Component Order Connectivity in Directed Graphs

07/14/2020 ∙ by J. Bang-Jensen, et al. ∙ SDU Royal Holloway, University of London 0

A directed graph D is semicomplete if for every pair x,y of vertices of D, there is at least one arc between x and y. Thus, a tournament is a semicomplete digraph. In the Directed Component Order Connectivity (DCOC) problem, given a digraph D=(V,A) and a pair of natural numbers k and ℓ, we are to decide whether there is a subset X of V of size k such that the largest strong connectivity component in D-X has at most ℓ vertices. Note that DCOC reduces to the Directed Feedback Vertex Set problem for ℓ=1. We study parametered complexity of DCOC for general and semicomplete digraphs with the following parameters: k, ℓ,ℓ+k and n-ℓ. In particular, we prove that DCOC with parameter k on semicomplete digraphs can be solved in time O^*(2^16k) but not in time O^*(2^o(k)) unless the Exponential Time Hypothesis (ETH) fails. The upper bound O^*(2^16k) implies the upper bound O^*(2^16(n-ℓ)) for the parameter n-ℓ. We complement the latter by showing that there is no algorithm of time complexity O^*(2^o(n-ℓ)) unless ETH fails. Finally, we improve (in dependency on ℓ) the upper bound of Göke, Marx and Mnich (2019) for the time complexity of DCOC with parameter ℓ+k on general digraphs from O^*(2^O(kℓlog (kℓ))) to O^*(2^O(klog (kℓ))). Note that Drange, Dregi and van 't Hof (2016) proved that even for the undirected version of DCOC on split graphs there is no algorithm of running time O^*(2^o(klogℓ)) unless ETH fails and it is a long-standing problem to decide whether Directed Feedback Vertex Set admits an algorithm of time complexity O^*(2^o(klog k)).

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1 Introduction

Motivated by various practical network applications, many different vulnerability measures of undirected graphs have been introduced and studied in the literature. The two most studied of such measures are vertex and edge connectivity of an undirected graph. However, these two measures often do not capture the more subtle vulnerability properties of networks that one might wish to consider, such as the number of vertices in the largest remaining connected component.

While both undirected and directed graphs are of great interest in graph theory and algorithms applications, undirected graphs have been studied much more than their directed counterparts arguably due to simpler structure of undirected graphs. In this paper, we study a number of parameterizations of a problem of interest from both theory and applications which was mainly studied for undirected graphs so far.

In many networks, the underlying graph is directed rather than undirected and the aim of this paper is to study an extension to directed graphs of the -component order connectivity of an undirected graph , which is the size of a minimum set such that where is the number of vertices in the largest connected component of (mco stands for maximum component order). By Component Order Connectivity will denote the following decision problem:

component order connectivity Input: A graph and a pair of natural numbers Question: Is there a subset of of size such that

For a survey on Component Order Connectivity, see Gross et al. [13]; for more recent research on the problem, see e.g. [11, 15, 16].

For a directed graph we define the -component order connectivity as the size of a minimum set such that where is the number of vertices in the largest strongly connected component of Using this definition of we state can state the following directed version of Component Order Connectivity.

directed component order connectivity Input: A digraph and a pair of natural numbers Question: Is there a subset of of size such that

In what follows, we will assume without loss of generality that (or, ). Indeed, if then our instance is a YES-instance since deleting any set of vertices implies

Clearly, Directed Component Order Connectivity is a generalization of Component Order Connectivity (each instance of Component Order Connectivity corresponds to an equivalent instance of Directed Component Order Connectivity, where is obtained from by replacing every edge of by a directed 2-cycle). For while Component Order Connectivity is equivalent to the Vertex Cover problem, Directed Component Order Connectivity is equivalent to the Directed Feedback Vertex Set problem. Unlike Vertex Cover whose fixed-parameter tractability is very easy to show, a fact that was known very early on in parameterized algorithmics [9], fixed-parameter tractability of Directed Feedback Vertex Set was a long-standing open problem until Chen et al. [7] in 2008 proved its fixed-parameter tractability by designing a -time algorithm. We provide basics on parameterized algorithms and complexity in the next section.

Since Component Order Connectivity is NP-complete (it remains NP-complete even for split, co-bipartite and chordal undirected graphs [11]), a number of researchers studied Component Order Connectivity using the framework of parameterized algorithmics, see e.g. [11, 15, 16]. Göke, Marx and Mnich [12] were the first to study the Directed Component Order Connectivity problem from the viewpoint of parameterized algorithms and complexity. They obtained an algorithm of running time which is close to the complexity of the algorithm of Chen et al. [7] when . Thus, Directed Component Order Connectivity parameterized by is fixed-parameter tractable (FPT).

