Vertex Cover Kernelization Revisited

An important result in the study of polynomial-time preprocessing shows that there is an algorithm which given an instance (G,k) of Vertex Cover outputs an equivalent instance (G′,k′) in polynomial time with the guarantee that G′ has at most 2k′ vertices (and thus \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mathcal{O}((k')^{2})$\end{document} edges) with k′≤k. Using the terminology of parameterized complexity we say that k-Vertex Cover has a kernel with 2k vertices. There is complexity-theoretic evidence that both 2k vertices and Θ(k2) edges are optimal for the kernel size. In this paper we consider the Vertex Cover problem with a different parameter, the size \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mathop{\mathrm{\mbox{\textsc{fvs}}}}(G)$\end{document} of a minimum feedback vertex set for G. This refined parameter is structurally smaller than the parameter k associated to the vertex covering number \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mathop{\mathrm{\mbox {\textsc{vc}}}}(G)$\end{document} since \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mathop{\mathrm{\mbox{\textsc{fvs}}}}(G)\leq\mathop{\mathrm{\mbox{\textsc{vc}}}}(G)$\end{document} and the difference can be arbitrarily large. We give a kernel for Vertex Cover with a number of vertices that is cubic in \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$\mathop{\mathrm{\mbox{\textsc{fvs}}}}(G)$\end{document}: an instance (G,X,k) of Vertex Cover, where X is a feedback vertex set for G, can be transformed in polynomial time into an equivalent instance (G′,X′,k′) such that |V(G′)|≤2k and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$|V(G')| \in\mathcal{O}(|X'|^{3})$\end{document}. A similar result holds when the feedback vertex set X is not given along with the input. In sharp contrast we show that the Weighted Vertex Cover problem does not have a polynomial kernel when parameterized by the cardinality of a given vertex cover of the graph unless NP ⊆ coNP/poly and the polynomial hierarchy collapses to the third level.


Introduction
The Vertex Cover problem is one of the six classic NP-complete problems discussed by Garey and Johnson in their famous work on intractability [22,GT1], and has played an important role in the development of parameterized algorithms [15,28,16].A parameterized problem is a language L ⊆ Σ * × N, and such a problem is (strongly uniform) fixed parameter tractable if membership of an instance (x, k) can be decided in f (k)|x| c time for some computable function f and constant c.Since the structure of Vertex Cover is so simple and elegant, it has proven to be an ideal testbed for new techniques in the context of parameterized complexity.The problem is also highly relevant from a practical point of view because of its role in bioinformatics [1] and other problem areas.
In this work we suggest a "refined parameterization" for the Vertex Cover problem using the feedback vertex number fvs(G) as the parameter, i.e. the size of a smallest vertex set whose deletion turns G into a forest.We give upper bounds on the kernel size for the unweighted version of Vertex Cover under this parameterization, and also supply a conditional superpolynomial lower bound on the kernel size for the variant of Vertex Cover where each vertex has a non-negative integral weight.But before we state our results we shall first survey the current state of the art for the parameterized analysis of Vertex Cover.
There has been an impressive series of ever-faster parameterized algorithms to solve k-Vertex Cover, which led to the current-best algorithm by Chen et al. that can decide whether a graph G has a vertex cover of size k in O(1.2738 k + kn) time and polynomial space [9,30,8,17].The Vertex Cover problem has also played an important role in the development of problem kernelization [24].A kernelization algorithm (or kernel) is a polynomial-time procedure that reduces an instance (x, k) of a parameterized decision problem to an equivalent instance (x , k ) such that |x |, k ≤ f (k) for some computable function f , which is the size of the kernel.We also use the term kernel to refer to the reduced instance (x , k ).
The k-Vertex Cover problem admits a kernel with 2k vertices and O(k 2 ) edges, which has been a subject of repeated study [6,8,10,2,11] and experimentation [1,13].There is some complexity-theoretic evidence that the size bounds for the kernel cannot be improved.Since practically all reduction-rules found to date are approximation-preserving [28], it appears that a kernel with less than 2k vertices would yield a polynomial-time approximation algorithm with a performance ratio smaller than 2 which would disprove the Unique Games Conjecture [25].A recent breakthrough result by Dell and Van Melkebeek [12] shows that there is no polynomial kernel which can be encoded into O(k 2− ) bits for any > 0 unless the polynomial hierarchy collapses to the third level (PH = Σ p 3 ), which suggests that the current bound of O(k 2 ) edges is tight up to logarithmic factors.This overview might suggest that there is little left to explore concerning kernelization for vertex cover, but this is far from true.All existing kernelization results for Vertex Cover use the requested size k of the vertex cover as the parameter.But there is no reason why we should not consider structurally smaller parameters, to see if we can preprocess instances of Vertex Cover such that their final size is bounded by a function of such a smaller parameter, rather than by a function of the requested set size k.We study kernelization for the Vertex Cover problem using the feedback vertex number fvs(G) as the parameter.Since every vertex cover is also a feedback vertex set we find that fvs(G) ≤ vc(G) which shows that the feedback vertex number of a graph is a structurally smaller parameter than the vertex covering number: there are trees with arbitrarily large values of vc(G) for which fvs(G) = 0. We call our parameter "refined" since it is structurally smaller than the standard parameter for the Vertex Cover problem.Related Work.The idea of studying parameterized problems using alternative parameters is not new (see e.g.[28]), but was recently advocated by Fellows et al. [19,20,29] in the call to investigate the complexity ecology of parameters.The main idea behind this program is to determine how different parameters affect the parameterized complexity of a problem.Some recent results in this direction include FPT algorithms for graph layout problems parameterized by the vertex cover number of the graph [21] and an algorithm to decide isomorphism on graphs of bounded feedback vertex number [26].We are aware of only two applications of this idea to give polynomial kernels using alternative parameters.Fellows et al. [20,18] show that the problems Independent Set, Dominating Set and Hamiltonian Circuit admit linear-vertex kernels on graphs G when parameterized by the maximum number of leaves in any spanning tree of G. Very recently Uhlmann and Weller [31] gave a polynomial kernel for Two-Layer Planarization parameterized by the feedback edge set number, which is a refined structural parameter for that problem since it is smaller than the natural parameter.
Our Results.We believe that we are one of the first to present a polynomial problem kernel using a non-standard but practically relevant refined parameter.We study the following parameterized problem: fvs-Weighted Vertex Cover Instance: A simple undirected graph G, a weight function w : V (G) → N + , a feedback vertex set X ⊆ V (G) such that G − X is a forest, an integer k ≥ 0. Parameter: The size |X| of the feedback vertex set.Question: Is there a vertex cover C of G such that v∈C w(v) ≤ k?
We also consider the unweighted variant fvs-Vertex Cover in which all vertices have a weight of 1.The problems fvs-Weighted Independent Set and fvs-Independent Set are defined similarly.Throughout this work k will always represent the total size or weight of the set we are looking for; depending on the context this is either a vertex cover or an independent set.
We prove that fvs-Vertex Cover has a kernel in which the number of vertices is bounded by min(O(|X| 3 ), 2k).This bound is at least as small as the current-best Vertex Cover kernel, but for graphs with small feedback vertex sets our bound is significantly smaller.We also study the weighted version of the problem, and obtain a contrasting result: we show that fvs-Weighted Vertex Cover does not admit a polynomial kernel unless PH = Σ p 3 .This is very surprising since both the weighted and unweighted versions of k-Vertex Cover admit polynomial kernels and can be attacked using similar reduction rules [10].To our knowledge we give the first example of a parameterized problem whose weighted and unweighted versions are both NP-complete and FPT, but for which the unweighted version allows a polynomial kernel but the weighted version does not.
When we present our results we will state them in terms of fvs-Independent Set and fvs-Weighted Independent Set since this simplifies the exposition.Because we are using the size of a feedback vertex set as the parameter, there are trivial parameterized reductions between these problems: an instance (G, X, k) of fvs-Vertex Cover is equivalent to an instance (G, X, |V (G)| − k) of fvs-Independent Set with the same parameter value |X|.Hence our kernelization bounds for Independent Set will immediately carry over to Vertex Cover.

Preliminaries
In this work we only consider undirected, finite, simple graphs.Let G be a graph and denote its vertex set by V (G) and the edge set by E(G).We denote the independence number of G by α(G), the vertex covering number by vc(G) and the feedback vertex number by fvs(G).
We will abbreviate maximum independent set as MIS, and feedback vertex set as FVS.  the vertices in X and their incident edges is denoted by G − X.The graph G[E(G) \ Y ] obtained from G by deleting the edges in Y but not their endpoints is denoted by G/Y .Carefully observe the difference between these two operators: if {u, v} is an edge in G, then G − {u, v} is the graph obtained from G by deleting the vertices u, v and their incident edges, whereas G/{{u, v}} is the graph obtained from G by removing the edge {u, v} while leaving the endpoints u and v intact.
We need the following proposition on the structure of maximum independent sets in trees by Zito [32, Theorem 2], which we re-state here in terms of forests: Proposition 1.Let F be a forest.Then there is a unique partition of the vertex set V (F ) into subsets A, N, S such that: 1. Any MIS for F contains all vertices of A and no vertices of N .

2.
For each vertex v ∈ S there is a MIS for F containing v and a MIS for F avoiding v.

3.
There is a perfect matching M in F [S], and any MIS for F contains exactly one endpoint of each edge in M .4. The matching M contains all the α-critical edges of F : for all e ∈ E(F ) it holds that α(F ) < α(F/{e}) ⇔ e ∈ M .This partition is uniquely characterized by adjacency relations.The sets A, N, S form the described partition if and only if: I.There is a matching M on the vertices of S. II.No vertex of A is adjacent to another vertex of A or to a vertex in S. III.Each vertex of N is adjacent to at least two vertices of A. We will refer to this decomposition of the vertex set of a forest F into the subsets A, N, S with the matching M as its independence decomposition (Figure 1).

Cubic Kernel for FVS-Independent Set
In this section we develop a cubic kernel for fvs-Independent Set.Consider an instance (G, X, k) of the problem, which asks whether graph G with the FVS X has an independent set of size k.Throughout this section F := G − X denotes the forest obtained by deleting the vertices in X.
We can ensure that G does not contain any vertices of degree ≤ 1.
To prove the correctness of Rule 4 we need the following lemma.

