(In)approximability of Maximum Minimal FVS
Louis Dublois, Tesshu Hanaka, Mehdi Khosravian Ghadikolaei, Michael Lampis, Nikolaos Melissinos
((In)approximability of Maximum Minimal FVS
Louis Dublois
Université Paris-Dauphine, PSL Research University, CNRS, UMR 7243, LAMSADE, Paris, [email protected]
Tesshu Hanaka
Chuo University, Tokyo, [email protected]
Mehdi Khosravian Ghadikolaei
Université Paris-Dauphine, PSL Research University, CNRS, UMR 7243, LAMSADE, Paris, [email protected]
Michael Lampis
Université Paris-Dauphine, PSL Research University, CNRS, UMR 7243, LAMSADE, Paris, [email protected]
Nikolaos Melissinos
Université Paris-Dauphine, PSL Research University, CNRS, UMR 7243, LAMSADE, Paris, [email protected]
Abstract
We study the approximability of the NP-complete
Maximum Minimal Feedback VertexSet problem. Informally, this natural problem seems to lie in an intermediate space between twomore well-studied problems of this type:
Maximum Minimal Vertex Cover , for which the bestachievable approximation ratio is √ n , and Upper Dominating Set , which does not admit any n − (cid:15) approximation. We confirm and quantify this intuition by showing the first non-trivial polynomialtime approximation for Max Min FVS with a ratio of O ( n / ), as well as a matching hardness ofapproximation bound of n / − (cid:15) , improving the previous known hardness of n / − (cid:15) . Along the way,we also obtain an O (∆)-approximation and show that this is asymptotically best possible, and weimprove the bound for which the problem is NP-hard from ∆ ≥ ≥ r , produces an r -approximate solution in time n O ( n/r / ) . Thistime-approximation trade-off is essentially tight: we show that under the ETH, for any ratio r and (cid:15) >
0, no algorithm can r -approximate this problem in time n O (( n/r / ) − (cid:15) ) , hence we preciselycharacterize the approximability of the problem for the whole spectrum between polynomial andsub-exponential time, up to an arbitrarily small constant in the second exponent. Mathematics of computing → Graph algorithms; Theory of Com-putation → Design and Analysis of Algorithms → Approximation algorithms analysis
Keywords and phrases
Approximation Algorithms, ETH, Inapproximability
Digital Object Identifier
Acknowledgements
This work is partially supported by PRC CNRS JSPS project PARAGA andby JSPS KAKENHI Grant Number JP19K21537.
In a graph G = ( V, E ), a set S ⊆ V is called a feedback vertex set (fvs for short) if thesubgraph induced by V \ S is a forest. Typically, fvs is studied with a minimization objective:given a graph we are interested in finding the best (that is, smallest) fvs. In this paper weare interested in an objective which is, in a sense, the inverse: we seek an fvs S which is as large as possible, while still being minimal. We call this problem Max Min FVS . © Louis Dublois, Tesshu Hanaka, Mehdi Khosravian Ghadikolaei, Michael Lampis, and NikolaosMelissinos;licensed under Creative Commons License CC-BY42nd Conference on Very Important Topics (CVIT 2016).Editors: John Q. Open and Joan R. Access; Article No. 23; pp. 23:1–23:21Leibniz International Proceedings in InformaticsSchloss Dagstuhl – Leibniz-Zentrum für Informatik, Dagstuhl Publishing, Germany a r X i v : . [ c s . CC ] S e p MaxMin and MinMax versions of many famous optimization problems have recentlyattracted much interest in the literature (we give references below) and
Max Min FVS canbe seen as a member of this framework. Although the initial motivation for studying suchproblems was a desire to analyze the worst possible performance of a naive heuristic, theseproblems have gradually been revealed to possess a rich combinatorial structure that makesthem interesting in their own right. Our goal in this paper is to show that
Max Min FVS displays an interesting complexity behavior with respect to its approximability.Our motivation for focusing on
Max Min FVS is the contrast between two of its morewell-studied cousins: the
Max Min Vertex Cover and
Upper Dominating Set problems,where the objective is to find the largest minimal vertex cover or dominating set respectively.At first glance, one would expect
Max Min VC to be the easier of these two problems:both problems can be seen as trying to find the largest minimal hitting set of a hypergraph,but in the case of
Max Min VC the hypergraph has a very restricted structure, while in
UDS the hypergraph is essentially arbitrary. This intuition turns out to be correct: while
UDS admits no n − (cid:15) -approximation [5], Max Min VC admits a √ n -approximation (but no n / − (cid:15) -approximation) [9].This background leads us to the natural question of the approximability of Max MinFVS . On an intuitive level, one may be tempted to think that this problem should be harderthan
Max Min VC , since hitting cycles is more complex than hitting edges, but easierthan
UDS , since hitting cycles still offers us more structure than an arbitrary hypergraph.However, to the best of our knowledge, no n − (cid:15) -approximation algorithm is currently knownfor Max Min FVS (so the problem could be as hard as
UDS ), and the best hardness ofapproximation bound known is n / − (cid:15) [36] (so the problem could be as easy as Max MinVC ).Our main contribution in this paper is to fully answer this question, confirming andprecisely quantifying the intuition that
Max Min FVS is a problem that lies “between”
Max Min VC and
UDS : We give a polynomial-time approximation algorithm with ratio O ( n / ) and a hardness of approximation reduction which shows that (unless P = NP) nopolynomial-time algorithm can obtain a ratio of n / − (cid:15) , for any (cid:15) >
0. This completelysettles the approximability of the problem in polynomial time. Along the way, we also givean approximation algorithm with ratio O (∆), show that no algorithm can achieve ratio ∆ − (cid:15) ,for any (cid:15) >
0, and improve the best known NP-completeness proof for
Max Min FVS from∆ ≥ ≥
6, where ∆ is the maximum degree of the input graph.One interesting aspect of our results is that they have an interpretation from extremalcombinatorics which nicely mirrors the situation for
Max Min VC . Recall that a corollaryof the √ n -approximation for Max Min VC [9] is that any graph without isolated verticeshas a minimal vertex cover of size at least √ n , and this is tight (see Remark 13). Hence,the algorithm only needs to trivially preprocess the graph (deleting isolated vertices) andthen find this set, which is guaranteed to exist. Our algorithms can be seen in a similarlight: we prove that if one applies two almost trivial pre-processing rules to a graph (deletingleaves and contracting edges between degree-two vertices), a minimal fvs of size at least n / (and Ω( n/ ∆)) is always guaranteed to exist, and this is tight (Corollary 11 and Remark 12).Thus, the approximation ratio of n / is automatically guaranteed for any graph where weexhaustively apply these very simple rules and our algorithms only have to work to constructthe promised set. This makes it somewhat remarkable that the ratio of n / turns out to bebest possible.Having settled the approximability of Max Min FVS in polynomial time, we consider thequestion of how much time needs to be invested if one wishes to guarantee an approximation . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:3 ratio of r (which may depend on n ) where r < n / . This type of time-approximationtrade-off was extensively studied by Bonnet et al. [8], who showed that Max Min VertexCover admits an r -approximation in time 2 O ( n/r ) and this is optimal under the randomizedETH.For Max Min FVS we cannot hope to obtain a trade-off with performance exponential in n/r , as this implies a polynomial-time √ n -approximation. It therefore seems more naturalto aim for a running time exponential in n/r / . Indeed, generalizing our polynomial-timeapproximation algorithm, we show that we can achieve an r -approximation in time n O ( n/r / ) .Although this algorithm reuses some ingredients from our polynomial-time approximation, itis significantly more involved, as it is no longer sufficient to compare the size of our solutionto n . We complement our result with a lower bound showing that our algorithm is essentiallybest possible under the randomized ETH for any r (not just for polynomial time), or moreprecisely that the exponent of the running time of our algorithm can only be improved by n o (1) factors. Related work
To the best of our knowledge,