We will continue the study of Directed Component Order Connectivity using parameterized algorithms and complexity. In particular, as in papers [11, 15, 16] which studied Component Order Connectivity, we study Directed Component Order Connectivity parameterized by three parameters: , and We will denote the corresponding parameterized problems by Directed Component Order Connectivity[], Directed Component Order Connectivity[] and Directed Component Order Connectivity[], respectively.

Moreover, we introduce and study a new parameterization of Directed Component Order Connectivity: parameter where is the number of vertices in One reason to introduce Directed Component Order Connectivity[] is that normally one requires the parameters to be relatively small compaired to the size of the problem under consideration. However, if is small it is possible that for every of size , is not much smaller than Then can be much smaller than

Since Component Order Connectivity is equivalent to the Vertex Cover problem for , Component Order Connectivity[] is para-NP-complete. Drange et al. [11, Theorem 8] proved that Component Order Connectivity[] is W[1]-hard even on split graphs. In their construction, Hence, Component Order Connectivity[] is also W[1]-hard. They also showed that Component Order Connectivity[] is FPT by obtaining an algorithm of running time The above mentioned results are written in the undirected graphs row of Table 1.

A directed graph is semicomplete if for every pair of distinct vertices of , there is an arc between and When we require that there is only one arc between and then we obtain a definition of a tournament. Clearly, the hardness results for the directed graphs row of Table 1 follow from the corresponding results in the undirected graphs row for columns and . Directed Component Order Connectivity[] is para-NP-complete for semicomplete digraphs as Directed Component Order Connectivity on semicomplete digraphs is NP-complete for . This follows from the fact that Directed Feedback Vertex Set is NP-complete even for tournaments, as proved by Bang-Jensen and Thomassen [3] and Speckenmeyer [18].

The FPT result in directed graphs row of Table 1 is first obtained by Göke et al. [12] as discussed above. The running time of their algorithm is By modifying their algorithm, we obtained an algorithm of complexity which decreases asymptotic dependence of the running time on 111We obtained this result independently from [17]; our approach is different from that in [17]. Our modification consists of replacing a branching algorithm in [12] with a randomized algorithm which can be derandomized without increasing the complexity upper bound. Note that Drange et al. [11, Theorem 14] proved that even for Component Order Connectivity on split graphs there is no algorithm of running time (here we assume that ) unless the Exponential Time Hypothesis (ETH) [14] fails and it is a long-standing problem to decide whether Directed Feedback Vertex Set admits an algorithm of time complexity

class of graphs
semicomplete digraphs FPT FPT para-NP-c. FPT
undirected graphs W[1]-hard W[1]-hard para-NP-c. FPT
directed graphs W[1]-hard W[1]-hard para-NP-c. FPT
Table 1: Parameterized Complexity of (Directed) Component Order Connectivity

The most interesting entry in the semicomplete digraphs row is a non-trivial result that Directed Component Order Connectivity[] on semicomplete digraphs is FPT. This FPT algorithm boils down to finding a shortest path in a suitably defined auxiliary weighted acyclic digraph. The running time of the algorithm is The other two FPT entries in this row follow from this result (for the parameter this is due to our assumption that ). We also prove the following lower bounds: no algorithm for Directed Component Order Connectivity[] on semicomplete digraphs can have time complexity unless ETH fails222Similarly, no algorithm for Directed Component Order Connectivity[] on semicomplete digraphs can have running time , unless ETH fails. and no such deterministic algorithm can run in time

Our paper is organised as follows. The next section is devoted to terminology and notation on directed and undirected graphs, and basics on parameterized algorithms and complexity. In Section 3, we describe our improvement on the algorithm of Göke et al. [12]. In Section 4, we prove that Directed Component Order Connectivity[] on semicomplete digraphs admits an algorithm of running time and show the lower bounds on running time with parameters and . We conclude the paper in Section 5.

2 Preliminaries

2.1 Directed and Undirected Graph Terminology and Notation

In this paper, all directed and undirected graphs are finite, without loops or parallel edges. As often the case in the directed graph theory, an edge of a digraph will be called an arc and the vertex and arc sets of a digraph will be denoted by and respectively. The out-neighbourhood and in-neighbourhood of a vertex of a digraph are denoted by and , respectively, and the subscript will be omitted if is clear from the context. The out-degree and in-degree of a vertex of is and respectively.

