Simultaneous Feedback Edge Set: A Parameterized Perspective^{1}
Abstract
In a recent article Agrawal et al. (STACS 2016) studied a simultaneous variant of the classic Feedback Vertex Set problem, called Simultaneous Feedback Vertex Set (SimFVS). In this problem the input is an vertex graph , an integer and a coloring function , and the objective is to check whether there exists a vertex subset of cardinality at most in such that for all , is acyclic. Here, and . In this paper we consider the edge variant of the problem, namely, Simultaneous Feedback Edge Set (SimFES). In this problem, the input is same as the input of SimFVS and the objective is to check whether there is an edge subset of cardinality at most in such that for all , is acyclic. Unlike the vertex variant of the problem, when , the problem is equivalent to finding a maximal spanning forest and hence it is polynomial time solvable. We show that for SimFES is NPhard by giving a reduction from Vertex Cover on cubicgraphs. The same reduction shows that the problem does not admit an algorithm of running time unless ETH fails. This hardness result is complimented by an FPT algorithm for SimFES running in time , where is the exponent in the running time of matrix multiplication. The same algorithm gives a polynomial time algorithm for the case when . We also give a kernel for SimFES with vertices. Finally, we consider the problem Maximum Simultaneous Acyclic Subgraph. Here, the input is a graph , an integer and, a coloring function . The question is whether there is a edge subset of cardinality at least in such that for all , is acyclic. Here, . We give an FPT algorithm for Maximum Simultaneous Acyclic Subgraph running in time . All our algorithms are based on parameterized version of the Matroid Parity problem.
1 Introduction
Deleting at most vertices or edges from a given graph , so that the resulting graph belongs to a particular family of graphs (), is an important research direction in the fields of graph algorithms and parameterized complexity. For a family of graphs , given a graph and an integer , the deletion (Edge deletion) problem asks whether we can delete at most vertices (edges) in so that the resulting graph belongs to . The deletion (Edge deletion) problems generalize many of the NPhard problems like Vertex Cover, Feedback vertex set, Odd cycle transversal, Edge Bipartization, etc. Inspired by applications, Cai and Ye introduced variants of deletion (Edge deletion) problems on edge colored graph [6]. Edge colored graphs are studied in graph theory with respect to various problems like Monochromatic and Heterochromatic Subgraphs [14], Alternating paths [5, 7, 19], Homomorphism in edgecolored graphs [2], Graph Partitioning in 2edge colored graphs [4] etc. One of the natural generalization to the classic deletion (Edge deletion) problems on edge colored graphs is the following. Given a graph with a coloring function , and an integer , we want to delete a set of at most edges/vertices in so that for each , belongs to . Here, is the graph with vertex set and edge set as . These problems are also called simultaneous variant of deletion (Edge deletion).
Cai and Ye studied the Dually Connected Induced subgraph and Dual Separator on 2edge colored graphs [6]. Agrawal et al. [1] studied a simultaneous variant of Feedback Vertex Set problem, called Simultaneous Feedback Vertex Set, in the realm of parameterized complexity. Here, the input is a graph , an integer , and a coloring function and the objective is to check whether there is a set of at most vertices in such that for all , is acyclic. Here, . In this paper we consider the edge variant of the problem, namely, Simultaneous Feedback Edge Set, in the realm of parameterized complexity.
In the Parameterized Complexity paradigm the main objective is to design an algorithm with running time , where is the parameter associated with the input, is the size of the input and is some computable function whose value depends only on . A problem which admits such an algorithm is said to be fixed parameter tractable parameterized by . Typically, for edge/vertex deletion problems one of the natural parameter that is associated with the input is the size of the solution we are looking for. Another objective in parameterized complexity is to design polynomial time preprocessing routines that reduces the size of the input as much as possible. The notion of such a preprocessing routine is captured by kernelization algorithms. The kernelization algorithm for a parameterized problem takes as input an instance of , runs in polynomial time and returns an equivalent instance of . Moreover, the size of the instance returned by the kernelization algorithm is bounded by , where is some computable function whose value depends only on . If is polynomial in , then the problem is said to admit a polynomial kernel. The instance returned by the kernelization is referred to as a kernel or a reduced instance. We refer the readers to the recent book of Cygan et al. [8] for a more detailed overview of parameterized complexity and kernelization.
A feedback edge set in a graph is such that is a forest. For a graph with a coloring function , simultaneous feedback edge set is a subset such that is a forest for all . Here, , where . Formally, the problem is stated below.
