Polynomial Kernels for Hitting Forbidden Minors under Structural Parameterizations

We investigate polynomial-time preprocessing for the problem of hitting forbidden minors in a graph, using the framework of kernelization. For a fixed finite set of connected graphs F, the F-Deletion problem is the following: given a graph G and integer k, is it possible to delete k vertices from G to ensure the resulting graph does not contain any graph from F as a minor? Earlier work by Fomin, Lokshtanov, Misra, and Saurabh [FOCS'12] showed that when F contains a planar graph, an instance (G,k) can be reduced in polynomial time to an equivalent one of size $k^{O(1)}$. In this work we focus on structural measures of the complexity of an instance, with the aim of giving nontrivial preprocessing guarantees for instances whose solutions are large. Motivated by several impossibility results, we parameterize the F-Deletion problem by the size of a vertex modulator whose removal results in a graph of constant treedepth $\eta$. We prove that for each set F of connected graphs and constant $\eta$, the F-Deletion problem parameterized by the size of a treedepth-$\eta$ modulator has a polynomial kernel. Our kernelization is fully explicit and does not depend on protrusion reduction or well-quasi-ordering, which are sources of algorithmic non-constructivity in earlier works on F-Deletion. Our main technical contribution is to analyze how models of a forbidden minor in a graph G with modulator X, interact with the various connected components of G-X. By bounding the number of different types of behavior that can occur by a polynomial in |X|, we obtain a polynomial kernel using a recursive preprocessing strategy. Our results extend earlier work for specific instances of F-Deletion such as Vertex Cover and Feedback Vertex Set. It also generalizes earlier preprocessing results for F-Deletion parameterized by a vertex cover, which is a treedepth-one modulator.


Introduction
How, and under which circumstances, can a polynomial-time algorithm prune the easy parts of an NP-hard problem input, without changing its answer?This question can rigorously be answered using the notion of kernelization [2, 24,30] which originated in parameterized complexity theory [9,13] where it can be naturally framed.After choosing a complexity parameter for the NP-hard problem of interest, which associates to every input x ∈ Σ * an integer k ∈ N that expresses its difficulty under the chosen type of measurement, the theory postulates that a good preprocessing algorithm can be captured by the notion of a polynomial kernelization: a polynomial-time algorithm that, given a parameterized instance (x, k) ∈ Σ * × N, outputs an instance (x , k ) with the same answer whose size is bounded polynomially in k.Not all parameterized problems admit polynomial kernelizations, and one can find meaningful ways to preprocess an NP-hard problem by studying those parameterizations for which it does.The study of kernelization has blossomed over the last decade, resulting in a myriad of interesting techniques for obtaining polynomial kernelizations [4,16,25,32,35], as well as frameworks for proving the non-existence of polynomial kernelizations under complexity-theoretic assumptions [2,3,12,14,21].
Originally, the study of kernelization focused on the natural parameterizations of (the decision variants of) search problems, where the complexity parameter k measures the size of the solution.A classic example [8,36] is that an instance (G, k) of the k-Vertex Cover problem, which asks whether an undirected graph G has a vertex cover of size k, can efficiently be reduced to an equivalent instance with at most 2k vertices.This guarantees that efficient pruning can be done on large inputs that have small vertex covers.However, such guarantees are meaningless when the smallest vertex cover contains more than half the vertices.By choosing a parameter that measures the structure of the input graph, rather than the size of the desired solution, one can hope to develop provably good preprocessing procedures even for inputs whose solutions are large.An early example of this approach was given by Jansen and Bodlaender [27], who showed that an instance of the Vertex Cover problem can efficiently be reduced to size O( 3 ), where is the size of a smallest feedback vertex set in G: Vertex Cover parameterized by the size of a feedback vertex set has a cubic-vertex kernel.The result effectively conveys that large instances of Vertex Cover that are vertex-deletions away from being acyclic, can be shrunk to size O( 3 ) in polynomial time.
Problem statement To understand the power of polynomial-time preprocessing algorithms over inputs to NP-hard problems that exhibit some structural regularities, but whose solutions are generally large, we set out to answer the following question: For which structural parameterizations of NP-hard graph problems is it possible to obtain polynomial kernelizations?Input: A graph G, integer k, and a modulator X ⊆ V (G) such that td(G − X) ≤ η.
Question: Is there a set Y ⊆ V (G) of size k such that G − Y is F-minor-free?
The restriction that F contains only connected graphs is needed to ensure that a solution on a disconnected graph can be formed from solutions on its connected components, which we require in some of our proofs.This restriction was also considered in previous work [19] on kernelization, but can be avoided when targeting single-exponential FPT algorithms [29].
For technical reasons, we assume that a modulator X is given in the input.If no modulator is known, one can compute an approximate modulator and use it as X.For example, Gajarský et al. [22,Lemma 4.2] showed that a modulator of size at most 2 η times the optimum can be found in quadratic time.Our problem setting is related to that of Gajarský et al. [22].They studied kernelization for a general class of graph problems that includes F-Deletion, parameterized by a constant-treedepth modulator, but under the 1 Bodlaender et al. [3, Theorem 1] show a superpolynomial kernelization lower bound for Independent Set parameterized by treewidth.Since the parameter is not related to the solution size, this is equivalent to Vertex Cover parameterized by treewidth.The lower bound holds under the assumption that NP ⊆ coNP/poly, which we implicitly assume when stating further lower bounds in this section. 2The lower bound is stated for distance to treewidth two, but the same proof works for pathwidth two.
if (G , X, k − ∆) is.This effectively shows that there is an optimal solution Y on G in which the non-essential components act in isolation: Y does not delete more vertices from such a component, than would be deleted by a solution on the graph G [C].
The overall kernelization follows straight-forwardly from this pruning of non-essential components by a recursive approach, similarly as in earlier work [5,22].The main challenge is therefore to understand which components are essential and which are not, and this is where our contribution lies.We present a stand-alone combinatorial lemma that captures our key insight in this direction.To state it, we introduce some terminology.
We work with a nonstandard notion of labeled graphs.For a finite set X, an X-labeled graph is a graph in which each vertex is assigned a (possibly empty) subset of X as its labelset; we stress that multiple vertices may carry the same label on their labelset.The minor relation on graphs extends to labeled graphs in a natural way: a labeled graph H is a minor of a labeled graph G, if H can be obtained from G by repeatedly deleting an edge, deleting a vertex, deleting a label from the labelset of a vertex, or contracting an edge.When contracting an edge {u, v} into a single vertex w, the labelset of w is formed as the union of the labelsets of u and v.
For a collection S of vertex subsets Y of an X-labeled graph C, and a set of X-labeled graphs Q, we say that all Y ∈ S leave a Q-minor in C, if for all Y ∈ S the graph C − Y contains some graph H ∈ Q as a labeled minor.We say that a set Q of X-labeled graphs is θ-saturated for an integer θ, if for each subset X ⊆ X of size θ, the graph consisting of one vertex with labelset X belongs to Q.Our main lemma states that if all optimal solutions to F-Deletion on C leave a Q-minor for some suitably saturated Q, then there is a small subset Q * for which the same holds.
Lemma 3 (Main lemma).Let F be a finite set of (unlabeled) connected graphs, let X be a set of labels, let Q be a (min H∈F |V (H)|)-saturated set of connected X-labeled graphs of at most max H∈F |E(H)| + 1 vertices each, and let C be an X-labeled graph.If all optimal solutions to F-Deletion on C leave a Q-minor, then there is a subset Q * ⊆ Q whose size depends only on (F, td(C)), such that all optimal solutions leave a Q * -minor.
In several aspects, the statement in the lemma is best-possible.In particular, we will show in Section 3 that the dependence of the size of Q * on td(G) rather than tw(G) is essential and that the precondition that Q is O(1)-saturated cannot be avoided.
Lemma 3 is the cornerstone in our understanding of which components of G − X are essential.In our applications of the lemma, the graph C consists of a connected component of G − X whose labels encode the adjacency of those vertices to the modulator X.The set Q contains potential fragments of models of forbidden F-minors, again labeled by adjacency to X, which we may be interested in destroying in C so that connections through X cannot form F-minors with fragments that remain in other components of G − X.The lemma then essentially says that if it is not possible to select a solution that deletes a minimum number of vertices from C while simultaneously destroying all fragments in Q, then there is a bounded-size subset of fragments Q * that cannot all be destroyed by such a solution.The full importance of Lemma 3 will become clear in Section 4.
Organization Section 2 provides basic preliminaries.In Section 3, we give some of the main ideas of the proof of Lemma 3. In Section 4 we show how Theorem 1 follows from a procedure that identifies relevant components.We give the procedure and its correctness proof later in the same section, while relying on Lemma 3. The proof of Lemma 3 is long and technical.In the appendix, we first develop a framework for boundaried labeled graphs and establish some useful auxiliary lemmata (Section B) and finally use these to prove the main lemma (Section C).Theorem 2 is proven in Section D in the appendix.The proofs of statements marked ( ) can be found in the appendix, Section A.

Preliminaries
For a positive integer n we use [n] as a shorthand for {1, . . ., n}.For a set S, let 2 S to denote the set of all subsets of S. All graphs we consider are finite, undirected, and simple.A graph G consists of a vertex set V (G) and edge set E(G) ⊆ V (G)

2
. The open neighborhood of a vertex v is denoted N G (v) assigning a (potentially empty) subset of labels to each vertex in G.The labeled graph G is θ-restricted if each vertex has at most θ labels.
If an edge {u, v} is contracted in a labeled graph to obtain a new vertex w, then the labelset of w is defined as , and if {u, v} ∈ E(H), then there exist u ∈ ϕ(u) and v ∈ ϕ(v) such that {u , v } ∈ E(G).The third condition implies that one can find an edge mapping ψ : E(H) → E(G) such that: For all {u, v} ∈ E(H), edge ψ({u, v}) has one endpoint in ϕ(u) and the other in ϕ(v).We will often use the existence of this edge mapping in our proofs.
For S ⊆ V (H) we define ϕ(S) := v∈S ϕ(v), and we define ϕ(V (H)) as the range of the minor model.A minor model ϕ of H in G is called minimal if no minor model ϕ exists with ϕ (V (H)) ϕ(V (H)).

Definition 7 (labeled minor model).
A labeled minor model of an X-labeled graph H in an X-labeled graph G is a mapping ϕ as in Definition 6, that additionally satisfies the following: If G contains a (labeled) minor model of H, then we say that G contains H as a (labeled) minor and denote this as H m G. Observe that G contains H as a (labeled) minor if and only if H can be obtained from G by deleting edges and vertices (and potentially labels), and contracting edges.

