Clustered Graph Coloring and Layered Treewidth∗

Chun-Hung Liu† David R. Wood‡ November 26, 2020

Abstract A graph coloring has bounded clustering if each monochromatic component has bounded size. This paper studies clustered coloring, where the number of colors depends on an excluded complete bipartite subgraph. This is a much weaker assumption than previous works, where typically the number of colors depends on an excluded minor. This paper focuses on graph classes with bounded layered treewidth, which include planar graphs, graphs of bounded Euler genus, graphs embeddable on a fixed surface with a bounded number of crossings per edge, amongst other examples. Our main theorem says that for fixed integers s, t, k, every graph with layered treewidth at most k and with no Ks,t subgraph is (s + 2)-colorable with bounded clustering. In the s = 1 case, which corresponds to graphs of bounded maximum degree, we obtain polynomial bounds on the clustering. This greatly improves a corresponding result of Esperet and Joret for graphs of bounded genus. The s = 3 case implies that every graph with a drawing on a fixed surface with a bounded number of crossings per edge is 5-colorable with bounded clustering. Our main theorem is also a critical component in two companion papers that study clustered coloring of graphs with no Ks,t-subgraph and excluding a fixed minor, odd minor or topological minor. arXiv:1905.08969v3 [math.CO] 25 Nov 2020

∗This material is based upon work supported by the National Science Foundation under Grant No. DMS-1664593, DMS-1929851 and DMS-1954054. †Department of Mathematics, Texas A&M University, Texas, USA, [email protected]. Partially supported by NSF under Grant No. DMS-1664593, DMS-1929851 and DMS-1954054. ‡School of Mathematics, Monash University, Melbourne, Australia, [email protected]. Research supported by the Australian Research Council.

1 Contents

1 Introduction 3 1.1 Main Results ...... 3 1.2 Applications ...... 5 1.3 Notation ...... 6 1.4 Organization of the Paper ...... 7

2 Bounded Layered Treewidth and Bounded Degree7

3 Bounded Neighborhoods 10

4 Fences in a Tree-decomposition 11 4.1 Parades and Fans ...... 13

5 Setup for Main Proof 14

6 Main Lemma 16

7 Allowing Apex Vertices 85

8 Open Problems 86

2 1 Introduction

This paper considers graph colorings where the condition that adjacent vertices are assigned distinct colors is relaxed. Instead, we require that every monochromatic component has bounded size (for a given graph class). More formally, a coloring of a graph G is a function that assigns one color to each vertex of G.A monochromatic component with respect to a coloring of G is a connected component of a subgraph of G induced by all the vertices assigned the same color. A coloring has clustering η if every monochromatic component has at most η vertices. Our focus is on minimizing the number of colors, with small monochromatic components as a secondary goal. The clustered chromatic number of a graph class G is the minimum integer k such that for some integer η, every graph in G is k-colorable with clustering η. There have been several recent papers on this topic [1,8, 13, 15, 16, 22–27, 31, 32, 35, 36]; see [46] for a survey. This paper is the ﬁrst of three companion papers [33, 34]. The unifying theme that distinguishes this work from previous contributions is that the number of colors is determined by an excluded subgraph. Excluding a subgraph is a much weaker assumption than excluding a minor, which is a typical assumption in previous results. In particular, we consider graphs with no Ks,t subgraph plus various other structural properties, and prove that every such graph is colorable with bounded clustering, where the number of colors only depends on s. All the dependence on t and the structural property in question is hidden in the clustering function. Note that the case s = 1 is already interesting, since a graph has no K1,t subgraph if and only if it has maximum degree less than t. Many of our theorems generalize known results for graphs of bounded maximum degree to the setting of an excluded Ks,t subgraph. Note that for s > 2, no result with bounded clustering is possible for graphs with no Ks,t subgraph, without making some extra assumption. In particular, for every graph H that contains a cycle, and for all k, η ∈ N, if G is a graph with chromatic number greater than kη and girth greater than |V (H)| (which exists [14]), then G contains no H subgraph and G is not k-colorable with clustering η, for otherwise G would be kη-colorable.

1.1 Main Results In this paper, the “other structural property” mentioned above is “bounded layered treewidth”. First we explain what this means. A tree-decomposition of a graph G is a pair (T, X = (Xx : x ∈ V (T )), where T is a tree, and for each node x ∈ V (T ), Xx is a non-empty subset of V (G) called a bag, such that for each vertex v ∈ V (G), the set {x ∈ V (T ): v ∈ Xx} induces a non- empty (connected) subtree of T , and for each edge vw ∈ E(G) there is a node x ∈ V (T ) such that {v, w} ⊆ Xx. The width of a tree-decomposition (T, X ) is max{|Xx| − 1 : x ∈ V (T )}. The treewidth of a graph G is the minimum width of a tree-decomposition of G. Treewidth is a key parameter in algorithmic and structural graph theory; see [3, 21, 39, 40] for surveys. A layering of a graph G is an ordered partition (V1,...,Vn) of V (G) into (possibly empty) sets such that for each edge vw ∈ E(G) there exists i ∈ [1, n − 1] such that {v, w} ⊆ Vi ∪ Vi+1. The layered treewidth of a graph G is the minimum nonnegative integer ` such that G has a tree-decomposition (T, X = (Xx : x ∈ V (T )) and a layering (V1,...,Vn), such that |Xx ∩ Vi| 6 ` for each bag Xx and layer Vi. This says that the subgraph induced by each layer has bounded treewidth, and moreover, a single tree-decomposition of G has bounded treewidth when restricted to each layer. In fact, these properties hold when considering a bounded sequence of consecutive layers. Layered treewidth was independently introduced by Dujmović, Morin, and Wood [12] and Shahrokhi [44]. Dujmović et al. [12] proved that every planar graph has layered treewidth at most

3 3; more generally, that every graph with Euler genus1 at most g has layered treewidth at most 2g + 3; and most generally, that a minor-closed class has bounded layered treewidth if and only if it excludes some apex graph as a minor. Layered treewidth is of interest beyond minor-closed classes, since as described below, there are several natural graph classes that have bounded layered treewidth but contain arbitrarily large complete graph minors. Consider coloring a graph G with layered treewidth w. Say (V1,...,Vn) is the corresponding S layering. Then G[ Vi] has treewidth at most w−1, and is thus properly w-colorable. Similarly, S i odd G[ i even Vi] is properly w-colorable. Thus G is properly 2w-colorable. This bound is best possible since K2w has layered treewidth w. In fact, for every integer d and integer w, there is a graph G with treewidth 2w − 1, such that every (2w − 1)-coloring of G has a vertex of monochromatic degree at least d, implying there is a monochromatic component with more than d vertices [46]. Bannister, Devanny, Dujmović, Eppstein, and Wood [2] observed that every graph with treewidth k+1 k has layered treewidth at most d 2 e (using two layers), so G has layered treewidth at most w. This says that the clustered chromatic number of the class of graphs with layered treewidth w equals 2w, and indeed, every such graph is properly 2w-colorable. On the other hand, we prove that for clustered coloring of graphs excluding a Ks,t-subgraph in addition to having bounded layered treewidth, only s + 2 colors are needed (no matter how large the upper bound on layered treewidth). This is the main result of the paper.

Theorem 1. For all s, t, w ∈ N there exists η ∈ N such that every graph with layered treewidth at most w and with no Ks,t subgraph is (s + 2)-colorable with clustering η. Theorem1 is proved in Section5. The case s = 1 in Theorem1 is of particular interest, since it applies for graphs of maximum degree ∆ < t. It implies that graphs of bounded Euler genus and of bounded maximum degree are 3-colorable with bounded clustering, which was previously proved by Esperet and Joret [15]. The clustering function proved by Esperet and Joret [15] is roughly O(∆32∆ 2g ), for graphs with Euler genus g and maximum degree ∆. While Esperet and Joret [15] made no eﬀort to reduce this function, their method will not lead to a sub-exponential clustering bound. In the case s = 1 we give a diﬀerent proof of Theorem1 with the following polynomial clustering bound.

Theorem 2. For every w, ∆ ∈ N, every graph with layered treewidth at most w and maximum degree at most ∆ is 3-colorable with clustering O(w19∆37). In particular, every graph with Euler genus at most g and maximum degree at most ∆ is 3-colorable with clustering O(g19∆37). Note that the proof of Theorem2 (presented in Section2) is relatively simpler than [15], avoiding many technicalities that arise when dealing with graph embeddings. This proof highlights the utility of layered treewidth as a general tool. The number of colors in Theorem2 is best possible since the Hex Lemma [20] says that every 2-coloring of the n × n planar triangular grid (which has layered treewidth 2 and maximum degree 6) contains a monochromatic path of length at least n. The planar triangular grid and its generalizations are a key lower bound in the theory of clustered colorings [46]. Further motivation for Theorem1 is that it is a critical ingredient in the proofs of results in our companion papers [33, 34] about clustered colorings of graphs excluding a minor, odd minor, or topological minor. For these proofs we actually need the following stronger result proved in Section7 (where the ξ = 0 case is Theorem1).

1The Euler genus of an orientable surface with h handles is 2h. The Euler genus of a non-orientable surface with c cross-caps is c. The Euler genus of a graph G is the minimum Euler genus of a surface in which G embeds (with no crossings).

4 Theorem 3. For all s, t, w ∈ N and ξ ∈ N0, there exists η ∈ N such that if G is a graph with no Ks,t subgraph and G − Z has layered treewidth at most w for some Z ⊆ V (G) with |Z| 6 ξ, then G is (s + 2)-colorable with clustering η.

Note that the number of colors in Theorem3 is best possible for s 6 2: Suppose that the theorem holds with s = 2, t = 7, w = 2 and ξ = 1, but with only 3 colors. Let n η2. Let G be obtained from the n × n triangular grid by adding one dominant vertex v. Then G contains no K2,7-subgraph, and G−v has layered treewidth 2. By assumption, G is 3-colorable with clustering η. Say v is blue. Since v is dominant, at most η rows or columns contain a blue vertex. The non-blue induced subgraph contains an η × η triangular grid (since n η2). This contradicts the Hex Lemma mentioned above. We remark that the class of graphs mentioned in Theorem3 is more general than the class of graphs with bounded layered treewidth. For example, the class of graphs that can be made planar (and hence bounded layered treewidth) by deleting one vertex contains graphs of arbitrarily large layered treewidth.

1.2 Applications We now give several examples of graph classes with bounded layered treewidth, for which Theorems1 and2 give interesting results.

(g, k)-Planar Graphs A graph is (g, k)-planar if it can be drawn in a surface of Euler genus at most g with at most k crossings on each edge (assuming no three edges cross at a single point). Such graphs can contain arbitrarily large complete graph minors, even in the g = 0 and k = 1 case [10]. On the other hand, Dujmović et al. [10] proved that every (g, k)-planar graph has layered treewidth at most (4g + 6)(k + 1). Theorem2 then implies:

Corollary 4. For all g, k, ∆ ∈ N, every (g, k)-planar graph with maximum degree at most ∆ is 3-colorable with clustering O(g19k19∆37).

Now consider (g, k)-planar graphs without the assumption of bounded maximum degree. Wood [46] proved that such graphs are 12-colorable with clustering bounded by a function of g and k. Os- sona de Mendez, Oum, and Wood [37] proved that every (g, k)-planar graph contains no K3,t subgraph for some t = O(kg2). Thus Theorem1 with s = 3 implies that this bound of 12 can be reduced to 5:

Corollary 5. For all g, k ∈ N0, there exists η ∈ N, such that every (g, k)-planar graph is 5-colorable with clustering η.

Corollary5 highlights the utility of excluding a Ks,t subgraph. It also generalizes a theorem of Esperet and Ochem [16] who proved the k = 0 case, which says that every graph with bounded Eu- ler genus is 5-colorable with bounded clustering. Note that Dvořák and Norin [13] recently proved that every graph with bounded Euler genus is 4-colorable (in fact, 4-choosable) with bounded clustering. It is open whether every (g, k)-planar graph is 4-colorable with bounded clustering.

5 Map Graphs

Map graphs are deﬁned as follows. Start with a graph G0 embedded in a surface of Euler genus g, with each face labelled a “nation” or a “lake”, where each vertex of G0 is incident with at most d nations. Let G be the graph whose vertices are the nations of G0, where two vertices are adjacent in G if the corresponding faces in G0 share a vertex. Then G is called a (g, d)-map graph.A (0, d)-map graph is called a (plane) d-map graph; such graphs have been extensively studied [5–7,9, 17]. The (g, 3)-map graphs are precisely the graphs of Euler genus at most g (see [7, 10]). So (g, d)-map graphs provide a natural generalization of graphs embedded in a surface that allows for arbitrarily large cliques even in the g = 0 case (since if a vertex of G0 is incident with d nations then G contains Kd). Dujmović et al. [10] proved that every (g, d)-map graph has layered treewidth at most (2g + 3)(2d + 1) and is (g, O(d2))-planar. Thus Corollary4 implis:

Corollary 6. For all g, d, ∆ ∈ N, every (g, d)-map graph with maximum degree at most ∆ is 3-colorable with clustering O(g19d38∆37).

Similarly, Corollary5 implies:

Corollary 7. For all g, d ∈ N, there exists η ∈ N, such that every (g, d)-map graph is 5-colorable with clustering η.

It is straightforward to prove that, in fact, if a (g, d)-map graph G contains a K3,t subgraph, then t 6 6d(g + 1). Thus Theorem1 can be applied directly with s = 3, slightly improving the clustering bounds. It is open whether every (g, d)-map graph is 4-colorable with clustering bounded by a function of g and d.

String Graphs A string graph is the intersection graph of a set of curves in the plane with no three curves meeting at a single point [18, 19, 30, 38, 42, 43]. For an integer k > 2, if each curve is in at most k intersections with other curves, then the corresponding string graph is called a k-string graph.A (g, k)-string graph is deﬁned analogously for curves on a surface of Euler genus at most g. Dujmović, Joret, Morin, Norin, and Wood [11] proved that every (g, k)-string graph has layered treewidth at most 2(k − 1)(2g + 3). By deﬁnition, the maximum degree of a (g, k)-string graph is at most k (and might be less than k since two curves might have multiple intersections). Thus Theorem2 implies:

Corollary 8. For all integers g > 0 and k > 2, there exists η ∈ N such that every (g, k)-string graph is 3-colorable with clustering O(g19k56).

1.3 Notation

Let G be a graph with vertex-set V (G) and edge-set E(G). For v ∈ V (G), let NG(v) := {w ∈ V (G): vw ∈ E(G)} be the neighborhood of v, and let NG[v] := NG(v) ∪ {v}. For X ⊆ V (G), let S NG(X) := v∈X (NG(v)\X) and NG[X] := NG(X)∪X. Denote the subgraph of G induced by X by G[X]. Let N := {1, 2,... } and N0 := {0, 1, 2,... }. For m, n ∈ N0, let [m, n] := {m, m + 1, . . . , n} and [n] := [1, n]. Denote the set of all positive real numbers by R+. For a coloring c of G, a c-monochromatic component is a monochromatic component with respect to c.

6 1.4 Organization of the Paper The proof of Theorem2 is self-contained and is presented in Section2. The proof of Theorem1 is more substantial and independent of the proof of Theorem2. Readers who are only interested in the more general theorems can safely skip Section2. We prove Theorem1 in Section5, assuming the key lemma, Lemma 21, which is then proved in Section6. Sections3 and4 are preparations for proving Lemma 21. In Section3, we show that for every subset X of vertices of a graph with no Ks,t subgraph, the number of vertices outside X but adjacent to at least s vertices in X is bounded. This helps us control the growth of monochromatic components each time we extend a precoloring. In Section4, we introduce the notions of fences, parades and fans, which are notions deﬁned on tree-decompositions that will be used in the proof of Lemma 21. Fences are a tool inspired by Robertson and Seymour’s separator theory for graphs of bounded treewidth. The proof of Lemma 21 is the most complicated part of this paper. We construct the desired coloring in the proof of Lemma 21 by an algorithm. The intuition of the algorithm is explained be- fore the formal statement. After stating the algorithm, we prove the correctness of the algorithm by a sequence of claims. These claims can be read in two parts. The ﬁrst part consists of Claims 21.1– 21.25, which show that one kind of monochromatic component has bounded size (Claim 21.22) and that the number of precolored vertices in the “interesting region” is always bounded (Claim 21.25). The remaining claims show that the other kind of monochromatic component has bounded size (Claim 21.59). In Section7, we prove Theorem3 using Theorem1. We conclude this paper by mentioning some open problems in Section8.

2 Bounded Layered Treewidth and Bounded Degree

This section proves Theorem2, which says that graphs of bounded layered treewidth and bounded maximum degree are 3-colorable with bounded clustering. We need the following analogous result by Alon et al. [1] for bounded treewidth graphs.

Lemma 9 ([1]). There is a function f : N × N → N such that every graph with treewidth w and maximum degree ∆ is 2-colorable with clustering f(w, ∆) ∈ O(k∆).

For a graph G with bounded maximum degree and bounded layered treewidth, if (V1,...,Vn) is the corresponding layering of G, then Lemma9 is applicable to G[Vi], which has bounded treewidth. The idea of the proof of Theorem2 is to use colors 1 and 2 for all layers Vi with i ≡ 1 (mod 3), use colors 2 and 3 for all layers Vi with i ≡ 2 (mod 3), and use colors 3 and 1 for all layers Vi with i ≡ 3 (mod 3). Then each monochromatic component is contained within two consecutive layers. The key to the proof is to control the growth of monochromatic components between consecutive layers. The next lemma is useful for this purpose.

Lemma 10. Let w, ∆, d, k, h ∈ N. Let G be a graph with maximum degree at most ∆. Let (T, X ) be a tree-decomposition of G with width at most w, where X = (Xt : t ∈ V (T )). For i > 1, let Yi be a subset of V (T ), T a subtree of T containing Y , and E a set of pairs of vertices in S X i i i x∈Yi x with |E | k. Let G0 be the graph with V (G0) = V (G) and E(G0) = E(G) ∪ S E . If every i 6 i>1 i vertex of G appears in at most d pairs in S E , and every vertex of T is contained in at most h i>1 i 0 members of {T1,T2,... }, then G has maximum degree at most ∆ + d and has a tree-decomposition (T, X 0) of width at most w + 2hk.

7 Proof. Since every vertex of G appears in at most d pairs in S E , G0 has maximum degree at i>1 i most ∆ + d. For every i > 1, let Zi be the set of the vertices appearing in some pair of Ei. Note that Z ⊆ S X and |Z | 2|E | 2k. For every t ∈ V (T ), let X0 := X ∪ S Z . Let i x∈Yi x i 6 i 6 t t {i:t∈V (Ti)} i 0 0 X := (Xt : t ∈ V (T )). 0 0 S 0 0 We claim that (T, X ) is a tree-decomposition of G . It is clear that t∈V (T ) Xt ⊇ V (G ). For 0 0 each i ∈ N, for every t ∈ V (Ti), since Xt ⊇ Zi, Xt contains both ends of each edge in Ei. For each v ∈ V (G0), 0 [ {t ∈ V (T ): v ∈ Xt} = {t ∈ V (T ): v ∈ Xt} ∪ V (Ti).

{i:v∈Zi}

Note that for every v ∈ V (G) and i > 1, if v ∈ Zi then v ∈ Xt for some t ∈ Yi ⊆ V (Ti). Hence 0 0 0 {t : v ∈ Xt} induces a subtree of T . This proves that (T, X ) is a tree-decomposition of G . Since for every t ∈ V (T ), |X0| |X | + P |Z | w + 1 + 2hk, the width of (T, X 0) is t 6 t {i:t∈V (Ti)} i 6 at most w + 2hk. We now prove Theorem2.

Theorem 11. Let ∆, w ∈ N. Then every graph G with maximum degree at most ∆ and with layered treewidth at most w is 3-colorable with clustering g(w, ∆), for some function g(w, ∆) ∈ O(w19∆37).

Proof. Let f be the function from Lemma9. Deﬁne

f1 := f(w, ∆) ∈ O(w∆) 2 3 ∆2 := ∆ + f1∆ ∈ O(w∆ ) 2 2 3 4 w2 := w + 2(w + 1)f1 ∆ ∈ O(w ∆ ), 4 7 f2 := f(w2, ∆2) ∈ O(w ∆ ) 2 4 9 ∆3 := ∆ + f2∆ ∈ O(w ∆ ) 2 2 11 20 w3 := w + 4(w2 + 1)f2 ∆ ∈ O(w ∆ ) 15 29 f3 := f(w3, ∆3) ∈ O(w ∆ ) 19 37 g(w, ∆) := (1 + f2∆)f3 ∈ O(w ∆ ).

Let G be a graph of maximum degree at most ∆ and layered treewidth at most w. Let (T, X ) and (Vi : i > 1) be a tree-decomposition of G and a layering of G such that |Xt ∩ Vi| 6 w for every S∞ t ∈ V (T ) and i > 1, where X = (Xt : t ∈ V (T )). For j ∈ [3], let Uj = i=0 V3i+j. By Lemma9, there exists a coloring c1 : U1 → {1, 2} such that every monochromatic component of G[U1] contains at most f1 vertices. For each i ∈ N0, let Ci be the set of c1-monochromatic components of G[U1] contained in V3i+1 with color 2. For each i ∈ N0 and C ∈ Ci, deﬁne the following:

• Let Yi,C be a minimal subset of V (T ) such that for every edge e of G between V (C) and NG(V (C)) ∩ V3i+2, there exists a node t ∈ Yi,C such that both ends of e belong to Xt.

• Let Ei,C be the set of all pairs of distinct vertices in NG(V (C)) ∩ V3i+2.

• Let Ti,C be the subtree of T induced by {t ∈ V (T ): Xt ∩ V (C) 6= ∅}.

Note that there are at most |V (C)|∆ 6 f1∆ edges of G between V (C) and NG(V (C)). So 2 2 |Yi,C | 6 f1∆ and |Ei,C | 6 f1 ∆ for every i ∈ N0 and C ∈ Ci. In addition, Yi,C ⊆ V (Ti,C ) for every i ∈ N0 and C ∈ Ci. Since (T, (Xt ∩ U1 : t ∈ V (T ))) is a tree-decomposition of G[U1] with width

8 at most w, for every t ∈ V (T ) and i ∈ N0, there exist at most w + 1 diﬀerent members C of Ci such that t ∈ V (T ). Furthermore, N (V (C)) ∩ V ⊆ S X , so each pair in E consists i,C G 3i+2 x∈Yi,C x i,C of two vertices in S X . Since every vertex v in U is adjacent in G to at most ∆ members x∈Yi,C x 2 of S C and every member C of S C creates at most f ∆ pairs in E involving v, where i∈N0 i i∈N0 i 1 iC ,C 2 iC is the index such that C ∈ Ci , every vertex in U2 appears in at most f1∆ · ∆ = f1∆ pairs in S C Ei,C . i∈N0,C∈Ci Let G2 be the graph with V (G2) := U2 and [ E(G2) := E(G[U2]) ∪ Ei,C .

i∈N0,C∈Ci

2 2 2 We have shown that Lemma 10 is applicable with k = f1 ∆ and d = f1∆ and h = w + 1. Thus (2) G2 has maximum degree at most ∆2 and a tree-decomposition (T, X ) of width at most w2. Say (2) (2) X = (Xt : t ∈ V (T )). By Lemma9, there exists a coloring c2 : U2 → {2, 3} such that every c2-monochromatic component of G2 contains at most f2 vertices. (2) Note that we may assume that (T, X ) is a tree-decomposition of G[U1 ∪ U2] by redeﬁning (2) (2) Xt to be the union of Xt and Xt ∩ U1, for every t ∈ V (T ). 0 For each i ∈ N0, let Ci be the set of the monochromatic components either of G[U1] with color 0 1 with respect to c1 or of G2 with color 3 with respect to c2. For each i ∈ N0 and C ∈ Ci, deﬁne the following:

0 • Let Yi,C be a minimal subset of V (T ) such that for every edge e of G between V (C) and 0 NG(V (C)) ∩ V3i+3, there exists a node t ∈ Yi,C such that both ends of e belong to Xt. 0 • Let Ei,C be the set of all pairs of distinct vertices of NG(V (C)) ∩ V3i+3. 0 (2) • Let Ti,C be the subtree of T induced by {t ∈ V (T ): Xt ∩ V (C) 6= ∅}.

Note that there are at most |V (C)|∆ 6 f2∆ edges of G between V (C) and NG(V (C)) ∩ V3i+3 for 0 0 0 2 2 0 every i ∈ N0 and C ∈ Ci. So |Yi,C | 6 f2∆ and |Ei,C | 6 f2 ∆ for every i ∈ N0 and C ∈ Ci. In 0 0 0 (2) addition, Yi,C ⊆ V (Ti,C ) for every i ∈ N0 and C ∈ Ci. Since (T, (Xt ∩ U2 : t ∈ V (T ))) is a tree- (2) decomposition of G2 with width at most w2 and (T, (Xt ∩U1 : t ∈ V (T ))) is a tree-decomposition of G[U1] with width at most w 6 w2, for every t ∈ V (T ) and i ∈ N0, there exist at most 2(w2 + 1) 0 0 diﬀerent members C ∈ Ci such that t ∈ V (Ti,C ). Furthermore, each pair in Ei,C consists of two S (2) vertices in 0 Xx . Since every vertex v in U3 is adjacent in G to at most ∆ members of x∈Yi,C S C0, and every member C of S C0 creates at most f ∆ pairs in E involving v, where i i∈N0 i i∈N0 i 2 iC ,C C 0 2 S is the index such that C ∈ C , every vertex in U3 appears in at most f2∆ pairs in 0 Ei,C . iC i∈N0,C∈Ci Let G3 be the graph with V (G3) := U3 and

[ 0 E(G3) := E(G[U3]) ∪ Ei,C . 0 i∈N0,C∈Ci

2 2 2 We have shown that Lemma 10 is applicable with k = f2 ∆ and d = f2∆ and h = 2(w2 + 1). (3) Hence G3 has maximum degree at most ∆3 and a tree-decomposition (T, X ) with width at most w3. By Lemma9, there exists a coloring c3 : U3 → {1, 3} such that every monochromatic component of G3 contains at most f3 vertices. Deﬁne c : V (G) → {1, 2, 3} such that for every v ∈ V (G), we have c(v) := cj(v), where j is the index for which v ∈ Uj. Now we prove that every c-monochromatic component of G contains at most g(w, ∆) vertices.

9 Let D be a c-monochromatic component of G with color 2. Since D is connected, for every pair of vertices u, v ∈ V (D) ∩ U2, there exists a path Puv in D from u to v. Since V (D) ⊆ U1 ∪ U2, for every maximal subpath P of Puv contained in U1, there exists i ∈ N0 and C ∈ Ci such that there exists a pair in Ei,C consisting of the two vertices in P adjacent to the ends of Puv. That is, there exists a path in G2[V (D) ∩ U2] connecting u, v for every pair u, v ∈ V (D) ∩ U2. Hence G2[V (D) ∩ U2] is connected. So G2[V (D) ∩ U2] is a c2-monochromatic component of G2 with color 2 and contains at most f2 vertices. Hence there are at most f2∆ edges of G between V (D) ∩ U2 and NG(V (D) ∩ U2). So D[V (D) ∩ U1] contains at most f2∆ components. Since each component of D[V (D) ∩ U1] is a c1-monochromatic component of G[U1], it contains at most f1 vertices. Hence D[V (D) ∩ U1] contains at most f2∆f1 vertices. Since D has color 2, V (D) ∩ U3 = ∅. Therefore, D contains at most (1 + f1∆)f2 6 g(w, ∆) vertices. Let D0 be a c-monochromatic component of G with color b, where b ∈ {1, 3}. Since D0 0 is connected, by an analogous argument to that in the previous paragraph, G3[V (D ) ∩ U3] is 0 connected. So G3[V (D )∩U3] is a c3-monochromatic component of G3 with color b and contains at 0 0 most f3 vertices. Hence there are at most f3∆ edges of G between V (D )∩U3 and NG(V (D )∩U3). 0 0 So D[V (D) ∩ Ub0 ] contains at most f3∆ components, where b = 1 if b = 1 and b = 2 if b = 3. Since each component of D[V (D) ∩ Ub0 ] is a cb0 -monochromatic component of G[Ub0 ], it contains at most f2 vertices. Hence D[V (D) ∩ Ub0 ] contains at most f3∆f2 vertices. Since D has color b, V (D) ∩ Ub+1 = ∅, where U4 = U1. Therefore, D contains at most (1 + f2∆)f3 6 g(w, ∆) vertices. This completes the proof.

3 Bounded Neighborhoods

We now set out to prove Theorem1 for all s. A key step in the above proof for the s = 1 case says that if X is a bounded-size set of vertices in a graph, then NG(X) also has bounded size (since G has bounded maximum degree). This fails for graphs with no Ks,t subgraph for s > 2, since vertices in X might have unbounded degree. To circumvent this issue we focus on those vertices with at least s neighbors in X, and then show that there are a bounded number of such vertices. The following notation formalizes this simple but important idea. For a graph G, a set X ⊆ V (G), and s ∈ N, deﬁne

>s NG (X) := {v ∈ V (G) − X : |NG(v) ∩ X| > s} and

~~>s >s~~

~~Lemma 12. For all s, t ∈ N, there exists a function fs,t : N0 → N0 such that for every graph G >s with no Ks,t subgraph, if X ⊆ V (G) then |N (X)| 6 fs,t(|X|).~~

x >s Proof. Deﬁne fs,t(x) = s (t − 1) for every x ∈ N0. For every y ∈ N (X), let Zy be a subset of |X| >s NG(y)∩X with size s. Since there are exactly s subsets of X with size s, if |N (X)| > fs,t(|X|), >s then there exists a subset Y of N (X) with size t such that Zy is identical for all y ∈ Y , implying S >s G[Y ∪ y∈Y Zy] contains a Ks,t subgraph, which is a contradiction. Thus |N (X)| 6 fs,t(|X|). The function f in Lemma 12 can be improved if we know more about the graph G, which helps to get better upper bounds on the clustering. A 1-subdivision of a graph H is a graph obtained from H by subdividing every edge exactly once. For a graph G, let ∇(G) be the maximum average

~~10 degree of a graph H for which the 1-subdivision of H is a subgraph of G. The following result is implicit in [37]. We include the proof for completeness.~~

+ ∇ ∇ Lemma 13. For all s, t ∈ N and positive ∇ ∈ R there is a number c := max{t−1, 2 +(t−1) s−1 }, such that for every graph G with no Ks,t-subgraph and with ∇(G) 6 ∇, if X ⊆ V (G) then s |N > (X)| 6 c|X|. Proof. In the case s = 1, the proof of Lemma 12 implies this lemma since c > t − 1. Now assume >s X >s that s > 2. Let H be the bipartite graph with bipartition {NG (X), 2 }, where v ∈ NG (X) is X adjacent in H to {x, y} ∈ 2 whenever x, y ∈ NG(v) ∩ X. Let M be a maximal matching in H. Let Q be the graph with vertex-set X, where xy ∈ E(Q) whenever {v, {x, y}} ∈ M for some >s vertex v ∈ NG (X). Thus, the 1-subdivision of every subgraph of Q is a subgraph of G. Hence ∇ ∇ |M| = |E(Q)| 6 2 |V (Q)| = 2 |X|. Moreover, Q is ∇-degenerate, implying Q contains at most ∇ >s s−1 |X| cliques of size exactly s. Exactly |M| vertices in NG (X) are incident with an edge in M. >s For each vertex v ∈ NG (X) not incident with an edge in M, by maximality, NG(v) ∩ X is a clique >s in Q of size at least s. Deﬁne a mapping from each vertex v ∈ NG (X) to a clique of size exactly >s s in Q[NG(v) ∩ X]. Since G has no Ks,t subgraph, at most t − 1 vertices v ∈ NG (X) are mapped >s ∇ ∇ to each ﬁxed s-clique in Q. Hence |NG (X)| 6 2 |X| + (t − 1) s−1 |X| 6 c|X|. Lemma 13 is applicable in many instances. In particular, every graph G with treewidth k has 0 0 ∇(G) 6 2k (since if a 1-subdivision of some graph G is a subgraph of G, then G has treewidth at most k, and every graph with treewidth at most k has average degree less than 2k). By an analogous argument, using bounds on the average degree independently due to Thomason [45] p and Kostochka [29], every H-minor-free graph G has ∇(G) 6 O(|V (H)| log |V (H)|). Similarly, using bounds on the average degree independently due to Komlós and Szemerédi [28] and Bollobás 2 and Thomason [4], every H-topological-minor-free graph G has ∇(G) 6 O(|V (H)| ). Finally, Dujmović et al. [12, Lemmas 8,9] proved that ∇(G) 6 5w for every graph with layered treewidth w. In all these cases, Lemma 13 implies that the function f in Lemma 12 can be made linear.

Corollary 14. For all s, t, w ∈ N, let fs,t,w : N0 → Q be the function deﬁned by fs,t,w(x) := 5w 5w ( 2 + (t − 1) s−1 )x for every x ∈ N0. Then for every graph G with no Ks,t-subgraph and with >s layered treewidth at most w, if X is a subset of V (G), then |N (X)| 6 fs,t,w(|X|) Proof. Dujmović et al. [12, Lemmas 8,9] showed that ∇(G) 6 5w for every graph with layered treewidth w. So this result immediately follows from Lemma 13.

~~4 Fences in a Tree-decomposition~~

~~This section introduces the notion of a fence, which will be used throughout. We start with a variant of a well-known result about separators in graphs of bounded treewidth.~~

Lemma 15. Let w ∈ N0. Let G be a graph and (T, X ) a tree-decomposition of G of width at most w, where X = (Xt : t ∈ V (T )). If Q is a subset of V (G) with |Q| > 12w + 13, then there exists ∗ 0 ∗ 2 S ∗ t ∈ V (T ) such that for every component T of T − t , |(Q ∩ ( t∈V (T 0) Xt)) ∪ Xt | < 3 |Q|. Proof. Suppose that there exists an edge xy of T such that |(Q ∩ (S X )) ∪ X ∪ X | 2 |Q| t∈V (Ti) t x y > 3 for each i ∈ [2], where T1,T2 are the components of T − xy. Then

~~2 X [ 4 |Q| + 2|X ∪ X | |(Q ∩ ( X )) ∪ X ∪ X | |Q|. x y > t x y > 3 i=1 t∈V (Ti)~~

11 1 13 Since the width of (T, X ) is at most w, 4(w + 1) > 2|Xx ∪ Xy| > 3 |Q| > 4w + 3 , a contradiction. First assume that there exists an edge xy of T such that |(Q ∩ (S X )) ∪ X ∪ X | < 2 |Q| t∈V (Ti) t x y 3 ∗ for each i ∈ [2], where T1,T2 are the components of T −xy. Let t := x. Then for every component 0 ∗ S S T of T − t , (Q ∩ ( 0 X )) ∪ X ∗ ⊆ (Q ∩ ( X )) ∪ X ∪ X for some i ∈ [2], and hence t∈V (T ) t t t∈V (Ti) t x y 2 S ∗ |(Q ∩ ( t∈V (T 0) Xt)) ∪ Xt | < 3 |Q|. So the lemma holds. Now assume that for every edge xy of T , there exists a unique r ∈ {x, y} such that |(Q ∩ (S X )) ∪ X ∪ X | 2 |Q|, where T ,T are the components of T − xy containing x, y, t∈V (Tr) t x y > 3 x y respectively. Orient the edge xy so that r is the head of this edge. We obtain an orientation of T . Note that the sum of the out-degree of the nodes of T equals |E(T )| = |V (T )| − 1. So some node t∗ has out-degree 0. 0 ∗ 0 ∗ For each component T of T − t , let tT 0 be the node in T adjacent in T to t . By the deﬁnition ∗ of the direction of tT 0 t , [ [ 2 |(Q ∩ ( X )) ∪ X ∗ | |(Q ∩ ( X )) ∪ X ∪ X ∗ | < |Q|. t t 6 t tT 0 t 3 t∈V (T ) t∈V (T ) tT 0 tT 0

0 0 ∗ Note that TtT 0 = T for every component T of T − t , which completes the proof. Let T be a tree and F a subset of V (T ). An F -part of T is an induced subtree of T obtained from a component of T − F by the adding nodes in F adjacent in T to this component. For an F -part T 0 of T , deﬁne ∂T 0 to be F ∩ V (T 0).

Lemma 16. Let (T, X ) be a tree-decomposition of a graph G of width at most w ∈ N0, where 1 X = (Xt : t ∈ V (T )). Then for every ∈ R with 1 > > w+1 and every Q ⊆ V (G), there 0 exists a subset F of V (T ) with |F | 6 max{(|Q| − 3w − 3), 0} such that for every F -part T of T , S S 1 |( t∈V (T 0) Xt ∩ Q) ∪ t∈∂T 0 Xt| 6 · (12w + 13).

1 Proof. We shall prove this lemma by induction on |Q|. If |Q| 6 (12w + 13), then let F := ∅; then 0 S S 1 for every F -part T of T , |( t∈V (T 0) Xt ∩ Q) ∪ t∈∂T 0 Xt| = |Q| 6 (12w + 13). So we may assume 1 that |Q| > (12w + 13) > 12w + 13 and the lemma holds for all Q with smaller size. By Lemma 15, there exists t∗ ∈ V (T ) such that for every component T 0 of T − t∗, |(Q ∩ 2 ∗ 0 S ∗ S S ( t∈V (T 0) Xt)) ∪ Xt | < 3 |Q|. That is, for every {t }-part T of T , |( t∈V (T 0) Xt ∩ Q) ∪ t∈∂T 0 Xt| < 2 3 |Q|. ∗ 0 2 0 S ∗ 0 For each {t }-part T of T , let QT := ( t∈V (T 0) Xt ∩Q)∪Xt , so |QT | < 3 |Q|. Let T1,T2,...,Tk ∗ 0 1 0 be the {t }-parts T of T with |QT | > (12w + 13). Let f be the function defined by f(x) := max{(x − 3w − 3), 0} for every x ∈ R. For each 2 i ∈ [k], since |QTi | < 3 |Q| < |Q|, the induction hypothesis implies that there exists Fi ⊆ V (Ti) with 0 S S 1 |Fi| 6 f(|QTi |) such that for every Fi-part T of Ti, |( t∈V (T 0) Xt ∩QTi )∪ t∈∂T 0 Xt| 6 (12w +13). ∗ Sk 0 0 ∗ Define F = {t } ∪ i=1 Fi. Note that for every F -part T of T , either T is a {t }-part 1 0 0 of T with |QT | 6 (12w + 13), or there exists i ∈ [k] such that T is an Fi-part of Ti. In 1 S S 0 the former case, |( t∈V (T 0) Xt ∩ Q) ∪ t∈∂T 0 Xt| = |QT | 6 (12w + 13). In the latter case, 1 S S S S ∗ |( t∈V (T 0) Xt ∩ Q) ∪ t∈∂T 0 Xt| 6 |( t∈V (T 0) Xt ∩ QTi ) ∪ t∈∂T 0 Xt| 6 (12w + 13) since Xt ⊆ QTi . S S 1 0 Hence |( t∈V (T 0) Xt ∩ Q) ∪ t∈∂T 0 Xt| 6 (12w + 13) for every F -part T of T . Pk To prove this lemma, it suffices to prove that |F | 6 f(|Q|). Note that |F | 6 1 + i=1|Fi| 6 Pk 1 Pk 1+ i=1 f(|QTi |). Since |QTi | > (12w +13) > 12w +13 for every i ∈ [k], |F | 6 1+ i=1 (|QTi |− 3w − 3).

12 If k = 0, then f(|Q|) = (|Q|−3w −3) = |Q|−(3w +3) > 12w +13−(3w +3) > 1 = |F | since 2 |Q| > 12w + 13 and 6 1. If k = 1, then |F | 6 1 + (|QT1 | − 3w − 3) 6 1 + ( 3 |Q| − 3w − 3) = (|Q| − 3w − 3) + 1 − 3 |Q| 6 f(|Q|) since |Q| > 12w + 13. Hence we may assume that k > 2. Then

~~k X [ |F | 6 1 + (|Q ∩ Xt| + |Xt∗ | − 3w − 3) ∗ i=1 t∈V (Ti)−{t }~~

~~6 1 + |Q| + k(|Xt∗ | − 3w − 3) = (|Q| − 3w − 3) + 1 + k|Xt∗ | − (k − 1)(3w + 3) 6 f(|Q|) + 1 + (k(w + 1) − 3(k − 1)(w + 1)) 6 f(|Q|) since k > 2. This completes the proof.~~

Lemma 17. Let (T, X ) be a tree-decomposition of a graph G of width at most w ∈ N0, where X = (Xt : t ∈ V (T )). Then for every Q ⊆ V (G), there exists a subset F of V (T ) with |F | 6 max{|Q| − 3w − 3, 0} such that:

0 S 1. for every F -part T of T , | t∈V (T 0) Xt ∩ Q| 6 12w + 13, and 2. if Q 6= ∅, then for every t∗ ∈ F , there exists at least two F -parts T 0 of T satisfying t∗ ∈ ∂T 0 S and Q ∩ t∈V (T 0) Xt − Xt∗ 6= ∅.

Proof. Let Q ⊆ V (G). By Lemma 16 with = 1, there exists F ⊆ V (T ) with |F | 6 max{|Q| − 0 S 3w − 3, 0} such that for every F -part T of T , | t∈V (T 0) Xt ∩ Q| 6 12w + 13. Assume further that F is minimal. Suppose that Q 6= ∅ and there exists t∗ ∈ F such that there is at most one F -part T 0 of T ∗ 0 S ∗ ∗ satisfying t ∈ ∂T and Q ∩ t∈∂T 0 Xt − Xt∗ 6= ∅. Let F := F − {t }. Note that for every F -part ∗ ∗ ∗ S W of T with t 6∈ ∂W , W is an F -part of T and ∂W ⊆ F − {t }, so | t∈V (W ) Xt ∩ Q| 6 12w + 13. In addition, there is exactly one F ∗-part T ∗ of T with t∗ ∈ V (T ∗), and every F ∗-part of T other than T ∗ is an F -part of T . Since there is at most one F -part T 0 of T satisfying t∗ ∈ ∂T 0 and S S S Q ∩ t∈∂T 0 Xt − Xt∗ 6= ∅, we know | t∈V (T ∗) Xt ∩ Q| = | t∈V (T 0) Xt ∩ Q| 6 12w + 13. This contradicts the minimality of F and proves the lemma. We call the set F mentioned in Lemma 17a (T, X ,Q)-fence.

4.1 Parades and Fans This subsection introduces parades and fans, and proves an auxiliary lemma that will be used in the proof of Theorem1. Let (T, X ) be a rooted tree-decomposition of a graph G of width w, where X = (Xt : t ∈ V (T )). 0 0 Let t, t be distinct nodes of T with t ∈ V (Tt). Let Tt be the subtree of T rooted at t. For 0 k ∈ [0, w + 1] and m ∈ N0, a (t, t , k)-fan of size m is a sequence (t1, t2, . . . , tm) of nodes such that:

~~0 (FAN1) for every j ∈ [m − 1], tj+1 ∈ V (Ttj ) − {tj} and t ∈ V (Ttj ),~~

~~(FAN2) for every j ∈ [m], |Xtj ∩ Xt| = k, and~~

~~(FAN3) Xtj − Xt are pairwise disjoint for all j ∈ [m].~~

13 Let T be a rooted tree. A parade in T is a sequence (t1, t2, . . . , tk) of nodes of T (for some k ∈ N) such that for every α ∈ [k − 1], tα+1 ∈ V (Ttα ) − {tα}. A sequence (a1, a2, . . . , aα) is a subsequence of a sequence (b1, b2, . . . , bβ) (for some positive integers α, β) if there exists an injection ι :[α] → [β] such that ι(1) < ι(2) < ··· < ι(α) and ai = bι(i) for every i ∈ [α].

~~Lemma 18. For every w ∈ N0 and k ∈ N, there exists N := N(w, k) ∈ N such that for every rooted tree-decomposition (T, X ) of a graph G, where X = (Xt : t ∈ V (T )), and for every parade~~

(t1, t2, . . . , tN ) in T with |Xti | 6 w + 1 for every i ∈ [N], if Xti 6⊆ Xtj for every i > j, then there 0 0 0 0 0 exist ` ∈ [0, w] and a (t1, tk, `)-fan (t1, t2, . . . , tk) of size k that is a subsequence of (t1, t2, . . . , tN ). Proof. Deﬁne N(0, k) := max{k, 2}, and for every x ∈ [w], deﬁne N(x, k) := N(x − 1, k) · (k − 1)(w + 1).

We proceed by induction on w. When w = 0, since Xti 6⊆ Xtj for every i > j, (t1, t2, . . . , tN ) is a (t1, tN , 0)-fan of size N > k. So we may assume that w > 1 and this lemma holds for every smaller w. 0 0 0 0 0 Suppose to the contrary that for every ` ∈ [0, w], there exists no (t1, tk, `)-fan (t1, t2, . . . , tk) of size k that is a subsequence of (t1, t2, . . . , tN ). In particular, there exists no k elements in {t1, t2, . . . , tk} with disjoint bags. Hence there exists a parade (q1, q2, . . . , qk−1) that is a subsequence 0 of (t1, t2, . . . , tN ) such that for every i ∈ [N], there exists i ∈ [k − 1] such that Xti ∩ Xqi0 6= ∅ and ti ∈ V (Tqi0 ). For each i ∈ [k − 1], let Si := {tj ∈ V (Tqi ): j ∈ [N],Xtj ∩ Xqi 6= ∅}. So there exists i ∈ [k−1] such that |S | N/(k−1). For every α ∈ [w+1], let S0 := {t ∈ S : |X ∩X | = α}. 1 i1 > α j i1 tj qi1 ∗ 0 N So there exists α ∈ [w + 1] such that |Sα∗ | > |Si1 |/(w + 1) > (k−1)(w+1) > 2. Since |Xti | 6 w + 1 ∗ for every i ∈ [N], and Xti 6⊆ Xtj for every i > j, α ∈ [w]. 0 Let (z , z , . . . , z 0 ) be the parade formed by the element of S ∗ . Since (T, X ) is a tree- 1 2 |Sα∗ | α ∗ 0 decomposition, there exists Z ⊆ X with |Z| = α such that for every i ∈ [|S ∗ |], X ∩ X = Z. qi1 α zi qi1 For every t ∈ V (T ), let X0 := X − Z. Let X 0 := (X0 : t ∈ V (T )). Then |X0 | = |X | − |Z| t t t zi zi 6 0 0 0 0 (w − 1) + 1 for every i ∈ [|S ∗ |]. If there exist i, j ∈ [|S ∗ |] with i > j such that X ⊆ X , then α α zi zj 0 0 0 0 0 X = X ∪ Z ⊆ X ∪ Z = X , a contradiction. So for any i, j ∈ [|S ∗ |] with i > j, X 6⊆ X . zi zi zj zj α zi zj 0 N Since |Sα∗ | > (k−1)(w+1) > N(w − 1, k), by the induction hypothesis, there exist ` ∈ [0, w − |Z|] 0 0 0 0 0 0 and a (t , t , `)-fan (t , t , . . . , t ) in (T, X ) of size k that is a subsequence of (z , z , . . . , z 0 ) and 1 k 1 2 k 1 2 |Sα∗ | 0 0 0 0 hence is a subsequence of (t1, t2, . . . , tN ). Since Z ⊆ Xzi for every i ∈ [|Sα∗ |], (t1, t2, . . . , tk) is a 0 0 (t1, tk, ` + |Z|)-fan in (T, X ). Note that ` + |Z| ∈ [0, w]. This proves the lemma.

~~5 Setup for Main Proof~~

This section proves Theorem1. The proof uses a list-coloring argument, where we assume that color i does not appear in the lists of vertices in layers Vj with j ≡ i (mod s + 2). This ensures that each monochromatic component is contained within at most s + 1 consecutive layers. For our purposes a color is an element of Z.A list-assignment of a graph G is a function L with domain containing V (G), such that L(v) is a non-empty set of colors for each vertex v ∈ V (G). For a list-assignment L of G, an L-coloring of G is a function c with domain V (G) such that c(v) ∈ L(v) for every v ∈ V (G). So an L-coloring has clustering η if every monochromatic component has at most η vertices. A list-assignment L of a graph G is an `-list-assignment if |L(v)| > ` for every vertex v ∈ V (G). Let G be a graph and Z ⊆ V (G).A Z-layering V of G is an ordered partition (V1,V2,... ) of V (G) − Z into (possibly empty) sets such that for every edge e of G − Z, there exists i ∈ N such

14 that both ends of e are contained in Vi ∪ Vi+1. Note that a layering is equivalent to an ∅-layering. For a tree-decomposition (T, X ) of G, with X = (Xt : t ∈ V (T )), the V-width of (T, X ) is

~~max max |Xt ∩ Vi|. i∈N t∈V (T )~~

Let G be a graph and let Z ⊆ V (G). For every s ∈ N and ` ∈ [s + 2], an s-segment of a Z- Sa+s layering (V1,V2,... ) of level ` is j=a Vj for some (possibly non-positive) integer a with a ≡ ` + 1 (mod s + 2), where Va = ∅ if a 6 0. When the integer s is clear from the context, we write segment instead of s-segment. Let G be a graph and let s ∈ N. Let Z ⊆ V (G) and V = (V1,V2,... ) be a Z-layering of G.A list-assignment L of G is (s, V)-compatible if the following conditions hold:

• L(v) ⊆ [s + 2] for every v ∈ V (G). S • i 6∈ L(v) for every i ∈ [s + 2] and v ∈ (Vj : j ≡ i (mod s + 2)). Note that there is no condition on L(v) for v ∈ Z. We remark that for every i ∈ [s + 2], if v ∈ V (G) with i ∈ L(v), then either v ∈ Z, or v belongs to a segment of V with level i. This leads to the following easy observation that we frequently use.

Proposition 19. Let G be a graph and let s ∈ N. Let V = (V1,V2,... ) be a layering of G, and let L be an (s, V)-compatible list-assignment. If k ∈ N and c is an L-coloring such that for every i ∈ [s+2] and every s-segment S with level i, every c-monochromatic component contained in G[S] with color i contains at most k vertices, then c has clustering at most k.

Proof. Since for every i ∈ [s + 2], if v ∈ V (G) with i ∈ L(v), then v belongs to a segment of V with level i, so every c-monochromatic component with color i is contained in some segment of V with level i. Let G be a graph, Z ⊆ V (G), s a positive integer, V a Z-layering of G, and L an (s, V)- compatible list-assignment of G. For Y1 ⊆ V (G), we say that (Y1,L) is a V-standard pair if the following conditions hold:

~~(L1) Y1 = {v ∈ V (G): |L(v)| = 1}.~~

~~and L(y) ∩ L(u) = ∅ for every u ∈ NG(y) ∩ Y1. (Note that |L(y)| > 2.)~~

~~(L3) For every v ∈ V (G) − NG[Y1], we have |L(v)| = s + 1.~~

Let G be a graph, Z ⊆ V (G), V a Z-layering, and (Y1,L) a V-standard pair. For a subset W S of V (G) and color i (not necessarily belonging to v∈V (G) L(v)), a (W, i)-progress of (Y1,L) is a 0 0 pair (Y1 ,L ) deﬁned as follows:

~~0 • Let Y1 := Y1 ∪ W . 0 • For every y ∈ Y1, let L (y) := L(y). 0 0 • For every y ∈ Y1 − Y1, let L (y) be a 1-element subset of L(y) − {i}, which exists by (L1).~~

~~[ 0 L(v) − {L (w): w ∈ NG(v) ∩ (W − Y1)}~~

~~0 of size |L(v)| − |NG(v) ∩ (W − Y1)|, which exists since |L (w)| = 1 for w ∈ NG(v) ∩ (W − Y1). 0~~

~~A(Y1,L)(W ) := {v ∈ V (G) − Y1 : v is a gate for some y ∈ W with respect to (Y1,L)}.~~

~~The next lemma implies Theorem1 by taking η = 0 and Y1 = ∅. Most of the work in proving Lemma 20 is done by Lemma 21 in the next section. So we prove Lemma 20 ﬁrst, assuming Lemma 21.~~

∗ Lemma 20. For every s, t, w ∈ N and η ∈ N0, there exists η ∈ N such that if G is a graph with no Ks,t-subgraph, V is a layering of G, L is an (s, V)-compatible list-assignment, Y1 is a subset of V (G), (Y1,L) is a V-standard pair, and (T, X ) is a tree-decomposition of G of V-width at most w such that |Y1 ∩ S| 6 η for every s-segment S of V, then there exists an L-coloring of G with clustering η∗.