Simultaneous Feedback Edge Set (SimFES) Input: An vertex graph , and a coloring function Parameter: Question: Is there a simultaneous feedback edge set of cardinality at most in
Feedback Vertex Set (FVS) is one of the classic NPcomplete [12] problems and has been extensively studied from all the algorithmic paradigms that are meant for coping with NPhardness, such as approximation algorithms, parameterized complexity and moderately exponential time algorithms. The problem admits a factor approximation algorithm [3], an exact algorithm with running time [11], a deterministic parameterized algorithm of running in time [15], a randomized algorithm running in time [9], and a kernel with vertices [23]. Agrawal et al. [1] studied Simultaneous Feedback Vertex Set (SimFVS) and gave an FPT algorithm running in time and a kernel of size . Finally, unlike the FVS problem, SimFES is polynomial time solvable when , because it is equivalent to finding maximal spanning forest.
Our results and approach
In Section 3 we design an FPT algorithm for SimFES by reducing to Matroid Parity on direct sum of elongated cographic matroids of , (see Section 2 for definitions related to matroids). This algorithm runs in time . Unlike the vertex counterpart, we show that for (2edge colored graphs) SimFES is polynomial time solvable. This follows from the polynomial time algorithm for the Matroid parity problem. In Section 4 we show that for , SimFES is NPhard. Towards this, we give a reduction from the Vertex Cover in cubic graphs which is known to be NPhard [21]. Furthermore, the same reduction shows that the problem cannot be solved in time unless Exponential Time Hypothesis (ETH) fails [13]. We complement our FPT algorithms by showing that SimFES is W[1]hard when parameterized by the solution size (Section 5). When , we give a parameter preserving reduction from the Hitting Set problem, a well known W[2]hard problem parameterized by the solution size [8]. However, SimFES remains W[1]hard even when . We show this by giving a parameter preserving reduction from Partitioned Hitting Set problem, a variant of the Hitting set problem, defined in [1]. In [1], Partitioned Hitting Set was shown to be W[1]hard parameterized by the solution size. In Section 6 we give a kernel with vertices. Towards this we apply some of the standard preprocessing rules for obtaining kernel for Feedback Vertex Set and use the approach similar to the one developed for designing kernelization algorithm for SimFVS [1]. In Section 7 we give an FPT algorithm for the problem, when parameterized by the dual parameter. Formally, this problem is defined as follows.
Maximum Simultaneous Acyclic Subgraph (MaxSimSubgraph) Input: An vertex graph , and a function . Parameter: Question: Is there a subset such that and for all , is acyclic?
For solving MaxSimSubgraph we reduce it to an equivalent instance of the Matroid Parity problem. As an immediate corollary we get an exact algorithm for SimFES running in time .
2 Preliminaries
We denote the set of natural numbers by . For , by we denote the set . For a set , by we denote the set of all subsets of . We use the term ground set/ universe to distinguish a set from its subsets. We will use to denote the exponent in the running time of matrix multiplication, the current best known bound for is [24].
2.1 Graphs
We use the term graph to denote undirected graph. For a graph , by and we denote its vertex set and edge set, respectively. We will be considering finite graphs possibly having loops and multiedges. In the following, let be a graph and let be a subgraph of . By , we denote the degree of the vertex in , i.e, the number of edges in H which are incident with . A selfloop at a vertex contributes to the degree of . For any nonempty subset , the subgraphs of induced by , are denoted by and respectively. Similarly, for , the subgraph of induced by is denoted by ; its vertex set is and its edge set is . For , by we denote the graph obtained by deleting the edges in . We use the convention that a double edge and a selfloop is a cycle. An edge colored graph is a graph with a color function . By we will denote the color (or color) graph of , where and . For an edge colored graph , the total degree of a vertex is . We refer the reader to [10] for details on standard graph theoretic notations and terminologies.
2.2 Matroids
A pair , where is a ground set and is a family of subsets (called independent sets) of , is a matroid if it satisfies the following conditions:

,

if and then , and

if and , then there is such that .
The axiom (I2) is also called the hereditary property and a pair satisfying only (I2) is called hereditary family. An inclusion wise maximal subset of is called a basis of the matroid. Using axiom (I3) it is easy to show that all the bases of a matroid have the same size. This size is called the rank of the matroid , and is denoted by rank(M). We refer the reader to [22] for more details about matroids.
Representable Matroids
Let be a matrix over an arbitrary field and let be the set of columns of . For , we define matroid as follows. A set is independent (that is ) if the corresponding columns are linearly independent over . The matroids that can be defined by such a construction are called linear matroids, and if a matroid can be defined by a matrix over a field , then we say that the matroid is representable over . A matroid is called representable or linear if it is representable over some field .
Direct Sum of Matroids
Uniform Matroid
A pair over an element ground set , is called a uniform matroid if the family of independent sets is given by , where is some constant. This matroid is also denoted as .
Graphic and Cographic Matroid
Given a graph , the graphic matroid is defined by taking the edge set as universe and is in if and only if is a forest. Let be a graph and be the number of components in . The cographic matroid of is defined by taking the the edge set as universe and is in if and only if the number of connected components in is .
Proposition 2 ([22]).