Lemma 4. Let T be a connected component of F and let X
Proof.Assume the conditions stated in the lemma hold.Consider the independence decomposition of T into the sets A, N, S, and let M be a perfect matching on T [S].We will try to construct a MIS I for T that does not use any vertices in N G (X I ); this must then also be a MIS for T − N G (X I ) of the same size.By the assumption that Conf T (X I ) > 0 any independent set in T must use at least one vertex in N G (X I ) in order to be maximum, hence our construction procedure must fail somewhere; the place where it fails will provide us with a set X as required by the statement of the lemma.
Construction of a MIS.By Proposition 1 any MIS for T must use all vertices in A, no vertices from N and exactly one endpoint of each edge in the matching M .It follows that if some vertex v ∈ A is adjacent in G to a vertex x ∈ X I , then α(T − {v}) < α(T ) and therefore α(T − N G (x)) < α(T ) which proves that Conf T ({x}) > 0; hence we can then use X := {x} as our desired subset to prove the claim.In the remainder of the proof we may therefore assume that no vertex of A is adjacent in G to a vertex in X I .
We now start building our independent set I for T that avoids vertices in N G (X I ).We start by taking all vertices of A in the independent set; we do not use any vertices in N G (X I ) here since A ∩ N G (X I ) = ∅ by assumption.To augment I into a MIS for T it remains to add one endpoint of each edge in the matching M .Since the endpoints of the matching are not adjacent to vertices in A by the adjacency rules of Proposition 1, we can now restrict ourselves to the subgraph T := T [S] induced by the matched vertices since no choice of independent vertices in T [S] will conflict with the choice of the vertices A. If there is a vertex v in T such that N T (v) = {u} and N G (v) ∩ X I = ∅, then the edge {v, u} must be in the matching M (since M is a perfect matching in T [S]).Because we must choose one of {u, v} in a MIS for T , and by Observation 2 choosing a degree-1 vertex will never conflict with choices that are made later on, we can add v to our independent set I while respecting the invariant that no vertex in I is adjacent in G to a vertex in X I .Since we have then chosen one endpoint of the matching edge {u, v} in I, we can delete u, v and their incident edges to obtain a smaller graph T (which again contains a perfect submatching of M ) in which we continue the process.As long as there is a vertex with degree 1 in T that has no neighbors in X I then we take it into I, delete it and its neighbor, and continue.If this process ends with an empty graph, then by Proposition 1 the set I must be a MIS for T , and since it does not use any vertices adjacent to X I it must also be a MIS for T − N G (X I ); but this proves that α(T ) = α(T − N G (X I )) which means Conf T (X I ) = 0, which is a contradiction to the assumption at the start of the proof.So the process must end with a non-empty graph T ⊆ T such that vertices with degree 1 in T are adjacent in G to a vertex in X I and for which the matching M restricted to T is a perfect matching on T .We use this subgraph T to obtain a set X as desired.
Using the subgraph to prove the claim.Consider a vertex v 0 in T with deg T (v 0 ) = 1, and construct a path P = {v 0 , v 1 , . . ., v 2p+1 } by following edges of T that are alternatingly in and out of the matching M , until arriving at a degree-1 vertex whose only neighbor was already visited.Since T is acyclic, M restricted to T is a perfect matching on T and we start the process at a vertex of degree 1, it is easy to verify that there must be such a path P (there can be many; any arbitrary such path will suffice), that P must contain an even number of vertices, that the first and last vertex on P have degree-1 in T and that the edges {v 2i , v 2i+1 } must be in M for all 0 ≤ i ≤ p.Since we assumed that all degree-1 vertices in T are adjacent in G to X I , there exist vertices and v 2p+1 ∈ N G (x 2 ).We now claim that X := {x 1 , x 2 } satisfies the requirements of the statement of the lemma, i.e. that Conf T ({x 1 , x 2 }) > 0. This fact is witnessed by considering the path P in the original tree T .Any MIS for T which avoids N G ({x 1 , x 2 }) must use one endpoint of the matched edge {v 0 , v 1 }, and since the choice of v 0 is blocked because v 0 is a neighbor to x 1 , it must use v 1 .But path P shows that v 1 is adjacent in T to v 2 , and hence we cannot choose v 2 in the independent set.But since {v 2 , v 3 } is again a matched edge, we must use one of its endpoints; hence we must use v 3 .Repeating this argument shows that we must use vertex v 2p+1 in a MIS for T if we cannot use v 0 ; but the use of v 2p+1 is also not possible if we exclude N G ({x 1 , x 2 }).Hence we cannot make a MIS for T without using vertices in . By the definition of conflicts this proves that Conf T (X ) > 0 for X = {x 1 , x 2 }, which concludes the proof.
Using this lemma we can prove the correctness of Rule 4.

Lemma 5. Rule 4 is correct
Proof.Assume the conditions in the statement of the lemma hold.It is trivial to see that α(G) ≤ α(G − T ) + α(T ).To establish the lemma we only need to prove that α(G) ≥ α(G − T ) + α(T ), which we will do by showing that any independent set I G−T in G−T can be transformed to an independent set of size at least |I G−T |+α(T ) in G.So consider such an independent set I G−T , and let X I := I G−T ∩ X be the set of vertices which belong to both I G−T and the feedback vertex set X. Suppose that α(T ) > α(T − N G (X I )).Then by Lemma 4 there is a subset X ⊆ X I with |X | ≤ 2 such that Conf T (X ) > 0. Since X I is an independent set, such a subset X would also be independent; but by the preconditions to this lemma such a set X does not exist and therefore we must have α(T ) = α(T − N G (X I )).We introduce the concept of blockability for the statement of the last two reduction rules.Definition 6.We say that the pair This can be interpreted as follows: any independent set in G that contains X cannot contain x or y, so the pair x, y is blocked from being in an independent set by choosing X .It follows directly from the definition that if x, y is not X-blockable, then for any combination of u ∈ N G (x) ∩ X and v ∈ N G (y) ∩ X we must have {u, v} ∈ E(G).
Reduction Rule 5.If there are distinct vertices u, v ∈ V (G) \ X which are adjacent in G and are not X-blockable such that deg F (u), deg F (v) ≤ 2 then reduce the graph as follows.Delete vertices u, v and decrease k by 1.If u has a neighbor t in F which is not v, then make all vertices of N G (v) ∩ X adjacent to t.If v has a neighbor w in F which is not u, then make all vertices of N G (u) ∩ X adjacent to w.If the vertices t, w exist then they must be unique; add the edge {t, w} to the graph.Reduction Rule 6.If there are distinct vertices t, u, v, w in V (G) \ X such that deg and {u, v} ∈ E(G) such that none of the pairs {u, t}, {v, w}, {t, w} are X-blockable, then reduce as follows.Let {p} = N F (u) \ {t, v} and let {q} = N F (v) \ {w, u}.Delete {t, u, v, w} and their incident edges from G, decrease k by 2, make p adjacent to all vertices of N G (t) ∩ X and make q adjacent to all vertices of N G (w) ∩ X.
See Figure 2 for an illustration of the final two reduction rules, which are meant to reduce the sizes of the trees in the forest F .The correctness of these rules can be proven by an exchange argument (Lemma 14, Lemma 15).Whereas Rule 4 deletes a tree T from the forest F when we can derive that for every independent set in G − T we can obtain an independent set in G which is α(T ) vertices larger, these last reduction rules act locally within one tree, but according to the same principle.Instead of working on an entire connected component of F , they reduce subtrees T ⊆ F in situations where we can derive that every independent set in X can be augmented with α(T ) vertices from T .In Rule 5 we reduce the subtree on vertices {u, v} which has independence number 1, and in Rule 6 we reduce the subtree on vertices {u, v, t, w} with independence number 2. Connections between the vertices adjacent to the reduced subtree are made to enforce that removal of the subtree does not affect the types of interactions between the neighboring vertices.
When no reduction rules can be applied to an instance, we call it reduced.In reduced instances the number of vertices in F must be bounded by a function of |X|, which can be proven using the following notion.Definition 7. Let F be a subforest of F , and define the number of active conflicts induced on F by the feedback set X as follows: Active F (X) := X ∈X Conf F (X ) using the abbreviation The number of active conflicts induced on F in a reduced instance is polynomially bounded in |X|.For every v ∈ X we have Conf F ({v}) < |X| by Rule 2, and every pair of distinct non-adjacent vertices {u, v} ⊆ X satisfies Conf F ({u, v}) < |X| by Rule 3. Hence for every reduced instance we have Active F (X) ≤ |X| 2 + |X| 2 |X|.A technical proof shows that in a reduced instance the number of active conflicts induced on the forest F is at least 1 83 |V (F )|.By combining this with the bound on the number of active conflicts we can obtain the following lemma, which is restated and proven in the appendix (Lemma 31).

Lemma 8. Let (G, X, k) be a reduced instance of fvs-Independent Set with forest
By applying the reduction rules and the lemma which bounds the size of reduced instances we can obtain the following kernelization results: (see Section B.7) Theorem 9. fvs-Independent Set has a kernel with a cubic number of vertices: there is a polynomial-time algorithm that transforms an instance (G, X, k) into an equivalent instance Corollary 10. fvs-Vertex Cover has a kernel with min(2k, |X| + 83|X| 3 ) vertices.

4
No Polynomial Kernel for FVS-Weighted Independent Set In this section we show that the introduction of vertex weights makes the parameterized Independent Set problem harder to kernelize, by proving that fvs-Weighted Independent Set does not have a polynomial kernel unless PH = Σ p 3 .To establish this result, we introduce a new parameterized problem called t-Paired Vector Agreement and show that it does not have a polynomial kernel unless PH = Σ p 3 .We then finish the proof by giving a polynomial-parameter transformation to fvs-Weighted Independent Set.

t-Paired Vector Agreement
Instance: A list L consisting of t pairs of vectors (a i , b i ) for 1 ≤ i ≤ t such that each vector is an element of {0, 1, #, ?} m , and an integer k ≥ 0. Parameter: The number of pairs t.Question: Is it possible to choose one vector from each pair, such that the chosen vectors S induce at most k conflict positions?A position 1 ≤ j ≤ m in a vector is a conflict position if some chosen vector v ∈ S has v j = #, or if we have chosen vectors u, v ∈ S such that u j = 0 and v j = 1.
The framework for proving that a parameterized problem does not have a polynomial kernel unless PH = Σ p 3 requires us to establish that the corresponding classical problem is NP-complete.Therefore we consider the classic variant Paired Vector Agreement, which is defined similarly as t-Paired Vector Agreement with the exception that there is no parameter.A reduction from Vertex Cover shows that Paired Vector Agreement is NP-complete (appendix, Lemma 32).By exploiting the fact that t-Paired Vector Agreement can be solved in O(2 t p(m)) time for some polynomial p (by trying all possible combinations of vectors), we can build an or-composition algorithm for the paired agreement problem using a bitmask selection strategy (appendix, Lemma 33); the techniques we use here are similar to those employed by Dom et al. [14].These two facts prove that t-Paired Vector Agreement does not have a polynomial kernel unless PH = Σ p 3 .To relate these results to fvs-Weighted Independent Set we use the following transformation.Lemma 11.There is a polynomial-parameter reduction from t-Paired Vector Agreement to fvs-Weighted Independent Set.
Proof.Let (L, t, m, k) be an instance of t-Paired Vector Agreement.We may assume that k < m otherwise the answer to the instance is trivially yes.We show how to build an equivalent instance (G , w , X , k ) of fvs-Independent Set in polynomial time such that |X | = 2t, which implies the existence of a polynomial-parameter reduction.
The graph G has 2(t + m) vertices, and is defined as follows.For each index 1 ≤ i ≤ t there is a pair of vertices v a i , v b i which are connected by an edge, and have weight 2(t + m).For each vector position 1 ≤ j ≤ m there are vertices p 0 j , p 1 j which are connected by an edge, and have weight 1.The vertices v a i and v b i correspond to the vectors a i , b i of the t-Paired Vector Agreement instance, and are connected to the position-vertices as follows.Let v be a vertex v a i or v b i corresponding to the vector vec(v) which is a i or b i , respectively.For 1 ≤ i ≤ t vertex v is adjacent in G to all p 0 j for which vector vec(v) has a 0 at position j; it is also adjacent to all p 1 j for which vector vec(v) has a 1 at position j, and finally vertex v is adjacent to all {p 0 j , p 1 j } for which vector vec(v) has a # at position j.This concludes the definition of the structure of graph G .
One may verify that a position vertex p x j is adjacent to exactly 1 other position vertex p 1−x j , which implies that the graph induced by the position vertices p 0,1 j has maximum degree 1 and is therefore a forest; this shows that the vector-vertices v a/b i form a feedback vertex set for G and thus we define the feedback vertex set for our instance as X := {v a i , v b i | 1 ≤ i ≤ t} which has size exactly 2t.We now ask for an independent set of total weight at least k := 2t(t + m) + (m − k), which completes the description of instance (G , w , X , k ).It is easy to see that this instance can be computed in polynomial time from the instance (L, t, m, k).The proof that these two instances are equivalent is not difficult, and has been moved to the appendix (Lemma 34) due to space restrictions.
Using standard kernelization lower-bound techniques (see [5,14]) the combination of Lemma 32, Lemma 33 (which are in the appendix) and Lemma 11 now allows us to derive the following results.
Theorem 12.The problems fvs-Weighted Independent Set and fvs-Weighted Vertex Cover do not admit polynomial kernels unless PH = Σ p 3 .
It is interesting to note that an instance (G , w , X , k ) of fvs-Independent Set resulting from the polynomial-parameter transformation of Lemma 11 has a very restricted graph structure: every connected component of the forest G − X is a path on two vertices.Hence our proof shows that even using the parameter "number of vertex deletions needed to turn the graph into a disjoint union of P 2 's" (a structurally larger parameter than the FVS size) there is no polynomial kernel unless PH = Σ p 3 .