Lemma 8 ( ).
Let G and H be unlabeled graphs, let X ⊆ V (G), and let ϕ be a minimal minor model of Two constructions of graphs and sets Q for n = 4, where no optimal F-deletion breaks Q, but for any Q ∈ Q there exists an optimal F-deletion breaking Q \ Q. Top: any solution breaking both F and Q (white vertices at the top) is larger than optF , but for any Q ∈ Q there is a solution of size optF breaking both F and Q \ {Q} (white vertices at the bottom).
We denote the size of an optimal F-Deletion solution on G by opt F (G), and the set of optimal solutions by optsol F (G).In our bounds, we use the notation O z (1) for some identifier(s) z to denote a constant that only depends on z.Lemma 9 ( ).Let F be a fixed set of (unlabeled) graphs, let η ≥ 1 be a constant, and let X be a set.For any set Q of X-labeled graphs and host graph C with td(C) ≤ η, one can: for some function f .Here L counts the number of elements of X that appear in the labelset of at least one vertex in at least one graph of Q.

Overview of the main lemma
In this section we discuss Lemma 3, whose long and technical proof is deferred to the appendix.The strength of the lemma comes from the fact that the bound on |Q * | is independent of the size of the graph C and of the number of labels |X| used on labelsets of vertices of C. The statement of Lemma 3 is best-possible in several ways.First of all, the dependence of |Q * | on td(G) instead of tw(G) is essential.In Figure 1 (left), a construction of a graph of treewidth 2 together with a set Q is shown.In this graph, no optimal {K 2 }-deletion (Vertex Cover) breaks all graphs in Q.However, for any Q ∈ Q there is an optimal vertex cover breaking Q \ {Q}.The example in Figure 1 can easily be extended to arbitrary n, showing that there is a set Q with |Q| = n such that no optimal vertex cover breaks Q, yet there is no Q * Q such that no optimal vertex cover breaks Q * .Since |Q| is not bounded in terms of tw(G) = 2 and F = {K 2 }, this shows that td(G) cannot be replaced by tw(G).
Secondly, the assumption that Q is (min H∈F |V (H)|)-saturated cannot be avoided already for F = {K 3 } (corresponding to Feedback Vertex Set).In Figure 1 (right) we show an example of a graph of treedepth 4 and a set Q of size 2n + 2 that consist of single vertices of two labels each, where we again cannot properly bound the size of Q * .The example is shown for n = 4 but can easily be generalized to arbitrary n, without increasing the treedepth.For any Q * Q there exists an optimal F-deletion breaking Q * , while |Q| is not bounded in terms of td(G) and F.
The proof of Lemma 3 follows an inductive strategy that mimics how a recursive algorithm would solve F-Deletion on a bounded-treedepth graph C. We pick a vertex v whose removal decreases the treedepth, and branch on whether v is part of the solution or not.If so, we remove v and recurse on a graph of smaller treedepth; if not, then we continue looking for solutions in which v is forbidden to be removed.The process builds up a set S with the property that removing S decreases the treedepth by |S|, and we are only interested in solutions disjoint from S. This proceeds while C − S remains connected; the branching depth is bounded since |S| ≤ td(C).When C − S becomes disconnected, we must take a more involved approach.We recurse on each of the connected components of C − S separately and find F-Deletion solutions there.But solutions for different components of C − S may not combine into a solution for C, since various fragments of F-minors left behind in different components of C − S, may be combined through their connections to S to form a forbidden minor.For this reason, when we recurse on connected components of C − S we place additional restrictions on the solutions chosen there, to ensure they also break fragments of F-minors in such a way that the solutions can be properly combined.
Our approach to bound the size of Q * is built on top of this inductive strategy.While branching over various ways to form an F-Deletion solution, we additionally branch on what fragments of labeled Q-minors are left behind by the solution in the various components of C − S. By exploiting the saturatedness of Q in a crucial way, we obtain the desired bound on |Q * |.The formalization of these ideas requires an extensive theory of how fragments of a forbidden minor in various components of C − S may combine to form a forbidden minor in C, which is developed in Appendix B.

Kernelization for F-Deletion
In this section we describe the recursive approach to kernelize the F-Deletion problem using a constant-treedepth modulator.The correctness of this strategy will crucially depend on Lemma 3. Lemma 10 identifies essential components in the input.
Lemma 10.Let F be a finite set of connected graphs and let η ≥ 1 be a constant.There is a polynomial-time algorithm that, given a graph G along with a modulator Before proving this lemma, we show how it implies Theorem 1.
Theorem 1.For every fixed finite set F of connected graphs and every constant η, the F-Deletion problem parameterized by a treedepth-η modulator has a polynomial kernelization.
Proof.Consider an input (G, X, k) to F-Deletion.The proof is by induction on η.
(η = 1) If td(G−X) = 1, then G−X is an independent set and any connected component of G − X contains one vertex.Apply Lemma 10 to find an induced subgraph G of G and integer ∆ such that opt (η > 1) For η > 1, we apply Lemma 10 on the input (G, X, k) and find G and ∆ as above.We will augment the modulator X into a superset X to ensure that td(G − X ) < η.To this end, we consider each connected component C of G − X.If C consists of a single vertex then its treedepth is already smaller than η > 1.Otherwise, C is a connected graph with more than one vertex, and by Definition 4 there is a vertex x C such that td(C − {x C }) < td(C).Since the Treedepth problem parameterized by the target width is fixed-parameter tractable [39], and η is a constant, we can find such a vertex x C by trying all options for x C and computing the treewidth of the resulting graph in f (η) • n O(1) time.(Alternatively, we can compute a treedepth-decomposition of C using the algorithm of Reidl et al. [39] and take its root as x C .)We initialize X as X.For each component C of G − X with treedepth larger than one, we add the corresponding treedepth-decreasing vertex x C to X .
Since Lemma 10 guarantees that the number of connected components of G − X is polynomial in |X| for fixed F and η, the resulting modulator X has size polynomial in |X|.Moreover, it guarantees that td(G − X ) < η.Hence we now have an instance (G , X , k − ∆) of F-Deletion parameterized by a treedepth-(η − 1) modulator, with the same answer as (G, X, k).We apply the kernel for the parameterization by a treedepth-(η − 1) modulator, which outputs an instance (G * , X * , k * ) with the same answer as (G , X , k − ∆) and therefore as (G, X, k).By induction, the size of G * is bounded by some polynomial in |X |, which in turn is bounded by a polynomial in |X|.Hence G * has size |X| O F ,η (1) for some suitably chosen constant, and we output (G * , X * , k * ) as the result of the kernelization.Now we prove Lemma 10.
Proof of Lemma 10.Let C be the connected components of G − X.To reduce their number, we have a single reduction rule stated in terms of labeled graphs.With each connected component C ∈ C, we naturally associate an X-labeled graph C L by assigning a vertex v ∈ V (C) the labelset N G (v) ∩ X.We are interested in which of these labeled graphs have optimal F-Deletion solutions that also hit certain fragments of potential F-minormodels.We therefore define a set H which is a superset of the relevant fragments.We use F as a shorthand for max H∈F |V (H)|.Let H consist of the connected F -restricted X-labeled graphs that have at most m F := max H∈F |E(H)| edges.We consider two X-labeled graphs to be identical if there is an isomorphism between them that respects the labelsets.
Proof.Graphs in H have at most m F + 1 vertices.There are less than 2 (m F +1) 2 distinct choices for the graph structure of a member of H, since there are less than 2 n 2 different n-vertex graphs.For each vertex, there are less than (|X| + 1) F choices for a labelset of size at most F .Hence each graph structure H can appear with less than ((|X| + 1) Choose γ ∈ O F ,η (1) such that Lemma 3 guarantees that for this choice of F and the treedepth bound η, one can always find (|X|).Consider the following marking procedure.
Procedure 12.For each set Q ⊆ H of size at most γ, do the following.Let Mark τ arbitrarily chosen components from C Q , or mark all of them if there are fewer than τ .
Let C ⊆ C denote the marked components, G := G[X ∪ C∈C C], and let ∆ := C∈C\C opt F (G[C]).The procedure can be executed in polynomial time, using variants of Courcelle's theorem to find the sets C Q .We explain how this is done in Lemma 9.
Since γ ∈ O F ,η (1), the number of subsets of H over which we iterate is polynomial in |H| and therefore in |X|.Since the graphs in Q are F -restricted, the number of labels involved is constant for fixed F and η, and therefore Lemma 9 guarantees a polynomial running time.
Proof.The procedure loops over |X| O F ,η (1) subsets Q.For each such set, we mark at The pair (G , ∆) is the desired outcome of Lemma 10.It remains to prove that opt F (G) = opt F (G ) + ∆.This follows from Claim 14 by induction.

Claim 14. For any unmarked component
Clearly, any solution for the graph G can be partitioned into a solution for G and a solution for We focus on proving the converse.Let Y ∈ optsol F ( G) be an optimal solution on G. Let X 0 := X \ Y and let H 0 ⊆ H contain those graphs for which the labelset of each vertex is contained in X 0 .Now define: Intuitively, one may think of Q as those labeled graphs (that represent potential fragments of forbidden F-minors) that can be realized in only few (ρ ∈ O F (|X|)) components of G − X after removing the solution Y .When lifting the solution Y in G to a solution in G by adding a solution in C * , it will be crucial to break all X-labeled minor models of Q in C * ; the fragments H 0 \ Q that remain in many different components turn out to be irrelevant.For a subset X ⊆ X 0 of labels, let I X be the labeled graph consisting of a single vertex with labelset X .Let n F := min H∈F |V (H)| and observe that n F ≤ ρ.We prove: (2) Suppose I X / ∈ Q for suitable X .Then there are ρ ≥ n F components of G − X that have I X as labeled minor after removing the solution Y .Take n F such components C 1 L , . . ., C n F L , and associate each one to a distinct vertex of X ⊆ V ( G) \ Y .The fact that I X is a labeled minor of C i L − Y for each i, implies that in each such component there is a connected vertex subset (3) This figure shows how to define HL based on H and G, and how to modify the minor model of H in G such that it uses fewer vertices of C * , in the proof of (4) in Claim 14.
To establish (3), assume that no solution of size opt F (G[C * ]) in G[C * ] breaks Q.We will use Lemma 3, together with our marking scheme, to argue for a contradiction.Observe that (2) implies that Q is an n F -saturated set of X 0 -labeled graphs.If no optimal solution on G[C * ] breaks Q, then by Lemma 3 there is a set Q * ⊆ Q of size at most γ such that no optimal solution on G[C * ] breaks Q * .Since the assumption that (3) does not hold means that the unmarked C * was eligible to be marked for the set C Q * in our procedure above, it has marked τ other components But this contradicts that Y is an optimal solution to F-Deletion on G: since F consists of connected graphs, we can form a solution Y by taking X together with a set of size opt So in the remainder we consider the case that the minor model contains at least one vertex of C * .We will build a minimal minor model of H in G using strictly fewer vertices of C * , thereby contradicting the choice of (ϕ, ψ).Consider the X 0 -labeled subgraph H L of G obtained by the following procedure, which is illustrated in Figure 2 As observed above, this subgraph is not empty.2. Remove all edges from this subgraph, except those in the range of ψ and those that connect two vertices that belong to a common branch set under ϕ.