∗ Proof of Lemma 20 assuming Lemma 21. Let s, t, w ∈ N and η ∈ N0. Let η be the number ∗ η (s, t, w + η) from Lemma 21. Let G be a graph with no Ks,t-subgraph, V a layering of G, L an (s, V)-compatible list-assignment, Y1 a subset of V (G), (Y1,L) a V-standard pair, and (T, X ) a tree-decomposition of G of V-width at most w such that |Y1 ∩ S| 6 η for every s-segment S of V. Say X = (Xt : t ∈ V (T )). ∗ ∗ ∗ ∗ For each t ∈ V (T ), let Xt := Xt ∪ Y1. Let X = (Xt : t ∈ V (T )). Let t be a node of T . Then ∗ ∗ (T, X ) is a tree-decomposition of G such that Xt∗ contains Y1. Since for every s-segment S of V, ∗ |Y1 ∩ S| 6 η, the V-width of (T, X ) is at most w + η. Therefore, by Lemma 21, there exists an L-coloring of G with clustering η∗.

~~6 Main Lemma~~

~~The next lemma is the heart of the proof of Theorem1.~~

Lemma 21. For every s, t, w ∈ N, there exists η∗ := η∗(s, t, w) ∈ N such that if G is a graph with no Ks,t-subgraph, V is a layering of G, L is an (s, V)-compatible list-assignment, Y1 is a subset of V (G), (Y1,L) is a V-standard pair, and (T, X ) is a tree-decomposition of G of V-width at most w ∗ such that some bag contains Y1, then there exists an L-coloring of G with clustering η .

~~Proof. Let s, t, w ∈ N. We start by deﬁning several values used throughout the proof:~~

~~• Let f be the function fs,t,w in Corollary 14.~~

~~• Let f0 be the identity function f on N0; for every i > 1, let fi be the function from N0 to N0 such that fi(x) = fi−1(x) + f(fi−1(x)) for every x ∈ N0.~~

~~16 • Let s∗ := 24(s + 2). ∗ • Let w0 := 12w s + 13.~~

~~• Let g0 : N0 → N0 be the function deﬁned by g0(0) := w0 and g0(x) := f1(g0(x − 1)) + 2w0 for every x ∈ N.~~

~~• Let g1 : N0 → N0 be the function deﬁned by g1(0) := g0(w0) and g1(x) := f1(g1(x − 1) + 3w0) for every x ∈ N.~~

~~• Let g2 : N0 × N0 → N0 be the function deﬁned by g2(0, y) := y + w0 for every y ∈ N0, and g2(x, y) := f2(g2(x − 1, y)) + 3w0 for every x ∈ N.~~

~~• Let g3 : N0 → N0 be the function deﬁned by g3(0) := g1(s+2), and g3(x) := g2(s+2, g3(x−1)) for every x ∈ N.~~

~~• Let η1 := w0g1(s + 2) + 3w0g3(4w0) + w0.~~

~~• Let g4 : N0 × N0 → N0 be the function deﬁned by g4(0, y) := y and g4(x, y) := (s + 2) · f1(g4(x − 1, y) + 2η1) for every x ∈ N and y ∈ N0,~~

~~• Let g5 : N0 → N0 be the function deﬁned by g5(x) := g4(s + 2, g4(w0, x)) + 2g3(4w0) for every x ∈ N0.~~

~~• Let g6 : N0 → N0 be the function deﬁned by g6(0) := w0, and g6(x) := g5(g6(x − 1)) + w0 for every x ∈ N.~~

~~• Let η2 := g6(5w0(2w0 + 1) + 9w0).~~

~~• Let η3 := (η1 + 2η2 + 1)w0.~~

~~• Let g7 : N0 × N0 → N0 be the function deﬁned by g7(0, y) := y and g7(x, y) := f1(g7(x − 1, y)) + w0 for every x ∈ N and y ∈ N0.~~

~~• Let g8 : N0 → N0 be the function deﬁned by g8(0) := η1 +2η2 and g8(x) := g7(s+2, g8(x−1)) for every x ∈ N.~~

~~• Let g9 : N0 → N0 be the function deﬁned by g9(0) := 2g8(w0) + η2 and g9(x) := f1(g9(x − 1)) + w0 for every x ∈ N.~~

~~• Let η4 := g9(s + 2).~~

~~• Let g10 : N0 → N0 be the function deﬁned by g10(0) := 2η4 and g10(x) := fs+3(g10(x − 1) + η3 + 2η4) for every x ∈ N.~~

• Let η5 := 3g10(η3). 2 2 • Let ψ1 : [0, w0] → N be the function such that for every (x1, x2) ∈ [0, w0] , ψ1(x1, x2) := 3w0+3 3 w0+1 (10(fw0 (η5)) w0 + w0) · (min{x1, x2}· (w0 + 1) · (f(η5) + 1) + 1). 2 2 • Let ψ2 : [0, w0] → N be the function such that for every (x1, x2) ∈ [0, w0] , ψ2(x1, x2) := ψ1(100x1, 100x2). 2 2 • Let ψ3 : [0, w0] → N be the function such that for every (x1, x2) ∈ [0, w0] , ψ3(x1, x2) := w (f(η5)) 0 ψ1(x1, x2) · . ψ1(0,0)

• Let κ0 := ψ2(w0, w0) + ψ3(w0, w0). 2 • Let κ1 := w0 · N18(w0, κ0 + w0 + 3), where N18 is the number N mentioned in Lemma 18. Px • Let φ1 : N → N be the function deﬁned by φ1(1) := κ1, and φ1(x+1) := κ1·(1+ i=1 φ1(i))+1 for every x ∈ N.

17 • Let φ2 : N0 → N0 be the function deﬁned by φ2(0) := 0, and φ2(x) := ((φ1(x) + 1)w0f(η5) + Px−1 1) · (1 + i=1 φ2(i)) for every x ∈ N. Px • Let φ3 : N → N be the function deﬁned by φ3(x) := (1 + i=1 φ2(i)) · (φ1(x) + 1) · x for every x ∈ N.

2w0+6 2 2 • Let h : N0 → N be the function deﬁned by h(0) := 2 (φ3(w0) · f(η5) · w0) (η5 + 3), and 2w0+7 2 2 for every x ∈ N0, h(x + 1) := (h(x) + 2) · 2 (φ3(w0) · f(η5) · w0) (η5 + 3).

~~• Let η6 := h(w0 − 1).~~

~~η6 • Let η7 := η5 · (f(η5)) . ∗ • Deﬁne η := η4 + η7.~~

Let G be a graph with no Ks,t-subgraph, V = (V1,V2,... ) a layering of G, L an (s, V)-compatible list-assignment, Y1 a subset of V (G), (Y1,L) a V-standard pair, and (T, X ) a tree-decomposition of G of V-width at most w such that some bag contains Y1, where X = (Xt : t ∈ V (T )). For every S R ⊆ V (T ), define XR := t∈R Xt. For every U ⊆ V (G), define X |U = (Xt ∩ U : t ∈ V (T )). For every integer i with i 6∈ [|V|], define Vi = ∅. ∗ ∗ Let r be the node of T such that Xr∗ contains Y1. Consider T to be rooted at r . Orient the ∗ edges of T away from r . For each node q of T , let Tq be the subtree of T induced by q and all the ∗ descendants of q. For each vertex v of G, let rv be the node of T closest to r with v ∈ Xrv . We construct the desired L-coloring of G by an algorithm. We now give an informal description of several notions that are used in the algorithm below. The layers are partitioned into pairwise disjoint belts, where each belt consists of a very large (but still bounded) set of consecutive layers. An interface consists of roughly the last one third of the layers within one belt, along with the first roughly one third of the layers in the next belt. Then the interior of an interface is the last few layers of the first belt, along with the first few layers of the next belt. The algorithm also uses a linear ordering of V (G) that never changes. We associate with each subgraph of G the first vertex in the ordering that is in the subgraph. We use this vertex to order subgraphs, whereby a subgraph that has a vertex early in the ordering is considered to be “old”. We now formalize these ideas. ∗ Define σT to be a depth-first-search order of T rooted at r . That is, σT (t) = 0, and if a node t 0 ∗ 0 of T is visited earlier than a node t of T in the depth-first-search starting at r then σT (t) < σT (t ). For every node t of T , we define it to be the integer such that σT (t) = it. Define σ to be a linear order of V (G) such that for any distinct vertices u, v, if σT (ru) < σT (rv), then σ(u) < σ(v). For every subgraph H of G, define σ(H) := min{σ(v): v ∈ V (H)}. Note that for any k ∈ N0, any set Z consisting of k consecutive layers, any t ∈ V (T ), any V-standard pair (Y 0,L0), any monochromatic component M with respect to any L0-coloring in G[Y 0] intersecting 0 Z ∩ Xt, there exist at most |Z ∩ Xt| 6 kw monochromatic components M with respect to any 0 0 0 L -coloring in G[Y ] intersecting Z ∩ Xt such that σ(M ) < σ(M). Sa+s∗ A belt of (T, X ) is a subset of V (G) of the form i=a+1 Vi for some nonnegative integer a with ∗ ∗ a ≡ 0 (mod s ). Note that each belt consists of s layers. So for every belt B of (T, X ), (T, X |B) is a tree-decomposition of G[B] of width at most s∗w. For every j ∈ [|V| − 1], let

~~(j+ 1 )s∗ (j+ 1 )s∗+(2s+5) [3 3 [ Ij := Vi and Ij := Vi. 1 ∗ 1 ∗ i=(j− 3 )s i=(j− 3 )s −(2s+5)~~

~~18 ∗ Ij is called the interface at j. (Note that s is a multiple of 3, so the indices in the deﬁnition of Ij and Ij are integers.) For every j ∈ [0, |V| − 1], deﬁne~~

js∗ js∗ [ [ Ij,0 := Vi and Ij,0 := Vi, i=js∗−(s+2)+1 i=js∗−2(s+2)+1 and define js∗+s+2 js∗+2(s+2) [ [ Ij,1 := Vi and Ij,1 := Vi. i=js∗+1 i=js∗+1 ◦ ◦ ◦ Also define Ij := Ij,0 ∪ Ij,1 and Ij := Ij,0 ∪ Ij,1. For every j ∈ [|V| − 1], define Sj to be the set ◦ ◦ S ◦ of all s-segments S intersecting I . Note that I ⊆ ◦ S ⊆ I ⊆ Ij for every j ∈ [|V| − 1]. In j j S∈Sj j addition, for every j ∈ [|V| − 1], Ij is contained in a union of two consecutive belts, so for every ∗ t ∈ V (T ), |Ij ∩ Xt| 6 2s w 6 w0. For a collection E of 2-element subsets of V (G) and a coloring c of G, a monochromatic E- pseudocomponent (with respect to c) is a component of the subgraph of G + W induced by a color class of c. We now give some intuition about the algorithm that follows. The input to the algorithm is a tree-decomposition of G with bounded layered width, a set Y1 of precolored vertices, and a list assignment L of G. Throughout the algorithm, Y (i,?,?) refers to the current set of colored vertices, where the superscript (i, ?, ?) indicates the stage of the algorithm. This vector is incremented in lexicographic order as the algorithm proceeds. The algorithm starts with stage (0, −1, 0), during which time we initialize several variables. Throughout the algorithm, a monochromatic component refers to a component of the subgraph of G induced by the current precolored set of vertices (or to be more precise, vertices with one color in their list). The algorithm does a depth-first-search on the tree T indexing the tree-decomposition, considering the nodes t of T with σT (t) = i in turn (for i = 0, 1, 2,... ). The algorithm first builds a fence around the subgraph of vertices in bags rooted at node t (relative to belts). At stage (i, −1,?), ◦ the algorithm tries to isolate the k-th oldest component intersecting some segment intersecting Ij . Then at stage (i, 0,?), the algorithm isolates the other monochromatic components intersecting Xt. In both these stages, ? refers to the color given to vertices around the component that we are trying to isolate. Then in stage (i, ?, ?) we isolate the fences associated with subtree Tj,t. Then we add “fakes edges” to the graph between precolored vertices to merge some monochromatic components according to certain rules. Finally, the algorithm moves to the next node in the tree, and builds new fences with respect to the next node in the tree. Now we give a more detailed intuition of the algorithm. We simplify some notations in this explanation for ease of presentation. The formal description of the algorithm is described later. During the algorithm, we construct the following. • Subsets Y (i,j,k) of V (G) for i ∈ [0, |V (T )|], j ∈ {−1} ∪ [0, |V (T )| + 1] and k ∈ [0, s + 2]: The set Y (i,j,k) is the set of precolored vertices at stage (i, j, k), where a vertex is said to be precolored if its list contains only one color. At each stage, we precolor more vertices, so these sets have the property that Y (i,j,k) ⊆ Y (i0,j0,k0) if (i, j, k) is lexicographically smaller than (i0, j0, k0).

• Subsets Fj,t of V (T ) for j ∈ [|V| − 1] and t ∈ V (T ): Each of these sets Fj,t is a union of 0 fences. For any ﬁxed j ∈ [|V| − 1] and node t of T , assuming Fj,p is given, where p is the parent of t, we do the following:

19 0 – We ﬁrst construct Fj,p as the following: We consider the Fj,p ∩ V (Tt)-parts of Tt containing t. Note that it is possible to have more than one such part since t could be in 0 0 Fj,p. Then for each such part T , we restrict ourselves to the subgraph of G induced by 0 XV (T 0) ∩ Ij. That is, the subgraph induced by the subset of vertices in the part T and in the closure of the interface at j. Note that this subgraph has a tree-decomposition naturally given by (T, X ). We construct a fence Fj,T 0 in this tree-decomposition based on the current set of precoloring vertices Y (i,−1,0) in this subgraph, where i = σ(t). Then 0 0 Fj,p is the union of Fj,p and Fj,T 0 among all such parts T . (i,−1,0) (i+1,−1,0) – When Fj,p is deﬁned, we extend our precolored set Y to Y by some pro- (i+1,−1,0) 0 cedure that is explained later. After Y is given, we construct Fj,t in a similar 0 0 (i,−1,0) way as we construct Fj,p from Fj,p, except we replace Fj,p by Fj,p and replace Y by Y (i+1,−1,0).

~~The key idea is to make sure that for every Fj,p ∩ V (Tt)-part containing t, the subgraph induced by this part only contains few precolored vertices.~~

• Subtrees Tj,t of Tt and the “boundary” ∂Tj,t of Tj,t, for j ∈ [|V| − 1] and t ∈ V (T ): Tj,t is the union of all Fj,p ∩ V (Tt)-parts containing t, and ∂Tj,t is the union of the boundary of those parts, where t is not included. Roughly speaking, Tj,t is a small portion of Tt containing t,

~~and XV (Tj,t) contains only few prepcolored vertices by the construction of the fences.~~

• Subsets Zt of XV (Tt) for every t ∈ V (T ): Zt consists of the vertices in XV (Tt) that are either not in the interior of any interface, or in the interior of the interface at j for some j but also

in XV (Tj,t). • Subsets D(∗,∗,∗) of V (G): These sets help to extend the current set of precolored vertices. We explain the usage of these sets later. (∗,∗) • Subsets Sj,t of V (G) for j ∈ [|V| − 1] and t ∈ V (T ): These sets play a minor role in the algorithm. Some vertices trigger the extension of the coloring during the algorithm and become “useless” afterwards. These sets collect such vertices and are used in the proof in order to bound the size of the monochromatic components. (i) (i,∗) • Sets Ej,t and Ej,t of “fake edges”, for i ∈ [0, |V (T )|], j ∈ [|V| − 1] and t ∈ V (T ): Each fake edge is a pair of two distinct vertices that are contained in the same s-segment in ◦ Sj . At any moment of the algorithm, the current set of precolored vertices form some monochromatic components, and it is possible that more than one of them will be merged into a bigger monochromatic component after we further color more vertices. These fake edges join current monochromatic components that are possibly to be merged into a bigger monochromatic component in the future to form “pseudocomponents”. Note that whether a pair of two vertices is a fake edge depends on j and t.

Now we give a sketch of the algorithm and explain the intuition. Recall that during the algorithm, (i,∗,∗) Y always contains all vertices of G contained in Xt with σT (t) 6 i. So at the very beginning (0,−1,0) of the algorithm, we color every uncolored vertex in Xr∗ to get Y . Then we proceed the algorithm for i = 0, 1, 2, ··· . Fix i ∈ N0 and t the node of T with σT (t) = i. Here we explain what the algorithm will do at this stage.

~~0 • First, for each j ∈ [|V| − 1], we build fences Fj,p and Fj,p, where p is the parent of t, and subtrees Tj,t of Tt. Note that it deﬁnes Zt. We further color vertices in Zt. The idea for~~

20 constructing those fences and Zt is to make sure that only a small number of precolor vertices in Zt are close to the boundary of two different belts. • Then we move to stage (i, −1, ∗). In this stage, we “isolate” the monochromatic pseudocom- ◦ ponents in Zt intersecting Xt and some segment in Sj for some j ∈ [|V| − 1]. That is, we color some vertices in Zt to stop the growth of those monochromatic components by coloring their gates by using a color that is different from the color of the monochromatic pseudocomponent. Note that we cannot isolate all such components at once, since some vertex can be a gate of two monochromatic pseudocomponents with different colors. The way we break the tie is according to the ordering σ of those monochromatic components. For each j ∈ [|V| − 1], we isolate the monochromatic pseudocomponent with minimum σ-order, and then isolate the monochromatic pseudocomponent with second minimum σ-order and so on. Say we are isolating the k-th monochromatic pseudocomponent among all those monochro- ◦ (i,−1,k) matic pseudocomponents intersecting Xt and some segment in Sj for each j. Then W0 consists of the vertices in the k-th monochromatic pseudocomponent for all j. Then for each j, there uniquely exists a color q + 1 such that this monochromatic pseudocomponent uses (i,−1,k,q+1) (i,−1,k) color q + 1. We write W1 to denote the vertices in W0 with color q + 1. Then (i,−1,k,q+1) we color the gates of W1 contained in Zt by using a color different from q + 1. Then we move to color the (k + 1)-th monochromatic pseudocomponent, and so on. Note that the (k + 1)-th monochromatic pseudocomponent can become larger due to the above process of coloring gates. • Now we move to stage (i, 0, ∗). In this stage we isolated all monochromatic pseudocomponents in Zt intersecting Xt. Note that all monochromatic pseudocomponents intersecting ◦ some s-segment in Sj for some j ∈ [|V| − 1] are isolated in the previous stage. So all the ◦ remaining monochromatic pseudocomponents are disjoint from the s-segments in Sj for every j ∈ [|V| − 1], and hence they are disjoint from all fake edges. Hence all monochromatic pseudocomponents that are dealt with at this stage are indeed monochromatic components. We isolate all such monochromatic components with color 1, and then the ones with color 2 and so on.

• Now we move to stage (i, > 1, ∗). In this stage we restrict to XV (Tj,t) ∩Ij, for each j ∈ [|V|−1]. (i) We denote the union of them by W4 . Now we ﬁx j to be an integer in [|V|−1]. The objective

~~is that at the end of this stage, the monochromatic components contained in XV (Tj,t) ∩ Ij~~

intersecting ∂Tj,t will not have more new vertices in XV (Tj,t) ∩ Ij in the future. The idea for 0 achieving this is to color Xt0 ∩Ij for some nodes t ∈ ∂Tj,t in the following way. Consider a set of “base nodes”. At the beginning, the only base node is t. If there exists a monochromatic 0 0 path P from a base node to a node t in ∂Tj,t, then we add t into the set of base nodes and (i,∗,∗) color Xt0 ∩ Ij, unless the vertex in V (P ) ∩ Xt0 belongs to D − Xt or in the intersection (i,∗,∗) 0 0 of D ∩ Xt and a bag of a some base nodes at stage (i , ∗, ∗) for some i < i, where D(i,∗,∗) is a special set of vertices roughly consisting of all vertices contained in the previous monochromatic paths for generating new base nodes. Then we isolate those newly colored vertices by coloring their gates. It might create new monochromatic paths from base nodes to other node in ∂Tj,t. We repeat the above process until no new base nodes can be added. The intuition of considering D(i,∗,∗) is to make sure that we only include “necessary” nodes in ∂Tj,t in the set of base nodes in order to limit the size of the set of precolored vertices. • Then we move to the stage for adding new fake edges. Recall that when we color more vertices, we always try to isolate some monochromatic pseudocomponents. But such isolation

21 is “incomplete” because of the definition of Zt. In addition, the isolation process depends on the order of the pseudocomponents. So we more or less require some extra information to indicate whether two monochromatic pseudocomponents will be joined into one in the future so that they should be treated the same when we order the pseudocomponents. This is the purpose of the fake edges. Whether a pair of vertices form a fake edge depends on where we “observe” it. That is, when we “stand on” a node t0 of T , we consider whether we should add a fake edge to connect two vertices if in the future there will be monochormatic paths 0 in XV (Tt) connecting these two vertices. So it is only relevant when t ∈ V (Tt). In addition, 0 (it) all nodes t ∈ V (Tt) − {t} will get the same new fake edges at this stage; and the sets Ej,t0 (it) 0 00 and Ej,t00 of fake edges for two different nodes t and t , respectively, are different only when 0 00 t is a descendant of the common ancestor of t and t with the largest σT -value. 0 We fix a node t ∈ V (Tt). Each fake edge is a pair of distinct vertices u, v in Xt00 ∩ S for 00 ◦ some j ∈ [|V| − 1], t ∈ ∂Tj,t and S ∈ Sj . Also, u and v must be already colored by the same color. So there exist monochromatic pseudocomponents Mu and Mv containing u and v, respectively. And Mu 6= Mv, otherwise u and v are already contained in the same monochromatic pseudocomponent and there is no need to add a fake edge to connected u and v. And the level of S is the color of u and v, so that no matter how Mu or Mv will grow, they must be contained in S. By symmetry, we may assume σ(Mu) < σ(Mv). Note that we want to add a fake edge to link u and v if it is possible that Mu and Mv will be joined into one monochormatic component in the future. So each Mu and Mv must have gates contained in X − X 00 . Recall that during the isolation process, the monochormatic V (Tt00 ) t psuedocomponent that with minimum σ-order will be isolated the first, so that it will not grow in the future. Hence Mu and Mv cannot be the monochromatic pseudocomponent with the minimum σ-order. In addition, the need of adding a fake edge is “triggered” by another monochromatic pseudocomponent intersecting Xt00 with smaller σ-value. That is, there exists a monochromatic pseudocomponent intersecting Xt00 with smaller σ-order than σ(Mu) and σ(Mv) so that isolating this monochromatic pseudocomponent might lead to possble needs for coloring more vertices to eventually make Mu and Mv be merged into one monochromatic pseudocomponent. Hence, there is no danger for merging Mu and Mv into one monochromatic pseudocomponent in the future if the gates of those monochromatic pseudocomponents with smaller σ-value than σ(Mu) and σ(Mv) are separated by a “thick buffer” from the gates of Mu and Mv. Here a “thick buffer” means a sequence of nodes that is similar to a fan defined in the previous section, so it is a sequence of many nodes of T such that they more or less have disjoint bags, and there exist a directed path in T passing through all of them. Moreover, due to the way we isolate monochromatic pseudocomponents in stages (6 i, ∗, ∗), there is no need to add the fake edge {u, v} if Tj,t ⊆ Tj,q for some node q with iq < it = i, and there exists some “evidence” that if we decide not to add a fake edge to join Mu and Mv at the stage (iq, ∗, ∗), then there is no need to add the fake now. 0 • Finally, color all vertices in the bag of the node with σT -value i + 1, build fences Fj,t and move to the next i.

~~Stage (0, −1, 0): Initialization~~

~~(0,−1,0) (0,−1,0) • Let (Y ,L ) be an (Xr∗ , 0)-progress of (Y1,L).~~

~~22 • Let D(0,0,0) := ∅.~~

• Let Fj,0 := ∅ for every j ∈ [|V| − 1]. (0,0) (0,1) (0,2) • Let Sj,t := ∅, Sj,t := ∅ and Sj,t := ∅ for every j ∈ [|V | − 1] and t ∈ V (T ). (0) • Let Ej,t := ∅ for every j ∈ [|V| − 1] and t ∈ V (T ).

~~For i = 0, 1, 2,... , let t be the node of T with σT (t) = i and perform the following steps: Stage: Building Fences~~

~~• Deﬁne the following:~~

– For every j ∈ [|V| − 1], 0 ∗ ∗ ∗ Let Fj,p := Fj,0 if t = r , and let p be the parent of t if t 6= r , 0 0 ∗ For every Fj,p ∩ V (Tt)-part T of Tt containing t, deﬁne Fj,T 0 to be a 0 (i,−1,0) (T , X | , (Y ∪ X∂T 0 ) ∩ XV (T 0) ∩ Ij)-fence. XV (T 0)∩Ij 0 S 0 0 ∗ Let Fj,p := Fj,p ∪ T 0 Fj,T 0 , where the union is over all Fj,p ∩ V (Tt)-parts T of Tt containing t. S 0 0 ∗ Let Tj,t := T 0 T , where the union is over all Fj,p ∩ V (Tt)-parts T of Tt containing t. S 0 0 ∗ Let ∂Tj,t := T 0 (∂T − {t}), where the union is over all Fj,p ∩ V (Tt)-parts T of Tt containing t. S|V|−1 ◦ S|V|−1 ◦ – Let Zt := (XV (Tt) − j=1 Ij ) ∪ ( j=1 (Ij ∩ XV (Tj,t))).

~~◦ Stage (i, −1,?): Isolate the k-th Oldest Component Intersecting a Segment in Sj :~~

~~• For each k ∈ [0, w0 − 1], deﬁne the following:~~

(t) (i) – For every j ∈ [|V| − 1], deﬁne Mj,k to be the monochromatic Ej,t -pseudocomponent in (i,−1,k) (t) G[Y ] such that σ(Mj,k ) is the smallest among all monochromatic (t) (i,−1,k) Ej,t -pseudocomponents M in G[Y ] intersecting Xt with AL(i,−1,k) (V (M))∩XV (Tt)− ◦ Xt 6= ∅ and contained in some s-segment S ∈ Sj whose level equals the color of M such Sk−1 (t) that V (M) ∩ `=0 V (Mj,` ) = ∅. (i,−1,k) S (t) – Let W0 := XV (Tt) ∩ j∈[|V|−1] V (Mj,k ). – Let (Y (i,−1,k,0),L(i,−1,k,0)) := (Y (i,−1,k),L(i,−1,k)). – For every q ∈ [0, s + 1], deﬁne the following:

(i,−1,k,q) (i,−1,k) (i,−1,k) ∗ Let W1 := {v ∈ W0 : L (v) = q + 1}. (i,−1,k,q) (i,−1,k,q) ∗ Let W2 := AL(i,−1,k,q) (W1 ) ∩ Zt. (i,−1,k,q+1) (i,−1,k,q+1) (i,−1,k,q) (i,−1,k,q) (i,−1,k,q) ∗ Let (Y ,L ) be a (W2 , q+1)-progress of (Y ,L ). (i,−1,k,q+1) (i,−1,k,q) (t) (t) ∗ If (Y − Y ) ∩ Ij ∩ Zt 6= ∅ and cj is undeﬁned, then let cj := q + 1. – Let (Y (i,−1,k+1),L(i,−1,k+1)) := (Y (i,−1,k,s+2),L(i,−1,k,s+2)).

~~Stage (i, 0,?): Isolate the other components intersecting Xt • Let (Y (i,0,0),L(i,0,0)) := (Y (i,−1,w0),L(i,−1,w0)). • For each k ∈ [0, s + 1], deﬁne the following:~~

23 (i,0,k) (i,0,k) (i,0,k) – Let W0 := {v ∈ Y ∩ XV (Tt) : L (v) = {k + 1} and (i,0,k) there exists a path P in G[Y ∩ XV (Tt)] from v to Xt such that L(i,0,k)(q) = {k + 1} for all q ∈ V (P )}. (i,0,k) (i,0,k) – Let W2 := AL(i,0,k) (W0 ) ∩ Zt. (i,0,k+1) (i,0,k+1) (i,0,k) (i,0,k) (i,0,k) – Let (Y ,L ) be a (W2 , k + 1)-progress of (Y ,L ).

Stage (i, > 1,?): Isolating the Fences of Tj,t (i,−1) S|V|−1 • Let W3 := j=1 (Xt ∩ Ij). (i) S|V|−1 • Let W4 := j=1 (XV (Tj,t) ∩ Ij). • For each ` ∈ [0, |V (T )|], deﬁne the following:

(i,`) (i,`−1) S|V|−1 S – Let W3 := W3 ∪ j=1 q(Xq ∩ Ij), where the last union is over all nodes q (i,`,s+2) (i) in ∂Tj,t for which there exists a monochromatic path in G[Y ∩ Ij ∩ W4 ] from (i,`−1) W3 to

(i,`,0) (i,`,0) 0 Xq∩Ij − ((D − Xt) ∪ {v ∈ D ∩ Xt ∩ Xq0 ∩ Ij : q ∈ V (Tt) − {t}, 0 (i0,`0) 0 0 q is a witness for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, i − 1] and ` ∈ [0, |V (T )|]})

internally disjoint from Xt ∪ X∂Tj,t . (i,`+1,0) (i,`+1,0) (i,`) (i,`,s+2) (i,`,s+2) – Let (Y ,L ) be a (W3 , 0)-progress of (Y ,L ). – For each k ∈ [0, s + 1], deﬁne the following: (i,`+1,k) (i,`+1,k) (i) (i,`+1,k) ∗ Let W0 := {v ∈ Y ∩ W4 : L (v) = {k + 1} (i,`+1,k) (i) (i,`) and there exists a path P in G[Y ∩ W4 ] from v to W3 such that L(i,`+1,k)(q) = {k + 1} for all q ∈ V (P )}. (i,`+1,k) (i,`+1,k) (i) ∗ Let W2 := AL(i,`+1,k) (W0 ) ∩ W4 . (i,`+1,k+1) (i,`+1,k+1) (i,`+1,k) (i,`+1,k) (i,`+1,k) ∗ Let (Y ,L ) be the (W2 , k+1)-progress of (Y ,L ). (i,`,k+1) (i,`,k) (i,`+1,k) ∗ Let D := D ∪ W0 . – Let D(i,`+1,0) := D(i,`,s+2).

Stage: Adding Fake Edges 0 (i+1) (i) • For each j ∈ [|V| − 1] and t ∈ V (T ) − (V (Tt) − {t}), let Ej,t0 := Ej,t0 . (i,0) (i) • For each j ∈ [|V| − 1], let Ej,t := Ej,t . 0 (i+1) (i,|V (G)|) • For each j ∈ [|V| − 1] and t ∈ V (Tt) − {t}, let Ej,t0 := Ej,t , where every ` ∈ (i,`+1) (i,`) [0, |V (G)| − 1], let Ej,t be the union of Ej,t and the set consisting of the 2-element sets {u, v} satisfying the following.

~~– {u, v} ⊆ Y (i,|V (T )|+1,s+2). – L(i,|V (T )|+1,s+2)(u) = L(i,|V (T )|+1,s+2)(v). ◦ – There exists an s-segment S in Sj whose level equals the color of u and v such that {u, v} ⊆ S.~~

~~24 00 – There exists t ∈ ∂Tj,t such that:~~

∗ {u, v} ⊆ Xt00 , (i,`) ∗ V (M) ∩ Xt00 6= ∅, where M is the monochromatic Ej,t -pseudocomponent in G[Y (i,|V (T )|+1,s+2)] such that σ(M) is the (` + 1)-th smallest among all monochro- (i,`) (i,|V (T )|+1,s+2) matic Ej,t -pseudocomponents in G[Y ], ◦ ∗ M is contained in some s-segment in Sj whose level equals the color of M, 00 (i,`0) 0 ∗ t is a witness for Xt00 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], ∗ A (i,|V (T )|+1,s+2) (V (M)) ∩ X − X 00 6= ∅, L V (Tt00 ) t ∗ A (i,|V (T )|+1,s+2) (V (M )) ∩ X − X 00 6= ∅, L u V (Tt00 ) t ∗ A (i,|V (T )|+1,s+2) (V (M )) ∩ X − X 00 6= ∅, and L v V (Tt00 ) t ∗ σ(M) < min{σ(Mu), σ(Mv)}, (i,`) (i,|V (T )|+1,s+2) where Mu and Mv are the monochromatic Ej,t -pseudocomponents in G[Y ] containing u and v, respectively. (i,`) – Let M, Mu and Mv be the monochromatic Ej,t -pseudocomponents as in the previous bullet. If (V (Mu) ∪ V (Mv)) ∩ Xt 6= ∅, then at least one of the following does not hold, (i,`) 0 (i,|V (T )|+1,s+2) ∗ for every monochromatic Ej,t -pseudocomponent M in G[Y ] with 0 0 ◦ σ(M ) 6 σ(M) such that M is contained in some s-segment in Sj whose level 0 0 0 equals the color of M , M satisﬁes that AL(i,|V (T )|+1,s+2) (V (M )) ∩ XV (Tt) − Xt ⊆ X − (X 00 ∪ Z ), V (Tt00 ) t t

∗ there exists a parade (t−βu,v,` , tβu,v,`+1, . . . , t−1, t1, t2, . . . , tαu,v,` ) for some αu,v,` ∈ N and βu,v,` ∈ N such that 00 00 · t−βu,v,` ∈ V (Tt ) − {t }, · αu,v,` = ψ2(αu,`, αv,`) and βu,v,` = ψ3(αu,`, αv,`), where αu,` and αv,` are the integers such that σ(Mu,`) and σ(Mv,`) are the αu,`-th (i,`) 0 and the αv,`-th smallest among all monochromatic Ej,t -pseudocomponents M (i,|V (T )|+1,s+2) ◦ in G[Y ] contained in some s-segment in Sj whose color equals the 0 color of M and M with A (i,|V (T )|+1,s+2) (V (M ))∩X −X 00 6= ∅, respectively, u v L V (Tt00 ) t · for any distinct ` , ` ∈ [−β , α ] − {0}, X ∩ I − X 00 and X ∩ I − X 00 1 2 u,v,` u,v,` t`1 j t t`2 j t are disjoint non-empty sets, S 00 ◦ · 00 (A (i,|V (T )|+1,s+2) (V (M ))∩XV (Tt) −Xt) ⊆ (XV (Tt ) −Xtα )∩I , where M L αu,v,` u,v,` j (i,`) 00 the union is over all monochromatic Ej,t -pseudocomponents M in (i,|V (T )|+1,s+2) 00 ◦ G[Y ] such that V (M ) is contained in some s-segment in Sj whose 00 00 00 level equals the color of M , σ(M ) 6 σ(M), V (M ) ∩ Xt00 6= ∅, and 00 A (i,|V (T )|+1,s+2) (V (M )) ∩ X − X 00 6= ∅, L V (Tt00 ) t

~~· there exists x` ∈ {u, v} such that AL(i,|V (T )|+1,s+2) (V (Mx` )) ∩ (XV (Tt) − Xt) ∩~~

XV (Tt ) − Xt1 = ∅, and AL(i,|V (T )|+1,s+2) (V (Mx` )) ∩ (XV (Tt) − Xt) ∩ XV (Tt ) − 1 −βu,v,` ◦ (Xt ∪ I ) = ∅, −βu,v,` j (i) ∗ (i,|V (T )|+1,s+2) ∗ there exists a monochromatic Ej,t -pseudocomponent M in G[Y ] with ∗ ∗ ◦ σ(M ) 6 σ(M) such that M is contained in some s-segment in Sj whose level ∗ ∗ equals the color of M satisfying that AL(i,|V (T )|+1,s+2) (V (M )) ∩ XV (Tt) − Xt ⊆ X − (X 00 ∪ Z ). V (Tt00 ) t t 0 – There do not exist q ∈ V (T ) with Tj,t ⊆ Tj,q and iq < it, a witness q ∈ ∂Tj,q for 0 (iq,` ) 0 (iq) Xq0 ∩Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic Ej,q -pseudocomponent

25 (iq,|V (T )|+1,s+2) in G[Y ] intersecting Xq0 and {u, v}. Stage: Moving to the Next Node in the Tree (i+1,−1,0) (i+1,−1,0) (i,|V (T )|+1,s+2) (i,|V (T )|+1,s+2) 0 • Let (Y ,L ) be a (Xt0 , 0)-progress of (Y ,L ), where t 0 is the node of T with σT (t ) = i + 1. • Let D(i+1,0,0) := D(i,|V (T )|,s+2). 0 0 • For every node t with t ∈ V (Tt) − {t} and j ∈ [|V | − 1],

0 – If t ∈ V (Tj,t) − ∂Tj,t, then define (i+1,0) (i,0) S|V (T )| Ss+1 (i,`+1,k) ∗ Sj,t0 := Sj,t0 ∪ `=0 k=0 W0 , (i+1,1) (i,1) S|V|−1 S ∗ Sj,t0 := Sj,t0 ∪ j=1 q(Xq ∩ Ij), where the last union is over all nodes q of 0 (i,`0) T such that q ∈ V (Tt0 ) − {t } and q is a witness for Xq ∩ Ij ⊆ W3 for some `0 ∈ [0, |V (T )|], and (i+1,2) (i,2) Sw0−1 (i,−1,k) Ss+1 (i,0,k) ∗ Sj,t0 := Sj,t0 ∪ k=0 W0 ∪ k=0 W0 ; (i+1,0) (i,0) (i+1,1) (i,1) (i+1,2) (i,2) – otherwise, define Sj,t0 := Sj,t0 , Sj,t0 := Sj,t0 and Sj,t0 := Sj,t0 . Stage: Building a New Fence • For every j ∈ [|V| − 1], define the following:

0 0 – For every Fj,p ∩ V (Tt)-part T of Tt containing t, let Fj,T 0 be a 0 (i+1,−1,0) (T , X | , (Y ∪ X∂T 0 ) ∩ XV (T 0) ∩ Ij)-fence. XV (T 0)∩Ij 0 S 0 0 – Let Fj,t := Fj,p ∪ T 0 Fj,T 0 , where the union is over all Fj,p ∩ V (Tt)-parts T of Tt containing t.

It is clear that if (i0, j0, k0) is a triple lexicographically smaller than a triple (i, j, k), then 0 0 0 0 0 0 (i,j,k) (i ,j ,k ) (i,j,k) (i ,j ,k ) (it,−1,0) L (v) ⊆ L (v) for every v ∈ V (G), so Y ⊇ Y . In addition, Xt ⊆ Y . (|V (T )|,−1,0) Recall that it is the number such that σT (t) = it. In particular, we have |L (v)| = 1 for every v ∈ V (G), so there exists a unique L(|V (T )|,−1,0)-coloring c of G. By construction, c is an L-coloring. We prove below that c has clustering η∗.

Claim 21.1. Let i ∈ N0, and let t ∈ V (T ) with it = i. Let j ∈ [|V| − 1] and ` ∈ [−1, |V (T )|]. If (i) k ∈ [0, s + 1] and P is a c-monochromatic path of color k + 1 contained in G[W4 ] intersecting (i,`) (i,`+1,k) (i) W3 , then V (P ) ⊆ Y and AL(i,`+1,k+1) (V (P )) ∩ W4 = ∅.

(i,`+1,k) (i,`) (i,`+1,0) (i,`) Proof. First suppose that V (P ) 6⊆ Y . Since W3 ⊆ Y , V (P ) ∩ W3 is a nonempty subset of Y (i,`+1,k). So there exists a longest subpath P 0 of P contained in G[Y (i,`+1,k)] inter- (i,`) 0 (i) 0 (i,`+1,k) 0 secting W3 . Since V (P ) ⊆ V (P ) ⊆ W4 , V (P ) ⊆ W0 . In addition, P 6= P , for otherwise V (P ) = V (P 0) ⊆ Y (i,`+1,k), a contradiction. So there exist v ∈ V (P 0) and u ∈ 0 (i,`+1,k) (i,`+1,k) NP (v)−V (P ). That is, c(u) = k+1 and u 6∈ Y . So u ∈ AL(i,`+1,k) ({v})∩V (P )−W0 ⊆ (i,`+1,k) (i) (i,`+1,k) (i,`+1,k+1) (i,`+1,k+1) (i,`+1,k) AL(i,`+1,k) (W0 )∩W4 ⊆ W2 . But (Y ,L ) is a (W2 , k+1)-progress, so k + 1 6∈ L(i,`+1,k+1)(u). Hence c(u) 6= k + 1, a contradiction. (i) (i) Now we suppose that AL(i,`+1,k+1) (V (P ))∩W4 6= ∅. So there exists z ∈ AL(i,`+1,k+1) (V (P ))∩W4 . (i,`+1,k) (i,`+1,k) Note that we have shown that V (P ) ⊆ Y , so V (P ) ⊆ W0 and AL(i,`+1,k+1) (V (P )) ⊆ (i,`+1,k+1) (i,`+1,k) (i,`+1,k) AL(i,`+1,k) (V (P )). In addition, z ∈ AL(i,`+1,k+1) (V (P )), so z 6∈ Y ⊇ Y ⊇ W0 . (i) (i,`+1,k) (i) (i,`+1,k) (i,`+1,k) So z ∈ AL(i,`+1,k+1) (V (P ))∩W4 −W0 ⊆ AL(i,`+1,k) (V (P ))∩W4 −W0 ⊆ W2 . Since

26 (i,`+1,k+1) (i,`+1,k+1) (i,`+1,k) (i,`+1,k) (i,`+1,k) (Y ,L ) is a (W2 , k + 1)-progress of (Y ,L ), we know k + 1 6∈ (i,`+1,k+1) L (z), so z 6∈ AL(i,`+1,k+1) (V (P )), a contradiction. This proves the claim.

0 0 (i0,`,s+2) (i0,`+1,s+2) 0 Proof. Let t be the node of T with it = i . Since Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ 0 XV (Tt) ∩ Ij − Xt, t is an ancestor of t. (i0,`) (i0,`−1) Suppose that (W3 −W3 )∩Ij = ∅. Then for every k ∈ [0, s+1] and every monochromatic (i0,`+1,k) (i0) (i0,`) path P of color k +1 in G[Y ∩W4 ] intersecting W3 ∩Ij, P is a c-monochromatic path in (i0) (i0,`) (i0,`−1) (i0,`,k) (i0) G[W4 ] intersecting W3 ∩ Ij ⊆ W3 ∩ Ij, so Claim 21.1 implies that V (P ) ⊆ Y ∩ W4 , (i0,`,k) (i0,`+1,k) (i0,`,k) and hence V (P ) ⊆ W0 . This implies that W0 ∩ Ij ⊆ W0 ∩ Ij for every k ∈ [0, s + 1]. 0 Since no edge of G is between Ij and Ij0 for any j 6= j, for every k ∈ [0, s + 1],

(i0,`+1,k) (i0,`+1,k) (i0) W2 ∩ Ij = AL(i0,`+1,k) (W0 ) ∩ W4 ∩ Ij (i0,`+1,k) (i0) = AL(i0,`+1,k) (W0 ∩ Ij) ∩ W4 ∩ Ij (i0,`,k) (i0) ⊆ AL(i0,`+1,k) (W0 ∩ Ij) ∩ W4 ∩ Ij (i0,`,k) (i0) ⊆ AL(i0,`+1,k) (W0 ) ∩ W4 ∩ Ij (i0,`,k) (i0) ⊆ AL(i0,`,k) (W0 ) ∩ W4 ∩ Ij (i0,`,k) = W2 ∩ Ij.

(i0,`+1,s+2) (i0,`,s+2) Ss+1 (i0,`+1,k) Ss+1 (i0,`,k) (i0,`,s+2) Hence (Y − Y ) ∩ Ij ⊆ k=0(W2 ∩ Ij) ⊆ k=0(W2 ∩ Ij) ⊆ Y ∩ Ij. (i0,`+1,s+2) (i0,`,s+2) So Y ∩ Ij ⊆ Y ∩ Ij, a contradiction. (i0,`) (i0,`−1) Hence (W3 − W3 ) ∩ Ij 6= ∅. (i0,`,s+2) (i0,`,0) (i0+1,0) Suppose to the contrary that ` = 0, |Xt ∩Ij ∩D | 6 |Xt ∩Ij ∩D |, |Xt ∩Ij ∩Sj,t | 6 (i0,0) (i0+1,1) (i0,1) (i0,`,s+2) (i0,`,0) (i0+1,0) (i0,0) |Xt∩Ij ∩Sj,t |, and |Xt∩Ij ∩Sj,t | 6 |Xt∩Ij ∩Sj,t |. Since D ⊇ D , Sj,t ⊇ Sj,t (i0+1,1) (i0,1) (i0,`,s+2) (i0,`,0) (i0+1,0) (i0,0) and Sj,t ⊇ Sj,t , we know Xt∩Ij ∩D = Xt∩Ij ∩D , Xt∩Ij ∩Sj,t = Xt∩Ij ∩Sj,t (i0+1,1) (i0,1) and Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t . (i0,`) (i0,`−1) Since (W3 − W3 ) ∩ Ij 6= ∅, there exists q ∈ ∂Tj,t0 such that q is a witness for Xq ∩ Ij ⊆ (i0,`) (i0,`−1) (i0,`) W3 and Xq ∩ Ij 6⊆ W3 . So Xq ∩ Ij is a nonempty subset of W3 ∩ Ij with Xq ∩ Ij 6⊆ (i0,`−1) (i0,`,s+2) (i0) (i0,`−1) W3 , and there exists a c-monochromatic path P in G[Y ∩ Ij ∩ W4 ] from W3 (i0,`,0) (i0,`,0) 0 0 0 to Xq ∩ Ij − ((D − Xt0 ) ∪ {v ∈ D ∩ Xt0 ∩ Xq0 ∩ Ij : q ∈ V (Tt0 ) − {t }, q is a witness (i00,`0) 00 0 0 for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, i − 1] and ` ∈ [0, |V (T )|]}) and internally disjoint from 0 Xt ∪ X∂Tj,t0 . Furthermore, choose such a node q such that q ∈ V (Tt) − {t} if possible. Note that there exists k ∈ [0, s + 1] such that the color of P is k + 1.