Conclusion
We have given upper and lower bounds on the size of kernels for the Vertex Cover and Independent Set problems using the parameter fvs(G).It would be very interesting to perform experiments with our new reduction rules to see whether they offer significant benefits over the existing Vertex Cover kernel on real-world instances.This result is one of the first examples of a polynomial kernel using a "refined" parameter which is structurally smaller than the standard parameterization.The contrasting result on the weighted problem shows that there is a rich structure waiting to be uncovered when studying kernelization using non-standard parameters.The kernel we have presented for fvs-Vertex Cover contains O(|X| 3 ) vertices and can therefore be encoded in O(|X| 6 ) bits using an adjacency matrix.
The results of Dell and Van Melkebeek [12] imply that it is unlikely that there exists a kernel which can be encoded in O(|X| 2− ) bits for any > 0. It might be possible to improve the size of the kernel to a quadratic or even a linear number of vertices, by employing new reduction rules.The current reduction rules can be seen as analogs of the traditional "high degree" rule for the Vertex Cover problem, and it would be interesting to see whether it is possible to find analogs of crown reduction rules when using fvs(G) as the parameter.
Although we have assumed throughout the paper that a feedback vertex set is supplied with the input, we can drop this restriction by applying the known polynomial-time 2approximation algorithm for FVS [3].Observe that the reduction algorithm does not require that the supplied set X is a minimum feedback vertex set; the kernelization algorithm works if X is any feedback vertex set, and the size of the output instance depends on the size of the FVS that is supplied.Hence if we compute a 2-approximate FVS and supply it to the kernelization algorithm, the bound on the number of vertices in the output instance is only a factor 2 worse than when running the kernelization using a minimum FVS.
This paper has focused on the decision version of the Vertex Cover problem, but the data reduction rules given here can also be translated to the optimization version to obtain the following result: given a graph G there is a polynomial-time algorithm that computes a graph G and a non-negative integer c such that vc 3 ); and a vertex cover S for G can be transformed back into a vertex cover of G of size |S | + c in polynomial time.
The approach of studying Vertex Cover parameterized by fvs(G) fits into the broad context of "parameterizing away from triviality" [28,7], since the parameter fvs(G) measures how many vertex-deletions are needed to reduce G to a forest in which Vertex Cover can be solved trivially in polynomial time.As there is a wide variety of restricted graph classes for which Vertex Cover is in P , this opens up a multitude of possibilities for non-standard parameterizations.For every graph class G which is closed under vertex deletion and for which the Vertex Cover problem is in P , the Vertex Cover problem is in FPT when parameterized by the size of a set X such that G − X ∈ G, assuming that X is given as part of the input.Recent work [4] into this direction shows that whenever G contains all cliques the resulting parameterized problem does not have a polynomial kernel unless PH = Σ p 3 .Examples of such classes G are chordal graphs, interval graphs and other types of perfect graphs.We conclude with two specific open problems.Is there a polynomial kernel using the deletion distance from a bipartite graph as the parameter?Does the Vertex Cover problem parameterized by the size of a minimum set X such that treewidth(G − X) ≤ i have a polynomial kernel for every fixed i, or is there some value of i for which this problem does not have a polynomial kernel?The classic Vertex Cover kernelizations can be interpreted as the case i = 0 whereas this paper supplies the result for i = 1.It appears that many interesting insights are waiting to be discovered in this direction.

A Additional Preliminaries
The degree of a vertex v in graph G is denoted by deg is a subset of edges then V (Y ) denotes the endpoints of the edges in Y .In particular, we will use V (M ) for a matching M to denote the endpoints of the matched vertices.A vertex of degree 1 is called a leaf.If v is a vertex in a tree and v is not a leaf, then it is an internal node of the tree.The leaf set of a graph G is the set of degree-1 vertices, denoted as Leaves(G) Proposition 2. Let T be a tree.The following must hold.

Lemma 13. Let T be a tree and consider its independence decomposition A, N, S, and let M be the corresponding matching on S. If e ∈ E(T ) \ M then α(T ) = α(T /{e}).
Proof.By Observation 1 we know that α(T ) ≤ α(T /{e}).By (4) of Proposition 1 we know that α(T ) ≥ α(T /{e}) and hence the two independence numbers must be equal.

B
Omitted Proofs: Cubic Kernel for FVS-Independent Set

B.1 Correctness of Reduction Rules
Lemma 14.Let (G, X, k) with F := G − X be an instance to which Rule 5 is applicable at vertices u, v, and let (G , X, k − 1) be the instance resulting from the reduction.Then it holds that α(G) Proof.Assume the conditions in the statement of the lemma hold.We prove the two directions separately.
(⇒) Let I G be an independent set for graph G of size at least k.We show how to obtain an independent set I G for graph G of size at least |I G | − 1 ≥ k − 1. Observe that no independent set in G can contain both {u, v} since they are adjacent.If I G does not contain any of the vertices {u, v} then we show how to obtain I G which is at least as large and does contain one of {u, v}; so assume I G avoids u and v. Since the pair u, v is not X-blockable by the preconditions for the reduction rule, we know that there is at least one vertex among u, v for which no neighbor in X is chosen in I G .Assume without loss of generality (by symmetry) that this holds for u, such that N G (u) ∩ X ∩ I G = ∅.Since v is not in I G by assumption, the only neighbor of u that can be in I G is its neighbor t in F unequal to v (if such a t exists; see Figure 2).If no such t exists then I G := I G ∪ {u} is a bigger independent set in G; otherwise I G := (I G \ {t}) ∪ {u} is an equally large independent set.So using this replacement argument and symmetry, we may assume that I G is an independent set of size at least k for G that contains u but not v.
We now claim that Since it is easy to see that I G has the desired size, it remains to show that it is an independent set in G .To establish this we need to show that the transformation to G does not add any edges between vertices of I G .This is ensured because all edges that are added by the transformation have at least one endpoint which is a neighbor of u: all added edges are either incident on w or a vertex in N G (u) ∩ X. Hence for each added edge one endpoint z is adjacent to u, and since we assumed u ∈ I G this implies that z cannot be in I G since I G is a subset of the independent set I G in G and having adjacent vertices u and z in I G would violate independence.Therefore I G is indeed an independent set of the required size in G .
(⇐) Let I G be an independent set for graph G of size at least k − 1.We show how to obtain an independent set I G for graph G of size at least |I G | + 1 ≥ k.The structure of I G determines how to augment to a larger independent set I G .From the structure of the reverse transformation of G to G it follows that I G is an independent set in G; hence for each case we will only show that the new vertex we add to the set will not violate independence in graph G.We now do a case analysis based on whether or not the neighbors t of u and w of v are present.
If vertex t exists and t ∈ I G , then define The correctness argument is symmetric to that of the previous case.
In the remaining case we know that {t, w} ∩ I G = ∅.There must be some z ∈ {u, v} Since N G (z) ∩ X ∩ I G = ∅ by our choice of z this proves that the addition of z to the independent set does not violate independence, because N G (z) ⊆ (N G (z)∩X)∪{t, u, v, w}.Since the case distinction is exhaustive this establishes the claim in this direction, which concludes the proof.Lemma 15.Let (G, X, k) with F := G − X be an instance to which Rule 6 is applicable at vertices t, u, v, w, and let (G , X, k − 2) be the instance resulting from the reduction.Then it holds that α(G) Proof.Assume the conditions in the statement of the lemma hold.We prove the two directions separately.
(⇒) Let I G be an independent set for graph G of size at least k.We show how to obtain an independent set We first show that without loss of generality we may assume that for one of the pairs {t, w}, {t, v}, {u, w} both vertices of the pair belong to I G .To see this, suppose that I G avoids at least one vertex in each pair.We then obtain an alternative independent set I G which is at least as large, and contains both vertices of at least one pair. If which is easily seen to be an independent set.Since no independent set can contain three or more vertices from {u, v, t, w} (because of the edges {u, t} and {v, w}) we now have for if both sets are non-empty, then taking one vertex from I G ∩ X ∩ N G (t) and one vertex from I G ∩ X ∩ N G (w) yields a pair which shows that {t, w} is X-blockable, which contradicts the preconditions to Rule 6.Using the same argument we must have that and we set I G := (I G \ {q, u, t, v, w}) ∪ {t, v}.The correctness argument is symmetric to that of the previous case.The argument above shows that we may assume without loss of generality that for one of the pairs {t, w}, {t, v}, {u, w} the independent set I G contains both vertices of the pair.Using this assumption we show how to obtain an independent know that all vertices in I G still exist in G .It remains to show that they form an independent set there.Because the reduction to G only adds edges incident on p and q, it suffices to show that for all edges incident on p or q which are added by the reduction there is at least one endpoint not in I G .The transformation from G to G adds edges from N G (t) ∩ X to p, and edges from N G (w) ∩ X to q.But since t, w ∈ I G we know that the independent set For all edges that we add, at least one endpoint is not in I G and therefore not in The proof of correctness is symmetric to that for the previous case.Since one of these cases must apply, the listing is exhaustive and it concludes the proof of this direction of the equivalence.
(⇐) Let I G be an independent set for graph G of size at least k − 2. We show how to obtain an independent set I G for graph G of size at least |I G | + 2 ≥ k.The structure of I G determines how to augment to a larger independent set I G by adding two vertices to I G .From the structure of the reverse transformation of G to G it follows that I G is an independent set in G; hence for each case we will only show that the new vertices we add to the set will not violate independence in graph G.
w}.Since vertices t, w are clearly non-adjacent in G, and because the vertices in I G form an independent set in G (as the transformation to G does not add edges between vertices in I G ) we now have that I G is an independent set in G of the required size.
would give a pair which shows that {t, w} is X-blockable in the original graph G, which contradicts the preconditions for Rule 6. Similarly we must have N G (u) ∩ X ∩ I G = ∅ by the assumption that {u, t} is not X-blockable in G. Since vertex p is adjacent in G to all vertices of N G (t) ∩ X, we know that by independence of We now set I G := I G ∪ {u, w} which must form an independent set in G because the established conditions show that none of the vertices of N G ({u, w}) can be in by the non-blockability of {w, t} and {w, v}.We assign I G := I G ∪ {t, v}.The correctness proof is symmetric to that of the previous case.Since the case distinction is exhaustive this establishes the claim in this direction, which concludes the proof.

B.2 Structure of Reduced Instances
The structure of reduced instances is captured by the following observations.Observation 3. Let (G, X, k) be a reduced instance with the corresponding forest F := G−X.
but degree-1 vertices are removed by Rule 1.This implies that a leaf v must be X-blockable in a reduced instance, or formally: the pair {v, v} must be X-blockable.
If there are distinct vertices t, u, v, w in F such that deg and {u, v} ∈ E(F ) then at least one of the pairs {u, t}, {v, w}, {t, w} is X-blockable by Rule 6.
For every tree T in the forest F there exists an independent subset of feedback vertices X ⊆ X with |X | ≤ 2 such that Conf T (X ) > 0 by Rule 4.