Graphic and cographic matroids are representable over any field of size and such a representation can be found in time polynomial in the size of the graph.
Elongation of Matroid
Let be a matroid and be an integer such that . A elongation matroid of is a matroid with the universe as and is a basis of if and only if, it contains a basis of and . Observe that the rank of the matroid is .
Proposition 3 ([17]).
Let be a linear matroid of rank , over a ground set of size , which is representable over a field . Given a number , we can compute a representation of the elongation of , over the field in field operations over .
Matroid Parity
In our algorithms we use a known algorithm for Matroid Parity. Below we define Matroid Parity problem formally and state its algorithmic result.
Matroid Parity Input: A representation of a linear matroid , a partition of into blocks of size and a positive integer . Parameter: Question: Does there exist an independent set which is a union of blocks?
3 FPT Algorithm for Simultaneous Feedback Edge Set
In this section we design an algorithm for SimFES by giving a reduction to Matroid Parity on the direct sum of elongated cographic matroids associated with graphs restricted to different color classes.
We describe our algorithm, AlgoSimFES, for SimFES. Let be an input instance to SimFES. Recall that for , is the graph with vertex set as and edge set as . Let . Note that for all . Let be the number of connected components in . To make acyclic we need to delete at least edges from . Therefore, if there is such that , then AlgoSimFES returns No. We let . Observe that for , . We need to delete at least edges from to make acyclic. Therefore, the algorithm AlgSimFES for each , guesses , where and computes a solution of SimFES such that . Let be the elongation of the cographic matroid associated with .
Proposition 5.
Let be a graph with connected components and be an elongation of the cographic matroid associated with , where . Then is a basis of if and only if the subgraph is acyclic and .
Proof.
In the forward direction let be a basis of . By Definition of it follows that and contains a basis of the cographic matroid of . Suppose has a cycle. This implies that has a cycle. But then, there is an edge whose removal from does not increase the number of connected components in . This contradicts that was a basis in the cographic matroid of .
In the reverse direction let such that and is acyclic. Consider a inclusion wise maximal subset such that the number of connected components in is . Observe that does not contain a cycle since is acyclic and is inclusion wise maximal. Therefore, it follows that is a basis in the cographic matroid of . But then contains a basis of the cographic matroid of and , therefore is a basis in . ∎
By Proposition 5, for any basis in , is acyclic. Therefore, our objective is to compute such that and the elements of restricted to the elements of form a basis for all . For this we will construct an instance of Matroid Parity as follows. For each and , we use to denote the corresponding element in . For each , by we denote the set of elements . For each edge , we define . Finally, for each edge , by we denote the set . Let . Furthermore, let and . Let be a uniform matroid over the ground set . That is, . By Propositions 1 to Proposition 3 we know that s are representable over , where is a prime number and their representation can be computed in polynomial time. Let be the linear representation of for all . Notice that for all . Let denote the direct sum with its representation matrix being . Note that the ground set of is . Now we define an instance of Matroid Parity, which is the linear representation of and the partition of ground set into , . Notice that for all , . Also for each , and . This implies that .
Now AlgoSimFES outputs Yes if there is a basis (an independent set of cardinality ) of which is a union of blocks in and otherwise outputs No. AlgoSimFES uses the algorithm mentioned in Proposition 4 to check whether there is an independent set of , composed of blocks. A pseudocode of AlgoSimFES can be found in Algorithm 1.
Lemma 3.1.
AlgoSimFES is correct.
Proof.
Let be a Yes instance of SimFES and let , where be a solution of . Let , where is the number of connected components in , for all . For all , let . Since is a solution, for all . This implies that AlgoSimFES will not execute Step 1. Consider the for loop for the choice . We claim that the columns corresponding to form a basis in and it is union of blocks. Note that by construction. For all , let , which is subset of ground set of . By Proposition 5, for all , is a basis for . This takes care of all the edges in . Now let . Observe that . Also, is a subset of ground set of and thus is a basis since . Hence is a basis of . Note that is the union of blocks corresponding to and hence is union of blocks. Therefore, AlgoSimFES will output Yes.
In the reverse direction suppose AlgoSimFES outputs Yes. This implies that there is a basis, say , that is the union of blocks. By construction corresponds to union of the sets for some edges in . Let these edges be . We claim that is a solution of . Clearly . Since is a basis of , for each , is a basis in . Let . Since is a basis of , by Proposition 5, is an acyclic graph. ∎
Lemma 3.2.
AlgoSimFES runs in deterministic time .
Proof.
Since Matroid Parity for can be solved in polynomial time [18] algorithm AlgoSimFES runs in polynomial time for . This gives us the following theorem.
Theorem 3.1.
SimFES is in FPT and when SimFES is in P.
