Diverse Collections in Matroids and Graphs
Fedor V. Fomin, Petr A. Golovach, Fahad Panolan, Geevarghese Philip, Saket Saurabh
DDiverse Collections in Matroids and Graphs
Fedor V. Fomin
University of Bergen, Norway [email protected]
Petr A. Golovach
University of Bergen, Norway https://folk.uib.no/pgo041/
Fahad Panolan
Department of Computer Science and Engineering, IIT Hyderabad, India https://iith.ac.in/~fahad/ [email protected]
Geevarghese Philip
Chennai Mathematical Institute, IndiaUMI ReLaX [email protected]
Saket Saurabh
Institute of Mathematical Sciences, IndiaUniversity of Bergen, Norway [email protected]
Abstract
We investigate the parameterized complexity of finding diverse sets of solutions to three fundamentalcombinatorial problems, two from the theory of matroids and the third from graph theory. Theinput to the
Weighted Diverse Bases problem consists of a matroid M , a weight function ω : E ( M ) → N , and integers k ≥ , d ≥
0. The task is to decide if there is a collection of k bases B , . . . , B k of M such that the weight of the symmetric difference of any pair of these bases is atleast d . This is a diverse variant of the classical matroid base packing problem. The input to the Weighted Diverse Common Independent Sets problem consists of two matroids M , M definedon the same ground set E , a weight function ω : E → N , and integers k ≥ , d ≥
0. The task is todecide if there is a collection of k common independent sets I , . . . , I k of M and M such that theweight of the symmetric difference of any pair of these sets is at least d . This is motivated by theclassical weighted matroid intersection problem. The input to the Diverse Perfect Matchings problem consists of a graph G and integers k ≥ , d ≥
0. The task is to decide if G contains k perfectmatchings M , . . . , M k such that the symmetric difference of any two of these matchings is at least d .The underlying problem of finding one solution (basis, common independent set, or perfectmatching) is known to be doable in polynomial time for each of these problems, and DiversePerfect Matchings is known to be NP -hard for k = 2. We show that Weighted Diverse Bases and
Weighted Diverse Common Independent Sets are both NP -hard. We show also that Diverse Perfect Matchings cannot be solved in polynomial time (unless P = NP ) even for thecase d = 1. We derive fixed-parameter tractable ( FPT ) algorithms for all three problems with ( k, d )as the parameter.The above results on matroids are derived under the assumption that the input matroids are givenas independence oracles . For
Weighted Diverse Bases we present a polynomial-time algorithmthat takes a representation of the input matroid over a finite field and computes a poly ( k, d )-sizedkernel for the problem. Theory of computation Fixed parameter tractability a r X i v : . [ c s . D S ] J a n Diverse Collections in Matroids and Graphs
Keywords and phrases
Matroids, Matching, Diverse solutions, Fixed-parameter tractable algorithms
Related Version
An extended abstract has been accepted for publication at STACS 2021.
Funding
Fedor V. Fomin : Supported by the Research Council of Norway via the project “MULTIVAL”(grant no. 263317).
Petr A. Golovach : Supported by the Research Council of Norway via the project “MULTIVAL”(grant no. 263317).
Fahad Panolan : Seed grant, IIT Hyderabad (SG/IITH/F224/2020-21/SG-79)
Saket Saurabh : European Research Council (ERC) under the European Union’s Horizon 2020research and innovation programme (grant no. 819416), and Swarnajayanti Fellowship grantDST/SJF/MSA-01/2017-18.
In this work we study the parameterized complexity of finding diverse collections of solutions to three basic algorithmic problems. Two of these problems arise in the theory of matroids.The third problem belongs to the domain of graph theory, and its restriction to bipartitegraphs can be rephrased as a question about matroids. Each of these is a fundamentalalgorithmic problem in its respective domain.
Diverse
FPT
Algorithms.
Nearly every existing approach to solving algorithmic problems focuses on finding one solutionof good quality for a given input. For algorithmic problems which are—eventually—motivatedby problems from the real world, finding “one good solution” may not be of much use forpractitioners of the real-world discipline from which the problem was originally drawn. Thisis primarily because the process of abstracting out a “nice” algorithmic problem from a“messy” real-world problem invariably involves throwing out a lot of “side information” whichis very relevant to the real-world problem, but is inconvenient, difficult, or even impossibleto model mathematically.The other extreme of enumerating all (or even all minimal or maximal) solutions to aninput instance is also usually not a viable solution. A third approach is to look for a fewsolutions of good quality which are “far away” from one another according to an appropriatenotion of distance. The intuition is that given such a collection of “diverse” solutions, anend-user can choose one of the solutions by factoring in the “side information” which isabsent from the algorithmic model.These and other considerations led Fellows to propose the Diverse X Paradigm [9]. Here“ X ” is a placeholder for an optimization problem, and the goal is to study the fixed-parametertractability of finding a diverse collection of good-quality solutions for X . Recall that the Hamming distance of two sets is the size of their symmetric difference. A natural measure ofdiversity for problems whose solutions are subsets of some kind is the minimum
Hammingdistance of any pair of solutions. In this work we study the parameterized complexity offinding diverse collections of solutions for three fundamental problems with this diversitymeasure and its weighted variant. . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 3Our problems.
Let M be a matroid on ground set E ( M ) and with rank function r (). The departure pointof our work is the classical theorem of Edmonds from 1965 [6] about matroid partition. Thistheorem states that a matroid M has k pairwise disjoint bases if and only if, for every subset X of E ( M ), k · r ( X ) + | E ( M ) − X | ≥ k · r ( M ) . An important algorithmic consequence of this result is that given access to an independenceoracle for a matroid M , one can find a maximum number of pairwise disjoint bases of M in polynomial time (See, e.g., [18, Theorem 42.5]). This in turn implies, for instance, thatthe maximum number of pairwise edge-disjoint spanning trees of a connected graph can befound in polynomial time.We take a fresh look at this fundamental result of Edmonds: what happens if we don’tinsist that the bases be pairwise disjoint, and instead allow them to have some pairwiseintersection? We work in the weighted setting where each element e of the ground set E ( M )has a positive integral weight ω ( e ) associated with it, and the weight of a subset X of E ( M )is the sum of the weights of the elements in X . The relaxed version of the pairwise disjointbases problem is then: Given an independence oracle for a matroid M and integers k, d asinput, find if M has k bases B , . . . , B k such that for every pair of bases B i , B j ; i = j theweight ω ( B i B j ) of their symmetric difference is at least d . We call this the WeightedDiverse Bases problem:
Input:
A matroid M , a weight function ω : E ( M ) → N , and integers k ≥ d ≥ Task:
Decide whether there are bases B , . . . , B k of M such that ω ( B i B j ) ≥ d holds for all distinct i, j ∈ { , . . . , k } . Weighted Diverse Bases
Due to the expressive power of matroids
Weighted Diverse Bases captures many interest-ing computational problems. We list a few examples; in each case the weight function assignspositive integral weights, k ≥ d ≥ diverse if the weight of the symmetric difference of each pair of objects in the collection isat least d . When M is a graphic matroid Weighted Diverse Bases corresponds to findingdiverse spanning trees in an edge-weighted graph. When M is a vector matroid then thisis the problem of finding diverse column (or row) bases of a matrix with column (or row)weights. And when M is a transversal matroid on a weighted ground set then this problemcorresponds to finding diverse systems of distinct representatives .Another celebrated result of Edmonds is the Matroid Intersection Theorem [7] whichstates that if M , M are matroids on a common ground set E and with rank functions r , r ,respectively, then the size of a largest subset of E which is independent in both M and M (a common independent set ) is given bymin T ⊆ E ( r ( T ) + r ( E − T )) . Edmonds showed that given access to independence oracles for M and M , a maximum-size common independent set of M and M can be found in polynomial time [7]. This iscalled the Matroid Intersection problem. Frank [12] found a polynomial-time algorithm
Diverse Collections in Matroids and Graphs for the more general
Weighted Matroid Intersection problem where the input hasan additional weight function ω : E → N and the goal is to find a common independent setof the maximum weight . The second problem that we address in this work is a “diverse”take on Weighted Matroid Intersection where we replace the maximality requirementon individual sets with a lower bound on the weight of their symmetric difference. Given M , M , ω as above and integers k, d , we ask if there are k common independent sets whosepairwise symmetric differences have weight at least d each; this is the Weighted DiverseCommon Independent Sets problem.
Input:
Matroids M and M with a common ground set E , a weight function ω : E → N , and integers k ≥ d ≥ Task:
Decide whether there are sets I , . . . , I k ⊆ E such that I i is independent inboth M and M for every i ∈ { , . . . , k } and ω ( I i I j ) ≥ d for all distinct i, j ∈ { , . . . , k } . Weighted Diverse Common Independent SetsWeighted Diverse Common Independent Sets also captures many interestingalgorithmic problems. We give a few examples ( cf. [18, Section 41.1a]). We use “diverse”here in the sense defined above. Given a bipartite graph G with edge weights, WeightedDiverse Common Independent Sets can be used to ask if there is a diverse collection of k matchings in G . A partial orientation of an undirected graph G is a directed graph obtainedby (i) assigning directions to some subset of edges of G and (ii) deleting the remaining edges.Given an undirected graph G = ( V, E ) with edge weights and a function ι : V → N , we saythat a partial orientation O of G respects ι if the in-degree of every vertex v in O is at most ι ( v ). We can use Weighted Diverse Common Independent Sets to ask if there is adiverse collection of k partial orientations of G , all of which respect ι . For a third example,let G = ( V, E ) be an undirected graph with edge weights, in which each edge is assigneda—not necessarily distinct— color . A colorful forest in G is any subgraph of G which is aforest in which no two edges have the same color. We can use Weighted Diverse CommonIndependent Sets to ask if there is a diverse collection of k colorful forests in G .Finding whether a bipartite graph has a perfect matching or not is a well-known applicationof Matroid Intersection ([18, Section 41.1a]). The third problem that we study in thiswork is a diverse version of the former problem, extended to general graphs. Note thatthere is no known interpretation of the problem of finding perfect matchings in (general)undirected graphs in terms of
Matroid Intersection . Input:
An undirected graph G on n vertices, and integers k ≥ d ≥ Task:
Decide whether there are perfect matchings M , . . . , M k of G such that | M i M j | ≥ d for all distinct i, j ∈ { , . . . , k } . Diverse Perfect Matchings
Our results.
We assume throughout that matroids in the input are given in terms of an independenceoracle . Recall that with this assumption, we can find one basis of the largest weight and one common independent set (of two matroids) of the largest weight, both in polynomial time.In contrast, we show that the diverse versions
Weighted Diverse Bases and
WeightedDiverse Common Independent Sets are both NP -hard, even when the weights are . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 5 expressed in unary . (cid:73) Theorem 1.
Both
Weighted Diverse Bases and
Weighted Diverse CommonIndependent Sets are strongly NP -complete, even on the uniform matroids U n . Given this hardness, we analyze the parameterized complexity of these problems with d, k as the parameters. Our first result is that
Weighted Diverse Bases is fixed-parametertractable (
FPT ) under this parameterization: (cid:73)
Theorem 2.
Weighted Diverse Bases can be solved in O ( dk (log k +log d )) · | E ( M ) | O (1) time. We have a stronger result if the input matroid is given as a representation over a finite field(and not just as a “black box” independence oracle): in this case we show that
WeightedDiverse Bases admits a polynomial kernel with this parameterization. (cid:73)
Theorem 3.
Given a representation of the matroid M over a finite field GF( q ) as input,we can compute a kernel of Weighted Diverse Bases of size O ( k d log q ) . We then show that our second matroid-related diverse problem is also
FPT under thesame parameterization. (cid:73)
Theorem 4.
Weighted Diverse Common Independent Sets can be solved in O ( k d log( kd )) · | E | O (1) time. We now turn to the problem of finding diverse perfect matchings.
Diverse PerfectMatchings is known to be NP -hard already when k = 2 and G is a 3-regular graph [16, 10].Since all perfect matchings of a graph have the same size the symmetric difference of twodistinct perfect matchings is at least 2. Setting d = 1 in Diverse Perfect Matchings is thus equivalent to asking whether G has at least k distinct perfect matchings. Since abipartite graph on n vertices has at most n ! perfect matchings and since log( n !) = O ( n log n )we get—using binary search—that there is a polynomial-time Turing reduction from theproblem of counting the number of perfect matchings in a bipartite graph to DiversePerfect Matchings instances with d = 1. Since the former problem is -complete [20]we get (cid:73) Theorem 5.