B.3 Outline of the proof of Lemma 8
The next subsections lead up to the proof of Lemma 8, which is re-stated and proven in this appendix as Lemma 31.As the proof is fairly substantial, we first sketch an outline of the proof.The size bound is obtained by showing that if T is a tree in the forest F of a reduced instance, then Active T (X) ≥ 1 83 |V (T )|.This claim is proven in two steps.Any tree must have at least a constant fraction of its vertices on degree-2 paths of length at least 8, or it must have at least a constant fraction of its vertices as leaves.We show that for both these quantities, the number of active conflicts induced by the set X on the tree T is at least a constant fraction of these quantities, i.e. at least a constant fraction of the total length of the degree-2 paths which have size ≥ 8 (proven in Section B.4), and at least a constant fraction of the total number of leaves (proven in Section B.5). Section B.6 then shows how these lemmas can be combined to give the complete proof of Lemma 8.

B.4 Conflicts from degree-2 paths in the forest
In this section we show that the total number of active conflicts induced on a tree T in the forest F must be at least a constant fraction of the total length of the paths containing at least 8 vertices of degree 2 in the tree T .To establish this result we make extensive use of the independence decomposition of a forest as described in Proposition 1.We show how this decomposition behaves when removing certain types of edges and vertices, and we also show that under certain conditions the decomposition allows us to obtain a direct bound on the decrease in the independence number of a tree that results from deleting specified vertices from that tree.Proof.Assume the conditions in the statement of the lemma hold.We will first show the relationship between the independence numbers of H and H .Because a MIS for H can use at most one vertex of u, v since they are adjacent, we can obtain an independent set in H by removing at most 1 vertex from an independent set in H. Hence α(H ) ≥ α(H) − 1.

From Observation 1 we know that α(H ) ≤ α(H). Now suppose that α(H ) = α(H).
Then the corresponding MIS in H is also a MIS in H that contains neither u nor v, contradicting (3) of Proposition 1 since the edge {u, v} is contained in M .Therefore we must have α(H ) < α(H) and combining this with the earlier derived fact that α(H We now show that the independence decomposition A , N , S of forest H is simply the decomposition A, N, S restricted to the vertex set of H .By Proposition 1 we know that the independence decomposition of a forest is uniquely determined by adjacency relations.We will show that the suggested restriction A , N , S satisfies these adjacency relations and must therefore be the independence decomposition of H . Property (I) must still hold because we have deleted a matched edge and its endpoints, which means that the matching restricted to the remaining vertices is still a perfect matching on H[S ].Since we do not give vertices in A extra neighbors, we do not violate (II).And finally (III) cannot be violated because we only delete vertices in S, which means that all vertices in N still have the same neighbors in A as they have in A. This concludes the proof.
Using the previous lemma as an induction step, the obtain the following.Lemma 18.Let (G, X, k) be a reduced instance with forest F := G − X, and let T be a tree in F .If there is a subset of edges Z ⊆ E(T ) such that the subgraph T := T /Z satisfies α(T ) = α(T ) and the matching M from the independence decomposition of T contains q edges whose endpoints are X-blockable in G, then Active T (X) ≥ q.
Proof.Assume the conditions in the statement of the lemma hold.Consider the matching M B containing the edges in M whose endpoints are X-blockable in G.By the definition of X-blockability we know that for every edge {u, v} in M B there is a set X ⊆ X which is independent in G such that |X | ≤ 2 and {u, v} ⊆ N G (X ).For such a set X define the edges in M B hit by X as H X (M B ) := {{u, v} | {u, v} ∈ M B ∧ {u, v} ⊆ N G (X )}.We will show that for every such independent set X we have Conf T (X ) ≥ |H X (M B )|.
Consider an independent set X ⊆ X in the graph G such that |X | ≤ 2, and the subgraph T − V (H X (M B )) obtained from T by removing all endpoints of edges in We can now bound the independence number of T − N G (X ) through the following derivation.
Observe that we are justified in applying Lemma 17 where T plays the role of the forest H, and the set H X (M B ) plays the role of the set edges Y , because all edges in H X (M B ) are contained in the matching M from the independence decomposition of T .By the definition of conflicts the derivation above shows that Conf T (X ) ≥ |H X (M B )|.By the definition of active conflicts, the independent set X ⊆ X contributes to the number of active conflicts since X is either a singleton or a pair of non-adjacent vertices from X.This implies that every independent set X ⊆ X in graph G with |X | ≤ 2 contributes at least |H X (M B )| to the number of active conflicts on T .Since there are q edges in the matching M that are X-blockable (by assumption in the statement of the lemma), each of these q edges occurs in the set H X (M B ) for at least one valid independent set X and hence the total number of conflicts for all such sets X must sum to at least q.By the definition of active conflicts this shows that Active T (X) ≥ q, which concludes the proof.

Lemma 19.
Let H be a forest, and consider the independence decomposition A, N, S of the vertex set of H with the corresponding matching M on the vertices S. If P is a simple path in H on vertices v 1 , v 2 , . . ., v l such that deg H (v i ) ≤ 2 for all 1 ≤ i ≤ l then either all vertices v i are in the set S, or the vertices on P are alternatingly contained in A and in N .
Proof.Assume the conditions in the statement of the lemma hold.We will give a proof by contradiction using the adjacency relations that define the independence decomposition, as given in Proposition 1.
We first show that if P contains one vertex in S, then all vertices on P must be in S.So assume for a contradiction that P contains a vertex from S and a vertex not in S; then there are adjacent vertices x, y on P such that x ∈ S and y ∈ S. Since the sets A, N, S partition the vertex set of H, the vertex y must be in A or in N .But by (II) of Proposition 1 no vertex in A can be adjacent to a vertex in S; since we assumed that vertices x, y are adjacent and x ∈ S this shows that y ∈ A. Hence y must be contained in N .But by (III) of Proposition 1 all vertices in N must be adjacent to at least two vertices in A. Because deg H (y) ≤ 2 by the definition of the path P , and because the neighbor x of y is in S, it follows that y cannot have two neighbors in A. Therefore y cannot be in N either, which is a contradiction to the fact that A, N, S partition V (H).This concludes this part of the proof; if P contains one vertex in S, then all vertices on P must be in S.
For the second part of the proof, assume that all vertices in P are from sets A or N , but that these do not alternate; then there must be two successive vertices v i , v i+1 on the path which are contained in the same subset of the partition.By the same argument as before, no adjacent vertices v i , v i+1 can be in the set A by (II).But the vertices v i , v i+1 cannot both be in the set N because vertices in N must have at least two neighbors in A. Since v i , v i+1 have degree at most 2 in H by definition of P this cannot hold if the adjacent vertices v i , v i+1 are in N .Hence we again arrive at a contradiction, which proves the claim.

v i+1 }} and let A , N , S be the independence decomposition of H . If one of the following conditions holds:
and all vertices v i+1 , . . ., v l are in the set S .
Proof.Assume the conditions in the statement of the lemma hold.We prove the two claims successively.
(1) Since the endpoints of the edge {v i , v i+1 } are not in S, this edge is not contained in the matching M and by Lemma 13 we must have α(H) = α(H ).We will show that v i+1 ∈ S by proving that there is a MIS for H which contains v i+1 , and that there is also a MIS for H which avoids v i+1 .
By assumption that v i+1 ∈ N a MIS for H does not contain v i+1 by Proposition 1, and is also an independent set in H ; since α(H ) = α(H) this shows there is a MIS for H avoiding v i+1 .Since deg H (v i+1 ) ≤ 2 by assumption, we must have deg H (v i+1 ) ≤ 1 because we deleted the edge to its old neighbor v i .By Observation 2 this means there is a MIS for H that contains v i+1 .By the properties of the sets A , N , S stated in Proposition 1 the fact that v i+1 is contained in some but not all maximum independent sets of H proves that v i+1 ∈ S .Now consider the path P := {v i+1 , . . ., v l } in H . Since the first vertex on this path v i+1 is in S, Lemma 19 shows that all vertices on this path must be in S .This proves the claim.
(2) Now consider the second case.Since it is explictly assumed that {v i , v i+1 } ∈ M we must have α(H) = α(H ) by Lemma 13.By Lemma 19 we derive that since v i ∈ S we must have v j ∈ S for all 1 ≤ j ≤ l.To prove that the vertices v i+1 , . . ., v l are in S we will show that the partitions A, N, S and A , N , S are actually identical.By Proposition 1 the independence decomposition is uniquely determined by adjacency relations.It is not hard to verify that since the sets A, N, S satisfy the adjacency relations on graph H, the same partition must satisfy the adjacency relations on graph H because we only remove a non-matching edge between vertices in S. Since the decomposition is unique, this shows that the partition A, N, S is also the independence decomposition of H . Hence we have v i+1 ∈ S .By considering the subpath P := {v i+1 , . . ., v l } and invoking Lemma 19 we now find that all vertices on P must be in the set S , which proves the last part of the claim.

Lemma 21.
Let H be a forest, and consider the independence decomposition A, N, S of the vertex set of H with the corresponding matching M on the vertices S. Let P be a set of mutually vertex-disjoint simple paths in H such that every P ∈ P contains at least 8 vertices, each of degree at most 2 in H. Then there is a set of edges Z such that α(H) = α(H/Z) and such that the matching M Z from the independence decomposition of H/Z contains at least Proof.Assume the conditions in the statement of the lemma hold.We will prove the statement by induction on |P|.If P = ∅ then the claim is trivially true.So assume that the claim holds for all sets of paths P with |P | < |P|.Consider a tree T in H containing at least one path in P; by the assumption that P = ∅ such a tree T must exist.Choose an arbitrary leaf r ∈ V (T ) of the tree as the root.For all paths P ∈ P which are contained in the tree T , define their upper endpoint to be the endpoint of the path closest to r and the lower endpoint is the one with largest distance to r.Now take the path P ∈ P for which the upper endpoint has maximum distance from r, taken over all paths from P that lie in T and breaking ties arbitrarily.Label the vertices on the path P by v 1 , v 2 , . . ., v l by increasing distance from the root r.By Lemma 19 we know that either all vertices on P are contained in S, or the vertices of P are alternatingly contained in A and N .Based on the local structure we select an edge e to go into the set Z.
If all vertices of P are contained in S, then consider the smallest i such that the edge {v i , v i+1 } is not contained in the matching M ; because M is a perfect matching in T [S] we must have i = 1 or i = 2. Set e := {v i , v i+1 }.
If the vertices of P are alternatingly contained in A and N , then consider the smallest i such that v i ∈ A. Because the vertices alternate we have i = 1 or i = 2. Set e := {v i , v i+1 }.Consider the forest H/{e} and its independence decomposition A , N , S with matching M .By Lemma 20 we know that α(H) = α(H/{e}) and that the vertices v i+1 , . . ., v l must be in S .Consider the connected component T P of H/{e} that contains the vertex v i+1 .By the choice of the path P the component T P cannot contain any vertices for paths P * ∈ P with P * = P ; such a path would have an upper endpoint which is further from the root r than the upper endpoint of P .Now look at the subpath P := {v i+1 , . . ., v l } in H/{e}.Since all vertices of P are in the set S of the independence decomposition of H/{e}, and because M is a perfect matching on the vertices of S , the edge {v i+1 , v i+2 } must be in M because v i+1 has degree 1 in H/{e} and cannot be matched to any other vertex.This shows that for all j in the range 0, 1, . . ., (l − i)/2 − 1, the edge {v i+2j+1 , v i+2j+2 } lies on P and is contained in M , and hence there are at least (l − i)/2 such edges.Because i ≤ 2 and l ≥ 8 this shows that the component T P contains at least l/3 edges which are contained in the matching M , and for which both endpoints lie on the same path in P.
To finish the proof we apply induction, and show that we can add the edge e that was defined above to the edge set resulting from the induction step.So apply the induction hypothesis on the forest H := H − V (T P ) with the set of paths P := P \ {P } to find a set of edges Z ⊆ E(H ) such that α(H ) = α(H /Z ), and such that the matching M Z from the independence decomposition of H contains at least 1
= α(H ) + α(T P ) Since H/{e} is the disjoint union of H and T P .