4 Hardness results for SimFES
In this section we show that when , SimFES is NPHard. Furthermore, from our reduction we conclude that it is unlikely that SimFES admits a subexponentialtime algorithm. We give a reduction from Vertex Cover (VC) in cubic graphs to the special case of SimFES where . Let be an instance of VC in cubic graphs, which asks whether the graph has a vertex cover of size at most . We assume without loss of generality that . It is known that VC in cubic graphs is NPhard [21] and unless the ETH fails, it cannot be solved in time
To construct , we first construct an instance of VC in subcubic graphs which is equivalent to . We set
That is, the graph is obtained from the graph by subdividing each edge in twice.
Lemma 4.1.
has a vertex cover of size if and only if has a vertex cover of size
Proof.
In the forward direction, let be a vertex cover in . We will construct a vertex cover in of size at most . Consider an edge . If both belongs to , then we arbitrarily add one of the vertices from to . If exactly one of belongs to , say then, we add to . If , then we add to . Clearly, is a vertex cover in and is of size at most .
In the reverse direction, given a vertex cover in . For each such that both and are in the vertex cover, we can replace by , and then, by removing all of the remaining vertices of the form (whose number is exactly ), we obtain a vertex cover of . ∎
Observe that in every path between two degree3 vertices contains an edge of the form . Thus, the following procedure results in a partition of such that for all , and , it holds that . Initially, . For each degree3 vertex , let , and be the edges containing . We insert into , into , and into (the choice of which edge is inserted into which set is arbitrary). Finally, we insert each edge of the form into a set that contains neither nor .
We are now ready to construct the instance . Let , where contains a copy of each vertex in . The set and coloring are constructed as follows. For each vertex , add an edge into and its colorset is . For each and for each , add the edges and into and its colorset is . We set . Clearly, the instance can be constructed in polynomial time, and it holds that .
Lemma 4.2 proves that is a Yes instance of VC if and only if is a Yes instance of SimFES. Observe that because of the above mentioned property of the partition of , we ensure that in , no vertex participates in two (or more) monochromatic cycles that have the same color. By construction, each monochromatic cycle in is of the form , where , and for each edge , where either or , contains exactly one monochromatic cycle of this form.
Lemma 4.2.
is a Yes instance of VC if and only if is a Yes instance of SimFES.
Proof.
In the forward direction, let be a vertex cover in of size at most . Define as the set of edges . We claim that is a solution to . Since , it holds that . Now, consider a monochromatic cycle in . Recall that such a cycle is of the form , where . Since is a vertex cover of , it holds that , which implies that .
In the reverse direction, let be a solution to . Recall that for each edge , where either or , contains exactly one monochromatic cycle of this form. Therefore, if contains an edge of the form or of the form , such an edge can be replaced by the edge . Thus, we can assume that only contains edges of the form . Define as the set of vertices . We claim that is a vertex cover of of size at most . Since , it holds that . Now, recall that for each edge , contains a monochromatic cycle of the form . Since is a solution to , it holds that , which implies that . ∎
Theorem 4.1.
SimFES where is NPhard. Furthermore, unless the Exponential Time Hypothesis (ETH) fails, SimFES when cannot be solved in time .
5 Tight Lower Bounds for SimFES
We show that SimFES parameterized by is hard when and hard when . Our reductions follow the approach of Agrawal et al. [1].
5.1 W[2] Hardness of SimFES when
We give a reduction from Hitting Set (HS) problem to SimFES where . Let be an instance of HS, where , which asks whether there exists a subset of size at most such that for all , . It is known that HS parameterized by is W[2]hard (see, e.g., [8]). Thus, to prove the result, it is sufficient to construct (in polynomial time) an instance of the form of SimFES that is equivalent to , where . We construct a graph such that and the number of colors used will be . The intuitive idea is to have one edge per element in the universe which is colored with all the indices of sets in the family that contains the element and for each creating a unique monochromatic cycle with color which passes through all the edges corresponding to the elements it contain. We explain the reduction formally in the next paragraph.
Without loss of generality we assume that each set in contains at least two elements from . The instance is constructed as follows. Initially, . For each element , insert two new vertices into , and , add the edge into and let be its colorset. Now, for all and for all such that and , perform the following operation: add a new vertex into , , add the edges and into and let their colorset be . Moreover, for each , let and be the elements with the largest and smallest index contained in , respectively, and perform the following operation: add a new vertex into , , add the edges and into , and let their colorset be . Observe that and that . Therefore, . It remains to show that the instances and are equivalent. By construction, each monochromatic cycle in is of the form , where , and for each set , contains exactly one such monochromatic cycle.
Lemma 5.1.
is a Yes instance of HS if and only if is a Yes instance of SimFES.
Proof.
In the forward direction, let be a solution to . Define as the set of edges . We claim that is a solution to . Since , it holds that . Now, consider a monochromatic cycle in . Recall that this cycle is of the form , where . In particular, observe that