Contract every edge between two vertices that belong to a common branch set of ϕ,
obtaining an X 0 -labeled graph H L .(Recall that labelsets merge during edge contraction.) Observe that H L has at most |E(H)| edges, since each edge remaining in H L corresponds to an edge in the range of ψ.We claim that H L is an n F -restricted graph: the labelset of each vertex has size less than n F .To see this, observe that if some vertex of H L has a labelset X ⊆ X 0 of size at least n F , then the pre-image of this vertex corresponds to a connected vertex subset To finish the argument, fix ] contains H as a minor.The role that vertices of T played in the minor model (ϕ, ψ) can be replaced by the vertices of C i L − Y : each edge of ψ that was realized between vertices of T yielded an edge of H L which is realized by a labeled and finally the connectivity of the branch sets is ensured because the labeling ensures that for all fragments of branch sets in T that were adjacent to vertices of X − Y = X 0 , the branch set of the H L -minor in C i − Y realizing that fragment is also adjacent to all those vertices of X 0 .Hence there is a minimal H-minor in G whose range is a subset of (ϕ(V (H)) \ T ) ∪ (C i − Y ).Since T ⊆ C * is not empty, this contradicts our choice of (ϕ, ψ) as a minimal H-model minimizing the intersection with C * .This concludes the proof of Lemma 10.

Conclusion
Our goal in this paper was to obtain polynomial kernelizations for a wide range of graph problems, in terms of a rich class of structural parameterizations.We obtained polynomial kernelizations for F-Deletion problems parameterized by a constant-treedepth modulator.
The kernelization algorithm as presented here is only of theoretical interest.While the kernel size is polynomial for fixed F and η, the degree of the polynomial grows very quickly with F and η.It would be desirable to have a uniformly polynomial kernel size, of the form f (F, η)|X| c for some constant c and function f .Unfortunately, Theorem 2 shows that even for the simplest choice of F, corresponding to the Vertex Cover problem, the degree of the polynomial must depend exponentially on η and no uniformly polynomial kernelization exists.The bad news also extends in the other direction: when taking the simplest choice for η and working with a treedepth-one modulator (a vertex cover), the degree of the polynomial in the kernel size for F-Deletion must depend on F [23, Theorem 1.1] and a uniformly-polynomial kernel does not exist.

A Omitted proofs from Section 2
Proof of Lemma 8. Consider a minimal minor model ϕ and let ψ : For each tree T v that consists of more than one vertex, all leaves of T v are incident on an edge in the range of ψ: if u ∈ V (G) is a leaf of T v not incident on an edge in the range of ψ, then replacing ϕ(v) by ϕ(v) \ {u} preserves connectivity of the branch set and validity of the edge mapping ψ.This yields a minor model of H in G of smaller range, contradicting the minimality of ϕ.
We call a connected component Observe that an edge ψ(e) cannot have endpoints in two different components of G − X, as the presence of such an edge would mean that they are connected and form a single component.Hence each edge of H contributes at most one terminal component, implying that the total number of terminal components is at most Consider the graph T v on vertex set X v obtained from T v by repeatedly contracting any edge that has at most one endpoint in X v , which is possible since no leaves in C by our observation above.Hence in the contraction process that turns T v into T v , the contraction of a nonterminal component C contributes at least one edge to T v .No other component can contribute this same edge, as that would contradict the fact that T v is acyclic.Hence the number of nonterminal components for v is bounded by the number of edges of T v .As any contraction of an acyclic graph is acyclic, it follows that T v is an acyclic graph on vertex set X v .Hence it has at most |X v | − 1 edges, yielding the desired bound on the number of nonterminal components for v.
Since each vertex of X belongs to at most one branch set, we have v∈V (H) |X v | ≤ |X| and hence the total number of nonterminal components is at most |X|.As each component of G−X that intersects the range of ϕ is a terminal or nonterminal component for some v ∈ V (H), this proves Lemma 8.
Proof of Lemma 9.This essentially follows from the fact that graphs of bounded treedepth have bounded treewidth, together with known meta-theorems for graphs of bounded treewidth.The second task is the more interesting one; it can be achieved by comparing the quantity opt F (C) to the smallest size of a vertex subset of C that breaks all F-minors and all Q-minors.These quantities can be computed in linear time for fixed F and η using the facts that C has constant treewidth, and that the questions can be phrased in the framework of linear Monadic Second Order Logic (MSOL) optimization extremum problems by Arnborg, Lagergren, and Seese [1, Theorem 5.6].The existence of a labeled minor in a labeled graph can be formulated in MSOL on labeled graphs.This allows the problem of finding a smallest vertex subsets that removes all such minors to be formulated as finding the smallest vertex subset that satisfies an MSOL formula whose length depends only on F and the total size of the graphs in Q, and where the number of labels in the instance depends only on the number L of labels used on graphs in Q; note that labels of C that do not occur on any graph in Q can be safely forgotten.

B Framework for boundaried labeled graphs
In this section we set up a careful framework of definitions for working with labeled boundaried graphs.We establish several of their properties, which will be used extensively in the (technical) proof of the main lemma in Section C.

B.1 Labeled and boundaried graphs
)} be the (possibly empty) set of boundary labels that are present in S. We use b G (v) as a shorthand for b G ({v}).
Note that by the above definition, every boundary vertex has a unique number between 1 and t, but not every integer between 1 and t corresponds to a vertex in the boundary of G. Furthermore, every t-boundaried graph, is by definition also a (t + 1)-boundaried graph.
For ease of presentation, we will often not define the boundary function of a t-boundaried graph explicitly.Only when relevant will we consider the values of b G (v), but it should be understood that all t-boundaried graphs have such a boundary function.Edges between boundary vertices cannot be contracted.When contracting an edge incident with exactly one boundary vertex, the vertex resulting from the contraction takes over the boundary role.
An X-labeled boundaried graph has both a label and a boundary function associated with it.As with the boundary function, we will only make the presence of the labeling function explicit when it is relevant to do so.When contracting an edge {u, v} in a labeled graph, the vertex resulting from the contraction is labeled by the union of the labelsets of u and v.
Definition 16 (isomorphism).We extend the definition of graph isomorphism to boundaried labeled graphs as follows.We say two t-boundaried X-labeled graphs G and G are isomorphic if there is an isomorphism f : V (G) → V (G ) for which the following additional conditions hold. For Intuitively, labeled boundaried graphs are isomorphic if there is an isomorphism that preserves the labelsets and the indices of the boundary vertices.
Definition 17 (universe of graphs).Let X be a finite set and let t ≥ 0 be an integer.We define the following sets of graphs, where two graphs are considered to be identical if they are isomorphic according to Definition 16. 1.Let G be the set of all (finite, undirected, simple) graphs.2. Let G con be the set of all connected graphs.3. Let G X be the set of all X-labeled graphs.4. Let G t be the set of all t-boundaried graphs.5. Let G t,att (for attached) be the subset of G t for which each connected component contains at least one boundary vertex.6.Let G t,def(i) be the set t-boundaried graphs, such that there exists a vertex v with b(v) = i.We will also consider combinations of these identifiers in the obvious way.For example, G X t is the set of all t-boundaried X-labeled graphs.
When convenient, we will interpret an unlabeled graph as an X-labeled graph, and a graph without boundary as a 0-boundaried graph.
Definition 18 (cc).Let G ∈ G X t be a graph, let cc (short for connected components) be defined as Note that contrary to Definition 17, if G contains multiple connected components that are isomorphic, then each of these appears as a distinct object in cc(G).

Definition 19 (boundaried minor model).
A boundaried minor model of a t-boundaried graph H in a t-boundaried graph G is a mapping ϕ as in Definition 6, that additionally satisfies the following for all v ∈ V (H): A boundaried labeled minor model simultaneously satisfies the conditions of Definitions 7 and 19.Refer to Figure 3 for an illustration of a boundaried labeled minor model.Using this notion of boundaried minors, we can define solutions to constrained versions of the F-Deletion problem.

Definition 20 (opt F ).
Let G ∈ G t with boundary set S, let F ⊆ G, and let Π be a set of t-boundaried graphs with boundary S. Define G − Y has no graph in Π as boundaried minor}.
Define opt F (G) := opt F (G, ∅, ∅), or simply the size of an optimal F-minor free deletion in G.
When analyzing the structure of F-Deletion problems, it will be convenient to argue about all possible graphs that can be obtained as a minor.The following notion captures this concept.Definition 21 (folio).For G ∈ G X t , the folio of G consist of all minors of G: The folio of an unlabeled graph, or an unboundaried graph, is defined analogously.Observe that the forget operation is only used to forget the boundary status of a vertex; it is not used to omit labels from a labelset.
We now introduce the ext (for extend) operation, which is illustrated in Figure 4.It is useful when translating the question whether H is an ordinary minor of G, into a question whether an extended version of H is a minor of a t-boundaried version of G in which t distinct vertices have been selected as the boundary.The presence of the boundary can then be used in a divide-and-conquer type of approach.

Definition 23 (ext). Let H ∈ G X
t for some t ≥ 0. Let ext +1 (H) (short for extend) be the set of all (t + 1)-boundaried graphs H that can be obtained from H by using exactly one of the following steps: Let H be equal to H, thereby forming H as a (t + 1)-boundaried graph in which there For each label on the labelset of u, either keep it on u or move it to the labelset of v. Define ext +t (H) as the set of (t + t )-boundaried graphs that can be obtained from H by applying exactly t of such operations in a row.The extend operation for unlabeled graphs is defined analogously, with the exception that there are no labels to be divided in the uncontract step.For a set of graphs Q, define ext +1 (Q) := Q∈Q ext +1 (Q), and ext +t (Q) analogously.