27 (i0,`,s+2) (i0) (i0,`+1,k) (i0) We first suppose that q ∈ V (Tt) − {t}. Since V (P ) ⊆ Y ∩ W4 ⊆ Y ∩ W4 , (i0,`+1,k) (i0,`,k+1) (i0,`,s+2) (i0,`,s+2) (i0,`,0) V (P ) ⊆ W0 ⊆ D ⊆ D . Since Xt ∩ Ij ∩ D = Xt ∩ Ij ∩ D , (i0,`,0) (i0,`,0) V (P ) ∩ Xt ⊆ D . Let z be the end of P belonging to Xq ∩ Ij − ((D − Xt0 ) ∪ {v ∈ (i0,`,0) 0 0 0 (i00,`0) D ∩ Xt0 ∩ Xq0 ∩ Ij : q ∈ V (Tt0 ) − {t }, q is a witness for Xq0 ∩ Ij ⊆ W3 for some 00 0 0 (i0,`−1) i ∈ [0, i − 1] and ` ∈ [0, |V (T )|]}), such that z 6∈ W3 if possible. (i0,`,0) (i0,`,0) Suppose z 6∈ D . Since Xt ∩ V (P ) ⊆ D , z 6∈ Xt. Since ` = 0, P has one end (i0,−1) 0 in W3 . Since t is an ancestor of t and q ∈ V (Tt), V (P ) intersects Xt, and there exists 00 (i0,`,0) (i0,`,0) a subpath P of P from Xt ∩ V (P ) ⊆ D to z ∈ XV (Tt) − (Xt ∪ D ) and internally 00 (i0,`,0) (i0,`,0) disjoint from Xt. So there exist a ∈ V (P ) ∩ D and b ∈ NP 00 (a) − D . Note that 00 (i0,`,0) (i0,`,0) (i0,`,0) (i0,`,0) b ∈ V (P ) − D ⊆ V (P ) − D . Since Xt ∩ V (P ) ⊆ D , b ∈ NP 00 (a) − (D ∪ Xt). ∗ 00 0 S|V (T )| (i00,`0+1,k) (i0,`,0) (i0,0,0) Let I = {i ∈ [0, i − 1] : a ∈ `0=0 W0 }. Since ` = 0 and a ∈ D , a ∈ D = (i0−1,|V (T )|,s+2) Si0−1 S|V (T )| (i00,`0+1,k) ∗ D . Since c(a) = k + 1, a ∈ i00=0 `0=0 W0 , so I 6= ∅. Note that for every 00 00 ∗ (i ,ì00 +1,k) i ∈ I , there exists ì00 ∈ [0, |V (T )|] such that either ì00 = 0 and a ∈ W0 , or ì00 > 0 00 00 (i ,ì00 +1,k) (i ,ì00 ,k) and a ∈ W0 − W0 , so there exists a c-monochromatic path Pi00 of color k + 1 in 00 00 00 00 (i ) (i ,` 00 ) (i ,` 00 ) 0 (i ,ì00 +1,k) i i (i ,`,0) G[Y ∩ W4 ] from a to W3 and internally disjoint from W3 . Since b 6∈ D , 00 00 00 (i ,` 00 +1,k) (i ) 00 ∗ 00 ∗ (i ) b 6∈ Y i ∩W for every i ∈ I . For every i ∈ I , since A (i00,` +1,k+1) (V (P 00 ))∩W = ∅ 4 L i00 i 4 (i00) (i00) 00 (i ,ì00 +1,k) by Claim 21.1, we have b 6∈ W4 (since b ∈ W4 implies that b 6∈ Y ⊇ V (Pi00 ) and hence (i00) 00 b ∈ A (i00,` +1,k+1) (V (P 00 )) ∩ W , a contradiction). Since P is from X to X and internally L i00 i 4 t q (i00) 00 ∗ 0 disjoint from Xt ∪ Xt ∪ X∂Tj,t0 , and since b 6∈ W4 , we know for every i ∈ I , q 6∈ V (Tj,ti00 ), where 00 00 ∗ 00 0 ti is the node of T with iti00 = i . Since q ∈ ∂Tj,t , for every i ∈ I , if some node in ∂Tj,ti00 belongs to the path in T from t0 to the parent of q, then this node must be t0. This together with the fact 00 ∗ 00 00 q 6∈ V (Tj,ti00 ) for every i ∈ I , where ti is the node of T with iti00 = i , we have t 6∈ V (Tj,ti00 ) 00 ∗ (i00+1,0) (i00,0) 00 0 S|V (T )| (i00,`0+1,k) for every i ∈ I . If a ∈ Sj,t − Sj,t for some i ∈ [0, i − 1], then a ∈ `0=0 W0 , so 00 ∗ |V (T )| (i00,`0+1,k) (i00) S 00 i ∈ I , but since a ∈ `0=0 W0 ⊆ W4 implies that t ∈ V (Tj,ti00 ), where ti is the node of 00 (i00+1,0) (i00,0) 00 0 (i0,0) T with iti00 = i , a contradiction. So a 6∈ Sj,t −Sj,t for every i ∈ [0, i −1]. Hence a 6∈ Sj,t . (i0,`+1,k) (i0+1,0) On the other hand, since t ∈ V (Tj,t0 ) − ∂Tj,t0 and a ∈ V (P ) ⊆ W0 , we know a ∈ Sj,t . (i0+1,0) (i0,0) (i00) 00 ∗ Since Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t , we know a 6∈ Xt. Recall that b 6∈ W4 for every i ∈ I . 00 ∗ Since ab ∈ E(G) and V (P ) ⊆ Ij, we know for every i ∈ I , a ∈ X∂T , where ti00 is the node of j,ti00 00 00 ∗ 00 00 T with iti00 = i . So for every i ∈ I , there exists qi ∈ ∂Tj,ti00 such that a ∈ Xqi00 ∩ Ij, where ti 00 00 ∗ 0 0 00 0 is a node of T with iti00 = i . If there exists i ∈ I with qi ∈ V (Tt ) − {t }, then t ∈ V (Tti00 ), 00 0 00 0 00 0 where ti is the node of T with iti00 = i , and since q ∈ ∂Tj,t and qi ∈ V (Tt ) − {t }, no node 00 ∗ 00 in ∂Tj,ti00 is contained in the path in T from ti to the parent of q for every i ∈ I , so we have (i00) (i0) 0 00 ∗ ∗ b ∈ W4 since b ∈ W4 , a contradiction. Hence qi00 6∈ V (Tt0 )−{t } for every i ∈ I . Since I 6= ∅, 0 00 ∗ 00 0 0 qi 6∈ V (Tt ) − {t } for some i ∈ I , so a ∈ Xt ∩ Xt, since a ∈ Xqi00 ∩ XV (Tt). This contradicts that a 6∈ Xt. (i0,`,0) (i0,`,0) 0 0 Hence z ∈ D . Since z 6∈ D − Xt0 , z ∈ Xt0 . So there does not exist q ∈ V (Tt0 ) − {t } 0 (i00,`0) 00 0 such that z ∈ Xq0 ∩ Ij and q is a witness for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, i − 1] and 0 (i0,`−1) (i0,−1) 0 ` ∈ [0, |V (T )|]. Since ` = 0, W3 = W3 ⊆ Xt0 . Since t is an ancestor of t and z ∈ Xq, (i0,`) z ∈ Xt0 ∩ Xt. Since z ∈ Xq ∩ Ij and q ∈ V (Tt) − {t} and q is a witness for Xq ∩ Ij ⊆ W3 , and (i0+1,1) (i0+1,1) (i0,1) (i0,1) t ∈ V (Tj,t0 ) − ∂Tj,t0 , we know z ∈ Sj,t . Since Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t , z ∈ Sj,t . So 0 there exist iz ∈ [0, i − 1], and a node qz ∈ V (Tt) − {t} such that z ∈ Xqz ∩ Ij and qz is a witness (iz,`z) 0 0 for Xqz ∩ Ij ⊆ W3 for some `z ∈ [0, |V (T )|]. Note that qz ∈ V (Tt) ⊆ V (Tt ) − {t }. So qz is a

28 0 0 0 (i00,`0) node q ∈ V (Tt0 ) − {t } such that z ∈ Xq0 ∩ Ij and q is a witness for Xq0 ∩ Ij ⊆ W3 for some i00 ∈ [0, i0 − 1] and `0 ∈ [0, |V (T )|], a contradiction. (i0,`) (i0,`−1) (i0,`) (i0,`−1) Therefore q 6∈ V (Tt) − {t}. So (W3 − W3 ) ∩ Ij ∩ XV (Tt) ⊆ (W3 − W3 ) ∩ Ij ∩ Xt (i0,`,s+2) (i0,`+1,0) by the choice of q. Hence Y ∩ XV (Tt) ∩ Ij − Xt = Y ∩ XV (Tt) ∩ Ij − Xt. Since (i0,`,s+2) (i0,`+1,s+2) 0 Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt, there exists k ∈ [0, s + 1] such that (i0,`+1,k0+1) (i0,`+1,k0) (i0,`+1,k0) Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt. So W2 ∩ XV (Tt) ∩ Ij − Xt 6= ∅. (i0,`+1,k0) Hence t ∈ V (Tj,t0 ) − ∂Tj,t0 , and there exists x ∈ W0 ∩ Xt ∩ Ij 6= ∅ such that there exists 0 (i0,`+1,k0) (i0) (i0,`) a monochromatic path Qx of color k + 1 in G[Y ∩ W4 ] from x to W3 and internally (i0,`) 0 0 0 disjoint from Xt ∪W3 , and there exists a monochromatic path Qx of color k +1 with Qx ⊆ Qx ⊆ (i0,`+1,k0) (i0,`+1,k0) (i0) (i0,`) 0 (i0,`) W0 in G[Y ∩ W4 ] from W3 to a vertex x and internally disjoint from W3 , 00 0 (i0,`+1,k0+1) (i0,`+1,k0) 00 0 0 0 and there exists x ∈ NG(x )∩(Y −Y )∩XV (Tt) −Xt 6= ∅ with x ∈ AL(i ,`+1,k ) (x ) 00 0 0 (i0,`+1,k0) (i0,`,k0+1) (i0,`,s+2) and c(x ) 6= c(x ) = c(x) = k + 1. Note that x ∈ W0 ⊆ D ⊆ D . Recall (i0,`,s+2) (i0,`,0) (i0,`,s+2) (i0,`,0) Xt ∩ Ij ∩ D = Xt ∩ Ij ∩ D . Since ` = 0, Xt ∩ Ij ∩ D = Xt ∩ Ij ∩ D = (i0,0,0) (i0−1,|V (T )|,s+2) (i0−1,|V (T )|,s+2) Xt ∩ Ij ∩ D = Xt ∩ Ij ∩ D . So x ∈ D . 0 00 0 S|V (T )| (i00,`0+1,k0) (i0−1,|V (T )|,s+2) 0 Let I = {i ∈ [0, i − 1] : x ∈ `0=0 W0 }. Since x ∈ D , I 6= ∅. So for 00 0 0 0 00 0 0 every i ∈ I , there exist t 00 ∈ V (T ) with σT (t 00 ) = i , ì00 ∈ [0, |V (T )|], a witness q 00 ∈ {t 00 }∪∂Tj,t0 i t i i i00 00 00 0 00 (i ,ì00 ) 0 0 (i ,` 00 +1,k ) (i ) for Xq0 ∩Ij ⊆ W , and a monochromatic path P 00 of color k +1 in G[Y i ∩W ] from i00 3 i 4 0 0 0 0 Xq0 ∩Ij to x. Note that if q 00 is in the path in T from t to t, then q 00 = t , since t ∈ V (Tt0,j)−∂Tt0,j. i00 i i (i00) 00 0 0 00 00 0 0 (i ,ì00 +1,k ) 0 0 Suppose V (Qx) ∪ {x } ⊆ W4 for some i ∈ I . If V (Qx) ⊆ Y , then Pi00 ∪ Qx is a 00 0 (i00) 0 (i ,ì00 +1,k ) connected monochromatic subgraph of color k + 1 contained in G[Y ∩W4 ] intersecting 00 00 0 00 0 0 (i ,ì00 ) 0 00 (i ,ì00 +1,k ) (i ,` 00 +1,k +1) (i ,0,0) Xq0 ∩ Ij ⊆ W and x , so x ∈ W ⊆ Y i ⊆ Y , contradicting that i00 3 2 0 0 0 0 00 0 00 (i ,`+1,k +1) (i ,`+1,k ) 0 (i ,ì00 +1,k ) x ∈ Y − Y . So V (Qx) 6⊆ Y . But this contradicts Claim 21.1, since 00 00 0 0 (i ) (i ,ì00 ) Pi00 ∪ Qx is a connected c-monochromatic subgraph contained in G[W4 ] intersecting W3 . 0 00 (i00) 00 0 Hence V (Qx) ∪ {x } 6⊆ W4 for every i ∈ I . (i0,0) 00 0 (i00+1,0) (i00,0) Suppose that x ∈ Sj,t . So there exists i ∈ [0, i − 1] such that x ∈ Sj,t − Sj,t ⊆ S|V (T )| (i00,`0+1,k0) 0 00 `0=0 W0 (since c(x) = k + 1) and t ∈ V (Tj,t00 ) − ∂Tj,t00 , where t is the node of T with 00 00 00 0 00 σT (t ) = i . Hence i ∈ I and no node in ∂Tj,t00 is in the path in T from t to the parent of t. 00 0 ∗ Since i < i < i and t ∈ V (Tj,t00 ) − ∂Tj,t00 , for every q ∈ ∂Tj,t0 , we know that ∂Tj,t00 is disjoint from 0 ∗ 00 the path in T from t to the parent of q , so ∂Tj,t00 is disjoint from the path in T from t to the ∗ (i0) (i00) 0 00 (i0) (i00) parent of q . So W4 ⊆ W4 . Hence V (Qx) ∪ {x } ⊆ W4 ⊆ W4 , a contradiction. (i0,0) (i0,`+1,k0) (i0+1,0) Hence x 6∈ Sj,t . However, x ∈ W0 and t ∈ V (Tj,t0 ) − ∂Tj,t0 , so x ∈ Sj,t . Therefore, (i0+1,0) (i0,0) Xt ∩ Ij ∩ Sj,t 6= Xt ∩ Ij ∩ Sj,t , a contradiction. This proves the claim. 0 0 0 0 0 Claim 21.3. Let i, i ∈ N0 with i < i. Let t, t be nodes of T with σT (t) = i and σT (t ) = i . Let j ∈ [|V | − 1]. Let ` ∈ [−1, |V (T )|] such that there exists a witness q ∈ ∂Tj,t0 ∩ V (Tt) − {t} for (i0,`+1) (i0,`) Xq ∩ Ij ⊆ W3 and Xq ∩ Ij 6⊆ W3 . Then either: (i0+1,0) (i0,0) • |Xt ∩ Ij ∩ Sj,t | > |Xt ∩ Ij ∩ Sj,t |, or (i0+1,1) (i0,1) • |Xt ∩ Ij ∩ Sj,t | > |Xt ∩ Ij ∩ Sj,t |, or (i0,`+1,s+2) (i0) (i0,`) • for every c-monochromatic path P in G[Y ∩ Ij ∩ W4 ] from W3 to (i0,`+1,0) (i0,`+1,0) 0 0 Xq ∩ Ij − ((D − Xt0 ) ∪ {v ∈ D ∩ Xt0 ∩ Xq0 ∩ Ij : q ∈ V (Tt0 ) − {t }, 0 (i00,`0) 00 0 0 q is a witness for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, i − 1] and ` ∈ [0, |V (T )|]})

~~29 0 and internally disjoint from Xt ∪ X∂Tj,t0 , we have V (P ) ⊆ XV (Tt) − Xt.~~

0 Proof. Note that t is an ancestor of t, since q ∈ ∂Tj,t0 ∩ V (Tt) − {t}. We may assume that (i0+1,0) (i0,0) (i0+1,1) (i0,1) |Xt ∩ Ij ∩ Sj,t | 6 |Xt ∩ Ij ∩ Sj,t | and |Xt ∩ Ij ∩ Sj,t | 6 |Xt ∩ Ij ∩ Sj,t |, for otherwise we (i0+1,0) (i0,0) (i0+1,1) (i0,1) are done. So Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t and Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t . (i0,`+1) (i0,`) Since there exists a witness q ∈ ∂Tj,t0 ∩ V (Tt) − {t} for Xq ∩ Ij ⊆ W3 and Xq ∩ Ij 6⊆ W3 , (i0,`+1,s+2) (i0) (i0,`) there exists a c-monochromatic path P in G[Y ∩ Ij ∩ W4 ] from W3 to a vertex

(i0,`+1,0) (i0,`+1,0) 0 0 z ∈ Xq ∩ Ij−((D − Xt0 ) ∪ {v ∈ D ∩ Xt0 ∩ Xq0 ∩ Ij : q ∈ V (Tt0 ) − {t }, 0 (i00,`0) 00 0 0 q is a witness for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, i − 1] and ` ∈ [0, |V (T )|]})

0 and internally disjoint from Xt ∪ X∂Tj,t0 . Furthermore, choose P such that V (P ) 6⊆ XV (Tt) − Xt if possible. We may assume that V (P ) 6⊆ XV (Tt) − Xt, for otherwise we are done. Let k ∈ [0, s+1] be the number such that P is of color k +1. By Claim 21.1, V (P ) ⊆ Y (i0,`+1,k). 0 Since V (P ) 6⊆ XV (Tt) − Xt and q ∈ V (Tt) − {t}, there exists a vertex z in V (P ) ∩ Xt such that the 0 subpath of P between z and z is contained in G[XV (Tt)] and internally disjoint from Xt. If ` 0, 0 0 > (i ,`+1,k) (i ,`1+1,k) then let `1 = `; if ` = −1, then let `1 = 0. Since `1 ` and V (P ) ⊆ Y , V (P ) ⊆ Y . 0 0 > 0 0 (i ,`1+1,k) 0 (i +1,0) (i +1,0) (i ,0) Since `1 > 0, V (P ) ⊆ W0 . So z ∈ Sj,t ∩Xt ∩Ij. Since Xt ∩Ij ∩Sj,t = Xt ∩Ij ∩Sj,t , 0 0 (i ,0) 0 0 (i1+1,0) (i1,0) z ∈ Sj,t ∩ Xt ∩ Ij. Hence there exists i1 ∈ [0, i − 1] such that z ∈ Sj,t − Sj,t . Note that (0,0) 0 (i1+1,0) (i1,0) such i1 exists since Sj,t = ∅. Let t1 be the node of T with σT (t1) = i1. Since z ∈ Sj,t −Sj,t , 0 (i1,`z0 +1,k) 0 t ∈ V (Tj,t1 )−∂Tj,t1 , and there exists `z ∈ [0, |V (T )|] such that z ∈ W0 . Hence there exists (i ) (i1,` 0 ) (i1,`z0 +1,k) 1 0 z a c-monochromatic path Pz0 in G[Y ∩ W4 ] from z to W3 . 0 0 Since t ∈ V (Tj,t1 )−∂Tj,t1 , ∂Tj,t1 is disjoint from the path in T from t1 to t . So V (Tj,t ) ⊆ V (Tj,t1 ) 0 (i1) as i1 < i . So P ∪ Pz0 is a c-monochromatic subgraph of color k + 1 contained in W4 containing (i1,`z0 ) (i1,`z0 ) z and intersecting W3 . By Claim 21.1, the path Pz contained in P ∪ Pz0 from W3 to z (i1,` 0 +1,k) 0 (i1,`z0 +1,k) z (i1,`z0 ,k+1) (i ,`+1,0) satisﬁes V (Pz) ⊆ Y . Hence V (Pz) ⊆ W0 . So z ∈ D ⊆ D . Since (i0,`+1,0) 0 z ∈ Xq ∩ Ij − (D − Xt0 ), z ∈ Xt0 . Since t belongs to the path in T between t and q, (i0+1,1) z ∈ Xt. Since q ∈ ∂Tj,t0 ∩ V (Tt) − {t}, t ∈ V (Tj,t0 ) − ∂Tj,t0 . Since z ∈ Xq ∩ Ij, z ∈ Sj,t . (i0+1,1) (i0,1) (i0,1) 0 0 Since Sj,t ∩ Xt ∩ Ij = Sj,t ∩ Xt ∩ Ij, z ∈ Sj,t . So there exists i1 ∈ [0, i − 1] such that 0 0 (i1+1,1) (i1,1) 0 (0,1) z ∈ Sj,t − Sj,t . Note that i1 exists since Sj,t = ∅. 0 0 0 0 0 (i1+1,1) (i1,1) Let t be the node of T with σ (t ) = i . Since z ∈ S − S , t ∈ V (T 0 ) − ∂T 0 and 1 T 1 1 j,t j,t j,t1 j,t1 0 0 (i1,`1) there exists qz ∈ V (Tt) − {t} such that z ∈ Xqz ∩ Ij and qz is a witness for Xqz ∩ Ij ⊆ W3 for 0 0 some `1 ∈ [0, |V (T )|]. Note that qz ∈ V (Tt0 ) − {t } since qz ∈ V (Tt) − {t}. However, the existence (i0,`+1,0) 0 0 0 of qz contradicts that z ∈ Xq ∩ Ij − {v ∈ D ∩ Xt0 ∩ Xq0 ∩ Ij : q ∈ V (Tt0 ) − {t }, q is a witness (i00,`0) 00 0 0 for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, i − 1] and ` ∈ [0, |V (T )|]}. This proves the claim.

0 0 0 0 0 Claim 21.4. Let i, i ∈ N0 with i < i. Let t, t be nodes of T with σT (t) = i and σT (t ) = i . Let j ∈ [|V | − 1]. Let ` ∈ [−1, |V (T )| − 1] such that there exists no witness q ∈ ∂Tj,t0 ∩ V (Tt) − {t} for (i0,`+1) (i0,`) (i0,`+2,s+2) (i0,`+1,s+2) Xq ∩ Ij ⊆ W3 and Xq ∩ Ij 6⊆ W3 . If Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt, then either:

~~(i0+1,0) (i0,0) • |Xt ∩ Ij ∩ Sj,t | > |Xt ∩ Ij ∩ Sj,t |, or (i0+1,1) (i0,1) • |Xt ∩ Ij ∩ Sj,t | > |Xt ∩ Ij ∩ Sj,t |.~~

30 (i0,`+2,s+2) (i0,`+1,s+2) 0 Proof. Since Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt, t is an ancestor of (i0+1,0) (i0,0) (i0+1,1) t. Suppose to the contrary that |Xt ∩ Ij ∩ Sj,t | 6 |Xt ∩ Ij ∩ Sj,t | and |Xt ∩ Ij ∩ Sj,t | 6 (i0,1) (i0+1,0) (i0,0) (i0+1,1) (i0,1) |Xt ∩ Ij ∩ Sj,t |. So Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t and Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t . (i0,`+1) (i0,`) Since there exists no witness q ∈ ∂Tj,t0 ∩V (Tt)−{t} for Xq ∩Ij ⊆ W3 and Xq ∩Ij 6⊆ W3 , (i0,`+1) (i0,`) (i0,`+2,0) W3 ∩XV (Tt) ∩Ij −Xt = W3 ∩XV (Tt) ∩Ij −Xt. This implies that Y ∩XV (Tt) ∩Ij −Xt = (i0,`+1,s+2) (i0,`+2,s+2) (i0,`+1,s+2) Y ∩ XV (Tt) ∩ Ij − Xt. Since Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt, (i0,`+2,0) (i0,`+2,s+2) we have Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt. Hence there exists k ∈ (i0,`+2,k+1) (i0,`+2,k) [0, s + 1] such that Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt. So there exist 0 (i0,`+2,k) (i0) (i0,`+2,k) 0 (i0,`+2,k) x ∈ W2 ∩W4 ∩XV (Tt) ∩Ij −(Xt ∪Y ), x ∈ NG(x )∩W0 and a monochromatic (i0,`+2,k) (i0) (i0,`+1) 0 path Q in G[Y ∩ W4 ] of color k + 1 from x to W3 , such that x ∈ AL(i0,`+2,k) (x). Note (i0,`+1) 0 that x ∈ XV (Tt). Let qQ ∈ ∂Tj,t be a witness for XqQ ∩ Ij ⊆ W3 such that one end of Q is in 0 0 XqQ ∩ Ij. Since x ∈ XV (Tt) − Xt, t 6∈ ∂Tj,t . Suppose that qQ ∈ V (Tt). Since qQ ∈ ∂Tj,t0 , qQ ∈ V (Tt) − {t}. Since there exists no witness (i0,`+1) (i0,`) (i0,`) 0 q ∈ ∂Tj,t ∩ V (Tt) − {t} for Xq ∩ Ij ⊆ W3 and Xq ∩ Ij 6⊆ W3 , we know XqQ ∩ Ij ⊆ W3 . (i0,`+1,k) 0 (i0,`+1,k) (i0,`+1,k) (i0,`+2,k) So by Claim 21.1, V (Q) ⊆ W0 and x ∈ W2 ⊆ Y ⊆ Y , a contradiction. ∗ ∗ So qQ 6∈ V (Tt). Hence there exists a vertex x ∈ Xt ∩ Ij such that the subpath of Q between x ∗ (i0,`+2,k) 0 0 and x is contained in G[XV (Tt)]. Since x ∈ Xt ∩V (Q) ⊆ Xt ∩Ij ∩W0 and t ∈ V (Tj,t )−∂Tj,t , ∗ (i0+1,0) (i0+1,0) (i0,0) 0 x ∈ Sj,t ∩ Xt ∩ Ij. Since Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t , there exists i2 ∈ [0, i − 1] such ∗ (i2+1,0) (i2,0) that x ∈ Sj,t − Sj,t . So the node t2 of T with σT (t2) = i2 satisﬁes that t ∈ V (Tj,t2 ) − ∂Tj,t2 , ∗ (i2,`x∗ +1,k) ∗ and there exists `x∗ ∈ [0, |V (T )|] such that x ∈ W0 , since c(x ) = k + 1. Hence there (i ) (i ,` ∗ ) (i2,`x∗ +1,k) 2 ∗ 2 x exists a monochromatic path Qx∗ in G[Y ∩ W4 ] from x to W3 . 0 0 Since t ∈ V (Tj,t2 ) − ∂Tj,t2 , ∂Tj,t2 is disjoint from the path in T from t2 to t . Since i2 < i , 0 ∗ V (Tj,t ) ⊆ V (Tj,t2 ). So Q ∪ Qx is a c-monochromatic subgraph of color k + 1 contained in (i2) (i2,`x∗ ) W4 containing x and intersects W3 . By Claim 21.1, the path Qx contained in Q ∪ Qx∗ (i ,` ∗ ) (i ) 2 x (i2,`x∗ +1,k) 2 from W3 to x satisﬁes V (Qx) ⊆ Y and AL(i2,`x∗ +1,k+1) (V (Qx)) ∩ W4 = ∅. But 0 0 (i ) (i2) 0 x ∈ AL(i ,`+2,k) (V (Qx)) ∩ W4 ⊆ AL(i2,`x∗ +1,k+1) (V (Qx)) ∩ W4 = ∅, a contradiction. This proves the claim.

(i0,0,s+2) (i0,|V (T )|+1,s+2) Proof. We may assume that Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt, (i0,0,s+2) (i0,1,s+2) for otherwise we are done. Since Y ∩ XV (Tt) ∩ Ij − Xt = Y ∩ XV (Tt) ∩ Ij − Xt, (i0,1,s+2) (i0,|V (T )|+1,s+2) Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt. So there exists a minimum (i0,`+1,s+2) (i0,`+2,s+2) ` ∈ [0, |V (T )| − 1] such that Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt. By the (i0,`,s+2) (i0,`+1,s+2) minimality of `, Y ∩ XV (Tt) ∩ Ij − Xt = Y ∩ XV (Tt) ∩ Ij − Xt. 0 0 0 (i0,0,s+2) (i0,|V (T )|+1,s+2) Let t be the node of T with σT (t ) = i . Since Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ 0 0 XV (Tt) ∩ Ij − Xt, t ∈ V (Tj,t ) − ∂Tj,t . (i0+1,0) (i0,0) (i0+1,1) Suppose to the contrary that |Xt ∩ Ij ∩ Sj,t | 6 |Xt ∩ Ij ∩ Sj,t | and |Xt ∩ Ij ∩ Sj,t | 6 (i0,1) (i0+1,0) (i0,0) (i0+1,1) (i0,1) |Xt ∩ Ij ∩ Sj,t |. So Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t and Xt ∩ Ij ∩ Sj,t = Xt ∩ Ij ∩ Sj,t .

31 We ﬁrst suppose that there exist `0 ∈ [−1, `] and a witness q ∈ ∂Tj,t0 ∩ V (Tt) for Xq ∩ Ij ⊆ 0 0 (i ,`0+1) (i ,`0) W3 and Xq ∩ Ij 6⊆ W3 . Choose `0 so that `0 is as small as possible. So there exists 0 0 0 (i ,`0+1,s+2) (i ) (i ,`0) a c-monochromatic path P in G[Y ∩ Ij ∩ W4 ] from W3 to a node z ∈ Xq ∩ Ij − 0 0 (i ,`0+1,0) (i ,`0+1,0) 0 0 0 ((D − Xt0 ) ∪ {v ∈ D ∩ Xt0 ∩ Xq0 ∩ Ij : q ∈ V (Tt0 ) − {t }, q is a witness for (i00,`0) 00 0 0 0 0 Xq ∩Ij ⊆ W3 for some i ∈ [0, i −1] and ` ∈ [0, |V (T )|]}) internally disjoint from Xt ∪X∂Tj,t0 . 0 Since t 6∈ ∂Tj,t , by Claim 21.3, V (P ) ⊆ XV (Tt) − Xt. In particular, `0 > 0. 0 (i ,`0−1) Let k ∈ [0, s + 1] be the number such that P is of color k + 1. If V (P ) intersects W3 , 0 0 (i ,`0,k) (i ,`0) then by Claim 21.1, V (P ) ⊆ Y , so q is a witness for Xq ∩ Ij ⊆ W3 , a contradiction. So 0 0 (i ,`0−1) 0 (i ,`0) V (P ) ∩ W3 = ∅. Then there exists a witness q ∈ V (Tt) − {t} for Xq0 ∩ Ij ⊆ W3 with 0 (i ,`0−1) Xq0 ∩ Ij 6⊆ W3 such that V (P ) is from Xq0 to Xq, contradicting the minimality of `0. 0 (i ,`0+1) Therefore, there do not exist `0 ∈ [−1, `] and a witness q ∈ ∂Tj,t0 ∩V (Tt) for Xq ∩Ij 6⊆ W3 0 (i ,`0) and Xq ∩ Ij 6⊆ W3 . In particular, there exists no witness q ∈ ∂Tj,t0 ∩ V (Tt) for Xq ∩ Ij ⊆ (i0,`+1) (i0,`) (i0+1,0) (i0,0) W3 and Xq ∩ Ij 6⊆ W3 . By Claim 21.4, either |Xt ∩ Ij ∩ Sj,t | > |Xt ∩ Ij ∩ Sj,t |, or (i0+1,1) (i0,1) |Xt ∩ Ij ∩ Sj,t | > |Xt ∩ Ij ∩ Sj,t |, a contradiction. This proves the claim.

Claim 21.6. Let i ∈ N0, and let t be the node of T with σT (t) = i. Let j ∈ [|V | − 1]. Let 0 (i0,|V (T )|+1,s+2) (i0,0,s+2) S := {i ∈ [0, i − 1] : Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt}. Then |S| 6 3w0. 0 (i0,1,s+2) (i0,0,s+2) Proof. Let S1 := {i ∈ [0, i − 1] : Y ∩ XV (Tt) ∩ Ij − Xt 6= Y ∩ XV (Tt) ∩ Ij − Xt}, 0 (i0+1,0) (i0+1,1) (i0,0,s+2) and let S2 := {i ∈ [0, i − 1] : |Xt ∩ Ij ∩ Sj,t | + |Xt ∩ Ij ∩ Sj,t | + |Xt ∩ Ij ∩ D | > (i0,0) (i0,1) (i0,0,0) |Xt ∩Ij ∩Sj,t |+|Xt ∩Ij ∩Sj,t |+|Xt ∩Ij ∩D |}. By Claim 21.5, S ⊆ S1 ∪S2. By Claim 21.2, S1 ⊆ S2. So S ⊆ S2. Since |Xt ∩ Ij| 6 w0, |S| 6 |S2| 6 3w0. Claim 21.7. Let t ∈ V (T ) and j ∈ [|V | − 1]. Let (Y 0,L0) be a V-standard pair and let Z ⊆ Y 0. s s 0 > 0 > 0 Then AL (Z) ⊆ NG [Z], AL (Z)∩XV (Tj,t) ⊆ NG (Z∩XV (Tj,t))∪X∂Tj,t ∪Xt, and |AL (Z)∩XV (Tj,t)| 6 f(|Z ∩ XV (Tj,t)|) + |X∂Tj,t | + |Xt|.

>s 0 0 0 0 Proof. It is obvious that AL0 (Z) ⊆ NG [Z] since (Y ,L ) is a V-standard pair. Since (Y ,L ) is a s s 0 > > V-standard pair, AL (Z) ∩ XV (Tj,t) ⊆ NG (Z) ∩ XV (Tj,t) ⊆ NG (Z ∩ XV (Tj,t)) ∪ (X∂Tj,t ∪ Xt). So s 0 > |AL (Z) ∩ XV (Tj,t)| 6 |NG (Z ∩ XV (Tj,t))| + |X∂Tj,t | + |Xt| 6 f(|Z ∩ XV (Tj,t)|) + |X∂Tj,t | + |Xt| by Corollary 14.

~~0 0 0 0 0 Claim 21.8. Let i, i ∈ N0 with i < i, and let t and t be nodes of T with σT (t) = i and σT (t ) = i . Let j ∈ [|V | − 1]. Then the following hold:~~

~~• |Y (i0,−1,0) ∩ I ∩ X ∩ X | w . j V (Tt) V (Tj,t0 ) 6 0 • |X ∩ X ∩ I | w . ∂Tj,t0 V (Tt) j 6 0~~

Proof. We may assume that t ∈ V (T 0 ), for otherwise, X ∩ X = ∅ and we are done. t V (Tt) V (Tj,t0 ) 0 0 ∗ 0 0 ∗ 0 Let Fj,p := Fj,0 if t = r ; let p be the parent of t if t 6= r . By deﬁnition, Fj,p = Fj,p ∪ S 0 0 0 T 0 Fj,T 0 , where the union is over all Fj,p ∩ V (Tt0 )-parts T of Tt0 containing t , and each Fj,T 0 is 0 (i0,−1,0) 0 a (T , X , (Y ∪ X∂T 0 ) ∩ XV (T 0) ∩ Ij)-fence. So for each Fj,p ∩ V (Tt0 )-part T of Tt0 , XV (T 0)∩Ij (i0,−1,0) S 0 |(Y ∩ Ij ∩ XV (T 0)) ∪ (X∂T 0 ∩ Ij)| 6 w0. By deﬁnition, Tj,t0 = T 0 T , where the union is 0 0 0 over all Fj,p ∩ V (Tt0 )-parts T of Tt0 containing t . Since t ∈ V (Tt0 ) − {t }, there exists at most one 0 Fj,p ∩V (Tt0 )-part of Tt0 containing both t and t. If there exists an Fj,p ∩V (Tt0 )-part of Tt0 containing

~~(t0) Proof. For every k ∈ [0, w0 − 1], let qk ∈ [0, s + 1] be the number such that the color of Mj,k is qk + 1. So for every k ∈ [0, w0 − 1],~~

(Y (i0,−1,k+1) − Y (i0,−1,k)) ∩ I ∩ X ∩ X − X j V (Tt) V (Tj,t0 ) t 0 (i ,−1,k,qk) ⊆ A (i0,−1,k,q ) (W ) ∩ Z 0 ∩ I ∩ X ∩ X − X L k 1 t j V (Tt) V (Tj,t0 ) t 0 >s (i ,−1,k,qk) ⊆ (N (W ∩ X ) ∪ X ∪ X 0 ) ∩ Z 0 ∩ I ∩ X ∩ X − X , G 1 V (Tj,t0 ) ∂Tj,t0 t t j V (Tt) V (Tj,t0 ) t where the last inclusion follows from Claim 21.7. Note that for every k ∈ [0, w0 − 1], since 0 0 (i ,−1,k,qk) S (i ,−1,k,qk) W ∩ Ij ⊆ ◦ S, we know NG[W ∩ Ij] ⊆ Ij. So for every k ∈ [0, w0 − 1], 1 S∈Sj 1

0 >s (i ,−1,k,qk) (N (W ∩ X ) ∪ X ∪ X 0 ) ∩ Z 0 ∩ I ∩ X ∩ X − X G 1 V (Tj,t0 ) ∂Tj,t0 t t j V (Tt) V (Tj,t0 ) t 0 >s (i ,−1,k,qk) ⊆ (N (W ∩ X ) ∩ Z 0 ∩ I ∩ X ∩ X − X ) ∪ (X ∩ I ∩ X ) G 1 V (Tj,t0 ) t j V (Tt) V (Tj,t0 ) t ∂Tj,t0 j V (Tt) 0 ⊆ (N >s(W (i ,−1,k,qk) ∩ X ∩ I ) ∩ I ∩ X ∩ X − X ) ∪ (X ∩ I ∩ X ) G 1 V (Tj,t0 ) j j V (Tt) V (Tj,t0 ) t ∂Tj,t0 j V (Tt) 0 ⊆ N >s(W (i ,−1,k,qk) ∩ X ∩ I ∩ X ) ∪ (X ∩ I ∩ X ) G 1 V (Tj,t0 ) j V (Tt) ∂Tj,t0 j V (Tt) 0 ⊆ N >s(Y (i ,−1,k) ∩ X ∩ I ∩ X ) ∪ (X ∩ I ∩ X ). G V (Tj,t0 ) j V (Tt) ∂Tj,t0 j V (Tt)

~~Hence for every k ∈ [0, w0 − 1],~~

~~33 • For every k ∈ [0, s + 1],~~

b+(s+3)−(k+1) 0 [ |Y (i ,0,k+1) ∩ ( V ) ∩ X ∩ X | α V (Tt) V (Tj,t0 ) α=a−(s+3)+(k+1) b+(s+3)−k 0 [ f (|Y (i ,0,k) ∩ ( V ) ∩ X ∩ X |) + |X ∩ X ∩ I | + 2w . 6 1 α V (Tt) V (Tj,t0 ) ∂Tj,t0 V (Tt) j 0 α=a−(s+3)+k

~~• If i0 < i, then |Y (i0,0,s+2) ∩ (Sb+1 V ) ∩ X ∩ X | g (s + 2). α=a−1 α V (Tt) V (Tj,t0 ) 6 1 • If i0 < i, then for every ` ∈ [0, |V (T )| + 1],~~

~~b+1 0 [ |Y (i ,`+1,s+2) ∩ ( V ) ∩ X ∩ X | α V (Tt) V (Tj,t0 ) α=a−1 b+1 0 [ g (s + 2, |Y (i ,`,s+2) ∩ ( V ) ∩ X ∩ X |). 6 2 α V (Tt) V (Tj,t0 ) α=a−1~~

~~• If i0 < i, then |Y (i0,|V (T )|+1,s+2) ∩ (Sb+1 V ) ∩ X ∩ X | g (4w ). α=a−1 α V (Tt) V (Tj,t0 ) 6 3 0~~

~~Proof. We may assume that t ∈ V (T 0 ), for otherwise X ∩ X = ∅ and we are done. t V (Tt) V (Tj,t0 ) We ﬁrst prove Statement 1. For every k ∈ [0, s + 1],~~

~~(i0,0,k+1) (i0,0,k) (i0,0,k) (Y − Y ) ∩ Ij ∩ XV (Tt) ⊆ W2 ∩ Ij ∩ XV (Tt) (i0,0,k) 0 0 ⊆ AL(i ,0,k) (Y ) ∩ Zt ∩ Ij ∩ XV (Tt).~~

~~(i0,0,k+1) (i0,0,k) (i0,0,k) 0 So (Y − Y ) ∩ Ij ∩ XV (Tt) ∩ XV (Tj,t) ⊆ AL(i ,0,k) (Y ) ∩ Ij ∩ XV (Tt) ∩ XV (Tj,t). By Claim 21.7,~~

b+(s+3)−(k+1) (i0,0,k) [ 0 AL(i ,0,k) (Y ) ∩ ( Vα) ∩ XV (Tt) − Xt α=a−(s+3)+(k+1) b+(s+3)−k b+(s+3)−(k+1) (i0,0,k) [ [ 0 ⊆ AL(i ,0,k) (Y ∩ XV (Tt) ∩ ( Vα)) ∩ ( Vα) ∩ XV (Tt) − Xt, α=a−(s+3)+k α=a−(s+3)+k+1 and

~~0 0 (i ,0,k) >s (i ,0,k) A (i0,0,k) (Y ∩ X ) ∩ X ⊆ N (Y ∩ X ∩ X ) ∪ X ∪ X 0 . L V (Tt) V (Tj,t0 ) G V (Tt) V (Tj,t0 ) ∂Tj,t0 t Therefore,~~

b+(s+3)−(k+1) 0 0 [ (Y (i ,0,k+1) − Y (i ,0,k)) ∩ ( V ) ∩ X ∩ X α V (Tt) V (Tj,t0 ) α=a−(s+3)+(k+1) b+(s+3)−(k+1) (i0,0,k) [ ⊆ (A (i0,0,k) (Y ) ∩ X ∩ ( V ) ∩ X − X ) ∪ (X ∩ I ) L V (Tj,t0 ) α V (Tt) t t j α=a−(s+3)+(k+1)

~~34 b+(s+3)−k b+(s+3)−(k+1) (i0,0,k) [ [ ⊆ A (i0,0,k) (Y ∩ X ∩ ( V )) ∩ ( V ) ∩ X ∩ X − X L V (Tt) α α V (Tj,t0 ) V (Tt) t α=a−(s+3)+k α=a−(s+3)+(k+1)~~

∪ (Xt ∩ Ij) b+(s+3)−k 0 >s (i ,0,k) [ ⊆ (N (Y ∩ ( V ) ∩ X ∩ X ) ∪ X ∪ X 0 ) G α V (Tt) V (Tj,t0 ) ∂Tj,t0 t α=a−(s+3)+k b+(s+3)−(k+1) [ ∩ ( Vα) ∩ XV (Tt) − Xt ∪ (Xt ∩ Ij), α=a−(s+3)+k+1 where the last inequality follows from Claim 21.7. Hence

~~35 This proves Statement 1. Now we prove Statement 2 of this claim. Assume i0 < i. By Claim 21.8 and Statement 1 of this claim, for every k ∈ [0, s + 1],~~

Then it is easy to verify that for every k ∈ [0, s+2], |Y (i0,0,k)∩(Sb+(s+3)−k V )∩X ∩X | α=a−(s+3)+k) α V (Tt) V (Tj,t0 ) 6 g1(k) by induction on k. (Note that the base case k = 0 follows from Claim 21.9.) Then Statement 2 follows from the case k = s + 2. Now we prove Statement 3 of this claim. Note that for every ` ∈ [0, |V (T )|],

b+1 (i0,`+1,k) [ W2 ∩ ( Vα) ∩ XV (Tt) α=a−1 b+1 (i0,`+1,k) (i0) [ 0 = AL(i ,`+1,k) (W0 ) ∩ W4 ∩ ( Vα) ∩ XV (Tt) α=a−1 (i0,`+1,k) ⊆ A (i0,`+1,k) (W ) ∩ X ∩ I L 0 V (Tj,t0 ) j (i0,`+1,k) ⊆ (A (i0,`+1,k) (W ∩ X ) ∪ X ∪ X 0 ) ∩ X ∩ I L 0 V (Tj,t0 ) ∂Tj,t0 t V (Tj,t0 ) j (i0,`+1,k) ⊆ (A (i0,`+1,k) (W ∩ I ∩ X ) ∪ X ∪ X 0 ) ∩ X ∩ I L 0 j V (Tj,t0 ) ∂Tj,t0 t V (Tj,t0 ) j b+1 0 >s (i ,`+1,k) [ ⊆ (N (W ∩ ( V ) ∩ X ) ∪ X ∪ X 0 ) ∩ X ∩ I . G 0 α V (Tj,t0 ) ∂Tj,t0 t V (Tj,t0 ) j α=a−1

~~36 So for every ` ∈ [0, |V (T )|] and k ∈ [0, s + 1],~~

b+1 (i0,`+1,k+1) (i0,`+1,k) [ (Y − Y ) ∩ ( Vα) ∩ XV (Tt) α=a−1 b+1 (i0,`+1,k) [ ⊆ W2 ∩ ( Vα) ∩ XV (Tt) α=a−1 b+1 b+1 0 >s (i ,`+1,k) [ [ ⊆ (N (W ∩ ( V ) ∩ X ) ∪ X ∪ X 0 ) ∩ X ∩ I ∩ ( V ) ∩ X G 0 α V (Tj,t0 ) ∂Tj,t0 t V (Tj,t0 ) j α V (Tt) α=a−1 α=a−1 b+1 0 >s (i ,`+1,k) [ ⊆ (N (W ∩ ( V ) ∩ X ∩ X ) ∪ X ∪ X ∪ X 0 ) ∩ X ∩ I ∩ X G 0 α V (Tj,t0 ) V (Tt) t ∂Tj,t0 t V (Tj,t0 ) j V (Tt) α=a−1 b+1 0 >s (i ,`+1,k) [ ⊆ N (Y ∩ ( V ) ∩ X ∩ X ) ∪ (X ∩ I ) ∪ (X ∩ I ∩ X ) ∪ (X 0 ∩ I ). G α V (Tj,t0 ) V (Tt) t j ∂Tj,t0 j V (Tt) t j α=a−1

~~Hence for every ` ∈ [0, |V (T )|] and k ∈ [0, s + 1],~~

b+1 (i0,`+1,k+1) (i0,`+1,k) [ |(Y − Y ) ∩ ( Vα) ∩ XV (Tt)| α=a−1 b+1 0 >s (i ,`+1,k) [ |N (Y ∩ ( V ) ∩ X ∩ X )| + |X ∩ I | + |X ∩ I ∩ X | + |X 0 ∩ I | 6 G α V (Tj,t0 ) V (Tt) t j ∂Tj,t0 j V (Tt) t j α=a−1 b+1 0 [ f (Y (i ,`+1,k) ∩ ( V ) ∩ X ∩ X ) + 3w 6 1 α V (Tj,t0 ) V (Tt) 0 α=a−1 by Claim 21.8. So for every ` ∈ [0, |V (T )|] and k ∈ [0, s + 1],

~~b+1 b+1 0 [ 0 [ |Y (i ,`+1,k+1) ∩ ( V ) ∩ X ∩ X | f (Y (i ,`+1,k) ∩ ( V ) ∩ X ∩ X ) + 3w . α V (Tj,t0 ) V (Tt) 6 2 α V (Tj,t0 ) V (Tt) 0 α=a−1 α=a−1~~

~~Then for every ` ∈ [0, |V (T )|], it is easy to verify by induction on k ∈ [0, s + 2] that~~

~~b+1 b+1 0 [ 0 [ |Y (i ,`+1,k) ∩ ( V ) ∩ X ∩ X | g (k, |Y (i ,`,s+2) ∩ ( V ) ∩ X ∩ X |). α V (Tj,t0 ) V (Tt) 6 2 α V (Tj,t0 ) V (Tt) α=a−1 α=a−1 Hence~~

b+1 b+1 0 [ 0 [ |Y (i ,`+1,s+2) ∩ ( V ) ∩ X ∩ X | g (s + 2, |Y (i ,`,s+2) ∩ ( V ) ∩ X ∩ X |). α V (Tj,t0 ) V (Tt) 6 2 α V (Tj,t0 ) V (Tt) α=a−1 α=a−1 Now we prove Statement 4 of this claim. Let

~~b+1 b+1 (i0,`+1,s+2) [ (i0,`,s+2) [ S := {` ∈ [0, |V (T )|]: |Y ∩ ( Vα) ∩ XV (Tt)| > |Y ∩ ( Vα) ∩ XV (Tt)|}. α=a−1 α=a−1~~

37 (i0,`+1,s+2) (i0,`,s+2) Sb+1 (i0,`+1,s+2) For every ` ∈ S, either (Y − Y ) ∩ Xt ∩ ( α=a−1 Vα) 6= ∅, or (Y − (i0,`,s+2) Sb+1 (i0,`+1,s+2) (i0,`,s+2) Sb+1 Y ) ∩ ( α=a−1 Vα) ∩ XV (Tt) − Xt 6= ∅. Note that (Y − Y ) ∩ α=a−1 Vα ⊆ (i0,`+1,s+2) (i0,`,s+2) (i0) Sb+1 (i0,`+1,s+2) (i0,`,s+2) (Y −Y )∩W4 ∩ α=a−1 Vα ⊆ (Y −Y )∩Ij. So for every ` ∈ S, either (i0,`+1,s+2) (i0,`,s+2) Sb+1 (i0,`+1,s+2) (i0,`,s+2) (Y − Y ) ∩ Xt ∩ ( α=a−1 Vα) 6= ∅, or (Y − Y ) ∩ Ij ∩ XV (Tt) − Xt 6= ∅. Let

~~b+1 (i0,`+1,s+2) (i0,`,s+2) [ S1 := {` ∈ [0, |V (T )|]:(Y − Y ) ∩ Xt ∩ ( Vα) 6= ∅} α=a−1 (i0,`+1,s+2) (i0,`,s+2) S2 := {` ∈ [0, |V (T )|]:(Y − Y ) ∩ Ij ∩ XV (Tt) − Xt 6= ∅}.~~

~~Sb+1 So S ⊆ S1 ∪ S2. Note that |S1| 6 |Xt ∩ ( α=a−1 Vα)| 6 w0. Let~~

~~(i0,`) (i0,`−1) S3 := {` ∈ [0, |V (T )|]:(W3 − W3 ) ∩ Ij ∩ XV (Tt) 6= ∅}.~~

(i0,`) (i0,−1) Since (W3 − W3 ) ∩ Ij ∩ XV (Tt) ⊆ X∂Tj,t0 ∩ Ij ∩ XV (Tt) for every ` ∈ [0, |V (T )|], we know |S | |X ∩ I ∩ X | w by Claim 21.8 since i0 < i. Let 3 6 ∂Tj,t0 j V (Tt) 6 0

~~(i0+1,0) (i0+1,1) (i0,0) (i0,1) S4 :={` ∈ [0, |V (T )|]: |Xt ∩ Ij ∩ Sj,t | + |Xt ∩ Ij ∩ Sj,t | > |Xt ∩ Ij ∩ Sj,t | + |Xt ∩ Ij ∩ Sj,t |}.~~

(i0,`) (i0,`−1) For every ` ∈ S2 − S3, we know (W3 − W3 ) ∩ Ij ∩ XV (Tt) = ∅, so there exists no witness (i0,`) (i0,`−1) q ∈ ∂Tj,t0 ∩V (Tt)−{t} for Xq ∩Ij ⊆ W3 and Xq ∩Ij 6⊆ W3 , and hence ` ∈ S4 by Claim 21.4. So S ⊆ S1 ∪ S2 ⊆ S1 ∪ S3 ∪ S4. Therefore, |S| 6 |S1| + |S3| + |S4| 6 w0 + w0 + 2w0 = 4w0. By Statements 2 and 3 of this claim, it is easy to verify by induction on k ∈ [|S|] that 0 b+1 |Y (i ,`k,s+2) ∩ (S V ) ∩ X ∩ X | g (k), where we denote the elements of S by α=a−1 α V (Tt) V (Tj,t0 ) 6 3 ` < ` < ··· < ` . Therefore, |Y (i0,|V (T )|+1,s+2) ∩ (Sb+1 V ) ∩ X ∩ X | g (|S|) 1 2 |S| α=a−1 α V (Tt) V (Tj,t0 ) 6 3 6 g3(4w0).

◦ Claim 21.11. Let j ∈ [|V|−1], and let M be a c-monochromatic component such that SM ∩Ij 6= ∅, where SM is the s-segment containing V (M) whose level equals the color of M. Let i ∈ N0 and ∗ t be a node of T with σT (t) = i such that V (M) ∩ Xt∗ 6= ∅ for some witness t ∈ ∂Tj,t ∪ {t} (i,`) (i,|V (T )|+1,s+2) ∗ for Xt ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|]. Then V (M) ∩ XV (Tj,t) ⊆ Y and

~~AL(i,|V (T )|+1,s+2) (V (M) ∩ XV (Tj,t)) ∩ XV (Tj,t) = ∅.~~

0 0 Proof. We may assume that there exists no t ∈ V (T ) with t ∈ V (Tt0 )−{t } and V (Tj,t0 )∩∂Tj,t 6= ∅ 00 0 (i0,`0) such that V (M) ∩ Xt00 6= ∅ for some witness t ∈ ∂Tj,t0 ∪ {t } for Xt00 ∩ Ij ⊆ W3 for some 0 0 0 0 ` ∈ [−1, |V (T )|], where i is the integer σT (t ), for otherwise we may replace t by t due to the (i0,|V (T )|+1,s+2) (i,|V (T )|+1,s+2) facts that Tj,t ⊆ Tj,t0 and Y ⊆ Y . (i,|V (T )|+1,s+2) Suppose to the contrary that either V (M)∩XV (Tj,t) 6⊆ Y , or AL(i,|V (T )|+1,s+2) (V (M)∩ ◦ (i) XV (Tj,t))∩XV (Tj,t) 6= ∅. Since SM ∩Ij 6= ∅, V (M)∩XV (Tj,t) ⊆ W4 . By Claim 21.1, some component (i,|V (T )|) ∗ Q of G[V (M) ∩ XV (Tj,t)] is disjoint from W3 . Since V (M) intersects Xt ⊆ X∂Tj,t ∪ Xt, V (Q) (i,−1) intersects X∂Tj,t ∪ Xt. Since Xt ∩ Ij ⊆ W3 , V (Q) ∩ Xt = ∅. So V (Q) intersects X∂Tj,t . (i,|V (T )|) 0 Since Xt∗ ∩ Ij ⊆ W3 , there exist a node t ∈ ∂Tj,t with Xt0 ∩ V (Q) 6= ∅ and Xt0 ∩ Ij 6⊆ (i,|V (T )|) (i,|V (T )|) W3 and a path P in M internally disjoint from W3 ∪ V (Q) passing through a vertex (i,|V (T )|) in V (M) ∩ W3 , a vertex in Xt0 − V (Q) and a vertex in Xt0 ∩ V (Q) in the order listed. Furthermore, choose Q such that P is as short as possible.

38 Let P0,P1,...,Pm (some some m ∈ N0) be the maximal subpaths of P contained in G[XV (Tj,t)] (i,|V (T )|) internally disjoint from X∂Tj,t ∪ Xt, where P0 intersects W3 and P passes through P0,P1,...,Pm in the order listed. So Pm intersects Q. Since P is internally disjoint from (i,|V (T )|) W3 ⊇ Xt, there exist (not necessarily distinct) t1, t2, . . . , tm+1 ∈ ∂Tj,t such that for every

` ∈ [m], P` is from Xt` to Xt`+1 . So V (Q) ∩ Xtm+1 6= ∅. (i,|V (T )|) ∗ (i,|V (T )|) Since V (Q)∩W3 = ∅, there exists a minimum ` ∈ [m+1] such that Xt`∗ ∩Ij 6⊆ W3 . (i,|V (T )|) 0 So Xt`∗−1 ∩ Ij ⊆ W3 . Let t0 = t . Note that if there exists q ∈ [−1, |V (T )|] such that (i,q) (i,q+1) (i,q) (i,|V (T )|) Ij ∩ W3 = Ij ∩ W3 , then Ij ∩ W3 = Ij ∩ W3 . This implies that if q ∈ [−1, |V (T )|] is (i,q) (i,|V (T )|) 00 the minimum such that Ij ∩W3 = Ij ∩W3 , then there exist at least q nodes t ∈ ∂Tj,t such (i,q) (i,|V (T )|) (i,|V (T )|−1) 00 that Xt ∩Ij ⊆ W3 . This together with Xt`∗ ∩Ij 6⊆ W3 imply that Xt`∗−1 ∩Ij ⊆ W3 . (i) (i,|V (T )|−1) ∗ Since P` −1 is contained in G[W4 ] and intersects Xt`∗−1 , and Xt`∗−1 ∩ Ij ⊆ W3 , (i,|V (T )|,k) Claim 21.1 implies that V (P`∗−1) ⊆ Y , where k+1 is the color of M. Let v be the vertex in (i,|V (T )|) (i,|V (T )|,0) (i,|V (T )|,0) ∗ 0 V (P` −1)∩Xt`∗ . Since Xt`∗ ∩Ij 6⊆ W3 , v ∈ (D −Xt)∪{u ∈ D ∩Xt ∩Xq ∩Ij : 0 0 (i00,`0) 00 0 q ∈ V (Tt) − {t}, q is a witness for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, i − 1] and ` ∈ [0, |V (T )|]}. 0 (i,|V (T )|,0) 0 Suppose there exists q ∈ V (Tt)−{t} such that v ∈ D ∩Xt ∩Xq0 ∩Ij and q is a witness (i0,`0) 00 0 00 for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, i − 1] and ` ∈ [0, |V (T )|]. Then there exists a node t of 00 00 00 0 T with σT (t ) = i ∈ [0, i − 1] and with t ∈ V (Tt00 ) − {t } such that q ∈ ∂Tj,t00 is a witness for (i00,`0) 0 0 Xq0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|]. Since q ∈ (V (Tt) − {t}) ∩ ∂Tj,t00 , we know ∂Tj,t00 is 00 00 disjoint from the path in T between t and t. So t`∗ ∈ ∂Tj,t ∩ V (Tj,t00 ). Hence t is a node of T 00 0 with t ∈ V (Tt00 ) − {t } and V (Tj,t00 ) ∩ ∂Tj,t 6= ∅ such that v ∈ V (M) ∩ Xq0 6= ∅ and q is a witness (i00,`0) 0 for Xq0 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], a contradiction. 0 (i,|V (T )|,0) (i0,`0+1,k) Therefore no such q exists, and hence v ∈ D − Xt. So v ∈ W0 for some i0 ∈ [0, i], `0 ∈ [0, |V (T )|] with (i0, `0) lexicographically at most (i, |V (T )|−1). We assume that (i0, `0) 0 (i0,`0+1,k) (i0) is lexicographically minimum. So there exists a monochromatic path P in G[Y ∩ W4 ] (i0,`0) (i0,`0) 0 0 (i0,`0,0) from v to W3 internally disjoint from W3 . The minimality of (i , ` ) implies that v 6∈ D . 0 (i,`0+1) (i,|V (T )|) 0 ∗ So if i = i, then t` is a witness for Xt`∗ ∩Ij ⊆ W3 ⊆ W3 , a contradiction. Hence i < i. 0 (i0,`0) 0 (i0,`0+1,k) (i0) (i0,`0) Let u be the end of P in W3 . Since P is in G[Y ∩ W4 ] from v to W3 internally (i0,`0) ∗ 00 00 00 0 disjoint from W3 , there exists q ∈ ∂Tj,t00 ∪{t }, where t is the node with σT (t ) = i , such that ∗ (i0,`0) 0 (i0) ∗ ∗ u ∈ Xq ∩ Ij and q is a witness for Xq ∩ Ij ⊆ W3 . Since i < i and v ∈ (XV (Tt) − Xt) ∩ W4 , 00 00 we know that t ∈ V (Tt00 ) − {t } and ∂Tj,t00 is disjoint from the path in T between t and t. So 0 V (Tj,t00 ) ∩ ∂Tj,t 6= ∅. Since v ∈ V (M) and P is between u and v, u ∈ V (M). So u ∈ V (M) ∩ Xq∗ ∗ 00 (i0,`0) 0 and q ∈ ∂Tj,t00 ∪ {t } is a witness for Xq∗ ∩ Ij ⊆ W3 with i < i, a contradiction. This proves the claim.

