Reducing the rank of a matroid

# Reducing the rank of a matroid

Gwenaël Joret
Département d’Informatique
Université Libre de Bruxelles
Brussels, Belgium
Department of Mathematics and Statistics, and School of Computer Science
McGill University
###### Abstract.

We consider the rank reduction problem for matroids: Given a matroid and an integer , find a minimum size subset of elements of whose removal reduces the rank of by at least . When is a graphical matroid this problem is the minimum -cut problem, which admits a -approximation algorithm. In this paper we show that the rank reduction problem for transversal matroids is essentially at least as hard to approximate as the densest -subgraph problem. We also prove that, while the problem is easily solvable in polynomial time for partition matroids, it is NP-hard when considering the intersection of two partition matroids. Our proof shows, in particular, that the maximum vertex cover problem is NP-hard on bipartite graphs, which answers an open problem of B. Simeone.

## 1. Introduction

Consider the well-known minimum -cut problem: Given a graph and an integer , find a minimum size subset of edges whose removal increases the number of connected components by at least . This problem is NP-hard, assuming is part of the input, and several -approximation algorithms have been developed for it over the years [14, 17, 18, 20]. Notice that the minimum -cut problem has a simple formulation in terms of matroids: Given a graph and an integer , find a minimum size subset of elements of the graphical matroid of whose removal reduces its rank by at least .

This observation motivates the study of the rank reduction problem in other classes of matroids. For example, is the rank reduction problem computationally hard and, if so, does it admit approximation algorithms with good approximation guarantees (as is the case for graphical matroids)? Moreover, as we will see, many fundamental problems can be formulated in this rank reduction framework.

In this paper, our focus is on the case of transversal matroids. First, we show that the rank reduction problem in transversal matroids is roughly at least as hard to approximate as the densest -subgraph problem: Given a graph and an integer , find a subset of vertices inducing a subgraph with a maximum number of edges. (Note that in our reduction the parameter is not necessarily the same one as in the rank reduction problem.) The densest -subgraph problem can be approximated to within a factor of due to a recent breakthrough result of [4]. Moreover, it is widely believed [5, 1, 10, 11] that its hardness is also close to this upper bound – indeed, Bhaskara et al. [5] present lower bounds for lift and project methods based upon the Sherali-Adams and the Lassere hierarchies. We will show in particular that an -approximation algorithm for the rank reduction problem on transversal matroids (where denotes the number of elements) implies an -approximation algorithm for the densest -subgraph (where is the number of vertices).

Secondly, we prove that while the rank reduction problem is easily solvable in polynomial time for partition matroids (a special class of transversal matroids), it is NP-hard when considering the intersection of two partition matroids. Our proof shows in particular that the maximum vertex cover problem—also known as the partial vertex cover problem—is NP-hard on bipartite graphs. Here, one is given a graph and a positive integer , and the goal is to find a set of vertices hitting as many edges as possible. The problem is obviously NP-hard on arbitrary graphs since it contains the vertex cover problem as a special case. Whether it remained NP-hard on bipartite graphs was an open problem of B. Simeone (see [15]). We note that we learned after finishing this paper that Apollonio and Simeone [2] independently obtained a proof of this result.

## 2. Preliminaries

In this section we give the necessary definitions and notations. All graphs and matroids in this paper are finite, and “graph” will always mean an undirected simple graph. We use the shorthands and for the number of vertices and edges of a graph , respectively. We denote by the maximum size of a matching in , which we call the matching number of .

A matroid is a pair where is a family of subsets, called the independent sets, of the ground set satisfying the following three axioms:

• the empty set is independent;

• every subset of an independent is again independent, and

• if and are two independent sets with then there exists such that is independent.

The inclusion-wise maximal independent sets are the bases of the matroid ; as follows from the third axiom the bases all have the same cardinality. The rank function of is the function that assigns to each subset of elements of the maximum size of an independent set contained in , called the rank of . In particular, is the cardinality of a basis of , which is called the rank of .

The rank reduction problem for matroids is defined as follows: Given a positive integer and a matroid on a set of elements with rank function , the goal is to find a minimum size subset such that .