= α(H/Z).
Since H/Z is the disjoint union of H /Z and T P .Now we prove that the matching M Z of the independence decomposition of H/Z must contain sufficiently many edges for which both endpoints belong to the same path in P. Since H/Z is just the disjoint union of H /Z and T P , all edges for which this condition holds in H /Z still satisfy this condition for H/Z.By induction it follows that there are at least P ∈P |V (P )| such edges.We saw earlier that the component T P contains at least l/3 of such edges, where l is the length of the path P we used for the induction step.Hence there are at least 1 To state the next lemma we need an additional definition.

Definition 22.
Let T be a tree, and let the set P(T ) contain all maximal simple paths in T containing only vertices which have degree at most 2 in T .Let P ≥8 (T ) be the subset obtained from P(T ) by removing paths which contain less than 8 vertices.
Note that the paths in P(T ) are mutually vertex-disjoint by definition.Proof.Assume the conditions in the statement of the lemma hold.From Observation 3 we know that all pairs {u, v} of adjacent vertices on paths in P(T ) are X-blockable, because these vertices all have degree at most 2 in T and hence in F .By applying Lemma 21 using the tree T as the forest H, and the set P ≥8 (T ) as the set of paths, we obtain a set of edges Z ⊆ E(T ) such that α(T ) = α(T /Z) and such that T /Z contains at least 1 3 P ∈P ≥8 (T ) |V (P )| edges that are in the independence matching of T /Z and for which both endpoints are on the same path in P(T ).Since all these edges must be X-blockable it follows from Lemma 18 that Active T (X) ≥ 1 3 P ∈P ≥8 (T ) |V (P )|.

B.5 Conflicts from leaves in the forest
In this section we show that if T is a tree in the forest F associated with a reduced instance, then the number of active conflicts induced on T must be proportional to the number of leaves in the tree.To prove this claim we first introduce five types of substructures in a tree, and show that they contribute to the number of conflicts.We then prove that the number of conflicts we can get from such structures in a reduced tree is proportional to the number of leaves of the tree, by giving a greedy algorithm to find such structures and proving that this algorithm must be able to find many such structures.Definition 24.We define five types of conflict structures for a tree T in the forest F of a reduced instance.For each such structure we also define its yield, which is a lower bound on the number of conflicts resulting from the presence of that structure.Given a subset X ⊆ X which is an independent set in G, we give conditions under which X hits a conflict structure, and the value for which the structure is hit by X .If X hits a conflict structure S for value q, then this means that if X is used in an independent set this will cause the number of vertices from S which can be used in the independent set to decrease by at least q, compared to the situation where no vertex from X is used in the independent set.Figure 3 shows examples of the structures.

1.
A conflict structure S of Type (1) consists of a parent vertex p ∈ V (T ) and all its adjacent leaves N T (p) ∩ Leaves(T ), under the condition that |N T (p) ∩ Leaves(T )| ≥ 2.
We define the yield of this structure to be |N T (p) ∩ Leaves(T )| − 1. Observe that all vertices Leaves(T ) are X-blockable by Observation 3. A subset X ⊆ X hits S for the value min(|N A conflict structure S of Type (2) consists of a leaf vertex l ∈ Leaves(T ) and its single neighbor {p} = N T (l) such that the pair {l, p} is X-blockable.The yield of this structure is 1.A subset X hits S for value 1 if {l, b} ⊆ N G (X ).
. By Observation 3 at least one of the pairs {u, t}, {v, w}, {t, w} is X-blockable in a reduced instance.The yield of this structure is 1.A subset X hits the structure for value The yield of this structure is 1.Observe that the condition deg T (t) = deg T (x) = 1 implies that both t and x are X-blockable by Observation 3. A subset X hits the structure for value Proof.Assume the conditions in the statement of the lemma hold.Consider some set X ⊆ X with |X | ≤ 2 which is independent in G.It follows from the definition of the conflict structures and Observation 3 that the sum over all such X of the values for which they hit the structures in S is at least the sum of the yields of these structures.We prove the main claim of the lemma by showing that for each set X satisfying the conditions stated above, the number of conflicts induced on T by the set X is at least the total value for which it hits conflict structures in S. Since the number of active conflicts that X induces on T is defined as the sum over all X of the number of conflicts induced by X , this will give a proof.
So consider a set X ⊆ X with |X | ≤ 2 such that X is independent in G. Let q be the total value for which X hits conflict structures in S. We prove that Conf T (X ) ≥ q, which is equivalent to showing that α(T ) − α(T − N G (X )) ≥ q and hence that α(T ) ≥ α(T − N G (X )) + q.Consider an independent set in T − N G (X ) and call this set I .We show how to construct an independent set I for tree T such that |I| ≥ |I | + q.We initialize  the independent set I as a copy of I , and then show that for each conflict structure S hit by X we can increase the size of I by the value for which X hits S. Let s be the number of conflict structures in S which are hit by X , and number these conflict structures as S 1 , . . ., S s .Let V (S i ) for 1 ≤ i ≤ s denote the vertices of the conflict structure S i .We show how to construct a series of independent sets I 0 , I 1 , . . ., I s starting with I 0 := I such that I i ∩ N G (X ) ∩ V (S j ) = ∅ for all i + 1 ≤ j ≤ s, and such that |I i+1 | − |I i | is at least the value for which X hits structure S i+1 .Since I 0 = I is an independent set in T − N G (X ) it is easy to see that I 0 satisfies the intersection requirement.Since the size of the independent set under construction will increase at each step by the value for which the corresponding conflict structure is hit by X , this ensures we will terminate with a sufficiently large independent set I s .Recall that for distinct conflict structures S i , S j we have V (S i ) ∩ V (S j ) = ∅ by the assumption that the conflict structures of S are vertex-disjoint.Given I i for 0 ≤ i < s we show how to construct I i+1 , based on the type of structure S i+1 .

1.
If S i+1 is a structure of Type (1) then we determine the contents of the set I i+1 based on whether or not p ∈ I i .If p ∈ I i then assign I i+1 := (I i \ {p}) ∪ (N T (p) ∩ Leaves(T )).It is easy to see that I i+1 is independent in this case since we explicitly remove the only neighbor p of all the leaves that we add to the set.Since p ∈ I i implies that none of the vertices in N T (p) ∩ Leaves(T ) are in I i , the size of the independent set increases by |N T (p) ∩ Leaves(T )| − 1 which is at least as much as the value for which X hits S i+1 , by definition of the value.If p ∈ I i then assign By the structure of the graph around p it is easy to see that I i+1 is independent, since the only neighbor p of the added vertices is not in I i by the case distinction.By the invariant of the construction procedure we know that I i does not contain any vertices of )) and therefore | which shows that the size increases by at least as much as the value for which X hits S i+1 .2. If S i+1 is a structure of Type (2) then assign I i+1 := I i ∪ {l}.Since {l, p} ∈ I i by the invariant since {l, p} = V (S i+1 ) ⊆ N G (X ), and because l is a leaf in T , the new set I i+1 is also an independent set in T and its size is 1 larger than the size of I i ; hence the size increase matches the value for which X hits S i+1 .