0 (i0,0,0) ◦ Proof. Assume that i is an integer in [0, i − 1] such that either |Y ∩ Ij ∩ XV (Tt) − Xt|= 6 0 0 0 (i ,0,s+2) ◦ (i ,−1,0) ◦ (i ,−1,w0) ◦ |Y ∩ Ij ∩ XV (Tt) − Xt| or |Y ∩ Ij ∩ XV (Tt) − Xt|= 6 |Y ∩ Ij ∩ XV (Tt) − Xt|. Let 0 0 0 (i0,−1,k,α) ◦ (i0,−1,k,α) t be the node of T with σT (t ) = i . Then either W2 ∩ Ij ∩ XV (Tt) − (Xt ∪ Y ) 6= ∅ (i0,0,α) ◦ (i0,0,α) for some k ∈ [0, w0 − 1] and α ∈ [0, s + 1], or W2 ∩ Ij ∩ XV (Tt) − (Xt ∪ Y ) 6= ∅ for some 0 α ∈ [0, s+1]. For the former, deﬁne `0 := −1, and let β(`0) := (−1, k, α) and β (`0) := (−1, k, α+1); 0 for the latter, deﬁne `0 := 0, and let β(`0) := (0, α) and β (`0) := (0, α + 1). So there exists a

39 0 0 (i ,`0,q) (i ) c-monochromatic path P of color α + 1 contained in G[W0 ] + Ej,t0 (for some q) intersecting (i0,β0(` )) (i0,β(` )) ◦ 0 0 0 0 0 Xt such that AL(i ,β(`0)) (V (P )) ∩ (Y − Y ) ∩ Zt ∩ Ij ∩ XV (Tt) − Xt 6= ∅. (i0) Note that t ∈ V (Tt0 ). So V (P ) ∩ Xt 6= ∅, since every element of Ej,t0 is contained in some ◦ 0 bag of (T, X ). Since AL(i ,β(`0)) (V (P )) ∩ Ij 6= ∅, V (P ) ⊆ Ij. So there exist v ∈ XV (Tt), u ∈ (i0,β0(` )) (i0,β(` )) ◦ 0 0 0 0 0 AL(i ,β(`0)) ({v})∩(Y −Y )∩Zt ∩Ij ∩XV (Tt) −Xt 6= ∅, and a subpath P of P contained 0 0 (i ,`0,q) (i ) 0 in G[W0 ] + Ej,t0 from Xt to v internally disjoint from Xt. Note that V (P ) ⊆ XV (Tt). 0 0 Suppose there exist i0 ∈ [0, i −1] with Tj,t ⊆ Tj,t0 and a c-monochromatic path Q0 containing v 00 (i0,`0) intersecting X 00 for some witness t ∈ ∂T ∪{t } for X 00 ∩I ⊆ W for some ` ∈ [−1, |V (T )|], t0 0 j,t0 0 t0 j 3 0 ◦ 0 where t0 is the node of T with σT (t0) = i0. Since u ∈ Ij ∩ AL(i ,β(`0)) ({v}), Q0 is contained in an ◦ s-segment in Sj whose level equals the color of Q0. By Claim 21.11, AL(i0,|V (T )|+1,s+2) (V (Q0)) ∩ ◦ 0 X = ∅. Since u ∈ Z 0 ∩ I , u ∈ X . Since i < i and T 0 ⊆ T , u ∈ A (i0,β(` )) ({v}) ∩ V (Tj,t0 ) t j V (Tj,t0 ) 0 j,t j,t0 L 0 X ⊆ A (i ,|V (T )|+1,s+2) (V (Q )) ∩ X = ∅, a contradiction. V (Tj,t0 ) L 0 0 V (Tj,t0 ) 0 0 Hence there do not exist i0 ∈ [0, i − 1] with Tj,t ⊆ Tj,t0 and a c-monochromatic path Q0 00 (i0,`0) containing v intersecting X 00 for some witness t ∈ ∂T ∪ {t } for X 00 ∩ I ⊆ W for some t0 0 j,t0 0 t0 j 3 `0 ∈ [−1, |V (T )|], where t0 is the node of T with σT (t0) = i0. 0 0 0 (i ,`0,q) 0 (i ) Suppose that P 6⊆ G[W0 ]. So E(P ) ∩ Ej,t0 6= ∅. Let P1 be the maximal subpath of 0 0 0 (i ) (i ,`0,q) P − Ej,t0 containing v. So P1 is a c-monochormatic path contained in G[W0 ], and one end (i0) of P1 is contained in an element e1 in Ej,t0 . So there exists the lexicographically minimal pair (i , ` ) such that e ∈ E(i1,`1), where t is the node of T with σ (t ) = i . Note that i < i0. 1 1 1 j,t1 1 T 1 1 1 (i1,`1) 00 00 By the deﬁnition of E , there exists t ∈ ∂T such that e ⊆ X 00 and t is a witness j,t1 1 j,t1 1 t1 1 0 (i1,` ) 0 0 0 for X 00 ∩ I ⊆ W for some ` ∈ [0, |V (T )|]. Since V (P ) ⊆ X and P is internally t1 j 3 V (Tt) 0 00 0 0 disjoint from Xt and e1 ∈ E(P ), we know t1 ∈ V (Tt) − {t} ⊆ V (Tt0 ) − {t }. Since i1 < i and 00 0 0 0 t1 ∈ ∂Tj,t1 ∩ V (Tt ) − {t }, Tj,t ⊆ Tj,t1 . But P1 is a c-monochromatic path containing v intersecting 0 00 (i1,` ) 0 X 00 for some witness t for X 00 ∩ I ⊆ W for some ` ∈ [0, |V (T )|], a contradiction. t1 1 t1 j 3 0 0 (i ,`0,q) 0 ◦ 0 So P ⊆ G[W0 ]. Let x be an end of P in Xt. Since Zt ∩ Ij ∩ XV (Tt) − Xt 6= ∅, t ∈ 0 0 0 (i ,`0,q) (i +1,2) V (Tj,t0 ) − ∂Tj,t0 . Since V (P ) ⊆ W0 , we know x ∈ Sj,t . (i0,2) 00 0 We may assume that x ∈ Sj,t , for otherwise we are done. So there exists i ∈ [0, i − 1] with 00 (i00+1,2) (i00,2) 00 00 t ∈ V (Tj,ti00 ) − ∂Tj,ti00 , where ti is the node of T with σT (ti ) = i , such that x ∈ Sj,t − Sj,t . 00 00 (i00,`00,k00) Hence there exist ` ∈ {0, −1} and k ∈ [0, w0 − 1] such that x ∈ W0 . So there exists a (i00,`00,k00) (i00) monochromatic path Q in G[Y ] + E from Xt to x internally disjoint from Xt . Since j,ti00 i00 i00 00 x ∈ V (P 0) ∩ Y (i00,`00,k00), there exists a maximal subpath P ∗ of P 0 contained in G[Y (i00,`00,k00)] + E(i ) j,ti00 containing x. ∗ 0 ∗ 0 If P = P , then let a = v and b = u; otherwise there exist a ∈ V (P ) and b ∈ NP 0 (a)∩V (P )− ∗ 0 V (P ). Note that for the former, b = u ∈ XV (Tt) − Xt; for the latter, b 6= x, so b ∈ V (P ) − Xt ⊆

XV (Tt) − Xt. 0 0 (i ,`0,q) 0 (i00,`00,k00) 00 0 Since P ⊆ G[W0 ], E(P ) ⊆ E(G). So b 6∈ Y by the fact i < i and the maximality 0 0 00 00 00 ∗ (i ,β (`0)) (i ,` ,k ) 000 00 0 of P . Hence b ∈ Y − Y . For each i ∈ [i , i ], let ti000 be the node of T with 000 ∗ (i00,`00,k00) (i00) σT (ti000 ) = i . Since Q ∪ P ⊆ G[Y ] + E intersects Xt00 and contains a, we know j,ti00 i (i00,`00,k00) (i00,−1,k00,c(a)−1) (i00,0,k00) a ∈ W0 . So if b ∈ Zti00 , then b ∈ W2 ∪ W2 , and hence c(a) 6= c(b) since 00 00 00 a ∈ W (i ,` ,k ), a contradiction. So b 6∈ Z . In particular, b ∈ I◦ − X . Since b ∈ X − X 0 ti00 j V (Tj,t00 ) V (Tt) t and t ∈ V (Tj,ti00 ) − ∂Tj,ti00 , there exists z ∈ ∂Tj,ti00 ∩ V (Tt) − {t} such that b ∈ XV (Tz) − Xz. Since 0 0 0 000 00 0 ◦ (i ,β (`0)) V (Tt) ⊆ V (Tt0 ) − {t }, b 6∈ XV (T ) for every i ∈ [i , i ]. However, since b ∈ I and b ∈ Y , j,ti000 j

~~40 ∗ 00 0 there exists a minimum i ∈ [i , i ] such that b ∈ XV (T ), a contradiction. This proves the j,ti∗ claim.~~

Claim 21.13. Let i ∈ N0 and let t be the node of T with σT (t) = i. Let j ∈ [|V| − 1]. Let 0 (i0,−1,0) ◦ (i0,0,s+2) ◦ S := {i ∈ [0, i − 1] : |Y ∩ Ij ∩ XV (Tt) − Xt|= 6 |Y ∩ Ij ∩ XV (Tt) − Xt|}. Then |S| 6 w0.

~~Claim 21.14. Let i ∈ N0 and let t be the node of T with σT (t) = i. Let j ∈ [|V| − 1]. Then (i,−1,0) ◦ |Y ∩ Ij ∩ XV (Tt)| 6 η1.~~

0 (i0,0,s+2) ◦ (i0,−1,0) ◦ Proof. Let S0 := {i ∈ [0, i − 1] : |Y ∩ Ij ∩ XV (Tt) − Xt| > |Y ∩ Ij ∩ XV (Tt) − Xt|}. 0 (i0,|V (T )|+1,s+2) ◦ By Claim 21.13, |S0| 6 w0. Let S2 := {i ∈ [0, i − 1] : |Y ∩ Ij ∩ XV (Tt) − Xt| > (i0,0,s+2) ◦ ◦ |Y ∩ Ij ∩ XV (Tt) − Xt|}. Since Ij ⊆ Ij, |S2| 6 3w0 by Claim 21.6. 0 0 For every i ∈ [0, i], let ti0 be the node of T with σT (ti0 ) = i . 0 (i0,0,s+2) (i0,−1,0) ◦ For every i ∈ S0, (Y − Y ) ∩ I ∩ XV (Tt) − Xt ⊆ XV (T ) ∩ Ij ∩ XV (Tt) − Xt. So for j j,ti0 0 (i0,0,s+2) (i0,−1,0) ◦ (i0,0,s+2) every i ∈ S0, |(Y −Y )∩I ∩XV (Tt)−Xt| |Y ∩Ij ∩XV (Tt)∩XV (T )| g1(s+2) j 6 j,ti0 6 by Statement 2 in Claim 21.10. 0 (i0,|V (T )|+1,s+2) (i0,0,s+2) ◦ For every i ∈ S2, (Y − Y ) ∩ I ∩ XV (Tt) − Xt ⊆ XV (T ) ∩ Ij ∩ XV (Tt) − Xt. j j,ti0 0 (i0,|V (T )|+1,s+2) (i0,0,s+2) ◦ (i0,|V (T )|+1,s+2) So for every i ∈ S2, |(Y − Y ) ∩ Ij ∩ XV (Tt) − Xt| 6 |Y ∩ Ij ∩

Claim 21.15. Let i ∈ N0, and let t be the node of T with σT (t) = i. Let j ∈ [|V | − 1]. Let 0 Zj be the set obtained from the j-th belt by deleting Ij−1,1 ∪ Ij,0. Let S := {i ∈ [0, i − 1] : (i0,−1,0) (i0,0,s+2) |Y ∩ XV (Tt) ∩ Zj − Xt| < |Y ∩ XV (Tt) ∩ Zj − Xt|}. Then |S| 6 5w0(2w0 + 1) + 2w0. ◦ ◦ 0 0 Proof. Let Zj := Zj ∪ Ij−1 ∪ Ij . For every i ∈ N0, let ti0 be the node of T with σT (ti0 ) = i . 0 (i0,−1,0) (i0,0,s+2) For every i ∈ S, since |Y ∩ XV (Tt) ∩ Zj − Xt| < |Y ∩ XV (Tt) ∩ Zj − Xt|, there 0 0 (i ,ì0 ,ki0 ) (i ,ì0 ,ki0 ) exist ` 0 ∈ {−1, 0} and k 0 ∈ [0, w − 1] such that W 6= ∅ and A (i0,` ,k ) (W ) ∩ Z ∩ i i 0 0 L i0 i0 0 j 0 0 (i ,0,s+2) (i ,ì0 ,ki0 ) Y ∩ X − X 6= ∅, so there exist v 0 ∈ W ∩ X and u 0 ∈ A (i0,` ,k ) ({v }) ∩ Z ∩ V (Tt) t i 0 V (Tt) i L i0 i0 i j (i0,0,s+2) 0 0 0 Y ∩ XV (Tt) − Xt 6= ∅. For every i ∈ S, since i < i, there exists a monochromatic path Pi in 0 0 (i ,` 0 ,k 0 ) (i ) i i 0 G[Y ] + R from Xti0 to vi ∈ XV (Tt) intersecting NG[Zj] internally disjoint from Xti0 , and (i0,0,s+2) (i0,−1,0) (i0) (i0) (i0) A (i0,`,k) (V (Pi0 )) ∩ (Y − Y ) ∩ Zj 6= ∅, where R = E if ì0 = −1, and R = ∅ if L j,ti0 (i0) 0 ì0 = 0. Since Pi0 is a monochromatic path, V (Pi0 ) ⊆ Zj by the definition of E . For every i ∈ S, j,ti0 0 let Pi0 be the subpath of Pi0 from vi0 to Xt internally disjoint from Xt, and let xi0 ∈ V (Pi0 ) ∩ Xt. 0 0 Note that V (Pi0 ) ⊆ XV (Tt) and xi ∈ Xt ∩ Zj. 0 0 (i0,−1,0) ◦ ◦ (i0,0,s+2) ◦ ◦ Let S = {i ∈ [0, i−1] : |Y ∩(Ij−1 ∪Ij )∩XV (Tt) −Xt|= 6 |Y ∩(Ij−1 ∪Ij )∩XV (Tt) − 0 Xt|}. By Claim 21.13, |S | 6 2w0. Suppose to the contrary that |S| > 5w0(2w0 + 1) + 2w0 + 1. Since |Xt ∩ Zj| 6 w0, there exist 0 i1, i2, i3 ∈ S with i1 < i2 < i3, ì1 = ì2 = ì3 and xi1 = xi2 = xi3 such that S ∩ [i1, i3] = ∅. Let

41 0 (iα−1,ìα−1 ,kiα−1 ) (iα−1) x = xi1 . For each α ∈ {2, 3}, let Qα be the maximal subpath of Piα ∩ G[Y ] + R containing x, and let aα be the end of Qα different from x if possible. For each α ∈ [3], if aα = viα , 0 then let bα = uiα ; otherwise, let bα be the neighbor of aα in Piα not in Qα. (iα0 ) Suppose that there exists α0 ∈ [2] such that {aα0+1, bα0+1} ∈ E(G) ∪ R . Then bα0+1 6∈ (iα ,` ,k ) (iα ) 0 iα0 iα0 (iα0 ) (iα0 ) 0 Y by the maximality of Qα0+1. If {aα0+1, bα0+1} ∈ R , then R = Ej,t and iα0 (iα ,` ,k ) (iα0 −1,|V (T )|+1,s+2) 0 iα0 iα0 {aα0+1, bα0+1} ⊆ Y ⊆ Y , a contradiction. So {aα0+1, bα0+1} ∈ E(G). (iα ,` ,k ) 0 iα0 iα0 (iα0 ) Note that Piα ∪ Qα +1 contains a monochromatic path in G[Y ] + R from Xt 0 0 iα0 (iα ,` ,k ) (iα ,` ,k ) 0 iα0 iα0 0 iα0 iα0 to aα +1. So aα +1 ∈ W and bα +1 ∈ A (iα ,` ,k ) (W ) ∩ XV (T ) − Xt. If 0 0 0 0 L 0 iα0 iα0 0 t (i +1,−1,0) (i ,` ,k ) b ∈ Z , then c(a ) 6= c(b ) and b ∈ Y α0 ⊆ Y α0+1 α0+1 α0+1 , so the former α0+1 tα0 α0+1 α0+1 α0+1 implies that b = u and the latter implies that b 6= u , a contradiction. So b 6∈ α0+1 iα0+1 α0+1 iα0+1 α0+1 ◦ ◦ 0 Ztα . So bα +1 ∈ I ∪ I . Note that ui ∈ Zj, so ui 6= bα +1. Hence bα +1 ∈ V (P ) ⊆ 0 0 j−1 j α0+1 α0+1 0 0 iα0+1 (iα +1,ì ,ki ) (iα +1,ì ,ki ) 0 α0+1 α0+1 0 α0+1 α0+1 (iα0 ,−1,0) ◦ ◦ Y . So bα0+1 ∈ (Y − Y ) ∩ (Ij−1 ∪ Ij ) ∩ XV (Tt) − Xt. Hence 0 [i1, iα0+1] ∩ S 6= ∅, a contradiction. (iα+1) (iα) (iα) (iα) Hence for each α ∈ [2], {aα+1, bα+1} ∈ R − R . In particular, ì = −1 and R = E α j,tiα (iα) (iα−1) for every α ∈ [3]. For each α ∈ {2, 3}, since {aα, bα} ∈ E −E , there exist βα ∈ [iα−1, iα −1] j,tiα j,tiα−1 0 (βα+1) (βα) (βα,` ) with {aα, bα} ∈ E − E and a witness qα ∈ ∂Tj,t for Xq ∩ Ij ⊆ W for some j,tiα j,tiα βα α 3 0 0 0 ` ∈ [0, |V (T )|] such that {aα, bα} ⊆ Xqα . For each α ∈ {2, 3}, since V (Piα ) ⊆ XV (Tt) and Piα is internally disjoint from X , q ∈ V (T ) − {t}. So T ⊆ T for every i0 ∈ [β , i ]. V (Tt) α t j,ti0 j,tβ2 2 3 Since P is a monochromatic path intersecting X and x ∈ X , V (P ) ∩ X 6= ∅. So i1 ti1 V (Tt) i1 tβ3 there exists a subpath P 00 of P from X to x ∈ X internally disjoint from X . Hence there i1 i1 tβ3 V (Tt) tβ3 ∗ exists k∗ ∈ [0, w − 1] such that V (P 00 ) ∩ X ⊆ W (β3,−1,k ). Let Q∗ be the maximal subpath of P 0 0 i1 tβ3 0 i3 ∗ (β ) contained in G[Y (β3,−1,k )] + E 3 containing x. Let a∗ be the end of Q∗ other than x if possible. j,tβ3 ∗ 0 ∗ ∗ ∗ 0 ∗ If Q = Pi3 , then a = vi3 , and we let b = ui3 ; otherwise, let b be the vertex in V (Pi3 ) − V (Q ) ∗ 0 adjacent to a in Pi3 . (β ) ∗ Suppose that {a∗, b∗} ∈ E(G) ∪ E 3 . Then b∗ 6∈ Y (β3,−1,k ) by the maximality of Q∗, since j,tβ3 ∗ ∗ ∗ (β3) ∗ ∗ (β3−1,|V (T )|+1,s+2) (β3,−1,k ) β3 < i3. If {a , b } ∈ E , then {a , b } ⊆ Y ⊆ Y , a contradiction. So j,tβ3 ∗ ∗ ∗ ∗ (β3,−1,k ) (β3) {a , b } ∈ E(G). Note that Pi1 ∪ Q contains a monochromatic path in G[Y ] + E j,tβ3 ∗ ∗ 00 ∗ ∗ (β3,−1,k ) ∗ (β3,−1,k ) from V (P ) ∩ X to a . So a ∈ W and b ∈ A (β ,−1,k∗) (W ) ∩ X − X . i1 tβ3 0 L 3 0 V (Tt) t Since β < i , if b∗ ∈ Z , then c(a∗) 6= c(b∗) and b∗ ∈ Y (β3+1,−1,0) ⊆ Y (i3,`3,k3), so the former 3 3 tβ3 implies that b∗ = u and the latter implies that b∗ 6= u , a contradiction. Hence b∗ 6∈ Z . i3 i3 tβ3 ∗ ◦ ◦ ∗ ∗ 0 (i3,−1,ki3 ) So b ∈ Ij−1 ∪ Ij . Note that ui3 ∈ Zj, so ui3 6= b . Hence b ∈ V (Pi3 ) ⊆ Y . So ∗ (i3,−1,ki3 ) (β3,−1,0) ◦ ◦ 0 0 b ∈ (Y − Y ) ∩ (Ij−1 ∪ Ij ) ∩ XV (Tt) − Xt. Hence [i1, i3] ∩ S ⊇ [β3, i3] ∩ S 6= ∅, a contradiction. ∗ ∗ (i3) (β3) (i3) (β3) Hence {a , b } ∈ E − E = E − E . So there exist γ ∈ [β3, i3 − 1] with {aα, bα} ∈ j,ti3 j,tβ3 j,ti3 j,ti3 (γ+1) (γ) ∗ (γ,`0) 0 ∗ E − E and a witness q ∈ ∂Tj,tγ for Xq ∩ Ij ⊆ W for some ` ∈ [0, |V (T )|] such that j,ti3 j,ti3 3 0 ∗ ∗ (β2,` ) {a , b } ⊆ X ∗ . However, T ⊆ T , β < γ, q ∈ ∂T is a witness for X ∩ I ⊆ W q j,tγ j,tβ2 2 2 j,tβ2 q2 j 3 0 ∗ (β3,|V (T )|+1,s+2) (β3) for some ` ∈ [0, |V (T )|], and Q2 ∪ Q is a monochromatic path in G[Y ] + E j,tβ3 ∗ ∗ (β3) (β3,|V (T )|+1,s+2) intersects Xq2 and {a , b }, so some E -pseudocomponent in G[Y ] intersects Xq2 j,tβ3 and {a∗, b∗}, contradicting that {a∗, b∗} ∈ E(γ+1). This proves the claim. j,ti3 0 0 Claim 21.16. Let i, i ∈ N0 with i < i, and let t be a node of T with σT (t) = i. Let j ∈ [|V |−1]. Let

~~42 (i0,|V (T )|+1,s+2) Zj be the set obtained from the j-th belt by deleting Ij−1,1∪Ij,0. Then |Y ∩Zj ∩XV (Tt)| 6 (i0,−1,0) g5(|Y ∩ Zj ∩ XV (Tt)|).~~

~~◦ ◦ Proof. Note that NG[Zj] ⊆ Zj ∪ Ij−1 ∪ Ij . For every k ∈ [0, w0 − 1] and q ∈ [0, s + 1],~~

~~(i0,0,k+1) |Y ∩ Zj ∩ XV (Tt)|~~

~~(i0,0,s+2) (i0,0,0) Therefore, |Y ∩ Zj ∩ XV (Tt)| 6 g4(s + 2, |Y ∩ Zj ∩ XV (Tt)|). Note that~~

(i0,|V (T )|+1,s+2) (i0,0,s+2) (Y − Y ) ∩ Zj ∩ XV (Tt) ⊆ (Y (i0,|V (T )|+1,s+2) − Y (i0,0,s+2)) ∩ ((I ∩ X ) ∪ (I ∩ X )) ∩ X j−1 V (Tj−1,t0 ) j V (Tj,t0 ) V (Tt) ⊆ Y (i0,|V (T )|+1,s+2) ∩ ((I ∩ X ) ∪ (I ∩ X )) ∩ X , j−1 V (Tj−1,t0 ) j V (Tj,t0 ) V (Tt)

0 0 0 (i0,|V (T )|+1,s+2) (i0,0,s+2) where t is the node of T with σT (t ) = i . So |(Y − Y ) ∩ Zj ∩ XV (Tt)| 6 |Y (i0,|V (T )|+1,s+2) ∩ ((I ∩ X ) ∪ (I ∩ X )) ∩ X | 2g (4w ) by Statement 4 in j−1 V (Tj−1,t0 ) j V (Tj,t0 ) V (Tt) 6 3 0 Claim 21.10. Therefore,

Claim 21.17. Let i ∈ N0, and let t be a node of T with σT (t) = i. Let j ∈ [|V | − 1]. Let Zj be the (i,−1,0) set obtained from the j-th belt by deleting Ij−1,1 ∪ Ij,0. Then |Y ∩ Zj ∩ XV (Tt)| 6 η2. Proof. Let

~~0 (i0,−1,0) (i0,0,s+2) S1 := {i ∈ [0, i − 1] : |Y ∩ XV (Tt) ∩ Zj − Xt| < |Y ∩ XV (Tt) ∩ Zj − Xt|}.~~

~~By Claim 21.15, |S1| 6 5w0(2w0 + 1) + 2w0. Let~~

~~0 (i0,0,s+2) (i0,|V (T )|+1,s+2) S2 := {i ∈ [0, i − 1] : |Y ∩ XV (Tt) ∩ Zj − Xt| < |Y ∩ XV (Tt) ∩ Zj − Xt|}.~~

0 (i0,|V (T )|+1,s+2) (i0,0,s+2) (i0,|V (T )|+1,s+2) Note that for every i ∈ S2, (Y − Y ) ∩ XV (Tt) ∩ Zj − Xt ⊆ (Y − (i0,0,s+2) Y ) ∩ XV (Tt) ∩ Zj ∩ (Ij−1 ∪ Ij) − Xt, where I0 = ∅. So

0 (i0,0,s+2) (i0,|V (T )|+1,s+2) |S2| 6 |{i ∈ [0, i − 1] : |Y ∩ XV (Tt) ∩ Ij−1 − Xt| < |Y ∩ XV (Tt) ∩ Ij−1 − Xt|}| 0 (i0,0,s+2) (i0,|V (T )|+1,s+2) + |{i ∈ [0, i − 1] : |Y ∩ XV (Tt) ∩ Ij − Xt| < |Y ∩ XV (Tt) ∩ Ij − Xt|}| 6 6w0 by Claim 21.6. Let

~~0 (i0,−1,0) (i0,|V (T )|+1,s+2) S3 := {i ∈ [0, i − 1] : |Y ∩ XV (Tt) ∩ Zj − Xt| < |Y ∩ XV (Tt) ∩ Zj − Xt|}.~~

~~So |S3| 6 |S1| + |S2| 6 5w0(2w0 + 1) + 8w0. Let~~

~~0 (i0,−1,0) (i0+1,−1,0) S := {i ∈ [0, i − 1] : |Y ∩ XV (Tt) ∩ Zj − Xt| < |Y ∩ XV (Tt) ∩ Zj − Xt|}.~~

44 0 (i0+1,−1,0) (i0,|V (T )|+1,s+2) Since for every i ∈ [0, i − 1], Y ∩ Zj ∩ XV (Tt) ⊆ (Y ∩ Zj ∩ XV (Tt)) ∪ (Xt ∩ Zj), we know |S| 6 |S3| + |Xt ∩ Zj| 6 5w0(2w0 + 1) + 9w0. 0 (i0+1,−1,0) (i0,|V (T )|+1,s+2) By Claim 21.16, for every i ∈ S, |Y ∩ Zj ∩ XV (Tt)| 6 |Y ∩ Zj ∩ XV (Tt)| + (i0,−1,0) (i0,−1,0) |Xt ∩ Zj| 6 g5(|Y ∩ Zj ∩ XV (Tt)|) + |Xt ∩ Zj| 6 g5(|Y ∩ Zj ∩ XV (Tt)|) + w0. (0,−1,0) Note that |Y ∩ Zj ∩ XV (Tt)| 6 |Xt ∩ Zj| 6 w0 = g6(0). Then it is easy to verify that (i,−1,0) |Y ∩Zj ∩XV (Tt)| 6 g6(|S|) by induction on the elements in S. Since |S| 6 5w0(2w0 +1)+9w0, (i,−1,0) |Y ∩ Zj ∩ XV (Tt)| 6 g6(5w0(2w0 + 1) + 9w0) = η2. ∗ Claim 21.18. Let i ∈ N0 and let t ∈ V (T ) with σT (t) = i. Let p be the parent of t if t 6= r ; ∗ otherwise let Fj,p be the set mentioned in the algorithm when t = r . Then

~~(i,−1,0) • for every q ∈ Fj,p, Y ∩ Ij ∩ XV (Tq) − Xq 6= ∅, and 0 (i+1,−1,0) • for every q ∈ Fj,t, Y ∩ Ij ∩ XV (Tq) − Xq 6= ∅.~~

∗ Proof. We shall prove this claim by induction on i. We first assume i = 0. So t = r . Since Fj,0 = ∅, (0,−1,0) T is the only Fj,0 ∩ V (Tt)-part containing t, and ∂T = ∅. Since Y = ∅, we know Fj,T = ∅. Hence the set Fj,p defined in the algorithm is ∅. So Statement 1 in this claim holds. Hence we may 0 0 0 0 assume that q ∈ Fj,t. Since Fj,p = ∅, q ∈ Fj,T 0 for some ∅-part T of Tt containing t. So T = T , and 0 (1,−1,0) F 0 is a (T, X | ,Y ∩I )-fence. By the definition of a fence (Statement 2 in Lemma 17), j,T XV (T )∩Ij j 0 00 00 (1,−1,0) there exist at least two Fj,T 0 -parts T satisfying q ∈ ∂T and XV (T 00) ∩ Y ∩ Ij − Xq 6= ∅. Note 00 (1,−1,0) that some such part T is contained in Tq. So Y ∩ Ij ∩ XV (Tq) − Xq 6= ∅. Hence the claim holds when i = 0. So we may assume i > 0, and the lemma holds for every smaller i. We first assume q ∈ Fj,p 0 0 0 0 and prove Statement 1. So either q ∈ Fj,p or q ∈ Fj,T 0 − Fj,p for some Fj,p ∩ V (Tt)-part T of Tt 0 0 0 (i,−1,0) containing t. If Fj,p ∩V (Tq) 6= ∅, then there exists q ∈ Fj,p ∩V (Tq) and Y ∩Ij ∩XV (Tq) −Xq ⊇ (i +1,−1,0) Y p ∩ I ∩ X − X 0 6= ∅ by the induction hypothesis, where i = σ (p). So we may j V (Tq0 ) q p T 0 0 0 0 assume Fj,p ∩V (Tq) = ∅. In particular, q ∈ Fj,T 0 −Fj,p for some Fj,p ∩V (Tt)-part T of Tt containing 00 t. By the definition of a fence (Statement 2 in Lemma 17), there exist at least two Fj,T 0 -parts T 00 (i,−1,0) 00 satisfying q ∈ ∂T and (Y ∪ X∂T 0 ) ∩ XV (T 00) ∩ Ij − Xq 6= ∅. So one such T is contained in Tq. 0 0 0 0 (i +1,−1,0) For every t ∈ ∂T , t ∈ F , so by the induction hypothesis, Y p ∩I ∩X −X 0 6= ∅. For j,p j V (Tt0 ) t 0 0 00 (i,−1,0) (i +1,−1,0) every t ∈ ∂T ∩V (T ), T 0 ⊆ T , so Y ∩I ∩X −X ⊇ Y p ∩I ∩X −X 0 6= ∅. t q j V (Tq) q j V (Tt0 ) t 0 00 (i,−1,0) 0 00 Hence if X∂T ∩ XV (T ) 6= ∅, then ∂T ∩ V (T ) 6= ∅, so Y ∩ Ij ∩ XV (Tq) − Xq 6= ∅; otherwise, (i,−1,0) (i,−1,0) 0 00 Y ∩ Ij ∩ XV (Tq) − Xq ⊇ (Y ∪ X∂T ) ∩ XV (T ) ∩ Ij ∩ −Xq 6= ∅. This proves Statement 1 of this claim. 0 0 Now we assume q ∈ Fj,t and prove Statement 2 of this claim. So either q ∈ Fj,p or q ∈ Fj,T 0 −Fj,p 0 (i,−1,0) for some Fj,p ∩ V (Tt)-part T of Tt containing t. If q ∈ Fj,p, then Y ∩ Ij ∩ XV (Tq) − Xq 6= ∅ 0 by Statement 1 of this claim. So we may assume q ∈ Fj,T 0 − Fj,p for some Fj,p ∩ V (Tt)-part 0 T of Tt containing t. By the definition of a fence (Statement 2 in Lemma 17), there exist at 0 00 00 (i+1,−1,0) least two Fj,T 0 -parts T satisfying q ∈ ∂T and (Y ∪ X∂T 0 ) ∩ XV (T 00) ∩ Ij − Xq 6= ∅. 00 0 0 0 So one such T is contained in Tq. For every t ∈ ∂T , t ∈ Fj,p, so by Statement 1 of this claim, (i,−1,0) 0 0 00 (i+1,−1,0) Y ∩I ∩X −X 0 6= ∅. For every t ∈ ∂T ∩V (T ), T 0 ⊆ T , so Y ∩I ∩X −X ⊇ j V (Tt0 ) t t q j V (Tq) q (i,−1,0) 0 00 (i+1,−1,0) Y ∩I ∩X −X 0 6= ∅. Hence if X 0 ∩X 00 6= ∅, then ∂T ∩V (T ) 6= ∅, so Y ∩I ∩ j V (Tt0 ) t ∂T V (T ) j (i+1,−1,0) (i+1,−1,0) 0 00 XV (Tq)−Xq 6= ∅; otherwise, Y ∩Ij ∩XV (Tq)−Xq ⊇ (Y ∪X∂T )∩XV (T )∩Ij ∩−Xq 6= ∅. This proves Statement 2 of this claim.

~~Claim 21.19. Let t ∈ V (T ). Let j ∈ [|V| − 1]. Then |X∂Tj,t ∩ Ij| 6 η3.~~

45 ∗ Proof. Let p be the parent of t if t 6= r ; otherwise, let Fj,p be the set mentioned in the algorithm ∗ (i,−1,0) when t = r . Note that ∂Tj,t ⊆ Fj,p, so by Claim 21.18, for every q ∈ ∂Tj,t, Y ∩ Ij ∩ 0 00 X − X 6= ∅. Since X − X 0 is disjoint from X − X 00 for any distinct q , q of V (Tq) q V (Tq0 ) q V (Tq00 ) q ∗ ∗ ∗ (i,−1,0) ∂Tj,t − {t }. where t is the node in Tj,t with σT (t ) minimum, we have |∂Tj,t| 6 |Y ∩ Ij|. (i,−1,0) (i,−1,0) (i,−1,0) Hence |∂Tj,t| 6 |Y ∩ Ij ∩ XV (Tt) − Xt| + 1 6 |Y ∩ Ij ∩ XV (Tt)| + 1 6 |Y ∩ (Zj ∪ ◦ Zj+1 ∪ Ij ) ∩ XV (Tt)| + 1 6 η1 + 2η2 + 1 by Claims 21.14 and 21.17, where Zj and Zj+1 are the sets obtained from the j-th belt by deleting Ij−1,1 ∪ Ij,0 and from the (j + 1)-th belt by deleting

~~Ij,1 ∪ Ij+1,0, respectively. Therefore, |X∂Tj,t ∩ Ij| 6 |∂Tj,t| · w0 6 (η1 + 2η2 + 1)w0 = η3.~~

~~|V|−1 |V|−1 |V|−1 (i,−1,k+1) (i,−1,k) (i,−1,k) [ [ [ ◦ [ 0 (Y − Y ) ∩ XV (Tt) ⊆ NG[W0 ] ∩ XV (Tt) ⊆ NG[ S] ⊆ Ij0 ⊆ Ij . j0=1 S∈S◦ j0=1 j0=1 j0~~

~~So for every k ∈ [0, w0 − 1] and q ∈ [0, s + 1],~~

(i,−1,k,q+1) (i,−1,k,q) (i,−1,k,q) ◦ (Y − Y ) ∩ Ij ∩ XV (Tt) − Xt ⊆ AL(i,−1,k,1) (Y1 ∩ Ij ) ∩ Ij ∩ XV (Tt) − Xt >s (i,−1,k,q) ◦ ⊆ NG (Y ∩ Ij ) ∩ Ij ∩ XV (Tt) − Xt >s (i,−1,k,q) ◦ ⊆ NG (Y ∩ Ij ∩ XV (Tt)).

46 Proof. Let a and b be the integers such that B = Sbj V . For every k ∈ [0, s + 2], let R := j j j α=aj α k Sbj +(s+2)−k V . Let Z := B −(I ∪I ). Note that B = R ⊆ R ⊆ I◦ ∪Z ∪I◦ = R α=aj −(s+2)+k α j j j−1,1 j,0 j s+2 k+1 j−1 j j 0 for every k ∈ [0, s + 1]. Note that for every k ∈ [0, s + 1],

(i,0,k+1) (i,0,k) (i,0,k) (Y − Y ) ∩ Rk+1 ∩ XV (Tt) − Xt ⊆ AL(i,0,k) (W0 ) ∩ Rk+1 ∩ XV (Tt) − Xt (i,0,k) ⊆ AL(i,0,k) (W0 ∩ Rk) ∩ Rk+1 ∩ XV (Tt) − Xt (i,0,k) ⊆ AL(i,0,k) (Y ∩ Rk) ∩ Rk+1 ∩ XV (Tt) − Xt >s (i,0,k) ⊆ NG (Y ∩ Rk ∩ XV (Tt)),

(i,0,0) |Y ∩ R0 ∩ XV (Tt)| (i,0,0) ◦ ◦ = |Y ∩ (Ij−1 ∪ Zj ∪ Ij ) ∩ XV (Tt)| (i,0,0) (i,0,0) (i,0,0) 6 |Y ∩ Ij−1 ∩ XV (Tt)| + |Y ∩ (Zj − (Ij−1 ∪ Ij)) ∩ XV (Tt)| + |Y ∩ Ij ∩ XV (Tt)| (i,−1,0) 6 g8(w0) + |Y ∩ Zj ∩ XV (Tt)| + g8(w0) 6 2g8(w0) + η2 =g9(0)

Proof. Let t be the node of T with minimum σT (t) such that V (M)∩XV (Tt) 6= ∅. Let i = σT (t). By (i,0,s+2) the minimality of i, V (M) ⊆ XV (Tt) and V (M)∩Xt 6= ∅. We claim that V (M) ⊆ Y ∩XV (Tt). Suppose to the contrary that V (M) 6⊆ Y (i,0,s+2). Let k ∈ [0, s + 1] such that c(v) = k + 1 for (i,−1,0) (i,−1,0) (i,0,k) every v ∈ V (M). Since Xt ⊆ Y , V (M) ∩ Y ∩ Xt 6= ∅. So V (M) ∩ Y ∩ Xt 6= ∅. 0 (i,0,k) Let M be the union of all components of M[Y ] intersecting Xt. Since V (M) ⊆ XV (Tt), 0 (i,0,k) (i,0,s+2) (i,0,k) 0 V (M ) ⊆ W0 . Since V (M) 6⊆ Y , V (M) 6⊆ Y , so there exists v ∈ V (M) − V (M ) 0 0 adjacent in G to V (M ). So v ∈ AL(i,0,k) (V (M )). Since v ∈ V (M), and V (M) is disjoint from S|V|−1 ◦ (i,0,k) (i,0,k+1) (i,0,k+1) (i,0,k) j=1 Ij , v ∈ Zt. So v ∈ W2 . Since (Y ,L ) is a (W2 , k + 1)-progress of (Y (i,0,k),L(i,0,k)), k + 1 6∈ L(i,0,k+1)(v). But c(v) = k + 1, a contradiction. (i,0,s+2) Therefore, V (M) ⊆ Y ∩ XV (Tt). Since M is a monochromatic component disjoint from S|V|−1 ◦ ∗ j=1 Ij , there exists j ∈ [|V| − 1] such that V (M) ⊆ Zj∗ , where Zj∗ is the set obtained from the ∗ (i,0,s+2) ∗ ∗ ∗ j -th belt by deleting Ij −1,1 ∪Ij ,0. So |V (M)| 6 |Y ∩XV (Tt) ∩Zj | 6 η4 by Claim 21.21.

47 Claim 21.23. Let i ∈ N0 and let t ∈ V (T ) with σT (t) = i. Let j ∈ [|V| − 1] and ` ∈ [0, |V (T )|]. (i,`+1,s+2) (i,`+1,0) Then |Y ∩ Ij ∩ XV (Tt)| 6 fs+2(|Y ∩ Ij ∩ XV (Tt)| + 2η4) − 2η4. (i,|V (T )|+1,s+2) (i,0,s+2) Proof. Let Bj be the j-th belt. Note that |Y ∩(Bj −(Ij−1 ∪Ij))∩XV (Tt)| = |Y ∩

(i,`+1,0) |Y ∩ ((Bj ∪ Bj+1) − (Ij−1 ∪ Ij+1)) ∩ XV (Tt)| (i,`,s+2) 6 |Y ∩ ((Bj ∪ Bj+1) − (Ij−1 ∪ Ij+1)) ∩ XV (Tt)| + |X∂Tj,t ∩ Ij| (i,`,s+2) 6 |Y ∩ ((Bj ∪ Bj+1) − (Ij−1 ∪ Ij+1) ∩ XV (Tt)| + η3.

(i,`+1,s+2) (i,`+1,0) For every ` ∈ [0, |V (T )|], (Y − Y ) ∩ ((Bj ∪ Bj+1) − (Ij−1 ∪ Ij+1)) ∩ XV (Tt) ⊆ (i,`+1,s+2) (i,`+1,0) (Y − Y ) ∩ Ij ∩ XV (Tt). So for every ` ∈ [0, |V (T )|], by Claim 21.23,

~~(i,|V (T )|+1,s+2) j+1 (i,|V (T )|+1,s+2) P 0 Proof. By Claim 21.24, |Y ∩ (Bj ∪ Bj+1) ∩ XV (Tt)| 6 j0=j−1|Y ∩ ((Bj ∪ 0 0 0 Bj +1) − (Ij −1 ∪ Ij +1)) ∩ XV (Tt)| 6 3 · η5/3 = η5.~~

Given any c-monochromatic component M, there uniquely exists a node rM of T such that V (M)∩X 6= ∅ and V (M) ⊆ X , and we deﬁne i := σ (r ). Recall that every monochro- rM V (TrM ) M T M matic component is contained in some s-segment. For every c-monochromatic component M, let SM be the s-segment containing V (M) whose level equals the color of M. Recall that for every node t of T , we denote σT (t) by it. ◦ Given any c-monochromatic component M with SM ∩ Ij 6= ∅ for some j ∈ [|V| − 1], deﬁne the following.

~~48 • Deﬁne K(M) to be the subset of V (T ) constructed by repeatedly applying the following process until no more nodes can be added:~~

~~– rM ∈ K(M). 0 – For every t ∈ K(M), if there exists t ∈ ∂T such that V (M) ∩ X − X 0 6= ∅, then j,t V (Tt0 ) t adding t0 into K(M).~~

∗ 0 0 • Deﬁne K (M) := {rM } ∪ {t ∈ K(M) − {rM } : t is the node in K(M) such that t ∈ ∂Tj,t, and |V (M) ∩ Y (it0 ,|V (T )|+1,s+2) ∩ X | > |V (M) ∩ Y (it,|V (T )|+1,s+2) ∩ X |}. V (Tt0 ) V (Tt0 ) Note that |V (M)| = |V (M) ∩ X | + P |V (M) ∩ X − X |. V (Tj,rM ) t∈K(M)−{rM } V (Tj,t) t ◦ Claim 21.26. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∩ Ij 6= ∅. Then ∗ |V (M)| 6 |K (M)| · η5.

(it,−1) Proof. For every t ∈ K(M), since t is a witness for Xt ∩ Ij ⊆ W3 , by Claim 21.11, |V (M) ∩ (it,|V (T )|+1,s+2) XV (Tj,t)| = |V (M) ∩ Y ∩ XV (Tj,t)|. So X |V (M)| = |V (M) ∩ X | + |V (M) ∩ X − X | V (Tj,rM ) V (Tj,t) t t∈K(M)−{rM } (i ,|V (T )|+1,s+2) X (i ,|V (T )|+1,s+2) = |V (M) ∩ Y rM ∩ X | + |V (M) ∩ Y t ∩ X − X |. V (Tj,rM ) V (Tj,t) t t∈K(M)−{rM }

Note that for every t ∈ K(M) − {rM }, there exists kt ∈ K(M) with t ∈ V (Tkt ) such that (ikt ,|V (T )|+1,s+2) (it,|V (T )|+1,s+2) |V (M) ∩ Y ∩ XV (Tj,t) − Xt| = |V (M) ∩ Y ∩ XV (Tj,t) − Xt| (since t is a ∗ candidate for kt). Choose such a node kt such that σT (kt) is as small as possible. Thus kt ∈ K (M) 0 0 0 for each t ∈ K(M) − {rM }. Let K = {t ∈ V (T ): t = kt for some t ∈ K(M) − {rM }}. Note that K0 ⊆ K∗(M), and for distinct t , t ∈ K(M) − {r }, X − X and X − X are 1 2 M V (Tj,t1 ) t1 V (Tj,t2 ) t2 disjoint. So

(i ,|V (T )|+1,s+2) X (i ,|V (T )|+1,s+2) |V (M)| = |V (M) ∩ Y rM ∩ X | + |V (M) ∩ Y t ∩ X − X | V (Tj,rM ) V (Tj,t) t t∈K(M)−{rM } (i ,|V (T )|+1,s+2) X (i ,|V (T )|+1,s+2) = |V (M) ∩ Y rM ∩ X | + |V (M) ∩ Y kt ∩ X − X | V (Tj,rM ) V (Tj,t) t t∈K(M)−{rM } (i ,|V (T )|+1,s+2) X (i ,|V (T )|+1,s+2) |V (M) ∩ Y rM ∩ X | + |V (M) ∩ Y q ∩ X | 6 V (TrM ) V (Tq) q∈K0

~~X (iq,|V (T )|+1,s+2) 6 |V (M) ∩ Y ∩ XV (Tq)| q∈K∗(M) ∗ 6 |K (M)| · η5 by Claim 21.25.~~

◦ Claim 21.27. Let j ∈ [|V|−1]. Let M be a c-monochromatic component with SM ∩Ij 6= ∅. Let t ∈ (it,−1,0) V (T ) with V (M)∩Xt 6= ∅. Then either V (M)∩XV (Tt) ⊆ Y , or there exists a monochromatic 0 (it,−1,0) 0 0 0 component M in G[Y ] with M ⊆ M, V (M )∩Xt 6= ∅ and AL(it,−1,0) (V (M ))∩XV (Tt) −Xt 6= ∅.

49 (it,−1,0) (it,−1,0) Proof. We may assume V (M)∩XV (Tt) 6⊆ Y , for otherwise we are done. Since Xt ⊆ Y , (it,−1,0) there exists a path P in M from Xt to a vertex v ∈ V (M) ∩ XV (Tt) − Y internally disjoint (it,−1,0) 0 (it,−1,0) from Xt such that V (P − v) ⊆ Y . Hence the monochromatic component M in G[Y ] 0 0 0 containing P −v satisﬁes that M ⊆ M, V (M )∩Xt 6= ∅ and v ∈ AL(it,−1,0) (V (M ))∩XV (Tt) −Xt 6= ∅.

◦ Claim 21.28. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∩ Ij 6= ∅. Let ∗ 0 t ∈ K(M). If K (M) ∩ V (Tt) − {t} 6= ∅, then there exists a monochromatic component M in (it,−1,0) 0 0 0 G[Y ] with M ⊆ M, V (M ) ∩ Xt 6= ∅ and AL(it,−1,0) (V (M )) ∩ XV (Tt) − Xt 6= ∅.

Proof. Suppose to the contrary. Since t ∈ K(M), V (M) ∩ Xt 6= ∅. So by Claim 21.27, V (M) ∩ (it,−1,0) ∗ ∗ XV (Tt) ⊆ Y . Since K (M) ∩ V (Tt) − {t}= 6 ∅, there exists t1 ∈ K (M) ∩ V (Tt) − {t}. Since ∗ t ∈ K(M) and t1 ∈ K (M) ∩ V (Tt) − {t}, t 6= rM . So there exists t2 ∈ K(M) with t1 ∈ ∂Tj,t2 such (i ,|V (T )|+1,s+2) (i ,|V (T )|+1,s+2) that V (M) ∩ Y t1 ∩ X 6= V (M) ∩ Y t2 ∩ X . Since t, t ∈ K(M) V (Tt1 ) V (Tt1 ) 2 and t ∈ ∂T ∩ X − X , t ∈ V (T ). So i > i i and X ⊆ X . Hence 1 j,t2 V (Tt) t 2 t t1 t2 > t V (Tt1 ) V (Tt) (i ,|V (T )|+1,s+2) (i ,−1,0) (i ,−1,0) (i ,|V (T )|+1,s+2) V (M)∩Y t1 ∩X ⊆ Y t ⊆ Y t2 . So V (M)∩Y t1 ∩X = V (Tt1 ) V (Tt1 ) (i ,|V (T )|+1,s+2) V (M) ∩ Y t2 ∩ X , a contradiction. V (Tt1 )

Recall that σ is a linear order of V (G) such that for any distinct vertices u, v, if σT (ru) < σT (rv), then σ(u) < σ(v); and for every subgraph H of G, σ(H) is deﬁned to be min{σ(v): v ∈ V (H)}. For any i ∈ [0, |V (T )| − 1], ` ∈ [−1, |V (T )| + 1], k ∈ [0, s + 2] and j ∈ [|V| − 1], we deﬁne the (i, `, k, j)-signature to be the sequence (a1, a2, . . . , a|V (G)|) such that for every α ∈ [|V (G)|],

~~(i,`,k) • if σ(M) = α for some monochromatic component M in G[Y ] intersecting Xt (where~~

~~t is the node of T with it = i) with AL(i,`,k) (V (M)) ∩ XV (Tt) − Xt 6= ∅ and contained in ◦ some s-segment in Sj whose level equals the color of M, then deﬁne aα = |AL(i,`,k) (V (M)) ∩~~

~~XV (Tt) − Xt|,~~

~~• otherwise, deﬁne aα = 0.~~

In the case aα > 0, we say that M defines aα. Claim 21.29. Let i ∈ [0, |V (T )| − 1], ` ∈ [−1, |V (T )| + 1], k ∈ [0, s + 2] and j ∈ [|V| − 1]. Then the (i, `, k, j)-signature has at most w0 nonzero entries, and every entry is at most f(η5). S Proof. Let t be the node of T with it = i. Since ◦ S ⊆ Ij and |Xt ∩Ij| w0, there are at most S∈Sj 6 (i,`,k) w0 monochromatic components M in G[Y ] intersecting Xt with AL(i,`,k) (V (M))∩XV (Tt) −Xt 6= ◦ ∅ and contained in some s-segment in Sj whose level equals the color of M. So the (i, `, k, j)- signature has at most w0 nonzero entries. Let aα be a nonzero entry of the (i, `, k, j)-signature. Let M be the monochromatic component >s (i,`,k) defining aα. So aα = |AL(i,`,k) (V (M)) ∩ XV (Tt) − Xt| 6 |NG (Y ∩ Ij ∩ XV (Tt)) ∩ XV (Tt)| 6 >s (i,|V (T )|+1,s+2) |NG (Y ∩ Ij ∩ XV (Tt)) ∩ XV (Tt)| 6 f(η5) by Claim 21.25. For any i ∈ [0, |V (T )| − 1], ` ∈ [−1, |V (T )| + 1], k ∈ [0, s + 2] and j ∈ [|V| − 1], we define the (i, `, k, j)-pseudosignature to be the sequence (p1, p2, . . . , p|V (G)|) such that for every α ∈ [|V (G)|], pα is a sequence of length |V (G)| such that

~~(i) (i,`,k) • if σ(M) = α for some monochromatic Ej,t -pseudocomponent M in G[Y ] intersecting Xt~~

~~(where t is the node of T with it = i) with AL(i,`,k) (V (M)) ∩ XV (Tt) − Xt 6= ∅ and contained ◦ in some s-segment in Sj whose level equals the color of M, then for every β ∈ [|V (G)|],~~

50 – the β-th entry of pα is the β-th entry of the (i, `, k, j)-signature if the β-th entry of the (i, `, k, j)-signature is positive and the monochromatic component deﬁning the β-th entry of the (i, `, k, j)-signature is contained in M,

~~– otherwise, the β-th entry of pα is 0,~~

~~• otherwise, pα is a zero sequence.~~

In the case pα is not a zero sequence, we say that M deﬁnes pα. Note that the (i, `, k, j)- pseudosignature has |V (G)|2 entries. Claim 21.30. Let i ∈ [0, |V (T )| − 1], ` ∈ [−1, |V (T )| + 1], k ∈ [0, s + 2] and j ∈ [|V| − 1]. Let (p1, p2, . . . , p|V (G)|) be the (i, `, k, j)-pseudosignature. Then for every α ∈ [|V (G)|], there exist at most one index β such that the α-th entry of pβ is nonzero. Furthermore, the (i, `, k, j)- pseudosignature has at most w0 nonzero entries, and every entry is at most f(η5). Proof. Every monochromatic component in G[Y (i,`,k)] is contained in at most one (i) (i,`,k) Ej,t -pseudocomponent in G[Y ]. So for every α ∈ [|V (G)|], there exist at most one index β such that the α-th entry of pβ is nonzero. Hence the number of nonzero entries and the sum of the entries of the (i, `, k, j)-pseudosignature equal the number of nonzero entries and the sum of the entries of the (i, `, k, j)-signature, respectively. Then this claim follows from Claim 21.29.

