reader to DBLP:books/daglib/0022205 ; Douglas2000 .

A bipartite graph is a graph  whose vertices can be divided into two disjoint sets and so that the edges of  are only between  and . A vertex  dominates another vertex  if there is an edge between them. For two disjoint subsets  and  of vertices, we say that  dominates  if every vertex in  is dominated by at least one vertex in .

Red-Blue Dominating Set (RBDS)
Input:

A bipartite graph  and an integer .

Question:

Is there a subset  of at most  vertices dominating ?

###### Problem 19

RBDS

is a well-known NP-hard problem, and from the parameterized complexity point of view it is W[2]-hard with respect to  DBLP:series/txcs/DowneyF13 . We will use this problem to establish our fixed-parameter intractability results.

Some of our problems are solved by reducing to a variant of the

###### Problem 20

Directed Steiner Tree

problem defined below. For a graph (resp. digraph) , we use  to denote its vertex set, and use  to denote its edge (resp. arc) set. For a subset  of edges (resp. arcs) of  is the set of vertices incident with edges (resp. arcs) in .

Directed Steiner Tree (
###### Problem 21
DST)
Input:

A digraph , a subset of vertices in  called terminals, a vertex called the root, a function assigning to each arc an integer weight, and an integer .

Question:

Is there a subset so that

(1) ; and

(2) for every , there is a directed path from  to  in the subgraph of  induced by ?

It is known that

###### Problem 22

DST is NP-hard and, moreover, it is FPT with respect to the number of terminals DBLP:journals/networks/DreyfusW71 ; DBLP:journals/siamdm/GuoNS11 ; DBLP:conf/icde/DingYWQZL07 . More precisely,

###### Problem 23

DST

can be solved in time DBLP:journals/networks/DreyfusW71 ; DBLP:journals/siamdm/GuoNS11 ; DBLP:conf/stoc/BjorklundHKK07 ; DBLP:journals/mst/FuchsKMRRW07  111 is the notion with the polynomial factor being ignored..

Directed Vertex Weighted Steiner Tree
(DVWST)
Input:

A directed graph , a subset of terminals, a vertex called the root, a function assigning to each vertex an integer weight, and an integer .

Question:

Is there a subset so that

(1) ; and

(2) for every , there is a directed path from  to  in the subgraph of  induced by ?

We shall use

###### Problem 24

DVWST

as an intermediate problem to established our FPT results, and as a byproduct, we show that

###### Problem 25

DVWST

is FPT with respect to the number of terminals (Theorem 1).

## 4 Our Results

We shall first study two FPT algorithms for

###### Problem 26

GCAI

in the general domain, and then we explore the complexity of

###### Problem 27

GCAI

restricted to the QC and DQC domains.

### 4.1 The General Domain

First, we resolve the open questions regarding

###### Problem 28

GCAI

for and in the affirmative, starting with the one for . To this end, we first show that the

DVWST

problem is FPT.

###### Problem 30

DVWST

can be solved in  time where  is the number of terminals.

The proof of Theorem 1 is differed to the Appendix. We start with a lemma which suggests that to solve

###### Problem 31

GCAI

for  we can make a guess on one of the individuals who are qualified by all individuals in the final profile. This enables us to split an instance of

###### Problem 32

GCAI

for  into polynomially many subinstances which are then solved via Theorem 1.

In order to state the lemma, we need the following notions. The incidence graph of a profile  over , denoted , is the digraph whose vertices are exactly the individuals in , and there is an arc from to if and only if  qualifies . Note that the incidence graph may contain loops.

###### Lemma 1.

Let  be a profile over a set  of individuals so that . Let be an individual qualified by all individuals in . Then, for every , it holds that  if and only if there is a directed path from  to  in .