For example consider the case of graphical matroids: Given a graph , the graphical matroid of is obtained by taking as ground set, and letting a subset of edges be independent if and only if the corresponding subgraph is acyclic. Here the rank reduction problem is the minimum -cut problem.

As stated, here we study transversal matroids. A bipartite graph with bipartition induces a matroid as follows: The matroid has as ground set, and is independent in if and only if there exists a matching of covering . The fact that this is indeed a matroid is well-known; see for instance [19]. Any matroid that can be obtained this way is called a transversal matroid, and the bipartite graph is said to be a model for . Observe that, letting denote the rank function of , the rank of is equal to . Also note that being a transversal matroid is a hereditary property, in the sense that for each set , taking the restriction of all independent sets yields a transversal matroid on ground set .

A special case of transversal matroids are partition matroids. Here we are given a collection of disjoint sets and integers such that for each . One can define a corresponding matroid with ground set by letting be independent if and only if for each . Such a matroid is called a partition matroid, with model . This corresponds to a transversal matroid on a bipartite graph with bipartition , where has vertices that are adjacent to all vertices in , and none other, for each set . Notice that partition matroids are also hereditary.

Throughout, since we restrict ourselves to specific families of matroids, we assume that the matroid is given concisely and not given explicitly as a set system in input. Specifically, a corresponding model of the matroid is provided: a bipartite graph for a transversal matroid, a graph for a graphical matroid, etc.

More generally, the rank reduction problem can be considered on the intersection of matroids. Given two matroids and with common ground set , the intersection of and is the pair where is the family of sets that are independent in both and , which are said to be the independent sets of . While the independence system is not necessarily a matroid anymore, it enjoys several of the nice properties of matroids (see [19]). In particular, letting as before the rank of be the maximum size of an independent set of contained in , the rank of can be computed in polynomial time given access to the rank functions and of and , respectively, by a classical result of Edmonds (see [19]). We examine the rank reduction problem for the intersection of two partition matroids in Section 4.

## 3. Transversal Matroids

We start our investigation of the rank reduction problem with an easy observation, namely that the problem can be solved in polynomial time if the input matroid is a partition matroid.

###### Theorem 3.1.

The rank reduction problem can be solved in polynomial time on partition matroids.

###### Proof.

Let be a given partition matroid with model and rank function . Let denote the ground set of . Observe that . Let be the given parameter for the rank reduction problem on . We may assume that . Let for each . Given , the rank of the set is equal to .

Let be such that . Moreover, assume is inclusion-wise minimal with this property. Then, for each , either or . Moreover, letting be the subset of indices such that , we have that and .

Conversely, suppose is such that . Then choosing arbitrarily elements of , for each , plus additional elements from gives a set with such that .

Therefore, computing an optimal solution to the rank reduction problem reduces to the problem of finding a subset such that and is minimum. Thus we obtain a knapsack problem. Moreover, as and are at most , they are of polynomial size when encoded in unary. Thus the knapsack problem can be solved easily in polynomial time using dynamic programming. ∎

While the rank reduction problem admits a simple polynomial-time algorithm on partition matroids, the problem turns out to be more difficult on the broader class of transversal matroids. In fact, up to some degree, the problem can be viewed as a generalization of the densest -subgraph problem. In the latter problem, one is given a graph and a positive integer , and the aim is to find a subgraph of with and maximum. Towards this goal, we consider a closely related problem, the minimum -edge subgraph problem: Given a graph and a positive integer , the goal is to find a subgraph of with and minimum.

We start by drawing a connection between the rank reduction problem on transversal matroids and the minimum -edge subgraph problem. Then we will extend the connection to the densest -subgraph problem.

###### Lemma 3.2.

For each constant with , every -approximation algorithm for the rank reduction problem on transversal matroids with elements can be turned into an -approximation algorithm for the minimum -edge subgraph problem on graphs with vertices.

###### Proof.

Let be an instance of the minimum -edge subgraph problem. Let . Let be disjoint copies of . Let be a disjoint copy of . Let be the bipartite graph with bipartition where and , and where is adjacent to if either corresponds to a vertex of that is incident to the edge corresponding to in , or if and correspond to the same edge of .