Max Min FVS was first considered by Mishraand Sikdar [36], who showed that the problem does not admit an n / − (cid:15) approximation(unless P = NP), and that it remains APX-hard for ∆ ≥
9. On the other hand,
UDS and
Max Min VC are well-studied problems, both in the context of approximation andin the context of parameterized complexity [1, 5, 9, 11, 13, 14, 19, 28, 33, 38, 40]. Manyother classical optimization problems have recently been studied in the MaxMin or MinMaxframework, such as
Max Min Separator [25],
Max Min Cut [21],
Min Max Knapsack (also known as the
Lazy Bureaucrat Problem ) [3, 23, 24], and some variants of
Max MinEdge Cover [35, 26]. Some problems in this area also arise naturally in other forms and havebeen extensively studied, such as
Min Max Matching (also known as
Edge DominatingSet [32]),
Grundy Coloring , which can be seen as a Max Min version of
Coloring [2, 6],and
Max Min VC in hypergraphs, which is known as
Upper Transversal [37, 29, 30, 31].The idea of designing super-polynomial time approximation algorithms which obtainguarantees better than those possible in polynomial time has attracted much attention inthe last decade [4, 10, 16, 18, 20, 22, 34]. As mentioned, the result closest to the time-approximation trade-off we give in this paper is the approximation algorithm for
MaxMin VC given by Bonnet et al. [8]. It is important to note that such trade-offs are onlygenerally known to be tight up to poly-logarithmic factors in the exponent of the runningtime. As explained in [8], current lower bound techniques can rule out improvements inthe running time that shave at least n (cid:15) from the exponent, but not improvements whichshave poly-logarithmic factors, due to the state of the art in quasi-linear PCP constructions.Indeed, such improvements are sometimes possible [4] and are conceivable for Max MinVC and
Max Min FVS . Lower bounds for this type of algorithm rely on the (randomized)Exponential Time Hypothesis (ETH), which states that there is no (randomized) algorithmfor running in time 2 o ( n ) . We use standard graph-theoretic notation and only consider simple loop-less graphs. For agraph G = ( V, E ) and S ⊆ V we denote by G [ S ] the graph induced by S . For u ∈ V , G − u isthe graph G [ V \ { u } ]. We write N ( u ) to denote the set of neighbors of u and d ( u ) = | N ( u ) | to denote its degree. For S ⊆ V , N ( S ) = ∪ u ∈ S N ( u ) \ S . We use ∆( G ) (or simply ∆) todenote the maximum degree of G . For uv ∈ E the graph G/uv is the graph obtained by
C V I T 2 0 1 6 contracting the edge uv , that is, replacing u, v by a new vertex connected to N ( u ) ∪ N ( v ).In this paper we will only apply this operation when N ( u ) ∩ N ( v ) = ∅ , so the result willalways be a simple graph.A forest is a graph that does not contain cycles. A feedback vertex set (fvs for short) is aset S ⊆ V such that G [ V \ S ] is a forest. An fvs S is minimal if no proper subset of S isan fvs. It is not hard to see that if S is minimal, then every u ∈ S has a private cycle , thatis, there exists a cycle in G [( V \ S ) ∪ { u } ], which goes through u . A vertex u of a feedbackvertex set S that does not have a private cycle (that is, S \ { u } is also an fvs), is called redundant . For a given fvs S , we call the set F := V \ S the corresponding induced forest. If S is minimal, then F is maximal.The main problem we are interested in is Max Min FVS : given a graph G = ( V, E ), finda minimal fvs of G of maximum size. Since this problem is NP-hard, we will be interested inapproximation algorithms. An approximation algorithm with ratio r ≥ n , the order of the graph) is an algorithm which, given a graph G , returns a solution ofsize at least mmfvs( G ) r , where mmfvs( G ) is the size of the largest minimal fvs of G .We make two basic observations about our problem: deleting vertices or contractingedges can only decrease the size of the optimal solution. (cid:73) Lemma 1.
Let G = ( V, E ) be a graph and u ∈ V . Then, mmfvs( G ) ≥ mmfvs( G − u ) .Furthermore, given any minimal feedback vertex set S of G − u , it is possible to construct inpolynomial time a minimal feedback vertex set of G of the same or larger size. Proof.
Let S be a minimal fvs of G − u . We observe that S ∪ { u } is an fvs of G . If S ∪ { u } is minimal, we are done. If not, we delete vertices from it until it becomes minimal. Wenow note that the only vertex which may be deleted in this process is u , since all vertices of S have a private cycle in G − u (that is, a cycle not intersected by any other vertex of S ).Hence, the resulting set is a superset of S . (cid:74)(cid:73) Lemma 2.
Let G = ( V, E ) be a graph, u, v ∈ V with N ( u ) ∩ N ( v ) = ∅ and uv ∈ E . Then mmfvs( G ) ≥ mmfvs( G/uv ) . Furthermore, given any minimal feedback vertex set S of G/uv ,it is possible to construct in polynomial time a minimal feedback vertex set of G of the sameor larger size. Proof.
Before we prove the Lemma we note that the contraction operation, under thecondition that N ( u ) ∩ N ( v ) = ∅ , preserves acyclicity in a strong sense: G is acyclic if andonly if G/uv is acyclic. Indeed, if we contract an edge that is part of a cycle, this cycle musthave length at least 4, and will therefore give a cycle in
G/uv . Of course, contractions nevercreate cycles in acyclic graphs.Let G := G/uv , w be the vertex of G which has replaced u, v , V = V ( G ), and S be aminimal fvs of G . We have two cases: w ∈ S or w S .In case w ∈ S , we start with the set S = ( S \ { w } ) ∪ { u, v } . It is not hard to see that S is an fvs of G . Furthermore, no vertex of S \ { u, v } is redundant: for all z ∈ S \ { w } ,there is a cycle in G [( V \ S ) ∪ { z } ], therefore there is also a cycle in G [( V \ S ) ∪ { z } ].Furthermore, we claim that S \ { u, v } is not a valid fvs. Indeed, there must be a cyclecontained (due to minimality) in G = G [( V \ S ) ∪ { w } ]. Therefore, if there is no cycle in G = G [( V \ S ) ∪ { u, v } ], we get a contradiction, as G can be obtained by G by contractingthe edge uv and contracting edges preserves acyclicity. We conclude that even if S is notminimal, if we remove vertices until it becomes minimal, we will remove at most one vertex,so the size of the fvs obtained is at least | S | .In case w S , we will return the same set S . Let F = V \ S, F = V \ S . By definition, G [ F ] is acyclic. To see that G [ F ] is also a forest, we note that G [ F ] is obtained from G [ F ] . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:5 by contracting uv , and as we noted in the beginning, the contractions we use strongly preserveacyclicity. To see that S is minimal, take z ∈ S and consider the graphs G = G [( V \ S ) ∪ { z } ]and G = G [( V \ S ) ∪ { z } ]. We see that G can be obtained from G by contracting uv . But G must have a cycle, by the minimality of S , so G also has a cycle. Thus, S is minimal in G . (cid:74) In this section we present a polynomial-time algorithm which guarantees an approximationratio of n / . As we show in Theorem 17, this ratio is the best that can be hoped for inpolynomial time. Later (Theorem 15) we show how to generalize the ideas presented hereto obtain an algorithm that achieves a trade-off between the approximation ratio and the(sub-exponential) running time, and show that this trade-off is essentially optimal.On a high level, our algorithm proceeds as follows: first we identify some easy cases inwhich applying Lemma 1 or Lemma 2 is safe , that is, the value of the optimal is guaranteedto stay constant, namely deleting vertices of degree at most 1, and contracting edges betweenvertices of degree 2. After we apply these reduction rules exhaustively, we compute a minimalfvs S in an arbitrary way. If S is large enough (larger than n / ), we simply return this set.If not, we apply some counting arguments to show that a vertex u ∈ S with high degree( ≥ n / ) must exist. We then have two cases: either we are able to construct a large minimalfvs just by looking at the neighborhood of u in the forest (and ignoring S \ { u } ), or u mustshare many neighbors with another vertex v ∈ S , in which case we construct a large minimalfvs in the common neighborhood of u, v .Because our algorithm is constructive (and runs in polynomial time), we find it interestingto remark an interpretation from the point of view of extremal combinatorics, given inCorollary 11. We begin by showing two safe versions of Lemmas 1, 2. (cid:73)
Lemma 3.