Diverse Perfect Matchings with d = 1 cannot be solved in time polyno-mial in n = | V ( G ) | even when graph G is bipartite, unless P = NP . Thus we get that
Diverse Perfect Matchings is unlikely to have a polynomial-timealgorithm even if one of the two numbers k, d is a small constant. We show that the problem does have a (randomized) polynomial-time algorithm when both these parameters are bounded;
Diverse Perfect Matchings is (randomized)
FPT with k and d as parameters: (cid:73) Theorem 6.
There is an algorithm that given an instance of
Diverse Perfect Match-ings , runs in time O ( kd ) n O (1) and outputs the following: If the input is a No -instance thenthe algorithm outputs No . Otherwise the algorithm outputs Yes with probability at least − e . Note that Theorem 6 implies, in particular, that
Diverse Perfect Matchings can besolved in (randomized) polynomial time when kd ≤ c + log log nc holds for some constants c , c which depend on the constant hidden by the O () notation. See Theorem 7 for an alternative hardness result for
Weighted Diverse Bases . Diverse Collections in Matroids and GraphsOur methods.
We prove the NP -hardness results (Theorem 1) by reduction from the 3 -Partition problem.To show that Weighted Diverse Bases is FPT (Theorem 2) we observe first that if theinput matroid M contains a set of size Ω( kd ) which is both independent and co-independentin M then the input is a Yes instance of
Weighted Diverse Bases (Lemma 10). We cancheck for the existence of such a set in time polynomial in | E ( M ) | , so we assume withoutloss of generality that no such set exists. We then show that starting with an arbitrarybasis of M and repeatedly applying the greedy algorithm (Proposition 9) poly ( k, d )-manytimes we can find, in time polynomial in ( | E ( M ) | + k + d ), (i) a subset S ∗ ⊆ E ( M ) of size poly ( k, d ) and (ii) a matroid f M on the ground set S ∗ such that ( f M , ω, k, d ) is equivalent tothe input instance (
M, ω, k, d ) (Lemma 11). We also show how to compute a useful partitionof E ( f M ) = S ∗ which speeds up the subsequent FPT -time search for a diverse set of bases in f M . The kernelization result for Weighted Diverse Bases (Theorem 3) follows directlyfrom Lemma 11. This “compression lemma” is thus the main technical component of ouralgorithms for
Weighted Diverse Bases .To show that
Weighted Diverse Common Independent Sets is FPT (Theorem 4)we observe first that if the two input matroids M , M have a common independent set ofsize Ω( kd ) then the input is a Yes instance of
Weighted Diverse Common IndependentSets (Lemma 16). So we assume that this is not the case, and then show (Lemma 17) thatwe can construct, in f ( k, d ) time, a collection F of common independent sets of M and M of size g ( k, d ) such that if the input is a Yes -instance then it has a solution I , . . . , I k with I i ∈ F for i ∈ { , . . . , k } . The FPT algorithm for
Weighted Diverse CommonIndependent Sets follows by a simple search in the collection F .Our algorithm for Diverse Perfect Matchings is based on two procedures. P1 Given an undirected graph G on n vertices, perfect matchings M , . . . , M r of G , and anon-negative integer s as input, this procedure (Lemma 23) runs in time 2 O ( rs ) n O (1) andoutputs a perfect matching M of G such that | M M i | ≥ s holds for all i ∈ { , . . . , r } (if such a matching exists), with probability at least e − rs . P2 Given an undirected graph G on n vertices, a perfect matching M of G , and non-negativeintegers r, d, s , this procedure(Lemma 25) runs in time 2 O ( r s ) n O (1) , and outputs r perfectmatchings M ? , . . . , M ?r of G such that | M M ?i | ≤ s holds for all i ∈ { , . . . , r } and | M ?i M ?j | ≥ d holds for all distinct i, j ∈ [ r ] (if such matchings exist), with probabilityat least e − rs . If no such perfect matchings exist, then the algorithm outputs No .Let ( G, k, d ) be the input instance of
Diverse Perfect Matchings . We use procedure P1 to greedily compute a collection of matchings which are “far apart”: We start withan arbitrary perfect matching M . In step i , we have a collection of perfect matchings M , . . . , M i − such that | M j M j | ≥ k − i d holds for any two distinct j, j ∈ { , . . . , i − } .We now run procedure P1 with r = i − s = 2 k − i d to find—if it exists—a matching M i such that | M i M j | ≥ k − i +1 d holds for all j ∈ { , . . . , i } . By exhaustively applying P1 weget a collection of perfect matchings M , . . . , M q such that (a) for any two distinct integers i, j ∈ { , . . . , q } , | M i M j | ≥ k − q +1 d , and (b) for any other perfect matching M / ∈ { M , . . . , M q } , | M M j | ≤ k − q d .Thus, if k ≤ q , then clearly { M , . . . , M k } is a solution. Otherwise, let M = { M ? , . . . , M ?k } be a hypothetical solution. Then for each M ?i there is a unique matching M j in { M , . . . , M q } such that | M j M ?i | < ( k − q ) d holds (Claim 28). For each i ∈ { , . . . , q } we guess thenumber r i of perfect matchings from M that are close to M i , and use procedure P2 to . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 7 compute a set of r i diverse perfect matchings that are close to M i . The union of all thematchings computed for all i ∈ { , . . . , q } form a solution.We use algebraic methods and color coding to design procedure P1 . The Tutte matrix A of an undirected graph G over the field F [ X ] is defined as follows, where F is the Galoisfield on { , } and X = { x e : e ∈ E ( G ) } . The rows and columns of A are labeled with V ( G ) and for each e = { u, v } ∈ E ( G ), A [ u, v ] = A [ v, u ] = x e . All other entries in A arezeros. There is a bijective correspondence between the set of monomials of det ( A ) and theset of perfect matchings of G . Procedure P1 extracts the required matching from det ( A )using color coding. Procedure P2 is realized using color coding and dynamic programming. Related work.
Recall that all bases of a matroid have the same size, and that the number of bases of amatroid on ground set E is at most 2 | E | . So using the same argument as for Theorem 5we get that Weighted Diverse Bases generalizes—via Turing reductions—the problemof counting the number of bases of a matroid. Each of these reduced
Weighted DiverseBases instances will have d = 1, and a weight function which assigns the weight 1 to eachelement in the ground set. Counting the number of bases of a matroid is known to be -complete even for restricted classes of matroids such as transversal [3], bircircular [14],and binary matroids [21]. Hence we have the following alternative hardness result for Weighted Diverse Bases (cid:73)
Theorem 7.
Weighted Diverse Bases cannot be solved in time polynomial in | E ( M ) | unless P = NP , even when d = 1 and every element of the ground set E ( M ) has weight . The study of the parameterized complexity of finding diverse sets of solutions is a veryrecent development, and only a handful of results are currently known. In the work whichintroduced this notion Baste et al. [1] showed that diverse variants of a large class of graphproblems which are
FPT when parameterized by the treewidth of the input graph, are also
FPT when parameterized by the treewidth and the number of solutions in the collection.In a second article [2] the authors show that for each fixed positive integer d , two diversevariants—one with the minimum Hamming distance of any pair of solutions, and the otherwith the sum of all pairwise Hamming distances of solutions—of the d - Hitting Set problemare
FPT when parameterized by the size of the hitting set and the number of solutions. Ina recent manuscript on diverse
FPT algorithms [10] the authors show that the problem offinding two maximum-sized matchings in an undirected graph such that their symmetricdifference is at least d , is FPT when parameterized by d . Note that our result on DiversePerfect Matchings generalizes this to k ≥ provided the input graph has aperfect matching.In a very recent manuscript Hanaka et al. [15] propose a number of results about findingdiverse solutions. We briefly summarize their results which are germane to our work. For acollection of sets X , . . . , X k let d sum ( X , . . . , X k ) denote the sum of all pairwise Hammingdistances of these sets and let d min ( X , . . . , X k ) denote the smallest Hamming distance ofany pair of sets in the collection. Hanaka et al. show that there is an algorithm which takesan independence oracle for a matroid M and an integer k as input, runs in time polynomialin ( | E ( M ) | + k ), and finds a collection B , B , . . . , B k of k bases of M which maximizes d sum ( B , B , . . . , B k ). This result differs from our work on Weighted Diverse Bases in Compare with Theorem 1.
Diverse Collections in Matroids and Graphs two key aspects. They deal with the unweighted (counting) case, and their diversity measureis the sum of the pairwise symmetric differences, whereas we look at the minimum (weightof the) symmetric difference. These two measures are, in general, not comparable.Hanaka et al. also look at the complexity of finding k matchings M , . . . , M k in a graph G where each M i is of size t . They show that such collections of matchings maximizing d min ( M , . . . , M k ) and d sum ( M , . . . , M k ) can be found in time 2 O ( kt log( kt )) ·| V ( G ) | O (1) . Thekey difference with our work is that their algorithm looks for matchings of a specified size t whereas ours looks for perfect matchings, of size t = | V ( G ) | ; note that this t does notappear in the exponential part of the running time of our algorithm (Theorem 6). Themanuscript [15] has a variety of other interesting results on diverse FPT algorithms as well.
Organization of the rest of the paper.
In the next section we collect together some definitions and preliminary results. In Section 3we prove that
Weighted Diverse Bases and
Weighted Diverse Common IndependentSets are strongly NP -hard. In Section 4 we derive our FPT and kernelization algorithms for
Weighted Diverse Bases , and in Section 5 we show that
Weighted Diverse CommonIndependent Sets is FPT . We derive our results for
Diverse Perfect Matchings inSection 6. We conclude in Section 7.
We use X Y to denote the symmetric difference ( X \ Y ) ∪ ( Y \ X ) of sets X and Y . Weuse N to denote the set of positive integers. Parameterized complexity.
A parameterized problem Π is a subset of Σ ∗ × N , where Σ is a finite alphabet. We say thata parameterized problem Π is fixed parameter tractable ( FPT ) , if there is an algorithm thatgiven an instance ( x, k ) of Π as input, solves in time f ( k ) | x | O (1) , where f is an arbitraryfunction and | x | is the length of x . A kernelization algorithm for a parameterized problemΠ is a polynomial time algorithm (computable function) A : Σ ∗ × N → Σ ∗ × N such that( x, k ) ∈ Π if and only if ( x , k ) = A (( x, k )) ∈ Π and | x | + k ≤ g ( k ) for some computablefunction g . When g is a polynomial function, we say that Π admits a polynomial kernel. Fora detailed overview about parameterized complexity we refer to the monographs [5, 4, 11] Matroids.