Let denote the rank function of the transversal matroid induced by on ; thus for , is the maximum size of a matching in . Obviously, , since every can be matched to its copy in . Let denote the number of elements of the transversal matroid. Now consider the rank reduction problem on this matroid with . Recall that a feasible solution is a subset such that .

As is well known, we have that for if and only if there exists such that , where denotes the set of vertices of that have a neighbor in .111This is a consequence of Hall’s Marriage Theorem, as we now explain for completeness. Add new vertices to the set , yielding a set , and make each of them adjacent to every vertex in . Let be the bipartite graph obtained from in this manner. Let . Then every matching in with can be extended to a matching of with . Conversely, every matching in with yields a matching of with by discarding the at most edges of incident to the vertices in . Hence, has a matching of size —or equivalently, —if and only if can be completely matched in . By Hall’s theorem, the latter happens if and only if for every , which is equivalent to for every . Therefore, if and only if there exists such that . Such a set is said to be a witness for . The set defines in turn a corresponding subgraph of consisting of all the edges of included in , and the vertices of incident to those edges. By definition of , the set consists of the copies of each vertex of , along with the copies in of each edge of . Observe that any set obtained by taking the copies in of each vertex of and arbitrarily chosen edges of in is such that . Moreover, since and , it follows that , that is, is a solution of size no greater than and having the same witness . Such a pair is called a canonical pair.

Now, if a canonical pair is such that , then consists of exactly edges of (or more precisely, their copies in ). For each with corresponding copy in , we have that is again a solution to the rank reduction problem, with witness , and of size smaller than .

To summarize the above discussion, given an arbitrary set such that , one can in polynomial time compute a canonical pair with and .

Conversely, for every subgraph with , there is a natural corresponding canonical pair , where contains the copies in of the edges of , and where . Letting and denote the size of an optimal solution for the rank reduction and minimum -edge subgraph problems, respectively, it follows that .

Now suppose that the rank reduction problem admits a -approximation algorithm, where and are absolute constants. Letting be a canonical pair with obtained using this algorithm, we have

 n|GY|+t=|X|⩽cmεx∗=c(n2+|E|)ε(nj∗+t)⩽c(2n2)ε(nj∗+t)

and hence

 |GY|⩽c(2n2)ε(nj∗+t)−tn⩽2cn2ε(j∗+tn)⩽2cn2ε(j∗+j∗)=4cn2εj∗.

(In the last inequality we used the fact that , and thus .) Therefore, is a -edge subgraph whose order is within a -factor of optimal. ∎

As pointed out to us by an anonymous referee, the following lemma is implicit in the recent work of Chlamtac, Dinitz, and Krauthgamer [8] (in [8], the minimum -edge subgraph problem is called the smallest -edge subgraph problem). We include a proof nevertheless, for completeness.

###### Lemma 3.3.

For each constant with , every -approximation algorithm for the minimum -edge subgraph problem can be turned into an -approximation algorithm for the densest -subgraph problem.

###### Proof.

Suppose that the minimum -edge subgraph problem admits a -approximation algorithm, which we denote , where and are absolute constants. Let be an instance of the densest -subgraph problem. As before, we let and denote the number of vertices and edges of , respectively. We may assume . Since and , there exists with such that is an integer. We will consider the approximation factor of to be in what follows, to avoid cumbersome floors and ceilings in the calculations.

Run algorithm on with . Let be the th subgraph returned by the algorithm. Clearly, we may suppose that for each .

Let denote the number of edges in an optimal solution to the densest -subgraph problem on . If then let , otherwise let be the index in such that and . Since algorithm is a -approximation algorithm, and since either or , it follows that every subgraph of with exactly vertices has at most edges, that is, .

Let . Observe that . Let be a partition of the vertex set of into subsets with and . Let be a pair with such that is maximized. By the pigeonhole principle,

 ||Ht′[Vi∪Vj]||⩾||Ht′||(q2)=t′(q2)⩾t′(3cnε)2⩾z∗9c2n2ε.