Proof Let , and  be as stipulated in the lemma. It is clear from the definition of that if there is a directed path from  to  then . It remains to show the other direction. Assume that . Due to the definition of , there must be an individual who is in the initial set of socially qualified individuals, and there is a directed path from  to  in the incidence graph  of . So,  is qualified by all individuals including , implying that there is a directed path from  to  in . ∎

Observe that if an individual in  qualifies another individual in , then if the former is socially qualified so is the latter. The following reduction rule implements this observation.

###### Reduction Rule 1.

If there are two distinct individuals such that  qualifies , move  from  into .

For a subset  of vertices in a digraph , let

 N−G(Y)={v∈V(G)∖Y∣∃(u∈Y)[(v,u)∈E(G)]}

be the set of inneighbors of vertices in , and let

 N+G(Y)={v∈V(G)∖Y∣∃(u∈Y)[(u,v)∈E(G)]}

be the set of outneighbors of vertices in . Merging  is the operation that creates one vertex  so that and , and removes all vertices of  from . We call  the merging vertex of .

Armed with Lemma 1 and Reduction Rule 1, we are ready to present our first FPT algorithm.

###### Problem 33

GCAI

for is FPT with respect to the number  of distinguished individuals. More precisely, it can be solved in time.

Proof Let be an instance of

###### Problem 34

GCAI

for . We first exhaustively apply Reduction Rule 1 to  so that in the resulting instance no individual in  qualifies another different individual in . Then, we split the instance into  subinstances, each of which takes  and an individual  as input, and determines if there exists of at most  individuals so that , , and all individuals in qualify . That is,  is our guessed individual who is in the initial set of socially qualified individuals in the final profile. Obviously, the original instance  is a YES-instance if and only if at least one of the subinstances is a YES-instance.

Now we focus on solving a subinstance with a guessed individual . We assume that  is already included in , since otherwise we simply move  from into  and decrease  by one. As  is supposed to be qualified by all individuals in the final profile, if there is an individual in  who disqualifies  we directly discard this subinstance and proceed to the next one. Otherwise, we remove from the subinstance all individuals in who disqualify  (this includes deleting them from both  and from ). We shall solve the subinstance by reducing it to

###### Problem 35

DVWST

. We create an instance of

###### Problem 36

DVWST

as follows. The digraph  in the

###### Problem 37

DVWST

instance is obtained from the incidence graph of  over  by merging . Obviously, , and so is nonempty. Let  denote the merging vertex of , and we let it be the given root of the

###### Problem 38

DVWST

instance. Furthermore, we let every individual in have weight , and all the other individuals have weight . The terminals are those in , and the weight upper bound is .

If there is a subset  of vertices of total weight at most  so that there is a directed path from  to every terminal , then  is socially qualified in due to Lemma 1. Moreover, as all individuals in have weight  and all the other individuals have weight , we know that  contains at most  individuals from , implying that is a certificate for the subinstance. For the reverse direction, if there is a subset of individuals so that for all it holds that , then due to Lemma 1 there is a directed path from  to  in the subgraph of  induced by . As the total weight of vertices in  is at most , the instance of

###### Problem 39

DVWST

is a YES-instance.

Regarding the running time, let be the number of distinguished individuals. As there are at most  subinstances to consider and each of them can be solved in  time (Theorem 1), the whole algorithm runs in  time. ∎

An analogous result for the liberal-start-respecting rule exists. In fact, the algorithm in this case is simpler. We first use a reduction rule to refine the structure of the instance.

###### Reduction Rule 2.

If there are such that  qualifies , move  from  into .

Notice that the difference between Reduction Rule 1 and Reduction Rule 2 is that in the second one  and  may be the same individual. The reason is that under  everyone qualifying herself is already socially qualified without needing other individuals qualifications.

###### Problem 40

GCAI

for is FPT with respect to the number  of distinguished individuals. More precisely, it can be solved in  time.