Let
G, u be as in Lemma 1 with d ( u ) ≤ . Then mmfvs( G − u ) = mmfvs( G ) . Proof.
We only need to show that mmfvs( G ) ≤ mmfvs( G − u ) (the other direction is givenby Lemma 1). Let S be a minimal fvs of G . Then, S is an fvs of G − u . Furthermore, u S ,as S is minimal in G . To see that S is also minimal in G − u , note that any cycle of G alsoexists in G − u (as no cycle contains u ). (cid:74)(cid:73) Lemma 4.
Let
G, u, v be as in Lemma 2 with d ( u ) = d ( v ) = 2 . Then mmfvs( G/uv ) =mmfvs( G ) . Proof.
Let G = G/uv , w be the vertex that replaced u, v in G , and V = V ( G ).We only need to show that mmfvs( G ) ≤ mmfvs( G ), as the other direction is given byLemma 2. Let S be a minimal fvs of G . We consider two cases:If u, v S , then we claim that S is also a minimal fvs of G . Indeed, G [ V \ S ] isobtained from G [ V \ S ] by contracting uv , so both are acyclic. Furthermore, for all z ∈ S , G [( V \ S ) ∪ { z } ] is obtained from G [( V \ S ) ∪ { z } ] by contracting uv , therefore both have acycle, hence no vertex of S is redundant in G .If { u, v } ∩ S = ∅ , we claim that exactly one of u, v is in S . Indeed, if u, v ∈ S , then G [( V \ S ) ∪ { u } ] does not contain a cycle going through u , as u has degree 1 in this graph. C V I T 2 0 1 6
Without loss of generality, let u ∈ S , v S . We set S := ( S \ { u } ) ∪ { w } and claim that S is a minimal fvs of G . It is not hard to see that S is an fvs of G , since it correspondsto deleting S ∪ { v } from G . To see that it is minimal, for all z ∈ S \ { w } we observe that G [( V \ S ) ∪{ z } ] obtained from G [( V \ S ) ∪{ z } ] by deleting v , which has degree 1. Therefore,this deletion strongly preserves acyclicity. Finally, to see that w is not redundant for S weobserve that G [( V \ S ) ∪ { u } ] has a cycle, and this cycle must be present in G [( V \ S ) ∪ { w } ],which is obtained from the former graph by contracting uv . (cid:74)(cid:73) Definition 5.
For a graph G = ( V, E ) we say that G is reduced if it is not possible to applyLemma 3 or Lemma 4 to G . We now present a counting argument which will useful in our algorithm and states,roughly, that if in a reduced graph we find a (not necessarily minimal) fvs, that fvs musthave many neighbors in the corresponding forest. (cid:73)
Lemma 6.
Let G = ( V, E ) be a reduced graph and S ⊆ V a feedback vertex set of G . Let F = V \ S . Then, | N ( S ) ∩ F | ≥ | F | . Proof.
Let n be the number of leaves of F , n the number of vertices of F with at least threeneighbors in F , n a the number of vertices of F with two neighbors in F and at least oneneighbor in S , and n b the number of remaining vertices of F . We have n + n a + n b + n = | F | .Furthermore, n ≤ n because the average degree of any forest is less than 2.We observe that all leaves of the tree have a neighbor in S (otherwise we would haveapplied Lemma 3). This gives | N ( S ) ∩ F | ≥ n + n a .Furthermore, none of the n b vertices which have degree two in the tree and no neighborsin S can be connected to each other, since then Lemma 4 would apply. Therefore, n b ≤ n + n a + n . Indeed, if n b > n + n a + n , then n b > | F | / n b verticesform an independent set, we would have | E ( F ) | ≥ n b > | F | , contradicting the assumptionthat F is a forest.Putting things together we get | F | = n + n a + n b + n ≤ n +2 n a +2 n ≤ n +2 n a ≤ | N ( S ) ∩ F | . (cid:74) We note that Lemma 6 immediately gives an approximation algorithm with ratio O (∆). (cid:73) Lemma 7.
In a reduced graph G with n vertices and maximum degree ∆ , every feedbackvertex set has size at least n . Proof.
Let S be a feedback vertex set of G and F the corresponding forest. If | S | < n then | N ( S ) ∩ F | < n so by Lemma 6 we have | F | < n . But then | V | = | S | + | F | < n , which is acontradiction. (cid:74)(cid:73) Remark 8.
Lemma 6 is tight.
Proof.
Take two copies of a rooted binary tree with n leaves and connect their roots. Theresulting tree has 2 n leaves and 2 n − u, v connected to every leaf. In the resulting graph S = { u, v } is an fvs.The corresponding forest has 8 n − n are connected to S . The graph isreduced. (cid:74) . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:7 u v v v v v v a u v w v w b u v w v w c Figure 1 ( a ) vertex u is a minimal fvs of the given graph and has 4 neighbors in G [ F ]. ( b ) acontracted form of G [ F ] with 4 vertices. ( c ) a new minimal fvs of the result graph of size 3. We begin with a final intermediate lemma that allows us to construct a large minimal fvs inany reduced graph that is a forest plus one vertex. (cid:73)
Lemma 9.
Let G = ( V, E ) be a reduced graph and u ∈ V such that G − u is acyclic.Then it is possible to construct in polynomial time a minimal feedback vertex set S of G with | S | ≥ d ( u ) / . Proof.
Let F = V \ { u } . Since the graph is reduced, all trees of G [ F ] contain at least twoneighbors of u . Indeed, since each tree T of G [ F ] has at least two leaves, both of them mustbe neighbors of u (otherwise Lemma 3 applies).As long as there exist v, w ∈ F with vw ∈ E and { v, w } 6⊆ N ( u ) we contract the edge vw .Note that we can apply Lemma 2 as v, w have no common neighbors ( u is not a commonneighbor by assumption, and they cannot have a common neighbor in the forest withoutforming a cycle). Furthermore, this operation does not change d ( u ). Therefore, it will besufficient to construct a minimal fvs in the resulting graph after applying this operationexhaustively.Suppose now that we have applied the contraction operation described above exhaustively.We eventually arrive at a graph where u is connected to all vertices of F , as all trees of F initially contain some neighbors of u and, after repeated contractions, all non-neighbors of u are absorbed into its neighbors (more precisely, each contraction decreases | F \ N ( u ) | ).Therefore, we arrive at a graph with d ( u ) = | F | . Furthermore, every component of F containsstrictly more than one vertex.Now, since G [ F ] is bipartite, there is a bipartition F = L ∪ R . Without loss of generality | L | ≤ | R | . We return the solution S = R . First, S does have the promised size, as | S | ≥ | F | / d ( u ) /
2. Second, S is an fvs, as L is an independent set, so L ∪ { u } induces astar. Finally, S is minimal, because all v ∈ S are connected to u , and also have at least oneneighbor w ∈ L , with w also connected to u . An illustration of the process is presented inFigure 1. (cid:74)(cid:73) Theorem 10.
There is a polynomial time approximation algorithm for
Max Min FVS with ratio O ( n / ) . C V I T 2 0 1 6
Proof.