For i ∈ N0, j ∈ [|V| − 1], ` ∈ [−1, |V (T )| + 1] and k ∈ [0, s + 2], we say a monochromatic (i,`,k) ◦ ◦ subgraph M in G[Y ] is Sj -related if V (M) is contained in some s-segment in Sj whose level equals the color of M.

~~Claim 21.31. Let i1, i2 ∈ N0. Let j ∈ [|V| − 1]. Let `1, `2 ∈ [−1, |V (T )| + 1]. Let k1, k2 ∈~~

[0, s + 2]. Let t1, t2 be nodes of T with σT (t1) = i1 and σT (t2) = i2. Assume that t2 ∈ V (Tt1 ) ◦ and (i1, `1, k1) is lexicographically smaller than (i2, `2, k2). Let M2 be an Sj -related monochromatic (i ) (i ) E 2 -pseudocomponent in G[Y (i2,`2,k2)] intersecting X . Then there exist a monochromatic E 1 - j,t2 t1 j,t1 (i1,`1,k1) pseudocomponent M1 in G[Y ] such that M1 ⊆ M2, σ(M1) = σ(M2) and V (M1) ∩ Xt1 6= ∅. Furthermore, if A (i ,` ,k ) (V (M ))∩X −X 6= ∅, then A (i ,` ,k ) (V (M ))∩X −X 6= ∅. L 2 2 2 2 V (Tt2 ) t2 L 1 1 1 1 V (Tt1 ) t1 Proof. Let v be the vertex of M such that σ(v ) = σ(M ). Since V (M )∩X 6= ∅, t ∈ V (T ). M 2 M 2 2 t1 1 rvM 0 (i1,`1,k1) So there exists a monochromatic component M in G[Y ] containing vM . Hence there exists (i ) a monochromatic E 1 -pseudocomponent M in G[Y (i1,`1,k1)] containing M 0. So σ(M ) σ(M 0) j,t1 1 1 6 6 σ(v ) = σ(M ). Since t ∈ V (T ), E(i1) ⊆ E(i1) ⊆ E(i2). Since (i , ` , k ) is lexicographically M 2 2 t1 j,t1 j,t2 j,t2 1 1 1 smaller than (i2, `2, k2), M1 ⊆ M2. Hence σ(M1) = σ(M2). Suppose V (M ) ∩ X = ∅. Since t ∈ V (T ) and v ∈ V (M ), V (M ) ∩ X = ∅. 1 t1 1 rvM M 1 1 V (Tt1 )

Since V (M2) ∩ Xt1 6= ∅, there exists a path P in M2 from Xt1 to V (M1) internally disjoint from (it2 ,`2,k2) V (M1) ∪ Xt1 . Let u be the vertex in V (P ) ∩ V (M1). Since V (P ) ⊆ Y and t2 ∈ V (Tt1 ) and P is internally disjoint from X , the neighbor of u in P is in Y (i1,−1,0) ⊆ Y (i1,`1,k1). Since M is V (Tt1 ) 1 a monochromatic E(i1)-pseudocomponent, the edge of P incident with u belongs to E(i2) − E(i1). j,t1 j,t2 j,t1 But u ∈ V (M ), so u 6∈ X . Hence there exists no element in E(i2) − E(i1) containing u, a 1 V (Tt1 ) j,t2 j,t1 contradiction. Therefore V (M1) ∩ Xt1 6= ∅. Now we assume that A (i ,` ,k ) (V (M )) ∩ X − X 6= ∅. Suppose that A (i ,` ,k ) (V (M )) ∩ L 2 2 2 2 V (Tt2 ) t2 L 1 1 1 1 X − X = ∅. Since (i , ` , k ) is lexicographically smaller than (i , ` , k ) and X − V (Tt1 ) t1 1 1 1 2 2 2 V (Tt2 ) X ⊆ X − X , V (M ) ⊃ V (M ). Since A (i ,` ,k ) (V (M )) ∩ X − X = ∅, there exists t2 V (Tt1 ) t1 2 1 L 1 1 1 1 V (Tt1 ) t1 e ∈ E(i2) − E(i1) such that e ∩ V (M ) 6= ∅ 6= e − V (M ). Since t ∈ V (T ), E(i1) = E(i1), so j,t2 j,t1 1 1 2 t1 j,t2 j,t1

51 E(i2) − E(i1) ⊆ E(i2) − E(i1). So there exists i ∈ [i , i − 1] and t ∈ V (T ) with σ (t ) = i and j,t2 j,t1 j,t2 j,t2 3 1 2 3 T 3 3 t ∈ V (T ) such that e ∈ E(i3+1) − E(i3). We may assume i is as small as possible. Hence there 2 t3 j,t2 j,t2 3 exist t ∈ ∂T and α ∈ [0, |V (G)|] such that t is the witness for e ∈ E(i3,α+1) − E(i3,α). 4 j,t3 4 j,t2 j,t2 Since A (i ,` ,k ) (V (M )) ∩ X − X = ∅, by the minimality of i , M is a monochromatic L 1 1 1 1 V (Tt1 ) t1 3 1 (i ,α) t3 (it ,|V (T )|+1,s+2) E -pseudocomponent in G[Y 3 ] and A (i ,|V (T )|+1,s+2) (V (M )) ∩ X − X 6= ∅. j,t2 L 3 1 V (Tt4 ) t4 Since i i , t ∈ V (T ). So A (i ,` ,k ) (V (M )) ∩ X − X ⊇ A (i ,|V (T )|+1,s+2) (V (M )) ∩ 3 > 1 4 t1 L 1 1 1 1 V (Tt1 ) t1 L 3 1 X − X 6= ∅, a contradiction. This proves the claim. V (Tt4 ) t4

(α) (α) (α) Claim 21.32. Let t1, t2, t3 be nodes of T such that Tt3 ⊆ Tt2 ⊂ Tt1 . For α ∈ [3], let (a1 , a2 , . . . , a|V (G)|) ∗ (1) be the (itα , −1, 0, j)-pseudosignature. Let α be the largest index such that aα∗ is a nonzero sequence. Then the following hold.

~~∗ (1) (2) • For every β ∈ [α − 1], if aβ is a zero sequence, then aβ is a zero sequence. ∗ (2) (3) • For every β ∈ [α ], if aβ is a zero sequence, then aβ is a zero sequence.~~

∗ (2) Proof. We first prove the first statement. Suppose there exists β ∈ [α − 1] such that aβ is a (it2 ) (i ,−1,0) nonzero sequence. Let M be the monochromatic E -pseudocomponent in G[Y t2 ] defining 2 j,t2 (2) (1) (it1 ) 0 a . Since a ∗ is a nonzero sequence, there exists a monochromatic E -pseudocomponent M in β α j,t1 1 (1) (it1 ,−1,0) ∗ 0 0 G[Y ] defining aα∗ . Since σ(M2) = β < α = σ(M1) and V (M1) ∩ Xt1 6= ∅ 6= V (M2) ∩ Xt2 , (i ,−1,0) V (M ) ∩ X 6= ∅. By Claim 21.31, there exists a monochromatic E t1 -pseudocomponent M 2 t1 j,t1 1 such that M ⊆ M , σ(M ) = σ(M ) = β, V (M )∩X 6= ∅, and A (i ,−1,0) (V (M ))∩X −X 6= 1 2 1 2 1 t1 L t1 1 V (Tt1 ) t1 (1) ∅. So aβ is a nonzero sequence, a contradiction. (2) Now we prove the second statement. Suppose to the contrary that aβ is a zero sequence and (3) ∗ (1) (1) ∗ aβ is a nonzero sequence. If β = α , then aβ = aα∗ is a nonzero sequence; if β ∈ [α − 1], then (1) (3) aβ is a nonzero sequence by the first statement of this claim. Since aβ is a nonzero sequence, (it3 ) (i ,−1,0) (3) (1) there exists a monochromatic E -pseudocomponent Q in G[Y t3 ] defining a . Since a j,t3 3 β β (it1 ) (i ,−1,0) is a nonzero sequence, there exists a monochromatic E -pseudocomponent Q in G[Y t1 ] j,t1 1 (1) defining aβ . Hence σ(Q1) = β = σ(Q3). So Q1 ⊆ Q3. Hence V (Q3) ∩ Xt1 6= ∅= 6 V (Q3) ∩ Xt3 . So (i ,−1,0) V (Q ) ∩ X 6= ∅. By Claim 21.31, there exists a monochromatic E t2 -pseudocomponent Q 3 t2 j,t2 2 such that Q ⊆ Q , σ(Q ) = σ(Q ) = β, V (Q )∩X 6= ∅ and A (i ,−1,0) (V (Q ))∩X −X 6= ∅. 2 3 2 3 2 t2 L t2 2 V (Tt2 ) t2 (2) So aβ is a nonzero sequence, a contradiction.

◦ (it) Claim 21.33. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M be an Sj -related monochromatic Ej,t - (it,−1,0) ∗ pseudocomponent in G[Y ] intersecting Xt. Let t ∈ V (T ) be the node with t ∈ V (Tj,t∗ ) (i ∗ ) t (it∗ ,|V (T )|+1,s+2) such that some monochromatic Ej,t∗ -pseudocomponent in G[Y ] intersecting V (M) 0 ∗ (it∗ ,`) intersects Xt0 for some witness t ∈ (∂Tj,t∗ ) ∪ {t } for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], (it∗ ) and subject to this, it∗ is minimum. Then for every monochromatic Ej,t∗ -pseudocomponent M1 in G[Y (it∗ ,|V (T )|+1,s+2)] intersecting V (M),

~~00 ∗ (it∗ ,`) • M1 intersects Xt00 for some witness t ∈ (∂Tj,t∗ ) ∪ {t } for Xt00 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and~~

~~52 ∗ (it∗ ,`) • for every node q ∈ (∂Tj,t∗ ) ∪ {t } with V (M1) ∩ Xq 6= ∅, q is a witness for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|].~~

∗ Proof. Note that t exists since t is an candidate. So it∗ 6 it. ∗ (it∗ ,`1) ∗ If there exist distinct nodes q1, q2 ∈ (∂Tj,t ) ∪ {t } where q1 is a witness for Xq1 ∩ Ij ⊆ W3 for some `1 ∈ [−1, |V (T )|] such that there exists a path P in M[V (M) ∩ XV (Tj,t∗ )] from Xq1 to Xq2 (it) ∗ ∗ internally disjoint from X∂Tj,t∗ ∪ Xt , then by the minimality of it and the construction of Ej,t , (i ) t (it∗ ,|V (T )|+1,s+2) E(P ) ∩ Ej,t = ∅, so P is a path in G, and hence by Claim 21.11, V (P ) ⊆ Y . Hence, (i ∗ ) t (it∗ ,|V (T )|+1,s+2) since |∂Tj,t∗ | 6 |V (T )|−1 and some monochromatic Ej,t∗ -pseudocomponent in G[Y ] 00 ∗ (it∗ ,`) intersecting V (M) intersects Xt00 for some witness t ∈ (∂Tj,t∗ ) ∪ {t } for Xt00 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], to prove this claim, it suﬃces to prove that if M1 is a component of 0 ∗ 0 ∗ 00 M[V (M) ∩ XV (Tj,t∗ )] − {e ∈ E(M): e ⊆ Xt for some t ∈ ∂Tj,t ∪ {t }} such that V (M1) ∩ Xt 6= ∅ 00 ∗ (it∗ ,`) for some witness t ∈ (∂Tj,t∗ ) ∪ {t } for Xt00 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], then for ∗ (it∗ ,`) every node q ∈ (∂Tj,t∗ ) ∪ {t } with V (M1) ∩ Xq 6= ∅, q is a witness for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|]. 0 ∗ 0 ∗ Let M1 be a component of M[V (M)∩XV (Tj,t∗ )]−{e ∈ E(M): e ⊆ Xt for some t ∈ ∂Tj,t ∪{t }} 00 ∗ (it∗ ,`) such that V (M1) ∩ Xt00 6= ∅ for some witness t ∈ (∂Tj,t∗ ) ∪ {t } for Xt00 ∩ Ij ⊆ W3 for some ` ∈ ◦ (it∗ ) (it) [−1, |V (T )|]. Since M is Sj -related, V (M1) ⊆ W4 . By the minimality of it∗ , E(M1) ∩ Ej,t = ∅. ∗ Suppose to the contrary that there exists q ∈ (∂Tj,t∗ ) ∪ {t } such that V (M1) ∩ Xq 6= ∅, (it∗ ,α) and q is not a witness for Xq ∩ Ij ⊆ W3 for any α ∈ [−1, |V (T )|]. So ` 6 |∂Tj,t∗ | − 1 6 00 |V (T )| − 1. We may choose q and t such that there exists v ∈ V (M1) ∩ Xq such that there exists ∗ 00 ∗ a monochromatic path P in M1 from Xt to Xq internally disjoint from Xt ∪ X∂Tj,t∗ . Recall that ∗ (i ∗ ,|V (T )|+1,s+2) P is a monochromatic path in G[Y t ] by the minimality of it∗ and Claim 21.11. ∗ (i ∗ ,`+1,k) Let k + 1 be the color of M1. By Claim 21.1, V (P ) ⊆ Y t . Since q is not a witness for (i ∗ ,`+1) t (it∗ ,`+1,0) (it∗ ,`+1,0) Xq ∩ Ij ⊆ W3 , either v ∈ D − Xt∗ , or v ∈ D ∩ Xt∗ ∩ Xq0 ∩ Ij for some witness 0 ∗ (i0,`0) 0 0 q ∈ V (Tt∗ ) − {t } for Xq0 ∩ Ij ⊆ W3 for some i ∈ [0, it∗ − 1] and ` ∈ [0, |V (T )|]. The latter cannot hold by the minimality of it∗ . (i ∗ ,`+1,0) (i ∗ ,0,0) Hence v ∈ D t − Xt∗ . By the minimality of it∗ and Claim 21.11, v 6∈ D t . So there ∗ ∗ (i ∗ ,`∗+1,k) ∗ (it∗ ,` +1,0) (it∗ ,` ,0) t exists ` ∈ [0, |V (T )| − 1] such that v ∈ D − D . So v ∈ W0 . Hence there ∗ (i ∗ ) (i ∗ ,`∗) (it∗ ,` +1,k) t t exists a path in G[Y ∩ W4 ] with color k + 1 from v to W3 . So q is a witness for 0 (it∗ ,` ) 0 ∗ Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, ` + 1], a contradiction. This proves the claim.

Claim 21.34. Let i ∈ N0 and t ∈ V (T ) with σT (t) = i. Let j ∈ [|V| − 1]. Let (p1, p2, . . . , p|V (G)|) be the (i, −1, 0, j)-pseudosignature. Let α ∈ [|V (G)|] be the smallest index such that pα is not (i) 0 a zero sequence. Let M be the monochromatic Ej,t -pseudocomponent deﬁning pα. Let i ∈ N0 0 0 0 0 (i0) and t ∈ V (Tt) − {t} with σT (t ) = i . Let M be the monochromatic Ej,t0 -pseudocomponent in G[Y (i0,−1,0)] containing M. Then V (M 0) = V (M) ⊆ Y (i,−1,0). Proof. Since V (M) ⊆ Y (i,−1,0), it suﬃces to show that V (M 0) = V (M). Suppose to the contrary that V (M 0) 6= V (M). We choose i, i0 such that i0 − i is minimum among all counterexamples. Since V (M 0) 6= V (M), there exists the minimum i00 ∈ [i, i0] such that V (M 00) 6= V (M), where 00 (i00) (i00,−1,0) 00 M is the monochromatic Ej,t00 -pseudocomponent in G[Y ] containing M and t is the node 00 00 00 00 with σT (t ) = i . Note that i > i. Let p be the parent of t . So p ∈ V (Tt). (iz) For every node z with z ∈ V (Tt), let Mz be the monochromatic Ej,z -pseudocomponent in G[Y (iz,−1,0)] containing M.

53 ◦ (iz) Suppose there exist z ∈ V (T ) with Tp ⊆ Tz ⊆ Tt and an Sj -related monochromatic Ej,z - 0 (iz,−1,0) 0 pseudocomponent Mz in G[Y ] with AL(iz,−1,0) (V (Mz)) ∩ XV (Tz) − Xz 6= ∅ intersecting Xz 0 00 such that σ(Mz) < σ(Mz). By the minimality of i , V (Mz) = V (M). Since Mz contains M, 0 0 0 V (Mz) ∩ Xt 6= ∅. Since σ(Mz) < σ(Mz) and V (Mz) ∩ Xz 6= ∅, V (Mz) ∩ Xt 6= ∅. By Claim 21.31, 0 there exists α ∈ [α − 1] such that pα0 is not a zero sequence, a contradiction. ◦ (iz) So for every z ∈ V (T ) with Tp ⊆ Tz ⊆ Tt, there exists no Sj -related monochromatic Ej,z - 0 (iz,−1,0) 0 pseudocomponent Mz in G[Y ] with AL(iz,−1,0) (V (Mz)) ∩ XV (Tz) − Xz 6= ∅ intersecting Xz 0 such that σ(Mz) < σ(Mz). (ip) Hence V (Q) = V (Mp) = V (M), where Q is the monochromatic Ej,p -pseudocomponent in (ip,|V (T )|+1,s+2) 00 00 G[Y ] containing Mp. Since V (M ) 6= V (M) and t ∈ V (Tj,p), by the minimality of 00 00 00 (i ) (ip) i and Claim 21.11, there exists e ∈ E(M ) ∩ Ej,t00 − Ej,p such that e ∩ V (Q) 6= ∅= 6 e − V (Q). 00 (ip) (ip) (i ) (ip) 00 Since Ej,t00 = Ej,p . So V (Q) intersects an element in Ej,t00 − Ej,t00 . Then since t is a ◦ (ip) 0 child of p, there exist q ∈ ∂Tj,p and an Sj -related monochromatic Ej,t00 -pseudocomponent Q (ip,|V (T )|+1,s+2) 0 in G[Y ] with AL(ip,|V (T )|+1,s+2) (V (Q )) ∩ XV (Tp) − Xp 6= ∅ intersecting Xq such that 0 0 σ(Q ) < σ(Q). Since V (Q) = V (M), σ(Q ) < σ(Q) = σ(M). Since V (Q) ∩ Xt 6= ∅ and 0 0 00 (i00) (i00,−1,0) V (Q ) ∩ Xq 6= ∅, V (Q ) ∩ Xt 6= ∅. Let Q be the Ej,t00 -pseudocomponent in G[Y ] con- 0 00 0 0 taining Q . Then σ(Q ) 6 σ(Q ) < σ(M). By Claim 21.31, there exists α ∈ [α − 1] such that 00 pα0 is not a zero sequence and is deﬁned by a subgraph of Q , a contradiction. This proves the claim.

~~◦ (it) For any j ∈ [|V| − 1], t ∈ V (T ) and Sj -related monochromatic Ej,t -pseudocomponent M in (it,−1,0) G[Y ] intersecting Xt, we say that (t, M) is a nice pair if the following hold:~~

∗ (it∗ ) • If t ∈ V (T ) is the node of T such that t ∈ V (Tj,t∗ ) and some Ej,t∗ -pseudocomponent in (i ∗ ,|V (T )|+1,s+2) ∗ G[Y t ] intersects V (M) and Xq for some witness q ∈ ∂Tj,t∗ ∪ {t } for Xq ∩ Ij ⊆ (it∗ ,`) W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ is minimum, then either: – t = t∗, or – t 6= t∗, and for every c-monochromatic path P in G intersecting V (M) with V (P ) ⊆

XV (Tt∗ ) internally disjoint from XV (Tt) and for every vertex uP ∈ V (P ) ∩ XV (Tj,t∗ ), there (i ∗ ,|V (G)|) t 0 (it∗ ,|V (T )|+1,s+2) exists a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] such that: 0 0 ∗ V (M ) ∩ V (M) 6= ∅= 6 V (M ) ∩ XV (Tt), 0 ∗ M contains uP , and 0 ∗ if AL(it,−1,0) (V (M))∩XV (Tt)−Xt 6= ∅, then AL(it∗ ,|V (T )|+1,s+2) (V (M ))∩XV (Tt)−Xt 6= ∅. ◦ Claim 21.35. Let j ∈ [|V| − 1]. Let t ∈ V (T ) and let M be an Sj -related monochromatic (it) (it,−1,0) ∗ Ej,t -pseudocomponent in G[Y ] intersecting Xt. Let t ∈ V (T ) be the node of T such that (i ∗ ) t (it∗ ,|V (T )|+1,s+2) t ∈ V (Tj,t∗ ) and some Ej,t∗ -pseudocomponent in G[Y ] intersects V (M) and Xq for ∗ (it∗ ,`) some witness q ∈ ∂Tj,t∗ ∪ {t } for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to ∗ ∗ this, it is minimum. If t 6= t, AL(it,−1,0) (V (M)) ∩ XV (Tt) − Xt 6= ∅ and (t, M) is a nice pair, then (i ∗ ,|V (G)|) t 0 (it∗ ,|V (T )|+1,s+2) every monochromatic Ej,t∗ -pseudocomponent M in G[Y ] intersecting V (M) 0 intersects XV (Tt) and satisﬁes AL(it∗ ,|V (T )|+1,s+2) (V (M )) ∩ XV (Tt) − Xt 6= ∅.

(i ∗ ,|V (G)|) t (it∗ ,|V (T )|+1,s+2) Proof. Let S1 = {Q : Q is a monochromatic Ej,t∗ -pseudocomponent in G[Y ] such that V (Q) intersects V (M) and XV (Tt), and AL(it∗ ,|V (T )|+1,s+2) (V (Q)) ∩ XV (Tt) − Xt 6= ∅}.

54 (i ∗ ,|V (G)|) t (it∗ ,|V (T )|+1,s+2) Let S2 = {Q 6∈ S1 : Q is a monochromatic Ej,t∗ -pseudocomponent in G[Y ] intersecting V (M)}. By Claims 21.11 and 21.33, since (t, M) is a nice pair, S1 6= ∅. (it) Suppose that the claim does not hold. That is, S2 6= ∅. By the construction of Ej,t and (it∗ ,|V (G)|) Claims 21.11 and 21.33, there exist monochromatic Ej,t∗ -pseudocomponents M1 and M2 in (i ∗ ,|V (T )|+1,s+2) G[Y t ] intersecting V (M) such that M1 ∈ S1, M2 ∈ S2 and there exist q ∈ ∂Tj,t∗ with

V (M1)∩Xq 6= ∅= 6 V (M2)∩Xq and a path P in M[V (M)∩XV (Tq)] from V (M1)∩Xq to V (M2)∩Xq internally disjoint from Xq ∪ V (M1) ∪ M2. By the minimality of it∗ , P is a c-monochromatic path in G. So V (P ) ∩ AL(it∗ ,|V (T )|+1,s+2) (V (M1)) ∩ XV (Tq) − Xq 6= ∅= 6 V (P ) ∩ AL(it∗ ,|V (T )|+1,s+2) (V (M2)) ∩

XV (Tq) − Xq. Since either V (M2) ∩ XV (Tt) = ∅ or AL(it∗ ,|V (T )|+1,s+2) (V (M2)) ∩ XV (Tt) − Xt = ∅, q 6∈ V (Tt) and P is internally disjoint from XV (Tt). Let u be the vertex in V (P ) ∩ V (M2) ∩ Xq. Since (t, M) is nice, there exists M3 ∈ S1 such that M3 contains u. Since u ∈ V (M2) ∩ V (M3), M3 = M2. Hence M2 = M3 ∈ S1, a contradiction.

◦ Claim 21.36. Let j ∈ [|V| − 1]. Let t ∈ V (T ) and let M be an Sj -related monochromatic (it) (it,−1,0) ∗ Ej,t -pseudocomponent in G[Y ] intersecting Xt. Let t ∈ V (T ) be the node of T such that (i ∗ ) t (it∗ ,|V (T )|+1,s+2) t ∈ V (Tj,t∗ ) and some Ej,t∗ -pseudocomponent in G[Y ] intersects V (M) and Xq for ∗ (it∗ ,`) some witness q ∈ ∂Tj,t∗ ∪ {t } for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, ∗ ∗ it is minimum. If t 6= t , AL(it,−1,0) (V (M)) ∩ XV (Tt) − Xt 6= ∅ and (t, M) is a nice pair, then there (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) exists a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] intersecting V (M) ∗ ∗ ∗ such that σ(M ) = σ(M), V (M )∩Xt 6= ∅ and satisﬁes AL(it∗ ,|V (T )|+1,s+2) (V (M ))∩XV (Tt) −Xt 6= ∅. ∗ Proof. Let vM be the vertex of M such that σ(vM ) = σ(M). If vM ∈ V (Tt∗ ) − {t }, then (it∗ ,|V (G)|) ∗ by Claims 21.11 and 21.33, there exists a monochromatic Ej,t∗ -pseudocomponent M in (i ∗ ,|V (T )|+1,s+2) ∗ G[Y t ] containing vM . If vM 6∈ V (Tt∗ ) − {t }, then by Claim 21.31, there exists (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] containing vM . Since t 6= t∗, M ∗ ⊆ M. Hence σ(M) = σ(M ∗) and V (M ∗) ∩ V (M) 6= ∅. So by Claim 21.35, ∗ ∗ V (M ) ∩ XV (Tt) 6= ∅ and AL(it∗ ,|V (T )|+1,s+2) (V (M )) ∩ XV (Tt) − Xt 6= ∅. Since V (M) ∩ Xt 6= ∅, ∗ ∗ ∗ ∗ t ∈ V (TrM ). Since vM ∈ V (M ), V (M )∩XrM 6= ∅. Since V (M )∩XV (Tt) 6= ∅, V (M )∩Xt 6= ∅.

◦ (iz) For any j ∈ [|V| − 1], z ∈ V (T ) and Sj -related monochromatic Ej,z -pseudocomponent M in (iz,−1,0) G[Y ] intersecting Xz, we say that (z, M) is an outer-safe pair if the following hold:

∗ (iz∗ ) • If z ∈ V (T ) is the node of T such that z ∈ V (Tj,z∗ ) and some Ej,z∗ -pseudocomponent in (i ∗ ,|V (T )|+1,s+2) ∗ G[Y z ] intersects V (M) and Xq for some witness q ∈ ∂Tj,z∗ ∪ {z } for Xq ∩ Ij ⊆ (iz∗ ,`) W3 for some ` ∈ [−1, |V (T )|], and subject to this, iz∗ is minimum, then there exist 0 (i ,−1,0) no t ∈ ∂T ∗ − V (T ), u, v ∈ X 0 , a monochromatic path in G[Y z ∩ X ] but j,z z t V (Tt0 ) (i ∗ ,|V (T )|+1,s+2) not in G[Y z ] from u to v internally disjoint from Xt0 such that there exists (i ∗ ,|V (G)|) z ∗ (iz∗ ,|V (T )|+1,s+2) ∗ a monochromatic Ej,z∗ -pseudocomponent M in G[Y ] with σ(M ) = σ(M) containing u but not v.

(Note that we do not require AL(iz,−1,0) (V (M)) ∩ XV (Tz) − Xz 6= ∅.) ◦ Claim 21.37. Let j ∈ [|V| − 1]. Let t ∈ V (T ) and let M be an Sj -related monochromatic (it) (it,−1,0) ∗ Ej,t -pseudocomponent in G[Y ] intersecting Xt. Let t ∈ V (T ) be the node of T such that (i ∗ ) t (it∗ ,|V (T )|+1,s+2) t ∈ V (Tj,t∗ ) and some Ej,t∗ -pseudocomponent in G[Y ] intersects V (M) and Xq for ∗ (it∗ ,`) some witness q ∈ ∂Tj,t∗ ∪ {t } for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this,

55 ∗ 0 it∗ is minimum. Assume that t 6= t . Assume that (z, M ) is a nice pair and an outer-safe pair ◦ (iz) 0 (iz,−1,0) for every z ∈ V (T ) and every Sj -related monochromatic Ej,z -pseudocomponent M in G[Y ] 0 intersecting Xz with σ(M ) < σ(M). 0 (it,−1,0) Then there exist no t ∈ ∂Tj,t∗ ∩ V (Tt), u, v ∈ Xt0 and a monochromatic path in G[Y ∩ (i ∗ ,|V (T )|+1,s+2) 0 X ] but not in G[Y t ] from u to v internally disjoint from X 0 such that t is V (Tt0 ) t (it∗ ,`) a witness for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and there exists a monochromatic (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) ∗ Ej,t∗ -pseudocomponent M in G[Y ] with σ(M ) = σ(M) containing u but not v.

0 0 0 Proof. Suppose to the contrary that there exist t ∈ ∂Tj,t∗ ∩ V (Tt), u , v ∈ Xt0 , a monochromatic 0 (i ,−1,0) 0 0 path P in G[Y t ∩ X ] from u to v internally disjoint from X 0 , and a monochromatic V (Tt0 ) t (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) ∗ 0 Ej,t∗ -pseudocomponent M in G[Y ] with σ(M ) = σ(M) such that t is a (i ∗ ,`) t 0 (it∗ ,|V (T )|+1,s+2) ∗ witness for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], V (P ) 6⊆ Y , and M contains u0 but not v0. We further assume that σ(M) is minimum among all counterexamples. 0 (i ,−1,0) 0 t0 Let t0 be the node of T such that it0 6 it and V (P ) 6⊆ Y , and subject to these, it0 is 0 (i ∗ ,|V (T )|+1,s+2) 0 (i ∗ +1,−1,0) maximum. Since it0 > it > it∗ and V (P ) 6⊆ Y t , we know V (P ) 6⊆ Y t , so 0 0 0 ∗ ∗ ∗ it < it0 . By the maximality of it0 , t ∈ V (Tt0 ) ⊆ V (Tt ). Since t ∈ ∂Tj,t , t 6∈ V (Tj,t0 )−∂Tj,t0 . Let 0 0 (it ,−1,0) 0 P0 be the maximal subpath of P [V (P ) ∩ Y 0 ] containing u . Let R0 be the monochromatic (it0 ) (i ,−1,0) ∗ 0 E -pseudocomponent in G[Y t0 ] containing P . So M ⊆ R . Let R be the monochromatic j,t0 0 0 0 (it0 ) (i ,−1,0) 0 E -pseudocomponent in G[Y t0 ] containing v . j,t0 0 (i ,−1,0) 0 0 0 t 0 Since V (P ) ⊆ Y and it 6 it , we know it0 < it and t0 6= t . Since t0 6= t and t 6∈ V (Tj,t0 )− (i ) ∂T , by the maximality of i , there exists an S◦-related monochromatic E t0 -pseudocomponent j,t0 t0 j j,t0 (it0 ,−1,0) 0 0 M0 in G[Y ] intersecting Xt0 such that there exist ` ∈ [−1, 0] and k ∈ [0, s + 2] such that 0 0 (it ,−1,0) σ(M ) = σ(Z ) < min{σ(R ), σ(R )} and A (i ,`0,k0) (V (Z ))∩V (P )∩N (V (P ))∩Z −Y 0 6= 0 0 0 0 L t0 0 G 0 t0 0 0 (it0 ) (i ,` ,k ) ∅, where Z is the monochromatic E -pseudocomponent in G[Y t0 ] containing M . So 0 j,t0 0 V (Z0) ∩ Xt0 6= ∅. ∗ ∗ ∗ ∗ Since M ⊆ R0, σ(R0) 6 σ(M ). Hence σ(M0) < σ(M ) = σ(M). Let z be the node of T such (i ∗ ) z (iz∗ ,|V (T )|+1,s+2) that t0 ∈ V (Tj,z∗ ) and some Ej,z∗ -pseudocomponent in G[Y ] intersects V (M0) and ∗ (iz∗ ,`) Xq for some witness q ∈ ∂Tj,z∗ ∪{z } for Xq ∩Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to ∗ ∗ this, iz∗ is minimum. Since t is a candidate for z , iz∗ 6 it∗ . Since σ(M0) < σ(M), (t0,M0) is a nice (iz∗ ,|V (G)|) pair by assumption. By Claim 21.36, there exists a monochromatic Ej,z∗ -pseudocomponent 0 (iz∗ ,|V (T )|+1,s+2) 0 0 M0 in G[Y ] intersecting V (M0) such that σ(M0) = σ(M0), V (M0) ∩ Xt0 6= ∅ and 0 A (i ∗ ,|V (T )|+1,s+2) (V (M )) ∩ X − X 6= ∅. L z 0 V (Tt0 ) t0 Since V (M )∩X 0 6= ∅ and A (i ,−1,0) (V (M ))∩X −X 0 6= ∅, by Claims 21.11 and 21.33 and 0 t L t0 0 V (Tt0 ) t 0 (iz∗ ,|V (G)|) the existence of M0, there exists a monochromatic Ej,z∗ -pseudocomponent M1 in (i ∗ ,|V (T )|+1,s+2) z 0 G[Y ] intersecting M0 such that V (M1) ∩ Xt 6= ∅ and AL(iz∗ ,|V (T )|+1,s+2) (V (M1)) ∩ 0 0 X − X 0 6= ∅. Suppose M 6⊆ M . Since M ∪ M ⊆ M , there exists a path Q in M from V (Tt0 ) t 1 0 1 0 0 0 (i ) V (M 0 ) to V (M ) internally disjoint from V (M 0 ). By the construction of E t0 and Claims 21.11 0 1 0 j,t0 (i ,−1,0) 00 00 00 t0 ∗ 00 and 21.33, Q is in G[Y ∩ XV (Tz∗ )]. Hence there exist t ∈ ∂Tj,z , u , v ∈ Xt , a monochro- 0 (i ,−1,0) 00 00 matic path Q in G[Y t0 ∩ X ] from u to v internally disjoint from X 00 but not in V (Tt00 ) t (iz∗ ,|V (T )|+1,s+2) 0 00 00 G[Y ] such that M0 contains u but not v . Since σ(M0) < σ(M), (t0,M0) is an 00 outer-safe pair, so t ∈ V (Tt0 ). Hence M0 is a counterexample of this claim, contradicting that σ(M) is minimum among all counterexamples. 0 0 (iz∗ ,|V (G)|) 0 So M0 contains M1 for every monochromatic Ej,z∗ -pseudocomponent M1 in

56 (i ∗ ,|V (T )|+1,s+2) 0 0 z 0 G[Y ] intersecting M0 such that V (M1) ∩ Xt 6= ∅ and AL(iz∗ ,|V (T )|+1,s+2) (V (M1)) ∩ (it∗ ,|V (G)|) ∗ X − X 0 6= ∅. Hence there exists a monochromatic E ∗ -pseudocomponent M in V (Tt0 ) t j,t 0 (i ∗ ,|V (T )|+1,s+2) 0 ∗ ∗ t 0 G[Y ] containing M0 such that V (M0 ) ∩ Xt 6= ∅ and AL(it∗ ,|V (T )|+1,s+2) (V (M0 )) ∩ X − X 0 6= ∅. V (Tt0 ) t (i ∗ ,|V (G)|) ∗ t (it∗ ,|V (T )|+1,s+2) Since M0 is an Ej,t∗ -pseudocomponent in G[Y ], there exists a nonnegative ∗ (it∗ ,β0) integer β0 such that M0 is the (β0+1)-smallest among all monochromatic Ej,t∗ -pseudocomponents (i ∗ ,β ) (it∗ ,|V (T )|+1,s+2) t 0 in G[Y ]. Let Ru and Rv be the monochromatic Ej,t∗ -pseudocomponents in (i ∗ ,|V (T )|+1,s+2) 0 0 0 G[Y t ] containing u and v , respectively. By the existence of P , Ru ∪ Rv ⊆ M, ∗ so σ(Ru) = σ(M ) = σ(M) 6 σ(Rv). ∗ 0 0 0 (it∗ ,|V (G)|) 0 0 Note that σ(M0 ) 6 σ(M0) = σ(M0) < σ(M) 6 σ(Ru). Since {u , v } 6∈ Ej,t∗ , {u , v } 6∈ (it∗ ,β0+1) ∗ (it∗ ,β0+1) Ej,t∗ . Since σ(M0 ) < σ(Ru), by the construction of Ej,t∗ and the minimality of it∗ we ◦ (it∗ ,β0) know (V (Ru)∪V (Rv))∩Xt∗ 6= ∅ and for every Sj -related monochromatic Ej,t∗ -pseudocomponent 0 (it∗ ,|V (T )|+1,s+2) 0 ∗ 0 M in G[Y ] with σ(M ) σ(M ), A (i ∗ ,|V (T )|+1,s+2) (V (M ))∩X −X ∗ ⊆ X − 6 0 L t V (Tt∗ ) t V (Tt0 ) (Xt0 ∪ Zt∗ ). 0 ◦ Since (V (Ru)∪V (Rv))∩Xt∗ 6= ∅, we know V (M )∩Xt∗ 6= ∅ for every Sj -related monochromatic (i ∗ ,β ) t 0 0 (it∗ ,|V (T )|+1,s+2) 0 ∗ 0 Ej,t∗ -pseudocomponent M in G[Y ] with σ(M ) 6 σ(M0 ). Since t ∈ ∂Tj,t∗ , and 0 ◦ A (i ∗ ,|V (T )|+1,s+2) (V (M ))∩X −X ∗ ⊆ X −(X 0 ∪Z ∗ ) for every S -related monochromatic L t V (Tt∗ ) t V (Tt0 ) t t j (i ∗ ,β ) t 0 0 (it∗ ,|V (T )|+1,s+2) 0 0 Ej,t∗ -pseudocomponent M in G[Y ] with σ(M ) 6 σ(M0), we know V (M0) = ∗ ∗ ∗ V (M ). Since σ(M ) = σ(M ) = σ(Z ) and Z ∩X ⊆ Z ∗ , we know V (Z ) = V (M ) = V (M ) 0 0 0 0 t0 V (Tt0 ) t 0 0 0 and

0 (it0 ,−1,0) ∅= 6 A (i ,`0,k0) (V (Z )) ∩ V (P ) ∩ N (V (P )) ∩ Z − Y L t0 0 G 0 t0 ∗ ⊆ A (i ,`0,k0) (V (M )) ∩ Z ∩ X − X 0 L t0 0 t0 V (Tt0 ) t ∗ ⊆ A (i ∗ ,|V (T )|+1,s+2) (V (M )) ∩ Z ∗ ∩ X − X 0 L t 0 t V (Tt0 ) t ∗ ⊆ A (i ∗ ,|V (T )|+1,s+2) (V (M )) ∩ (X − X ∗ ) ∩ Z ∗ ∩ X − X 0 L t 0 V (Tt∗ ) t t V (Tt0 ) t ⊆ (X − (X 0 ∪ Z ∗ )) ∩ Z ∗ ∩ X − X 0 = ∅, V (Tt0 ) t t t V (Tt0 ) t a contradiction.

(it) Claim 21.38. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M be an Sj-related monochromatic Ej,t - (it,−1,0) ∗ pseudocomponent in G[Y ] intersecting Xt. Let t be the node of T with t ∈ V (Tj,t∗ ) such (i ∗ ) t (it∗ ,|V (T )|+1,s+2) that some Ej,t∗ -pseudocomponent in G[Y ] intersecting V (M) intersects Xt0 for some 0 ∗ (it∗ ,`) witness t ∈ ∂Tj,t∗ ∪ {t } for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ is minimum. Assume that (z, Mz) is a nice pair and an outer-safe pair for every z ∈ V (T ) and ◦ (iz) (iz,−1,0) Sj -related monochromatic Ej,z -pseudocomponent Mz in G[Y ] intersecting Xz with σ(Mz) < σ(M). If (t, M) is an outer-safe pair, then the following hold.

(i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) • There uniquely exists a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] intersecting V (M). ∗ ∗ • V (M ) ∩ Xt 6= ∅ and σ(M ) = σ(M). ∗ ∗ • If t 6= t and AL(it,−1,0) (V (M))∩XV (Tt)−Xt 6= ∅, then AL(it∗ ,|V (T )|+1,s+2) (V (M ))∩XV (Tt)−Xt 6= ∅.

57 (it∗ ,|V (G)|) Proof. By Claims 21.11, 21.31 and 21.33, there exists a monochromatic Ej,t∗ -pseudocomponent ∗ (i ∗ ,|V (T )|+1,s+2) ∗ ∗ M in G[Y t ] with σ(M ) = σ(M) such that V (M ) intersects Xq for some witness ∗ (it∗ ,`) q ∈ ∂Tj,t∗ ∪ {t } for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|]. (i ∗ ,|V (G)|) t 0 (it∗ ,|V (T )|+1,s+2) Suppose there exists a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] intersecting V (M) such that M 0 6= M ∗. Then t 6= t∗, so M 0 ∪ M ∗ ⊆ M, and hence there exists a monochromatic path P in G[Y (it,−1,0)] from V (M ∗) to V (M 0) internally disjoint from V (M ∗). By ∗ 0 (it,−1,0) tracing P from V (M ), there exist t ∈ ∂Tj,t∗ , u, v ∈ Xt0 , a monochromatic path Q in G[Y ∩ ∗ ∗ X ] from u to v such that M contains u but not v, and V (Q) ∩ A (i ∗ ,|V (T )|+1,s+2) (V (M )) ∩ V (Tt0 ) L t (i ∗ ,|V (T )|+1,s+2) ∗ X − X 0 6= ∅. Note that V (Q) 6⊆ Y t , since M contains u but not v. By V (Tt0 ) t ∗ ∗ 0 (it∗ ,`) Claim 21.33, since V (M ) ∩ Xt0 6= ∅ 6= V (M ) ∩ V (M), t is a witness for Xt0 ∩ Ij ⊆ W3 for 0 some ` ∈ [0, |V (T )|]. Since (t, M) is an outer-safe pair, t ∈ V (Tt). However, by Claim 21.37, 0 t 6∈ V (Tt), a contradiction. (i ∗ ,|V (G)|) ∗ t (it∗ ,|V (T )|+1,s+2) Hence M is the unique monochromatic Ej,t∗ -pseudocomponent in G[Y ] ∗ intersecting V (M). Recall that σ(M ) = σ(M). Since V (M) ∩ Xt 6= ∅ and t ∈ V (Tj,t∗ ), by ∗ Claims 21.11 and 21.33, V (M ) ∩ Xt 6= ∅. ∗ ∗ If t 6= t and AL(it,−1,0) (V (M))∩XV (Tt)−Xt 6= ∅, then since V (M)∩Xt ⊆ V (M ) by Claim 21.11, (i ∗ ,|V (G)|) ∗ t (it∗ ,|V (T )|+1,s+2) and M is the unique monochromatic Ej,t∗ -pseudocomponent in G[Y ] inter- ∗ secting V (M), AL(it∗ ,|V (T )|+1,s+2) (V (M )) ∩ XV (Tt) − Xt 6= ∅. ◦ Claim 21.39. Let j ∈ [|V| − 1]. Let t ∈ V (T ) and let M be an Sj -related monochromatic (it) (it,−1,0) ∗ Ej,t -pseudocomponent in G[Y ] intersecting Xt. Let t ∈ V (T ) be the node of T such that (i ∗ ) t (it∗ ,|V (T )|+1,s+2) t ∈ V (Tj,t∗ ) and some Ej,t∗ -pseudocomponent in G[Y ] intersects V (M) and Xq for ∗ (it∗ ,`) some witness q ∈ ∂Tj,t∗ ∪ {t } for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, ∗ 0 it∗ is minimum. Assume that t 6= t . Assume that (z, M ) is a nice pair and an outer-safe pair ◦ (iz) 0 (iz,−1,0) for every z ∈ V (T ) and every Sj -related monochromatic Ej,z -pseudocomponent M in G[Y ] 0 intersecting Xz with σ(M ) < σ(M). If (t, M) is an outer-safe pair and there exists a c-monochromatic path P in G from V (M) to (it) (it,−1,0) a monochromatic Ej,t -pseudocomponent M0 in G[Y ] other than M internally disjoint from (it) (it,−1,0) every monochromatic Ej,t -pseudocomponent in G[Y ] such that V (P ) ∩ AL(it,−1,0) (V (M)) ∩ 0 ∗ 0 XV (Tt) − Xt 6= ∅, then there exist t ∈ ∂Tj,t ∩ V (Tt), u, v ∈ Xt and a c-monochromatic path in (i ,−1,0) 0 G[X ] but not in G[Y t ] from u to v internally disjoint from X 0 such that t is a witness V (Tt0 ) t (it∗ ,`) (it∗ ,|V (G)|) for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and there exists a monochromatic Ej,t∗ - pseudocomponent M ∗ in G[Y (it∗ ,|V (T )|+1,s+2)] with σ(M ∗) = σ(M) containing u but not v.

(it) (it,−1,0) Proof. Since P is internally disjoint from every monochromatic Ej,t -pseudocomponent in G[Y ] and V (P ) ∩ AL(it,−1,0) (V (M)) ∩ XV (Tt) − Xt 6= ∅, V (P ) ⊆ XV (Tt). 0 0 Since M 6= M0, there exist t ∈ ∂Tj,t∗ ∩ V (Tt), u, v ∈ Xt0 , a subpath P of P ∪ M ∪ M0 from (i ,−1,0) u to v contained in G[X ] internally disjoint from X 0 but not contained in G[Y t ], and a V (Tt0 ) t (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) monochromatic Ej,t∗ -pseudocomponent M in G[Y ] intersecting V (M)∩V (P ) 0 (it∗ ,`) containing u but not containing v. Note that t is a witness for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|] by Claims 21.11 and 21.33. To prove this claim, it suﬃces to prove that σ(M ∗) = σ(M). (it∗ ,|V (G)|) By Claims 21.11, 21.31 and 21.33, there exists a monochromatic Ej,t∗ -pseudocomponent Q∗ in G[Y (it∗ ,|V (T )|+1,s+2)] with σ(Q∗) = σ(M). Suppose σ(M ∗) 6= σ(M). So M ∗ 6= Q∗. Since ∗ ∗ (it) 00 M ∪ Q ⊆ M, by the construction of Ej,t and Claims 21.11 and 21.33, there exist t ∈ ∂Tj,t∗ ,

58 0 (it,−1,0) ∗ (it∗ ,|V (G)|) a monochromatic path Q in G[Y ∩ X ] from V (Q ) to a monochromatic E ∗ - V (Tt00 ) j,t pseudocomponent in G[Y (it∗ ,|V (T )|+1,s+2)] intersecting V (M) other than Q∗ internally disjoint from 0 ∗ 00 X 00 such that V (Q )∩A (i ∗ ,|V (T )|+1,s+2) (V (Q ))∩X −X 00 6= ∅, and t is a witness for X 00 ∩I ⊆ t L t V (Tt00 ) t t j (it∗ ,`) 00 W3 for some ` ∈ [0, |V (T )|]. By Claim 21.37, t 6∈ V (Tt). So (t, M) is not an outer-safe pair, a contradiction. Hence σ(M ∗) = σ(M). This proves the claim.

0 For any t ∈ V (T ), t ∈ V (Tt), j ∈ [|V|−1], α0 ∈ [0, w0], α1, α2 ∈ [w0] with α0 < α1 6 α2, k ∈ N, 0 0 0 k ∈ N0, ` ∈ [−1, |V (T )|+1], ξ ∈ [0, s+2] and ξ ∈ [0, |V (G)|], we deﬁne an (α0, α1, α2, k, k )-blocker 0 0 for (j, t, `, ξ, ξ , t ) to be a parade (t−k0 , t−k0+1, . . . , t−1, t1, t2, . . . , tk) such that the following hold:

0 ◦ (it,ξ ) • For every β ∈ [w0], let Mβ be the Sj -related monochromatic Ej,t -pseudocomponent in 0 (it,`,ξ) ◦ (it,ξ ) G[Y ] such that σ(Mβ) is the β-th smallest among all Sj -related monochromatic Ej,t - (it,`,ξ) pseudocomponents M in G[Y ] intersecting X 0 with A (i ,`,ξ) (V (M)) ∩ X − X 0 6= ∅ t L t V (Tt0 ) t (if there are less than β such M, then let Mβ = ∅). 0 0 • If k 6= 0, then t−k0 ∈ V (Tt0 ) − {t }. 0 • t1 ∈ V (Tt0 ) − {t }. 0 • For every distinct ` , ` ∈ [−k , k] − {0}, X ∩ I − X 0 and X ∩ I − X 0 are disjoint 1 2 t`1 j t t`2 j t non-empty sets. Sα0 ◦ • A (i ,`,ξ) (V (M )) ∩ X − X ⊆ (X − X ) ∩ I . β=1 L t β V (Tt) t V (Ttk ) tk j ∗ • There exists α ∈ {α , α } such that A (i ,`,ξ) (V (M ∗ )) ∩ (X − X ) ∩ X − X = ∅, 1 2 L t α V (Tt) t V (Tt1 ) t1 0 ◦ and if k 6= 0, then A (it,`,ξ) (V (Mα∗ )) ∩ (XV (T ) − Xt) ∩ XV (T ) − (Xt 0 ∪ I ) = ∅. L t t−k0 −k j

0 0 Claim 21.40. Let j ∈ [|V| − 1]. Let t ∈ V (T ) and let t ∈ ∂Tj,t such that t is a witness for (it,`) 0 0 0 0 ◦ Xt0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|]. Let M1,M2,...,Mk0 (for some k ∈ N) be the Sj -related (i 0 ) t (it0 ,−1,0) 0 0 monochromatic Ej,t0 -pseudocomponents in G[Y ] such that for every α ∈ [k ], V (Mα)∩Xt0 6= 0 0 0 0 ∅ and A (i ,−1,0) (V (M )) ∩ X − X 0 6= ∅. Assume that σ(M ) < σ(M ) < ··· < σ(M 0 ). Let L t0 α V (Tt0 ) t 1 2 k 0 0 0 0 0 0 V (Mγ) = ∅ for every γ > k + 1. Let α1 ∈ [w0] and α2 ∈ [α1 + 1, k ]. (0) (0) (0) (0) ◦ Let M1 ,M2 ,...,Mk(0) (for some k ∈ N) be the Sj -related monochromatic (it) (it,|V (T )|+1,s+2) (0) Ej,t -pseudocomponents in G[Y ] such that for every α ∈ [k ], we have (0) (0) (0) (0) AL(it,|V (T )|+1,s+2) (V (Mα )) ∩ XV (Tt) − Xt 6= ∅. Assume that σ(M1 ) < σ(M2 ) < ··· < σ(Mk(0) ). (0) 0 Let β0 be the minimum such that M ⊆ M 0 . Let β1 ∈ [|V (G)|] be the integer such that the β0 α1 (i ,β −1) (0) monochromatic E t 1 -pseudocomponent in G[Y (it,|V (T )|+1,s+2)] containing M has the β -th j,t β0 1 (it,β1−1) (it,|V (T )|+1,s+2) smallest σ-value among all monochromatic Ej,t -pseudocomponents in G[Y ]. Let (it,β2−1) β2 be the largest element in [0, β1 − 1] such that the monochromatic Ej,t -pseudocomponent (it,|V (T )|+1,s+2) (it,β2−1) M in G[Y ] with the β2-th smallest σ-value among all monochromatic Ej,t - (it,|V (T )|+1,s+2) ◦ pseudocomponents in G[Y ] satisﬁes that M is Sj -related, V (M) ∩ Xt0 6= ∅ and ◦ A (i ,|V (T )|+1,s+2) (V (M)) ∩ X − X 0 6= ∅. Let M ,M ,...,M (for some k ∈ ) be the S -related L t V (Tt0 ) t 1 2 k N j (it,β2−1) (it,|V (T )|+1,s+2) monochromatic Ej,t -pseudocomponents in G[Y ] such that for every α ∈ [k], V (M ) ∩ X 0 6= ∅ and A (i ,|V (T )|+1,s+2) (V (M )) ∩ X − X 0 6= ∅. Assume that σ(M ) < σ(M ) < α t L t α V (Tt0 ) t 1 2 ··· < σ(Mk). Let V (Mγ) = ∅ for every γ > k + 1. 0 For γ ∈ {1, 2}, let Q = {M : M ⊆ M 0 }. For every Q ∈ Q ∪ Q , let α be the index such γ ` ` αγ 1 2 Q that Q = MαQ .