We give a brief description of the matroid-related notions that we need. See the book ofOxley [17] for a detailed introduction to matroids. A pair M = ( E, I ), where E is a finite ground set and I is a family of subsets of the ground set, called independent sets of E , is a matroid if it satisfies the following conditions, called independence axioms : (I1) ∅ ∈ I . (I2) If A ⊆ B ⊆ E ( M ) and B ∈ I then A ∈ I . (I3) If A, B ∈ I and | A | < | B | , then there is e ∈ B \ A such that A ∪ { e } ∈ I .We use E ( M ) and I ( M ) to denote the ground set and the set of independent sets, respectively.As is standard for matroid problems, we assume that each matroid M that appears in theinput is given by an independence oracle , that is, an oracle that in constant (or polynomial)time replies whether a given A ⊆ E ( M ) is independent in M or not. An inclusion-wise . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 9 maximal independent set B is called a basis of M . We use B ( M ) to denote the set of basesof M . The bases satisfy the following properties, called basis axioms : (B1) B ( M ) = ∅ . (B2) If B , B ∈ B ( M ), then for every x ∈ B \ B , there is y ∈ B \ B such that( B \ { x } ) ∪ { y } ∈ B ( M ).All the bases of M have the same size that is called the rank of M , denoted rank ( M ). The rank of a subset A ⊆ E ( M ), denoted rank ( A ), is the maximum size of an independent set X ⊆ A ; the function rank : 2 E ( M ) → Z is the rank function of M . A set A ⊆ E ( M ) spans an element x ∈ E ( M ) if rank ( A ∪ { x } ) = rank ( A ). The closure (or span ) of A is the set cl ( A ) = { x ∈ E ( M ) | A spans x } . Closures satisfy the following properties, called closureaxioms : (CL1) For every A ⊆ E ( M ), A ⊆ cl ( A ). (CL2) If A ⊆ B ⊆ E ( M ), then cl ( A ) ⊆ cl ( B ). (CL3) For every A ⊆ E ( M ), cl ( A ) = cl ( cl ( A )). (CL4) For every A ⊆ E ( M ) and every x ∈ E ( M ) and y ∈ cl ( A ∪ { x } ) \ cl ( A ), x ∈ cl ( A ∪ { y } ).The dual of a matroid M = ( E, I ), denoted M ∗ , is the matroid whose ground set is E and whose set of bases is B ∗ = { B | B ∈ B ( M ) } . That is, the bases of M ∗ are exactly thecomplements of the bases of M . A basis (independent set, rank, respectively) of M ∗ is a cobasis ( coindependent set , corank , respectively) of M . We use I ∗ ( M ) to denote the set ofcoindependent sets of M . Also, corank ( M ) denotes the corank of M and corank ( A ) denotesthe corank of a set A ⊆ E ( M ); corank ( A ) is the rank of set A in the dual matrix M ∗ , and corank ( A ) = | A | − rank ( M ) + rank ( E \ A ). Given an independence oracle for M we canconstruct—using the augmentation property (I3) and with an overhead which is polynomialin | E | —a rank oracle for M , and thence corank and coindependence oracles for M .For e ∈ E ( M ), the matroid M = M − e is obtained by deleting e if E ( M ) = E ( M ) \ { e } and I ( M ) = { X ∈ I ( M ) | e / ∈ X } . It is said that M = M/e is obtained by contracting e if M = ( M ∗ − e ) ∗ . In particular, if e is not a loop (i.e., if { e } is independent) in M , then I ( M ) = { X \ { e } | e ∈ X ∈ I ( M ) } . Notice that deleting an element in M is equivalent tocontracting it in M ∗ and vice versa. Let X ⊆ E ( M ). Then M − X denotes the matroidobtained from M by the deletion of the elements of X and M/X is the matroid obtainedby consecutive contractions of the elements of X . Note that an independence oracle for M can itself act as an independence oracle for M − X if we restrict our queries to subsetsof E ( M ) \ X . Let rank M/X denote the rank function of the matroid
M/X . Then for any Y ⊆ ( E ( M ) \ X ) we have that rank M/X ( Y ) = rank ( X ∪ Y ) − rank ( X ) [17, 3.1.7]. Given anindependence oracle for M we can thus easily construct an independence oracle for M/X .Let M be a matroid and let F be a field. An n × m -matrix A over F is a representationof M over F if there is one-to-one correspondence f between E ( M ) and the set of columnsof A such that for any X ⊆ E ( M ), X ∈ I ( M ) if and only if the columns f ( X ) are linearlyindependent (as vectors of F n ); if M has such a representation, then it is said that M has a representation over F . In other words, A is a representation of M if M is isomorphic to the linear matroid of A , i.e., the matroid whose ground set is the set of columns of A and a setof columns is independent if and only if these columns are linearly independent. Observethat, given a representation A of M , we can verify whether a set is independent by checkingthe linear independence of the corresponding columns of A . Hence, we don’t need an explicitindependence oracle in this case.Let 1 ≤ r ≤ n be integers. We use U rn to denote the uniform matroid, that is, the matroidwith the ground set of size n such that the bases are all r -element subsets of the ground set. We use the classical results of Edmonds [7] and Frank [12] about the
Weighed MatroidIntersection problem. The task of this problem is, given two matroids M and M withthe same ground set E and a weight function ω : E → N , find a set X of maximum weightsuch that X is independent in both matroids. Edmonds [7] proved that the problem canbe solved in polynomial time for the unweighted case (that is, the task is to find a commonindependent set of maximum size; we refer to this variant as Matroid Intersection ) andthe result was generalized for the variant with the weights by Frank in [12]. (cid:73)
Proposition 8 ([7, 12]) . Weighted Matroid Intersection can be solved in polynomialtime.
We also need another classical result of Edmonds [8] that a basis of maximum weight canbe found by the greedy algorithm. Recall that, given a matroid M with a weight function ω : E ( M ) → N , the greedy algorithm finds a basis B of maximum weight as follows. Initially, B := ∅ . Then at each iteration, the algorithm finds an element of x ∈ E ( M ) \ B of maximumweight such that B ∪ { x } is independent and sets B := B ∪ { x } . The algorithms stops whenthere is no element that can be added to B . (cid:73) Proposition 9 ([8]) . The greedy algorithm finds a basis of maximum weight of a weightedmatroid in polynomial time.
We need the following observation(See [17, Lemma 2.1.10]). (cid:73)
Observation 1.
Let X and Y be disjoint sets such that X is independent and Y iscoindependent in a matroid M . Then there is a basis B of M such that X ⊆ B and Y ∩ B = ∅ . Observe that for any sets X and Y that are subsets of the same universe, X Y = X Y .This implies the following. (cid:73) Observation 2.
For every matroid M , every weight function ω : E ( M ) → N , and allintegers k ≥ and d ≥ , the instances ( M, ω, k, d ) and ( M ∗ , ω, k, d ) of Weighted DiverseBases are equivalent.
We show that
Weighted Diverse Bases and
Weighted Diverse Common IndependentSets are NP -complete in the strong sense even for uniform matroids. (cid:73) Theorem 1.
Both
Weighted Diverse Bases and
Weighted Diverse CommonIndependent Sets are strongly NP -complete, even on the uniform matroids U n . Proof.
We prove the claim for
Weighted Diverse Bases by a reduction from the 3 -Partition problem. The input to 3 -Partition consists of a positive integer b and amultiset S = { s , . . . , s n } of 3 n positive integers such that (i) b < s i < b holds for each i ∈ { , . . . , n } and (ii) P ni =1 s i = nb . The task is to decide whether S can be partitionedinto n multisets S , . . . , S n such that P s ∈ S i s = b holds for each S i . Note that each multiset S i in such a partition must contain exactly three elements from S . This problem is knownto be NP -complete in the strong sense, i.e., it is NP -complete even if the input integers areencoded in unary [13, SP15].Let ( b, S = { s , . . . , s n ) } be an instance of 3 -Partition with n ≥
3. We set M to bethe uniform matroid U n on the ground set { , . . . , n } , and define the weight function to be . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 11 ω ( i ) = s i for i ∈ { , . . . , n } . We set d = 2 b . We will now show that ( b, S ) is a yes-instanceof 3 -Partition if and only if ( M, ω, n, d ) is a yes-instance of
Weighted Diverse Bases .In the forward direction, suppose that S , . . . , S n is a partition of S into triples of integerssuch that the sum of elements of each S i is b . Let B , . . . , B n be the corresponding partitionof { , . . . , n } , that is, B i = { i , i , i } if and only if S i = { s i , s i , s i } for each i ∈ { , . . . , n } .Clearly, B , . . . , B n are pairwise disjoint bases of M . Then for every distinct i, j ∈ { , . . . , n } , ω ( B i B j ) = ω ( B i ) + ω ( B j ) = 2 b . Therefore, ( M, ω, n, d ) is a yes-instance of
WeightedDiverse Bases .In the reverse direction, assume that (
M, ω, n, d ) is a yes-instance of
Weighted DiverseBases . Let B , . . . , B n be bases of M such that ω ( B i B j ) ≥ d = 2 b for distinct i, j ∈{ , . . . , n } .We claim that B , . . . , B n are pairwise disjoint. For the sake of contradiction, assume thatthere are distinct i, j ∈ { , . . . , n } such that B i ∩ B j = ∅ . Let X = B i \ B j and X = B j \ B i .Note that | X | ≤ | X | ≤
2. We have that ω ( X ) = P h ∈ X ω ( h ) = P h ∈ X s h < | X | b/ < b . Similarly, ω ( X ) < b . Therefore, ω ( B i B j ) = ω ( X ) + ω ( X ) < b ; acontradiction. We conclude that the bases B , . . . , B n are pairwise disjoint. This impliesthat B , . . . , B n is a partition of { , , . . . , n } .Next we show that ω ( B i ) = b holds for every i ∈ { , . . . , n } . Suppose that there is an h ∈ { , . . . , n } such that ω ( B h ) > b . Let I = { , . . . , n } \ B h and J = { , . . . , n } \ { h } . Wehave that P i ∈ I ω ( i ) < P ni =1 ω ( i ) − b = b ( n − B , . . . , B h − , B h +1 . . . , B n form apartition of I , we get that P i ∈ J ω ( B i ) < b ( n −
1) holds as well. Recall that n ≥
3. Then X { i,j } s.t. i,j ∈ J, i = j ( ω ( B i ) + ω ( B j )) ≤ ( n − X i ∈ J ω ( B i ) < b ( n − n − . The first inequality above comes from the fact that since | J | = n −
1, for each index i ∈ J theterm ω ( B i ) appears in at most n − ω ( B i ) + ω ( B j )) in the summation onthe left hand side. Now suppose ( ω ( B i ) + ω ( B j )) ≥ b holds for all pairs i, j ∈ J, i = j . Thenthe sum on the left hand side would be at least (cid:0) | J | (cid:1) · b = ( n − n − b , a contradiction.Therefore, there must exist distinct i, j ∈ J such that ω ( B i ) + ω ( B j ) < b holds. And thiscontradicts our assumption that ω ( B i B j ) ≥ b holds for all such i, j . We conclude that ω ( B i ) ≤ b holds for every i ∈ { , . . . , n } . And since P ni =1 ω ( B i ) = bn , we get that ω ( B i ) = b holds for every i ∈ { , . . . , n } .Finally, we consider the partition S , . . . , S n of S corresponding to B , . . . , B i , that is, foreach B i = { i , i , i } , we define S i = { s i , s i , s i } . Clearly, s i + s i + s i = ω ( B i ) = b . Thuswe get that ( b, S ) is a yes-instance of 3 -Partition . This concludes the proof for WeightedDiverse Bases .The reduction for
Weighted Diverse Common Independent Sets is also from3 -Partition , and is nearly identical to the above reduction for
Weighted Diverse Bases .Given an instance ( b, S = { s , . . . , s n ) } of 3 -Partition with n ≥
3, we set each of M , M to be the uniform matroid U n on the ground set { , . . . , n } , and define the weight functionto be ω ( i ) = s i for i ∈ { , . . . , n } . We set d = 2 b . We will now show that ( b, S ) is ayes-instance of 3 -Partition if and only if ( M , M , ω, n, d ) is a yes-instance of WeightedDiverse Common Independent Sets .In the forward direction, suppose that S , . . . , S n is a partition of S into triples of integerssuch that the sum of elements of each S i is b . Let I , . . . , I n be the corresponding partition of { , . . . , n } , that is, I i = { i , i , i } if and only if S i = { s i , s i , s i } for each i ∈ { , . . . , n } .Clearly, I , . . . , I n are pairwise disjoint common independent sets of M and M , and forevery distinct i, j ∈ { , . . . , n } , ω ( I i I j ) = ω ( I i ) + ω ( I j ) = 2 b . Therefore, ( M , M , ω, n, d )is a yes-instance of Weighted Diverse Common Independent Sets . In the reverse direction, assume that ( M , M , ω, n, d ) is a yes-instance of WeightedDiverse Common Independent Sets , and let I , . . . , I n be common independent setsof M and M such that ω ( I i I j ) ≥ d = 2 b for distinct i, j ∈ { , . . . , n } . Since everyindependent set in the matroids M , M has at most three elements, and since s i < b holdsfor each i ∈ { , . . . , n } , we get that the sets I , . . . , I n are pairwise disjoint. If two of thesesets, say I i , I j have at most two elements each then ω ( I i I j ) < · b = 2 b , a contradiction.So at most one of these sets has at most two elements; every other set in the collection hasexactly three elements.If all the sets I , . . . , I n have three elements each then they are a pairwise disjoint collectionof n bases of M , and the argument that we used for the reverse direction in the proof for Weighted Diverse Bases tells us that ( b, S ) is a yes-instance of 3 -Partition . In theremaining case there is exactly one set of size two among I , . . . , I n ; without loss of generality,let this smaller set be I . Then | S ni =1 I i | = 3 n −
1. Let x = { , , . . . , n } \ S ni =1 I i be theunique element which is not in any of these independent sets. Then ( I ∪ { x } ) , . . . , I n is apairwise disjoint collection of n bases of M such that the weight of the symmetric differenceof any pair of these bases is at least d = 2 b , and the argument that we used for the reversedirection in the proof for Weighted Diverse Bases tells us that ( b, S ) is a yes-instance of3 -Partition . (cid:74) In this section, we show that
Weighted Diverse Bases is FPT when parameterized by k and d . Moreover, if the input matroid is representable over a finite field and is given by sucha representation, then Weighted Diverse Bases admits a polynomial kernel.We start with the observation that if the input matroid has a sufficiently big set that issimultaneously independent and coindependent, then diverse bases always exist. (cid:73)
Lemma 10.