We are given a graph G = ( V, E ). We begin by applying Lemmas 3,4 exhaustively inorder to obtain a reduced graph G = ( V , E ). Clearly, if we obtain a | V | / approximationin G , since the reductions we applied do not change the optimal, and we can construct asolution of the same size in G , we get a | V | / ≤ | V | / approximation ratio in G . So, inthe remainder, to ease presentation, we assume G is already reduced and has n vertices.Our algorithm begins with an arbitrary minimal fvs S . This can be constructed, forexample, by starting with S = V and removing vertices from S until it becomes minimal. If | S | ≥ n / then we return S . Since the optimal solution cannot have size more than n , weachieve the claimed ratio.Suppose then that | S | < n / . Let F be the corresponding forest. We have | F | >n − n / > n/ n . By Lemma 6, | N ( S ) ∩ F | ≥ n/
8. Since | S | < n / there must exist u ∈ S such that u has at least n / neighbors in F .Let w ∈ F ∩ N ( u ). We say that w is a good neighbor of u if there exists w ∈ F ∩ N ( u )with w = w and w is in the same tree of G [ F ] as w . Otherwise w is a bad neighbor of u .By extension, a tree of G [ F ] that contains a good (resp. bad) neighbor of u will be called agood (resp. bad) tree. Every vertex of N ( u ) ∩ F is either good or bad.We have argued that | N ( u ) ∩ F | ≥ n / . We distinguish two cases: either u has at least n / good neighbors in F , or it has at least that many bad neighbors in F .In the former case, we delete from the graph the set S \ { u } and apply Lemmas 3, 4exhaustively again. We claim that the number of good neighbors of u does not decrease inthis process. Indeed, two good neighbors of u cannot be contracted using Lemma 4, sincethey have a common neighbor (namely u ). Furthermore, suppose w is the first good neighborof u to be deleted using Lemma 3. This would mean that w currently has no other neighborexcept u . However, since w is good, initially there was a w ∈ N ( u ) in the same tree of G [ F ]as w . The vertex w has not been deleted (since we assumed w is the first good neighborto be deleted). Furthermore, Lemmas 3, 4 cannot disconnect two vertices which are in thesame component, so we get a contradiction. We therefore have a reduced graph, where { u } is an fvs, and d ( u ) ≥ n / . By Lemma 9 we obtain a minimal fvs of size at least n / , whichis an O ( n / ) approximation.In the latter case, u has at least n / bad neighbors in F . Consider a bad tree T . Weclaim that T must have a neighbor in S \ { u } , because T has at least two leaves, at mostone of which is a neighbor of u (since T is bad). If the second leaf is not connected to S , itwill be deleted by Lemma 3. Furthermore, since u is connected to one vertex in each badtree, u is connected to at least n / bad trees.We now find the vertex v ∈ S \ { u } such that v is connected to the maximum number ofbad trees connected to u . Since | S | ≤ n / , v must be connected to at least n / bad treesconnected to u . We now delete from the graph the set S \ { u, v } as well as all trees of G [ F ],except the bad trees connected to u, v . Furthermore, in each bad tree T connected to both u, v let u ∈ T ∩ N ( u ) and v ∈ T ∩ N ( v ) such that u , v are as close as possible in T (notethat perhaps v = u ). We delete all vertices of the tree T except those on the path from v to u . Then, we contract all internal edges of this path (note that internal vertices of thepath are not connected to { u, v } by the selection of u , v ). It is not hard to verify that, byusing Lemmas 1, 2, if we are able to produce a large minimal fvs in the resulting graph, weobtain a solution for G . Furthermore, in the resulting graph, every bad tree T connected to u, v has been reduced to a single vertex connected to u, v . So the graph is now either a K ,s ,with s ≥ n / , or the same graph with the addition of the edge uv . In either case, it is nothard to see that starting with the fvs that contains all vertices except { u, v } , and making itminimal, we obtain a solution of size at least s − . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:9 O ( n / ). (cid:74)(cid:73) Corollary 11.
For any reduced graph G on n vertices we have mmfvs( G ) = Ω( n / ) . Proof.
We simply note that the algorithm of Theorem 10 always constructs a solution ofsize at least n / c , where c is a small constant, assuming that the original n -vertex graph G was reduced. (cid:74)(cid:73) Remark 12.
Corollary 11 is tight.
Proof.
Take a K n and for every pair of vertices u, v in the clique, add 2 n new verticesconnected only to u, v . The graph has order n + 2 n (cid:0) n (cid:1) = n + n ( n −
1) = n − n + n ≥ n / n − n − n ≤ n . We have mmfvs( G ) | V ( G ) | ≤ nn = O ( n ) thereforemmfvs( G ) = O ( | V ( G ) | / ). (cid:74) Finally, we remark that a similar combinatorial point of view can be taken for therelated problem of
Max Min VC , giving another intuitive explanation for the difference inapproximability between the two problems. (cid:73)
Remark 13.
Any graph G = ( V, E ) without isolated vertices, has a minimal vertex cover ofsize at least p | V | , and this is asymptotically tight. Proof.
We will prove the statement under the assumption that G is connected. If not, wecan treat each component separately. If the components of G have sizes n , . . . , n k , thenwe rely on the fact that P ki =1 √ n i ≥ qP ki =1 n i and that the union of the minimal vertexcovers of each component is a minimal vertex cover of G .If G = ( V, E ) has a vertex u of degree at least √ n , then we begin with the vertex cover V \ { u } and remove vertices until it becomes minimal. In the end, our solution contains asuperset of N ( u ), therefore we have a minimal vertex cover of size at least √ n as promised.If, on the other hand, ∆( G ) < √ n , then any vertex cover of G must have size at least √ n .Indeed, a vertex cover of size at most √ n − √ n − √ n < n − G is connected we have | E ( G ) | ≥ n −
1. So, in this case, any minimal vertex coverhas the promised size.To see that the bound given is tight, take a K n and attach n leaves to each of its vertices.This graph has n + n vertices, but any minimal vertex cover has size at most 2 n . (cid:74) In this section we give an approximation algorithm that generalizes our n / -approximationand is able to guarantee any desired performance, at the cost of increased running time. Ona high level, our initial approach again constructs an arbitrary minimal fvs S and if S isclearly large enough, returns it. However, things become more complicated from then on,as it is no longer sufficient to consider vertices of S individually or in pairs. We thereforeneed several new ideas, one of which is given in the following lemma, which states that wecan find a constant factor approximation in time exponential in the size of a given fvs. Thiswill be useful as we will use the assumption that S is “small” and then cut it up into evensmaller pieces to allow us to use Lemma 14. (cid:73) Lemma 14.
Given a graph G = ( V, E ) on n vertices and a feedback vertex set S ⊆ V ofsize k , it is possible to produce a minimal fvs S of G of size | S | ≥ mmfvs( G )3 in time n O ( k ) . C V I T 2 0 1 6
Proof.