59 0 Assume V (M 0 ) ∩ Xt 6= ∅, and for every Q1 ∈ Q1 and Q2 ∈ Q2, there exists an (β2, αQ1 , αQ2 , α1 0 ψ2(αQ1 , αQ2 ), ψ3(αQ1 , αQ2 ))-blocker for (j, t, |V (T )| + 1, s + 2, β2 − 1, t ). If either

~~0 (i) α1 6 αQ for every Q ∈ Q1 ∪ Q2, or 0 0 0 (ii) for every β ∈ [α1 − 1], (t ,Mβ) is a nice pair,~~

0 0 0 0 0 0 0 0 0 then there exists an (α1 − 1, α1, α2, ψ2(α1, α2), ψ3(α1, α2) − ψ3(0, 0))-blocker for (j, t , −1, 0, 0, t ). Proof. We ﬁrst prove that condition (ii) implies condition (i). 0 0 Suppose that condition (ii) holds but condition (i) does not hold. So α1 > 2. Let β ∈ [α1 − 1]. ∗ 0 (it∗ ) Let t ∈ V (T ) be the node of T such that t ∈ V (Tj,t∗ ) and some Ej,t∗ -pseudocomponent in (it∗ ,|V (T )|+1,s+2) 0 ∗ G[Y ] intersects V (Mβ) and Xq for some witness q ∈ ∂Tj,t∗ ∪ {t } for Xq ∩ Ij ⊆ (it∗ ,`) W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ is minimum. Note that t is a candidate ∗ 0 ∗ 0 0 for t , so t 6= t . Since V (M 0 ) ∩ Xt 6= ∅, V (M ) ∩ Xt 6= ∅. By Claim 21.31, there exists α1 β (0) (0) 0 (0) 0 αβ ∈ [β0 − 1] such that M (0) ⊆ Mβ and σ(M (0) ) = σ(Mβ). By Claims 21.11 and 21.33, there αβ αβ (i ∗ ,|V (G)|) (0) t (it∗ ,|V (T )|+1,s+2) exists a monochromatic Ej,t∗ -pseudocomponent Rβ in G[Y ] intersecting M (0) αβ with σ(R ) = σ(M (0) ). Since R intersects V (M 0 ), by Claim 21.35, R intersects X and β (0) β β β V (Tt0 ) αβ satisﬁes A (i ∗ ,|V (T )|+1,s+2) (V (R )) ∩ X − X 0 6= ∅. Since R intersects X and σ(R ) = L t β V (Tt0 ) t β V (Tt0 ) β (0) (0) (0) σ(M (0) ) and V (M (0) ) ∩ Xt 6= ∅, we have Rβ intersects Xt0 . Since αβ < β0, Rβ is contained αβ αβ (it,β1−2) 0 (it,|V (T )|+1,s+2) 0 (0) in a monochromatic Ej,t -pseudocomponent Rβ in G[Y ]. So Rβ ⊇ M (0) . Since αβ

A (i ∗ ,|V (T )|+1,s+2) (V (R )) ∩ X − X 0 6= ∅ and A (i ∗ ,|V (T )|+1,s+2) (V (R )) ∩ (X − X 0 ) ∩ Z ∗ = ∅ L t β V (Tt0 ) t L t β V (Tt0 ) t t 0 0 0 and t ∈ V (T ∗ ), A (i ,|V (T )|+1,s+2) (V (R ))∩X −X 0 6= ∅. Since V (R )∩X 0 ⊇ V (R )∩X 0 6= ∅ j,t L t β V (Tt0 ) t β t β t 0 and A (i ,|V (T )|+1,s+2) (V (R )) ∩ X − X 0 6= ∅, by the maximality of β , there exists α ∈ [β − 1] L t β V (Tt0 ) t 2 β 2 0 (0) such that Rβ = Mαβ . Since αβ < β0, αβ < αQ for every Q ∈ Q1 ∪ Q2. Hence there exists a 0 mapping ι that maps each element β ∈ [α1 − 1] to αβ. Note that ι is an injection. So there are at 0 0 least α1 − 1 diﬀerent numbers strictly smaller than minQ∈Q1∪Q2 αQ. Hence α1 6 minQ∈Q1∪Q2 αQ and condition (i) holds, a contradiction. Therefore, to prove this claim, it suﬃces to assume the case that condition (i) holds.

For every Q1 ∈ Q1 and Q2 ∈ Q2, let κQ1,Q2 = ψ2(αQ1 , αQ2 ) and ξQ1,Q2 = ψ3(αQ1 , αQ2 ), and Q1,Q2 Q1,Q2 Q1,Q2 Q1,Q2 Q1,Q2 let t = (t , . . . , t−1 , t1 , . . . , tκ ) be a (β2, αQ1 , αQ2 , κQ1,Q2 , ξQ1,Q2 )-blocker for −ξQ1,Q2 Q1,Q2 0 P (j, t, |V (T )| + 1, s + 2, β2 − 1, t ) such that x ix is maximum, where the sum is over all entries of tQ1,Q2 . We also denote κ by κ , and denote ξ by ξ . For every Q ∈ Q Q1,Q2 αQ1 ,αQ2 Q1,Q2 αQ1 ,αQ2 1 1 ∗ ∗ and Q2 ∈ Q2, let α ∈ {αQ1 , αQ2 } such that AL(it,|V (T )|+1,s+2) (V (Mα )) ∩ (XV (Tt) − Xt) ∩ Q1,Q2 Q1,Q2 ◦ X −(Xt ∪I ) = ∅, and A (it,|V (T )|+1,s+2) (V (Mα∗ ))∩(X −Xt)∩X −Xt = V (Tt−ξ ) −ξQ ,Q j L Q ,Q V (Tt) V (Tt1 ) 1 Q1,Q2 1 2 1 2 ∗ ∅, and subject to this, αQ1,Q2 = αQ2 if possible. 00 00 00 00 ◦ Let M1 ,M2 ,...,Mk (for some k ∈ N) be the Sj -related monochromatic (it,|V (G)|) (it,|V (T )|+1,s+2) 00 00 Ej,t -pseudocomponents in G[Y ] such that for every α ∈ [k ], V (Mα) ∩ Xt0 6= ∅ 00 00 00 0 and A (i ,−1,0) (V (M )) ∩ X − X 0 6= ∅. Assume that σ(M ) < σ(M ) < ··· < σ(M 00 ). Let L t0 α V (Tt0 ) t 1 2 k 00 00 00 V (Mγ ) = ∅ for every γ > k + 1. Let b be the sequence

~~00 ◦ 00 ◦ (|A (i ,|V (T )|+1,s+2) (V (M ))∩X −(X 0 ∪I )|,..., |A (i ,|V (T )|+1,s+2) (V (M ))∩X −(X 0 ∪I )|). L t 1 V (Tt0 ) t j L t w0 V (Tt0 ) t j~~

60 0 Let m1 := minQ∈Q1 αQ. Let m2 := minQ∈Q2 αQ. Since α1 6 αQ for every Q ∈ Q1 ∪ Q2, 0 0 0 α1 6 min{m1, m2}. So ψ2(α1, α2) 6 ψ2(m1, m2). ∗ We ﬁrst assume that for every Q2 ∈ Q2, α = αQ . Mm1 ,Q2 2 M ,M Note that for all Q ∈ Q , κ + ξ κ + ξ , and by the choices of t m1 m2 2 2 m1,m2 m1,m2 6 m1,αQ2 m1,αQ2 M ,Q Mm1 ,Mm2 Mm1 ,Mm2 Mm1 ,Mm2 Mm1 ,Mm2 Mm1 ,Q2 Mm1 ,Q2 and t m1 2 , (t , . . . , t , t , . . . , t ) is a suﬃx of (t , . . . , t , ξm ,m −1 1 κm1,m2 −ξm ,α −1 1 2 1 Q2 Mm1 ,Q2 Mm1 ,Q2 ∗ t , . . . , tκm ,α ). Since for every Q2 ∈ Q2, α = αQ2 , we know 1 1 Q2 Mm1 ,Q2 [ 0 A (it,|V (T )|+1,s+2) (V (Q2)) ∩ (XV (T 0 ) − Xt ) ∩ XV (T ) − X Mm1 ,Mm2 = ∅ and L t Mm1 ,Mm2 t t1 1 Q2∈Q2 [ ◦ A (i ,|V (T )|+1,s+2) (V (Q )) ∩ (X − X 0 ) ∩ X − (X Mm ,Mm ∪ I ) = ∅. L t 2 V (Tt0 ) t V (T Mm ,Mm ) 1 2 j t 1 2 t −ξ −ξm1,m2 Q2∈Q2 m1,m2

(i 0 ,−1,0) (i ,|V (T )|+1,s+2) ◦ Since for each vertex in (Y t − Y t ) ∩ I ∩ X , it is in X ∩ I − (X 0 ∪ I ), j V (Tt0 ) V (Tt0 ) j t j and when it gets colored, it decreases the lexicographic order of b00. So by Claim 21.25, we loss at w0−1 w0 most η5 · (f(η5)) 6 (f(η5)) 6 ψ3(0, 0) nodes in the blocker. That is,

0 0 A (i 0 ,−1,0) (V (M 0 )) ∩ (XV (T 0 ) − Xt ) ∩ (XV (T ) − X Mm1 ,Mm2 ) = ∅ and L t α2 t Mm1 ,Mm2 t t1 1 0 ◦ 0 A (i 0 ,−1,0) (V (M 0 )) ∩ (XV (T 0 ) − Xt ) ∩ XV (T ) − (X Mm1 ,Mm2 ∪ I ) = ∅. L t α2 t Mm1 ,Mm2 t j t −ξm ,m +ψ3(0,0) −ξm1,m2 +ψ3(0,0) 1 2

Mm1 ,Mm2 Mm1 ,Mm2 Mm1 ,Mm2 Mm1 ,Mm2 0 So (t , . . . , t−1 , t1 , . . . , tκm ,m ) contains a subsequence that is an (α1 − −ξm1,m2 +ψ3(0,0) 1 1 0 0 0 0 0 0 0 0 1, α1, α2, ψ2(α1, α2), ψ3(α1, α2) − ψ3(0, 0))-blocker for (j, t , −1, 0, 0, t ). ∗ ∗ Hence we may assume that there exists Q ∈ Q2 such that m1 = α ∗ 6= αQ∗ . Since for every 2 Mm1 ,Q2 2 ∗ Q1 ∈ Q1 − {Mm }, ψ2(αQ , αQ∗ ) > ψ2(m1, αQ∗ ) + ψ3(m1, αQ∗ ), we know α ∗ = αQ , for oth- 1 1 2 2 2 Q1,Q2 1 ∗ ∗ ∗ ∗ ∗ ∗ Mm1 ,Q2 Mm1 ,Q2 Mm1 ,Q2 Mm1 ,Q2 ∗ erwise α ∗ = αQ by the choice of α ∗ . So (t , . . . , t , t , . . . , tκm ,α ∗ ) Mm1 ,Q2 2 Mm1 ,Q2 −ξm1,α ∗ −1 1 1 Q Q2 2 ∗ ∗ ∗ ∗ Q1,Q2 Q1,Q2 Q1,Q2 Q1,Q2 is a suﬃx of (t , . . . , t−1 , t1 , . . . , tκα ,α ∗ ) for every Q1 ∈ Q1. Note that for each −ξαQ ,α ∗ Q1 Q 1 Q2 2 (i 0 ,−1,0) (i ,|V (T )|+1,s+2) ◦ vertex in (Y t − Y t ) ∩ I ∩ X , it is in X − (X 0 ∪ I ), and when it j V (Tt0 ) V (Tt0 ) t j gets colored, it decreases the lexicographic order of b00. So by Claim 21.25, we loss at most M ,Q∗ M ,Q∗ w0−1 w0 m1 2 m1 2 η5 · (f(η5)) (f(η5)) ψ3(0, 0) nodes in the blocker. Hence (t , . . . , tκm ,α ∗ ) 6 6 −ξm1,α ∗ +ψ3(0,0) 1 Q Q2 2 0 0 0 0 0 0 0 contains a subsequence that is an (α1 − 1, α1, α2, ψ2(α1, α2), ψ3(α1, α2) − ψ3(0, 0))-blocker for (j, t0, −1, 0, 0, t0). This proves the claim.

Claim 21.41. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M1,M2,...,Mk (for some k ∈ N) be ◦ (it) (it,−1,0) the Sj -related monochromatic Ej,t -pseudocomponents in G[Y ] such that for every α ∈ [k],

V (Mα) ∩ Xt 6= ∅ and AL(it,−1,0) (V (Mα)) ∩ XV (Tt) − Xt 6= ∅. Assume that σ(M1) < σ(M2) < ··· < σ(Mk). Let V (Mγ) = ∅ for every γ > k + 1. Let α1, α2 ∈ [2, k] with α1 < α2. Assume that the following hold.

~~0 • for every α ∈ [2, α1] and α ∈ [α + 1, k], if:~~

~~0 – either α 6 α1 − 1 or (α, α ) = (α1, α2), – Mα and Mα0 have the same color, and ◦ – the s-segment in Sj containing V (Mα) whose level equals the color of Mα contains V (Mα0 ),~~

~~61 then either:~~

0 0 0 – there exists an (α−1, α, α , ψ2(α, α ), ψ3(α, α )−ψ3(0, 0))-blocker for (j, t, −1, 0, 0, t), or ∗ 0 – there exist t ∈ V (T ) with it∗ < it and t ∈ V (Tj,t∗ ), a witness q ∈ ∂Tj,t∗ for Xq0 ∩ (it∗ ,`) (it∗ ) Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic Ej,t∗ -pseudocomponent in (i ∗ ,|V (T )|+1,s+2) G[Y t ] intersecting Xq0 and V (Mα) ∪ V (Mα0 ).

(iz) • (z, Mz) is a nice pair and an outer-safe pair for every z ∈ V (T ) and monochromatic Ej,z - (iz,−1,0) pseudocomponent Mz in G[Y ] with V (Mz) ∩ Xz 6= ∅ and σ(Mz) < σ(Mα1 ). Then either:

• there exists no c-monochromatic path P in G from V (Mα1 ) ∩ V (P ) to V (Mα2 ) ∩ V (P ) in- Sw0 ternally disjoint from γ=1 V (Mγ) such that V (P ) ∩ AL(it,−1,0) (V (Mα1 )) ∩ XV (Tt) − Xt 6= ∅, or

~~• (t, Mα1 ) is not an outer-safe pair.~~

~~Proof. Suppose that this claim does not hold. Among all counterexamples, we choose (t, Mα1 ) such that σ(Mα1 ) is as small as possible, and subject to this, it is as large as possible.~~

~~Hence (t, Mα1 ) is an outer-safe pair, and there exists a c-monochromatic path P in G from Sw0 V (Mα1 ) ∩ V (P ) to V (Mα2 ) ∩ V (P ) internally disjoint from γ=1 V (Mγ) such that~~

V (P ) ∩ AL(it,−1,0) (V (Mα1 )) ∩ XV (Tt) − Xt 6= ∅. Since for every α ∈ [2, α1 − 1], (t, Mα) is an outer-safe pair, applying this claim to Mα implies 0 0 that for every α ∈ [2, α1 − 1] and α ∈ [α + 1, k], there exists no c-monochromatic path P 0 0 Sw0 in G from V (Mα) ∩ V (P ) to V (Mα0 ) ∩ V (P ) internally disjoint from γ=1 V (Mγ) such that 0 V (P ) ∩ AL(it,−1,0) (V (Mα)) ∩ XV (Tt) − Xt 6= ∅. So we have the following, where the case α = 1 follows from Claim 21.34.

0 0 (a) For every α ∈ [α1 − 1] and α ∈ [α + 1, k], there exists no c-monochromatic path P in 0 0 Sw0 G from V (Mα) ∩ V (P ) to V (Mα0 ) ∩ V (P ) internally disjoint from γ=1 V (Mγ) such that 0 V (P ) ∩ AL(it,−1,0) (V (Mα)) ∩ XV (Tt) − Xt 6= ∅.

0 0 Suppose there exist α ∈ [2, α1] and α ∈ [α + 1, k] with either α 6 α1 − 1 or (α, α ) = (α1, α2), ◦ such that Mα and Mα0 have the same color, and the s-segment in Sj containing V (Mα) whose 0 0 0 level equals the color of Mα contains V (Mα0 ), and there exists no (α−1, α, α , ψ2(α, α ), ψ3(α, α )− ∗ ψ3(0, 0))-blocker for (j, t, −1, 0, 0, t), and for every t ∈ V (T ) with it∗ < it and t ∈ V (Tj,t∗ ), if there 0 (it∗ ,`) (it∗ ) exist a witness q ∈ ∂Tj,t∗ for Xq0 ∩Ij ⊆ W3 for some ` ∈ [0, |V (T )|] and a monochromatic Ej,t∗ - (i ∗ ,|V (T )|+1,s+2) pseudocomponent in G[Y t ] intersecting Xq0 and V (Mα) ∪ V (Mα0 ), then t ∈ ∂Tj,t∗ . ∗ ∗ By our assumption, such t exists, and we choose t such that it∗ is as small as possible. So (it∗ ,`) t ∈ ∂Tj,t∗ . By Claims 21.11 and 21.33, t is a witess for Xt ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|]. (i ∗ ) ∗ t1 Let t be the node of T with t ∈ V (Tj,t∗ ) such that some monochromatic E ∗ -pseudocomponent 1 1 j,t1 (it∗ ,|V (T )|+1,s+2) 0 in G[Y 1 ] intersecting X 0 and V (M ) for some witness q ∈ ∂T ∗ for X 0 ∩ I ⊆ q 1 j,t1 q j (i ∗ ,`) t1 W for some ` ∈ [0, |V (T )|], and subject to this, i ∗ is minimum. Since V (M ) ∩ X 6= ∅, 3 t1 1 t ∗ ∗ ∗ t is a candidate of t . So i ∗ i ∗ and hence t 6= t. Since σ(M ) < σ(M ), (t, M ) is a 1 t1 6 t 1 1 α1 1 (i ∗ ,|V (G)|) t1 ∗ nice pair, so by Claim 21.36, there exists a monochromatic E ∗ -pseudocomponent M in j,t1 (it∗ ,|V (T )|+1,s+2) ∗ ∗ G[Y 1 ] intersecting V (M1) such that σ(M ) = σ(M1), V (M ) ∩ Xt 6= ∅ and satisﬁes

62 ∗ 0 ∗ A (i ∗ ,|V (T )|+1,s+2) (V (M )) ∩ XV (Tt) − Xt 6= ∅. Since it < it and α 6= α , there exists no element L t1 (it∗ ,|V (G)|) in Ej,t∗ intersecting both Mα and Mα0 . Since Mα ∩ Xt 6= ∅= 6 Mα0 ∩ Xt and t ∈ ∂Tj,t∗ is a (it∗ ,`) (it∗ ,|V (G)|) witness for Xt ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], by the construction of Ej,t∗ and the ∗ existence of M1 and the minimality of t , we know V (Mα ∪ Mα0 ) ∩ Xt∗ 6= ∅ (so V (Mα) ∩ Xt∗ 6= ∅ since α < α0) and there exist blockers for (j, t∗, |V (T )| + 1, s + 2, β, t) (for some β ∈ [0, |V (G)|]) 0 0 as stated in the assumption of Claim 21.40 (where the M 0 and M 0 in Claim 21.40 are replaced α1 α2 by Mα and Mα0 , respectively). Since α 6 α1, the assumption of this claim implies that condition 0 0 0 (ii) in Claim 21.40 holds. So there is an (α − 1, α, α , ψ2(α, α ), ψ3(α, α ) − ψ3(0, 0))-blocker for (j, t, −1, 0, 0, t) by Claim 21.40, a contradiction. Therefore, we have

~~0 (b) for every α ∈ [2, α1] and α ∈ [α + 1, k], if:~~

0 – either α 6 α1 − 1 or (α, α ) = (α1, α2), – Mα and Mα0 have the same color, and ◦ – the s-segment in Sj containing V (Mα) whose level equals the color of Mα contains V (Mα0 ), then either:

0 0 0 – there exists an (α −1, α, α , ψ2(α, α ), ψ3(α, α )−ψ3(0, 0))-blocker for (j, t, −1, 0, 0, t), or ∗ 0 – there exist t ∈ V (T ) with it∗ < it and t ∈ V (Tj,t∗ )−∂Tj,t∗ , a witness q ∈ ∂Tj,t∗ for Xq0 ∩ (it∗ ,`) (it∗ ) Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic Ej,t∗ -pseudocomponent in (i ∗ ,|V (T )|+1,s+2) G[Y t ] intersecting Xq0 and V (Mα) ∪ V (Mα0 ). Now we prove that:

∗ 0 (c) there exist no t ∈ V (T ) with it∗ < it and t ∈ V (Tj,t∗ ) − ∂Tj,t∗ , a witness q ∈ ∂Tj,t∗ for (it∗ ,`) (it∗ ) Xq0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic Ej,t∗ -pseudocomponent in (i ∗ ,|V (T )|+1,s+2) t 0 G[Y ] intersecting Xq and V (Mα1 ) ∪ V (Mα2 ).

∗ Suppose to the contrary that (c) does not hold. So there exist t ∈ V (T ) with it∗ < it and 0 (it∗ ,`) t ∈ V (Tj,t∗ ) − ∂Tj,t∗ , a witness q ∈ ∂Tj,t∗ for Xq0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and (it∗ ) (i ∗ ,|V (T )|+1,s+2) t 0 a monochromatic Ej,t∗ -pseudocomponent in G[Y ] intersecting Xq and V (Mα1 ) ∪ ∗ ∗ ∗ V (Mα2 ). We choose t such that it is as small as possible. For each γ ∈ [2], let tαγ be the node of T (i ∗ ) tα (i ∗ ,|V (T )|+1,s+2) γ tαγ with t ∈ V (Tj,t∗ ) such that some monochromatic E ∗ -pseudocomponent in G[Y ] αγ j,tαγ (i ∗ ,`) 0 tαγ intersecting X 0 and V (M ) for some witness q ∈ ∂T ∗ for X 0 ∩ I ⊆ W for some q αγ j,tαγ q j 3 ∗ ∗ ` ∈ [0, |V (T )|], and subject to this, i ∗ is minimum. Note that t is a candidate for t for some tαγ αγ ∗ ∗ γ ∈ [2]. So it∗ min{it∗ , it∗ }. Let t0 ∈ {t , t } with it = min{it∗ , it∗ }. If it < it∗ , then > α1 α2 α1 α2 0 α1 α2 0 ∗ ∗ ∗ ∗ ∗ since t ∈ V (Tj,t ) − ∂Tj,t , t ∈ V (Tj,t0 ) − ∂Tj,t0 , so t0 is a candidate for t . Hence t = tαγ for some ∗ ∗ ∗ γ ∈ [2]. By Claims 21.11 and 21.33 and by tracing along P , we have t = tα1 = tα2 . Since α1 6= α2 0 and (t, Mα ) is outer-safe, by (b) and Claim 21.39, there exist t ∈ ∂Tj,t∗ ∩ V (Tt), u, v ∈ Xt0 , 1 α1 a c-monochromatic path P 0 in G[X ] but not in G[Y (it,−1,0)] from u to v internally disjoint V (Tt0 ) 0 (it∗ ,`) from Xt0 such that t is a witness for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and there exists (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) ∗ a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] with σ(M ) = σ(Mα1 )

63 0 containing u but not containing v. Since t ∈ V (Tj,t∗ ) − ∂Tj,t∗ , t ∈ V (Tt) − {t}. Let M0 be (i 0 ) t (it0 ,−1,0) ∗ 0 the monochromatic Ej,t0 -pseudocomponent in G[Y ] containing M . Since t ∈ ∂Tj,t∗ , by ∗ 0 00 Claims 21.11 and 21.33, t is the node of T with t ∈ V (Tj,t∗ ) such that there exist a witness q ∈ (it∗ ,`) (it∗ ) ∂Tj,t∗ for Xq00 ∩Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic Ej,t∗ -pseudocomponent (i ∗ ,|V (T )|+1,s+2) in G[Y t ] intersecting Xq00 and V (M0), and subject to this, it∗ is minimum. ∗ ∗ Note that there exists a tuple (q, q , Q, uq, vq,Pq,Q ) satisfying the following statement, since 0 ∗ 0 ∗ ∗ ∗ (t , t ,M0, u, v, P ,M ) is a candidate for (q, q , Q, uq, vq,Pq,Q ): q ∈ ∂Tj,q∗ ∩ V (Tt) − {t}, Q is (iq) (iq,−1,0) a monochromatic Ej,q -pseudocomponent in G[Y ] intersecting Xq, uq, vq ∈ Xq, Pq is a c- (iq∗ ,|V (T )|+1,s+2) monochromatic path in G[XV (Tq)] but not in G[Y ] from uq to vq internally disjoint (i ∗ ,|V (G)|) ∗ q (iq∗ ,|V (T )|+1,s+2) from Xq, and Q is a monochromatic Ej,q∗ -pseudocomponent in G[Y ] with ∗ ∗ ∗ ∗ σ(Q ) 6 σ(Mα1 ) containing uq but not containing vq such that Q ⊇ Q , Tj,t ⊆ Tj,q and q 0 (iq∗ ,` ) 0 ∗ is a witenss for Xq ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and q is the node with iq∗ < iq 00 (iq∗ ,`) and q ∈ V (Tj,q∗ ) such that there exist a witness q ∈ ∂Tj,q∗ for Xq00 ∩ Ij ⊆ W3 for some (i ∗ ) q (iq∗ ,|V (T )|+1,s+2) ` ∈ [0, |V (T )|] and a monochromatic Ej,q∗ -pseudocomponent in G[Y ] intersecting ∗ Xq00 and V (Q), and subject to this, iq∗ is minimum. We further choose them such that σ(Q ) is ∗ as small as possible. Note that σ(Q) 6 σ(Q ) 6 σ(Mα1 ). (iq,−1,0) Suppose V (Pq) ⊆ Y . Since σ(Q) 6 σ(Mα1 ), we can apply Claim 21.37 by taking (t, M) to be (q, Q). By applying Claim 21.37 by taking (t, M) to be (q, Q), the non-existence of the node t0 mentioned in the conclusion of Claim 21.37 and the existence of q here imply that σ(Q∗) > σ(Q). (iq∗ ) 00 So by Claims 21.11, 21.31 and 21.33, there exists a monochromatic Ej,q∗ -pseudocomponent Q in G[Y (iq∗ ,|V (T )|+1,s+2)] with σ(Q00) = σ(Q) < σ(Q∗). Since Q00 and Q∗ are disjoint subgraphs of Q, 00 by Claims 21.11 and 21.33, there exist q ∈ ∂Tj,q∗ , uq00 , vq00 ∈ Xq00 , a monochromatic path Pq00 in (i ,−1,0) (i ∗ ,|V (T )|+1,s+2) G[Y q ∩ X ] but not in G[Y q ] from u 00 to v 00 internally disjoint from X 00 V (Tq00 ) q q q 00 ∗ 00 00 such that Q contains uq but not vq . Since σ(Q) < σ(Q ) 6 σ(Mα1 ), (q, Q) is outer-safe, so 00 00 00 ∗ 00 ∗ q ∈ V (Tq). Since q ∈ ∂Tj,q∗ , q = q. Since σ(Q ) < σ(Q ), Q contradicts the choice of Q . (iq,−1,0) So V (Pq) 6⊆ Y . By the assumption of this claim, (q, Q) is a nice pair. Hence, by (iq,−1,0) Claim 21.36 (with taking (t, M) = (q, Q)) and Claim 21.34, since V (Pq) 6⊆ Y , there exists a (i ∗ ,|V (G)|) q ∗ (iq∗ ,|V (T )|+1,s+2) ∗ ∗ monochromatic Ej,q∗ -pseudocomponent Q1 in G[Y ] such that σ(Q1) < σ(Q ), ∗ ∗ ∗ (iq) V (Q )∩X 6= ∅ and A (i ∗ ,|V (T )|+1,s+2) (V (Q ))∩X −X 6= ∅. Let R be the monochromatic E - 1 q L q 1 V (Tq) q j,q (iq,−1,0) ∗ ∗ ∗ pseudocomponent in G[Y ] containing Q . Let R1,...,Rβ (for some β ∈ N) be the monochro- (iq) 0 (iq,−1,0) 0 matic Ej,q -pseudocomponents M in G[Y ] intersecting Xq with AL(iq,−1,0) (V (M ))∩XV (Tq) − 0 ∗ ∗ ∗ ∗ ∗ ∗ Xq 6= ∅ and σ(M ) < σ(R ). Assume σ(R1) < σ(R2) < ··· < σ(Rβ). Let Rβ+1 = R . ∗ ∗ (iq∗ ,|V (G)|) ∗ Since uq ∈ V (Q ) and vq 6∈ V (Q ), {uq, vq} 6∈ Ej,q∗ . By the existence of Q1 and the (iq∗ ,|V (G)|) ∗ ∗ ∗ construction of Ej,q∗ , V (Q ) ∩ Xq 6= ∅. Since σ(Q ) 6 σ(Mα1 ), for every α ∈ [β], ∗ Rα contains Mα for some α ∈ [α1 − 1] by Claim 21.31. So by the assumption of this claim and Claim 21.40, the assumptions of this claim hold when t is replaced by q and Mα1 is re- ∗ ∗ ∗ placed by Rβ+1 = R . Since σ(R ) 6 σ(Mα1 ) and iq > it, this claim is applicable to q and ∗ ∗ R , and by applying it, the existence of Pq implies that (q, R ) is not an outer-safe pair. So ∗ ∗ ∗ ∗ ∗ ∗ ∗ σ(R ) = σ(Mα1 ) > σ(Q ). So σ(R ) = σ(Q ) and R = Q ⊇ Mα1 . Hence t = q by ∗ 00 Claims 21.11 and 21.33. Since (q, R ) is not an outer-safe pair, there exist t ∈ ∂Tj,q∗ − V (Tq), 00 00 00 (i ,−1,0) (i ∗ ,|V (T )|+1,s+2) u , v ∈ X 00 , a monochromatic path P in G[Y q ∩ X ] but not in G[Y q ] t V (Tt00 ) 00 00 (q∗,|V (G)|) 00 from u to v internally disjoint from Xt00 , and a monochromatic Ej,q∗ -pseudocomponent M (iq∗ ,|V (T )|+1,s+2) 00 ∗ 00 00 in G[Y ] with σ(M ) = σ(R ) = σ(Mα1 ) containing u but not v . Since q ∈ V (Tt)

64 and V (P 00) ⊆ Y (iq,−1,0) ∩ X , if t00 6∈ V (T ) − {t}, then V (P 00) ⊆ Y (it,−1,0), so it contradicts that V (Tt00 ) t 00 00 ∗ (t, Mα1 ) is an outer-safe pair. So t ∈ ∂Tj,q ∩ (V (Tt) − {t}) − V (Tq). In particular, t 6= q. (iq∗ ,|V (G)|) 00 (it00 ,−1,0) 00 00 If V (P ) 6⊆ Y , then by Claims 21.34 and 21.36, the fact that {u , v } 6∈ Ej,q∗ , the (iq∗ ,|V (G)|) ∗ existence of P and P 00 , and the construction of E ∗ , we know A (i ∗ ,|V (T )|+1,s+2) (V (Q )) ∩ q t j,q L q 1 00 00 X − X ∗ ⊆ (X − X ) ∩ (X − X 00 ), so t = q, a contradiction. Hence V (P ) ⊆ V (Tq∗ ) q V (Tq) q V (Tt00 ) t (i 00 ) (it00 ,−1,0) 00 ∗ ∗ 00 ∗ 000 t Y . Since σ(M ) = σ(R ) = σ(Q ), M = Q . Let M be the monochromatic Ej,t00 - (it00 ,−1,0) 00 00 ∗ 000 pseudocomponent in G[Y ] containing M . Since σ(M ) = σ(Q ) = σ(Mα1 ), M contains ∗ ∗ 00 0 ∗ ∗ Mα1 , so t = q is the node in T with t ∈ ∂Tj,t such that there exist a witness q ∈ ∂Tj,t (it∗ ,`) (it∗ ) for Xq0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|] and a monochromatic Ej,t∗ -pseudocomponent in (i ∗ ,|V (T )|+1,s+2) 000 G[Y t ] intersecting Xq0 and V (M ). 00 ∗ 00 000 Since σ(M ) = σ(Q ) = σ(Mα1 ), we can apply Claim 21.37 by taking (t, M) to be (t ,M ). By applying Claim 21.37 by taking (t, M) to be (t00,M 000), the non-existence of the node t0 mentioned in Claim 21.37 and the existence of t00 here imply that σ(Q∗) = σ(M 00) > σ(M 000). So by (iq∗ ,|V (G)|) 000 Claims 21.11, 21.31 and 21.33, there exists a monochromatic Ej,q∗ -pseudocomponent Q (i ∗ ,|V (T )|+1,s+2) 000 000 00 ∗ 000 in G[Y q ] with σ(Q ) = σ(M ) < σ(M ) = σ(Q ). Hence there exist q ∈ ∂Tj,q∗ , (i 00 ,−1,0) (i ∗ ,|V (T )|+1,s+2) u 000 , v 000 ∈ X 000 , a monochromatic path P 000 in G[Y t ∩X ] but not in G[Y q ] q q q q V (Tq000 ) 000 from uq000 to vq000 internally disjoint from Xq000 such that Q contains uq000 but not vq000 . Since 000 ∗ 00 000 000 000 00 00 σ(Q ) < σ(Q ) 6 σ(Mα1 ), (t ,M ) is outer-safe, so q ∈ V (Tt ). Hence q = t . Since σ(Q000) < σ(Q∗), Q000 contradicts the choice of Q∗. This proves (c). ◦ By the existence of P , Mα1 and Mα2 have the same color, and the s-segment in Sj containing

V (Mα1 ) whose level equals the color of Mα1 contains V (Mα2 ). By (b) and (c), there exists an (α1 − 1, α1, α2, κ, ξ)-blocker (t−ξ, t−ξ+1, . . . , t−1, t1, t2, . . . , tκ) for (j, t, −1, 0, 0, t), where κ = ψ2(α1, α2) Sα1−1 ◦ (i ,−1,0) and ξ = ψ3(α1, α2)−ψ3(0, 0). So α=1 AL t (V (Mα))∩XV (Tt) −Xt ⊆ (XV (Ttκ ) −Xtκ )∩Ij . Let ∗ ◦ α be the element in {α , α } such that A (i ,−1,0) (V (M ∗ ))∩(X −(X ∪I ))∩X −X = ∅ 1 2 L t α V (Tt) t j V (Tt−ξ ) t−ξ (if ξ 6= 0), and A (i ,−1,0) (V (M ∗ )) ∩ (X − X ) ∩ X − X = ∅. L t α V (Tt) t V (Tt1 ) t1 Now we prove

~~(d) t 3 6∈ V (T ). η5 j,t~~

Suppose to the contrary that t 3 ∈ V (T ). η5 j,t (it,−1,α1−1) (it,−1,0) Sα1−1 Suppose (Y −Y ) 6⊆ X . By Claim 21.25, since A (i ,−1,0) (V (M ))∩ V (Tt 2 ) α=1 L t α κ−α1η5 ◦ XV (Tt)−Xt ⊆ (XV (Ttκ )−Xtκ )∩Ij and {(Xtγ −Xt)∩Ij : γ ∈ [−ξ, κ]−{0}} consists of pariwise disjoint 0 0 (it,−1,α1−1) members, there exist α ∈ [α1 − 1], α ∈ [α + 1, w0] and a monochromatic path P in G[Y ] Sw0 0 from V (Mα) to V (Mα0 ) internally disjoint from γ=1 V (Mγ) such that all internal vertices of P (it,−1,0) ◦ 0 are in Y ∪ (I ∩ X − X ) and V (P ) ∩ A (i ,−1,0) (V (M )) ∩ X − X 6= ∅, j V (Tt 2 ) tκ−α η2 L t α V (Tt) t κ−α1η5 1 5 contradicting (a). (it,−1,α1−1) (it,−1,0) Hence (Y − Y ) ⊆ X . Since A (i ,−1,0) (V (M ∗ )) ∩ (X − X ) ∩ V (Tt 2 ) L t α V (Tt) t κ−α1η5 2 (it) (X − X ) = ∅ and κ − α η 2, M ∗ is a monochromatic E -pseudocomponent in V (Tt1 ) t1 1 5 > α j,t (it,−1,α1−1) (it,−1,α1−1) 0 (it) G[Y ]. So V (P ) 6⊆ Y . Let M be the monochromatic Ej,t -pseudocomponent in (it,−1,α1−1) (it,−1,α1−1) G[Y ] containing Mα1 . Since V (P ) 6⊆ Y , there exists 0 (it,−1,α1) u ∈ V (P ) ∩ AL(it∗ ,−1,α1−1) (V (M )) ∩ XV (Tt) − Xt. Then either u is colored in Y by a color diﬀerent from c(P ), or u 6∈ Y (it,|V (T )|+1,s+2). The former cannot hold, so the latter holds. So ◦ u 6∈ Zt. Hence there exists qu ∈ ∂Tj,t with u ∈ Ij ∩ XV (Tqu ) − Xqu .

65 00 (it) (it,|V (T )|+1,s+2) Let M be the monochromatic Ej,t -pseudocomponent in G[Y ] containing Mα∗ . (it,|V (T )|+1,s+2) 00 Since u 6∈ Y , there exists v ∈ V (P ) ∩ AL(it,|V (T )|+1,s+2) (V (M )) ∩ XV (Tt) − Xt. We choose v such that σ(v) is as small as possible. Since P is internally disjoint from Xt, and A (i ,−1,0) (M ∗ )∩X −X ⊆ V (G)−(X −X ), by Claim 21.25, v ∈ V (G)−X . Note L t α V (Tt) t V (Tt1 ) t1 V (Tt 2 ) η5 that v 6∈ Zt. Hence there exists qv ∈ ∂Tj,t with v ∈ XV (Tqv ) − Xqv . (it,|V (T )|+1,s+2) 2 By Claim 21.25, |Y ∩ I | η . Since t 3 ∈ V (T ), there exist η + 1 β < β j 6 5 η5 j,t 5 6 1 2 6 3 2 3 2 η5 −(α1+1)η5 (it,|V (T )|+1,s+2) η5 −α1η5 with β2 −β1 > η5 such that Y ∩XV (Tt ) −XV (Tt ) = ∅. Since > η5+1 β1 β2 v ∈ V (G) − X , q ∈ V (T ) − V (T ). V (Tt 2 ) v tβ1 η5 ∗ (it,|V (T )|+1,s+2) Suppose that qu ∈ V (Ttβ ). So α = α2. Since Y ∩ XV (Tt ) − XV (Tt ) = ∅ and 1 β1 β2 X ∩ I ⊆ Y (it,|V (T )|+1,s+2), q ∈ V (T ). By the existence of P and Claims 21.11 and 21.33, there qu j u tβ2 (it,|V (T )|+1,s+2) exist subpaths of P contained in G[Y ] such that for each β ∈ [β1, β2], some of those subpaths intersects Xtβ − Xt. But it contradicts Claim 21.25 since β2 − β1 > η5. Hence q ∈ V (T ) − V (T ). Let M ∗ be the monochromatic E(iqu )-pseudocomponent in u tβ1 j,qu (iq ,−1,0) ∗ u (i ,−1,0) G[Y ] containing Mα1 . Note that u ∈ V (P ) ∩ AL qu (V (M )) ∩ XV (Tqu ) − Xqu since ◦ ∗ u ∈ Ij ∩ XV (Tqu ) − Xqu . By Claim 21.34, M does not have the smallest σ-value among all (i ) ◦ qu 000 (iqu ,−1,0) Sj -related monochromatic Ej,qu -pseudocomponents M in G[Y ] intersecting Xqu with 000 ◦ (iqu ) (i ,−1,0) AL qu (V (M )) ∩ XV (Tqu ) − Xqu 6= ∅. So there exists an Sj -related monochromatic Ej,qu - ∗ (iq ,−1,0) ∗ u (i ,−1,0) pseudocomponents Q in G[Y ] intersecting Xqu with AL qu (V (Q ))∩XV (Tqu ) −Xqu 6= ∅ ∗ ∗ ∗ such that σ(Q ) < σ(M ). By Claim 21.31, there exists α ∈ [α1 − 1] such that Mα ⊆ Q . (it,|V (G)|) By Claims 21.11 and 21.33, there exists a monochromatic Ej,t -pseudocomponent Q in (it,|V (T )|+1,s+2) (i ,|V (T )|+1,s+2) G[Y ] containing Mα and a vertex u0 ∈ Xqu such that AL t ({u0})∩XV (Tqu ) − ◦ (i ,−1,0) Xqu 6= ∅. Since AL t (V (Mα))∩XV (Tt)−Xt ⊆ (XV (Ttκ )−Xtκ )∩Ij , u0 6∈ V (Mα). Since Q contains M and u , and Y (it,−1,α1−1) −Y (it,−1,0) ⊆ X , V (Q)∩X 6= ∅ for every β ∈ [β , κ−α η2]. α 0 V (Tt 2 ) tβ 1 1 5 κ−α1η5 2 But κ − α1η5 − β1 > η5, contradicting Claim 21.25. This proves (d) (That is, t 3 6∈ V (T )). η5 j,t 0 0 So there exists t ∈ ∂T such that t 3 ∈ V (T 0 ) − {t }. Since j,t η5 t

~~α1−1 [ ◦ (i ,−1,0) AL t (V (Mα)) ∩ XV (Tt) − Xt ⊆ (XV (Ttκ ) − Xtκ ) ∩ Ij , α=1~~

(it) (it,|V (T )|+1,s+2) Mα1 is a monochromatic Ej,t -pseudocomponent in G[Y ] and ◦ V (P ) ∩ AL(it,|V (T )|+1,s+2) (V (Mα1 )) ∩ XV (Tt) − Xt is a nonempty subset of Ij . Since the process for adding fake edges in one interface is irrelevant with the process for adding fake edges in other interfaces, to simplify the notation, we may without loss of generality assume (it,`) 0 (it,|V (T )|+1,s+2) that for every ` ∈ [0, |V (G)|], the monochromatic Ej,t -pseudocomponent M in G[Y ] 0 (it,`) in which σ(M ) is the (` + 1)-th smallest among all monochromatic Ej,t -pseudocomponents in (it,|V (T )|+1,s+2) ◦ ◦ (it,`) G[Y ] is Sj -related, unless there are less than ` + 1 Sj -related monochromatic Ej,t - pseudocomponents in G[Y (it,|V (T )|+1,s+2)]. 0 0 0 (γ ) ◦ (it,γ ) For every γ ∈ [w0] and γ ∈ [0, |V (G)|], let Mγ be the Sj -related monochromatic Ej,t - (it,|V (T )|+1,s+2) ◦ pseudocomponent in G[Y ] with the γ-th smallest σ-value among all Sj -related monochro- 0 (it,γ ) 0 (it,|V (T )|+1,s+2) 0 matic Ej,t -pseudocomponents M in G[Y ] with V (M ) ∩ Xt0 6= ∅ and 0 0 (γ0) A (i ,|V (T )|+1,s+2) (V (M )) ∩ X − X 0 6= ∅ (if there are at most γ − 1 such M , then let M = ∅). L t V (Tt0 ) t γ

66 (γ0) 00 By Claim 21.31, each Mγ either contains Mγ00 for some γ , or is disjoint from Xt. We shall prove the following statements (i) and (ii) by induction on α0: for every α0 ∈ [0, α1 −2] and α ∈ [α0 + 1, α1 − 1],

~~(it,α0) (it) (α0) (i) there exists no e ∈ Ej,t − Ej,t such that e ∩ V (Mα ) 6= ∅, and~~

0 (α0−1) (α0−1) (ii) for every α ∈ [α + 1, w0], if α0 > 1, and Mα and Mα0 have the same color, and ◦ (α0−1) (α0−1) the s-segment in Sj containing V (Mα ) whose level equals the color of Mα contains (α0−1) V (Mα0 ), then either:

0 0 0 0 – there exists an (α0, α, α , ψ2(α, α ), ψ3(α, α ))-blocker for (j, t, |V (T )|+1, s+2, α0 −1, t ), or ∗ 0 0 – there exist t ∈ V (T ) with it∗ < it0 and t ∈ V (Tj,t∗ ), a witness q ∈ ∂Tj,t∗ for Xq0 ∩ (it∗ ,`) (it∗ ) Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic Ej,t∗ -pseudocomponent in (α −1) (α −1) (it∗ ,|V (T )|+1,s+2) 0 0 G[Y ] intersecting Xq0 and V (Mα ) ∪ V (Mα0 ).

Note that (i) and (ii) obviously hold for α0 = 0. And for every α0 > 1, if (i) holds for all smaller (it,α0) α0, then (ii) for α0 implies (i) for α0 by the construction for Ej,t and Claims 21.11 and 21.33. So it suﬃces to show that if (i) holds for every smaller α0, then (ii) holds for α0. (it,α0−1) (it) Assume (i) holds for every smaller α0. That is, there exist no e ∈ Ej,t − Ej,t and 00 (α0−1) (α0−1) 0 α ∈ [0, α1 − 1] such that e ∩ V (Mα00 ) 6= ∅. So Mγ = Mγ for every γ ∈ [α1 − 1]. Let α be a 0 0 w0+3 ﬁxed element in [α + 1, w0]. Since α 6 α1 − 1, ψ2(α, α ) + ψ3(α, α ) 6 ψ2(α1, α2) − 4f1(η5) . If (α0−1) w +3 0 V (Mα0 ) ∩ Xt = ∅, then since t(f1(η5)) 0 ∈ V (Tt ) and (t1, t2, . . . , tκ) is an (α1 − 1, α1, α2, κ, 0)- (α0−1) blocker for (j, t, −1, 0, 0, t), we know A (i ,|V (T )|+1,s+2) (V (M 0 )) ∩ X = ∅ by L t α V (Tt w +3 ) (f1(η5)) 0 +2η5 0 0 0 w +3 Claim 21.25, so (t(f1(η5)) 0 +3η5+1, . . . , tκ) contains a subsequence that is an (α0, α, α , ψ2(α, α ), ψ3(α, α ))- 0 (α0−1) blocker for (j, t, |V (T )| + 1, s + 2, α0 − 1, t ). So we may assume that V (Mα0 ) ∩ Xt 6= ∅. Hence (α0−1) 00 0 (α0−1) Mα0 contains Mα00 for some α ∈ [α , k] by Claim 21.31. Let S = {β ∈ [w0]: Mβ ⊆ Mα0 }. 0 Note that for every β ∈ S, β > α > α + 1. ∗ 0 If there exist β ∈ S and t ∈ V (T ) with it∗ < it and t ∈ V (Tj,t∗ ) − ∂Tj,t∗ , a witness q ∈ ∂Tj,t∗ (it∗ ,`) (it∗ ) for Xq0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic Ej,t∗ -pseudocomponent in (i ∗ ,|V (T )|+1,s+2) 0 G[Y t ] intersecting Xq0 and V (Mα) ∪ V (Mβ), then it∗ < it0 and t ∈ V (Tj,t∗ ), so the second case for (ii) holds and we are done. So by (b), we may assume that fore every β ∈ S, there exists an (α , α, β, κ , ξ )-blocker qβ = (qβ , qβ , . . . , qβ , qβ, qβ, . . . , qβ ) for (j, t, −1, 0, 0, t), 0 β β −ξβ −ξβ +1 −1 1 2 κβ β where κβ = ψ2(α, β) and ξβ = ψ3(α, β) − ψ3(0, 0). We further choose each q such that the sequence (i β , i β , . . . , i β , i β , i β , . . . , i β ) is lexicographically maximal. q q q q q qκ −ξβ −ξβ +1 −1 1 2 β 0 Since β > α > α + 1 for every β ∈ S, we know κβ is identical for all β ∈ S, and ξβ is identical β β for all β ∈ S. Since q are chosen such that the sequence of the σT -value of the nodes in q is lexicographically maximal, qβ is identical for all β ∈ S. Since α 6 α1 − 1, for every β ∈ S,

~~w0+3 κβ + ξβ 6 ψ2(α1, α2) − (2f1(η5) + 3η5 + ψ1(20, 20)) w0+3 = κ − (2f1(η5) + 3η5 + ψ1(20, 20)),~~

β β w +3 β w +3 so if t2f1(η5) 0 +3η5+ψ1(20,20) ∈ V (T ), then (q , . . . , q−1, t2f1(η5) 0 +4η5+ψ1(20,20)+1, . . . , tκ) contains q1 −ξβ a subsequence that is an (α0, α, β, κβ, ξβ)-blocker for (j, t, −1, 0, 0, t), contradicting the maximality

67 β of the σT -value of the nodes. Hence for every β ∈ S, we have q1 ∈ V (Tt w +3 ) − 2f1(η5) 0 +3η5+ψ1(20,20) w +3 {t2f1(η5) 0 +3η5+ψ1(20,20)}. Note that for every β ∈ S, ψ3(α, β) 6 ψ1(3, 3). So if there exists β ∈ S w +3 β such that t2f1(η5) 0 +3η5+ψ1(10,10) ∈ V (Tq ), then −ξβ β β (t w +3 , . . . , t w +3 , q , . . . , q ) 2f1(η5) 0 +3η5+ψ1(10,10)+1 2f1(η5) 0 +3η5+ψ1(20,20) 1 κβ contains a subsequence that is an (α0, α, β, κβ, ξβ)-blocker for (j, t, −1, 0, 0, t), contradicting the lexicographical maximality. Hence for every β ∈ S, β w +3 q−ξ ∈ V (Tt w +3 ) − {t2f1(η5) 0 +3η5+ψ1(10,10)}. β 2f1(η5) 0 +3η5+ψ1(10,10)

Sw0 (α0−1) Note that for each v ∈ γ=0 V (Mγ ), if there exists a monochromatic path in Sw0 (α0−1) (it) ◦ G[ V (M )] + E from v to X , then A (i ,|V (T )|+1,s+2) ({v}) ∩ X − (X 0 ∪ I ) = ∅; γ=0 γ j,t t L t V (Tt0 ) t j ◦ if A (i ,|V (T )|+1,s+2) ({v}) ∩ X − (X 0 ∪ I ) 6= ∅ and there exists no monochromatic path in L t V (Tt0 ) t j Sw0 (α0−1) (it) G[ V (Mγ )] + E from v to X , then since t 3 ∈ V (T 0 ) and (t , t , . . . , t ) is a blocker γ=0 j,t t η5 t 1 2 κ for (j, t, −1, 0, 0, t), we know v ∈ V (G) − (X ∪ X ) and A (i ,|V (T )|+1,s+2) ({v}) ∩ X − V (Tt 3 ) t L t V (Tt0 ) η5+η5+1 ◦ 0 (Xt ∪ I ) ⊆ V (G) − XV (Tt ) by Claim 21.25. Hence j 3 w0 η5+η5·f1(η5)

w0 [ (α0−1) ◦ A (i ,|V (T )|+1,s+2) ( V (M )) ∩ X − (X 0 ∪ I ) ⊆ V (G) − X L t γ V (Tt0 ) t j V (Tt 3 w ) η +η5·f1(η5) 0 γ=0 5 ⊆ V (G) − X . V (Tt w +3 ) 2f1(η5) 0 +3η5+ψ1(10,10) ∗ For every β ∈ S, let α ∈ {α, β} such that A (i ,−1,0) (M ∗ ) ∩ (X − X ) ∩ X − X β = β L t αβ V (Tt) t V (T β ) q q1 1 ◦ ∗ β ∅ and AL(it,−1,0) (Mα ) ∩ (XV (Tt) − Xt) ∩ XV (T β ) − (X ∪ Ij ) = ∅. So for every β ∈ S, β q q−ξ −ξβ β β β (t w0+3 , . . . , t w0+3 , q , . . . , q ) is an (α , α, β, ψ (α, β), ψ (α, β)− 2f1(η5) +3η5+ψ1(5,5)+1 2f1(η5) +3η5+ψ1(10,10) −ξβ κβ 0 2 3 0 ψ3(0, 0) + ψ1(5, 5))-blocker for (j, t, |V (T )| + 1, s + 2, t ). Note that for every β ∈ S, β > 0 0 ∗ α and ψ3(α, β) − ψ3(0, 0) + ψ1(5, 5) > ψ3(α, α ). Hence, if αβ = α for some β ∈ S, then β β (t w0+3 , . . . , t w0+3 , q , . . . , q ) contains a subsequence that is 2f1(η5) +3η5+ψ1(5,5)+1 2f1(η5) +3η5+ψ1(10,10) −ξβ κβ 0 0 0 0 an (α0, α, α , ψ2(α, α ), ψ3(α, α ))-blocker for (j, t, |V (T )|+1, s+2, t ), so we are done. So we may as- ∗ β sume that α = β for every β ∈ S. Then (t w0+3 , . . . , t w0+3 , q , β 2f1(η5) +3η5+ψ1(5,5)+1 2f1(η5) +3η5+ψ1(10,10) −ξβ . . . , qβ ) contains a subsequence that is an (α , α, α0, ψ (α, α0), ψ (α, α0))-blocker for (j, t, |V (T )| + κβ 0 2 3 1, s + 2, t0), so we are done. Therefore, (i) and (ii) hold for all α0 ∈ [0, α1 − 2]. In particular, there exists no e ∈ (it,α1−2) (it) 00 (α1−2) Ej,t − Ej,t and α ∈ [α1 − 1] such that e ∩ V (Mα00 ) 6= ∅. Hence there exists no (it,|V (G)|) (it) Sα1−1 S 0 e ∈ Ej,t −Ej,t such that e∩ γ=1 V (Mγ) 6= ∅. So M 0 AL(it,|V (T )|+1,s+2) (V (M ))∩XV (Tt)−Xt ⊆ Sα1−1 ◦ ◦ (i ,−1,0) α=1 AL t (V (Mα)) ∩ XV (Tt) − Xt ⊆ XV (Ttκ ) ∩ Ij , where the ﬁrst union is over all Sj -related (it,|V (G)|) 0 (it,|V (T )|+1,s+2) 0 monochromatic Ej,t -pseudocomponents M in G[Y ] with σ(M ) < σ(Mα1 ). (i 0 ) 00 ◦ t (it0 ,−1,0) For each γ ∈ [w0], let Mγ be the Sj -related monochromatic Ej,t0 -pseudocomponent in G[Y ] ◦ (it0 ) 0 with the γ-th smallest σ-value among all Sj -related monochromatic Ej,t0 -pseudocomponents M (i 0 ,−1,0) 0 0 in G[Y t ] with V (M ) ∩ X 0 6= ∅ and A (i ,−1,0) (V (M )) ∩ X − X 0 6= ∅ (if there are less t L t0 V (Tt0 ) t 0 00 than γ such M , then let V (Mγ ) = ∅). Since (i) and (ii) hold for α0 ∈ [0, α1 − 2], by Claim 21.40, for every α ∈ [2, α1 − 1] and 0 00 00 ◦ 00 α ∈ [α + 1, w0], if Mα and Mα0 have the same color, and the s-segment in Sj containing V (Mα) 00 00 whose level equals the color of Mα contains V (Mα0 ), then either:

~~68 0 0 0 0 0 (iii) there exists an (α − 1, α, α , ψ2(α, α ), ψ3(α, α ) − ψ3(0, 0))-blocker for (j, t , −1, 0, 0, t ), or~~

∗ 0 0 (it∗ ,`) (iv) there exist t ∈ V (T ) with it∗ < it0 and t ∈ V (Tj,t∗ ), a witness q ∈ ∂Tj,t∗ for Xq0 ∩Ij ⊆ W3 (i ∗ ) t (it∗ ,|V (T )|+1,s+2) for some ` ∈ [0, |V (T )|], and a monochromatic Ej,t∗ -pseudocomponent in G[Y ] 00 00 intersecting Xq0 and V (Mα) ∪ V (Mα0 ).