Let M be a matroid, and let k ≥ and d ≥ be integers. If there is X ⊆ E ( M ) of size at least k d d e such that X is simultaneously independent and coindependent, then ( M, ω, k, d ) is a yes-instance of Weighted Diverse Bases for any weight function ω . Proof.
Let X ⊆ E ( M ) be a set of size at least k d d e such that X is simultaneously independentand coindependent. Then there is a partition X , . . . , X k of X such that | X i | ≥ d d e for every i ∈ { , . . . , k } . Let i ∈ { , . . . , k } . Since X is independent, X i is independent, and since X iscoindependent, then X \ X i is coindependent. Then by Observation 1, there is a basis B i of M such that X i ⊆ B i and B i ∩ ( X \ X i ) = ∅ . The latter property means that B i ∩ X j = ∅ for every j ∈ { , . . . , k } such that j = i . We consider the bases B i defined in this manner forall i ∈ { , . . . , k } . Then for every distinct i, j ∈ { , . . . , k } , X i ∪ X j ⊆ B i B j . Therefore, ω ( B i B j ) ≥ ω ( X i ∪ X j ) ≥ | X i ∪ X j | = | X i | + | X j | ≥ d d e ≥ d for any ω : E ( M ) → N .Hence, ( M, ω, k, d ) is a yes-instance of
Weighted Diverse Bases . (cid:74) Our results are based on the following lemma. (cid:73)
Lemma 11.
There is an algorithm that, given an instance ( M, ω, k, d ) of WeightedDiverse Bases , runs in time polynomial in ( | E ( M ) | + k + d ) and either correctly decidesthat ( M, ω, k, d ) is a yes-instance or outputs an equivalent instance ( f M , ω, k, d ) of WeightedDiverse Bases such that E ( f M ) ⊆ E ( M ) and | E ( f M ) | ≤ d d e k . In the latter case, thealgorithm also computes a partition ( L, L ∗ ) of E ( f M ) with the property that for every basis B of f M , | B ∩ L | ≤ d d e k and | L ∗ \ B | ≤ d d e k , and the algorithm outputs an independenceoracle for f M that answers queries for f M in time polynomial in | E ( M ) | . Moreover, if M is . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 13 representable over a finite field F and is given by such a representation, then the algorithmoutputs a representation of f M over F . Proof.
Let (
M, ω, k, d ) be an instance of
Weighted Diverse Bases . Recall that M isgiven as an independence oracle. We construct an independence oracle for the dual matroid M ∗ , and then solve Matroid Intersection for M and M ∗ using Proposition 8. Let X be the set computed by the Matroid Intersection algorithm. Then X ⊆ E ( M ) is a setof maximum size that is both independent and coindependent in M . If | X | ≥ k d d e , then( M, ω, k, d ) is a yes-instance of
Weighted Diverse Bases by Lemma 10; the problem issolved and we return the answer.Assume from now on that this is not the case, and that | X | ≤ k d d e − B bean arbitrary basis of M , and let B = ( E ( M ) \ B ). If rank ( B ) ≥ k d d e then there exists anindependent set Y ⊆ B of size at least k d d e . But Y is also a coindependent set of size atleast k d d e , which contradicts our assumption. Thus we get that rank ( B ) ≤ k d d e − corank ( B ) ≤ k d d e − B of M .Let ‘ = d d e k . Fix an arbitrary basis B of M . We construct sets S , . . . , S ‘ iteratively.We set S = B . For i ≥ S i from S ( i − as follows. If E ( M ) \ S i − = ∅ , we set X i = ∅ . Otherwise we set X i to be a basis of maximum weight in the matroid M − S i − ; wefind X i using the greedy algorithm (see Proposition 9). Finally, we set S i = S i − ∪ X i .Let S = S ‘ and L = S \ B . Since rank ( B ) ≤ k d d e −
1, we get that every independent setcontained in the set B = ( E ( M ) \ B ) has size at most k d d e −
1. And since L is a disjointunion of ‘ such independent sets we get that | L | = | S \ B | ≤ ‘ ( k d d e − ≤ d d e k . We showthe following crucial claim. (cid:66) Claim 12.
If (
M, ω, k, d ) is a yes-instance of
Weighted Diverse Bases , then there isa solution, that is, a family of bases B , . . . , B k such that ω ( B i B j ) ≥ d for all distinct i, j ∈ { , . . . , k } , with the property that B i ⊆ S for every i ∈ { , . . . , k } . Proof.
Let (
M, ω, k, d ) be a yes-instance, and let the family of bases B , . . . , B k be a solutionwhich maximizes the size of the set (( S ki =1 B i ) ∩ S ) of vertices in the bases which arealso in the set S . We show that B i ⊆ S holds for every i ∈ { , . . . , k } . The proof isby contradiction. Assume that there is an h ∈ { , . . . , k } such that B h \ S = ∅ . Recallthat rank ( M − B ) = rank ( B ) ≤ k d d e −
1. Therefore, | B i \ B | ≤ k d d e − i ∈ { , . . . , k } , and | ∪ ki =1 ( B i \ B ) | ≤ k ( k d d e − < ‘ . Let X , . . . , X ‘ be the independentsets used to construct the sets S , . . . , S ‘ . Since ( X j ∩ B ) = ∅ holds for all j ∈ { , . . . , ‘ } we get that ( X j ∩ B i ) = ( X j ∩ ( B i \ B )) holds for all i ∈ { , . . . , k } , j ∈ { , . . . , ‘ } . So thenumber | ∪ ki =1 ( B i \ B ) | of elements from the bases B , . . . , B k which could potentially bepart of any of the sets X , . . . , X ‘ is strictly less than the number of these latter sets. Hencefrom the pigeonhole principle we get that there is a t ∈ { , . . . , ‘ } such that X t ∩ B i = ∅ holds for all i ∈ { , . . . , k } . Let A = B h ∩ S ( t − and Y = B h \ S ( t − . We show that there is Z ⊆ X t such that(i) B h = A ∪ Z is a basis, and(ii) ω ( Z ) ≥ ω ( Y ).We construct Z by greedily augmenting A with elements of X t . Let σ be the order inwhich the greedy algorithm picks elements from the set E ( M ) \ S ( t − to add them to the set X t . Initially we set Z := ∅ . Then we select the first x ∈ X t \ Z in σ such that A ∪ Z ∪ { x } is independent, and we set Z := Z ∪ { x } . We stop when there is no x ∈ X t \ Z such that A ∪ Z ∪ { x } is independent. We prove that (i) and (ii) are fulfilled for Z .First we show that (i) holds. From the construction we get that X t is a basis of the matroid M − S ( t − . This implies that ( E ( M ) \ S ( t − ) ⊆ cl ( X t ) holds. Now since Y = ( B h \ S ( t − ) is a subset of ( E ( M ) \ S ( t − ) we get that Y ⊆ cl ( X t ) holds. Since ( A ∪ Z ) is independent, andthere is no x ∈ X t \ Z such that A ∪ Z ∪ { x } is independent, we get that X t ⊆ cl ( A ∪ Z ) holds.Now by (CL2) and (CL3) , Y ⊆ cl ( cl ( A ∪ Z )) = cl ( A ∪ Z ). And by (CL1) , A ⊆ cl ( A ∪ Z )and we conclude that A ∪ Y ⊆ cl ( A ∪ Z ) holds. But ( A ∪ Y ) = B h is a basis of M , andso [17, Proposition 1.4.9] cl ( A ∪ Y ) = E ( M ). Applying (CL2) and (CL3) we get that E ( M ) ⊆ cl ( A ∪ Z ) which implies that cl ( A ∪ Z ) = E ( M ). Now since A ∪ Z is independentwe get (See, e.g., [17, Section 1.4, Exercise 2]) that B h = A ∪ Z is a basis.Now we show that (ii) holds. Let Z = { z , . . . , z s } , where the elements are indexedaccording to the order in which they are added to Z by the greedy augmentation describedabove. Note that ω ( z ) ≥ · · · ≥ ω ( z s ). Since B h and B h are bases, Y = ( B h \ S ( t − ), and Z = ( B h \ S ( t − ), we get that | Y | = | Z | . Observe also that Y ∩ Z = ∅ . We define (i) Z = ∅ and (ii) Z i = { z , . . . , z i } for i ∈ { , . . . , s } . We show that there is an ordering h y , . . . , y s i ofthe elements of Y such that the set A ∪ Z i − ∪ { y i } is independent for every i ∈ { , . . . , s } .We define this order inductively, starting with y s and proceeding in decreasing order of thesubscript.We set y s to be an element y ∈ ( A ∪ Y ) \ ( A ∪ Z s − ) = ( Y \ Z s − ) such that A ∪ Z s − ∪ { y } is independent. Since | A ∪ Z s − | < | A ∪ Y | we know from (I3) such an element must exist.For the inductive step, assume that for some fixed i ∈ { , . . . , s − } distinct elements y i +1 , . . . , y s ∈ Y have been defined such that A ∪ Z i ∪ { y i +1 , . . . , y s } is independent. Notethat | A ∪ Z i ∪ { y i +1 , . . . , y s }| = | A ∪ Y | . Then R = A ∪ Z i − ∪ { y i +1 , . . . , y s } is independentby (I2) , and by (I3) , there must exist an element y ∈ ( A ∪ Y ) \ R such that R ∪ { y } = A ∪ Z i − ∪ { y, y ( i +1) , . . . , y s } is independent. We set y i to be this element y . Observe thatdue to (I2) , A ∪ Z i − ∪ { y i } is indeed independent for every i ∈ { , . . . , s } .We claim that ω ( y i ) ≤ ω ( z i ) holds for every i ∈ { , . . . , s } . For the sake of contradiction,assume that this is not the case and let i ∈ { , . . . , s } be the first index such that ω ( y i ) > ω ( z i )holds. Recall that X t is constructed by the greedy algorithm. Denote by W ⊂ X t the set ofelements that are prior z i in the ordering σ . Suppose that y i ∈ cl ( W ). By the constructionof Z , W ⊆ cl ( A ∪ Z i − ), because z i is the first element in σ such that A ∪ Z i − ∪ { z i } is independent. By (CL2) and (CL3) , we have that y i ∈ cl ( cl ( A ∪ Z i − )) = cl ( A ∪ Z i − ).However, this contradicts the property that A ∪ Z i − ∪{ y i } is independent. Hence, y i / ∈ cl ( W ).This implies that W ∪ { y i } is independent. But this means that the greedy algorithm wouldhave y i over z i in the construction of X t , because ω ( y i ) > ω ( z i ); a contradiction. This provesthat ω ( y i ) ≤ ω ( z i ) holds for every i ∈ { , . . . , s } . Therefore, ω ( Z ) ≥ ω ( Y ) and (ii) is fulfilled.This completes the proof of the existence of a set Z ⊆ X t satisfying (i) and (ii).We replace the basis B h in the solution by B h = A ∪ Z = ( B h \ Y ) ∪ Z . We show thatthe resulting family of bases is a solution to the instance ( M, ω, k, d ). Clearly, it is sufficientto show that for every i ∈ { , . . . , k } such that i = h , ω ( B h B i ) ≥ d , as the other pairsof bases are the same as before. By the choice of t ∈ { , . . . , ‘ } we have that X t ∩ B i = ∅ holds for all i ∈ { , . . . , k } . And since Z ⊆ X t we have that Z ⊆ B h B i holds. Then, ω ( B h B i ) = ω ((( B h \ Y ) ∪ Z ) B i )) ≥ ω ( B h B i ) − ω ( Y ) + ω ( Z ) ≥ ω ( B h B i ) ≥ d asrequired. We have that the replacement of B h by B h gives a solution. However B h ⊆ S t ⊆ S whereas B h \ S = ∅ , and this contradicts the assumption that B , . . . , B k is a solution suchthat the number of vertices of the bases in S is the maximum. This concludes the proof ofthe claim. (cid:67) Let c M = M − ( E ( M ) \ S ). Then E ( c M ) = S and the set B is a basis of c M as well.Claim 12 immediately implies the following property. . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 15 (cid:66) Claim 13.