If , then let . If, on the other hand, , then let where is an arbitrary subset of of size . Thus in both cases and . Hence, is a solution to the densest -subgraph problem on whose number of edges is within a -factor of the optimum. ∎

Combining Lemma 3.2 and 3.3 gives:

###### Theorem 3.4.

For each constant with , every -approximation algorithm for the rank reduction problem on transversal matroids with elements can be turned into an -approximation algorithm for the densest -subgraph problem on graphs with vertices.

As discussed in the introduction, the best approximation algorithm for the densest -subgraph problem currently known has an approximation ratio of for any fixed  [4] and it is conjectured that the inapproximability of the problem is of a similar magnitude. It would be nice to obtain strong inapproximability bounds for the rank reduction problem that do not rely on this conjecture. One approach may be to analyze hypergraphs as the rank reduction problem in transversal matroids incorporates the hypergraph version of the minimum -edge subgraph problem. That is, we wish to select as few vertices as possible that induce at least hyperedges. Perhaps surprisingly, little is known about this problem. As far as we are aware, the only specific hardness result is NP-hardess due Vinterbo [21] who studied the problem in the context of making medical databases anonymous.

We conclude this section with a remark about the approximability of the minimum -edge subgraph problem itself. Given the existence of a -approximation algorithm for the densest -subgraph problem, in view of Lemma 3.3 it is perhaps natural to wonder whether one could achieve a -approximation for the former problem. While this is still open as far as we know, Chlamtac et al. [8] recently made progress in that direction by describing an algorithm for the minimum -edge subgraph problem with an approximation ratio of for fixed .

## 4. The Maximum Vertex Cover Problem in Bipartite Graphs

As we have seen, the rank reduction problem admits a fairly simple polynomial-time algorithm on partition matroids but becomes much harder on transversal matroids, in the sense that approximation algorithms offering good guarantees seem unlikely to exist. Another interesting generalization of the case of partition matroids is to consider the intersection of two partition matroids.

As is well-known, the set of matchings of a bipartite graph with bipartition can be modeled as the family of common independent sets of two partition matroids and defined on : Take to be the partition matroid with model and the partition matroid with model , where , , and for the set denotes the set of edges incident to . Hence, in this specific case the rank reduction problem on amounts to finding a subset of edges of of minimum size such that . In this section we show that this problem is NP-hard. More accurately, we show that a problem polynomially equivalent to it, the maximum vertex cover problem on bipartite graphs, is NP-hard; see Theorem 4.1. This solves an open problem of B. Simeone (see [15]).

The maximum vertex cover problem (also known as the partial vertex cover problem) is defined as follows: Given a graph and a positive integer , find a subset of vertices of with such that the number of edges covered by is maximized. (An edge of is covered by if has at least one endpoint in .)

Now, if is bipartite, is a positive integer with , and is a subset of edges of such that , then by Kőnig’s theorem has a vertex cover of size , and hence covers at least edges of . (We remark that could cover some edges of too, and that can be computed in polynomial time given .). Conversely, for every set with , the set of edges of not covered by is such that . Therefore, for bipartite graphs, the maximum vertex cover problem is polynomially equivalent to that of finding a minimum-size set of edges decreasing the matching number by a prescribed amount.

It should be noted that two recent works [9, 6] with an overlapping set of authors claim that the NP-hardness of the maximum vertex cover problem on bipartite graphs can be derived directly from the reduction of Corneil and Perl [7] showing that the densest -subgraph problem is NP-hard on bipartite graphs. However, the argument relating the latter reduction to the maximum vertex cover problem, described explicitly in [9, Lemma 4], is flawed.222As mentioned in [9, Lemma 3], the maximum vertex cover problem in bipartite graphs is polynomially equivalent to the densest -subgraph problem in complements of bipartite graphs. Thus one may equivalently consider the complexity of the latter problem. In the proof of Lemma 4 in [9], the authors point out that the reduction of [7] implies that the problem of finding a densest -subgraph in the complement of a bipartite graph with bipartition with and with the extra requirement that it contains exactly vertices from and vertices from is NP-hard. From this they wrongly conclude that the densest -subgraph problem, without this extra constraint, is also NP-hard on complements of a bipartite graphs. (In fact, the instances obtained via the reduction in [7] satisfy , and thus a densest -subgraph is trivially obtained by taking vertices in the clique .) We also mention that the proof of Theorem 1 in [6], showing that a related problem called the maximum quasi-independent set problem is NP-hard on bipartite graphs, relies on the assumption that the maximum vertex cover problem is NP-hard on bipartite graphs. Thus our result also fills a gap in that proof.