Sα1−1 (α1−1) ◦ 00 Since A (i ,−1,0) (V (M ) ∩ X − X ⊆ (X − X 0 ) ∩ I , M = M for every γ=1 L t γ V (Tt) t V (Tt0 ) t j γ γ γ ∈ [α1 − 1]. (|V (G)|) (it,−1,0) For every v0 ∈ V (P ) ∩ AL(it,−1,0) (V (Mα1 )) − Y , since c(v0) = c(P ) and α1−1 ◦ 0 (i ,|V (T )|+1,s+2) S 0 t α=1 AL(it,−1,0) (V (Mα)) ∩ XV (Tt) − Xt ⊆ Ij and tκ ∈ V (Tt ) − {t }, we know v0 6∈ Y , ◦ ∗ (iq∗ ,−1,0) ∗ v0 so v0 ∈ Ij and there exists qv0 ∈ ∂Tj,t such that v ∈ XV (Tq∗ ) − Xqv and v 6∈ Y . Let v be v0 0 (|V (G)|) (it,−1,0) ∗ 0 a vertex in V (P ) ∩ AL(it,−1,0) (V (Mα1 )) − Y such that qv 6= t if possible. Now we prove

~~(|V (G)|) (it,−1,0) ∗ 0 (i ,−1,0) (e) For every v0 ∈ V (P ) ∩ AL t (V (Mα1 )) − Y , qv0 = t .~~

∗ 0 Suppose (e) does not hold. Then qv 6= t by the choice of v. By Claim 21.34, the monochro- (i ∗ ) qv (iq∗ ,−1,0) matic E ∗ -pseudocomponent in G[Y v ] containing M does not have the smallest σ- j,qv α1 (i ∗ ) ◦ qv 0 (iq∗ ,−1,0) value among all S -related monochromatic E ∗ -pseudocomponents M in G[Y v ] with j j,qv 0 (it,|V (G)|) A (i ∗ ,−1,0) (V (M )) ∩ XV (T ∗ ) − Xq∗ 6= ∅. By Claim 21.31, since there exists no e ∈ E − L qv qv v j,t (it) Sα1−1 (it) E such that e ∩ V (M ) 6= ∅, there exists a c-monochromatic path R in G + E ∗ j,t α=1 α j,qv Sα1−1 Sα1−1 Sα1−1 ∗ (i ,−1,0) from α=1 V (Mα) to Xqv internally disjoint from α=1 V (Mα). Since AL t ( α=1 V (Mα)) ∩ ◦ 0 00 0 00 0 XV (Tt) − Xt ⊆ XV (Tt ) ∩ I , there exists a subpath R from V (M 0 ) ∩ V (R ) to V (M 0 ) ∩ V (R ) κ j α1 α2 Sw0 00 0 00 internally disjoint from V (M ) such that V (R ) ∩ A (i 0 ,−1,0) (V (M 0 )) ∩ XV (T 0 ) − Xt0 6= ∅ α=1 α L t α1 t 0 0 0 0 00 for some α1 ∈ [α1 − 1] and α2 ∈ [α1 + 1, w0]. By Claim 21.34, α1 ∈ [2, α1 − 1]. Let α1 be the 00 minimum γ ∈ [2, α1 − 1] such that there exists α2 ∈ [γ + 1, w0] such that there exists a sub- 00 00 00 00 00 Sw0 00 path R from V (M ) ∩ V (R ) to V (M 00 ) ∩ V (R ) internally disjoint from V (M ) such that γ α2 α=1 α 00 00 00 0 V (R ) ∩ A (i ,−1,0) (V (M )) ∩ X − X 0 6= ∅. Note that α α < α , so (iii) and (iv) hold L t0 γ V (Tt0 ) t 1 6 1 1 00 0 00 00 for every α ∈ [2, α ] and α ∈ [α + 1, w0]. Since σ(M 00 ) σ(M 0 ) < σ(Mα1 ), the minimality of 1 α1 6 α1 0 00 γ implies that this claim is applicable when t is replaced by t and Mα1 is replaced by M 00 , so α1 0 00 00 00 0 00 (t ,M 00 ) is not an outer-safe pair by the existence of R . But σ(M 00 ) < σ(Mα1 ), so (t ,M 00 ) is an α1 α1 α1 outer-safe pair by the assumption of this claim, a contradiction. 00 (|V (G)|) 00 Hence (e) holds. So M ⊇ M ⊇ M and V (P )∩A (i ,−1,0) (V (M ))∩X −X 0 6= ∅. α1 α1 α1 L t0 α1 V (Tt0 ) t ∗ 0 00 (|V (G)|) Note that by an argument similar with the one for showing that qv = t , Mα1 = Mα1 . Since v 6∈ Y (it0 ,−1,0), V (P ) 6⊆ Y (it0 ,−1,0). Suppose there exist β(0) ∈ [w ] with M (0) 6⊆ M 00 and a subpath P 0 of P from V (M 00 ) to V (M (0)) 2 0 β2 α1 α1 β2 Sw0 (0) 0 00 internally disjoint from V (M ) such that V (P ) ∩ A (i ,−1,0) (V (M )) ∩ X − X 0 6= ∅. γ=1 γ L t0 α1 V (Tt0 ) t (i ,|V (T )|+1,s+2) 0 0 Since X 0 ∩ I ⊆ Y t , V (P ) ⊆ X and P is internally disjoint from X 0 . So t j V (Tt0 ) t 00 00 (0) 0 00 there exists α ∈ [w0] such that M 00 contains M . Since P is a subpath of P , α ∈ [α1, w0]. 2 α2 β2 2 00 00 00 (α1−1) (0) Since M 6⊆ Mα1 , α2 ∈ [α1 + 1, w0]. Let α1 = α1. For each β ∈ [2], let Qβ = {Mγ : β2 (α1−1) 00 00 (it,α1) (it,α1−1) γ ∈ [w0],Mγ ⊆ M 00 }. Since α2 6= α1, there exists no e ∈ Ej,t − Ej,t such that αβ e ∩ S V (Q) 6= ∅= 6 e ∩ S V (Q). By the construction of E(it,α1), for every Q ∈ Q and Q∈Q1 Q∈Q2 j,t 1 1 ∗ 0 0 Q2 ∈ Q2, either there exists t ∈ V (T ) with it∗ < it and t ∈ V (Tj,t∗ ), a witness q ∈ ∂Tj,t∗

69 (it∗ ,`) (it∗ ) for Xq0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic Ej,t∗ -pseudocomponent (it∗ ,|V (T )|+1,s+2) 00 00 in G[Y ] intersecting Xq0 and V (M 0 ) ∪ V (M 0 ), or there exists a blocker for Q1 α1 α2 0 and Q2. If the former holds, then by Claims 21.11 and 21.33 and the existence of P , there ex- (it∗ ) (i ∗ ,|V (T )|+1,s+2) t 0 ists a monochromatic Ej,t∗ -pseudocomponent in G[Y ] intersecting Xq and V (Mα1 ), contradicting (c). So desired blockers for Q1 ∈ Q1 and Q2 ∈ Q2 exist. But (iii) and (iv) hold for every α ∈ [2, α1 − 1], so Claim 21.40 implies that the conditions of this claim hold, when t is replaced by t0 and M replaced by M 00 . Since σ(M 00 ) σ(M ) and i0 > i , the choice α1 α1 α1 6 α1 t t 0 0 00 of (t, Mα1 ) for this claim and the existence of P implies that (t ,Mα1 ) is not outer-safe. But 00 (|V (G)|) (it,|V (T )|+1,s+2) V (Mα1 ) = V (Mα1 ) ⊆ Y , a contradiction. 00 (|V (G)|) 00 (|V (G)|) This together with (e) and the fact that Mα1 = Mα1 imply that Mα2 ⊆ Mα1 = Mα1 . (|V (G)|) 00 (it0 ,−1,0) 00 Since Mγ = Mγ for every γ ∈ [α1 − 1], V (P ) − Y 6= ∅, so V (P ) 6⊆ Mα1 = Mα1 . Recall ∗ that there exists α ∈ {α , α } such that A (i ,−1,0) (V (M ∗ )) ∩ (X − X ) ∩ X − X = ∅. 1 2 L t α V (Tt) t V (Tt1 ) t1 Since V (P ) is internally disjoint from Xt, by (d) and Claim 21.25, there exists a vertex o in V (P )∩ (|V (G)|) 00 A (i ,|V (T )|+1,s+2) (V (Mα )) ∩ X − (X 0 ∪ X )). So o ∈ V (P ) ∩ A (i 0 ,−1,0) (V (M )) ∩ L t 1 V (Tt0 ) t V (Tt 3 ) L t α1 2η5 X − (X 0 ∪ X )). V (Tt0 ) t V (Tt 3 ) 2η5 0 0 Let z be the node of T with iz > it such that either o ∈ XV (Tj,z) or there exists z ∈ ∂Tj,z such that o ∈ X − X 0 and t 6∈ V (T 0 ), and subjec to this, i is minimum. Since o ∈ V (Tz0 ) z κ z z 00 A (i 0 ,−1,0) (V (M )) − X , an augument similar to the one for showing (d) shows that t 3 6∈ L t α1 V (Tt 3 ) 3η5 2η5 00 V (T ). If o ∈ X , then since t 3 6∈ V (T ) and o ∈ A (i 0 ,−1,0) (V (M )) − X , o ∈ j,z V (Tj,z) 3η5 j,z L t α1 V (Tt 3 ) 2η5 (i ,0,α ) 0 Y z 1 with c(o) 6= c(P ), a contradiction. So there exists z ∈ ∂T such that o ∈ X − X 0 j,z V (Tz0 ) z and tκ 6∈ V (Tz0 ). Then an argument similar to the one for showing (e) leads to a contradiction. This proves the claim.

(it) Claim 21.42. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M be an Sj-related monochromatic Ej,t - (it,−1,0) ∗ pseudocomponent in G[Y ] intersecting Xt. Let t ∈ V (T ) be the node of T such that t ∈ (i ∗ ) t (it∗ ,|V (T )|+1,s+2) V (Tj,t∗ ) and some Ej,t∗ -pseudocomponent in G[Y ] intersects V (M) and Xq for some ∗ (it∗ ,`) witness q ∈ ∂Tj,t∗ ∪ {t } for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ 0 is minimum. Assume there exist t ∈ ∂Tj,t∗ − V (Tt), u, v ∈ Xt0 , a c-monochromatic path Pt0 in (i ∗ ,|V (T )|+1,s+2) G[X ] but not in G[Y t ] from u to v internally disjoint from X 0 such that there V (Tt0 ) t (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) ∗ exists a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] with σ(M ) = (i 0 ) 0 t (it0 ,−1,0) σ(M) containing u but not v. Let M be the monochromatic Ej,t0 -pseudocomponent in G[Y ] containing M ∗. ◦ Assume that (z, Mz) is a nice pair and an outer-safe pair for every z ∈ V (T ) and Sj -related (iz) (iz,−1,0) monochromatic Ej,z -pseudocomponent Mz in G[Y ] intersecting Xz with σ(Mz) < σ(M). 0 0 (i 0 ,−1,0) If (t ,M ) is an outer-safe pair, then V (Pt0 ) ⊆ Y t .

(i 0 ,−1,0) Proof. Suppose V (Pt0 ) 6⊆ Y t . 0 ∗ 0 0 ∗ 0 ∗ Since t ∈ ∂Tj,t∗ − V (Tt), t 6= t 6= t . Since M contains M , σ(M ) 6 σ(M ) = σ(M). Since (it0 ,−1,0) ◦ Pt0 is a c-monochromatic path and V (Pt0 ) 6⊆ Y , by Claim 21.34, there exists an Sj -related (it0 ) 0 (i 0 ,−1,0) 0 t 0 monochromatic Ej,t0 -pseudocomponent Q in G[Y ] intersecting Xt with AL(it0 ,−1,0) (V (Q ))∩ 0 0 0 0 ∗ 0 0 X − X 0 6= ∅ and σ(Q ) < σ(M ). Since σ(Q ) < σ(M ) σ(M ) = σ(M), (t ,Q ) is V (Tt0 ) t 6 0 (it∗ ,`) an outer-safe pair and a nice pair. By Claim 21.33, t is a witness for Xt0 ∩ Ij ⊆ W3 for ∗ 0 (iq∗ ) some ` ∈ [0, |V (T )|]. Let q ∈ V (T ) be the node of T such that t ∈ V (Tj,q∗ ) and some Ej,q∗ -

70 (i ∗ ,|V (T )|+1,s+2) 0 ∗ pseudocomponent in G[Y q ] intersects V (Q ) and Xq for some witness q ∈ ∂Tj,q∗ ∪{q } (iq∗ ,`) ∗ for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, iq∗ is minimum. Since t is a ∗ ◦ (it∗ ,|V (G)|) candidate for q , iq∗ 6 it∗ . So by Claim 21.36, there exists an Sj -related monochromatic Ej,t∗ - (it∗ ,|V (T )|+1,s+2) pseudocomponent Q in G[Y ] intersecting X 0 with A (i ∗ ,|V (T )|+1,s+2) (V (Q))∩X − t L t V (Tt0 ) 0 ∗ Xt0 6= ∅ and σ(Q) = σ(Q ) < σ(M ). (`) (it∗ ,`) For every ` ∈ [0, |V (G)|−1] and x ∈ {u, v}, let Qx be the monochromatic Ej,t∗ -pseudocomponent in G[Y (it∗ ,|V (T )|+1,s+2)] containing x. For every ` ∈ [0, |V (G)| − 1], let Q(`) be the monochromatic (i ∗ ,`) t (it∗ ,|V (T )|+1,s+2) (`) Ej,t∗ -pseudocomponent in G[Y ] such that σ(Q ) is the (` + 1)-th smallest among (i ∗ ,`) t (it∗ ,|V (T )|+1,s+2) all monochromatic Ej,t∗ -pseudocomponents in G[Y ]. (`) ◦ (`) (`) 0 Let L = {` ∈ [0, |V (G)| − 1] : Q is Sj -related,V (Q ) ∩ Xt 6= ∅,AL(it∗ ,|V (T )|+1,s+2) (V (Q )) ∩ (`) (`) (`) X − X 0 6= ∅, σ(Q ) < min{Q ,Q }}. By the existence of Q, L= 6 ∅. V (Tt0 ) t u v (`) (`) (`) ◦ For each ` ∈ L, let M1 ,M2 ,...,Mβ (for some β ∈ N) be the Sj -related monochromatic (it∗ ,`) 00 00 E ∗ -pseudocomponents M intersecting X 0 with A (i ∗ ,|V (T )|+1,s+2) (V (M )) ∩ X − X 0 6= ∅ j,t t L t V (Tt0 ) t (`) (`) (`) (`) such that σ(M1 ) < σ(M2 ) < ··· < σ(Mβ ) and V (Mγ ) = ∅ for every γ > β + 1. Let mL = max L. 0 0 0 0 ◦ (it0 ) Let M1,M2,...,Mβ (for some β ∈ N) be the Sj -related monochromatic Ej,t0 -pseudocomponents 00 00 0 0 M intersecting X 0 with A (i ,−1,0) (V (M )) ∩ X − X 0 6= ∅ such that σ(M ) < σ(M ) < ··· < t L t0 V (Tt0 ) t 1 2 0 0 0 σ(Mβ0 ). Let V (Mγ) = ∅ for every γ > β + 1. 0 ∗ Let α1 ∈ [w0] be the minimum index such that V (Mα1 ) ∩ {u, v}= 6 ∅. Since M contains u but (it∗ ,|V (G)|) not v, there exists no monochromatic Ej,t∗ -pseudocomponent containing both u and v. So by (it∗ ,|V (G)|) (mL) (mL) 0 ∗ S the construction of Ej,t∗ , (V (Qu )∪V (Qv ))∩Xt 6= ∅ and M 0 AL(it∗ ,|V (T )|+1,s+2) (V (M ))∩ ◦ ◦ (it∗ ,mL) X −X ∗ ⊆ (X −X 0 )∩I , where the union is over all S -related E ∗ -pseudocomponents V (Tt∗ ) t V (Tt0 ) t j j j,t 0 (i ∗ ,|V (T )|+1,s+2) 0 0 t 0 (i ,|V (T )|+1,s+2) ∗ M in G[Y ] with V (M ) ∩ Xt 6= ∅, AL t∗ (V (M )) ∩ XV (Tt∗ ) − Xt 6= ∅ and 0 (mL) 0 (|V (G)|) σ(M ) 6 σ(Q ). So by Claim 21.31, we know for every α ∈ [α1 −1], Mα = Mα by induction 0 (|V (G)|) 0 on α. This together with Claim 21.31 imply that Mα1 = Mα1 . So |Mα1 ∩ {u, v}| = 1. Hence 0 ∗ 0 there exists α2 ∈ [α1 + 1, w0] such that Mα2 ∩ {u, v} 6= ∅. Note that M is contained in Mα1 or M 0 . So σ(M 0 ) σ(M ∗) and M 0 ∈ {M 0 ,M 0 }. α2 α1 6 α1 α2 (mL) (mL) 0 For ` ∈ [2], let Q = {M : γ ∈ [w ],M ⊆ M }. By the existence of P 0 , the α γ 0 γ α` t (it∗ ,|V (G)|) (mL) (mL) minimality of it∗ and the construction of Ej,t∗ , for every Mγ1 ∈ Q1 and Mγ2 ∈ Q2, we (mL) (mL) have (V (Mγ1 ) ∪ V (Mγ2 )) ∩ Xt∗ 6= ∅ and there exists an (mL + 1, γ1, γ2, ψ2(γ1, γ2), ψ3(γ1, γ2))- ∗ 0 0 ∗ blocker for (j, t , |V (T )| + 1, s + 2, mL, t ). Hence, by Claim 21.31, V (Mα1 ) ∩ Xt 6= ∅. Since σ(M 0 ) σ(M ∗) = σ(M), for every β ∈ [α − 1], (t0,M 0 ) is a nice pair. Hence by Claim 21.40, α1 6 1 β 0 0 there exists an (α1 − 1, α1, α2, ψ2(α1, α2), ψ3(α1, α2) − ψ3(0, 0))-blocker for (j, t , −1, 0, 0, t ). (it∗ ,mL) 0 Similarly, by the construction of Ej,t∗ and Claim 21.40, for every α ∈ [2, α1−1] and α ∈ [α+ 0 0 ◦ 0 1, w0], if Mα and Mα0 have the same color, and the s-segment in Sj containing V (Mα) whose level 0 0 0 0 0 equals the color of Mα contains V (Mα0 ), then either there exists an (α−1, α, α , ψ2(α, α ), ψ3(α, α )− 0 0 000 ψ3(0, 0))-blocker for (j, t , −1, 0, 0, t ), or there exist t ∈ V (T ) with it000 < it∗ and Tj,t∗ ⊆ Tj,t000 , 0 (it000 ,`) a witness q ∈ ∂Tj,t000 for Xq0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|], and a monochromatic (i 000 ) t (it000 ,|V (T )|+1,s+2) 0 0 Ej,t000 -pseudocomponent in G[Y ] intersecting V (Mα) ∪ V (Mα0 ) and Xq0 . (i 0 ,−1,0) 0 0 0 0 0 0 0 t Since V (Pt ) 6⊆ Y , by Claim 21.34, α1 6= 1. If Mα1 = M , then (t ,Mα1 ) = (t ,M ) is an outer-safe pair; if M 0 6= M 0, then M 0 = M 0 , so σ(M 0 ) < σ(M 0 ) = σ(M 0) σ(M ∗) = σ(M), α1 α2 α1 α2 6 0 0 and hence (t ,Mα1 ) is an outer-safe pair. By Claim 21.41, there exists no c-monochromatic path

71 ∗ 0 ∗ 0 ∗ Sw0 0 O in G from V (Mα1 ) ∩ V (O ) to V (Mα2 ) ∩ V (O ) internally disjoint from γ=1 V (Mγ) such that ∗ 0 V (O ) ∩ A (i ,−1,0) (V (M )) ∩ X − X 0 6= ∅, contradicting the existence of P 0 . This proves the L t0 α1 V (Tt0 ) t t claim.

(it) Claim 21.43. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M be an Sj-related monochromatic Ej,t - (it,−1,0) pseudocomponent in G[Y ] intersecting Xt. Assume that (z, Mz) is a nice pair and an ◦ (iz) outer-safe pair for every z ∈ V (T ) and Sj -related monochromatic Ej,z -pseudocomponent Mz in (iz,−1,0) G[Y ] intersecting Xz with σ(Mz) < σ(M). Then (t, M) is an outer-safe pair.

Proof. Suppose to the contrary that (t, M) is not an outer-safe pair, and subject to this, it is as small as possible. ∗ (it∗ ) Let t ∈ V (T ) be the node of T such that t ∈ V (Tj,t∗ ) and some Ej,t∗ -pseudocomponent in (i ∗ ,`) (it∗ ,|V (T )|+1,s+2) ∗ t G[Y ] intersects V (M) and Xq for some witness q ∈ ∂Tj,t∗ ∪{t } for Xq ∩Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ is minimum. Since (t, M) is not an outer-safe 0 (i ,−1,0) pair, there exist t ∈ ∂T ∗ − V (T ), u, v ∈ X 0 , a monochromatic path P 0 in G[Y t ∩ X ] j,t t t t V (Tt0 ) (i ∗ ,|V (T )|+1,s+2) but not in G[Y t ] from u to v internally disjoint from Xt0 such that there exists (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) ∗ a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] with σ(M ) = σ(M) ∗ 0 (it∗ ,`) containing u but not v. In particular, t 6= t . By Claim 21.33, t is a witness for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|]. (i 0 ) 0 t (it0 ,−1,0) ∗ Let M be the monochromatic Ej,t0 -pseudocomponent in G[Y ] containing M . By ∗ 0 (it∗ ) Claims 21.11 and 21.33, t ∈ V (T ) is the node of T such that t ∈ V (Tj,t∗ ) and some Ej,t∗ - (i ∗ ,|V (T )|+1,s+2) 0 ∗ pseudocomponent in G[Y t ] intersects V (M ) and Xq for some witness q ∈ ∂Tj,t∗ ∪{t } (it∗ ,`) for Xq ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ is minimum. Note that 0 ∗ 0 ∗ t 6= t and σ(M ) 6 σ(M ). (i 0 ,−1,0) 0 ∗ 0 Suppose V (Pt0 ) ⊆ Y t . If σ(M ) = σ(M ), then since t ∈ ∂Tj,t∗ ∩ V (Tt0 ), it is a contradiction by Claim 21.37. So σ(M 0) < σ(M ∗) = σ(M). Hence (t0,M 0) is an outer-safe pair by (it∗ ,|V (G)|) assumption. By Claim 21.38, there uniquely exists a monochromatic Ej,t∗ -pseudocomponent Q∗ in G[Y (it∗ ,|V (T )|+1,s+2)] intersecting V (M 0). Since M ∗ ⊆ M 0, Q∗ = M ∗ by the uniqueness of Q∗. By Claim 21.38, σ(Q∗) = σ(M 0). So σ(M 0) = σ(M ∗), a contradiction. (i 0 ,−1,0) 0 0 Hence V (Pt0 ) 6⊆ Y t . By Claim 21.42, (t ,M ) is not an outer-safe pair. By the assumption 0 0 ∗ 0 ∗ 0 of this claim, σ(M ) > σ(M). Since M ⊇ M , σ(M ) 6 σ(M ) = σ(M). So σ(M ) = σ(M). 0 0 0 Since (t ,M ) is not outer-safe and σ(M ) = σ(M), by the maximality of it, it0 > it. But V (Pt0 ) ⊆ (it,−1,0) (i 0 ,−1,0) Y and V (Pt0 ) 6⊆ Y t . So it > it0 , a contradiction. This proves the claim.

(it) Claim 21.44. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M be an Sj-related monochromatic Ej,t - (it,−1,0) pseudocomponent in G[Y ] intersecting Xt. Assume that (z, Mz) is a nice pair and an ◦ (iz) outer-safe pair for every z ∈ V (T ) and Sj -related monochromatic Ej,z -pseudocomponent Mz in (iz,−1,0) G[Y ] intersecting Xz with σ(Mz) < σ(M). Then (t, M) is a nice pair.

∗ Proof. By Claim 21.43, (t, M) is an outer-safe pair. Let t be the node of T with t ∈ V (Tj,t∗ ) such (i ∗ ) t (it∗ ,|V (T )|+1,s+2) that some Ej,t∗ -pseudocomponent in G[Y ] intersecting V (M) intersects Xt0 for some 0 ∗ (it∗ ,`) witness t ∈ ∂Tj,t∗ ∪ {t } for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ is (it∗ ,|V (G)|) minimum. By Claim 21.38, there uniquely exists a monochromatic Ej,t∗ -pseudocomponent ∗ (i ∗ ,|V (T )|+1,s+2) ∗ ∗ ∗ M in G[Y t ] intersecting V (M); V (M ) ∩ Xt 6= ∅ and σ(M ) = σ(M); if t 6= t and ∗ AL(it,−1,0) (V (M)) ∩ XV (Tt) − Xt 6= ∅, then AL(it∗ ,|V (T )|+1,s+2) (V (M )) ∩ XV (Tt) − Xt 6= ∅.

72 Suppose to the contrary that (t, M) is not a nice pair. So t 6= t∗, and there exist a c- monochromatic path P in G intersecting V (M) with V (P ) ⊆ XV (Tt∗ ) internally disjoint from (it∗ ,|V (G)|) XV (Tt) and a vertex uP ∈ V (P ) ∩ XV (Tj,t∗ ) such that there exists no monochromatic Ej,t∗ - 0 (it∗ ,|V (T )|+1,s+2) 0 0 0 pseudocomponent M in G[Y ] such that V (M ) ∩ V (M) 6= ∅= 6 V (M ) ∩ XV (Tt), M 0 contains uP , and if AL(it,−1,0) (V (M))∩XV (Tt)−Xt 6= ∅, then AL(it∗ ,|V (T )|+1,s+2) (V (M ))∩XV (Tt)−Xt 6= ∗ (it∗ ,|V (G)|) ∅. So M does not contain uP . By Claims 21.11 and 21.33, some monochromatic Ej,t∗ - (i ∗ ,|V (T )|+1,s+2) ∗ pseudocomponent Q in G[Y t ] contains uP . Hence Q 6= M . ∗ ∗ Since V (P )∩V (M) 6= ∅ and V (P ) ⊆ XV (Tt∗ ), V (P )∩V (M ) 6= ∅ by the uniqueness of M . Since ∗ 0 ∗ P is internally disjoint from XV (Tt), by tracing along P from M , there exists t ∈ ∂Tj,t − V (Tt), (i ∗ ,|V (T )|+1,s+2) u, v ∈ X 0 , a c-monochromatic path P 0 in G[X ] but not in G[Y t ] from u to t t V (Tt0 ) ∗ ∗ v internally disjoint from Xt0 such that M contains u but not v. Since M does not contain ∗ v, M does not contain v by Claims 21.11 and 21.33 and the uniqueness of M . So V (Pt0 ) 6⊆ (i 0 ) (it,−1,0) 0 t (it0 ,−1,0) ∗ Y . Let M be the monochromatic Ej,t0 -pseudocomponent in G[Y ] containing M . 0 ∗ 0 0 So σ(M ) 6 σ(M ) = σ(M). Hence (t ,M ) is an outer-safe pair by Claim 21.43. By Claim 21.42, (i 0 ,−1,0) V (Pt0 ) ⊆ Y t . (it,−1,0) 0 ∗ Since (t, M) is outer-safe, V (Pt0 ) 6⊆ Y . So it0 > it. Since t 6∈ V (Tt) and V (M ) ∩ Xt 6= ∅ ∗ ∗ 0 and u ∈ V (M ), V (M ) intersects the bag at the common ancestor of t and t . So V (Pt0 ) ⊆ (i 0 ,−1,0) (it,−1,0) Y t implies V (Pt0 ) ⊆ Y . This proves the claim.

◦ (it) Claim 21.45. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M be an Sj -related monochromatic Ej,t - (it,−1,0) ∗ pseudocomponent in G[Y ] intersecting Xt. Let t be the node of T with t ∈ V (Tj,t∗ ) such (i ∗ ) t (it∗ ,|V (T )|+1,s+2) that some Ej,t∗ -pseudocomponent in G[Y ] intersecting V (M) intersects Xt0 for some 0 ∗ (it∗ ,`) witness t ∈ ∂Tj,t∗ ∪ {t } for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ is minimum. Then • (t, M) is a nice pair and an outer-safe pair, and

• if P is a c-monochromatic path in G[XV (Tt∗ )] intersecting V (M)∩Xt internally disjoint from (it∗ ,|V (G)|) XV (Tt) and uP is a vertex in V (P ) ∩ XV (Tj,t∗ ), then there exists a monochromatic Ej,t∗ - (i ∗ ,|V (T )|+1,s+2) pseudocomponent in G[Y t ] containing V (M) ∩ V (P ) ∩ Xt and containing uP . Proof. Suppose that (t, M) is either not a nice pair or not an outer-safe pair, and subject to this, σ(M) is as small as possible. By Claim 21.43, (t, M) is an outer-safe pair. By Claim 21.44, (t, M) is a nice pair. Hence the ﬁrst statement of this claim holds. Now we prove the second statement of this claim. By Claim 21.38, there exists uniquely (i ∗ ,|V (G)|) t ∗ (it∗ ,|V (T )|+1,s+2) a monochromatic Ej,t∗ -pseudocomponent M in G[Y ] intersecting V (M). If ∗ t = t , then since V (P ) ⊆ XV (Tt∗ ) and P is internally disjoint from XV (Tt), P is a c-monochromatic ∗ ∗ path in G[Xt], so M containing P by Claims 21.11 and 21.33. Hence we may assume that t 6= t . ∗ ∗ ∗ Since (t, M) is nice, the uniqueness of M implies that V (M ) ∩ V (M) 6= ∅ 6= V (M ) ∩ XV (Tt) ∗ ∗ ∗ and M contains uP . By Claims 21.11, 21.33 and 21.38, V (M ) contains V (M) ∩ Xt, so V (M ) contains V (M) ∩ V (P ) ∩ Xt. This proves the claim.

(it) Claim 21.46. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M be an Sj-related monochromatic Ej,t - (it,−1,0) ∗ pseudocomponent in G[Y ] intersecting Xt. Let t be the node of T with t ∈ V (Tj,t∗ ) such (i ∗ ) t (it∗ ,|V (T )|+1,s+2) that some Ej,t∗ -pseudocomponent in G[Y ] intersecting V (M) intersects Xt0 for some 0 ∗ (it∗ ,`) witness t ∈ ∂Tj,t∗ ∪ {t } for Xt0 ∩ Ij ⊆ W3 for some ` ∈ [−1, |V (T )|], and subject to this, it∗ is minimum. Let P be a c-monochromatic path in G intersecting V (M) ∩ Xt internally disjoint from

~~XV (Tt). Let uP be a vertex of P . If either uP ∈ XV (Tj,t∗ ) or σ(uP ) = σ(P ), then M contains uP .~~

73 Proof. Suppose this claim does not hold. Among all counterexamples, we choose (t, M, P, uP ) such that it is as small as possible. ∗ Suppose t = t . If V (P ) ⊆ XV (Tt), then since P is internally disjoint from XV (Tt), P is a c-monochromatic path in G[Xt], so P ⊆ M, a contradiction. So t is not the root of T . Let p be ∗ ∗ the parent of t. So V (P ) ∩ Xp 6= ∅. Hence p is a candidate for t , so t 6= t , a contradiction. ∗ So t 6= t . We may assume that uP is an end of P . Let v0 be a vertex in V (P ) ∩ V (M). We may assume that v0 6= uP , for otherwise we are done. (it∗ ,|V (G)|) By Claims 21.38 and 21.45, there uniquely exists a monochromatic Ej,t∗ -pseudocomponent ∗ (Y (it∗ ,|V (T )|+1,s+2)) ∗ M in G[Y ] intersecting V (M). By Claims 21.11 and 21.33, M contains v0.

Let C be the collection of the maximal subpaths of P contained in G[XV (Tj,t∗ )]. Denote C by {Pα : α ∈ [|C|]}. So for each α ∈ [|C|], there exist distinct vertices vα−1, uα ∈ V (P ) such that Pα is from vα−1 to uα. By renaming, we may assume that P passes through v0, u1, v1, u2, v2, . . . , u|C|, uP ∗ in the order listed. Note that for every α ∈ [|C| − 1], there exist tα ∈ ∂Tj,t∗ ∪ {t } and a subpath

Qα of P from uα to vα internally disjoint from Xtα such that V (Qα) − XV (Tj,t∗ ) 6= ∅. Since P is internally disjoint from XV (Tt), tα 6∈ V (Tt). ∗ We shall prove that for every α ∈ [|C| − 1], {uα, vα, u|C|} ⊆ V (M ) by induction on α. For each ∗ α ∈ [|C|], since Qα is in G[XV (Tj,t∗ )], we know if M contains vα−1, then by Claims 21.11 and 21.33, ∗ ∗ M contains uα. In particular, u1 ∈ V (M ). So it suffices to show that for every α ∈ [|C| − 1], if ∗ ∗ M contains uα, then M contains vα. ∗ Now we fix α to be an element in [|C| − 1], and assume that M contains uα. We first assume (i ∗ ) ∗ t (it∗ ,−1,0) tα = t . Let Zu and Zv be the monochromatic Ej,t∗ -pseudocomponents in G[Y ] containing uα and vα, respectively. Let vQα be the vertex of Qα with σ(vQα ) = σ(Qα). For each x ∈ {uα, vα}, let Qx be the subpath of Qα between x and vQα . Note that each Qx is internally disjoint from ∗ XV (Tt∗ ). Since it < it and we choose it to be minimal among all couterexamples of this claim, we ∗ ∗ know that both Zu and Zv contain vQα , so Zu = Zv. Since M contains uα, M contains Zu = Zv, ∗ so M contains vα. ∗ So we may assume tα ∈ ∂Tj,t − V (Tt). Hence Qα is a c-monochromatic path in G[XV (Tt∗ )] ∗ internally disjoint from XV (Tt), and vα is a vertex of P in XV (Tj,t∗ ). And Qα contains uα ∈ V (M ) ⊆ (it∗ ,|V (G)|) 0 V (M). Since (t, M) is a nice pair, there exists a monochromatic Ej,t∗ -pseudocomponent M (i ∗ ,|V (T )|+1,s+2) 0 0 in G[Y t ] such that V (M ) ∩ V (M) 6= ∅ and M contains vα. By the uniqueness of ∗ 0 ∗ ∗ M , M = M . So M contains vα. ∗ This shows that for every α ∈ [|C| − 1], {uα, vα, u|C|} ⊆ V (M ). Note that if uP ∈ XV (Tj,t∗ ), ∗ then uP = u|C| ∈ V (M ) ⊆ V (M), a contradiction. So uP 6∈ XV (Tj,t∗ ). Hence σ(uP ) = σ(P ), and ∗ there exists a subpath P of P from u|C| to uP internally disjoint from XV (Tt∗ ). Let M0 be the (i ∗ ) t (it∗ ,−1,0) ∗ monochromatic Ej,t∗ -pseudocomponent in G[Y ] containing u|C|. So V (P )∩V (M0)∩Xt∗ 6= ∗ ∅ and σ(uP ) = σ(P ). Since it∗ < it and we choose it to be minimal among all couterexamples of this claim, M0 contains uP . Since M contains u|C|, M ⊇ M0. So M contains uP . This proves the claim.

~~Claim 21.47. Let j ∈ [|V| − 1]. Let t ∈ V (T ). Let M1 and M2 be Sj-related monochromatic (it) (it,−1,0) Ej,t -pseudocomponents in G[Y ] intersecting Xt. Let P be a c-monochromatic path from~~

~~V (M1) to V (M2) internally disjoint from XV (Tt). Then M1 = M2.~~

~~Proof. Let vP be the vertex of P such that σ(vP ) = σ(P ). By Claim 21.46, both M1 and M2 contain vP . So M1 = M2.~~

74 Claim 21.48. Let j ∈ [|V| − 1]. Let t1, t2 ∈ V (T ) with t2 ∈ V (Tt1 ). For each α ∈ [2], let Mα be (i ) tα (itα ,−1,0) an Sj-related monochromatic Ej,tα -pseudocomponent in G[Y ] intersecting Xtα . Let P be a c-monochromatic path in G from V (M ) ∩ X to V (M ) internally disjoint from X . Then 2 t2 1 V (Tt2 ) M2 ⊇ M1.

~~Proof. Suppose to the contrary that M2 6⊇ M1. Among all counterexamples, we choose (t1, t2,M1,~~

M2,P ) such that it2 is as small as possible. By Claim 21.47, t2 ∈ V (Tt1 ) − {t1}. So by taking a subpath, we may assume that P is internally disjoint from V (M1). Let v1 be the end of P in M1, and let v2 be the end of P in V (M2) ∩ Xt2 . Note that v1 6∈ V (M2), for otherwise M ⊇ M . By Claim 21.47, v ∈ X . 2 1 1 V (Tt1 ) (i ∗ ) ∗ t2 Let t be the node of T with t2 ∈ V (Tj,t∗ ) such that some monochromatic E ∗ -pseudocomponent 2 2 j,t2 (it∗ ,|V (T )|+1,s+2) 0 ∗ in G[Y 2 ] intersecting V (M ) intersects X 0 for some witness t ∈ ∂T ∗ ∪ {t } for 2 t j,t2 2 (i ∗ ,`) t2 X 0 ∩ I ⊆ W for some ` ∈ [−1, |V (T )|], and subject to this, i ∗ is minimum. t j 3 t2 Suppose v1 ∈ XV (T ∗ ). Since v1 6∈ V (M2), by Claim 21.46, v1 6∈ XV (T ∗ ). So there exists t2 j,t2 q ∈ ∂T ∗ − V (T ) such that v ∈ X − X , and there exists a path Q in M from v to X j,t2 t2 1 V (Tq) q 1 1 q internally disjoint from Xq. If Q is a path in G, then by applying Claim 21.46 to the path P ∪ Q, we know M2 contains V (Q) ∩ Xq, so M2 ⊇ M1, a contradiction. So Q is not a path in G. Hence ∗ 0 (iq∗ ,|V (G)|) there exist q ∈ V (T ) with T ∗ ⊂ T ∗ , q ∈ ∂T ∗ ∩ X − X and e ∈ E(Q) ∩ E ∗ 6= ∅ j,t2 j,q j,q V (Tq) q j,q 0 (iq∗ ,`) such that e ⊆ Xq0 and q is a witness for Xq0 ∩ Ij ⊆ W3 for some ` ∈ [0, |V (T )|]. We may 0 0 assume that there exists a subpath Q of Q from V (Q)∩Xq to an element in e with E(Q ) ⊆ E(G). So i ∗ < i ∗ . Since v ∈ X − X , there exists a subpath P of P ∪ Q from a vertex u in X q t2 1 V (Tq) q 1 1 q 0 to an element of e internally disjoint from XV (Tq) with E(P1) ⊆ E(G). So V (P1) ∩ Xq 6= ∅. Since u1 ∈ XV (T ∗ ), u1 ∈ V (M2) by Claim 21.46. By the existence of P1 and Claims 21.11 and 21.33, j,t2 ∗ ∗ V (M ) ∩ X 0 6= ∅, so q is a candidate for t . But i ∗ < i ∗ , a contradiction. 2 q 2 q t2 ∗ ∗ Hence v1 6∈ XV (T ∗ ). By the existence of P , the parent of t2 is a candidate for t2, so t2 6= t2. Let t2 0 0 P be the maximal subpath of P from v to X ∗ . Let u be the end of P other than v if possible. 2 t2 2 2 Since u2 ∈ Xt∗ ⊆ XV (T ∗ ), by Claim 21.46, M2 contains u2. Let Mu be the monochromatic 2 j,t2 (i ∗ ) t2 (it∗ ,−1,0) 00 E ∗ -pseudocomponent in G[Y 2 ] containing u2. So there exists a subpath P of P from u2 j,t2 ∗ ∗ to v1 internally disjoint from XV (T ∗ ). Since v1 ∈ XV (Tt ) − XV (T ∗ ), t2 ∈ V (Tt1 ). Since it < it2 , t2 1 t2 2 the minimality of it2 among all counterexamples implies that Mu ⊇ M1. Since M2 contains u2 ∈ V (Mu), M2 ⊇ Mu ⊇ M1, a contradiction. This proves the claim.

~~A parade (t1, t2, . . . , tk) is tamed if for every α ∈ [k − 1], tα+1 ∈ ∂Tj,tα .~~

Claim 21.49. Let j ∈ [|V| − 1]. Let κ = κ1. Let (t1, t2, . . . , tκ) be a parade that is a subse- (β) (β) (β) (β) quence of a tamed parade. For every β ∈ [κ], let p = (p1 , p2 , . . . , p|V (G)|) be the (itβ , −1, 0, j)- (1) pseudosignature. Let α ∈ [|V (G)|] such that pα is not a zero sequence. Let M1 be the monochro- (it1 ) (i ,−1,0) (1) matic E -pseudocomponent in G[Y t1 ] deﬁning p . For every β ∈ [2, κ], let M be the j,t1 α β (itβ ) (i ,−1,0) monochromatic E -pseudocomponent in G[Y tβ ] containing M . For every α0 ∈ [α − j,tβ 1 (1) (1) (it1 ) 1] in which p 0 is a nonzero sequence, let Q 0 be the monochromatic E -pseudocomponent α α j,t1 (it ) (it ,−1,0) (1) (β) β in G[Y 1 ] deﬁning p 0 , and for every β ∈ [2, κ], let Q 0 be the monochromatic E - α α j,tβ (1) (1) (itβ ,−1,0) 0 pseudocomponent in G[Y ] containing Qα0 . Assume that for every α ∈ [α − 1] in which pα0 (1) (κ) is a nonzero sequence, V (Qα0 ) = V (Qα0 ). If V (Mβ+1) 6= V (Mβ) for every β ∈ [κ − 1], then there

75 ∗ (κ) (1) (κ) (1) exists α ∈ [α − 1] such that pα∗ is lexicographically smaller than pα∗ , and pβ = pβ for every β ∈ [α∗ − 1] Proof. Suppose to the contrary that this claim does not hold. We may assume that among all counterexamples, α is minimal. 0 (1) (1) (κ) Since for every α ∈ [α−1] in which pα0 is a nonzero sequence, V (Qα0 ) = V (Qα0 ), we know for (β) (1) 0 every β ∈ [κ], V (Qα0 ) = V (Qα0 ). So, together with Claim 21.32, for every α ∈ [α − 1], β ∈ [κ − 1] (β+1) (β) and γ ∈ [|V (G)|], the γ-th entry of pα0 is at most the γ-th entry of pα . Hence if there exist 0 (β) (1) (κ) (1) β ∈ [2, κ] and α ∈ [α − 1] such that pα0 6= pα0 , then pα0 is lexicographically smaller than pα0 and we are done if we further choose α0 to be minimum. 0 (β) (1) 0 So for every β ∈ [κ] and α ∈ [α − 1], pα0 = pα0 . Hence for every β ∈ [κ] and α ∈ [α − 1] (1) (β) (β) (1) in which p 0 is a nonzero sequence, Q 0 deﬁnes p 0 , and A (i ,−1,0) (V (Q 0 )) ∩ X − X = α α α L t1 α V (Tt1 ) t1 (β) 0 (1) A (it ,−1,0) (V (Q 0 ))∩XV (T ) −Xt ⊆ XV (T ) −Xt . For every α ∈ [α−1], since V (Q 0 )∩Xt 6= ∅, L β α tβ β tκ κ α 1 (1) (κ) we know A (i ,−1,0) (V (Q 0 )) ∩ X − X = A (i ,−1,0) (V (Q 0 )) ∩ X − X ⊆ X − L t1 α V (Tt1 ) t1 L tκ α V (Ttκ ) tκ V (Ttκ ) ◦ (Xtκ ∪ Zt1 ) ⊆ (XV (Ttκ ) − Xtκ ) ∩ Ij . Since (t1, . . . , tκ) is a subsequence of a tamed parade, for every 0 (β) β ∈ [κ − 1] and α ∈ [α − 1], (A (it ,−1,0) (V (Q 0 )) ∩ XV (T ) − Xt ) ∩ Zt = ∅. L β α tβ β β 0 (β) ◦ Let κ = κ0 + w0 + 3. Since A (it ,−1,0) (V (Q 0 )) ∩ XV (T ) − Xt ⊆ (XV (T ) − Xt ) ∩ I for every L β α tβ β tκ κ j β ∈ [κ − 2] and α0 ∈ [α − 1], we know that for any distinct β , β ∈ [κ] with t ∈ V (T ) − {t }, 1 2 β2 tβ1 β1 if X ∩ I ⊆ X ∩ I , then since M 6= M , some element in E(M ) − E(M ) is a fake edge tβ2 j tβ1 j β2 β1 β2 β1 contained in X ∩ I by the rules of adding fake edges. So for every β ∈ [κ], there are at most tβ1 j w0 elements β0 ∈ [β + 1, κ] such that X ∩ I ⊆ X ∩ I . Hence there exists a subsequence 2 tβ0 j tβ j 2 2 (tj1 , tj2 , ...) of (t1, t2, ..., tκ) of length at least κ/w0 such that for any distinct β1, β2 ∈ [dκ/w0e] with 2 2 0 β1 < β2, Xtj ∩ Ij 6⊆ Xtj ∩ Ij. Since κ/w0 κ1/w0 N18(w0, κ ) and (t1, t2, . . . , tκ) is a parade β2 β1 > > 0 0 ∗ 0 0 0 with |Xtβ ∩ Ij| 6 w0 for every β ∈ [κ], by Lemma 18, there exists a (t1, tκ0 , ` )-fan (t1, t2, . . . , tκ0 ) 0 ∗ in (T, X |G[Ij ]) of size κ (for some ` ∈ [0, w0]) that is a subsequence of (t1, t2, . . . , tκ) such that 0 X 0 ∩ I − X 0 6= ∅ for every β ∈ [κ ]. tβ j t1 0 0 0 Since (t1, t2, . . . , tκ0 ) is a subsequence of (t1, t2, . . . , tκ), for simplicity, by removing some nodes 0 0 0 in (t1, t2, . . . , tκ), we may assume (t1, t2, . . . , tκ0 ) = (t1, t2, . . . , tκ0 ). ∗ 0 (β∗) Let β ∈ [κ −κ0 −2]. Since V (Mβ∗+1) 6= V (Mβ∗ ) and A (it ,−1,0) (V (Q 0 ))∩XV (T ) −Xt ∗ = L β∗ α tβ∗ β (β∗+1) ◦ 0 A (it ,−1,0) (V (Q 0 )) ∩ XV (T ) − Xt ∗ ⊆ (XV (T ) − Xt ) ∩ I for every α ∈ [α − 1], there L β∗+1 α tβ∗+1 β +1 tκ κ j (i ) tβ∗+1 exists e ∗ ∈ E(M ∗ ) ∩ E − E(M ∗ ) such that |e ∩ V (M ∗ )| = 1. So there exists z ∈ V (T ) β β +1 j,tβ∗+1 β β (iz,|V (G)|) (iz) ∗ with Ttβ∗+1 ⊆ Tz ⊆ Ttβ∗ and itβ∗ 6 iz < itβ∗+1 such that eβ ∈ Ej,z − Ej,z . Hence there exists (iz,`+1) (iz,`) 0 ` ∈ [0, |V (G)| − 1] such that eβ∗ ∈ Ej,z − Ej,z . Let t ∈ ∂Tj,z be the witness for Xt0 ∩ Ij ⊆ 0 (iz,` ) 0 ∗ 0 ∗ ∗ 0 W3 for some ` ∈ [0, |V (T )|] such that eβ ⊆ Xt . Since V (Mβ ) ∩ Xt1 6= ∅ 6= V (Mβ ) ∩ Xt , V (Mβ∗ ) ∩ Xz 6= ∅. (iz,`) Denote eβ∗ by {u, v}. Let Mu,Mv be the monochromatic Ej,z -pseudocomponents in (iz,|V (T )|+1,s+2) ◦ (iz,`) G[Y ] containing u, v, respectively. Let M be the Sj -related monochromatic Ej,z - pseudocomponent in G[Y (iz,|V (T )|+1,s+2)] such that σ(M) is the (` + 1)-th smallest among all (iz,`) (iz,|V (T )|+1,s+2) monochromatic Ej,z -pseudocomponents in G[Y ]. So σ(M) < min{σ(Mu), σ(Mv)} ◦ and M is Sj -related. Since Mβ∗ contains u or v, Mβ∗ is contained in one of Mu and Mv. So σ(M) < σ(Mβ∗ ). ∗ 0 0 0 Since β 6 κ − κ0 − 2 and (t1, . . . , tκ ) is a subsequence of a tamed parade, tκ −κ0 6∈ V (Tj,z).

76 0 (1) (z) (iz) For every α ∈ [α − 1] in which pα0 is a nonzero sequence, let Qα0 be the monochromatic Ej,z - (iz,−1,0) (1) 0 pseudocomponent in G[Y ] containing Q 0 . Since iz < it 0 , for every α ∈ [α − 1], α κ −κ0−1 (z) (1) A (iz,|V (T )|+1,s+2) (V (Q 0 )) ∩ XV (T ) − Xz ⊆ XV (T ) − (Xt 0 ∪ Zz). By Claim 21.31, since Q 0 = L α z tκ0 κ α (z) (κ) 0 0 (z) 0 (1) 0 0 Qα0 = Qα0 for every α ∈ [α − 1], we know tκ −κ0 ∈ V (Tt ) − {t } and {Qα0 : α ∈ [α − 1], pα0 (iz,`) 0 is a nonzero sequence} contains the set of the monochromatic Ej,z -pseudocomponents M in (iz,|V (T )|+1,s+2) 0 (iz,`) (iz) G[Y ] with σ(M ) 6 σ(M). So Ej,z − Ej,z ⊆ Xt0 . ∗ 0 0 Since (t , t , . . . , t 0 ) is a (t , t 0 , ` )-fan in (T, X | ), for every κ −κ β < β κ , X ∩I − 1 2 κ 1 κ G[Ij ] 0 6 1 2 6 tβ1 j X 0 and X ∩I −X 0 are disjoint non-empty sets. Since (V (M )∪V (M ))∩X ⊇ V (M ∗ )∩X 6= ∅ t tβ2 j t u v z β z (iz,`+1) (iz,`) ∗ 0 0 0 and eβ ∈ Ej,z − Ej,z , (tκ −(ψ2(w0,w0)+ψ3(w0,w0))+1, tκ −(ψ2(w0,w0)+ψ3(w0,w0))+2, ..., tκ ) is not an (` + 1, w0, w0, ψ2(w0, w0), ψ3(w0, w0))-blocker, so for every x ∈ {u, v},

~~(i ,|V (T )|+1,s+2) AL z (V (Mx)) ∩ (XV (Tz) − Xz) ∩ XV (Tt 0 ) − Xtκ0−κ +1 κ −κ0+1 0~~

~~(i ,|V (T )|+1,s+2) = AL z (V (Mx)) ∩ (XV (Tz) − Xz) ∩ XV (Tt 0 ) − Xtκ0−(ψ (w ,w )+ψ (w ,w ))+1 κ −(ψ2(w0,w0)+ψ3(w0,w0))+1 2 0 0 3 0 0 6= ∅,~~

~~and hence V (Mx) ∩ Xt 0 ∩ Ij 6= ∅. Since |eβ∗ ∩ V (Mβ∗ )| = 1, there exists x ∈ {u, v} such that κ −κ0~~

(V (Mβ∗+1) − V (Mβ∗ )) ∩ Xt 0 ∩ Ij ⊇ V (Mx) ∩ Xt 0 ∩ Ij 6= ∅. κ −κ0 κ −κ0 ∗ 0 0 Since β is an arbitrary element in [κ − κ0 − 2], we know |Xt 0 ∩ Ij| κ − κ0 − 2 w0 + 1, κ −κ0 > > a contradiction. This proves the claim.