The instances (
M, ω, k, d ) and ( c M , ω, k, d ) of
Weighted Diverse Bases areequivalent.We now repeat the argument that preceded Claim 12, this time with the dual matroid c M ∗ and starting with its basis b B = ( E ( c M ) \ B ) = L . Recall that ‘ = d d e k . We constructsets S ∗ , . . . , S ∗ ‘ iteratively. We set S ∗ = b B . For i ≥ S ∗ i from S ∗ ( i − asfollows. If E ( c M ) \ S ∗ i − = ∅ , we set X ∗ i = ∅ . Otherwise we set X ∗ i to be a basis of maximumweight in the matroid c M ∗ − S ∗ i − , which we find using the greedy algorithm. Finally, we set S ∗ i = S ∗ i − ∪ X ∗ i .Let S ∗ = S ∗ ‘ and L ∗ = S ∗ \ b B = S ∗ ∩ B . Since corank ( B ) ≤ k d d e −
1, we get that every coindependent set contained in the set B has size at most k d d e −
1. And since L ∗ is a disjointunion of ‘ such coindependent sets we get that | L ∗ | = | S ∗ ∩ B | ≤ ‘ ( k d d e − ≤ d d e k .Restating Claim 12 for c M ∗ , we get that if ( c M ∗ , ω, k, d ) is a yes-instance of WeightedDiverse Bases , then there is a solution, that is, a family of bases B ∗ , . . . , B ∗ k of c M ∗ suchthat ω ( B ∗ i B ∗ j ) ≥ d holds for all distinct i, j ∈ { , . . . , k } , with the property that B ∗ i ⊆ S ∗ for every i ∈ { , . . . , k } . In terms of c M , the same property can be stated as follows. (cid:66) Claim 14.
If ( c M , ω, k, d ) is a yes-instance of
Weighted Diverse Bases , then thereis a solution, that is, a family of bases B , . . . , B k such that ω ( B i B j ) ≥ d for alldistinct i, j ∈ { , . . . , k } with the property that B i ⊆ S ∗ for every i ∈ { , . . . , k } , where B i = ( E ( c M ) \ B i ).Since E ( c M ) = ( B ∪ b B ) and b B ⊆ S ∗ ⊆ E ( c M ) we have that E ( c M ) = ( B ∪ S ∗ ). Hencefrom Claim 14 we get that if ( c M , ω, k, d ) is a yes-instance, then it has a solution B , . . . , B k such that ( B \ S ∗ ) ⊆ B i holds for every i ∈ { , . . . , k } . That is, elements from the set( B \ S ∗ ) do not contribute to the weight ω ( B i B j ) for any distinct i, j ∈ { , . . . , k } . So atransformation that removes the subset ( B \ S ∗ ) from the ground set of c M is safe, providedthat (i) B i \ ( B \ S ∗ ) = ( B i ∩ S ∗ ) is a basis of the resulting matroid for all i ∈ { , . . . , k } ,and (ii) for any basis B of the resulting matroid, B ∪ ( B \ S ∗ ) is a basis of c M .We now show that the operation of contracting the set ( B \ S ∗ ) has both these properties.Let f M = c M / ( B \ S ∗ ). Then E ( f M ) = ( B ∪ S ∗ ) \ ( B \ S ∗ ) = S ∗ . Let rank ( c M ) , rank ( f M )be the ranks and d rank , g rank be the rank functions of the two matroids c M , f M , respectively.Recall that g rank ( X ) = d rank (( B \ S ∗ ) ∪ X ) − d rank ( B \ S ∗ ) holds for all X ⊆ E ( f M ) = S ∗ .Now rank ( f M ) = g rank ( S ∗ ) = d rank ( B ∪ S ∗ ) − d rank ( B \ S ∗ ) = rank ( c M ) − | B \ S ∗ | , where thelast equation holds because B is a basis of c M . And for any i ∈ { , . . . , k } , g rank ( B i ∩ S ∗ ) = d rank (( B \ S ∗ ) ∪ ( B i ∩ S ∗ )) − d rank ( B \ S ∗ ) = rank ( c M ) − | B \ S ∗ | = rank ( f M ), where thesecond equation holds because B, B i are bases of c M and B i ⊆ (( B \ S ∗ ) ∪ ( B i ∩ S ∗ )). Thus( B i ∩ S ∗ ) is a basis of f M . Finally, let B be an arbitrary basis of f M . Then g rank ( B ) = rank ( f M ) = rank ( c M ) − | B \ S ∗ | . Rearranging the expression for g rank ( B ) in terms of d rank weget: d rank (( B \ S ∗ ) ∪ B ) = g rank ( B ) + d rank ( B \ S ∗ ) = rank ( c M ) − | B \ S ∗ | + | B \ S ∗ | = rank ( c M )where the second equation holds because B is a basis of c M . Thus ( B \ S ∗ ) ∪ B is a basis of c M , and we have (cid:66) Claim 15.
The instances (
M, ω, k, d ) and ( f M , ω, k, d ) of
Weighted Diverse Bases areequivalent.Recall the sets L = b B ⊆ S ∗ and L ∗ = ( S ∗ \ L ) from the construction. ( L, L ∗ ) is thusa partition of E ( f M ) = S ∗ . From the construction we get L = b B ⊆ B and L ∗ ⊆ B . Nowsince rank ( B ) ≤ d d e k and corank ( B ) ≤ d d e k in M , we have that for every basis B of f M , | B ∩ L | ≤ d d e k and | L ∗ \ B | ≤ d d e k hold. This completes the description of the algorithm that returns the instance ( f M , ω, k, d )and the partition (
L, L ∗ ) of E ( f M ). Since | L | ≤ d d e k and | L ∗ | ≤ d d e k , we have that | E ( f M ) | ≤ d d e k . It is straightforward to verify that given an independence oracle for M we can construct the following in polynomial time: (i) the set E ( f M ), (ii) an independenceoracle for f M that in time polynomial in | E ( M ) | answers queries for f M , and (iii) the sets L and L ∗ . To see this, note that A ⊆ E ( f M ) is independent in f M if and only if A = A ∪ ( B \ S ∗ )is independent in M .To show the second claim of the lemma, assume that we are given representation A of M over a finite field F . It is well-known that M ∗ also is representable over F and, given A , therepresentation of M ∗ over F can be computed in polynomial time by linear algebra tools(see, e.g., [17]). Taking into account that contraction of a set is equivalent to the deletion ofthe same set in the dual matroid and vice versa, we obtain that the representation ˜ A of f M can be constructed in polynomial time from A . This concludes the proof of the lemma. (cid:74) Using Lemma 11 we can prove that
Weighted Diverse Bases is FPT when parameter-ized by k and d . (cid:73) Theorem 2.
Weighted Diverse Bases can be solved in O ( dk (log k +log d )) · | E ( M ) | O (1) time. Proof.
Let (
M, ω, k, d ) be an instance of
Weighted Diverse Bases . We run the algorithmfrom Lemma 11. If the algorithm solves the problem, then we are done. Otherwise, thealgorithm outputs an equivalent instance ( f M , ω, k, d ) of
Weighted Diverse Bases suchthat E ( f M ) ⊆ E ( M ) and | E ( f M ) | ≤ d d e k . Moreover, the algorithm computes the partition( L, L ∗ ) of E ( f M ) with the property that for every basis B of f M , | B ∩ L | ≤ d d e k and | L ∗ \ B | ≤ d d e k . Then we check all possible k -tuples of bases by brute force and verifywhether there are k bases forming a solution. By the properties of L and L ∗ , f M has( d k ) O ( dk ) distinct bases. Therefore, we check at most ( d k ) O ( dk ) k -tuples of bases. Weconclude that this checking can be done in 2 O ( dk (log k +log d )) · | E ( M ) | O (1) time, and the claimfollows. (cid:74) If the input matroid is given by a representation over a finite field, then
WeightedDiverse Bases admits a polynomial kernel when parameterized by k and d . (cid:73) Theorem 3.
Given a representation of the matroid M over a finite field GF( q ) as input,we can compute a kernel of Weighted Diverse Bases of size O ( k d log q ) . Proof.
Let (
M, ω, k, d ) be an instance of
Weighted Diverse Bases . Let also A be itsrepresentation over GF( q ) . We run the algorithm from Lemma 11. If the algorithm solvesthe problem and reports that ( M, ω, k, d ) is a yes-instance, we return a trivial yes-instanceof the problem. Otherwise, the algorithm outputs an equivalent instance ( f M , ω, k, d ) of
Weighted Diverse Bases such that E ( f M ) ⊆ E ( M ) and | E ( f M ) | ≤ d d e k . Moreover,the algorithm computes a representation ˜ A of f M over GF( q ) . Clearly, it can be assumedthat the number of rows of the matrix ˜ A equals rank ( f M ). Since rank ( f M ) ≤ | E ( f M ) | , thematrix ˜ A has O ( k d ) elements. Because ˜ A is a matrix over GF( q ) , it can be encoded by O ( k d log q ) bits. Finally, note that the weights of the elements can be truncated by d , thatis, we can set ω ( e ) := min { ω ( e ) , d } for every e ∈ E ( f M ). Then the weights can be encodedusing O ( d k log d ) bits. This concludes the construction of our kernel. (cid:74) . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 17 FPT algorithm for Weighted Diverse Common Independent Sets
In this section we show that
Weighted Diverse Common Independent Sets is FPT when parameterized by k and d .We use a similar win-win approach as for Weighted Diverse Bases and observe thatif the two matroids from an instance of
Weighted Diverse Common Independent Sets have a sufficiently big common independent set, then we have a yes-instance of
WeightedDiverse Common Independent Sets . (cid:73) Lemma 16.
Let M and M be matroids with a common ground set E , and let k ≥ and d ≥ be integers. If there is an X ⊆ E of size at least k d d e such that X is acommon independent set of M and M , then ( M , M , ω, k, d ) is a yes-instance of WeightedDiverse Common Independent Sets for any weight function ω : E → N . Proof.
Let X ⊆ E be a set of size at least k d d e such that X is a common independentset of M and M . Then there is a partition I , . . . , I k of X such that | I i | ≥ d d e for every i ∈ { , . . . , k } . Clearly, I , . . . , I k are common independent sets of M and M . Also wehave that ω ( I i I j ) = ω ( I i ) + ω ( I j ) ≥ d for all distinct i, j ∈ { , . . . , k } and every weightfunction ω which assigns positive integral weights. This means that I , . . . , I k is a solutionfor ( M , M , ω, k, d ); that is, ( M , M , ω, k, d ) is a yes-instance. (cid:74) Lemma 16 implies that we can assume that the maximum size of a common independentset of the input matroids is bounded. We prove the following crucial lemma. (cid:73)
Lemma 17.
Let ( M , M , ω, k, d ) be an instance of Weighted Diverse Common In-dependent Sets such that the maximum size of a common independent set of M and M is at most s . Then there is a set F of common independent sets of M and M , of size |F| = 2 O ( s log( ks )) · d , such that if ( M , M , ω, k, d ) is a yes-instance of Weighted DiverseCommon Independent Sets then the instance has a solution I , . . . , I k with I i ∈ F for i ∈ { , . . . , k } . Moreover, F can be constructed in O ( s log( ks )) · d · | E | O (1) time where E isthe (common) ground set of M and M . Proof.