Before we begin, let us point out that for k = 1, Max Min FVS can be solvedoptimally in time O ( n ), using standard arguments from parameterized complexity, namelythe fact that in this case G has treewidth 2, and invoking Courcelle’s theorem, since theproperties “ S is an fvs” and “ S is minimal” are MSO-expressible [15]. Unfortunately, thistype of argument is not good enough for larger values of k , as the running time guaranteedby Courcelle’s theorem could depend super-exponentially on k . We could try to avoid this byformulating a treewidth-based DP algorithm to obtain a better running time, but we preferto give a simpler more direct branching algorithm, since this is good enough for Theorem 15.We will assume that S is minimal (if not, we can remove vertices from it to make itminimal and this only decreases the available running time of our algorithm). As a result,we assume that mmfvs( G ) ≥ k , as otherwise S is already a 3-approximation.Let S OP T be a maximum minimal fvs of G , and F OP T = V \ S OP T . We formulatean algorithm that maintains two disjoint sets of vertices S SOL , F
SOL which, intuitively,correspond to vertices we have decided to place in the fvs or the induced forest, respectively.We will denote U := V \ ( S SOL ∪ F SOL ) the set of undecided vertices. Our algorithm will benon-deterministic, that is, it will sometimes “guess” some vertices of U that will be placed in S SOL or F SOL . We will bound the total number of guessing possibilities by n O ( k ) , whichwill imply that the algorithm can be made deterministic by trying all possibilities for everyguess and returning the best returned solution.Throughout the algorithm, we will work to maintain the following invariants: S SOL ∪ F SOL is an fvs of G . S SOL ⊆ S OP T and F SOL ⊆ F OP T . G [ F SOL ] is acyclic and has at most 2 k components. All vertices of S SOL have at least two neighbors in F SOL .To begin, we guess a set F ⊆ S such that G [ F ] is acyclic and set F SOL = F and S SOL = S \ F . Property 1 is satisfied as F SOL ∪ S SOL = S . Property 2 is satisfied for theguess F = F OP T ∩ S . If there exists u ∈ S SOL which does not satisfy Property 4, we guessone or two vertices from N ( u ) ∩ U and place them into F SOL so that u has two neighbors in F SOL . Since u has a private cycle in G [ F OP T ], if the vertices we guessed are the neighbors of u in that cycle, we maintain Property 2. We continue in this way until Property 4 is satisfied.We now observe that F SOL is acyclic (as F SOL ⊆ F OP T ), and that since we have added atmost two vertices for each vertex of S SOL , it contains at most 2 k vertices, hence at most2 k components, so we satisfied Property 3. So far, the total number of possible guesses isupper-bounded by 2 k n k : 2 k for guessing F and n k for guessing at most two neighbors foreach u ∈ S SOL .We will now say that a “connector” is a path P ⊆ F OP T \ F SOL , such that G [ F OP T ∪ P ]has strictly fewer components that G [ F SOL ]. Our algorithm will now repeateadly guess ifa connector exists, and if it does it will guess the first and last vertex u, v of P . Note that u, v ∈ U and if we guess u, v correctly we can infer all of P , as G [ U ] is acyclic, so there is atmost one path from u to v in G [ U ]. We set F SOL := F SOL ∪ P and continue guessing, untilwe guess that no connector exists. Observe that guessing the endpoints of a connector gives n possibilities, and that adding a connector to F SOL decreases the number of connectedcomponents of F SOL , which can happen at most 2 k times by Property 3. So we have a totalof n O ( k ) possible guesses and for the correct guess Property 2 is maintained.We now consider every vertex of u ∈ U that has at least two neighbors in F SOL and placeall such vertices in S SOL . Properties 1, 3, and 4 are trivially still satisfied. Furthermore,if our guesses so far are correct, all such vertices u belong in S OP T , as they either already . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:11 have a private cycle in F OP T , or if they have neighbors in distinct components of F SOL , theywould function as connectors in F OP T (and we assume we have correctly guessed that nomore connectors exist).We are now in a situation where every vertex of U has at most one neighbor in F SOL .We construct a new graph H by deleting from G all of S SOL and replacing F SOL by a singlevertex f that is connected to N ( F SOL ). Note that H is a simple graph (it has no paralleledges) with an fvs of size 1 (as H − w is acyclic). We therefore use the aforementionedalgorithm implied by Courcelle’s theorem to produce a maximum minimal fvs of H which,without loss of generality, does not contain w . Let S ∗ ⊆ U be this set. In G , we check if S SOL ∪ S ∗ is an fvs. If it is we delete vertices from it (if necessary) to make it redundantand return the resulting set S ∗∗ , which is a minimal fvs.To see that the resulting solution has the desired size we focus on the case where allguesses were correct and therefore Properties 1-4 were maintained throughout the executionof the algorithm. As mentioned, since the total number of possibilities considered in n O ( k ) , adeterministic algorithm can simply try out all possible choices and return the best solution.We first observe that mmfvs( H ) ≥ mmfvs( G ) − | S SOL | , where S SOL is the set of verticeswe deleted from G to obtain H . Indeed, S := S OP T \ S SOL is a minimal fvs of H . Tosee that S is an fvs, suppose that H contains a cycle after deleting S . This cycle mustnecessarily go through w . Let P be the vertices of this cycle except w . We have P ⊆ U \ S OP T therefore, P ⊆ F OP T . However, this means either that P forms a cycle with a component of F SOL (which contradicts the acyclicity of F OP T by Property 2), or that P is a connector,which contradicts our guess that no other connector exists. Therefore, S must be an fvsof H . To see that it is minimal we note that for all u ∈ S there exists a private cycle in G [ U ∪ F SOL ∪ { u } ], and this cycle is not destroyed by contracting the vertices of F SOL into w .We now have that | S ∗ ∪ S SOL | ≥ | S OP T | , because | S ∗ | ≥ | S OP T \ S SOL | . We argue thatin the process of making S ∗ minimal to obtain S ∗∗ we delete at most 2 k vertices. Indeed,every time a vertex u of S SOL is removed from S ∗ ∪ S SOL as redundant, since u has at leasttwo neighbors in F SOL by Property 4, the number of components of G [ F SOL ] must decrease.Similarly, if we remove a vertex u ∈ S ∗ as redundant, we consider the private cycle of u in H \ S ∗ . All of the vertices of this cycle are present in G after we delete S ∗ , except w ,therefore, this cycle forms a path between two distinct components of G [ F SOL ]. We concludethat, since removing a vertex from our fvs decreases the number of connected componentsof G [ F SOL ], by Property 3 we have | S ∗∗ | ≥ | S OP T | − k . But recall that we have assumedthat k ≤ S OPT (otherwise S was already a sufficiently good approximation), so we have | S ∗∗ | ≥ mmfvs( G )3 . (cid:74)(cid:73) Theorem 15.
There is an algorithm which, given an n -vertex graph G = ( V, E ) and avalue r , produces an r -approximation for Max Min FVS in G in time n O ( n/r / ) . Proof.
First, let us note that we may assume that r is ω (1), because if r is bounded by aconstant, then we can solve the problem exactly in the given time. To ease presentation, wewill give an algorithm with approximation ratio O ( r ). A ratio of exactly r can be obtainedby multiplying r with an appropriate (small) constant.Our algorithm borrows several of the basic ideas from Theorem 10, but requires somenew ingredients (including Lemma 14). The first step is, again, to construct a minimal fvs S in some arbitrary way, for example by setting S = V and then removing vertices from S untilit becomes minimal. If | S | ≥ n/r we are done, as we already have an r -approximation, so wesimply return S . From this point, this algorithm departs from the algorithm of Theorem10, because it is no longer sufficient to compare the size of the returned solution with a C V I T 2 0 1 6 function of n (we need to compare it to the actual optimal in order to obtain a ratio of r ),and because we need to partition S into non-trivial parts that contain more than one vertex.The algorithm proceeds as follows:Let k = d√ r e and partition S into k parts of (almost) equal size S , . . . , S k . Ouralgorithm proceeds as follows: for each i, j ∈ { , . . . , k } (not necessarily distinct) considerthe graph G i,j obtained by deleting all vertices of S \ ( S i ∪ S j ). Compute, using Lemma 14a solution for G i,j , taking into account that S i ∪ S j is a feedback vertex set of this graph.Output the largest of the solutions found, using Lemma 1 to transform them into solutionsof G (or output S if it is larger than all solutions).The algorithm clearly runs in the promised time: | S i ∪ S j | ≤ nrk , so the algorithm ofLemma 14 takes time n O ( n/r / ) and is executed a polynomial number of times.Let us now analyze the approximation ratio of the produced solution. Let S OP T bean optimal solution and let F := V \ S and F OP T = V \ S OP T be the induced forestscorresponding to S and to the optimal solution. We would like to argue