In this paper all paths and cycles in digraphs are directed, so we will omit the adjective ‘directed’ when referring to paths and cycles in digraphs. If is a digraph and , then we denote by the subdigraph induced by the vertices in . A digraph is strongly connected (or, just strong) if there is a path from to

for every ordered pair

of distinct vertices. A strong component of a digraph is a maximal strong induced subgraph of Strong components of do not share vertices and can be ordered such that there is no arc in from to when Such an ordering is called an acyclic ordering. Note that if is a semicomplete digraph, then the strong components of have a unique acyclic ordering and we have for every

Basic digraph terminology not introduced in this section can be found in [1, 2].

2.2 Parameterized Complexity

An instance of a parameterized problem is a pair where is the main part and is the parameter; the latter is usually a non-negative integer. A parameterized problem is fixed-parameter tractable (FPT) if there exists a computable function such that instances can be solved in time where denotes the size of  and is an absolute constant. The class of all fixed-parameter tractable decision problems is called FPT and algorithms which run in the time specified above are called FPT algorithms. As in other literature on FPT algorithms, we will sometimes omit the polynomial factor in and write instead.

While FPT is a parameterized complexity analog of P in classic complexity, there are many hardness classes in parameterized complexity and they form a nested sequence starting from W[1]. It is well known that if the Exponential Time Hypothesis holds then FPTW[1]. Due to this and other complexity results, it is widely believed that FPTW[1] and hence W[1] is viewed as a parameterized analog of NP in classical complexity.

para-NP is the class of parameterized problems which can be solved by a nondeterministic algorithm in time where is a computable function and is an absolute constant. It is well-known that if a problem with parameter is NP-complete when equals to some constant, then is para-NP-complete. It is also well known that FPT=para-NP if and only if P=NP.

For more information on parameterized algorithms and complexity, see recent books [8, 10].

3 Directed Component Order Connectivity[] on General Digraphs

Göke, Marx and Mnich [12] showed that Directed Component Order Connectivity[] is FPT with a running time given as

The core of their algorithm is as follows. Begin with the iterative compression version of the problem, where in addition to the input also contains a solution with , which can be used to guide the search for a smaller solution. This is a standard ingredient in FPT algorithms; see, e.g., [8]. At the cost of a simple branching step, we may also assume that we are looking for a solution with . Next, they observe that if we knew the strongly connected components of that the vertices of are contained in, then the problem reduces to a previously studied, simpler problem known as Skew Separator [7], which occurs in the design of the FPT algorithm for Directed Feedback Vertex Set (DFVS) of Chen et al. [7]. Indeed, if the precise strong components containing the vertices of are known, then the problem can be solved in time using a strategy much like that for DFVS [7, 12]. Hence the bottleneck in Directed Component Order Connectivity[] is the guessing of the strong components of in .

Göke et al. [12] solve this via a branching algorithm that they analyse as taking time at most . We show a simpler randomized method solving this problem with an improved time bound of

(1)

The method can be derandomized by standard means.

Lemma 3.1.

Let be an instance of Directed Component Order Connectivity[], and let be a solution with . Let be an unknown solution with such that

. There is a randomized procedure that with success probability at least

computes a set such that for every , the strong components containing in and in are identical.

Proof.

Initialize , then for every vertex place in independently at random with probability . We declare a guess a success if the following conditions apply:

  1. For every we have , where is the strong component of containing

Let . Our guess is successful if and only if for every , and for every . Since these are independent events, this clearly happens with probability precisely

hence the worst case occurs when all sets are disjoint and have , and , i.e., and . Let . We bound the probability of success carefully in two steps:

  1. We estimate the probability that

    , without caring about the precise intersection (i.e., success in this stage includes cases where ).

  2. We estimate the probability of success, conditional on the previous event.

Note by assumption.

For the first step, note that the expected number of vertices of not in is

Also note that in a successful guess, this value is precisely . Hence the expected value differs from the intended value by less than 1. Since

is a binomial distribution, due to the guesses being independent, this clearly happens with probability at least inverse polynomial in

.

Subject to this event, the set

is uniformly distributed among all subsets of

of size by independence, hence the conditional probability of success is one in . We conclude that the success probability matches the bound in the lemma.

Finally, assume that the guess was successful for some set and consider the strong component of in for some . Let be this strong component. Since is strongly connected and , we have . On the other hand, by assumption is an induced subgraph of , and since is a strongly connected component in we must have . We conclude for each , as required. ∎

For the derandomization, we employ a cover-free family construction of Bshouty and Gabizon [4]. We get the following:

Lemma 3.2.

There is a deterministic procedure that produces a set with

in time , such that there is a set such that for every , the strong components containing in and in are identical.

Proof.

Let be integers. Bshouty and Gabizon (in a slightly non-standard definition) define an -cover free family as a set such that for every disjoint pair of sets with and there is a set such that and . Bshouty and Gabizon [4] show how to compute an -cover free family of size

in time .