Consider ( M , M , ω, k, d ). Let E = E ( M ) = E ( M ). It is convenient to assumethat the weights of the elements are bounded by d . For this, we set ω ( e ) := min { d, ω ( e ) } for every e ∈ E . It is straightforward to see that by this operation we obtain an equivalentinstance of Weighted Diverse Common Independent Sets . Notice that for everycommon independent set I of M , M , we now have ω ( I ) ≤ ds .For every w ∈ { , . . . , ds } , we use a recursive branching algorithm to construct a family F w of size 2 O ( s log( ks )) of common independent sets of M and M with the followingproperties: (i) each set in F w has weight at least w , and (ii) if S = { I , . . . , I k } is a solutionto the instance ( M , M , ω, k, d ) such that ω ( I i ) = w for some i ∈ { , . . . , k } , then there isan I i ∈ F w such that ( S \ I i ) ∪ I i is also a solution to ( M , M , ω, k, d ).The algorithm, denoted by A , takes as its input a common independent set X of M and M , and two matroids M and M such that M i = ( M i − W ) /X for i = 1 , W ⊆ E \ X . For the very first call to A we set X := ∅ and M i = M i for i = 1 , W := ∅ ). Algorithm A outputs at most ks common independent sets of M and M of the form X ∪ Y , where Y ⊆ E = E ( M ) = E ( M ) is a common independent setof M and M . Note that E = E \ ( X ∪ W ). Algorithm A performs the following steps. Step 1. If ω ( X ) ≥ w , then output X and return. Step 2.
Greedily compute at most ks disjoint common independent sets Y , . . . , Y ‘ of M and M , each of weight at least w = w − ω ( X ), as follows.(a) Set i = 1 , Ys = {} .(b) If | Ys | = ( i −
1) = ks then set ‘ = ( i −
1) and go to
Step 3 .(c) Set M h = M h − ( [ Y j ∈ Ys Y j ), for h = 1 , Z of M and M of the maximum weight.(e) If ω ( Z ) < w , then set ‘ = ( i −
1) and go to
Step 3 . Otherwise, set Y i = Z, Ys =( Ys ∪ { Y i } ) and i = i + 1, and go to Step 2(b) . Step 3.
At this point we have Ys = { Y , . . . , Y ‘ } .If ‘ = 0 then return.If ‘ = ks then output the sets X ∪ Y , . . . , X ∪ Y ‘ and return.If neither of the above holds then:Set R = [ Y j ∈ Ys Y j .For each nonempty common independent set Z ⊆ R of M and M , set W = R \ Z and recursively invoke A ( X ∪ Z, ( M − W ) /Z, ( M − W ) /Z ).This completes the description of A . To construct F w , we call A ( ∅ , M , M ). Thenthe set F w includes all the sets output by A . Note that in every recursive step we call A ( X ∪ Z, ( M − W ) /Z, ( M − W ) /Z ) only if Z = ∅ . So the size of the first argument ( X ∪ Z )to a recursive call of A is strictly larger than the size of the first argument X of the parentcall to A . Moreover, since Z is a common independent set of M and M , we have that X ∪ Z is a common independent set of M and M . Because the maximum size of the commonindependent set of M and M is at most s , we obtain that the depth of the recursion isbounded by s , that is, the algorithm is finite. We show the crucial property of F w mentionedabove. (cid:66) Claim 18. If S = { I , . . . , I k } is a solution to the instance ( M , M , ω, k, d ) such that ω ( I i ) = w for some i ∈ { , . . . , k } , then there is an I i ∈ F w such that ( S \ I i ) ∪ I i is also asolution to ( M , M , ω, k, d ). Proof.
Fix a set I i ∈ S ; ω ( I i ) = w . Recall that an arbitrary invocation of A has the form A ( X, ( M − W ) /X, ( M − W ) /X ) where X is a common independent set of M , M and W ⊆ ( E \ X ). For the very first invocation of A these sets are X = ∅ , W = ∅ , and these setstrivially satisfy the viability condition ( X ∪ W ) ∩ I i = X ; that is: I i contains all of X ,and none of W . We show that any invocation of A whose arguments satisfy the viabilitycondition either outputs a set I i that can be used to replace I i in S , or makes at leastone recursive call to A such that the arguments to this recursive call satisfy the viabilitycondition. Since the size of the first argument ( X ∪ Z ) to a recursive call of A is strictlylarger than the size of the first argument X of the parent call to A , we get that some call to A will output a set I i with the desired property.Assume inductively that A ( X, M = ( M − W ) /X, M = ( M − W ) /X ) is an invocationof A whose arguments satisfy the viability condition. If ω ( X ) ≥ w , then X = I i , because ω ( I i ) = w . In this case, the algorithm outputs X = I i in Step 1 . Clearly, we can set I i = I i ,and we are done. So let us assume that this is not the case, and that ω ( X ) < w . In this case X (cid:40) I i holds and the algorithm goes to Step 2 .Let E = E \ ( X ∪ W ) be the common ground set of M and M . Let Y = ( I i \ X ) and w = ω ( Y ) = w − ω ( X ). Then since Y is a common independent set of M and M , thegreedy computation of Step 2 produces a nonempty family Y , . . . , Y ‘ of disjoint common . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 19 independent sets of M and M of weight at least w each. Note that X ∪ Y , . . . , X ∪ Y ‘ arecommon independent sets of M and M . We consider two cases depending on the value of ‘ in Step 3 . Note that by the above reasoning the case ‘ = 0 does not arise here. Case 1. ‘ = ks . In this case the algorithm outputs the sets X ∪ Y , . . . , X ∪ Y ‘ where( X ∩ Y i ) = ∅ holds for all i ∈ { , . . . , ‘ } . Recall that every common independent set of M and M has size at most s . Therefore, the set J = S j ∈{ ,...,k } , j = i I j has size at most( k − s < ‘ . Hence, by the pigeonhole principle, there is an h ∈ { , . . . , ‘ } such that Y h ∩ I j = ∅ holds for all j ∈ { , . . . , k } ; j = i .Let I i := X ∪ Y h . Then I i is a common independent set of M and M and ω ( I i ) = ω ( X ) + ω ( Y h ) ≥ ω ( X ) + w = ω ( X ) + ω ( I i \ X ) = ω ( I i ), where the last equationfollows from the fact that X ⊆ I i holds. From this chain of relations we also get that ω ( I i \ X ) ≤ ω ( Y h ) holds. Consider an arbitrary index j ∈ { , . . . , k } ; j = i . Then ω ( I i I j ) = ω ( I j \ I i ) + ω ( I i \ I j ) ≤ ω ( I j \ X ) + ω ( I i \ I j ) = ω ( I j \ X ) + ω ( X \ I j ) + ω (( I i \ X ) \ I j ) ≤ ω ( I j X ) + ω ( I i \ X ) ≤ ω ( I j X ) + ω ( Y h ). But since I i = X ∪ Y h ,( X ∩ Y h ) = ∅ and ( I j ∩ Y h ) = ∅ we get that ω ( I j I i ) = ω ( I j X ) + ω ( Y h ) holds. Thus ω ( I j I i ) ≥ ω ( I i I j ), and so replacing I i by I i in the solution S indeed gives us asolution to the instance ( M , M , ω, k, d ). Case 2. < ‘ < ks . In this case we set R := S ‘i =1 Y i . From the construction we get that thematroids M − R and M − R have no common independent set of weight at least w . Since( I \ X ) \ R is such a common independent set we have that ω (( I \ X ) \ R ) < w , and since ω ( I i \ X ) = w we get that Z = ( I i \ X ) ∩ R = ∅ . Clearly, Z is a common independentset of M and M . Our algorithm considers all such sets. Hence there is a recursive call A ( X , ( M − W ) /Z, ( M − W ) /Z ) where Z = (( I i \ X ) ∩ R ) , X = X ∪ Z, W = R \ Z .By the choice of Z and W we get that ( X ∪ W ) ∩ I i = X , so that this recursive callsatisfies the viability condition. Moreover, we have that | X | > | X | . This completes thesecond case and the proof of the claim. (cid:74) We already observed that the algorithm A is finite. Now we evaluate its running timeand the size of F w . (cid:66) Claim 19.
The set F w has size 2 O ( s log( ks )) and can be constructed in 2 O ( s log( ks )) · | E | O (1) time. Proof.
To give an upper bound on the size of F w , observe that in each recursive call, thealgorithm A either outputs some sets, or performs some recursive calls, or simply returnswithout outputting anything. Notice that in Step 1, A can output at most one set, and A may output ks sets in Step 3. The number of recursive calls is upper bounded by thenumber of nonempty common independent sets Z ⊆ R of M and M . Since ‘ < ks and | Y i | ≤ s for i ∈ { , . . . , ‘ } , | R | ≤ ks . Because for each Z , | Z | ≤ s , the branching factor is atmost ( ks ) s = 2 O ( s log( ks )) . Since the depth of the recursion is at most s , the search tree has2 O ( s log( ks )) leaves. This implies that the size of F w is 2 O ( s log( ks )) .To evaluate the running time, note that in Step 2, the algorithm greedily constructs thesets Y , . . . , Y ‘ that are common independent sets of M and M . By Proposition 8, this canbe done in polynomial time, because in each iteration we find a common independent set ofmaximum weight. Because the search tree has 2 O ( s log( ks )) leaves, the total running time is2 O ( s log( ks )) · | E | O (1) . (cid:67) We construct F = S dsw =0 F w . By Claim 19, |F| ≤ ( ds + 1) max w ∈{ ,...,ds } |F w | =2 O ( s log( ks )) · d and F can be constructed in total 2 O ( s log( ks )) · d ·| E | O (1) time. Claim 18 implies that if ( M , M , ω, k, d ) is a yes-instance of Weighted Diverse Common IndependentSets , then the instance has a solution I , . . . , I k with I i ∈ F for i ∈ { , . . . , n } . (cid:74) Combining Lemma 16 and Lemma 17, we obtain the main result of the section. (cid:73)
Theorem 4.
Weighted Diverse Common Independent Sets can be solved in O ( k d log( kd )) · | E | O (1) time. Proof.
Let ( M , M , ω, k, d ) be an instance of Weighted Diverse Common IndependentSets . First, we use Proposition 8 to solve
Matroid Intersection for M and M andfind a common independent set X of maximum size. If | X | ≥ k d d e , then by Lemma 16, weconclude that ( M , M , ω, k, d ) is a yes-instance. Assume that this is not the case. Thenthe maximum size of a common independent set of M and M is s < k d d e . We applyLemma 17 and construct the set F of size 2 O (( kd ) log( kd )) in 2 O (( kd ) log( kd )) · | E | O (1) time.By this lemma, if ( M , M , ω, k, d ) is a yes-instance, it has a solution I , . . . , I k such that I i ∈ F for i ∈ { , . . . , k } . Hence, to solve the problem we go over all k -tuples of the elementsof F , and for each k -tuple, we verify whether these common independent sets of M and M give a solution. Clearly, we have to consider 2 O ( k d log( kd )) tuples. Hence, the total runningtime is 2 O ( k d log( kd )) · | E | O (1) . (cid:74) In this section we prove that
Diverse Perfect Matchings is fixed parameter tractablewhen parameterized by k and d . We need the following simple observations later in thissection. (cid:73) Observation 3.
The cardinality of symmetric differences of perfect matchings in a graphobeys the triangle inequality. That is, for a graph G and perfect matchings M , M , M in G , | M M | + | M M | ≥ | M M | . Observation 3 follows from the fact that Hamming distance is a metric and hence obeystriangular inequality. (cid:73)
Observation 4.