that one of theconsidered subproblems contains at least a r fraction of S OP T and that most (though notall) of these vertices form part of a minimal fvs of that subgraph.To be more precise, we will define the notion of “type“ for each u ∈ S OP T ∩ F . For eachsuch u there must exist a cycle in the graph G [ F OP T ∪ { u } ] (if not, this would contradict theminimality of S OP T ). Call this cycle c ( u ) (select one such cycle arbitrarily if several exist).The cycle c ( u ) must intersect S , as S is an fvs. Let v be the vertex of c ( u ) ∩ S closest to u onthe cycle. Let v be the vertex of c ( u ) ∩ S that is closest to u if we traverse the cycle in theopposite direction (note that v, v are not necessarily distinct). Suppose that v ∈ S i , v ∈ S j and without loss of generality i ≤ j . We then say that u ∈ S OP T ∩ F has type ( i, j ). Inthis way, we define a type for each u ∈ S OP T ∩ F . Note that according to our definition, allinternal vertices of the path in c ( u ) from u to v (and also from u to v ) belong in F OP T ∩ F .According to the definition of the previous paragraph, there are k ( k + 1) / ≤ r possibletypes of vertices in S OP T ∩ F . Therefore, there must be a type ( i, j ) such that at least | S OPT ∩ F | r vertices have this type. We now concentrate on the graph G i,j , for the type ( i, j )which satisfies this condition. Our algorithm constructed G i,j by deleting all of S except S i ∪ S j . We would like to claim that this graph has a minimal feedback vertex set of sizecomparable to | S OPT ∩ F | r .For the sake of the analysis, construct a minimal feedback vertex set S ∗ of G i,j as follows:we begin with the fvs S ∗ = S OP T ∩ ( F ∪ S i ∪ S j ) and the corresponding induced forest F ∗ = F OP T ∩ ( F ∪ S i ∪ S j ). The set S ∗ is a feedback vertex set as it contains all vertices of S OP T found in G i,j and S OP T is a feasible feedback vertex set of all of G . We then make S ∗ minimal by arbitrarily removing redundant vertices. Call the resulting set S ∗∗ ⊆ S ∗ and thecorresponding induced forest F ∗∗ ⊇ F ∗ .Our main claim now is that the number of vertices of S ∗ ∩ F of type ( i, j ) which were“lost” in the process of making S ∗ minimal, is upper-bounded by | S i ∪ S j | . Formally, weclaim that |{ u ∈ ( S ∗ ∩ F ) \ S ∗∗ | u has type ( i, j ) }| ≤ | S i ∪ S j | . Indeed, consider such avertex u ∈ ( S ∗ ∩ F ) \ S ∗∗ of type ( i, j ), let c ( u ) be the cycle that defines its type and v, v the vertices of S i ∪ S j which are closest to u on the cycle in either direction. All vertices of c ( u ) in the paths from u to v and from u to v belong to F OP T ∩ F , therefore also to F ∗ .If u was removed as redundant, this means that v, v must have been in distinct connectedcomponents at the moment u was removed from the feedback vertex set (and also that v, v are distinct). However, the addition of u to the induced forest creates a path from v to v in the induced forest and hence decreases the number of connected components (that is,trees in the induced forest) containing vertices of S i ∪ S j . The number of such connected . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:13 components cannot decrease more than | S i ∪ S j | times, therefore, during the process ofmaking S ∗ minimal we may have removed at most | S i ∪ S j | vertices of type ( i, j ) from S ∗ ∩ F .Using the above analysis and the assumption that S ∗ contains at least | S OPT ∩ F | r verticesof type ( i, j ), we conclude that mmfvs( G i,j ) ≥ | S ∗∗ | ≥ | S OPT ∩ F | r − | S i ∪ S j | . We nownote that if | S OP T ∩ S | ≥ | S OPT | r , then S is already an r -approximation, so it is safe toassume | S OP T ∩ F | ≥ ( r − | S OPT | r . Furthermore, | S i ∪ S j | ≤ | S |√ r ≤ | S OPT | r √ r , where againwe are assuming that S is not already an r -approximation. Putting things together weget mmfvs( G i,j ) ≥ ( r − | S OPT | r − | S OPT | r √ r ≥ | S OPT | r , for sufficiently large r . Hence, sincethe algorithm will return a solution that is at least as large as mmfvs( G i,j )3 , we obtain an O ( r )-approximation. (cid:74) In this section we establish lower bound results showing that the approximation algorithmsgiven in Theorems 10 and 15 are essentially optimal, under standard complexity assumptions.
We begin by showing that the best approximation ratio achievable in polynomial time isindeed (essentially) n / . For this, we rely on the celebrated result of Håstad on the hardnessof approximating Max Independent Set , which was later derandomized by Zuckerman,cited below. (cid:73)
Theorem 16. [27, 39] For any (cid:15) > , there is no polynomial time algorithm whichapproximates Max Independent Set with a ratio of n − (cid:15) , unless P = NP.
Starting from this result, we present a reduction to
Max Min FVS . (cid:73) Theorem 17.
For any (cid:15) > , Max Min FVS is inapproximable within a factor of n / − (cid:15) unless P = NP.
Proof.
We give a gap-preserving reduction from
Max Independent Set , which cannot beapproximated within a factor of n − (cid:15) , unless P = NP. We are given a graph G = ( V, E ) on n vertices as an instance of Max Independent Set . Recall that α ( G ) denotes the size ofthe maximum independent set of G .We transform G into an instance of Max Min FVS as follows: For every pair of u, v ∈ V ,we add n vertices such that they are adjacent only to u and v . We denote by I uv the set ofsuch vertices. Then I uv is an independent set. Let G = ( V , E ) be the constructed graph.We now make the following two claims: (cid:66) Claim 18. mmfvs( G ) ≥ ( n − (cid:0) α ( G )2 (cid:1) Proof.
We construct a minimal fvs of G as follows: let C be a minimum vertex cover of G .Then we begin with the set that contains C and the union of all I uv (which is clearly an fvs)and remove vertices from it until it becomes minimal. Let S be the final minimal fvs. Weobserve that for all u, v ∈ V \ C , S contains at least n − I uv . Since C is aminimum vertex cover of G , there are (cid:0) α ( G )2 (cid:1) pairs u, v ∈ V \ C . (cid:74)(cid:66) Claim 19. mmfvs( G ) ≤ n (cid:0) α ( G )2 (cid:1) + n C V I T 2 0 1 6
Proof.
Let S be a minimal fvs of G and F be the corresponding forest. It suffices to showthat | S \ V | ≤ n (cid:0) α ( G )2 (cid:1) , since | S ∩ V | ≤ n . Consider now a set I uv . If u ∈ S or v ∈ S ,then I uv ∩ S = ∅ , because all vertices of I uv have at most one neighbor in F , and aretherefore redundant. So, I uv contains (at most n ) vertices of S only if u, v ∈ F . However, | F ∩ V | ≤ α ( G ), because F is bipartite, so F ∩ V induces two independent sets, both ofwhich must be at most equal to the maximum independent set of G . So the number of pairs u, v ∈ F ∩ V is at most (cid:0) α ( G )2 (cid:1) and since each corresponding I uv has size n , we get thepromised bound. (cid:74) The two claims together imply that there exist constants c , c such that (for suffi-ciently large n ) we have c n ( α ( G )) ≤ mmfvs( G ) ≤ c n ( α ( G )) . That is, mmfvs( G ) =Θ( n ( α ( G )) ).Suppose now that there exists a polynomial-time approximation algorithm which, givena graph G , produces a minimal fvs S with the property mmfvs( G ) r ≤ | S | ≤ mmfvs( G ),that is, there exists an r -approximation for Max Min FVS . Running this algorithm onthe instance we constructed, we obtain that c n ( α ( G )) r ≤ | S | ≤ c n ( α ( G )) . Therefore, α ( G ) √ rc /c ≤ q | S | c n ≤ α ( G ). As a result, we obtain an O ( √ r ) approximation for the value of α ( G ). We therefore conclude that, unless P = NP, any such algorithm must have √ r > n − (cid:15) ,for any (cid:15) >
0, hence, r > n − (cid:15) , for any (cid:15) >
0. Since the graph G has N = Θ( n ) vertices,we get that no approximation algorithm can achieve a ratio of N / − (cid:15) . (cid:74) We notice that in the construction of the previous theorem, the maximum degree of thegraph is approximately equal to the approximation gap. Thus, the following corollary alsoholds. (cid:73)
Corollary 20.
For any positive constant (cid:15) , Max Min FVS is inapproximable within afactor of ∆ − (cid:15) unless P = NP.
In this section we extend Theorem 17 to the realm of sub-exponential time algorithms. Werecall the following result of Chalermsook et al. (cid:73)
Theorem 21. [12] For any (cid:15) > and any sufficiently large r , if there exists an r -approximation algorithm for Max Independent Set running in ( n/r ) − (cid:15) , then the ran-domized ETH is false. We remark that Theorem 21, which gives an almost tight running time lower bound for
Max Independent Set , has already been used as a starting point to derive a similarlytight bound for the running time of any sub-exponential time approximation for
Max MinVC . Here, we modify the proof of Theorem 17 to obtain a similarly tight result for
MaxMin FVS . Nevertheless, the reduction for
Max Min FVS is significantly more challenging,because the ideas used in Theorem 17 involve an inherent quadratic (in n ) blow-up of thesize of the instance. As a result, in addition to executing an appropriately modified versionof the reduction of Theorem 17, we are forced to add an extra “sparsification” step, and usea probabilistic analysis with Chernoff bounds to argue that this step does not destroy theinapproximability gap. (cid:73) Theorem 22.