Polynomial Kernels for Hitting Forbidden Minors under Structural Parameterizations
Figure 5 This figure illustrates the reason for having an ext operation, that can split vertices.We show graph H, and 2-boundaried graph G.Here H m G, but H cannot be obtained from graph G using minor operations (and adding a boundary vertex), since the edge between boundary vertices 1 and 2 cannot be contracted.
Observe that, in terms of graph structure, the operations above can be reversed by minor operations that contract an edge (in the third case).This implies that, for all H ∈ G X t and H ∈ ext +t (H) we have that H m forget(H , t): after forgetting the boundary status of the last t boundary vertices, the resulting t-boundaried labeled graph has H as a minor.
The next lemma follows quite easily from the above definitions.It shows that a tboundaried graph H can be obtained from an unboundaried graph by t + 1 extend operations if and only if it can be obtained by t extend operations.It is later used to justify a technical step in the proof of the main lemma.The labelsets of all vertices of H match the labelsets of the corresponding vertices in H, with the exception of possibly u.In addition, we define a labelset for v.For each label on the labelset of u, if C 1 contains a vertex carrying that label, then add to the labelset of u; otherwise, one of the vertices in {b G (t)} ∪ k j=2 C j carries label and we add to the labelset of v.
This concludes the construction of H .We have H ∈ ext +((t−1−t )+1) (H ) = ext +(t−t ) (H ) since it was obtained from H by the reverse of an edge contraction; note that contracting the edge {u, v} recovers H.It remains to prove that H m G. Towards that end, we build a model (ϕ , ψ ) of H in G, as follows. For To see that a vertex x as required in the last step exists, observe that vertex x is adjacent to at least one vertex of C i for all i ∈ [k].This follows from the fact that G[ϕ(u)] is connected by definition, and C 1 , . . ., C k are the components that result from that connected graph by removing vertex x.The same argument shows that G[ϕ (v)] is connected.We trivially satisfy the requirement that the branch set of v contains x, and have y ∈ C 1 = ϕ (u).Using these facts, it is easy to verify that (ϕ , ψ ) is a valid labeled boundaried minor model of H in G, which concludes the proof.

B.2 Summing pieces of a boundaried graph
We will now give some useful properties of boundaried graphs.Boundaried graphs can be summed together using the following notion.

Definition 26 (⊕
We stress that no parallel edges are introduced in this step.
For a set P = {p 1 , . . ., p k } of t-boundaried graphs, define p∈P p as p 1 ⊕ p 2 ⊕ . . .⊕ p k .For P = ∅, we define p∈P p as the empty graph.
The following notion will be instrumental to analyze how a large t-boundaried graph can be formed by gluing together pieces of smaller t-boundaried graphs.Refer to Figure 6  From these definitions, it follows that p∈pcs(G) p = G for all G ∈ G X t .The following lemma shows the key property of this definition of pcs.
Lemma 28.Let H, G 1 and G 2 be t-boundaried X-labeled graphs.

Proof. Suppose ∃P ⊆ pcs(H) :
p∈P p m G 1 and p∈pcs(H)\P p m G 2 .Let H 1 := p∈P p and let H 2 := p / ∈P p.We start by showing Let ϕ 1 be a minor model of H 1 in G 1 and let ϕ 2 be a minor model of H 2 in G 2 .For any non-boundary vertex v ∈ V (H 1 ⊕ H 2 ) that was originally in H x for x ∈ {1, 2}, it is easy to see that we may define ϕ(v) := ϕ x (v).Similarly, for any boundary vertex v ∈ δ(H) that only occurs in H x for x ∈ {1, 2}, we define ϕ(v) := ϕ x (v).For any other boundary vertex ). Verify that this branch set remains connected, since branch set ϕ(v x ) for x ∈ [2] contains the vertex in G x with boundary label b H1 (v 1 ) = b H2 (v 2 ) by definition, and in G these vertices were identified.
Suppose H m G 1 ⊕ G 2 .Find a minor model ϕ with edge mapping ψ of H in G 1 ⊕ G 2 .We now show how to define P .Let p ∈ pcs(H).We consider four options: and all properties follow.Else, if ϕ 1 (v) is not connected, there are at least two parts in G 1 , that were connected in G.But this is not possible, since then ϕ(v) contained more than one boundary vertex.Non-intersecting -This follows immediately from the definition.Edges -Let {u, v} ∈ E(H 1 ).If u, v / ∈ δ(H), the result follows.Suppose u, v ∈ δ(H), then this edge exists because the part p consisting of the two boundary vertices u and v connected by an edge (namely, {u, v}) was added to P .Thereby, ψ({u, v}) = {u , v } with {u , v } ∈ E(G 1 ) and by u , v ∈ V (G 1 ) and the definition of ϕ 1 it follows that u ∈ ϕ 1 (u) and v ∈ ϕ 1 (v) as required.Suppose u / ∈ δ(H) and v ∈ δ(H).Consider ψ({u, v}) = {u , v }.Clearly, {u , v } is an edge in G.It remains to show that u , v ∈ V (G 1 ), then it follows by definition that u ∈ ϕ 1 (u) and v ∈ ϕ 1 (v).Clearly, u ∈ V (G 1 ) as ϕ 1 (u) ⊆ V (G 1 ) since u is not a boundary vertex.Furthermore, v ∈ V (G 1 ) as {u , v } is an edge, and no edges go from V (G 1 ) to G \ V (G 1 ).Thereby, u and v are in V (G 1 ) as required.Boundary -Let u ∈ δ(H 1 ).Let u ∈ G be the vertex with the same boundary label, by definition u ∈ ϕ(u).Since u ∈ δ(H 1 ), it follows from the choice of H 1 that u ∈ V (G 1 ), and thereby u ∈ ϕ 1 (u).The other properties follow immediately from ∈ δ(H 1 ) the result follows.Suppose u ∈ δ(H 1 ).By point three of the construction, there exists u ∈ ϕ(u) such that L G (u) = {x} and u ∈ V (G 1 ), thereby u ∈ ϕ 1 (u).
In combination with Lemma 28, the above lemma shows that if we are given graphs  7, where we see that the extend of H is really necessary.Let mpcs +t (Q) := mpcs(ext +t (Q)).
By these definitions, the set pcs(Q) may contain graphs that are isomorphic to each other, but the set mpcs(Q) can not.
The following observation follows immediately from the definition of mpcs.

Remove an isolated boundary vertex.
The following lemma says that if a (t + 1)-boundaried graph can be obtained by first extending a (ordinary) graph in F for t times, then restricting the result to the sum of a subset of its pieces, and extending that sum once more to obtain H , then one can also obtain H by first extending (t + 1)-times and then restricting to a subset of the pieces.Note that it applies to both labeled and unlabeled graphs, by using an empty label set for X.
Proof.By Definition 29 we have H ∈ mpcs(Q) for some Q ∈ ext +t (Q).Consider the operation that extends H to H ∈ ext +1 (H).We will mimic this operation in Q to find some Q ∈ ext +(t+1) (Q) that forms a supergraph of H respecting the boundary and labelsets; then we will argue that H can be obtained as the sum of a subset of pieces of Q and therefore H ∈ mpcs(Q ) ⊆ mpcs +(t+1) (Q).
If H was simply a copy of H, then obtain let Q equal Q.If H was obtained by setting b H (v) = t + 1 for some v ∈ V (H), then since H is the sum of some pieces of Otherwise, H was obtained by the reverse of an edge contraction operation on some boundary vertex b −1 ), and finally changing some edges that were incident on b −1 H (i) to now be incident on b −1 H (t + 1) instead; let T be the set of vertices that are adjacent to b −1 and make those vertices adjacent to b −1 Q (t + 1) instead.In all cases, we find Q ∈ ext +1 (Q) such that Q is a supergraph of H with the same boundary function and the same labelsets.By Observation 30 and H ∈ mpcs(Q), it follows H can be obtained from Q by a sequence of the described removal operations.We show that H can be obtained from Q by a similar sequence of operations, implying H ∈ mpcs(Q ) by Observation 30.This is easy to see for operations 2, 3, and 4. If H was obtained from Q by (amongst others) removing the vertices C of a connected component of Q − δ(Q), then no vertex of C was chosen as b −1 H (t + 1) and we claim that C is also the vertex set of a connected component of Q − δ(Q ).Observe that H being a component of Q − δ(Q) implies that all neighbors of C in Q belong to the boundary δ(Q).Therefore, the operation that took Q to Q was splitting a boundary vertex (causing all neighbors of C to lie in the new boundary), or turning a vertex not in C into a boundary vertex (which again does not affect the neighborhood of C).Hence the neighborhood of C in Q is a subset of the boundary, and all such C are connected components of Q − δ(Q ).They therefore correspond to pieces of Q that can be omitted from the sum of pieces if desired.It follows that H ∈ mpcs(Q ), which concludes the proof.
For the opposite direction, suppose H ∈ mpcs +t+1 (G).By definition, we can find 8 for an illustration.We will show that H ∈ mpcs(Q), to conclude the proof.To show this, we will use the alternative definition for mpcs, as given in Observation 30.We do a case distinction on the extend operation applied to obtain Q from  The next lemma shows that when starting from a set of connected graphs Q, any multi piece that is obtained by extending a graph Q ∈ Q is either attached, or consists of a real extension of Q (rather than a subset of its pieces).
that H is obtained as p∈P p for some subset P ⊆ pcs(Q ).Observe that if all connected components of Q contain a boundary vertex, then the same is true for all pieces of Q and hence for each multi piece of Q .So if Q ∈ G X t,att , then H ∈ G X t,att and we are done.Suppose Q / ∈ G X t,att , it follows that Q was obtained from Q by simply copying Q in every step of ext.Thereby, Q = Q, δ(Q ) = ∅ and it is easy to see that mpcs(Q ) = {Q } = {Q}.Thus, H = Q and thereby H ∈ ext +t (Q).

B.3 Splitting and merging families of boundaried graphs
The definitions in this section will be useful when analyzing how the task of breaking the models of a certain family Π of boundaried minors in a graph G 1 ⊕ G 2 , can be divided into breaking certain minor models Π 1 in G 1 and breaking Π 2 in G 2 .Definition 34 (split).Let Π ⊆ G t .Define split F (Π) as the set of all pairs (Π 1 , Π 2 ) with Π 1 , Π 2 ⊆ G t that can be obtained by the following procedure.Initialize Π 1 := Π and Π 2 := Π. Continue as follows.
For each G ∈ Π, consider each set Y ⊆ pcs(G), let Y := pcs(G) \ Y .Add y∈Y y to Π 1 or add y∈Y y to Π 2 (or both).
For each i ∈ {1, 2} and G ∈ mpcs +t (F), if some minor of G is in Π i , then add G to Π i .
The subscript of split F (Π) will be omitted when it is clear from the context.Observe that by the above definition Π 1 ⊇ Π and Π 2 ⊇ Π for any (Π 1 , Π 2 ) ∈ split(Π).The following operation acts as an inverse to split.