Let G be a graph and M and M be two perfect matchings in G . Then | M M | = 2 · | M \ M | = 2 · | M \ M | . For an undirected graph G , the Tutte matrix A of G over the field F [ X ] is defined asfollows, where F is the Galois field on { , } and X = { x e : e ∈ E ( G ) } . The rows andcolumns of A are labeled with V ( G ) and for each e = { u, v } ∈ E ( G ), A [ u, v ] = A [ v, u ] = x e .All other entries in the matrix are zeros. That is, for any pair of vertices u, v ∈ V ( G ), ifthere is no edge between u and v , A [ u, v ] = 0. It is well known that det ( A ) = 0 if and only if G has a perfect matching. As the characterstic of F is a 2, the determinant of A coincideswith the permanent of A . That is, det ( A ) = perm ( A ) = X σ ∈ S V ( G ) Π v ∈ V ( G ) A [ v, σ ( v )] . (1)Here, S V ( G ) is the set of all permutations of V ( G ). Let PM ( G ) be the set of perfect matchingson G . Then, one can show that det ( A ) = P M ∈ PM ( G ) Π e ∈ M x e .Let Y be a set of variables disjoint from X . For each edge e , let L ( e ) ⊆ Y be a subsetof variables. Let A be the matrix obtained from A by replacing each entry of the form x e with x e · Π y ∈ L ( e ) y . Then, . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 21 det ( A ) = perm ( A ) = X σ ∈ S V ( G ) Π v ∈ V ( G ) A [ v, σ ( v )]= X σ ∈ S V ( G ) Π v ∈ V ( G ) (cid:0) A [ v, σ ( v )] · Π y ∈ L ( { v,σ ( v ) } ) y (cid:1) , (2)where Π y ∈ L ( { v,σ ( v ) } ) y = 1 if { v, σ ( v ) } / ∈ E ( G ) or L ( e ) = ∅ . (cid:73) Lemma 20.
Let G be an undirected graph and let X = { x e : e ∈ E ( G ) } and Y = { y , . . . , y ‘ } be two sets of variables such that X ∩ Y = ∅ . For each edge e ∈ E ( G ) , weare also given a subset L ( e ) ⊆ Y . Let A be the matrix defined as above. For any perfectmatching M , Π e ∈ M x e Π y ∈ L ( e ) y is a monomial in det ( A ) . Moreover, for any monomial m in det ( A ) , M = { e : x e is a variable in m } is perfect a matching in G and for each e ∈ M , L ( e ) is a subset of variables in the monomial m . Proof. A cycle-matching cover of G is a subset of edges F ⊆ E ( G ) such that V ( F ) = V ( G )and each connected component of G [ F ] is either a cycle or an edge. Each non-zero term inthe summation of (2), there is a cycle-matching cover defined as follows. Let σ ∈ S V ( G ) suchthat Π v ∈ V ( G ) A [ v, σ ( v )] · Π y ∈ L ( { v,σ ( v ) } ) y is non-zero. Then, Π v ∈ V ( G ) A [ v, σ ( v )] is non-zero.As G is a simple graph, A [ v, v ] = 0. Therefore, since Π v ∈ V ( G ) A [ v, σ ( v )] = 0, there is no1-cycle in σ . Moreover any ‘ -cycle in σ corresponds to a cycle in G and any 2-cycle in σ corresponds to an edge in G , where the vertices covered in the cycle are the vertices presentthe cycle of the permutation. That is, for each cycle ( u , u , . . . , u ‘ in σ , u , u , . . . , u ‘ , u is acycle in G if ‘ > u u is a matching edge if ‘ = 2. Therefore, there is a cycle-matchingcover corresponding to the non-zero term Π v ∈ V ( G ) A [ v, σ ( v )] · Π y ∈ L ( { v,σ ( v ) } ) y .Let F be a cycle-matching cover. Let { C , . . . , C r } be the set of cycles in G [ F ] and { e , . . . , e s } be the set of the edges in F \ ( S i E ( C i )). Let F = S i ∈ [ r ] E ( C i ). For each cycle C = u , u , . . . , u ‘ , u in G [ F ], where ‘ > σ and σ on V ( C ) as follows: ( u , u , . . . , u ‘ ) and ( u , u ‘ , u ‘ − , . . . , u ). That is,Π v ∈ V ( C ) A [ v, σ ( v )] = Π v ∈ V ( C ) A [ v, σ ( v )] = Π e ∈ E ( C ) (cid:0) x e · Π y ∈ L ( e ) y (cid:1) . This implies that there are 2 r terms in (2) which are equal to P e ∈ F \ F (cid:0) x e · Π y ∈ L ( e ) y (cid:1) + P e ∈ F Π e ∈ F (cid:0) x e · Π y ∈ L ( e ) y (cid:1) (which we call the terms corresponding to F ). In other wordsif F is a perfect matching then Π e ∈ F (cid:0) x e · Π y ∈ L ( e ) y (cid:1) is a unique term in (2), and if F is hascycle then the terms corresponding to F will cancel each other, because the characteristic ofthe F [ X ∪ Y ] is 2. Therefore, for any perfect matching M , Π e ∈ M x e Π y ∈ L ( e ) y is a monomialin det ( A ).As any non-zero term in (2) corresponds to a cycle-matching cover, and for any cycle match-ing cover that contains at least one cycle all the terms corresponding to it cancels each otherwe have the following. For any monomial m in det ( A ), M = { e : x e is a variable in m } isperfect a matching in G . Also, from the construction of A , it follows that for each e ∈ M , L ( e ) is subset of variables in m . (cid:74) We use the following two known results. (cid:73)
Proposition 21 (Schwartz-Zippel Lemma [19, 23]) . Let P ( x , . . . , x n ) be a multivariatepolynomial of total degree at most d over a field F , and P is not identically zero. Let r , . . . , r n be the elements in F choses uniformly at random with repetition. Then Pr( P ( r , ..., r n ) =0) ≤ d | F | . For a multivariate polynomial P and a monomial m , we let P ( m ) denote the coefficientof m in P . (cid:73) Proposition 22 ([22]) . Let P ( x , . . . , x n ) be a polynomial over a field of characteristictwo, and T ⊆ [ n ] be a set of target indices. For a set I ⊆ [ n ] , define P − I ( x , . . . , x n ) = P ( y , . . . , y n ) where y i = 0 for i ∈ I and y i = x i otherwise. Define Q ( x , . . . , x n ) = X I ⊆ T P − I ( x , . . . , x n ) . Then, for any monomial m such that t := Π i ∈ T x i divides m we have Q ( m ) = P ( m ) , and forevery other monomial we have Q ( m ) = 0 . (cid:73) Lemma 23.
There is an algorithm that given an undirected graph G , perfect matchings M , . . . , M r , and a non-negative integer s , runs in time O ( rs ) n O (1) , and outputs a perfectmatching M such that | M \ M i | ≥ s for all i ∈ { , . . . , r } (if such a matching exists) withprobability at least e − rs . Proof.
For each i ∈ { , . . . , r } , we color each edge in E ( G ) \ M i uniformly at randomusing colors { c i, , . . . , c i,s } . Now we label each edge with a subset of the variable set Y = { y i,j : i ∈ [ r ] , j ∈ [ s ] } . For each edge e , we label it with L ( e ) = { y i,j : i ∈ [ r ] and e is colored with c i,j in the random coloring for i } . Let A be the Tutte matrix of G over the field F [ X ], where X = { x e : e ∈ E ( G ) } . Let A bethe matrix obtained from A by replacing each entry of the form x e with x e · Π y ∈ L ( e ) y .Suppose there is a matching M such that | M \ M i | ≥ s for all i ∈ { , . . . , r } . Then, foreach i ∈ [ r ], let { e i, , . . . , e i,s } ⊆ ( M \ M i ) be an arbitrary subset. We say that { e i, , . . . , e i,s } is colorful if the edges in { e i, , . . . , e i,s } gets distinct colors from { c i, , . . . , c i,s } in the randomcoloring for i . Then for each i ∈ [ r ], the probability that { e i, , . . . , e i,s } is colorful is s ! s s ≥ e − s . For each q ∈ [ r ], let E q be the even that { e q, , . . . , e q,s } is colorful. As therandom coloring for i ∈ [ r ] is different from the random coloring for j ∈ [ r ] \ { i } , theevents E i and E j are independent. That is, E , . . . , E r are independent events and hencePr[ T ri =1 E i ] ≥ e − rs . Therefore, there is a monomial m in det ( A ) with probability at least e − rs such that M = { e ∈ E ( G ) : x e is a variable in m } and Y is a subset of variables in m .Now, suppose there is a monomial m in det ( A ) such that Y is a subset of variables in m .Therefore, since for each i ∈ [ r ] only the edges in E ( G ) \ M i are colored and { y i,j : j ∈ [ s ] } ⊆ Y , we have that | M \ M i | ≥ s . Moreover, by Lemma 20, { e ∈ E ( G ) : x e is a variable in m } is a perfect matching in G .Thus, it is enough to check whether there exists a monomial m in det ( A ) such that allthe variables in Y are present in m . (cid:66) Claim 24.
There is an algorithm, that runs in time 2 O ( | Y | ) n O (1) and it outputs thefollowing. If there is no monomial in det ( A ) that contains Y , then the algorithm outputsNo. If there is a monomial in det ( A ) that contains Y , then the algorithm outputs Z ⊆ X with probability at least 2 / m in det ( A ) with variables in m is exactly equal to Z ∪ Y . Proof.
Let { e , . . . , e m } = E ( G ). Let P ( x e , . . . , x e m , y , , . . . , y r,s ) = det ( A ). For each I ⊆ Y , we define P − I ( x e , . . . , x e m , y , , . . . , y r,s ) = P ( x e , . . . , x e m , z , , . . . , z r,s ), where forall i ∈ [ r ] and j ∈ [ s ], z i,j = 0 if y i,j ∈ I and z i,j = y i,j otherwise. Let Q = Q ( x e , . . . , x e m , y , , . . . , y r,s ) = X I ⊆ Y P − I ( x e , . . . , x e m , y , , . . . , y r,s ) . . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 23 By Proposition 22, the set of monomials in Q are the set of monomial is det ( A ) thatcontains Y . Our objective is to find out the variables in such a monomial if it exists. Towardsthat we consider Q be a polynomial in a field extension F of F such that the number ofelements in the field F is at least t , which we fix later. From the construction of det ( A ),we know that the degree of det ( A ) and Q is at most d = n + 2 rs . By Proposition 21, theexistence of a required monomial can be tested in polynomial time with failure probabilityat most dt . But, recall that we need to find out the variables in such a monomial. Towardsthat we do the following. Notice that if Q is a polynomial identically zero, then our answeris No. This can be checked using Proposition 21. Assume that Q is not identically zero.Let Q = Q (0 , x e , . . . , x e m , y , , . . . , y r,s ). If Q is not identically equal to zero, then thereis a monomial satisfying the property mentioned in the statement of the claim is presentin Q . Otherwise we know that every monomial with the required property contains thevariable x e . Again this can be checked using Proposition 21. So if Q is not identically zero,then set Q = Q . Next, we let Q = Q ( x e , , x e , . . . , x e m , y , , . . . , y r,s ). Again, Q is notidentically equal to zero, then there is a monomial satisfying the property mentioned in thestatement of the claim, is present in Q . Otherwise we know that every monomial with therequired property contains the variable x e . By repeating this process at most m times, wewill be able to obtain all the variables present in the required monomial. Our algorithm willsucceed if all the m + 1 application of Proposition 21 do not fail. Thus, by union boundthe failure probability of is at most ( m + 1) dt . We set ( m + 1) dt = and this implies that t = ( m + 1)( n + rs ). Hence the success probability of our algorithm is at least .Towards the running time analysis, notice that the construction of the polynomial Q takes time 2 | Y | n O (1) and each application of Proposition 21 takes time polynomial in thesize of Q . This implies that the total running time is bounded by 2 O ( | Y | ) n O (1) . (cid:67) Now we run the algorithm in Claim 24, and get a subset Z ⊆ { x e : e ∈ E ( G ) } (if itexists) such that there is a monomial m in det ( A ) and the variables in m is exactly equal to Z ∪ Y with probability at least 2 /
3. As the initial random coloring of edges succeeds withprobability at least e − rs , the success probability of out algorithm is at least e − rs . If nosuch monomial exists then the algorithm outputs No. By Lemma 20, M = { e : x e ∈ Z } is aperfect matching and it is the required output. The running time of the algorithm followsfrom Claim 24. This completes the proof of the lemma. (cid:74)(cid:73) Lemma 25.
There is an algorithm that given an undirected graph G , a perfect matching M ,and non-negative integers r, d, s , runs in time O ( r s ) n O (1) , and outputs r perfect matchings M ? , . . . , M ?r such that | M M ?i | ≤ s for all i ∈ { , . . . , r } and | M ?i M ?j | ≥ d for alldistinct i, j ∈ [ r ] (if such matchings exist) with probability at least e − rs . If no such perfectmatchings exist, then the algorithm outputs No Proof.
Suppose there exist perfect matchings M , . . . , M r such that | M M i | ≤ s forall i ∈ { , . . . , r } and | M i M j | ≥ d for all distinct i, j ∈ [ r ]. Then we know that P ri =1 | M M i | ≤ rs . Let S i = M M i for all i ∈ [ r ]. Notice that, as M and M i areperfect matchings S i forms a collection of alternating cycles (i.e., edges in the cycles alternatebetween M and M i ). Let S = S ri =1 S i .We do a random coloring on the edges of G using rs colors. That is, we color each edgeof G uniformly at random with a color from { , . . . , rs } . We say that the random coloring is good if all the edges in S gets distinct colors. The probability that the random coloring isgood is e − rs .Now on assume that the random coloring is good. For each i ∈ [ r ], let C i be the set ofcolors on the edges S i . Notice that | C i | = | S i | . (cid:66) Claim 26.
For any two distinct integers i, j ∈ [ r ], | C i C j | ≥ d . Proof.
We know that | M i M j | ≥ d . Let E i = M i \ M j and E j = M j \ M i . Notice that | E i | + | E j | ≥ d and E i ∩ E j = ∅ .Let E i, = E i ∩ M , E i, = E i \ E j, , E j, = E j ∩ M , and E j, = E j \ E j, . As E i, ⊆ M \ M j and E i, ⊆ M ∩ M i , we have that E i, ⊆ S j \ S i . Similarly E j, ⊆ S i \ S j . As E i, ⊆ M i \ ( M j ∪ M ), we have that E i, ⊆ S i \ S j . Similarly, we have E j, ⊆ S j \ S i . Thatis, we prove that E i, ∪ E j, ⊆ S j \ S i and E j, ∪ E i, ⊆ S i \ S j . Also, since all the edges in S gets distinct colors and the colors on the edges in S i and S j are C i and C j , respectively,we have that | C i C j | ≥ | E i ∪ E j | ≥ d . This completes the proof of the claim. (cid:67) Next we prove the reverse direction of the above claim. (cid:66)
Claim 27.
Let Q and Q be two collections of alternating cycles in G (i.e., the edges in Q i are alternating between M and E ( G ) \ M for each i ∈ { , } ) such that following hold: | E ( Q ) | = | C | , | E ( Q ) | = | C | , the edges in E ( Q ) uses distinct colors from C , and theedges in E ( Q ) uses distinct colors from C . Let P = E ( Q ) M and P = E ( Q ) M .Then, P and P are perfect matchings and | P P | ≥ d . Proof. As M is a perfect matching and Q is a collection of alternating cycles, we havethat P = E ( Q ) M is a perfect matching. By similar arguments, we have that P isa perfect matching. Now we prove that E ( Q ) E ( Q ) ⊆ P P . Consider an edge e ∈ E ( Q ) \ E ( Q ). We have two cases based on whether e ∈ M or not. In the first case,assume that e ∈ M . Since e ∈ E ( Q ), e ∈ M , and P = E ( Q ) M , we have that e / ∈ P .Also since e / ∈ E ( Q ), e ∈ M , and P = E ( Q ) M , we have that e ∈ P . Therefore, e ∈ P P .For the second case, we have that e / ∈ M . Since e ∈ E ( Q ), e / ∈ M , and P = E ( Q ) M ,we have that e ∈ P . Also, since e / ∈ E ( Q ) and e / ∈ M , we have that e / ∈ P . Therefore, e ∈ P P .By arguments, similar to above, one can prove that an edge e ∈ E ( Q ) \ E ( Q ) alsobelongs to P P . Thus, we proved that E ( Q ) E ( Q ) ⊆ P P . Since | E ( Q ) | = | C | , | E ( Q ) | = | C | , the edges in E ( Q ) uses distinct colors from C , and the edges in E ( Q )uses distinct colors from C , we have that | E ( Q ) E ( Q ) | ≥ | C C | ≥ d . Therefore, | P P | ≥ | E ( Q ) E ( Q ) | ≥ d . (cid:67) Thus, to prove the lemma, it is enough to find a collection Q i of alternating cycles suchthat | E ( Q i ) | = | C i | , the edges in E ( Q i ) uses distinct colors from C i , for each i ∈ [ r ]. That is,our algorithm guesses C , . . . C r and computes Q , . . . , Q r . The cost of guessing C , . . . , C r is 2 r s . Now, given C i , to compute Q i with desired property ( | E ( Q i ) | = | C i | , the edges in E ( Q i ) are colored with distinct colors from C i ), we design a simple dynamic programming(DP) algorithm. We give a brief outline of this algorithm below. For each subset L ⊆ C i and pair of vertices u, v we have table entries D [ L, u, v ] and D [ L, ⊥ , ⊥ ] which stores thefollowing. If there is a collection Q of alternating cycles and an alternating path with u and v as endpoints such that | E ( Q ) | = | L | and the edges in E ( Q ) are colored with distinctcolors from L , then we store one such collection in D [ L, u, v ]. Otherwise, we store ⊥ in D [ L, u, v ]. If there is a collection Q of alternating cycles such that | E ( Q ) | = | L | and theedges in E ( Q ) are colored with distinct colors from L , then we store one such collectionin D [ L, ⊥ , ⊥ ]. Otherwise, we store ⊥ in D [ L, ⊥ , ⊥ ]. We compute the DP table entries inthe increasing order of the size of L . The base case is when L = ∅ . That is, D [ ∅ , ⊥ , ⊥ ] = ∅ and D [ ∅ , u, v ] = ⊥ for any two vertices u and v . Now, for any ∅ 6 = L ⊆ C i , and two distinctvertices u, v ∈ V ( G ), we compute D [ L, u, v ] and D [ L, ⊥ , ⊥ ] as follows. If there is an edge . V. Fomin and P. A. Golovach and F. Panolan and G. Philip and S. Saurabh 25 ( x, y ) ∈ E ( G ) such that the color c of ( x, y ) belongs to L , and D [ L \ { c } , x, y ] = Q = ⊥ ,then we store the graph induced on E ( Q ) ∪ { ( x, y ) } in D [ L, ⊥ , ⊥ ]. Otherwise we store ⊥ in D [ L, ⊥ , ⊥ ]. Also, if there a vertex w adjacent to v such that the color c of ( w, v ) belongs to L , and D [ L \ { c } , u, w ] = Q = ⊥ , then we store the graph induced on E ( Q ) ∪ { ( w, v ) } in D [ L, u, v ]. Otherwise we store ⊥ in D [ L, u, v ]. At the end we output D [ C i , ⊥ , ⊥ ].By using standard induction, one can prove that the computation of D [ C i , ⊥ , ⊥ ] is correctand the details is omitted here. As the number of table entries for D [ ., ., . ] is upper boundedby 2 rs +1 n , the running time to compute Q i is 2 rs n O (1) . We have already mentioned thatthe cost of guessing C , . . . , C r is 2 r s . Therefore, the total running time to compute therequired r perfect matchings is 2 O ( r s ) n O (1) . (cid:74) Finally, we put together both the lemmas and prove the main theorem of the section. (cid:73)
Theorem 6.
There is an algorithm that given an instance of
Diverse Perfect Match-ings , runs in time O ( kd ) n O (1) and outputs the following: If the input is a No -instance thenthe algorithm outputs No . Otherwise the algorithm outputs Yes with probability at least − e . Proof.
Let (
G, k, d ) be the input instance. Our algorithm A has two steps. In the firststep of A we compute a collection of matchings greedily such that they are far apart usingLemma 23. Towards that first we run an algorithm to compute a maximum matching in G and let M be the output. If M is not a perfect matching we output No and stop. Nextwe iteratively apply Lemma 23 to compute a collection of perfect matchings that are farapart. Formally, at the beginning of step i , where ≤ ≤ i < k , we have perfect matchings M , . . . , M i such that | M j \ M j | ≥ k − i d for any two distinct j, j ∈ { , . . . , i } . Now, weapply Lemma 23 with r = i and s = 2 k − i − d and it will either output a matching M i +1 suchthat | M i +1 \ M j | ≥ k − i − d for all j ∈ { , . . . , i } , or not. If no such matching exists, thenthe first step of the algorithm A is complete. So at the end of the first step of the algorithm A , we have perfect matchings M , . . . , M q , where q ∈ { , . . . , k } such that (i) for any two distinct integers i, j ∈ { , . . . , q } , | M i \ M j | ≥ k − q d , and (ii) if q = k , then for any other perfect matching M / ∈ { M , . . . , M q } , | M \ M j | ≤ k − q − d .If q = k , then { M , . . . , M k } is a solution to the instance ( G, k, d ), and hence our algorithm A outputs Yes. Now on, we assume that q ∈ { , . . . , k − } . Statements ( i ) and ( ii ), andObservation 4 imply that (iii) for any two distinct integers i, j ∈ { , . . . , q } , | M i M j | ≥ k − q +1 d , and (iv) for any perfect matching M / ∈ { M , . . . , M q } , | M M j | < k − q d .Statements ( ii ) and ( iv ), and Observation 3 imply the following claim. (cid:66) Claim 28.
For any perfect matching M , there exists a unique i ∈ { , . . . , q } such that | M M i | < k − q d .Let M = { M ? , . . . , M ?k } is a solution to the instance ( G, k, d ). Then, by Claim 28, thereis a partition of M into M ] . . . ] M q (with some blocks possibly being empty) such thatfor each i ∈ { , . . . , q } , and each M ∈ M i , | M M i | ≤ k − q d . Thus, in the second stepof our algorithm A , we guess r = |M | , . . . , r q = |M q | and apply Lemma 25. That is,for each i ∈ { , . . . , q } such that r i = 0, we apply Lemma 25 with M = M i , r = r i , and s = 2 k − q d . Then for each i ∈ , . . . , q , let the output of Lemma 25 be N i, , . . . , N r i . Clearly | N i,j N i,j | ≥ d for any two distinct j, j ∈ { , . . . , r i } . Observation 3 and statement ( iii ) implies that for any two distinct i, j ∈ { , . . . , q } , the cardinality of the symmetric differencebetween a matching in { N i, , . . . , N i,r i } and a matching in { N j, ,...,N j,rj } is at least d .If algorithm A computes a solution in any of the guesses for r , . . . , r d , then we outputYes. Otherwise we output No. As the number of choices for r , . . . r k is upper boundedby k O ( k ) , from Lemmas 23 and 25 we get that the running time of A is 2 O ( kd ) n O (1) andthe success probability is at least 2 − ckd for some constant c . To get success probability1 − /e , we do 2 ckd many executions of A and output Yes if we succeed in at least oneof the iterations and output No otherwise. Thus, running time of the overall algorithm is2 O ( kd ) n O (1) . (cid:74) We took up weighted diverse variants of two classical matroid problems and the unweighteddiverse variant of a classical graph problem. We showed that the two diverse matroidproblems are NP -hard, and that the diverse graph problem cannot be solved in polynomialtime even for the smallest sensible measure of diversity. We then showed that all threeproblems are FPT with the combined parameter ( k, d ) where k is the number of solutionsand d is the diversity measure.We conclude with a list of open questions:We showed that the unweighted, counting variant of Weighted Diverse Bases doesnot have a polynomial-time algorithm unless P = NP (Theorem 7). This is the case whenall the weights are 1 and d = 1 or d = 2. Both the weighted and unweighted variants canbe solved in polynomial time when k = 1 (the greedy algorithm) and k = 2 ((weighted)matroid intersection). What happens for larger, constant values of d and/or k ? Till whatvalues of d, k does the problem remain solvable in polynomial time? These questions areinteresting also for special types of matroids. For instance, is there a polynomial-timealgorithm that checks if an input graph has three spanning trees whose edge sets havepairwise symmetric difference at least d , or is this already NP -hard?A potentially easier question along the same vein would be: we know from Theorem 7that Weighted Diverse Bases is unlikely to have an
FPT algorithm parameterized by d alone. Is Weighted Diverse Bases
FPT parameterized by k alone?Unlike for the other two problems, we don’t have hardness results for Weighted DiverseCommon Independent Sets for small values of k or d . Is Weighted Diverse CommonIndependent Sets
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