Proof Let be an instance of

###### Problem 41

GCAI

for . We first apply Reduction Rule 2 to  iteratiely untill it does not apply. Then, we solve the instance by reducing it to a

###### Problem 42

DVWST

instance as follows. The digraph  of the

###### Problem 43

DVWST

instance is obtained from the incidence graph of  over  by creating one new vertex  and creating arcs from  to everyone in  qualifying herself (i.e., ). We set  as the root, and set  as the set of the terminals. Finally, we let the weight of all individuals in be , and those of others be . Similar to the analysis of the proof of Theorem 2, we can show that the two instances are equivalent. The running time of the algorithm follows from Theorem 1. ∎

Next, we strengthen the NP-hardness of

###### Problem 44

GCAI

for the consensus-start-respecting rule by showing its W[2]-hardness with respect to the number of added individuals. We also have a W[2]-hardness result for the liberal-start-respecting rule, but we will present it in the next section because this result holds even in a specific domain which is not the focus of this section.

###### Problem 45

GCAI

for is W[2]-hard with respect to the number of added individuals.

Proof We prove the theorem via a reduction from the

###### Problem 46

RBDS

problem. Let be an

###### Problem 47

RBDS

instance where is a bipartite graph with the vertex partition and  is an integer. We create an instance of

###### Problem 48

GCAI

for as follows. First, we create for each vertex in  an individual denoted by the same symbol for notational brevity. Let and . We define a profile  over  so that:

• each qualifies all individuals in  and disqualifies all individuals in ; and

• each qualifies all individuals in  and, moreover, for each individual it holds that  qualifies  if and only if  and  are adjacent in .

The instance of

###### Problem 49

GCAI

for is . The reduction can be done in polynomial time. We show the correctness of the reduction as follows.

Assume that there is a subset of at most  vertices dominating . According to the definition of , individuals in  are qualified by all individuals in . Therefore, it holds that . Moreover, as  dominates , for every  there is at least one which dominates . By the definition of , qualifies , implying that . As this holds for all , we know that the instance of

###### Problem 50

GCAI

for constructed above is a YES-instance.

Assume that there is a subset (recall that ) of cardinality at most  so that . Let  be any arbitrary individual in . According to the definition of  is qualified only by individuals in  who dominate  in . As , this implies that  contains at least one vertex dominating . As this holds for all , we conclude that  dominates . Given , we conclude that the

###### Problem 51

RBDS

instance is a YES-instance. ∎

### 4.2 The Consecutive Domains

Now we explore the complexity of

###### Problem 52

GCAI

for the two procedural rules restricted to the consecutive domains.

###### Lemma 2.

Let  be a profile over  which is QC with respect to a linear order  of . Then, all individuals in are consecutive in the order .

Proof If , the lemma vacuously holds. Otherwise, there is an individual who is qualified by all individuals. As  is QC with respect to , all individuals only qualify individuals consecutive in , and they all qualify the same individual . It follows that all individuals in are consecutive in . ∎

Based on Lemma 2, we can derive a polynomial-time algorithm for

GCAI

for .

###### Problem 54

GCAI

for is polynomial-time solvable when restricted to QC profiles.

Proof Let be an instance of

###### Problem 55

GCAI

for  where  is QC with respect to a linear order  over . Let  and  be respectively the left-most and the right-most individuals in  that are from . Due to Lemma 2, the question of  is equivalent to making  and  socially qualified in  by adding at most  individuals from into . By light of this fact, we move all individuals except  and  from  into . After this operation,  contains at most two individuals ( is a singleton when ). Then, we solve the instance in polynomial time by Theorem 2. ∎

Now we move on to the liberal-start-respecting rule. Unlike , we show that

###### Problem 56

GCAI

for remains computationally hard even when restricted to QC profiles.

###### Problem 57

GCAI

for is NP-hard and W[2]-hard with respect to the number of added individuals even when restricted to QC profiles.

Proof We prove the theorem by giving a reduction from the

RBDS

problem to

###### Problem 59

GCAI

for  restricted to QC profiles. Let be an instance of

###### Problem 60

RBDS

, where is a bipartite graph, and  is an integer. For each , we construct an individual denoted still by  for notational simplicity. For each , let  be the degree of  in . For each , we construct individuals . Let for each , and let . In addition, let  denote the set of the above constructed individuals, let , and let . We define a profile  over  as follows.

• For each red vertex , the individual  qualifies , and each  qualifies exactly one neighbor of  in  so that every neighbor of  is qualified by exactly one of these  individuals.

• For each where is not specified above, we define .

The instance of

###### Problem 61

GCAI

is .

It is easy to see that the profile  is QC. In fact, except those in , all the other individuals qualify at most one individual. Moreover, as every  where qualifies exactly the individuals created for , the profile is QC with respect to any linear order over  where for every the individuals created for  are consecutive.

The construction takes polynomial time. In the following, we show the correctness of the reduction.

Assume that there is a subset of at most  vertices dominating . Let . We show that . Note that as for all , and  qualifies also all the other individuals created for , we know that for every , the individuals  are all socially qualified in at . Let  be an individual in . As  dominates  and has at least one neighbor in . Then, due to the above construction, there exists an individual  where who qualifies . As , it follows that . As this holds for all , the above constructed instance of

###### Problem 62

GCAI

for  is a YES-instance.

Assume that there is a such that and . Let . Clearly, . We claim that  dominates . Let  be a vertex (individual) in . By the definition of  is only qualified by individuals in such that and dominates . Then, as , there exists an individual where and such that and . Note that the only individual who qualifies  is the individual  who qualifies herself. This means that , and hence , further implying that  is dominated by . As the above argument holds for all , we conclude that  dominates . ∎

When restricted to DQC, we can show that

###### Problem 63

GCAI

for both procedural rules are polynomial-time solvable. A crucial observation is that if the given instance is a YES-instance, we need at most two individuals to bring all distinguished individuals into the set of socially qualified individuals.

GCAI

for and

###### Problem 65

GCAI

for are polynomial-time solvable when restricted to DQC profiles.

Proof Let be an instance of

###### Problem 66

GCAI

for  (resp. ), where  is DQC with respect to a linear order over . For each individual , let  be the left-most individual  qualifies, and let  be the right-most individual  qualifies in . More precisely, (resp. ) such that and, moreover, for all such that it holds that (resp. ). Let

 {a∈N∣φ(a,an)=1,∀(a′∈N,φ(a′,an)=1)[L(a)⊳L(a′)]}

and let

 {a∈N∣φ(a,a1)=1,∀(a′∈N,φ(a′,a1)=1)[R(a′)⊳R(a)]}.
###### Observation 1.

Let and . Then, for every subset , the individuals who are qualified by at least one in  is a subset of individuals qualified by at least one of  and .

Based on Observation 1, we can obtain an equivalent instance of  by resetting . The new instance can be solved in polynomial time (precisely in time) by a brute-force search. ∎

## 5 Concluding Remarks

We have proved that

###### Problem 67

GCAI

for both the consensus-start-respecting rule () and the liberal-start-respecting rule () are FPT with respect to the number of distinguished candidates (Theorems 2 and 3), resolving two open questions left in DBLP:journals/aamas/YangD18 . Additionally, we studied these problems restricted to the qualifying consecutive (QC) domain and the disqualifying consecutive (DQC) domain. We showed that both problems become polynomial-time solvable when restricted to the DQC domain (Theorem 7). However, when restricted to the QC domain,

###### Problem 68

GCAI

for  is polynomial time solvable (Theorem 5), while

###### Problem 69

GCAI

for  turned out to be NP-hard (Theorem 6).

Given the fixed-parameter tractability of

###### Problem 70

GCAI

with respect to the number of distinguished individuals (Theorems 2 and 3), one may wonder whether the two problems admit polynomial kernels. The reader may already observed that both reductions in the proofs of Theorems 4 and 6 are in fact polynomial parameter transformations with respect to the combined parameter of the number of distinguished candidates and the number of added individuals. Then, by the lower bound technique developed by Dom, Lokshtanov, and Saurabh DBLP:journals/talg/DomLS14 , we have the following two corollaries refuting the possibility of the existence of polynomial kernels for the two problems.

###### Problem 71

GCAI

for does not admit any polynomial kernel with respect to the parameter unless the polynomial hierarchy collapses to the third level .

###### Problem 72

GCAI

for does not admit any polynomial kernel with respect to the parameter unless the polynomial hierarchy collapses to the third level. Moreover, this holds even when restricted to QC profiles.

###### Problem 73

GCAI

is FPT with respect to because it can be solved in time by a brute-force search. Because

###### Problem 74

RBDS

is unlikely to admit any polynomial kernel with respect to  Cygan2015 , our reductions in Theorems 4 and 6 respectively lead to the following two corollaries.

###### Problem 75

GCAI

for does not admit any polynomial kernel with respect to the parameter unless the polynomial hierarchy collapses to the third level.

###### Problem 76

GCAI

for does not admit any polynomial kernel with respect to the parameter unless the polynomial hierarchy collapses to the third level. Moreover, this holds even when restricted to QC profiles.

Finally, observe that

###### Problem 77

GCAI

can be also solved in time by a brute-force search. As

###### Problem 78

RBDS

cannot be solved in time assuming the Strong Exponential Time Hypothesis (SETH) Cygan2015 , and it cannot be solved in time either assuming ETH DBLP:journals/jcss/ChenHKX06 , our reduction in Theorems 4 and 6 imply that these brute-force based algorithms are essentially optimal.

###### Corollary 5.

Unless SETH fails,

###### Problem 79

GCAI

for cannot be solved in time, and unless ETH fails

###### Problem 80

GCAI

for cannot be solved in  time.

###### Corollary 6.

Unless SETH fails,

###### Problem 81

GCAI

for cannot be solved in time, and unless ETH fails

###### Problem 82

GCAI

for cannot be solved in  time. Moreover, this holds even when restricted to QC profiles.

## Appendix

Proof of Theorem 1 Let be an instance of

###### Problem 83

DVWST

. We create an instance of

###### Problem 84

DST

equivalent to  as follows.

We first create an arc-weighted digraph  obtained from  by performing the following operations:

1. Replace every vertex with two vertices  and , add an arc from to  with weight , and add some arcs so that the inneighbors of are exactly the inneighbors of in , and the outneighbors of are exactly the outneighbors of  in .

2. Set the weight of all the arcs whose weights are not yet specified to be .

Let be the function corresponding to the weights specified for the arcs in  above. The instance of

###### Problem 85

DST

is . The reduction clearly can be done in polynomial time. It remains to show the correctness.

Assume that the

###### Problem 86

DVWST

instance is a YES-instance, i.e., there is a subset such that , and for every terminal there is a directed path from  to  in the subgraph of  induced by . Let . Due to the above construction every original arc in  has weight  under , and every arc has weight  under . It follows that

 ∑(vin,vout)∈J′w′((vin,vout))=∑v∈Jw(v)≤p.

Moreover, if is a directed path from the root  to some terminal in the digraph , by the definition of  we know that is a directed path in . Therefore, the constructed

###### Problem 87

DST

instance is a YES-instance.

Assume that the constructed instance of

###### Problem 88

DST

is a YES-instance, i.e., there is a subset  of arcs in  so that and, moreover, for every terminal there is a directed path from  to  in the subgraph of  induced by . Let

 J′={v∈V(G)∖(X∪{u})∣(vin,vout)∈J}.

Similar to the above analysis, we know that and for every terminal we can change any directed path from  to  in the subgraph of  induced by  into a directed path from  to  in the subgraph of  induced by . In particular, observe that each  has a unique outneighbor . So, in any - directed path containing a vertex , the next vertex after  must be . Therefore, from a directed path from  to  in , we can obtain a directed path from  to  in  by replacing every arc  in the path with the vertex .

The theorem follows from the above reduction and the fact that

DST