Definition 36 ( ).
Let Π 1 , Π 2 ⊆ G t .Let F ⊆ G, and define: We omit the subscript from F when it is clear from the context.
Proof.Suppose for contradiction that G has a boundaried Π 1 Π 2 -minor H.By Lemma 28, there exist P ⊆ pcs(H) such that H 1 := p∈P p m G 1 and H 2 := p∈pcs(H)\P p m G 2 .But then no minor of H 1 can be in Π 1 , no minor of H 2 is in Π 2 and H 1 and H 2 share no edges, and Lemma 39.Let Π A1 , Π A2 , Π B ⊆ mpcs +t (F) be three sets of boundaried graphs, then Proof.It is easy to see that It is easy to see from the definition that is commutative, so that it is both associative and commutative.Hence, reordering or parenthesizing an expression of the form Π 1 Π 2 . . .Π n , will not change the result.
and only if for all partitions of pcs(H) into P 1 , P 2 , P 3 , there exists i ∈ [3] such that Π i contains a minor of p∈Pi p.
Proof.This follows from Lemma 28 and Definition 36.
The following technical statement is used in the inductive proof of the main lemma.There, we will be considering F-Deletion solutions in a graph G that is decomposed as ) is connected, we progress in the induction by selecting a vertex of G A − δ(G A ) whose removal decreases the treedepth of G A − δ(G A ) and using it as the t + 1'th boundary vertex.This turns G A into a (t + 1)-boundaried graph G A , and we interpret G B and G C also as (t + 1)boundaried graphs G B , G C in which the (t + 1)'th boundary vertex is undefined.The next lemma shows that if breaking the t-boundaried graphs Π . By Lemma 40, to show that F ∈ Π 1 Π 2 Π 3 it suffices to prove that for all partitions P 1 , P 2 , P 3 of pcs(F ), there exists i ∈ [3] such that Π i contains a minor of p∈P i p.So consider an arbitrary partition of pcs(F ).Define F i := p∈P i p for all i ∈ [3]; by Definition 29 it follows that F i ∈ mpcs +(t+1) (F) if P i = ∅.Since the P i partition the pieces, we have F 1 ⊕ F 2 ⊕ F 3 = F and these three graphs are edge-disjoint.We consider the status of the (t + 1)'th boundary vertex in these graphs.
If F j ∈ G t+1,def(t+1) for some j ∈ {2, 3}, then we have F j ∈ mpcs +(t+1) (F) ∩ G t+1,def(t+1) and therefore F j ∈ Π j , by assumption.Hence Π j contains a minor of F j and we are done.In the remainder of the proof, we consider the case that the (t + 1)'th boundary vertex is left undefined in both F 2 and F 3 .
Since F ∈ ext +(t+1) (F), there is some graph F ∈ ext +t (F) such that F ∈ ext +1 (F ).Following Definition 23, we consider three cases of how F was extended to form F .In the first case, the proof is trivial.In the second and third case, we will show that there exists some i ∈ [3] and t-boundaried graph F i such that F i ∈ ext +1 (F i ) and such that Π i contains a t-boundaried minor H i of F i .This will later be shown to prove the lemma.
(No change: F := F) Suppose F equals F because the (t + 1)'th boundary vertex is left undefined in F .It follows that pcs(F ) = pcs(F ) and thereby P 1 , P 2 , P 3 form a partition of the pieces of F .It follows from Lemma 40, that there exists an i ∈ [3] such that Π i contains a minor π of F i .By definition of extend, π ∈ Π i , which concludes the proof.
(Boundary status for existing vertex) Next, consider the case that F was obtained from F by assigning b F (v) := t + 1 for some v ∈ V (F ) \ δ(F ).Recall that F 2 , F 3 do not define a (t + 1)'th boundary vertex.Since F 1 ⊕ F 2 ⊕ F 3 = F , this means F 1 contains a vertex v 1 that acts as the t + 1'th boundary vertex.Define F i := forget(F i , t), i.e., by forgetting the boundary status of the (t + 1)'th boundary vertex.It follows that F = F 1 ⊕ F 2 ⊕ F 3 , since boundary vertex t + 1 is not defined in F j for j ∈ {2, 3}.As these operations invert the extend operation, we have Since F is an unlabeled graph and F is the sum of the edge-disjoint graphs F 1 ⊕ F 2 ⊕ F 3 , there exists a partition P 1 , P 2 , P 3 of pcs(F ) such that F i = p∈Pi p for all i ∈ [3].As F ∈ ext +t (F), we have F ∈ Π 1 Π 2 Π 3 by assumption.By Lemma 40, these last two facts imply that there exists i ∈ [3] such that Π i contains a t-boundaried minor H i of F i .
(Splitting a boundary vertex) Finally, we consider the case that F was obtained from F by splitting the boundary vertex b −1 F (i) for some i ∈ [t], by the reverse of an edge contraction.This resulted in the edge {b −1 F (i), b −1 F (t + 1)} in F , and contracting this edge recovers F .Since {b F (i), b F (t + 1)} is an edge of F = F 1 ⊕ F 2 ⊕ F 3 , while the (t + 1)'th boundary vertex is undefined in F 2 and F 3 , this edge is contained in F 1 .Obtain F 1 from forget(F 1 , t) by contracting this edge.Let F j equal F j for j ∈ {2, 3}.By construction, F i ∈ ext(F i ) for each i ∈ [3].Similarly as before we find F = F 1 ⊕ F 2 ⊕ F 3 and there exists a partition P 1 , P 2 , P 3 of pcs(F ) such that F i = p∈Pi p for all i ∈ [3].Just as in the previous case, this implies that there exists i ∈ [3] such that Π i contains a t-boundaried minor H i of F i .
(Concluding the proof ) In the last two cases, we have found some i ∈ [3] and F i such that F i ∈ ext +1 (F i ) and H i m F i for some H i ∈ Π i .By Definition 23 and the fact that F i ∈ ext +1 (F i ), we have F i m forget(F i , t).Hence H i m forget(F i , t), which implies by Lemma 25 applied for X = ∅ that there exists and therefore Π i contains a minor of F i , as desired.This concludes the proof.

B.4 Starred folio
Often, we will be interested in whether graphs of a certain kind appear in the folio.We therefore introduce the following abbreviation.
Definition 42 (folio * ).For a set Q ⊆ G X con , an integer t, and G ∈ G X t , define Intuitively, folio * Q,t (G) for a t-boundaried X-labeled graph G gives the set of fragments of Q-minors that can be formed in G, so that different F-Deletion solutions leaving behind the same folio are interchangeable.The following lemma states that if we have three tboundaried labeled graphs G α , G β1 , G β2 such that the set of Q-minor fragments that can be formed in G β1 is a subset of the fragments that can be formed in G β2 , then when we glue on a third graph G α , the set of fragments that can be made in ). Then: Proof.We first prove the ⊆-part of the statement, from which we will later derive the statement about equality.So assume that folio . By Lemma 28 there is a subset of pieces P ⊆ pcs(H) such that H 1 := ( p∈P p) is a minor of G α , and As the given argument applies to arbitrary ).Now we prove the statement about equality.
), then by applying the above argument once we find folio * Applying it once more with the roles of G β1 and G β2 reversed, we obtain the containment in the other direction, and conclude that folio

C Main lemma
In this section we prove our main lemma, Lemma 3, by stating an inductive version of the lemma in Lemma 46.We use this result to prove our main lemma at the end of this section.
To state the inductive version of our main lemma, we need the following additional definition.
we will call these prohibitions), set Q of X-labeled graphs, and R B ⊆ G X t , define optsol F st (for opt.solution such that) as: has no boundaried Π i -minor for any i ∈ {A, B, C}, and The prohibitions introduced above will be used to keep track of how minor models for F are broken in graph G A ⊕ G B ⊕ G C , by keeping track of which pieces are broken in each of the three graphs G A , G B , and , can be completely determined from the folios of G A , G B , and G C .In particular, using Lemma 28, we only need to consider minors in mpcs +t (F).Similarly, to keep track of which minor models for graphs in Q are broken by optimal solutions in G, we only need to keep track of which mpcs +t (Q) remain in G A , G B , and G C , meaning that we only keep track of folio * Q,t of these graphs after removing a solution.In the inductive application, we will prescribe a certain folio * Q,t for G B and we will query for the folio * Q,t of G A and leave the solution within G C unrestrained.
Definition 45.Let two graphs be distinct when they are not isomorphic (see Definition 16).Define N ( , t, n, θ) as the number of distinct θ-restricted t-boundaried [ ]-labeled graphs on at most n vertices.
We can now state the version of our main lemma that we will use to do induction.Recall that F is defined as max H∈F |V (H)|.For a graph G, define isCon(G) = 1 if G is connected and 0 otherwise.

Lemma 46 (Main lemma (inductively)). Let X be a finite set, let G
We also say that Y is the corresponding solution for a remainder R in R 0 .Let R be defined as the set of ⊆-minimal elements from R 0 .Let R Q ⊆ R (the remainders that leave a Q-minor) be defined as Then there exist functions f and g (that do not depend on and for each R ∈ R Q there exist q ∈ Q * , r ∈ R with q m forget(r).
Before giving the proof, we will explain the lemma statement and some of the main ideas of the proof.The above statement is a generalization of Lemma 3 that is needed in order to be able to do induction.Initially, G A should be the entire graph and G B and G C should be

23:32
Polynomial Kernels for Hitting Forbidden Minors under Structural Parameterizations empty.When inductively using the lemma, parts from G A will be moved to G B and G C .We use Π A , Π B and Π C to keep track of which parts of graphs in F are broken in G A , G B , and G C by the solutions we are considering.A solution breaking Π X in G X for all X ∈ {A, B, C} is guaranteed to break F in G, by the requirement that Π A Π B Π C ⊇ ext +t (F).Since the graphs in F are unlabeled, the graphs in Π A , Π B , and Π C are also unlabeled.Graph G A is the "query" part, i.e., we want the lemma to tell us what folio * 's may remain in G A after removing a solution.Then G B is a "prescribed" part; we only want to know folio * 's in G A for solutions that have a prescribed effect in G B .The behavior of G C is unrestrained, but not part of the output; we want to know all possible folio * 's that can be generated within G A by global solutions that do exactly as prescribed in G B and are free in G C .
To see which decisions we made for G B , remainder R B is introduced.Informally speaking, this set contains all fragments of graphs in Q that solutions for G that we are currently considering, will leave behind in G B .This is relevant, since combined with graph G A these parts may form a graph in Q. R 0 will then contain sets of graphs that we call remainders.Each remainder set corresponds to the parts of Q that are left behind after deleting some optimal solution from G (one remainder may be generated by more than one optimal solution).Intuitively, if some remainder in R 0 is not minimal with respect to ⊆, the corresponding minimal remainder is preferable: if no F-Deletion solution breaks all Q-minors, then any minimal remainder contains a graph from Q. Thereby, will only consider the ⊆-minimal remainders from this set, which form R. We then promise that there exists a small subset Q * ⊆ Q that covers all remainders that have some Q-minor, which will give us Lemma 3. Furthermore we promise that the set of remainders with no Q-minor, called R N , is appropriately bounded.This last claim is only needed for the induction, as these remainders can later be combined into bigger remainders that contain Q-minors, and we need the number of options to consider to be sufficiently small.
The general proof strategy is to consider graph G A − S. If it is connected, we find a vertex s whose removal from G A − S reduces its treedepth.We consider two types of optimal solutions, namely those containing s and those not containing s.For solutions containing s, we obtain remainders by removing s from the graph and applying the induction hypothesis.For solutions avoiding s, we obtain remainders by adding s to S and applying induction.We then take the union of both.
If G A − S is not connected, we split off one connected component C. We show how to combine each remainder obtained from C with the remainders obtained from what remains of G A , when C is moved to G B and R B is adapted correctly.This requires some careful analysis, as the size bounds are not allowed to depend on the number of connected components of G A − S.
We bound |Q | and |R N | by f and g respectively, that depend on a number of parameters introduced for the induction.isCon(G A − S) is used to denote if G A − S is connected, since this case is "easier" than the disconnected case, where we need to carefully combine remainders from all different connected components, without breaking the size bounds.In the case that G A − S is not connected, we split off a component C from G A − S. It can happen that a globally optimal solution, is not locally optimal in C.However, this should not be the case for too many components, because otherwise removing S and taking an optimal solution in each component gives a smaller solution.We will use µ to keep track of the difference between the size of a globally optimal solution in G A , and sizes of the locally optimal solutions in all components of G A − S, and by the reasoning above µ should be bounded by |S|.We use ν to keep track of how restrictive prohibition Π A is, the idea is that as we split G A further, smaller bits of F should be broken in each component and thus ν will decrease.Furthermore, ξ(R B ) takes care of the complexity of R B .At some point you can add nothing new to R B without introducing Q-minors, which will help to bound ξ(R B ).The bounds depend on |S| but the size of this set will be bounded by the treedepth of G, by the precondition that td(G A − S) + |S| ≤ td(G).
Proof of Lemma 46.We will construct sets R N and Q ⊆ Q such that and for all R ∈ R Q there exist graphs q ∈ Q , r ∈ R such that q m forget(r).
We will then show that R N and Q satisfy the required size bounds.We do this by induction on the number of vertices of G A − S.

Base cases
We have a number of different base cases.
are required to always equal R B and no vertex from G A may be removed (as it only contains vertices in S).Thus, it holds that Since this holds for any two solutions, there is in fact at most one remainder R ∈ R. If R has no Q-minor, R N has size one, Q * is empty and we are done.If R does have a Q-minor, R N is empty and we add one such Q-minor to Q * such that |Q * | = 1.

µ(G
) is empty and therefore the statement is vacuously true.
Then by Definition 44, the set R 0 we consider only contains remainders corresponding to solutions Y in which q ∈ folio * Q,t (G B − Y ).Let q ∈ Q such that q ∈ ext +t (Q).It follows that q m forget(q ) and therefore that q is a minor of forget(q ) with q ∈ R for all R ∈ R 0 .Thereby, R N = ∅ and if we define Q * := {q}, it is easy to see that every remainder in R has a Q * -minor and A sketch of G and its treedepth decomposition, where GA, GB, and GC are indicated, as they could be when the lemma is applied inductively, starting from GA = G and both other graphs empty.The 4-vertex set S is the common boundary of the three subgraphs.Let X be any subset of (s) of size min H∈F |V (H)|.We show that the single-vertex graph with labelset X is a minor of G − Y for any ) and thus we can use a Q * of size one and R N = ∅.Since for each x ∈ X there is a graph in R B that has a vertex with label x in the same component as s, in G − Y vertex s is in the same component as one vertex with label x, for all x ∈ X .If we contract this component to s and delete all other components in the graph, we get exactly the desired (unboundaried) minor.Furthermore, since G − Y always has a Q * -minor, any remainder in R will contain a graph from ext +t (Q * ) by Lemma 25.

Step
Having covered the base cases, we continue by describing the induction step.Suppose the lemma statement holds for all cases where |V (G A − S)| is smaller.We do a case distinction on whether G A − S is connected or not.Refer to Figure 9 for a sketch of the situation in the graph.

S G
Figure 10 On the left the graph in which R Π A 2 ,Π A 1 is obtained is shown, on the right the graph where RΠ A 1 ,Π A 2 ,R is obtained is given.

G A − S disconnected
Suppose G A − S consists of multiple components.Order them arbitrarily, and let C be the last component.Define G A2 as the X-labeled t-boundaried subgraph of G A induced by S ∪ C, from which all edges between vertices of S have been removed.Let By this definition, G A1 ⊕ G A2 equals G A , and all edges between vertices in S are in G A1 .
In the following procedures and proofs, for ) denote the set R (respectively R 0 , R N ) obtained by applying the lemma inductively to the graph G A2 ⊕ G B ⊕ (G C ⊕ G A1 ), with G A := G A2 and the new G C as G C ⊕ G A1 and with prohibitions Π A2 and Π B and Π A1 Π C and remainder R B .Let Q Π A 2 ,Π A 1 ⊆ Q be a set Q * of bounded size as described by the lemma applied to the stated parameters.
Let RΠ A 1 ,Π A 2 ,R (respectively and with prohibitions Π A1 , Π A2 Π B , and Π C and remainder R. Let QΠ A 1 ,Π A 2 ,R ⊆ Q be a set Q * obtained with the same parameters, that has suitably bounded size.It can be found by the induction hypothesis.
See Figure 10 for a sketch of the graphs in which R Π A 2 ,Π A 1 and RΠ A 1 ,Π A 2 ,R are obtained.Using these inductively acquired objects, we define a set R N ⊆ 2 G X t by the following procedure.An intuitive description is provided below its definition.

5.
for each pair (Π A1 , Π A2 ) ∈ split(Π A ), such that there exists It is easy to verify that when the lemma is used inductively in the above procedure, all preconditions for the lemma are satisfied.
Let us try to give some intuition behind this procedure.Let R ∈ R be a remainder.We want to make sure that if R ∈ R N , it is added to R N and otherwise some Q-minor of R is added to Q .The idea is that there exists a solution ) that corresponds to R, such that we can do the following.
Clearly, there exist (Π A1 , Π A2 ) ∈ split(Π A ) such that Y breaks Π A1 in G A1 and Π A2 in G A2 .We try all such pairs in line 5.To do induction it is needed to decrease the number of vertices in G A .To achieve this, we "split off" one connected component of G A − S, namely G A2 .The idea is that when G A2 −Y has a Q-minor, then it is added to Q in line 6 and we are done.Otherwise, we move G A2 to G B , by considering the graph In line 7, we "guess" this R B .Then, the remainder R should be a remainder of G A1 ⊕(G A2 ⊕G B )⊕G C , with respect to R B and the new prohibitions (note that We will use that the induction hypothesis is applied to situations with "smaller" parameters to bound the sizes of Q and R N . At this point, one might wonder why the if-part of the above procedure (line 1 to 3) exists.The point is that in this case, we cannot guarantee that the parameters for which we apply the induction hypothesis will indeed be strictly "smaller".Luckily, in this specific situation we can "copy" all remainders from a single application of the induction hypothesis, whose parameters are not worse than the current parameters, which is enough to get the appropriate size bounds.When the if-condition does not hold and the else-part (starting in line 4) is executed, we can guarantee that the induction hypothesis is applied to "smaller" parameters than the current ones, and use this to argue that their combination will satisfy the required size bound.
The following two claims show that R N and Q satisfy the requirements given in (5) and (6).Claim 52 proves that they satisfy the desired size bound.
Claim 47. Suppose G A −S is not connected, and that there exists Before giving the proof, let us give an outline.This claim considers the situation that a solution exists in G A2 of size opt F (G A2 ) that does not intersect S and has a very large effect in breaking Π A in G A , namely by breaking Π A2 in G A2 such that Π A2 Π A ⊇ Π A .The second condition R B ∈ R Π A 2 ,Π A states that there exists a minimum F-deletion in G whose restriction to G A2 has these desired properties, and such that the remainder it leaves in G A2 ⊕ G B is exactly the same as the remainder R B (which equals the remainder that is made in G B by the solutions we are currently considering).The main idea is that under these conditions, the remainders in R can all be generated by solutions that break Π A2 in G A2 and leave remainder and we added set QΠ A 2 ,Π A ,R B to Q .To actually show these properties, let R ∈ R and let Y be its corresponding solution, such that folio We combine these solutions to obtain a solution Ŷ , that equals Y in G A1 and G C and equals Y in G A2 and G B .We denote the corresponding 7) in the proof.Since we do not necessarily know if R is minimal for this set, we replace it by minimal remainder R .For this remainder, we have added elements to R N and Q as needed.We then show that R ∈ R 0 in (8), in order to conclude that either R is smaller than R implying R is not minimal (which contradicts that R ∈ R), or R equals R and we are done.
We regularly use Lemma 38, saying that if a solution breaks Π A1 in G A1 and Π A2 in G A2 , it breaks Π A1 Π A2 in G A1 ⊕ G A2 .Furthermore, we regularly use that if a set breaks Π A Π B Π C in G, it breaks all occurrences of F, by Π A Π B Π C ⊇ ext +t (F) together with Lemma 25.We proceed by formalizing these ideas into a rigorous proof.

Proof of Claim
This is a valid solution, since it gives the same prohibitions, and it is smaller than Y , which is a contradiction.Thereby, R ∈ RΠ A ,Π A 2 ,R B 0 , which establishes (7).By (7), there exists a corresponding minimal remainder R ∈ RΠ A ,Π A 2 ,R B such that R ⊆ R. We now show that: R ∈ R 0 . (8)

23:38 Polynomial Kernels for Hitting Forbidden Minors under Structural Parameterizations
Let Y be a solution corresponding to R , thus and it was added to R N in line 2. It follows from (8) and (9) that either R is not minimal, which is a contradiction, or R ∈ R N .
Finally, we prove If R ∈ R Q , and R has no Q-minor, it follows from (8) and (9) that R was not minimal in R 0 , which is a contradiction with R ∈ R. If R does have a Q-minor, one of these minors was added to Q in line 3, and this minor is also a minor of some graph in R.
Claim 48.Suppose G A − S is not connected, and furthermore that there exists no Π A2 such that (Π A , Π A2 ) ∈ split(Π A ), opt F (G A2 , Π A2 , S) = opt F (G A2 ), and R B ∈ R Π A 2 ,Π A .Then R N ⊆ R N and for each R ∈ R Q , there exist q ∈ Q , r ∈ R such that q m forget(r).
Before stating the proof, we will first give an outline.The precondition for this claim that there exists no Π A2 such that (Π A , Π A2 ) ∈ split(Π A ), opt F (G A2 , Π A2 , S) = opt F (G A2 ), and R B ∈ R Π A 2 ,Π A , is essential to show that we perform lines 4 to 8 of the procedure, but is not needed for the proof of Claim 48; it will be used later when bounding the size of R N and Q .The claim then states that R N and Q satisfy requirements (5) and (6).Let R ∈ R be a remainder in G A ⊕ G B ⊕ G C corresponding to some solution Y , such

8.
Let G A be G A with boundary label t + 1 assigned to vertex s, thus b G A (s) = t + 1.

9.
Let (Observe that forget(G , t) = G and recall that any t-boundaried graph can be interpreted as being (t + 1)-boundaried)

10.
Compute R + N , Q + for Π A , Π B , Π C , R B on graph G with S as the boundary of G 11. Let We start by arguing that in every step where the induction hypothesis is applied, all preconditions are satisfied.In Step 3, this is trivially true, Step 10 is more interesting.Claim 49.Suppose G A − S is connected, then R N ⊆ R N and for each R ∈ R Q there exists q ∈ Q and r ∈ R such that q m forget(r).
Before giving the proof, we give a short outline.Given a remainder R, we will do a case distinction.If a corresponding solution of R contains s, it is easy to see that R is a remainder for G − {s}, thereby R ∈ R − and the result follows.If a corresponding solution of R avoids s, the proof is a bit more involved.The idea is to show that there exists a minimal remainder R in R + , such that if R has a Q-minor then R also has a Q-minor.Furthermore if R has no Q-minor, we show that R ∈ R + N , such that in line 11 of the procedure we add R to R + N .In the proof we will use implicitly that G B and G C contain no G t+1,def(t+1) -minors, which follows trivially from the definitions: boundary vertex t + 1 is not defined in G B and

23:42
Polynomial Kernels for Hitting Forbidden Minors under Structural Parameterizations G C as these are t-boundaried graphs, so any graph in which the t + 1'th boundary vertex is defined cannot be obtained from G B or G C by minor operations.
Proof of Claim 49.Let R ∈ R be a remainder with corresponding solution Y ∈ optsol F st(G A ⊕ G B ⊕ G C , Π A , Π B , Π C , R B ). Suppose s ∈ Y , it is easy to see that thereby, we applied line 3 of the procedure.Let Y := Y \ {s}.It is easy to see that R is a remainder for Y in graph G − {s}.We show R ∈ R − 0 , by showing that Y has the required properties.The only property it does not inherit from Y is that it has optimal size.However, if there exists a solution in G A ⊕ G B ⊕ G C − {s} that is smaller than |Y | − 1, we could use this solution and extend it with s to obtain a solution in the entire graph that is strictly smaller than Y , which is a contradiction.Thus, Y ∈ optsol F st(( It is easy to see that if R is not minimal for R − 0 , then it is not minimal for R 0 and then R / ∈ R, which is a contradiction.Thereby if R ∈ R N , it follows R ∈ R − N and if R ∈ R Q , then there exists q ∈ Q − ⊆ Q such that q m forget(r) for some r ∈ R.
Suppose s / ∈ Y , it follows that we applied Step 10, as Y is an example of such a solution avoiding s.
It is easy to see that Y is a solution of optimal size, as forget(G , t) = G by definition and graphs in F are not boundaried.
For a vertex set S ⊆ V (G), its open neighborhood is N G (S) := v∈S N G (v) \ S. For an edge {u, v} in a graph G, contracting {u, v} results in the graph G obtained from G by removing u and v, and replacing them by a new vertex w with N G (w) = N G ({u, v}).For a vertex set S ⊆ V (G), we use G − S to denote the graph obtained from G by deleting all vertices in S and their incident edges.The subgraph of G induced by vertex set S is denoted G[S].Definition 4 (treedepth).Treedepth is defined as follows.The trivial one-vertex graph has treedepth 1.The treedepth of a disconnected graph G with connected components C 1

Figure 3
Figure 3 Illustration of Definition 19, showing 2-boundaried {a, b}-labeled graphs H and G such that H m G, together with a boundaried labeled minor model of H in G and an edge model (the red edges).

Figure 6
Figure 6This figure shows a 2-boundaried {a, b}-labeled graph H, together with its set of pieces pcs(H).
for an illustration.Definition 27 (pcs).Let G ∈ G X t .Let pcs(G) (for pieces) contain the following tboundaried graphs.For all vertices in δ(G), pcs(G) contains a graph P consisting of a single vertex u with L P (u) := ∅ and b P (u) := b G (v).For all v ∈ δ(G), for all x ∈ L G (v), pcs(G) contains a graph P consisting of a single vertex u with L P (u) := {x} and b P (u) := b G (v).For every edge {u, v} ∈ E(G) with u, v ∈ δ(G), pcs(G) contains a graph P with vertices x and y and edge {x, y}.Define b P (x) := b P (u), b P (y) := b P (v), and L P (u) := L P (v) = ∅.For every connected component C of G − δ(G), define C as the set C together with all vertices in δ(G) that are adjacent to C. Let pcs(G) contain a graph P that is equal to G[C ] after removing all edges between boundary vertices.Remove all labels from the vertices in δ(P ).For unlabeled graphs, pcs(G) is defined analogously by treating it as a ∅-labeled graph.

Figure 7
Figure 7 This figure illustrates how the ext and mpcs operations combine.It shows G = G1 ⊕ G2 and H, the choice of H ∈ ext+2(H) such that H m G, and the choice of H1, H2 ∈ mpcs+2(H) such that H1 m G1 and H2 m G2.Minor models are shown for all the claimed minor relations.

Figure 8
Figure 8 A figure showing an example of G and H ∈ mpcs+t+1(G) for Lemma 32 and the graphs Q and Q used in the proof for t = 1, together with the relations between them.
A , Π A , S) > |S|.As F only contains connected graphs, C∈cc(G A −S) opt F (C) = opt F (G A − S) and there is anF-Deletion solution Y * in G of size |S| + opt F (G A − S) + opt F (G B − S) + opt F (G C − S)that removes all of S together with optimal solutions in the three subgraphs.If µ(G A , Π A , S) > |S|, then any F-Deletion solution in G A that breaks Π A and is disjoint from S is strictly larger than Y * , since it contains opt F

R
B has more than min H∈F |V (H)| labels.Then R N = ∅, since every remainder R ∈ R satisfies R B ⊆ R, and R B contains a graph with at least min H∈F |V (H)| labels on one vertex.Since Q is min H∈F |V (H)|-saturated, the vertex with the first min H∈F |V (H)| of these labels is in Q.Let Q * contain only this graph, it is easy to see that every remainder in R has a Q * -minor and |Q * | = 1.Suppose more than t • min H∈F |V (H)| labels are used in total.For a vertex s ∈ S, let (s) := {x ∈ X | ∃r ∈ R B : ∃v ∈ V (r) : x ∈ L r (v) and v is in the same connected component as s in graph r}.Thus, (s) contains all labels that occur in the same connected component as s in at least one graph in R B .By the assumption that R B ∩ ext +t (Q) = ∅, together with the precondition that R B ⊆ mpcs +t (Q) and Lemma 33, we find that R B ⊆ G X t,att : all connected components of all graphs in R B contain a boundary vertex.Hence every label occurring on a vertex in R B appears in a common component with some boundary vertex, which implies | s∈S (s)| > t • min H∈F |V (H)|.By the pigeonholeprinciple, there exists s ∈ S such that | (s)| > min H∈F |V (H)|.
and G A1 − Y = G A1 − Ŷ , it follows from Lemma 43 that R is a remainder corresponding to Ŷ inG A ⊕ G B .Now we show that Ŷ ∈ optsol F st(G A ⊕ G B ⊕ G C , Π A , Π B , Π C , R B ), such that R ∈ R 0 follows.By definition, Ŷ ∩ S = ∅.Furthermore, Ŷ breaks Π A2 in G A2 since Ŷ ∩ V (G A2 ) = Y ∩ V (G A2 ) which breaks Π A2 in G A2 by definition.Also, Ŷ breaks Π A in G A1 , since Ŷ ∩ V (G A1 ) = Y ∩ V (G A1 ) which breaks Π A in V (G A1 ) by definition.We conclude Ŷ breaks Π A in G A .Furthermore Ŷ breaks Π B in G B , since Ŷ ∩ V (G B ) = Y ∩ G B which breaks Π B by definition.Similarly, Ŷ ∩ V (G C ) = Y ∩ V (G C ), which breaks Π C in G C by definition.The requirement for R B is satisfied since Ŷ ∩ V (G B ) = Y ∩ V (G B) and Y satisfies the requirement for R B .That Ŷ is a solution follows from the fact that Ŷ gives the required prohibitions in G A , G B and G C .It remains to show that Ŷ has minimal size, we will use that|Y | = |Y | = opt F (G A ⊕ G B ⊕ G C ). Suppose | Ŷ | > |Y |, then |Y ∩ V (G A2 ⊕ G B )| > |Y ∩ V (G A2 ⊕ G B )|.But then Y is not minimal, as a smaller solution can be obtained by taking Y in G B and a solution breaking Π A2 in G A2 of size opt F (G A2 ).This solution breaks Π A2 Π B in G A2 ⊕ G B and Π A Π C in G A1 ⊕ G C , thus it is a valid solution.Thus, | Ŷ | = opt F (G A ⊕ G B ⊕ G C ). Thereby, R ∈ R 0 , which proves (8).Furthermore, we proveR ⊆ R ⊆ R. (9) R ⊆ R is true by definition.Since G A1 − Y = G A1 − Ŷ and folio * Q,t ((G A2 ⊕ G B ) − Ŷ ) = R B ⊆ folio * Q,t ((G A2 ⊕ G B ) − Y ), it follows that R ⊆ R from Lemma 43.We can now show that R ∈ R N ⇒ R ∈ R N .(10)If R ∈ R N , it follows that R has no Q-minor, thus R ∈ RΠ A ,Π A 2 ,R B N
Furthermore, since s / ∈ Y and Y ∩ S = ∅ it follows that Y ∩ S = ∅.Then suppose G A − Y has a Π A -minor.Since G A was obtained from G A by assigning a boundary label to an existing vertex it is easy to see that this implies G A has a Π A -minor, which is a contradiction.Furthermore, suppose G B − Y has a π -minor for π ∈ Π B .Since G B has no G t+1,def(t+1) -minors, π ∈ ext +1 (Π B ). However since G B has no G t+1,def(t+1) -minors, it follows from the definition of ext that π ∈ Π B , which contradicts the choice of Y .The same reasoning implies that G C has no Π C -minors.It remains to show the property of R B :R B = folio * Q,t+1 (G B − Y ).(17) Let r ∈ R B , hereby r ∈ folio(G B − Y ) and r ∈ mpcs +t (Q).It follows from Lemma 32 that r ∈ mpcs +t+1 (Q).Thereby, r ∈ folio * Q,t+1 (G B − Y ).Suppose r ∈ folio * Q,t+1 (G B −Y ), it follows that r ∈ folio(G B −Y ) and r ∈ mpcs +t+1 (Q).By Lemma 32, it follows that r ∈ mpcs +t (Q).Since we know that R B = folio * Q,t (G B − Y ), it follows that r ∈ R B .This shows 17, which was the last step towards proving (16).LetR := folio * Q,t+1 ((G A ⊕G B )−Y ).It follows from (16) that R ∈ R + 0 .Thereby, there exists R ⊆ R such that R ∈ R + .Let Y ∈ optsol F st(G A ⊕G B ⊕G C , Π A , Π B , Π C , R B ) be its corresponding solution, hence R = folio * Q,t+1 ((G A ⊕ G B ) − Y ).Let R := folio * Q,t ((G A ⊕ G B ) − Y ).We show that R ⊆ R. (18)Let r ∈ R, then there exists r ∈ ext +1 (r) such that r ∈ folio(G A ⊕G B −Y ) by Lemma 25.We know that r ∈ mpcs +t (Q) by definition, it follows from Lemma 31 that r ∈ mpcs +t+1 (Q).Thereby r∈ folio * Q,t+1 (G A ⊕G B −Y ) = R .Since R ⊆ R = folio * Q,t+1 (G A ⊕G B −Y ), it follows that r m G A ⊕ G B − Y .But since r m forget(r , t), it follows that r m forget(G A ⊕ G B − Y, t) = G A ⊕ G B − Y .Sincer ∈ mpcs +t (Q) by definition, it follows that r ∈ folio * Q,t ((G A ⊕ G B ) − Y ) = R, which establishes (18).We now prove R ∈ R 0 , (19) each label of X appears at least once on a vertex of S i .Considering the corresponding vertex subset in G − Y and taking into account that the labeling of C i L represents adjacency to X in G, this implies that we can contract each S i into a single vertex s i that becomes adjacent to all vertices of X .Then contract each s i into a distinct vertex of X : these minor operations on graph G − Y turn X into a clique of size n F .Hence any graph on n F vertices is a minor of G − Y , contradicting that G − Y is F-minor-free since F has a graph on n F vertices.So (2) holds.
1) we have for each graph H ∈ Q * that there are fewer than ρ components C i among C 1 , . . ., C τ for which C i L − Y contains H as a labeled minor.Since |Q * | ≤ γ, it follows that there are at most γ • ρ indices i ∈ [τ ] for which C i L − Y contains some graph from Q * as a labeled minor.But since τ = |X| + 1 + γ • ρ, there are at least |X|+1 components C i L in which all Q * -minors are broken by Y .Since no optimal solution breaks Q * in the marked components, we have | with strict inequality for at least |X| + 1 components, we have | Y | < | Y |.
* | = opt F ( G) + opt F (G[C * ]).Assume for a contradiction that G := G − ( Y ∪ Y * )contains some graph H ∈ F as a minor.Consider a minimal minor model of H in G, which is given by a vertex mapping ϕ : V (H) → 2 V ( G) , and let ψ : E(H) → E( G) be a corresponding edge mapping.Out of all possible minimal minor models of H in G, select a model (ϕ, ψ) that minimizes the quantity |ϕ Definition 15 (boundaried graph).A t-boundaried graph G is a graph with boundary set B ⊆ V (G) together with an injective boundary function b Then by definition, there exists H ∈ ext +t (G) such that H ∈ ext +1 (H ).Since the H is a t-boundaried graph, it follows that the (t + 1)'th boundary vertex is undefined and thus since H ∈ ext +1 (H ), the only operation that leaves this vertex undefined is letting H := H .It follows that H ∈ ext +t (G).The following lemma gives a key property of the extend operation.It shows that if a labeled graph H with a boundary of size t (possibly 0) is a minor of the labeled tboundaried graph forget(G, t ) obtained from G ∈ G X t by forgetting the boundary status of its last (t − t ) boundary vertices, then there is an extension H ∈ ext +(t−t ) (H) that appears as a labeled t-boundaried minor in G.To see that this extend operation is really necessary with our definition of boundaried minor models, consider Figure5.Then there existsH ∈ ext +(t−t ) (H) ⊆ G X t such that H m G.We prove the statement by induction on t − t .If t = t then the statement is trivial since forget(G, t) = G and H ∈ ext +0 (H), so the base case t − t = 0 holds.For the induction step, suppose that t > t .Recall that forget(G, t − 1) is the (t − 1)boundaried graph obtained by omitting the t'th vertex from the boundary, but keeping it in the graph.By induction, there is a graph H ∈ ext +(t−1−t ) (H) such that H m forget(G, t−1); let (ϕ, ψ) be a corresponding labeled boundaried minor model.If boundary vertex t is not defined in G, thus if there exists no vertex x ∈ δ(G) with b G (x) = t, then trivially forget(G, t − 1) = G and (ϕ, ψ) is a boundaried labeled minor model of H in G. Thereby, we can choose H := H.If x is not in the range of ϕ, then let H be H.Then H ∈ ext +1 ( H), and since H ∈ ext +(t−1−t ) (H) it follows that H ∈ ext +(t−t ) (H).A minor model of H in G is given by (ϕ, ψ). that is, x is part of the branch set of a non-boundary vertex of H, then obtain H from H by setting b H (u) = t so that u becomes the t'th boundary vertex of H . Model (ϕ, ψ) shows that H m G.(Part of a boundary branch set) Finally, if x ∈ ϕ(u) for some u ∈ δ( H), that is, x is part of the branch set of one of the at most t − 1 boundary vertices of H, then the procedure is somewhat more delicate.Let y ∈ δ(G) such that b G (y) = b H (u). Consider the connected components C 1 , . .., C k of G[ϕ(u) \ {x}]; we may have k = 1.By definition of a boundaried minor model, we know that y ∈ ϕ(u).Since the boundary vertices of G are all distinct, we have y = x.Hence y is contained in one of the connected components of G[ϕ(u) \ {x}].Assume without loss of generality that y ∈ C 1 .We obtain H from H by the reverse of an edge contraction operation, as follows.
Lemma 25.Let H ∈ G Xt for some integer t ≥ 0 and let G ∈ G X t with t ≥ t such that H m forget(G, t ).Otherwise, let x ∈ δ(G) such that b G (x) = t be the t'th boundary vertex of G, which does not belong to the boundary of forget(G, t − 1).We consider the role of x.(Not in range)(Part of a non-boundary branch set) If x ∈ ϕ(u) for some u / ∈ δ( H), Initialize H as a copy of H, into which a new boundary vertex v with b H (v) = t is inserted.Add the edge {u, v}.For each edge {u, w} ∈ E( H), we may re-attach it to v instead of u, based on the following distinction.By Definition 6, we know that ψ({u, w}) has exactly one endpoint b in ϕ(u).If w / ∈ C 1 , then replace the edge {u, w} in H by the edge {v, w} in H .If w ∈ C 1 , then the edge is preserved in H .
, add p to P .p consists of boundary vertices u and v connected by edge {u, v}.Let {u , v } := ψ({u, v}).If u , v ∈ V (G 1 ) and {u , v } ∈ E(G 1 ), add p to P .Else, {u , v } ∈ E(G 2 ) and we do not add p to P .p contains some non-boundary vertex v / ∈ δ(H).Note that ϕ(v) is either completely contained in G 1 or in G 2 .If ϕ(v) ⊆ V (G 1 ), add p to P .Else, do nothing.Let H 1 := p∈P p and let H 2 := p / ∈P p, we show how to find boundaried minor models of then there exists H ∈ ext +t (G) such that H m G. Furthermore, there exists P ⊆ pcs(H ) such that p∈P m G 1 and p / ∈P p m G 2 .An example is shown in Figure the result follows.Q was obtained from Q by setting b Q (v) := t + 1 for an existing vertex v.To obtain H from Q, start by removing the connected component of Q − δ(Q) containing v and its edges to the boundary.Since H is a t-boundaried graph, v is not a vertex of H and all components of Q − δ(Q ) that contain a neighbor of v, were indeed removed from H in the process of obtaining Q .Furthermore, v itself and edges to other boundary vertices were indeed removed.If a connected component S of Q − δ(Q ) was removed to obtain H that does not have connections to v, it is also a component of Q − δ(Q) and we remove it.Furthermore we forget any edges between boundary vertices, labels within the boundary, and (now isolated) boundary vertices as needed.
Q was obtained from Q by splitting vertex u ∈ δ(Q) into u and v (such that b Q (v) = t+1)and redistributing labels and edges.Consider the steps used to obtain H from Q .Suppose component S of Q − δ(Q ) was removed, then we also remove this component from Q to does not have any graph in Π as a minor if and only if there exist (Π 1 , Π 2 ) ∈ split(Π) such that G i has no graph in Π i as a minor for i ∈ [2].Proof.Suppose there exist (Π 1 , Π 2 ) ∈ split(Π) such that G i has no graph in Π i as a minor for i ∈ [2].Suppose for contradiction that there exists H ∈ Π such that H m G.By Lemma 28, this implies that there exists P ⊆ pcs(H) such that p∈P p m G 1 and p / ∈P p m G 2 .But by the definition of split, either p∈P p ∈ Π 1 or p / ∈P p ∈ Π 2 , which is a contradiction.Suppose G does not have any graph in Π as a minor.We show how to define Π 1 and Π 2 .Initialize Π i := Π for i ∈ [2].Consider every graph H ∈ Π, and all P ⊆ pcs(H).If p∈P p m G 1 , add p∈P p to Π 1 .If p / ∈P p m G 2 , add p / ∈P p to Π 2 .Finally, for any H ∈ mpcs +t (F) such that Π i contains a minor of H , add H to Π i for i ∈ [2].It remains to show that (Π 1 , Π 2 ) ∈ split(Π), since by definition, G 1 has no minors in Π 1 and G 2 has no minors in Π 2 .We need to verify that for all H ∈ Π and all P ⊆ pcs(H), either p∈P p ∈ Π 1 or p / ∈P p ∈ Π 2 .Suppose for contradiction that there exists such P with p∈P p / ∈ Π 1 and p / ∈P p / ∈ Π 2 .By the construction of Π 1 and Π 2 , this implies that p∈P p m G 1 and p / ∈P p m G 2 .By Lemma 28, it follows that p∈pcs(H) p m G 1 ⊕ G 2 = G.Since p∈pcs(H) p = H, it follows that H m G. Since H ∈ Π, this contradicts that G has no graph in Π as a minor.
H 1 , H 2 such that Π 1 contains no minor of H 1 and Π 2 contains no minor of H 2 , and furthermore H 1 ⊕ H 2 = H and the two graphs share no edges.These must exist by the definition of .By definition, H m H 1 ⊕ H 2 and by Lemma 28 there exists P ⊆ pcs(H) such that H 1 := p∈P p m H 1 and H 2 := p∈pcs(H)\P p m H 2 and H 1 This means that |R B | > N (t • min H∈F |V (H)|, t, t + Q , min H∈F |V (H)|) (recall Definition 17 for the meaning of N ( , t, n, θ)).Since R B is a subset of mpcs +t (Q), it follows that all graphs in R B have at most t + Q vertices.Furthermore, they are t-boundaried.So, if |R B | > N (t • min H∈F |V (H)|, t, t + Q , min H∈F |V (H)|), either some graph has more than min H∈F |V (H)| labels on a single vertex, or more than t • min H∈F |V (H)| labels are used in total.Suppose some vertex in