∗ Claim 21.50. Let j ∈ [|V| − 1]. Let γ ∈ [w0] and κ ∈ N. Let (t1, t2, . . . , tκ) be a parade that (β) (β) (β) (β) is a subsequence of a tamed parade. For every β ∈ [κ], let p = (p1 , p2 , . . . , p|V (G)|) be the (1) (itβ , −1, 0, j)-pseudosignature. Let α ∈ [|V (G)|] such that pα is not a zero sequence, and α is the (i ) γ∗-th smallest index ` such that p(1) is a nonzero sequence. Let M be the monochromatic E t1 - ` 1 j,t1 (it ,−1,0) (1) pseudocomponent in G[Y 1 ] deﬁning pα . For every β ∈ [2, κ], let Mβ be the monochromatic (it ) (1) β (itβ ,−1,0) 0 E -pseudocomponent in G[Y ] containing M . For every α ∈ [α − 1] in which p 0 is a j,tβ 1 α (i ) (1) t1 (it ,−1,0) nonzero sequence, let Q 0 be the monochromatic E -pseudocomponent in G[Y 1 ] deﬁning α j,t1 (1) (β) (it ) β (itβ ,−1,0) p 0 , and for every β ∈ [2, κ], let Q 0 be the monochromatic E -pseudocomponent in G[Y ] α α j,tβ (1) containing Qα0 . (β) (β) Assume (p1 , . . . , pα−1) is identical for every β ∈ [κ]. If V (Mβ+1) 6= V (Mβ) for every β ∈ [κ], ∗ then κ 6 φ1(γ ). ∗ ∗ Proof. We shall prove this claim by induction on γ . When γ = 1, since φ1(1) = κ1, this claim ∗ ∗ follows from Claim 21.49. So we may assume γ > 2 and this claim holds for smaller γ . ∗ Suppose to the contrary that κ > φ1(γ ). Let Z = {z1, z2, . . . , zγ∗ } be the set of elements in ∗ 0 (1) [α] such that for each ` ∈ [γ ], z` is the `-th smallest index ` such that p`0 is a nonzero sequence. 0 ∗ (β+1) (β) 0 For every α ∈ [γ ], let Sα = {i ∈ [κ − 1] : V (Qzα0 ) 6= V (Qzα0 )}. 0 ∗ 0 Suppose that for every α ∈ [γ − 1], |Sα0 | 6 φ1(α ). Then there exists a set R ⊆ [κ − 1] κ−1 κ−1 of consecutive positive integers with |R| γ∗−1 γ∗−1 κ1 such that R ∩ > 1+P |S 0 | > 1+P φ (α0) > α0=1 α α0=1 1 Sγ∗−1 α0=1 Sα0 = ∅. But since |R| > κ1 and V (Mβ+1) 6= V (Mβ) for every β ∈ R, Claim 21.49 implies ∗ (a) (b) that there exists α ∈ [α − 1] such that pα∗ 6= pα∗ for some distinct a, b ∈ R ⊆ [κ], a contradiction.

77 ∗ 0 So there exists the smallest element ξ in [γ −1] such that |Sξ| > φ1(ξ)+1. Since for every α ∈ 0 ∗ ∗ Pξ−1 0 [ξ −1], |Sα0 | 6 φ1(α ), there exists a set R ⊆ Sξ ⊆ [κ−1] with |R | > |Sξ|/(1+ α0=1 φ1(α )) > κ1 ∗ Sξ−1 ∗ Sξ−1 such that R ∩ α0=1 Sα0 = ∅ and there exist no β1 < β2 < β3 with β1, β3 ∈ R and β2 ∈ α0=1 Sα0 . ∗ (β+1) (β) ∗ But since |R | > κ1 and V (Qzξ ) 6= V (Qzξ ) for every β ∈ R , Claim 21.49 implies that there ∗ (a) (b) ∗ exists α ∈ [α − 1] such that pα∗ 6= pα∗ for some distinct a, b ∈ R , a contradiction. This proves the claim.

∗ Claim 21.51. Let j ∈ [|V| − 1]. Let γ ∈ [w0] and κ ∈ N. Let (t1, t2, . . . , tκ) be a parade that (β) (β) (β) (β) is a subsequence of a tamed parade. For every β ∈ [κ], let p = (p1 , p2 , . . . , p|V (G)|) be the (1) (itβ , −1, 0, j)-pseudosignature. Let α ∈ [|V (G)|] such that pα is not a zero sequence, and α is the ∗ (1) 0 (1) γ -th smallest index ` such that p` is a nonzero sequence. For every α ∈ [α] in which pα0 is a (i ) (1) t1 (it ,−1,0) nonzero sequence, let Q 0 be the monochromatic E -pseudocomponent in G[Y 1 ] deﬁning α j,t1 (1) (β) (it ) β (itβ ,−1,0) p 0 , and for every β ∈ [2, κ], let Q 0 be the monochromatic E -pseudocomponent in G[Y ] α α j,tβ (1) (β+1) (β) containing Qα0 . If for every β ∈ [κ], there exists αβ ∈ [α] such that V (Qαβ ) 6= V (Qαβ ), then ∗ κ 6 φ3(γ ). ∗ Proof. Suppose to the contrary that κ > φ3(γ ) + 1. Let Z = {z1, z2, . . . , zγ∗ } be the subset of [α] 0 ∗ 0 (1) such that for each α ∈ [γ ], zα0 is the α -th smallest index ` such that p` is a nonzero sequence. 0 ∗ (i) (i−1) 0 For every α ∈ [γ ], let Sα = {i ∈ [2, κ]: pzα0 6= pzα0 }. ∗ Suppose there exists ξ ∈ [γ ] such that |Sξ| > φ2(ξ) + 1. We choose ξ to be as small as |Sξ| φ2(ξ)+1 possible. So there exists a subset R of Sξ with |R| ξ−1 ξ−1 (φ1(ξ) + 1) · > 1+P |S 0 | > 1+P φ (α0) > α0=1 α α0=1 2 Sξ−1 w0f(η5) + 1 such that R ∩ α0=1 Sα0 = ∅ and there exist no distinct elements a, b ∈ R such that ξ−1 0 (β) (β−1) ξ−1 S 0 S 0 [a, b] ∩ α0=1 Sα 6= ∅. Let Sξ = {β ∈ R : V (Qzξ ) 6= V (Qzξ )}. Since R ∩ α0=1 Sα = ∅ and there Sξ−1 0 exist no distinct elements a, b ∈ R such that [a, b] ∩ α0=1 Sα0 6= ∅, |Sξ| 6 φ1(ξ) by Claim 21.50. So 0 0 |R| 0 0 there exists a subset R of R with |R | 0 w0f(η5) + 1 such that R ∩ Sξ = ∅ and for any > |Sξ|+1 > 0 0 0 0 (β) (β−1) distinct elements a < b of R , [a, b]∩Sξ = ∅. Since for every β ∈ R , β ∈ R−Sξ, so pzξ 6= pzξ and (β) (β−1) (β) (β−1) 0 V (Qzξ ) = V (Qzξ ), and hence pzξ < pzξ . So |R | 6 w0f(η5) by Claim 21.30, a contradiction. 0 ∗ 0 00 Therefore, for every α ∈ [γ ], |Sα0 | 6 φ2(α ). Hence there exists a subset R of [2, κ] with

~~∗ 00 κ − 1 κ − 1 φ3(γ ) |R | ∗ ∗ ∗ > Pγ > Pγ 0 > Pγ 0 1 + α0=1|Sα0 | 1 + α0=1 φ2(α ) 1 + α0=1 φ2(α )~~

00 00 Sγ∗ ∗ 00 such that R consists of consecutive integers and R ∩ α0=1 Sα0 = ∅. So there exists R ⊆ R ∗ 00 ∗ ∗ ∗ ∗ with |R | > |R |/γ > φ1(γ ) + 1 such that there exists x ∈ Z such that for every x ∈ R , (x+1) (x) V (Qx∗ ) 6= V (Qx∗ ). But it contradicts Claim 21.50. This proves the claim.

~~◦ Claim 21.52. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∈ Sj . Let ∗ ∗ t ∈ K (M) such that K (M) ∩ V (Tt) − {t}= 6 ∅. Then:~~

~~(it) 0 (it,−1,0) 0 • There exists a monochromatic Ej,t -pseudocomponent M in G[Y ] such that V (M ) ∩ 0 0 Xt 6= ∅= 6 V (M ) ∩ V (M) and AL(it,−1,0) (V (M )) ∩ XV (Tt) − Xt 6= ∅.~~

~~(it) 0 (it,−1,0) 0 • There exists a monochromatic Ej,t -pseudocomponent M in G[Y ] such that V (M ) ∩ 0 Xt 6= ∅ and M contains vM , where vM is the vertex of M with σ(vM ) = σ(M).~~

~~78 (it) 0 (it,−1,0) 0 • For every monochromatic Ej,t -pseudocomponent M in G[Y ] with V (M ) ∩ Xt 6= ∅= 6 0 0 V (M ) ∩ V (M), AL(it,−1,0) (V (M )) ∩ XV (Tt) − Xt 6= ∅.~~

∗ 0 ∗ Proof. Since K (M) ∩ V (Tt) − {t}= 6 ∅, there exists t ∈ K (M) ∩ XV (Tt) − {t}. So there exists v ∈ V (M) ∩ Y (it0 ,|V (T )|+1,s+2) ∩ X − Y (it,|V (T )|+1,s+2). Hence v 6∈ Y (it,−1,0). Let P be a path in V (Tt0 ) ∗ M from v to vM , where vM is the vertex of M with σ(vM ) = σ(M). Since t ∈ K (M), t ∈ V (TrM ). 0 So V (P ) ∩ Xt 6= ∅ and there exists a subpath P of P from v to Xt internally disjoint from (it,−1,0) (it) 0 (it,−1,0) Xt. Since v 6∈ Y , there exists a monochromatic Ej,t -pseudocomponent M in G[Y ] 0 0 0 such that V (M ) ∩ Xt 6= ∅= 6 V (M ) ∩ V (P ) and AL(it,−1,0) (V (M )) ∩ XV (Tt) − Xt 6= ∅. Since P ⊆ M, V (M 0) ∩ V (M) 6= ∅. This proves the ﬁrst statement of this claim. In addition, since 00 00 V (P ) ∩ Xt 6= ∅, there exists a subpath P from vM to a vertex u in Xt internally disjoint from 00 (it) (it,−1,0) 00 XV (Tt). Let M be the monochromatic Ej,t -pseudocomponent in G[Y ] containing u . So 00 00 P is a c-monochromatic path in G intersecting V (M ) ∩ Xt internally disjoint from XV (Tt). Since 00 00 σ(vM ) = σ(M) and P ⊆ M, σ(vM ) = σ(P ). By Claim 21.46, M contains vM . This proves the second statement of this claim. S 000 Now we prove the third statement of this claim. Let Z1 = M 000 V (M ), where the union (it) (it,−1,0) 000 is over all monochromatic Ej,t -pseudocomponents in G[Y ] such that V (M ) ∩ Xt 6= ∅= 6 000 000 V (M )∩V (M) and AL(it,−1,0) (V (M ))∩XV (Tt)−Xt 6= ∅. By the ﬁrst statment of this claim, Z1 6= ∅. S 000 (it) Let Z2 = M 000 V (M ), where the union is over all monochromatic Ej,t -pseudocomponents in (it,−1,0) 000 000 000 G[Y ] such that V (M )∩Xt 6= ∅= 6 V (M )∩V (M) and AL(it,−1,0) (V (M ))∩XV (Tt) −Xt = ∅. 0 Suppose Z2 6= ∅. There exists a path P in M from Z2 to Z1 internally disjoint from Z1 ∪ Z2. Note 0 0 that the neighbor u of the vertex in V (P ) ∩ Z2 in P belongs to AL(it,−1,0) (Z2). So u 6∈ XV (Tt) − Xt. 0 Hence there exists a subpath of P from Z2 to Z1 internally disjoint from Z1 ∪ Z2 ∪ XV (Tt). Let (it) (it,−1,0) Q1 be the monochromatic Ej,t -pseudocomponents in G[Y ] such that V (Q1) ⊆ Z1 and Q1 0 0 contains the vertex in Z1 ∩ V (P ). Since P is internally disjoint from XV (Tt), by Claim 21.47, (it) (it,−1,0) Q1 equals the monochromatic Ej,t -pseudocomponents in G[Y ] containing V (P ) ∩ Z2. So V (Q1) ⊆ Z1 ∩Z2, a contradiction. Hence Z2 = ∅. This proves the third statement of this claim.

~~◦ For any i ∈ [0, |V (T )| − 1], j ∈ [|V| − 1] and c-monochromatic component M with SM ∈ Sj , the (i, j, M)-truncation is the sequence (b1, . . . , b|V (G)|) deﬁned as follows.~~

• Let (p1, p2, . . . , p|V (G)|) be the (i, −1, 0, j)-pseudosignature. ∗ • Let α be the largest index γ such that pγ is a nonzero sequence and the monochromatic (it) (i,−1,0) Ej,t -pseudocomponent (where t is the node of T with it = i) in G[Y ] defining pγ intersects M. (If no such index γ exists, then define α∗ = 0.) ∗ • For every α ∈ [α ], define bα := pα. ∗ • For every α ∈ [|V (G)|] − [α ], define bα := 0.

◦ Claim 21.53. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∈ Sj . Let ∗ (t1, t2, . . . , tη5+3) be a parade such that tα ∈ K (M) for every α ∈ [η5 + 3]. Then there exists ∗ α ∈ [2, η5 + 2] such that either

• the (itα∗ , j, M)-truncation is diﬀerent from the (it1 , j, M)-truncation, or (i ) ∗ tα (itα ,−1,0) • for each α ∈ {1, α }, there exists a monochromatic Ej,tα -pseudocomponent Mα in G[Y ] (i ,−1,0) with V (Mα) ∩ Xtα 6= ∅ 6= V (Mα) ∩ V (M) and AL tα (V (Mα)) ∩ XV (Ttα ) − Xtα 6= ∅ such that M1 ⊆ Mα∗ and V (M1) 6= V (Mα∗ ).

79 (itα ) 0 Proof. For every α ∈ [η5 + 2], let Cα = {V (Q): Q is a monochromatic Ej,tα -pseudocomponent M (it ,−1,0) 0 0 0 α (i ,−1,0) in G[Y ] with V (M )∩Xtα 6= ∅= 6 V (M )∩V (M) and AL tα (V (M ))∩XV (Ttα )−Xtα 6= ∅}.

Suppose to the contrary that this claim does not hold. So for every α ∈ [η5 + 2], the (itα , j, M)- truncation equals the (it1 , j, M)-truncation. Hence, for every α ∈ [η5 +2] and every Qα ∈ Cα, there 0 0 0 exists Q1 ∈ C1 such that σ(Qα) = σ(Q1), so Qα ⊇ Q1. Therefore for every α ∈ [η5 + 2], Cα = C1, for otherwise second outcome of this claim holds. ∗ (itα ,|V (T )|+1,s+2) For every α ∈ [2, η5+2], since tα ∈ K (M), there exists vα ∈ V (M)∩XV (Ttα )∩Y − (it ,|V (T )|+1,s+2) Y α−1 . So {vα : α ∈ [2, η5 + 2]} is a set of η5 + 1 diﬀerent vertices. By Claim 21.25, ∗ ∗ there exists α ∈ [2, η5 + 2] such that vα 6∈ XV (Tt ). η5+2 Let vM be the vertex of M such that σ(vM ) = σ(M). Let P be a path in M from vα∗ to vM . ∗ 0 ∗ ∗ Since tα −1 ∈ K (M), V (P )∩Xtα∗−1 6= ∅. Let P be the subpath of P from vα to Xtα∗−1 internally disjoint from Xtα∗−1 . (it ∗ ) (i ,−1,0) Let Q be the monochromatic E α −1 -pseudocomponent in G[Y tα∗−1 ] containing the j,tα∗−1 (i ,−1,0) 0 tα∗−1 0 vertex in V (P ) ∩ Xt ∗ . Since vα∗ 6∈ Y , A (it ,−1,0) (V (Q)) ∩ V (P ) − Xt ∗ 6= α −1 L α∗−1 α −1 0 ∗ ∗ ∅. Since P is internally disjoint from Xtα∗−1 , V (Q) ∈ Cα −1. Since Cη5+2 = C1 = Cα −1, V (Q) ∈ C . Since both the (i , j, M)-truncation and (i , j, M)-truncation equal the η5+2 tη5+2 tα∗−1

(it1 , j, M)-truncation, A (it ∗ ,−1,0) (V (Q)) ∩ XV (Tt ) − Xt ∗ ⊆ XV (Tt ) − Xtη +2 . Hence L α −1 α∗−1 α −1 η5+2 5 0 00 0 ∗ V (P ) ∩ XV (Tt ) − Xtη +2 6= ∅. Since vα 6∈ XV (Tt ), there exists a subpath P of P from η5+2 5 η5+2 v ∗ to X internally disjoint from X . α tη5+2 tη5+2 (i ) tη +2 (it ,−1,0) 0 5 η5+2 Let Q be the monochromatic E -pseudocomponent in G[Y ] containing Xtη +2 ∩ j,tη5+2 5 00 ∗ 0 V (P ). Since tη5+3 ∈ K (M) ∩ XV (Tt ) − {tη5+2}, by Claim 21.52, A (it ,−1,0) (V (Q )) ∩ η5+2 L η5+2 0 (it ∗ ,−1,0) 0 ∗ α −1 ∗ XV (Tt ) − Xtη +2 6= ∅. So V (Q ) ∈ Cη5+2 = Cα −1. Since vt ∗ 6∈ Y and V (Q ) ∈ Cα −1, η5+2 5 α 0 0 00 0 00 vt ∗ 6∈ V (Q ). So A (it ,−1,0) (V (Q )) ∩ V (P ) 6= ∅. Let u ∈ A (it ,−1,0) (V (Q )) ∩ V (P ). Since α L α∗−1 L α∗−1 (i ,−1,0) u 6∈ Y tα∗−1 , u 6∈ X . Since both the (i , j, M)-truncation and (i , −1, 0)-truncation tα∗−1 tη5+2 tα∗−1 0 equal the (it1 , j, M)-truncation, A (it ∗ ,−1,0) (V (Q )) ∩ XV (Tt ) − Xt ∗ ⊆ XV (Tt ) − Xtη +2 . L α −1 α∗−1 α −1 η5+2 5 00 0 Since V (P ) ⊆ XV (Tt ), u ∈ A (it ∗ ,−1,0) (V (Q )) ∩ XV (Tt ) − Xt ∗ ⊆ XV (Tt ) − Xtη +2 . α∗−1 L α −1 α∗−1 α −1 η5+2 5 00 But P is internally disjoint from XV (Tt ), a contradiction. This proves the claim. η5+2 ◦ Claim 21.54. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∈ Sj . Let t , t ∈ K∗(M) such that t ∈ V (T ) − {t } and K∗(M) ∩ X − X 6= ∅. Let α∗ be the largest 1 2 2 t1 1 V (Tt2 ) t2 index ` such that the `-th entry of the (it1 , j, M)-truncation is not a zero sequence. For α ∈ [2], let (α) (α) ∗ (1) (2) (q1 , . . . , q|V (G)|) be the (itα , j, M)-truncation. Let α ∈ [0, α − 1]. If qβ = qβ for every β ∈ [α], then either

∗ ∗ (2) • there exists β ∈ [α + 1, α ] such that qβ∗ is not a zero sequence, or ∗ (1) (2) • there exists β ∈ [α] such that qβ∗ and qβ∗ are nonzero sequences and the vertex-set of (i ) t2 (it ,−1,0) (2) the monochromatic E -pseudocomponent in G[Y 2 ] defining q ∗ is different from the j,t2 β (i ) t1 (it ,−1,0) (1) vertex-set of the monochromatic E -pseudocomponent in G[Y 1 ] defining q ∗ . j,t1 β

∗ (2) Proof. Suppose this claim does not hold. So for every β ∈ [α + 1, α ], qβ is a zero sequence. And ∗ (1) (2) for every β ∈ [α], if qβ∗ and qβ∗ are nonzero sequences, then the vertex-set of the monochromatic (i ) t2 (it ,−1,0) (2) E -pseudocomponent in G[Y 2 ] deﬁning q ∗ equals the vertex-set of the monochromatic j,t2 β (i ) t1 (it ,−1,0) (1) E -pseudocomponent in G[Y 1 ] deﬁning q ∗ . j,t1 β

80 Let vM be the vertex of M such that σ(vM ) = σ(M). For every k ∈ [2], by Claim 21.52, (itk ) (i ,−1,0) there exists a monochromatic E -pseudocomponent M in G[Y tk ] with V (M ) ∩ X 6= ∅ j,tk k k tk containing v . By Claim 21.52, for each k ∈ [2], A (i ,−1,0) (V (M )) ∩ X − X 6= ∅, so there M L tk k V (Ttk ) tk ∗ (k) (k) exists γ ∈ [σ(vM )] such that q ∗ is not a zero sequence and Mk defines q ∗ . k γk γk ∗ ∗ ∗ ∗ ∗ By the definition of α , γ1 ∈ [α ]. Since vM ∈ V (M1) ∩ V (M2), M1 ⊆ M2. So γ2 6 γ1 . Since (2) ∗ ∗ ∗ ∗ ∗ ∗ q ∗ is not a zero sequence, γ 6∈ [α + 1, α ]. Since γ γ α , γ ∈ [α]. In particular, α 1. γ2 2 2 6 1 6 2 > (1) Since vM ∈ V (M2) and V (M2) ∩ Xt 6= ∅, V (M2) ∩ Xt 6= ∅. So by Claim 21.31, q ∗ is not a zero 2 1 γ2 sequence. (1) For every γ ∈ [|V (G)|], if qγ is not a zero sequence, then let Qγ be the monochromatic (it1 ) (i ,−1,0) (1) E -pseudocomponent in G[Y t1 ] defining q ; otherwise, let V (Q ) = ∅. j,t1 γ γ Sα (1) ∗ ∗ Let Z1 = V (Qγ). Since q ∗ is not a zero sequence and γ α, Z1 6= ∅. Since γ ∈ [α], γ=1 γ2 2 6 2 V (Q ∗ ) = V (M ). Since M contains v , Q ∗ contains v . Hence Z ∩ V (M) 6= ∅. γ2 2 2 M γ2 M 1 α∗ (1) (1) S ∗ Let Z2 = γ=α+1(V (Qγ)). Since (q1 , . . . , q|V (G)|) is the (it1 , j, M)-truncation, V (Qα ) ∩ V (M) 6= ∅. So Z2 ∩ V (M) 6= ∅. Since Z1 ∩ V (M) 6= ∅= 6 Z2 ∩ V (M), there exists a path P in M from Z1 to Z2 internally disjoint from Z ∪ Z . By Claims 21.47 and 21.52, P is contained in G[X ] and is internally 1 2 V (Tt1 ) ∗ ∗ ∗ disjoint from Xt1 . Let β be the index in [α] such that Qβ contains V (P ) ∩ Z1. Since β 6 α, (2) (1) (it2 ) ∗ q ∗ = q ∗ is not a zero sequence. So there exists a monochromatic E -pseudocomponent Q in β β j,t2 (it ,−1,0) (2) ∗ ∗ (1) (2) G[Y 2 ] defining q ∗ . Since β ∈ [α], V (Q ) = V (Q ∗ ). Since q ∗ = q ∗ , A (i ,−1,0) (V (Q ∗ )) ∩ β β β β L t1 β ∗ X − X = A (i ,−1,0) (V (Q )) ∩ X − X ⊆ X − X . Since P is contained in V (Tt1 ) t1 L t2 V (Tt2 ) t2 V (Tt2 ) t2 G[X ] and is internally disjoint from X , V (P ) ∩ A (i ,−1,0) (V (Q ∗ )) ∩ X − X 6= ∅. So V (Tt1 ) t1 L t1 β V (Tt1 ) t1 V (P ) ∩ X − X 6= ∅. V (Tt2 ) t2

Let v be the vertex in Z2 ∩ V (P ). Let βv be the index in [|V (G)|] such that Qβv contains Sα∗ ∗ v ∈ V (P ) ∩ Z2 ⊆ V (M) ∩ γ=α+1(V (Qγ)). So βv ∈ [α + 1, α ]. 00 (2) Suppose there exists β ∈ [|V (G)|] such that qβ00 is not a zero sequence and the monochromatic (i ) t2 0 (it ,−1,0) (2) 0 E -pseudocomponent Q in G[Y 2 ] defining q 00 contains Q . Since Q contains Q , j,t2 β βv βv 00 00 ∗ (2) 00 0 β ∈ [βv]. Since β 6 βv 6 α and qβ00 is a nonzero sequence, β ∈ [α], so V (Q ) = V (Qβ00 ). Hence 00 00 V (Qβ ) contains V (Qβv ). So Z1 ∩ Z2 ⊇ V (Qβ ) ∩ V (Qβv ) 6= ∅, a contradiction. 00 (2) Hence there exists no β ∈ [|V (G)|] such that qβ00 is not a zero sequence and the monochromatic (i ) t2 (it ,−1,0) (2) E -pseudocomponent in G[Y 2 ] defining q 00 contains Q . j,t2 β βv Suppose v ∈ X . Since V (Q ) ∩ X 6= ∅ and Q contains v, V (Q ) ∩ X 6= ∅. So by V (Tt2 ) βv t1 βv βv t2 00 (2) Claim 21.52, there exists β ∈ [βv] such that qβ00 is not a zero sequence and the monochromatic (i ) t2 (it ,−1,0) (2) E -pseudocomponent in G[Y 2 ] defining q 00 contains Q , a contradiction. j,t2 β βv Hence v ∈ V (G) − X . Since V (P ) ∩ X − X 6= ∅, there exists v0 ∈ V (P ) ∩ X such V (Tt2 ) V (Tt2 ) t2 t2 that the subpath P 0 of P between v and v0 is internally disjoint from X . So there exists a V (Tt2 ) (it2 ) (i ,−1,0) 0 monochromatic E -pseudocomponent R in G[Y t2 ] containing v . By Claim 21.52, there j,t2 2 exists γ ∈ [|V (G)|] such that q(2) is not a zero sequence, and R defines q(2) . But by Claim 21.48, R2 γR2 2 γR2

~~R2 ⊇ Qβv , a contradiction. This proves the claim.~~

~~We say a sequence (a1, a2, . . . , am) is a substring of another sequence (b1, b2, . . . , bn) (for some m, n ∈ N) if there exists α ∈ [0, n − m] such that ai = bα+i for every i ∈ [m].~~

81 ◦ Claim 21.55. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∈ Sj . Let ∗ κ ∈ N. Let (t1, t2, . . . , tκ) be a parade in T such that tα ∈ K (M) for every α ∈ [κ]. For each (α) (α) ∗ α ∈ [κ], let (a1 , . . . , a|V (G)|) be the (itα , j, M)-truncation. Let α be an index such that:

~~(1) • aα∗ is a nonzero sequence, and ∗ (1) (2) (κ) • for every α ∈ [α − 1], aα = aα = ··· = aα .~~

~~∗ (1) Let n0 be the number of indices γ ∈ [α ] such that aγ is a nonzero sequence. Then κ 6 h(w0 −n0).~~

Proof. Note that w0 − n0 > 0 by Claim 21.30. Suppose to the contrary that this claim does not hold. That is, κ > h(w0 − n0) + 1. We further assume that w0 − n0 is as small as possible among all counterexamples. (t) (β) For every t ∈ {t1, t2, . . . , tκ} and α ∈ [|V (G)|], let aα = aα , where β is the integer such (t) that t = tβ. For every α ∈ [|V (G)|] and t ∈ {t1, t2, . . . , tκ}, if aα is not a zero sequence, then (t) (it) (it,−1,0) (t) let Mα be the monochromatic Ej,t -pseudocomponent in G[Y ] deﬁning aα ; otherwise, let (t) ∗ V (Mα ) = ∅. Since each tα is in K (M), every subsequence of (t1, t2, . . . , tκ) is a subsequence of a ∗ (t) tamed parade. For every t ∈ {t1, t2, . . . , tκ}, let αt be the largest index γ such that aγ is a nonzero sequence. 0 0 0 Let τ1 = (κ − 1)/(η5 + 3). By Claim 21.53, there exists a subsequence (t1, t2, . . . , tτ1 ) of (t1, t2, . . . , tκ−1) such that for every α ∈ [τ1 − 1],

~~(i) either the (i 0 , j, M)-truncation is diﬀerent from the (i 0 , j, M)-truncation, or there exists tα tα+1 0 0 (tα+1) (tα) γα ∈ [|V (G)|] such that V (Mγα ) 6= V (Mγα ).~~

0 0 0 Since (t1, t2, . . . , tτ1 ) satisﬁes (i), by Claim 21.51 and the assumption of this claim, there exists ∗ ∗ 0 0 β ∈ [φ3(w0) + 1] such that a 0 α . For each γ ∈ [τ1 − φ3(w0)], we redeﬁne tγ to be t . So tβ > β+γ−1 ∗ ∗ a 0 α . t1 > w0 00 00 00 Let τ2 = (τ1 − φ3(w0))/2 . By Claim 21.32, there exists a substring (t1, t2, . . . , tτ2 ) of 0 0 0 ∗ (t , t , . . . , t ) such that for each α ∈ [α 0 ], 1 2 τ1−φ3(w0) t1

~~(t) 00 00 00 (t) (ii) either aα is a zero sequence for every t ∈ {t1, t2, . . . , tτ2 }, or aα is a nonzero sequence for 00 00 00 every t ∈ {t1, t2, . . . , tτ2 }.~~

00 00 00 0 0 0 00 00 00 Since (t1, t2, . . . , tτ2 ) is a substring of (t1, t2, . . . , tτ1 ), (t1, t2, . . . , tτ2 ) satisﬁes (i). 000 000 000 00 00 00 Let τ3 = τ2/(φ3(w0)+1). By Claim 21.51, there exists a substring (t1 , t2 , . . . , tτ3 ) of (t1, t2, . . . , tτ2 ) such that

~~∗ 0 000 000 000 (t) (t0) (iii) for every α ∈ [α 0 ] and t, t ∈ {t , t , . . . , t }, V (Mα ) = V (Mα ). t1 1 2 τ3~~

000 000 000 00 00 00 000 000 000 Since (t1 , t2 , . . . , tτ3 ) is a substring of (t1, t2, . . . , tτ2 ), (t1 , t2 , . . . , tτ3 ) satisﬁes (i)-(iii). τ3 000 000 000 ∗ 0 Let τ4 = 2 . Since (t1 , t2 , . . . , tτ ) satisﬁes (iii), for every α ∈ [α 0 ] and 1 β < β τ3, f(η5)w0+1 3 t1 6 6 (t000 ) 000 β0 (tβ ) the sum of the entries of aα is at most the sum of the entries of aα . By Claim 21.30, the 000 (tβ ) sum of the entries of aα is at most f(η5)w0. So there exists a substring (q0, q1, q2, . . . , qτ4−1) of 000 000 000 (t1 , t2 , . . . , tτ3 ) such that

~~∗ 0 (t) (t0) (t) (t0) (iv) for every α ∈ [α 0 ] and t, t ∈ {q0, q1, . . . , qτ4−1}, V (Mα ) = V (Mα ) and aα = aα . t1~~

82 000 000 000 Since (q0, q1, . . . , qτ4−1) is a substring of (t1 , t2 , . . . , tτ3 ), (q0, q1, . . . , qτ4−1) satisfies (i)-(iv). ∗ ∗ ∗ Since (q0, q1, . . . , qτ4−1) satisfies (i) and (iv), there exists β ∈ {0, 1} such that α ∗ α 0 + 1. qβ > t1 ∗ For each γ ∈ [τ4 − 1], redefine qγ to be qβ +γ−1. Then (q1, q2, . . . , qτ4−1) satisfies (i)-(iv) and ∗ ∗ ∗ ∗ ∗ α α 0 + 1. For each β ∈ [2, τ4 − 1], by taking (t1, t2, α , α) = (q1, qβ, α , α 0 ) in Claim 21.54, q1 > t1 q1 t1 ∗ ∗ ∗ (qβ ) since (q1, . . . , qτ4−1) satisfies (iv), there exists γ ∈ [α 0 +1, α ] such that a ∗ is a nonzero sequence. β t1 q1 γβ ∗ ∗ For each β ∈ [2, τ4 − 1], we choose γβ 6= αq1 if possible. ∗ ∗ (qβ ) Let τ5 = (τ4 − 1)/2. By Claim 21.32, if γτ 6= αq , then aγ∗ is a nonzero sequence for every 5 1 τ5 ∗ ∗ ∗ ∗ β ∈ [τ5]. Since we choose γβ 6= αq1 for each β ∈ [2, τ4 − 1] if possible, we know if γτ5 = αq1 , then ∗ ∗ 0 0 0 γβ = αq1 for every β ∈ [τ5, τ4 − 1] by Claim 21.32. Hence there exists a substring (q1, q2, . . . , qτ5 ) of (q1, q2, . . . , qτ4−1) such that

~~∗ ∗ ∗ (t) 0 0 0 (v) there exists γ ∈ [α 0 +1, α ] such that a ∗ is a nonzero sequence for every t ∈ {q , q , . . . , q }. t1 q1 γ 1 2 τ5~~

0 0 0 0 0 0 Since (q1, q2, . . . , qτ5 ) is a substring of (q1, q2, . . . , qτ4−1), (q1, q2, . . . , qτ5 ) satisﬁes (i)-(v). Note that ∗ ∗ α 0 γ . q1 > τ5 Let τ6 = w0 2 . By Claims 21.30, 21.32 and 21.51, there exists a substring 2 ·(φ3(w0)+1)·(f(η5)w0+1) 00 00 00 0 0 0 (q1 , q2 , . . . , qτ6 ) of (q1, q2, . . . , qτ5 ) satisfying (i)-(v) and

~~∗ 0 00 00 00 (t) (t0) (t) (t0) (vi) for every α ∈ [α 0 ] and t, t ∈ {q , q , . . . , q }, V (Mα ) = V (Mα ) and aα = aα . q1 1 2 τ6~~

00 00 00 ∗ ∗ ∗ ∗ Since (q , q , . . . , q ) satisifies (i) and (vi), there exists β ∈ {1, 2} such that a 00 α 0 +1 γ +1. 1 2 τ6 1 q ∗ > q1 > β1 00 00 ∗ ∗ For every γ ∈ [τ6−1], redefine q = q ∗ . So there exists the smallest integer ξ ∈ [γ +1, |V (G)|] γ γ−1+β1 00 00 (q1 ) ∗ (qk ) such that aξ∗ is a nonzero sequence. By (vi) and Claim 21.32, for every α ∈ [ξ − 1], aα is identical for every k ∈ [τ6 − 1]. 00 ∗ (1) 0 ∗ (q1 ) Let Z0 = {γ ∈ [α ]: aγ is a nonzero sequence}. So |Z0| = n0. Let Z0 = {γ ∈ [ξ ]: aγ is 0 0 ∗ (1) (2) (κ) a nonzero sequence}. Let n0 = |Z0|. Recall that for every α ∈ [α − 1], aα = aα = ··· = aα . ∗ ∗ ∗ ∗ 0 ∗ ∗ 0 0 Since ξ > γ + 1 > α + 1, Z0 − {α } ⊆ Z0. By (v), {γ , ξ } ⊆ Z0 − Z0. So n0 > n0 + 1. In 0 particular, w0 − n0 > 1. By the minimality of w0 − n0, τ6 − 1 6 h(w0 − n0) 6 h(w0 − n0 − 1). But τ6 − 1 > h(w0 − n0 − 1) + 1 since κ > h(w0 − n0) + 1, a contradiction. This proves the claim. ◦ Claim 21.56. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∈ Sj . Let ∗ τ ∈ N. Let t1, t2, . . . , tτ be elements in K (M) such that ti+1 ∈ V (Tti ) − {ti} for every i ∈ [τ − 1]. Then τ 6 η6. (α) (α) (α) Proof. Suppose τ > η6 +1. For each α ∈ [τ], let (a1 , a2 , . . . , a|V (G)|) be the (itα , j, M)-truncation. ∗ (1) Since η6 + 1 > 2, by Claim 21.52, there exists γ such that aγ∗ is a nonzero sequence. We ∗ (1) ∗ may assume that γ is as small as possible. So aγ is a zero sequence for every γ ∈ [γ − 1]. ∗ (α) By Claim 21.32, for every γ ∈ [γ − 1] and α ∈ [τ], aγ is a zero sequence. By Claim 21.55, τ 6 h(w0 − 1). So η6 6 h(w0 − 1) − 1, a contradiction. ◦ For j ∈ [|V| − 1] and a c-monochromatic component M with SM ∈ Sj , we define TM to be the rooted tree obtained from TrM by contracting each maximal subtree of TM rooted at a node in ∗ ∗ ∗ K (M) containing exactly one node in K (M) into a node. Note that |V (TM )| = |K (M)|. ◦ Claim 21.57. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∈ Sj . If P is a directed path in TM , then |V (P )| 6 η6.

~~83 ∗ Proof. By the deﬁnition of TM , there exist nodes t1, t2, . . . , t|V (P )| in K (M) such that for each~~

α ∈ [|V (P )| − 1], tα+1 ∈ V (Ttα ) − {tα}. By Claim 21.56, |V (P )| 6 η6. ◦ Claim 21.58. Let j ∈ [|V| − 1]. Let M be a c-monochromatic component with SM ∈ Sj . Then the maximum degree of TM is at most f(η5).

~~∗ Proof. For every node t ∈ K (M) − {rM }, let qt be the node in K(M) such that t ∈ ∂Tj,qt . Let x be a node of TM . It suﬃces to show that the degree of x in TM is at most f(η5).~~

Let Rx be the subtree of TrM in T such that x is obtained by contracting Rx. We say a node ∗ q in Rx is important if q = qt for some t ∈ K (M) − {rM }. Note that the degree of x equals the ∗ number of nodes t in K (M) − {rM } such that there exists an important node q in Rx with q = qt. ∗ Let rx be the root of Rx. Since rx is the only element in V (Rx) ∩ K (M), for every important (iq,|V (T )|+1,s+2) (irx ,|V (T )|+1,s+2) node q, V (M) ∩ Y ∩ XV (Tq) = V (M) ∩ Y ∩ XV (Tq). ∗ ∗ ∗ ∗ Suppose there exists an important node q and node t ∈ K (M) − {rM } with q = qt∗ such (iq∗ ,|V (T )|+1,s+2) ∗ ∗ that A (i ∗ ,|V (T )|+1,s+2) (Y ∩ V (M)) ∩ X − X ∗ = ∅. Since t ∈ K (M) − {r }, L q V (Tt∗ ) t M (it∗ ,|V (T )|+1,s+2) (iq∗ ,|V (T )|+1,s+2) there exists v ∈ V (M) ∩ (Y − Y ) ∩ XV (Tt∗ ). By Claim 21.11, since (iq∗ ,−1) ∗ V (M) ∩ Xq∗ 6= ∅ and Xq∗ ∩ Ij ⊆ W3 and t ∈ ∂Tj,q∗ , we have v 6∈ Xt∗ . Since M is connected, there exists a path P in M from v to V (M) ∩ Y (iq∗ ,|V (T )|+1,s+2) internally disjoint from Y (iq∗ ,|V (T )|+1,s+2). Let u be the neighbor of the end of P in V (M) ∩ Y (iq∗ ,|V (T )|+1,s+2). So (iq∗ ,|V (T )|+1,s+2) (iq∗ ,|V (T )|+1,s+2) u ∈ A (i ∗ ,|V (T )|+1,s+2) (V (M) ∩ Y ). Since A (i ∗ ,|V (T )|+1,s+2) (Y ∩ V (M)) ∩ L q L q ∗ ∗ ∗ ∗ ∗ XV (Tt∗ ) − Xt = ∅, u 6∈ XV (Tt∗ ) − Xt . Since u ∈ V (M) and t ∈ ∂Tj,q , u 6∈ Xt by Claim 21.11. ∗ ∗ ∗ Since v ∈ XV (Tt∗ ) − Xt , some internal node of P other than u belongs to Xt . But V (P ) ∩ Xt ⊆ (i ∗ ,|V (T )|+1,s+2) ∗ q V (M) ∩ Xt ⊆ V (M) ∩ XV (Tj,q∗ ) = V (M) ∩ Y ∩ XV (Tj,q∗ ) by Claim 21.11. Hence P is not internally disjoint from Y (iq∗ ,|V (T )|+1,s+2), a contradiction. ∗ Therefore, for every important node q and node t ∈ K (M) − {rM } with q = qt, we have (iq,|V (T )|+1,s+2) ∗ AL(iq,|V (T )|+1,s+2) (Y ∩ V (M)) ∩ XV (Tt) − Xt 6= ∅. Hence for every t ∈ K (M) − {rM } in which there exists an important node q in Rx with q = qt,

~~(irx ,|V (T )|+1,s+2) AL(irx ,|V (T )|+1,s+2) (Y ∩ V (M)) ∩ XV (Tt) − Xt~~

(irx ,|V (T )|+1,s+2) = AL(irx ,|V (T )|+1,s+2) (Y ∩ V (M) ∩ XV (Tq)) ∩ XV (Tt) − Xt (iq,|V (T )|+1,s+2) = AL(irx ,|V (T )|+1,s+2) (Y ∩ V (M) ∩ XV (Tq)) ∩ XV (Tt) − Xt (iq,|V (T )|+1,s+2) ⊇ AL(iq,|V (T )|+1,s+2) (Y ∩ V (M) ∩ XV (Tt)) ∩ XV (Tt) − Xt (iq,|V (T )|+1,s+2) = AL(iq,|V (T )|+1,s+2) (Y ∩ V (M)) ∩ XV (Tt) − Xt 6= ∅. Note that

(irx ,|V (T )|+1,s+2) (i ,|V (T )|+1,s+2) |AL rx (Y ∩ V (M)) ∩ XV (Trx ) − Xrx | >s (irx ,|V (T )|+1,s+2) 6 |N (Y ∩ V (M) ∩ XV (Trx ))| (irx ,|V (T )|+1,s+2) 6 f(|Y ∩ V (M) ∩ XV (Trx )|) 6 f(η5) ∗ by Claim 21.25. So there are at most f(η5) nodes t ∈ K (M) − {rM } in which there exists an important node q in Rx with q = qt. Therefore, the degree of x is at most f(η5).

~~S|V|−1 ◦ Claim 21.59. If M is a c-monochromatic component with V (M) ∩ j=1 Ij 6= ∅, then |V (M)| 6 η7.~~

84 S|V|−1 ◦ ◦ Proof. Since V (M) ∩ j=1 Ij 6= ∅, there exists j ∈ [|V| − 1] such that SM ∈ Sj . By Claims 21.57 and 21.58, TM is a rooted tree with maximum degree at most f(η5) and the longest directed path in ∗ Pη6−1 k η6 TM from the root contains at most η6 nodes. So |K (M)| = |V (TM )| 6 k=0 (f(η5)) 6 (f(η5)) . ∗ η6 By Claim 21.26, |V (M)| 6 |K (M)| · η5 6 η5 · (f(η5)) = η7. Now we are ready to complete the proof of this lemma. Let M be a c-monochromatic compo- ◦ ∗ ◦ nent. If SM ∩ Ij = ∅ for every j ∈ [|V| − 1], then |V (M)| 6 η4 6 η by Claim 21.22. If SM ∩ Ij 6= ∅ ∗ for some j ∈ [|V| − 1], then |V (M)| 6 η7 6 η∗ by Claim 21.59. Therefore, |V (M)| 6 η . This proves Lemma 21.

~~7 Allowing Apex Vertices~~

~~The following lemma implies Theorem3 by taking k = 1 and Y1 = ∅.~~

Lemma 22. For all s, t, w, k, ξ ∈ N, there exists η∗ ∈ N such that for every graph G containing no Ks.t-subgraph, if Z is a subset of V (G) with |Z| 6 ξ, V = (V1,V2,...,V|V|) is a Z-layering of G, (T, X ) is a tree-decomposition of G − Z with V-width at most w, Y1 is a subset of V (G) with |Y1| 6 k, L is an (s, V)-compatible list-assignment of G such that (Y1,L) is a V-standard pair, then there exists an L-coloring c of G such that every c-monochromatic component contains at most η∗ vertices.

Proof. Let s, t, w, k, ξ ∈ N. Let f be the function fs,t in Lemma 12. Let h0 be the identity function. For i ∈ N, let hi be the function defined by hi(x) := x + f(hi−1(x)) for every x ∈ N0. ∗ Let η1 := hs+1(k + ξ). Let η2 be the number η in Lemma 20 taking s = s, t = t, w = w + ξ and ∗ η = hs+2(k + ξ) + (s + 1)ξ. Define η := max{η1, η2}. Let G be a graph with no Ks,t-subgraph, Z a subset of V (G) with |Z| 6 ξ, V = (V1,V2,...,V|V|) a Z-layering of G, (T, X ) a tree-decomposition of G − Z with V-width at most w, Y1 a subset of V (G) with |Y1| 6 k, L an (s, V)-compatible list-assignment of G such that (Y1,L) is a V-standard pair. Say X = (Xp : p ∈ V (T )). (0) (0) We may assume that |V| > 2, since we may add empty layers into V. Let (Y ,L ) be a (i) (i) >s (i−1) (Z ∪ Y1, 0)-progress of (Y1,L). For each i ∈ [s + 2], define (Y ,L ) to be an (NG (Y ), i)- progress of (Y (i−1),L(i−1)). Note that every L(s+2)-coloring of G is an L-coloring of G. It is clear (i) that |Y1 | 6 hi(|Z ∪ Y1|) 6 hi(ξ + k) for every i ∈ [0, s + 2] by Lemma 12 and induction on i. (s+2) Claim 22.1. For every L -coloring of G, every monochromatic component intersecting Z ∪ Y1 (s+1) is contained in G[Y1 ]. Proof. We shall prove that for each i ∈ [s+2] and for every L(i)-coloring of G, every monochromatic (i−1) component colored i and intersecting Z ∪ Y1 is contained in G[Y ]. Suppose to the contrary that there exist i ∈ [s+2], an L(i)-coloring of G, and a monochromatic (i−1) component M colored i and intersecting Z ∪ Y1 such that V (M) 6⊆ Y . Since M intersects (i−1) (i−1) (i−1) Z ∪ Y1 ⊆ Y , we have V (M) ∩ Y 6= ∅. So there exist y ∈ V (M) ∩ Y and v ∈ (i−1) (i−1) (i) (i−1) (i−1) (i−1) NG(y) ∩ V (M) − Y . In particular, L (y) = {i} ⊆ L (v) ⊆ L (v). Since (Y ,L ) s (i−1) (i) (i) >s (i−1) (i) v ∈ NG (Y ). But since (Y ,L ) is an (NG (Y ), i)-progress, i 6∈ L (v), a contradiction. This proves that for each i ∈ [s + 2] and for every L(i)-coloring of G, every monochromatic (i−1) (s+2) component colored i and intersecting Z ∪ Y1 is contained in G[Y ]. Since every L -coloring of G is an L(i)-coloring of G for every i ∈ [s + 2], every monochromatic component with respect to (s+2) Ss+2 (i−1) (s+1) L intersecting Z ∪ Y1 is contained in G[ i=1 Y ] ⊆ G[Y ].

85 0 Let G be the graph obtained from G − Z by adding a copy zi of z into Vi for every z ∈ Z and i ∈ [|V|] and adding an edge ziv for every edge zv ∈ E(G) with z ∈ Z and v ∈ Vi. For i ∈ [|V|], 0 0 0 0 0 0 0 0 deﬁne Vi := Vi ∪ {zi : z ∈ Z}. Let V := (V1 ,V2 ,...,V|V|). Then V is a layering of G . Let (T, X ) 0 0 0 0 be the tree-decomposition of G , where X := (Xp : p ∈ V (T )) and Xq := Xq ∪{zi : z ∈ Z, i ∈ [|V|]} for every q ∈ V (T ). Then (T, X 0) is a tree-decomposition of G0 with V0-width at most w + ξ. 0 (s+2) 0 (s+2) Let Y1 := Y ∪ {zi : z ∈ Z, i ∈ [|V|]}, let L (v) := L (v) for every v ∈ V (G) − Z, and let 0 (s+2) 0 0 0 0 L (zi) := L (z) for every z ∈ Z and i ∈ [|V|]. Note that (Y1 ,L ) satisﬁes (L1)–(L3). But (Y1 ,L ) 0 0 is possibly not (s, V )-compatible since some L (zi) might contain a color that is not allowed for 0 0 0 0 0 Vi. For each i ∈ [|V |] and z ∈ Z with L (zi) = {i } for some i ∈ [s + 2] with i ≡ i (mod s + 2), 0 0 0 delete zi from G and add edges zi−1v for each edge ziv ∈ E(G ) if i > 1 (or delete zi from G 0 0 and add edges zi+1v for each ziv ∈ E(G ) if i = 1). Note that if zi is deleted from G , then zi−1 0 0 0 0 00 and zi+1 are not deleted from G (since L (zi−1) = L (zi)L (zi+1)). Let G be the resulting graph. 00 0 00 00 00 00 00 00 00 For each i ∈ [|V|], let Vi := Vi ∩ V (G ). Let V := (V1 ,V2 ,...,V|V|). So V is a layering of G . 00 0 00 00 0 00 00 Let Y1 := Y1 ∩ V (G ) and L := L |V (G00). Then L is an (s, V )-compatible list-assignment of 00 00 00 00 00 00 0 00 G , and (Y1 ,L ) is a V -standard pair of G . Let Xq := Xq ∩ V (G ) for every q ∈ V (T ), and let 00 00 00 00 00 X := (Xq : q ∈ V (T )). Then (T, X ) is a tree-decomposition of G with V -width at most w + ξ. 00 Claim 22.2. G does not contain Ks,t as a subgraph.

00 00 Proof. Suppose to the contrary that some subgraph H of G is isomorphic to Ks,t. Note that 00 for every z ∈ Z, NG0 (zi) ∩ NG0 (zj) = ∅ for all distinct i, j ∈ [|V |]. So for every z ∈ Z, no vertex 00 00 00 of G is adjacent to two vertices in {zi ∈ V (G ): i ∈ [|V |]}. Hence for every z ∈ Z, each part 00 00 of the bipartition of H contains at most one vertex in {zi : z ∈ Z, i ∈ [|V |]}. Furthermore, if 0 00 0 00 0 00 zizj ∈ E(H ) for some z, z ∈ Z and i, j ∈ [|V |], then z 6= z . Therefore, for every z ∈ Z, H 00 00 contains at most one vertex in {zi : i ∈ [|V |]} Note that for every z ∈ Z and i ∈ [|V |] with 00 zi ∈ V (G ), we have NG00 (zi) ⊆ NG(z). 00 00 Deﬁne H to be the subgraph of G obtained from H by replacing each vertex zi ∈ V (H ) (for 00 00 some z ∈ Z and i ∈ [|V |]) by z. Then H is isomorphic to Ks,t, which is a contradiction. So G does not contain Ks,t as a subgraph. Note that for every s-segment S of V00,

00 (s+2) |Y1 ∩ S| 6 |Y1 ∩ S| + (s + 1)|Z| 6 hs+2(k + ξ) + (s + 1)ξ. By Lemma 20, there exists an L00-coloring c00 of G00 such that every c00-monochromatic component 00 contains at most η2 vertices. Let c be the function deﬁned by c(v) := c (v) for every v ∈ V (G) − Z 00 00 00 and c(z) := c (zi) for every z ∈ Z and some i ∈ [|V |] with zi ∈ V (H ). