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
Problem 7
Group Control By Adding/Deleting Individuals
(
Problem 8
GCAI
/
Problem 9
GCDI
) problems. In particular,
Problem 10
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 consensusstartrespecting rule and the liberalstartrespecting rule, and showed that
Problem 12
GCAI
for both rules are NPhard, while
Problem 13
GCDI
for both rules turned out to be polynomialtime solvable. However, it is left open whether
Problem 14
GCAI
for these two rules are fixedparameter 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 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 liberalstartrespecting rule holds even when restricted to a very special case where the qualifications of individuals satisfy the socalled consecutive ones property. However, for the consensusstartrespecting rule, the problem is polynomialtime 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
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 singlepeaked 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 NPhard in general become polynomialtime 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 socalled 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 consensusstartrespecting rule and the liberalstartrespecting rule. The reader is referred to DBLP:journals/mss/DimitrovSX07 for axiomatic characterizations of these rules.
 Consensusstartrespecting 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
For each , , …, let
Then, for some such that .
 Liberalstartrespecting 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
For each integer , , …, let
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
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 polynomialtime (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.
Group Control by Adding Individuals  

(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 YESinstance in time , where is a computable function in the parameter
. The following hierarchy has been developed to classify parameterized problems:
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 FPTalgorithms 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 YESinstance if and only if is a YESinstance, 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 .
RedBlue 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 wellknown NPhard 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 fixedparameter 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 21DST) 


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 NPhard 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 ^{1}^{1}1 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
Problem 29
DVWST
problem is FPT.
Theorem 1.
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
be the set of inneighbors of vertices in , and let
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 .
Theorem 2.
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 YESinstance if and only if at least one of the subinstances is a YESinstance.
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 YESinstance.
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 liberalstartrespecting 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.
Theorem 3.
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 NPhardness of
Problem 44
GCAI
for the consensusstartrespecting 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 liberalstartrespecting 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.
Theorem 4.
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 YESinstance.
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 YESinstance. ∎
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 polynomialtime algorithm for
Problem 53
GCAI
for .
Theorem 5.
Problem 54
GCAI
for is polynomialtime 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 leftmost and the rightmost 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 liberalstartrespecting rule. Unlike , we show that
Problem 56
GCAI
for remains computationally hard even when restricted to QC profiles.
Theorem 6.
Problem 57
GCAI
for is NPhard 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
Problem 58
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 YESinstance.
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 polynomialtime solvable. A crucial observation is that if the given instance is a YESinstance, we need at most two individuals to bring all distinguished individuals into the set of socially qualified individuals.
Theorem 7.
Problem 64
GCAI
for and
Problem 65
GCAI
for are polynomialtime 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 leftmost individual qualifies, and let be the rightmost individual qualifies in . More precisely, (resp. ) such that and, moreover, for all such that it holds that (resp. ). Let
and let
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 bruteforce search. ∎
5 Concluding Remarks
We have proved that
Problem 67
GCAI
for both the consensusstartrespecting rule () and the liberalstartrespecting 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 polynomialtime 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 NPhard (Theorem 6).
Given the fixedparameter 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.
Corollary 1.
Problem 71
GCAI
for does not admit any polynomial kernel with respect to the parameter unless the polynomial hierarchy collapses to the third level .
Corollary 2.
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.
Additionally, note that
Problem 73
GCAI
is FPT with respect to because it can be solved in time by a bruteforce 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.
Corollary 3.
Problem 75
GCAI
for does not admit any polynomial kernel with respect to the parameter unless the polynomial hierarchy collapses to the third level.
Corollary 4.
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 bruteforce 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 bruteforce 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.
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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 arcweighted digraph obtained from by performing the following operations:

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 .

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 YESinstance, 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
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 YESinstance.
Assume that the constructed instance of
Problem 88
DST
is a YESinstance, 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
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
Problem 89
DST
can be solved in time DBLP:journals/networks/DreyfusW71 ; DBLP:journals/siamdm/GuoNS11 ; DBLP:conf/stoc/BjorklundHKK07 ; DBLP:journals/mst/FuchsKMRRW07 . ∎