If
. Assume without loss of generality (by symmetry) that t ∈ N G (X ).By the invariant this implies that t ∈ I i , and because G contains the edges {u, v} and {w, x} we know that from each of these edges at most one endpoint is in I i .This shows that I i contains at most 2 vertices from {t, u, v, w, x}, as required.5.If S i+1 is a structure of Type (5) then assign I i+1 := (I i \{l, p, u, v})∪{l, u}.Once again it is easy to see that I i+1 is independent, and we show that I i contains at most 1 vertex from the set {l, p, u, v}.By the invariant we know that {u, v} By the existence of the edge {l, p} the set I i contains at most 1 vertex from {l, p} which proves that |I i+1 | ≥ |I i | + 1.The fact that these induction steps preserve the invariant that I i ∩ N G (X ) ∩ V (S j ) = ∅ for all i + 1 ≤ j ≤ s can be seen from the fact that the vertices which are added to I i by some update step are a subset of V (S i+1 ).Since the conflict structures in S are vertex-disjoint this shows that vertices of some set V (S j ) ∩ N G (X ) can only be part of sets I j for j ≥ j which implies that the invariant is maintained.
The incremental process shows that |I s | ≥ |I | + q, where q is the total value for which X hits the conflict structures in S and where I s is an independent set in T .As we have shown that we can transform any independent set for T − N G (X ) into an independent set for T containing at least q more vertices, it follows that α(T ) ≥ α(T − N G (X )) + q.By the definition of conflicts this implies that Conf T (X ) ≥ q.Because the sum over all subsets X satisfying the requirements must be at least as large as the total yield of the conflict structures S, this proves by the definition of active conflicts that Active T (X) ≥ S∈S Yield(S).Lemma 26.Let (G, X, k) be a reduced instance with forest F := G − X, and let T be a tree in F such that |V (T )| > 1.There is a set S of vertex-disjoint conflict structures in T such that S∈S Yield(S) ≥ 1  5 | Leaves(T )|.
Proof.We use a proof by construction which finds a set of conflict structures.The procedure grows a subtree T ⊆ T and set S incrementally, and during each augmentation step of the tree we enforce an incremental inequality which shows that the number of leaves in T which are contained in T , is proportional to the total yield of the conflict structures found so far.This proof strategy is inspired by the method of "amortized analysis by keeping track of dead leaves" which is used in extremal graph theory [23].
We use the following properties in the analysis.We say that the vertices Leaves(T ) ∩ Leaves(T ) are the closed branches of the tree T .The vertices Leaves(T ) \ Leaves(T ) are the open branches of T .Observe that when we have grown the tree T until it equals T , then for T = T the number of open branches is 0 and the number of closed branches of T is equal to the number of leaves in T .For the subtree T ⊆ T let O T denote the number of open branches of T , and let C T denote the number of closed branches.We also construct a set S of conflict structures, and enforce the invariant that at each stage of the construction for all conflict structures S ∈ S we have V (S) ⊆ V (T ) \ (Leaves(T ) \ Leaves(T )), i.e. the vertices we use in conflict structures are contained in T and are not open branches of T .Define the total current yield Y S as S∈S Yield(S).We will ensure that our augmentation operations respect the following incremental inequality: ( The ∆ values in the incremental inequality refer to the difference between the new status of T and S compared to their old status; if T has 5 open branches at a given moment, and we perform an augmentation after which it has 4 open branches then ∆O T = −1 for that step.We define the augmentations to the tree T by adding vertices to it; it will be understood implicitly that the subtree T we are considering is the subtree of T induced by the vertices we have added to T .We will show that the subtree T and the set S can be initialized and grown such that each augmentation satisfies this incremental inequality, until T = T .At that stage we will have O T = 0, and by summing the incremental inequality over all augmentation steps we then find Y S ≥ C T /5 = | Leaves(T )|/5 which will prove the claim.Hence to establish the claim all that remains is to give the initialization and augmentation operations.Figure 4 illustrates some augmentations.
We ensure that for all vertices which are internal to the tree T we have N T (v) = N T (v); i.e. if we make a vertex internal in T then we add all its neighbors to T , and not just a selected few.
Initialization.There are three possible ways to initialize the tree T , depending on the structure of T .
and add the Type (1) conflict structure S consisting of the vertices {p} ∪ (N T (p) ∩ Leaves(T )) to the set S. The number of closed branches of T is equal to It is straightforward to verify that this combination of values satisfies the incremental inequality.We leave this verification as an exercise to the reader, and do not explicitly re-state that all further augmentation operations satisfy the incremental inequality.Operation 2. If there is no vertex p as described by Operation 1 then every vertex in T is adjacent to at most 1 leaf.If T contains a leaf l with N T (l) = {p} such that the pair {l, p} is X-blockable, then initialize T := N T [p] and add the Type (2) conflict structure S Figure 4 Illustrations of some augmentation operations.The subtree T is visualized by drawing the edges in E(T ) ∩ E(T ) using thick lines, whereas the edges E(T ) \ E(T ) are thin lines.on vertices {l, p} to the set S, which results in a yield of 1. Observe that all ∆ values must be non-negative, since we are initializing the tree.Because p is adjacent to at most 1 leaf (p itself can be a leaf if T = K 2 ) we have ∆C T ≤ 2 which satisfies the incremental inequality.Operation 3. If both of the previous operations are not applicable then every vertex in T is adjacent to at most 1 leaf, and no leaf of T is X-blockable with its parent; this implies by Observation 3 that all leaves of T are adjacent to vertices of degree ≥ 3. Consider a leaf l with N T (l) = {p} and initialize T := N T [p].It follows that deg T (p) ≥ 3, and since l is the only neighbor of p which is a leaf we know that we have created at least 2 open branches and exactly 1 closed branch.
Since we assume |V (T )| > 1 one of the three given operations must be applicable to any tree T .
Augmentation.We now show that if the subtree T and set S are initialized and T = T then we can augment T by adding more vertices to it, and possibly adding more conflict structures to S, while respecting the incremental inequality.If T = T then there is some open branch of T , i.e. there is a vertex v ∈ Leaves(T ) \ Leaves(T ).We show how to augment the tree T near this vertex v.We say that a vertex u ∈ N T (v) \ V (T ) is a neighbor of u outside T , and a vertex u ∈ N T (v) ∩ V (T ) is a neighbor inside T .Since we initialize tree T to contain at least two vertices, we know that v has exactly one neighbor inside T .
It can be verified from the description of the initialization procedure that for this vertex v its single neighbor inside T cannot be a leaf of T or T ; this fact will be used later on.
When describing the augmentation steps of the subtree T we will use T a to refer to the status of the tree before the augmentation, and T b to refer to its status after the augmentation.When the intended meaning is clear from the context we will just write T .Now consider an open branch vertex v of tree T as described above.In the remaining cases we know that v is adjacent to at most 1 leaf of T .In the remaining situations we have deg T (v) = 3 and v is adjacent to at most 1 leaf of T .Operation 7. If v is not adjacent to any leaves of T , then add N T (v) to T .We have ∆C T = 0, ∆O T ≥ 0 and ∆Y S = 0.
If none of the cases above hold then we know that deg T (v) = 3 and v is adjacent to exactly 1 leaf l ∈ Leaves(T ).From the definition of the initialization procedure we know that l is a neighbor of v outside T .Hence vertex v has exactly one non-leaf neighbor w outside T .Consider the path v, w, . . . in T obtained by following degree-2 vertices until the first vertex u of degree = 2 is encountered; this implies that all the vertices which are in the interior of the unique path P uv between u, v in T have degree-2 in T , the endpoints u, v of P uv have degree = 2 in T , and v is the only vertex on this path which is in T .Observe that we could have u = w if u is a neighbor of v outside T of degree at least 3. Operation 8.If deg T (u) = 1 then the path P uv must contain at least one vertex of degree 2 in its interior, by the assumption that v is adjacent to only 1 leaf and we started tracing the path in the direction of the non-leaf neighbor w of v. Let {p} = N T (u) be the predecessor of u on P uv .Add N T (v) and all vertices on P to the tree T , and add the conflict structure of type Type (2) on vertices {p, u} to S for a yield of 1.If we consider this augmentation operation in its entirety then the leaf-neighbor l of v and the vertex u become closed branches of T .The number of open branches decreases by 1 since vertex v becomes an internal node in T b .Hence we find that ∆C T = 2, ∆O T = −1 and ∆Y S = 1.
In the remainder we may assume that deg T (u) ≥ 3. Operation 9.If u is adjacent to at least 2 leaves of T , then these must be neighbors outside T by the structure of the initialization.Add N T (v) and all vertices on the path P uv and the vertices N T (u) to the tree T , and create a conflict structure of Type ( 1 In the remainder we know that u is adjacent to at most 1 leaf of T .Operation 10.If u is not adjacent to any leaves or deg T (u) ≥ 4 then add all vertices N T (v), all vertices on P uv and the vertices N T (u) to T .We find that ∆C T ≤ 2, ∆O T = |N T (u) \ Leaves(T )| − 2 ≥ 1 and ∆Y S = 0. Operation 11.In the final situation we know that u is adjacent to exactly 1 leaf and deg T (u) = 3.We now augment and create a conflict structure based on the number of degree-2 vertices on the path P uv between u and v in T .
If there are no degree-2 vertices between u and v (i.e.vertices u and v are adjacent in T ) then we add all vertices N T (v) and N T (u) to T .Consider the unique leaf w adjacent to v and the leaf t adjacent to u.By Observation 3 we know that at least one of the pairs {v, w}, {u, t}, {t, w} must be X-blockable, and hence we can add a conflict structure of Type (3) to S for a yield of 1.For this augmentation we find ∆C T = 2 since the leaves t, w are added to T , we have ∆O T = 0 because one neighbor of u will be a leaf in T b but not in T (which compensates for the fact that v ceases to be an open branch), and ∆Y S = 1.If there is exactly 1 degree-2 vertex between u and v then call this vertex w.Add all vertices N T (v), all vertices on P uv and the set N T (u) to T .Let t be the unique leaf adjacent to u and let x be the unique leaf adjacent to v. By Observation 3 both vertices t and x are X-blockable, which means we may add a conflict structure of Type (4) consisting of vertices t, u, w, v, x to S for a total yield of 1.We have ∆C T = 2, ∆O T = 0 and ∆Y S = 1 which suffices.
In the remaining situations there must be at least two degree-2 vertices between u and v on the path P uv .Let x, y be the first two degree-2 vertices succeeding v on this path, which means that N T (x) = {v, y}.We add the vertices N T (v) ∪ {x, y} ∪ N T (y) to T .Since x, y have degree-2 in T we know from Observation 3 that the pair {x, y} is X-blockable.Therefore we may add a conflict structure of Type ( 5): let l be the unique leaf of T adjacent to u, and add the conflict structure on vertices {l, u, x, y} to S for a yield of 1.We find that ∆C T = 1, ∆O T = 0 and ∆Y S = 1.This concludes the description of the augmentation operations.Although we have not explicitly argued that each operation maintains the invariant that for all confict structures S that we add to S we have V (S) ⊆ V (T ) \ (Leaves(T ) \ Leaves(T )), this is straight-forward to verify.From this invariant it also follows that the resulting set of conflict structures is vertex-disjoint: the conflict structures that we add consist of vertices which were either open branches of T before the augmentation or were not yet contained in T prior to the augmentation.Because the case distinction is exhaustive we have shown that whenever T is not yet equal to T we can augment the tree T while respecting the incremental inequality.By the arugment given above this proves that the resulting set of conflict structures S satisfies S∈S Yield(S) ≥

B.6 Putting It Together
In this section we use the results from the previous sections to obtain a proof of Lemma 8.

Lemma 28.
If T is a tree then for all constants 0 < c < Proof.Let T be a tree.We will show that Consider the subtree T obtained from T by repeatedly deleting a degree-2 vertex v with N (v) = {u, v} and replacing it by a direct edge {u, v}.From this definition it follows that Leaves(T ) = Leaves(T ) and V ≥3 (T ) = V ≥3 (T ).Since T does not contain any degree-2 vertices we know that V (T ) = Leaves(T ) ∪ V ≥3 (T ).We partition the edge set of T into E ≥8 and E <8 for a counting argument, based on how many degree-2 vertices were deleted to form the edge.
An edge {u, v} ∈ E(T ) belongs to the set E ≥8 if the unique path between u and v in T contains at least 8 vertices of degree 2, and it belongs to E <8 otherwise.Let P T (u, v) for {u, v} ⊆ V (T ) denote the unique simple path in T between u and v, and let V (P T (u, v)) be the vertices on that path, including the endpoints.Now define: Since every degree-2 vertex in T must lie on the interior of some path P T (u, v) for vertices u, v of degree unequal to 2, it follows that the sets V (E ≥8 ) and V (E <8 ) partition the set of degree-2 vertices in T .Hence the vertex set of T is partitioned into Leaves(T ), V ≥3 (T ), V (E ≥8 ) and V (E <8 ).Since all the vertices in V (E ≥8 ) lie on a path consisting of at least 8 degree-2 vertices, we know that all paths containing these vertices are in the set P ≥8 (T ) and therefore |V (E ≥8 )| ≤ P ∈P ≥8 (T ) |V (P )|.Since every edge in E <8 is the result of deleting at most 7 degree-2 vertices we have |V (E <8 )| ≤ 7|E(T )|.We now have sufficient information to complete the proof.
Proof.The first claim follows from the fact that a MIS in a forest is simply the union of maximum independent sets for the trees in the forest.The second claim follows directly from the first by the definition of active conflicts.

B.7 The kernelization algorithm
We now give the proof of Theorem 9 which was omitted from the main text.
Proof of Theorem 9. We sketch the algorithm and then prove that it is correct.When given an input (G, X, k) the algorithm exhaustively applies the reduction rules, until these no longer change the graph.Each reduction rule can be applied and tested for applicability in polynomial time.For Rule 1 this follows from Theorem 1, and for the other reduction rules this is easy to verify since it involves finding and manipulating bounded-size bounded-degree substructures in the graph, and computing the independence numbers of forests.Each application of a reduction rule either deletes vertices and their incident edges, or adds edges to the graph.We may assume that the target size k for the vertex cover is less than n, otherwise it is a trivial yes instance.We now describe the vector pairs in the Paired Vector Agreement instance.Each vector pair corresponds to a vertex, so t := n.The first vector of a pair will represent the choice of taking the vertex in the vertex cover, and the second vector represents the choice to avoid it.We set k := k.There are vectors (a i , b i ) for 1 ≤ i ≤ n.The pairs of vectors in the list L have length m := n 2 + n each.

C
Vector a i has a # at position n 2 + i, and ?elsewhere.Vector b i has a 0 at positions [(i − 1)n + 1 . . .(i − 1)n + n], has a 1 at all positions [(j − 1)n + 1 . . .(j − 1)n + n] for which vertex j is a neighbor in G of vertex i, and has ?elsewhere.This concludes the description of the Paired Vector Agreement instance (L , t , m , k ).It is trivial to see that the instance can be constructed from (G, k) in polynomial time.We now prove that the two instances are equivalent: we can select one vector from each pair for a total of at most k conflict positions ⇔ graph G has a vertex cover of size at most k.We prove the two directions separately.
(⇒) Suppose that we can select one vector from each pair such that the chosen vectors induce at most k conflict positions.Look at two vertices i, j which are adjacent in G, and consider the effect of choosing the b-vectors b i and b j for both these vertices.By the definition of the b-vectors we know that all positions [(i − 1)n + 1 . . .(i − 1)n + n] are a 1 for vector b i , but since j is adjacent to i these positions are set to 0 in vector b j ; hence if we choose b i and b j simultaneously then all these positions are conflict positions.Similarly all positions [(j − 1)n + 1 . . .(j − 1)n + n] must then be conflict positions.This shows that if we choose the b-vectors for two adjacent vertices, then this induces at least 2n conflicts.By our assumption that k < n this shows that whenever the chosen vectors induce at most k conflict positions, then the a-vectors which are chosen must form a vertex cover of G: if we take the b-vectors for adjacent vertices then this induces > k conflict positions.Every a-vector, say a i , that is chosen induces a unique conflict position through the # symbol at position n 2 + i, and therefore a choice of vectors with at most k conflict positions contains at most k vectors of type a.Since these vectors form a vertex cover, this proves that G has a vertex cover of size at most k.
(⇐) Now suppose that graph G has a vertex cover of size k.By the structure of the Paired Vector Agreement instance sketched previously, it follows that if we take the a-vectors of all vertices in the cover, and the b-vectors for all other vertices, then we have a vector set which induces at most k conflicts; one conflict for each chosen a-vector.This concludes the proof.

Lemma 33. t-Paired
Vector Agreement is or-compositional: there is an algorithm that receives s instances of t-Paired Vector Agreement with equal value t of the parameter, and outputs an instance (L * , t * , m * , k * ) in time polynomial in the total size of the input instances, such that (L * , t * , m * , k * ) is a yes-instance if and only if one of the input instances is a yes-instance, and such that t * ≤ 2t.
Proof.Suppose we are given s instances of t-Paired Vector Agreement that all have the same parameter value t.It is easy to see that we can solve an instance of t-Paired Vector Agreement in O * (2 t ) time by exhaustively trying all ways of choosing one vector from each pair.This means that if s ≥ 2 t then the composition algorithm can successively solve each instance exactly, for a total running time of O * (s2 t ) which is polynomial in the total input size if s ≥ 2 t .We can then output a constant-sized yes instance if any of the input instances is a yes-instance, and otherwise output a constant-sized no instance.So in the remainder we may assume that s < 2 t .We now pad the instances to make it easier to compose them together.If the number of input instances s is not a power of 2, then we may duplicate some instances in the sequence of inputs to increase s to the next power of 2. Next we ensure that all instances look for the same number of conflicts k.We find the instance with the largest value of k, and if there is some other instance x with k < k then we append k − k elements to each vector of instance x , and give each vector a value of # in these positions; this ensures that the original version of x has a set with k conflict positions if and only if the padded version of x has a set with k conflict positions.As the final step we unify the vector lengths.If some instances use differing vector lengths then we pad vectors by appending ?-entries to the end, until all vectors have the same length.It is easy to verify that these steps do not increase the size of the input by more than a polynomial factor.
So now we may assume that we are given 2 r instances (L j , t, m, k) indexed by 0 ≤ j < 2 r for some integer r ≤ t, where all instances use the same values for t, m and k.We show how to compose all these instances into one new instance (L * , t * , m * , k * ).We have r + t pairs of vectors in our output instance, so t * = r + t ≤ 2t.The first t pairs are formed by concatenating vectors from the input instances.Pair i for 1 ≤ i ≤ t is defined as (a * i , b * i ) where a * i is obtained by concatenating all vectors a i from the sets L j for 0 ≤ j < 2 r , and similarly b * i is the concatenation of all vectors b i from L j for 0 ≤ j < 2 r .Hence all vectors have length m * := 2 r m.This construction implies that if we restrict the resulting vectors to positions [jm + 1 . . .jm + m] then we re-obtain the vectors corresponding to instance j.It remains to describe the last r pairs of vectors; these will enforce the selection of exactly 1 active instance in any solution set.So we have exactly r instance selection pairs defined as follows: The a-vector of the i-th instance selection pair has a ?at positions [qm + 1 . . .qm + m] for which the i-th bit in the binary notation of the value q is a 0, and has a # elsewhere.
The b-vector of the i-th instance selection pair has a ?at positions [qm + 1 . . .qm + m] for which the i-th bit in the binary notation of the value q is a 1, and has a # elsewhere.This gives a complete description of the instance selection pairs.The total vector set for the output instance consists of the union of the r instance selection pairs, and the t pairs that hold concatenated values for the input instances.We define the target number of conflicts as k * := (2 r − 1)m + k, which concludes the description of the output instance.It is not hard to verify that the output instance can be computed in time which is polynomial in the total size of the input instances.The new parameter value t * is clearly bounded polynomially in the old parameter value t since t * ≤ 2t.All that remains is to prove that the answer to the output instance is yes if and only if the answer to one of the input instances is yes.We will separately prove both directions of this equivalency.
(⇐) First assume that input instance j (with 0 ≤ j < 2 r ) is a yes instance; we show that the output instance is a yes instance.Consider the choice of vectors for instance j that induce at most k conflicts.We show how to choose one vector from each pair in the output instance such that the resulting set induces at most (2 r − 1)m + k conflicts.Consider the binary notation of the number j, which can be specified by r bits b r , b r−1 , . . ., b 1 .We choose vectors from the instance selection pairs according to the values of these bits.For 1 ≤ i ≤ r we choose the a-vector of the i-th instance selection pair if b i = 0, and we choose the b-vector if b i = 1.Now consider the effect of these choices on the conflict positions in the chosen vectors.For all values j = j with 0 ≤ j < 2 r , there is some bit in which the binary notation of j differs from that of j; since the vector we chose at that position corresponds to the value of j instead of the value of j , this implies that all positions [j m + 1 . . .j m + m] must be conflict positions since we have chosen a vector which has a # there.Since there are 2 r − 1 such values j = j these choices already induce (2 r − 1)m conflicts, and only the positions [jm + 1 . . .jm + m] can be conflict-free -the instance selection vectors that we chose all have ?values there.We have yet to specify a choice of vectors in the pairs containing concatenated input vectors.The choices for these pairs match the choice made in the solution for the yes instance j: if the solution to instance j chooses the a-vector of pair i, then we also choose the a-vector of concatenated pair i; otherwise we choose the b-vector there.Since the concatenated vectors, when restricted to positions [jm + 1 . . .jm + m] are equal to the input vectors of instance j, these vectors induce at most k conflicts in positions [jm + 1 . . .jm + m] (by the assumption that we started from a valid solution for yes instance j).Since the instance-selection vectors only have ?-elements in the range [jm + 1 . . .jm + m] these do not induce additional conflicts, and thus the total number of conflicts induced by this choice of vectors is at most (2 r − 1)m + k.
(⇒) Now assume that the output instance is a yes instance.Similar as before, the choice of (a, b) vectors in the instance selection pairs correspond to the binary representation of a number 0 ≤ j < 2 r and for all j = j with 0 ≤ j < 2 r we must have conflicts at positions [j m + 1 . . .j m + m] by the definition of the instance selection vectors.Hence we have (2 r − 1)m conflicts in these positions, and if the answer to the output instance is yes then it has at most k conflicts in the positions [jm + 1 . . .jm + m].But since the concatenated set of vectors restricted to the positions [jm + 1 . . .jm + m] forms exactly the vector set of instance j, it follows that instance j is a yes instance.concludes the proof of correctness of the composition algorithm.
Lemma 34.The instances (L, t, m, k) and (G , w , X , k ) in the polynomial parameter transformation of Lemma 11 are equivalent: we can choose one vector out of each pair in L for a total of ≤ k induced conflict positions ⇔ graph G has an independent set with total weight ≥ k .
Proof.We prove the two directions separately.
(⇒) Suppose S is a set containing exactly one vector from each pair in L such that the number of conflict positions induced by S is at most k.Now construct an independent set I in G as follows.Each vector corresponds 1-1 with a vertex in the feedback vertex set X of the graph G .Choose vertex v a i in I if S contains the a-vector for pair i, and choose v b i in I if S contains the b-vector of pair i.By the adjacency structure of G the set we obtain in this way is independent, and it has weight exactly t(2(t + m)) = 2t(t + m) since we have chosen one vertex from each pair for a total of t pairs, and all such vertices have weight 2(t + m) by definition.Now consider how the selected vertices in I are adjacent to position-vertices p 0 j and p 1 j .If the vectors of S have a conflict at position j, then our current set I contains neighbors of both p 0 j and p 1 j (either because we have selected a vertex corresponding to a vector with a # at position j, or because we have selected one vector with a 0 there and one with a 1).But if the vectors of S have no conflict at position j then for at least one vertex of {p 0 j , p 1 j } we have not chosen any of its neighbors in I and we may add such a vertex to I.This choice will not cause position vertices for positions j = j to obtain neighbors in I, because of the adjacency structure of the graph.This implies that for every position where the vectors S induce no conflict, we gain an additional vertex of weight-1 in the independent set.Since there are at least m − k such positions, we add at least m − k extra weight to our independent set which proves that we find an independent set of weight at least 2t(t + m) + (m − k) = k which establishes the claim in this direction.
(⇐) Suppose I is an independent set in G of weight at least k = 2t(t + m) + (m − k).We first show that I must contain exactly one vertex from each pair {v a i , v b i } for 1 ≤ i ≤ t.Observe that no independent set can contain both vertices of a pair, as the vertices are adjacent in G .Now suppose that we have an independent set for which there is a pair i such that I contains no vertex of the pair.Then this independent set has weight at most 2(t − 1)(t + m) + m because it contains at most 1 vector vertex with weight 2(t + m) of each pair, and the remaining m vertices all have weight 1.But since we assumed that k < m we then find that 2(t − 1)(t + m) + m < 2t(t + m) + (m − k), a contradiction to the assumption that the weight of I is at least k .So we know that I contains exactly one vertex from each pair {v a i , v b i }.Since the remaining vertices have weight 1, the set I must contain exactly m−k of such degree-1 position vertices.But as we showed earlier, an independent set in G can only contain one vertex from a position pair, and only if the vector-vertices that are selected do not induce a conflict there.So for the selected vector-vertices in I there are at least m − k positions where they do not induce a conflict, which implies that those vectors induce at most k conflicts.This proves the claim.
For v ∈ V (G) we denote the open and closed neighborhoods of v by NG (v) and N G [v], respectively.For a set S ⊆ V (G) we have N G (S) := v∈S N G (v) \ S, and N G [S] := v∈S N G [v].We write G ⊆ G if G is a subgraph of G.The graph G[V (G) \ X] obtained from G by deleting(a) Graph G and its independence decomposition.(b) Decomposition for graph G/{{u, v}}.(c) Decomposition for graph G − {u, v}.

Figure 1
Figure 1 Examples of the independence decomposition of a graph.Black vertices are in A, white vertices are in N , gray vertices are in S and the edges in M are drawn with thick lines.
Now we show how to transform I G−T into an independent set for G of the requested size.Let I T be a MIS in T − N G (X I ), which has size α(T − N G (X I )) = α(T ).It is easy to verify that I G−T ∪ I T must be an independent set in G because vertices of T are only adjacent to vertices of G − T which are contained in X. Hence the set I G−T ∪ I T is independent in G and it has size |I G−T | + α(T ).Since this argument applies to any independent set I G−T in graph G − T it holds in particular for a MIS in G − T , which proves that α(G) ≥ α(G − T ) + α(T ) which proves the claim.

(a) Rule 5 :Figure 2
Figure 2 Illustrations of two reduction rules.The original structure is shown on the left, and the image on the right shows the structure after the reduction.Feedback vertices X are drawn in the bottom container, whereas the forest G − X is visualized in the top container.
contradicting the precondition to the reduction rule.We now assign I G := I G ∪ {z}.Since N G (z) ∩ F ⊆ {t, u, v, w} and these vertices either do not exist in G or are not in I G by the case distinction, we know {t, u, v, w} ∩ I G = ∅.
and hence the defined set I G is an independent set in G .If t, v ∈ I G then define I G := I G \ {t, v}.All vertices in I G must exist in G since u, w cannot be in I G because their neighbors t, v are in I G .The edges we add in the transformation to G do not violate independence: because t ∈ I G we have N G (t) ∩ I G = ∅, and similarly because v ∈ I G we have N G (v) ∩ I G = ∅ which in particular means q ∈ I G .

Lemma 16 .
Let H be a forest, and consider the independence decomposition A, N, S of the vertex set of H with the corresponding matching M on the vertices S. Let {u, v} ∈ M and consider the subforest H := H − {u, v} obtained by deleting u, v and their incident edges.Then α(H) = α(H ) + 1 and the independence decomposition A , N , S of H can be obtained from A, N, S by restriction to the vertex set of H .

Lemma 17 .
Let H be a forest, and consider the independence decomposition A, N, S of the vertex set of H with the corresponding matching M on the vertices S. Let Y ⊆ M ⊆ E(H) consist of a subset of the edges matched under M , and consider the graph H := H − V (Y ) obtained from H by deleting all vertices which are endpoint of an edge in Y .Then α(H) = α(H )+|Y | and the independence decomposition A , N , S of H can be obtained from A, N, S by restriction to the vertex set of H . Proof.Assume the conditions in the statement of the lemma hold.We will prove the statement by induction on |Y |.For Y = ∅ the statement is clearly true since then H = H and |Y | = 0. Now consider the case that |Y | ≥ 1, and assume the statement has been proven for all smaller values of |Y |.Consider some edge {u, v} ∈ Y .Since {u, v} is an edge matched under M (by definition of Y ) it follows from Lemma 16 that α(H) = α(H −{u, v})+1.By the second property of the Lemma we know that the independence decomposition of H − {u, v} is just the decomposition of H restricted to the vertex set of H and that all edges in Y := Y \{{u, v}} are matched under the restriction M .Hence we may apply induction to the graph H −{u, v} with the edge set Y to establish that α(H −{u, v}) = α(H −{u, v}−V (Y ))+|Y |, which shows that α(H) = α(H − {u, v} − V (Y )) + |Y | + 1 = α(H − V (Y )) + |Y |.This proves the claim.

1 3 P
∈P |V (P )| edges for which both endpoints lie on the same path in P.

3 P 3 P
∈P |V (P )| + l/3 = 1 ∈P |V (P )| edges in H/Z which are contained in the matching M Z of the forest H/Z and for which both endpoints lie on the same path in P.This concludes the proof.

Lemma 23 . 3 P
Let (G, X, k) be a reduced instance with forest F := G − X, and let T be a tree in F .Then Active T (X) ≥ 1 ∈P ≥8 (T ) |V (P )|.

5 .
A conflict structure S of Type (5) consists of vertices l, p, u, v in V (T ) such that deg T (p) = 3, deg T (u) = deg T (v) = 2 with N T (l) = {p} and N T (u) = {p, v}.The yield of this structure is 1.Observe that the condition deg T (u) = deg T (v) = 2 implies that the pair {u, v} is X-blockable by Observation 3. A subset X hits the structure for value 1 if {u, v} ⊆ N G (X ).If S is a conflict structure then Yield(S) represents the yield value of that structure.Lemma 25.Let (G, X, k) be a reduced instance with forest F := G − X, and let T be a tree in F .If S is a set of vertex-disjoint conflict structures in T then Active T (X) ≥ S∈S Yield(S).

( a )
Structure Type (1) hit at vertices {u, v} by set X = {a} for value 2. (b) Structure Type (2) hit at vertices {l, p} by set X = {a, b} for value 1. (c) Structure Type (3) hit at vertices {t, w} by set X = {a} for value 1.(d) Structure Type (4) hit at vertex t by set X = {a} for value 1.(e) Structure Type (5) hit at vertices {u, v} by set X = {a, b} for value 1.

Figure 3
Figure 3 Illustrations of conflict structures which are hit by subsets X ⊆ X.The left image shows the original situation, whereas the right image shows the tree T := T \ NG(X ).
Figure 4 Illustrations of some augmentation operations.The subtree T is visualized by drawing the edges in E(T ) ∩ E(T ) using thick lines, whereas the edges E(T ) \ E(T ) are thin lines.(a) Tree T in the forest F .(b).Initialized subtree T by Operation 1 around p, which involves adding a conflict structure of Type (1) on vertices {p} ∪ (NT (p) ∩ Leaves(T )) = {p, a, b, c}.Vertices {d, e} are now open branches of T and {a, b, c} are closed branches.∆C T = 3, ∆O T = 2 and ∆YS = 2. (c) The vertex v is an open branch of T , with its neighbor u outside T .(d) Added {u} to T through Operation 4, which involves adding a conflict structure of Type (2) on vertices {u, v} for yield 1. ∆C T = 1, ∆O T = −1 and ∆YS = 1.(e) Result of considering the next open branch (re-labeled v) with its neighbor u, and adding {u} to T through Operation 4. ∆C T = ∆O T = ∆YS = 0. (f) Result of applying the first case of Operation 11 to the vertex drawn as v in this picture, which became an open branch after the previous operation.This operation involved adding a conflict structure of Type (3) on vertices {v, u, t, w} for yield 1. ∆C T = 2, ∆O T = 0, ∆YS = 1.(g) After applying Operation 4 for a final time, again adding a conflict structure of Type (2) for yield 1, the process is complete since T = T .∆C T = 1, ∆O T = −1 and ∆YS = 1.

Operation 4 .Operation 5 .
If deg T (v) = 2, then consider the unique neighbor u of v outside T .If u ∈ Leaves(T ) then add {u} to T ; the numbers of open and closed branches do not change.If u ∈ Leaves(T ) then add {u} to T and add the conflict structure S on vertices {v, u} of Type (2) to tree T .By Observation 3 the pair {v, u} must be X-blockable and therefore S is a valid conflict structure of yield 1. We have ∆O T = −1, ∆C T = 1 and ∆Y S = 1.So in the remaining cases the open branch of T at vertex v has deg T (v) ≥ 3.If v is adjacent to at least 2 leaves of T then add N T (v) to T and add a conflict structure of Type (1) on vertices {v} ∪ (N T (v) ∩ Leaves(T )) to S. Since the neighbor of v which is inside T a cannot be a leaf of T or T by the definition of the initialization, all leaf-neighbors of v are not in T a but are added to the tree to obtain T b .Let us consider how the number of open branches is affected by this operation.Vertex v is no longer an open branch vertex in T b , but its non-leaf neighbors which are not in T a have become open branch vertices.So we find that ∆O T = |N T (v) \ Leaves(T )| − 2, ∆C T = |N T (v) ∩ Leaves(T )| and ∆Y S = |N T (v) ∩ Leaves(T )| − 1, which satisfies the incremental inequality since |N T (v) ∩ Leaves(T )| ≥ 2.

1 5 |Lemma 27 . 5 | 5 | 5 | 5 |
Leaves(T )|.Let (G, X, k) be a reduced instance with forest F := G − X, and let T be a tree in F .Then Active T (X) ≥ 1 Leaves(T )|.Proof.Consider such a tree T .If |V (T )| = 1 then it follows from Observation 3 that there is an independent subset X ⊆ X with |X | ≤ 2 such that Conf T (X ) > 0, from which it follows that Active T (X) ≥ 1 ≥ 1 Leaves(T )| =1  5 .If |V (T )| ≥ 2 then apply Lemma 26 to tree T to obtain a set S of vertex-disjoint conflict structures in T with a total yield of at least 1 Leaves(T )|.We apply Lemma 25 with this set S which proves that Active T (X) ≥ S∈S Yield(S) ≥ 1 Leaves(T )|.
Since we can delete at most |V (G)| vertices, and since we can add only O(|V (G)| 2 ) edges, the procedure must stop after a polynomial number of applications of the reduction rules.The resulting instance is output as (G , X , k ) if k > 0; otherwise we output a constant-size yes instance.The reduction rules do not increase the values |X|, k or |V (G)| − k which proves the bounds on |X |, k and |V (G )| − k .The fact that |V (G )| ≤ 2(|V (G )| − k ) follows from Theorem 1, whereas the bound |V (G )| ≤ |X| + 83|X| 3 follows directly from Lemma 8.The correctness proofs of the reduction rules show that (G, X, k) and (G , X , k ) are indeed equivalent instances.Using the previous proof, Corollary 10 follows easily from the dual nature of Independent Set and Vertex Cover.Proof of Corollary 10.Given an instance (G, X, k) of fvs-Vertex Cover we transform it into an instance (G, X, |V (G)| − k) of fvs-Independent Set, which is an equivalent instance because the complement of an independent set is a vertex cover.We apply the kernelization algorithm from Theorem 9 to (G, X, |V (G)| − k) to compute in polynomial time an equivalent instance (G , X , |V (G )| − k ).By adjusting the target value we transform this back to an instance (G , X , k ) of fvs-Vertex Cover and use it as the output which shows that |V (G )| ≤ 2k and |V (G )| ≤ |X | + 83|X | 3 .
There is a polynomial-time algorithm that takes an instance (G, k) of Vertex Cover as input, and computes in polynomial time an equivalent instance ( Our starting point is the current-best Vertex Cover kernelization [8, Theorem 2.2] which exploits a theorem by Nemhauser and Trotter [27].Theorem 1. , v, w}) ∪ {u, w}.The neighborhood conditions show that no neighbors of u, w in X are contained in I G (and hence in I G ), and because we explicitly delete any neighbors that u, w might have in F when forming I G we see that I G is also an independent set in G.If I G ∩ X ∩ N G (t) = ∅ as specified by the precondition for this case, then we cannot have t ∈ I G because then I G would not be independent.The edges {p, u} and {v, w} in G show that of the set {p, u, v, w} at most two vertices are in an independent set; hence in this situation I G contains at most two vertices from {p, u, t, v, w} and therefore we have S i+1 is a structure of Type (3) then assign I i+1 := (I i \{u, v})∪{t, w}.Because the only neighbors of t, w in T are u, v and these vertices are explicitly removed from I i+1 it is easily seen that I i+1 is independent in T .It remains to show that |I i+1 | ≥ |I i | + 1.To establish this we will show that I i contains at most 1 vertex from {t, u, v, w}.By the definition of X hitting the conflict structure S i+1 of Type (3) we know that {u, t} ⊆ N G (X ) or {v, w} ⊆ N G (X ) or {t, w} ⊆ N G (X ).But in each case the set {t, u, v, w} \ N G (X ) contains at most two vertices, and if it contains two vertices then these are connected by an edge.Hence the independent set I i which does not contain any vertices of N G (X ) ∩ V (S i+1 ) (by the invariant) can contain at most 1 vertex from {t, u, v, w}, and thus|I i+1 | ≥ |I i | + 1. 4. If S i+1 is a structure of Type (4) then assign I i+1 := (I i \ {t, u, v, w, x}) ∪ {t, v,w}.It is easy to see that I i+1 is an independent set in T ; we will prove that |I i+1 | ≥ |I i | + 1 by showing that I i contains at most 2 vertices from {t, u, v, w, x}.By the definition of X hitting the conflict structure S i+1 we know that t

Omitted Proofs: No Polynomial Kernel for FVS-Weighted Independent Set Lemma 32
It is trivial to see that Paired Vector Agreement is in NP.To prove it is NPhard, we give a polynomial-time many-one reduction from Vertex Cover.Let (G, k) be an instance of Vertex Cover on n vertices; we construct an equivalent instance (L , t , m , k ) of Paired Vector Agreement.For simplicity we identify the vertices of G with numbers [1 . . .n].
. Paired Vector Agreement is NP-complete.Proof.