###### Theorem 4.1.

The maximum vertex cover problem is NP-hard on bipartite graphs.

Before proving Theorem 4.1, we need to introduce a technical lemma.

###### Lemma 4.2.

Let be an integer with . Then the integer program

 \em\bf minimizex+2y+3z\em s.\ t.x+y+z−s=(ℓ2)−ℓx⩽(s2)x,y,z,s∈N

has a unique optimal solution given by .

###### Proof.

The proof is a straightforward case analysis. Consider an optimal solution to the integer program and, arguing by contradiction, assume it differs from the solution described above. Let .

Case 1: . We have and thus

 y+z=(ℓ2)−ℓ+s−x⩾(ℓ2)−ℓ.

It follows that . But since , contradicting the optimality of the solution.

Case 2: .

Here, the last inequality follows from the fact that .

Now, increment by , by , and decrement and in such a way that they remain non-negative integers and that the sum decreases by exactly . The modified solution is still feasible and decreases by at least , a contradiction.

Case 3: . Then , since otherwise we would have the solution described in the lemma statement. It follows that . Thus we can increment by and decrement by a positive variable among . This strictly decreases , a contradiction.

Case 4: . Then . But , otherwise the solution cannot be minimum. Therefore . Thus we improve the solution by incrementing by and decrementing by a positive variable among . ∎

Now we may turn to the proof of Theorem 4.1.

###### Proof of Theorem 4.1..

The reduction is from the NP-complete problem Clique: Given a graph and an integer , decide whether contains a clique on vertices or not. We may assume (otherwise, we simply check the existence of an -clique by brute force). We may also suppose that has minimum degree at least . Indeed, a vertex with degree at most cannot be part of an -clique, and thus those vertices can iteratively be removed from the graph. Finally, we assume that . This last assumption can also be made without loss of generality. Indeed, if is too small then one can simply consider the disjoint union of with a large enough -regular graph; since no vertex from this new -regular component can be part of an -clique.

We build an instance of the maximum vertex cover problem as follows. First, create two adjacent vertices and for every vertex , and similarly two adjacent vertices and for every edge . Next, for every edge , add the edges , and , . Finally, let

 k:=|H|+||H||−(ℓ2)+ℓ.

Observe that is bipartite with bipartition

 ({ax:x∈V(H)∪E(H)},{bx:x∈V(H)∪E(H)}).

See Figure 1 for an illustration of the construction.

A feasible solution for this instance of the maximum vertex cover problem is a subset of vertices of with , which we call a partial vertex cover for short. We let denote the number of edges covered by such a set . Let denote the maximum of over every partial vertex cover of .

A partial vertex cover of is nice if

 X∩{au,bu}∈{{au},{au,bu}}

for every and

 X∩{ae,be}∈{∅,{ae}}

for every .

###### Claim 4.3.

Given a partial vertex cover of , one can find a nice partial vertex cover of with .

###### Proof.

First we define a partial vertex cover based on which is close to being nice: Let

 ~X:= ∪{ae,be:e∈E(H),ae,be∈X}∪{ae:e∈E(H),|{ae,be}∩X|=1}.

By construction . Clearly, an edge with is covered by if and only if it is covered by . Also, given a pair of vertex and edge such that is incident to in , the set covers at least as many edges in as (though not necessarily the same ones). It follows that .

A useful property of the set is that if for some then necessarily . For simplicity we call this property the -property of .

We need to introduce an additional definition. An element is said to be bad in a partial vertex cover of if either and ( is a bad vertex), or and ( is a bad edge). Observe that is nice if and only if has the -property and there is no bad element.

Suppose is an edge of which is bad in . If or is also bad in , say , then let

 ~X′:=(~X−{be})∪{au}.

We have , thus the edge is still covered by . Since covers also , there is at most one edge incident to in (namely, ) which is not covered by . On the other hand, covers the previously uncovered edge . Hence, is a partial vertex cover with . Observe that still has the -property, and the edge is no longer bad in .

If, on the other hand, none of is bad in , then by the -property. Since , it follows that . There exists an element such that , because (since ). Let then

 ~X′:=(~X−{be})∪{ax}.

The set is a partial vertex cover with the -property and with . Moreover, the edge is no longer bad in .

Now apply iteratively the above modifications on as long as there exists a bad edge. This results in a partial vertex cover with the -property, without bad edges, and with .

Next we deal with bad vertices in . Suppose is such a vertex, that is, . Consider two edges incident to in . (Recall that has minimum degree at least .) Since and by our assumption on , we have . Together with the -property of , it follows that for some edge (possibly or ). Note that , because otherwise would be a bad edge for . Let

 ˆX′:=(ˆX−{ae′})∪{au}.

Since has degree in we have . Furthermore, as there are no bad edges in . Thus, the three edges of were not covered by but are covered by , so we have . Similarly as before, the partial vertex cover has the -property and one less bad vertex than . Therefore, iterating this procedure as long as there is a bad vertex, we eventually obtain a partial vertex cover with having the -property and no bad element, as desired. ∎

Consider a nice partial vertex cover of . Let be the set of vertices such that , and let . An edge of satisfies exactly one of the following three conditions:

1. ;

2. exactly one of belongs to ,

3. .

We say that edge is of type () if satisfies the th condition above and moreover . (We will focus on edges of such that in what follows, which is why the other ones do not get assigned a type.) Let be the set of edges of with type , and let .

###### Claim 4.4.

Let be a nice partial vertex cover. Then

 c(X)=||G||−e1(X)−2e2(X)−3e3(X).
###### Proof.

As is nice, for all . Therefore every edge of the form or is covered. Also, for each edge we have that are all covered. Thus the only uncovered edges are incident to vertices where . Suppose and let us consider which edges among the three edges are covered by . If , then covers and but not . If , then covers exactly one of , , and avoids . If , then covers none of the three edges. Hence, the total number of edges not covered by is exactly . ∎

###### Claim 4.5.

Let be a nice partial vertex cover. Then

 c(X)⩽||G||−(ℓ2),

with equality if and only if , , .

###### Proof.

Let , , and . Then , where by the previous claim.

Every edge in has its two endpoints in ; hence,

 (1) (s2)⩾x.

Also,

 |H|+||H||−(x+y+z−s)=|X|=k=|H|+||H||−((ℓ2)−ℓ),

and thus

 (2) x+y+z−s=(ℓ2)−ℓ.

Since and are non-negative integers satisfying (1) and (2), by Lemma 4.2 we have , or equivalently,

 c(X)⩽||G||−(ℓ2).

Moreover, equality holds if and only if by the same lemma. ∎

It follows from Claims 4.3 and 4.5 that

 OPT⩽||G||−(ℓ2).

If has an -clique , then the subset defined by

 X:= {au,bu:u∈V(K)}∪{au:u∈V(H)−V(K)} ∪{ae:e∈E(H)−E(K)}

is a partial vertex cover of with , implying .

Conversely, if , then there exists a partial vertex cover of with , and by Claim 4.3 we may assume that is nice. From Claim 4.5 we then have , , , implying that induces an -clique in .

Therefore, we can decide in polynomial time if has an -clique by checking if . This concludes the proof. ∎

The NP-hardness of the maximum vertex cover problem on bipartite graphs motivates the search for non-trivial approximation algorithms for this class of graphs. A recent result in this direction is due to Apollonio and Simeone [3], who gave an LP-based -approximation algorithm for bipartite graphs.

## Acknowledgments

We are grateful to Attila Bernath and Tamás Kiraly for their comments on the proof of Theorem 4.1, and to the two anonymous referees for their helpful remarks and suggestions. We also thank Mohit Singh, Bill Cunningham and Anupam Gupta for interesting discussions, and Nicola Apollonio for providing us with a preliminary version of [2].

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