By Lemma 3.1, it suffices to construct a cover-free family with parameters , and . Here , but we can simply compute an -cover free family and take the complement of every member. Hence we get a family of size

computed in output-linear time. ∎

The two lemmas of this section and (1) imply the following:

Theorem 3.1.

There is a randomized FPT algorithm that solves Directed Component Order Connectivity[] in time with probability at least . The algorithm can be derandomized in the same time, up to a lower-order overhead factor.

4 Directed Component Order Connectivity on Semicomplete Digraphs

Let us first summarize main ideas behind our FPT algorithm, before providing more technical details. Let be a semicomplete digraph, and let of size such that . The vertices of can be partitioned into such that each is the vertex set of a strong component of and

  1. for every is , and

  2. for every with and every , we have and .

Figure 1: An example of a valid triple . A semicomplete digraph , the set is such that and are strong components of . and , where , . The arcs , , for are omitted as well as the arcs between and . The set is the set of the three red vertices, one in each of , , and , is a minimal vertex cover of the red arcs from to . Note that the vertex in is not in as the arc incident to it with the tail in is already covered by . Note also the blue vertex in , the only reason it is in is to reduce the size of and as such it will not appear in any , , in the set of valid triples defining these components.

In our algorithm, we would like to discover the strong components one by one in the ascending order from to . Now let be a partition of into (possibly empty) parts and let, for each , and , where , . Moreover, let be a subset of such that for each and we have and . See also Figure 1. Note that, given , it suffice to solve our problem in subgraphs and separately. Moreover, the set is basically the strong component up to few vertices in that are not incident to any arc with tail in or head in . Such vertices can actually be replaced in by any vertex in . It follows, that if we are given , then we can easily reconstruct a solution of size as plus some arbitrary vertices of to have at most vertices in each strong component of .

Therefore, our goal will be to search for triples , , where is a partition of and is a minimal subset of such that there is no arc in with and . The first step of our proof is to show that there are at most triples we need to consider (Lemma 4.4). We will call these important triples valid and we postpone the precise definition for later. The main reason for the bound is that we only need to consider triples for which and that if we fix (and hence also ), then vertices with out-degree at least (resp. in-degree at least ) have to be in (resp. in ) or in and we can fix these vertices in (resp. in ). Once we bound the number of the triples we need to consider, we can define compatible pairs of triples , for which and these triples, loosely speaking can define a strong component of with at most vertices as and the arcs from to are all hit by a vertex in . This allows us to create an auxiliary acyclic “state” digraph whose vertices are valid triples and arcs are the compatible pairs of triples. The paths from to in this graph then define a solution for . Note that our algorithm can be equivalently seen as a dynamic programming which computes for each valid triple a minimum size set such that .

The following lemma allows us to show that if we fix in a triple , then only vertices of could potentially be in both and and all other vertices are fixed. The lemma is an easy consequence of the fact that every semicomplete digraph on at least , , vertices has a vertex of out-degree at least . We give the proof for the convenience of the reader.

Lemma 4.1.

Let be a semicomplete digraph and let be a partition of such that for every and every we have . Then for every (1) there are at most vertices in with and (2) there are at most vertices in with .

Proof.

We will first prove Part (1). Let be the set of vertices in with out-degree at most in . Since for every and every is , it follows that all vertices in have out-degree at most in . Hence , i.e., the sum of out-degrees of vertices in in , is at most . Hence,

Since is a semicomplete digraph,

It follows that Part (2) follows directly from Part (1) applied to a digraph obtained from by reversing all the arcs i.e. . ∎

Let be a semicomplete digraph and . We will call a triple -valid if

  1. is a partition of with ,

  2. is a minimal (w.r.t. inclusion) set such that for all and , if , then ,

  3. ,

  4. for all , if , then ,

  5. for all , if , then .

We will say a triple is valid, if it is -valid for some . The following simple observation will help us bound the number of partitions that could lead to a -valid triple .

Lemma 4.2.

For any -valid triple all vertices with are in and all vertices with are in .

Proof.

If , the lemma follows directly from the definition of a -valid triple. If and , then has an out-neighbour in , because , and follows by property 2. Similarly, if and , then has an in-neighbour in and by property 2. ∎

Lemma 4.3.

Let be a semicomplete digraph, , and let . If there exists a -valid triple, then there are at most vertices in with and .

Proof.

Let us assume that there is at least one -valid triple and let us denote it . Note that for all and it holds that . Since is a semicomplete digraph, it follows that . Due to Lemma 4.1 applied to there are at most vertices in with and there are at most vertices in with . Let By the above,

Thus,

Lemma 4.4.

Let be a semicomplete digraph, , and let . There are at most -valid triples . Moreover, if we are given the in- and out-degrees of all vertices in on the input, then we can enumerate all such triples in time .

Proof.

Let By Lemma 4.3, . If the out- and in-degrees of all vertices in are given on the input, we can construct the set in time By Lemma 4.2, there are at most possible partitions that could lead to a -valid triple for some , each such partition is uniquely determinate by fixing .

For the rest of the proof, we assume that we computed the set of vertices in with and , . Let be one of partitions that could lead to a -valid triple. We show that we can enumerate all minimal sets , , such that for all and , if , then . Let be an undirected bipartite graph such that , the partite sets of are and and for every , it holds if and only if . Then is a minimal vertex cover in . Moreover, every minimal vertex cover in leads to a -valid triple . It is well known and easy to show that we can enumerate all minimal vertex covers of size at most in in time . This is done by including all vertices with degree at least in every vertex cover. If the resulting graph has more than edges, then there is no vertex of size at most  [5]. Then we can enumerate all vertex covers, by using simple search-tree algorithm that picks an edge, say , and recursively enumerate all minimal vertex covers of size at most that includes or , respectively. Given the algorithm, it is also easy to see that there are at most distinct minimal vertex covers of size at most .

It follows that there are at most -valid triples and we can enumerate all of them in time . ∎

We are now ready to present our algorithm.

Theorem 4.1.

There is an FPT algorithm that solves Directed Component Order Connectivity[] on semicomplete digraphs in time .

Proof.

Let be a semicomplete digraph and let be an instance of Directed Component Order Connectivity[].

Algorithm.

Our algorithm boils down to finding a shortest path in an auxiliary weighted acyclic digraph whose vertex set consists of all the valid triples. The main idea is to find a sequence of valid triples such that is a solution for and the strongly connected components of are subsets of , where and for all , , it holds that .

We define the weighted directed acyclic state graph as follows. The set of vertices is the set of all -valid triples for all . The set of arcs contains an arc from a -valid triple to a -valid triple if and only if the following conditions holds:

  • (and ),

  • if and , then ,

  • if , then , and

  • .

We let the weight of an arc from to be

This finishes the description of the auxiliary weighted acyclic digraph. In the remainder of the proof we first show that is YES-instance if and only if the cost of the shortest path in from to is at most . Afterwards, we bound by and prove that we can construct the auxiliary digraph in time. We can then find a shortest path from to in linear time, that is, in time since is acyclic (by dynamic programming using an acyclic ordering of the vertices), which finishes the proof.

Correctness of the Algorithm.

Suppose first that is a YES-instance of Directed Component Order Connectivity[] such that is a semicomplete digraph. Let be a minimum size solution for , that is, a minimum size set such that . Since is YES-instance, and , the vertices of can be partitioned in sets such that

  1. for every is , and

  2. for every with and every , we have and .

Our goal is to define a sequence of valid triples , , such that the arc is in and the cost of the path in defined by this sequence is . We will construct these triples from and with some additional restrictions that makes it easier to show that they indeed define a path in of cost at most . Namely, we will define them such that for all , the triples satisfy the following properties:

  1. is -valid for some ,

  2. ,

  3. ,

  4. ,

  5. and ,

  6. if and , then ,

  7. if , then .

It is straightforward to verify that, given the above properties, the arc

We first show that a sequence with the above properties indeed exists and defer the computation of the cost of the path defined by this sequence to later.

To obtain this sequence, we need to discuss how to distribute the vertices of in the sets ’s and ’s and how to compute , (Note that the partition of the vertices in is fixed by properties 2 and 3).

To distribute the vertices of between and , we put all with in and all with in . The remaining vertices in we can distribute arbitrarily, we only have to make sure that for all , , it holds that and . Now and for all and we have . The set is defined to be those vertices such that one of the following holds:

  1. and there exists such that ,

  2. and there is an arc , such that .

Note that all arcs from to are covered by and for each there is an arc from to with . Note that if , then for all . On the other hand, if , then there is such that . Moreover, for all , . Therefore, if , then . From the above two properties it follows that if , then . This finishes the proof of the existence of a sequence of valid triples with properties 1-7.

We claim that the cost of path following this sequence is . First note that if , then and for all it holds , hence every vertex in is counted in at most one of the sets . Now the set is precisely . If is in some set , then from the properties 5, 6 and 7 of the sequence of triples it follows that is in . Hence is precisely the number of vertices in that are in and in none of the sets , . Note that for such vertex and a vertex