For any (cid:15) > and any sufficiently large r , if there exists an r -approximationalgorithm for Max Min FVS running in ( n/r / ) − (cid:15) , then the randomized ETH is false. . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:15 Proof.
We recall some details about the reduction used to prove Theorem 21. The reductionof [12] begins from a instance φ on n variables, and for any (cid:15), r , constructs a graph G with n (cid:15) r (cid:15) vertices which (with high probability) satisfies the following properties: if φ is satisfiable, then α ( G ) ≥ n (cid:15) r ; otherwise α ( G ) ≤ n (cid:15) r (cid:15) . Hence, any approximationalgorithm with ratio r − (cid:15) for Max Independent Set would be able to distinguish betweenthe two cases (and solve the initial instance). If, furthermore, this algorithm runs in2 ( | V | /r ) − (cid:15) , we get a sub-exponential algorithm for .Suppose we are given (cid:15), r , and we want to prove the claimed lower bound on the runningtime of any algorithm that r -approximates Max Min FVS . To ease presentation, we willassume that r is the square of an integer (this can be achieved without changing the value of r by more than a small constant). We will also perform a reduction from to show thatan algorithm that achieves this ratio too rapidly would give a sub-exponential (randomized)algorithm for . We begin by executing the reduction of [12], starting from a instance φ on n variables, but adjusting their parameter r appropriately so we obtain a graph G with the following properties (with high probability): | V ( G ) | = n (cid:15) r / (cid:15) If φ is satisfiable, then α ( G ) ≥ n (cid:15) r / If φ is not satisfiable, then α ( G ) ≤ n (cid:15) r (cid:15) We now construct a graph G as follows: for each pair u, v ∈ V ( G ), we introduce anindependent set I uv of size √ r connected to u, v . We claim that G has the followingproperties (assuming G has the properties cited above): | V ( G ) | = Θ( n (cid:15) r / (cid:15) )If φ is satisfiable, then mmfvs( G ) = Ω( n (cid:15) r / )If φ is not satisfiable, then mmfvs( G ) = O ( n (cid:15) r / (cid:15) )Before proceeding, let us establish the properties mentioned above. The size of | V ( G ) | is easy to bound, as for each of the (cid:0) | V ( G ) | (cid:1) pairs of vertices of G we have constructed anindependent set of size √ r . If φ is satisfiable, we construct a minimal fvs of G by startingwith a minimum vertex cover of G to which we add all vertices of all I uv . We then makethis fvs minimal. We claim that for each I uv for which u, v ∈ V \ C , our set will in the endcontain all of I uv , except maybe at most one vertex. Furthermore, if one vertex of I uv isremoved from the fvs as redundant, this decreases the number of components of the inducedforest that contain vertices of V (as u, v are now in the same component). This cannothappen more than | V ( G ) | times. The number of I uv with u, v ∈ V \ C is (cid:0) α ( G )2 (cid:1) = Ω( n (cid:15) r ).So, mmfvs( G ) = Ω( n (cid:15) r / − | V ( G ) | ).For the third property, take any minimal fvs S of G and let F be the correspondingforest. We have | F ∩ V | ≤ α ( G ), because F is bipartite. It is sufficient to bound | S \ V | to obtain the bound (as | S ∩ V | is already small enough). To do this, we note that in a set I uv where u, v are not both in F , we have I uv ∩ S = ∅ , as all vertices of I uv are redundant.So, the number of sets I uv which contribute vertices to S is at most (cid:0) | F ∩ V | (cid:1) = O ( n (cid:15) r (cid:15) ).Each such set has size √ r , giving the claimed bound.We have now constructed an instance where the gap between the values for mmfvs( G ),depending on whether φ is satisfiable, is almost r (in fact, it is r − (cid:15) , but we can make itequal to r by adjusting the parameters accordingly). The problem is that the order of thenew graph depends quadratically on n . This blow-up makes it impossible to obtain a runningtime lower bound, as a fast approximation algorithm for Max Min FVS (say with running
C V I T 2 0 1 6 time 2 n/r ) would not result in a sub-exponential algorithm for . We therefore needto “sparsify” our instance.We construct a graph G by taking G and deleting every vertex of V ( G ) \ V ( G ) withprobability n − n . That is, every vertex of the independent sets I uv we added survives(independently) with probability 1 /n . We now claim the following properties hold with highprobability: | V ( G ) | = Θ( n (cid:15) r / (cid:15) )If φ is satisfiable, then mmfvs( G ) = Ω( n (cid:15) r / )If φ is not satisfiable, then mmfvs( G ) = O ( n (cid:15) r / (cid:15) )Before we proceed, let us explain why if we establish that G satisfies these properties,then we obtain the theorem. Indeed, suppose that for some sufficiently large r and (cid:15) > Max Min FVS with ratio r − (cid:15) running intime 2 ( N/r / ) − (cid:15) for graphs with N vertices. The algorithm has sufficiently small ratioto distinguish between the two cases in our constructed graph G , as the ratio betweenmmfvs( G ) when φ is satisfiable or not is Ω( r − (cid:15) ) (and r is sufficiently large), so we canuse the approximation algorithm to solve . Furthermore, to compute the running timewe see that N/r / = Θ( n (cid:15) r (cid:15) ) = O ( n (cid:15) ). Therefore, ( N/r / ) − (cid:15) = o ( n ) and we geta sub-exponential time algorithm for . We conclude that for any sufficiently large r and any (cid:15) >
0, no algorithm achieves ratio r − (cid:15) in time 2 ( N/r / ) − (cid:15) . By adjusting r, (cid:15) appropriately we get the statement of the theorem.Let us therefore try to establish that the three claimed properties all hold with highprobability. We will use the following standard Chernoff bound: suppose X = P ni =1 X i isthe sum of n independent random 0 / X i and that E [ X ] = P ni =1 E [ X i ] = µ . Then,for all δ ∈ (0 ,
1) we have
P r [ | X − µ | ≥ δµ ] ≤ e − µδ / The first property is easy to establish: we define a random variable X i for each vertexof each I uv of G . This variable takes value 1 if the corresponding vertex appears in G and 0 otherwise. Let X be the sum of the X i variables, which corresponds to thenumber of such vertices appearing in G . Suppose that the number of vertices in sets I uv of G is cn (cid:15) r / (cid:15) , where c is a constant. Then, E [ X ] = cn (cid:15) r / (cid:15) . Also, P r [ | X − E [ X ] | ≥ E [ X ]2 ] ≤ e − E [ X ] / = o (1). So with high probability, | V ( G ) | is of thepromised magnitude.The second property is also straightforward. This time we consider a maximum minimalfvs S of G of size cn (cid:15) r / . Again, we define an indicator variable for each vertex of thisset in sets I uv . The expected number of such vertices that survive in G is cn (cid:15) r / . As inthe previous paragraph, with high probability the actual number will be close to this bound.We now need to argue that (almost) the same set is a minimal fvs of G . We start in G with (the surviving vertices of) S , which is clearly an fvs of G , and delete vertices until theset is minimal. We claim that the size of the set will decrease by at most | V ( G ) | = n (cid:15) r (cid:15) .Indeed, if S ∩ I uv = ∅ , then u, v S . The two vertices u, v are (deterministically) included in G and start out in the corresponding induced forest in our solution. If a vertex of S ∩ I uv isdeleted as redundant, placing that vertex in the forest will put u, v in the same component,reducing the number of components of the forest with vertices from | V ( G ) | . This can happenat most | V ( G ) | times. Since | V ( G ) | < c ( n (cid:15) r / ) (for n, r sufficiently large), deletingthese redundant vertices will not change the order of magnitude of the solution.Finally, in order to establish the third property we need to consider every possible minimalfvs of G and show that none of them end up being too large. Consider a set F ⊆ V ( G )that induces a forest in G . Our goal is to prove that any minimal fvs S of G that satisfies . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:17 u
The edge gadget of e = ( u, v ) in the constructed graph G . V ( G ) \ S = F has a probability of being “too large” (that is, violating our claimed bound)much smaller than 2 −| V ( G ) | . If we achieve this, then we can take a union bound over all sets F and conclude that with high probability no minimal fvs of G is too large.Suppose then that we have fixed an acyclic set F ⊆ V ( G ). We have | F | ≤ α ( G ) ≤ n (cid:15) r (cid:15) . Any minimal fvs with V ( G ) \ S = F can only contain vertices from a set I uv if u, v ∈ F . The total number of such vertices in G is at most O ( n (cid:15) r / (cid:15) ). The expectednumber of such vertices that survive in G is (for some constant c ) at most µ = cn (cid:15) r / (cid:15) .Now, using the Chernoff bound cited above we have P r [ | X − µ | ≥ µ ] ≤ e − µ/ . We claim2 e − µ/ = o (2 −| V ( G ) | ). Indeed, this follows because | V ( G ) | = n (cid:15) r / (cid:15) = o ( µ ). As a result,the probability that a large minimal fvs exists for a fixed set F ⊆ V ( G ) exists is low enoughthat taking the union bound over all possible sets F we have that with high probability nominimal fvs exists with value higher than 3 µ/
2, which establishes the third property. (cid:74) ∆ = 6 (cid:73)
Theorem 23.
Max Min FVS is NP-hard on planar bipartite graphs with ∆ = 6 . Proof.
We give a reduction from
Max Min VC , which is NP-hard on planar bipartitegraphs of maximum degree 3 [40]. Note that the NP-hardness in [40] is stated for
MinimumIndependent Dominating Set , but any independent dominating set is also a maximalindependent set (and vice-versa) and the complement of the minimum maximal independentset of any graph is a maximum minimal vertex cover. Thus, we also obtain NP-hardness for
Max Min VC on the same instances.We are given a graph G = ( V, E ). For each edge e = ( u, v ) ∈ E , we add a path oflength three from u to v going through two new vertices e (1) , e (2) (see Figure 2). Note that u, e (1) , e (2) , v form a cycle of length 4. Then we add two cycles of length 4, e ( i ) , c ( i ) e , c ( i ) e , c ( i ) e and e ( i ) , c ( i ) e , c ( i ) e , c ( i ) e for i ∈ { , } . Let G = ( V , E ) be the constructed graph. Because∆( G ) = 3, we have ∆( G ) = 6. Moreover, since G is planar and bipartite, G is also planarand bipartite. We will show that there is a minimal vertex cover of size at least k in G ifand only if there is a minimal feedback vertex set of size at least k + 4 | E | in G .Given a minimal vertex cover S of size at least k in G , we construct the set S = S ∪ S e ∈ E { c (1) e , c (1) e , c (2) e , c (2) e } . Then | S | ≥ k + 4 | E | . Let us first argue that S is an fvs of G . For each e = ( u, v ) ∈ E we have at least one of u, v ∈ S , without loss of generalitylet u ∈ S . Now in G [ V \ S ] the edges ( e (1) , e (2) ) and ( e (2) , v ) are bridges and thereforecannot be part of any cycle. The remaining cycles going through e (1) , e (2) are handled by { c (1) e , c (1) e , c (2) e , c (2) e } . Furthermore, since G [ V \ S ] is an independent set, it is also acyclic.To see that S is a minimal fvs, we remark that for each c ( i ) e , c ( i ) e contained in S there is aprivate cycle in G [ V \ S ]. We also note that since S is a minimal vertex cover of G , for C V I T 2 0 1 6 each u ∈ S , there exists v S with e = ( u, v ) ∈ E . This means that u has the private cycleformed by { u, v, e (1) , e (2) } in G [ V \ S ]. Therefore, S is a minimal fvs.Conversely, suppose we are given a minimal fvs S of G with | S | ≥ k + 4 | E | . We willedit S so that is contains only vertices in V \ S e ∈ E { e (1) , e (2) } , without decreasing its size.First, suppose e (1) , e (2) ∈ S , for some e ∈ E . We construct a new minimal fvs S = S \ { e (2) } ∪ { c (2) e , c (2) e } which is larger that S , since by minimality we have c (2) ei S for i ∈ { , . . . , } . It is not hard to see that S is indeed an fvs, as no cycle can go through e (2) in G [ V \ S ]. The two vertices we added have a private cycle, while all vertices of S ∩ S retain their private cycles, so S is a minimal fvs. As a result in the remainder we assumethat S contains at most one of { e (1) , e (2) } for all e ∈ E .Suppose now that for some e = ( u, v ) ∈ E , we have S ∩{ u, v } 6 = ∅ and S ∩{ e (1) , e (2) } 6 = ∅ .Without loss of generality, let e (1) ∈ S . We set S = S \ { e (1) } ∪ { c (1) e , c (1) e } and claim that S is a larger minimal fvs than S . Indeed, no cycle goes through e (1) in G [ V \ S ], the newvertices we added to S have private cycles, and all vertices of S ∩ S retain their privatecycles in G [ V \ S ]. Therefore, we can now assume that if for some e = ( u, v ) ∈ E we have S ∩ { e (1) , e (2) } 6 = ∅ then u, v S .For the remaining case, suppose that for some e = ( u, v ) ∈ E we have u, v S and(without loss of generality) e (1) ∈ S . We construct the set S = S \ { e (1) } ∪ { c (1) e , c (2) e , u } .Note that | S | ≥ | S | + 2. It is not hard to see that S is an fvs, since by adding c (1) e , c (1) e , v to our set we have hit all cycles containing e (1) in G . The problem now is that S is notnecessarily minimal. We greedily delete vertices from S to obtain a minimal fvs S ∗ . Weclaim that in this process we cannot delete more than two vertices, that is | S ∗ \ S | ≤ c (1) e , c (2) e , u cannot be removed from S as they have privatecycles in G [ V \ S ]. Suppose now that w ∈ S \ S ∗ is the first vertex we removed from S ,so G [( V \ S ) ∪ { w } ] is acyclic. This vertex must have had a private cycle in G [ V \ S ],which was necessarily going through u . Therefore, G [( V \ S ) ∪ { w } ] has a path connectingtwo neighbors of u and this path does not exist in G [( V \ S )]. With a similar reasoning,removing another vertex w ∈ S from the fvs will create a second path between neighbors of u in the induced forest. We conclude that this cannot happen a third time, since | N ( u ) | ≤ u , this will create a cycle. As a result, | S ∗ | ≥ | S | . We assume in the remainder that S does not contain e (1) , e (2) for any e ∈ E .Now, given a minimal fvs S of G with | S | ≥ k + 4 | E | and S ∩ ( ∪ e ∈ E { e (1) , e (2) } ) = ∅ we set S = S ∩ V and claim that S is a minimal vertex cover of G with | S | ≥ k . Indeed S isa vertex cover, as for each e = ( u, v ) ∈ E , if u, v S then we would get the cycle formed by { u, v, e (1) , e (2) } . To see that S is minimal, suppose N G [ u ] ⊆ S . We claim that in that case u has no private cycle in G [ V \ S ] (this can be seen by deleting all bridges in G [ V \ S ],which leaves u isolated). This contradicts the minimality of S as an fvs of G . Finally, weargue that | S \ V | ≤ | E | , which gives the desired bound on | S | . Consider an e = ( u, v ) ∈ E . S cannot contain more than one vertex among c (1) e , c (1) e , c (1) e , since any of these vertices hitsthe cycle that goes through the others. With similar reasoning for the three other length-fourcycles we conclude that S contains at most 4 vertices for each edge e ∈ E . (cid:74) We have essentially settled the approximability of
Max Min FVS for polynomial and sub-exponential time, up to sub-polynomial factors in the exponent of the running time. It wouldbe interesting to see if the running time of our sub-exponential approximation algorithm canbe improved by poly-logarithmic factors in the exponent, as in [4]. In particular, improving . Dublois, T. Hanaka, M. Khosravian Ghadikolaei, M. Lampis, and N. Melissinos 23:19 the running time to 2 O ( n/r / ) seems feasible, but would likely require a version of Lemma14 which uses more sophisticated techniques, such as Cut&Count [7, 15, 17].Another problem of similar spirit which deserves to be studied is Max Min OCT , wherean odd cycle transversal (OCT) is a set of vertices whose removal makes the graph bipartite.This problem could also potentially be “between”
Max Min VC and
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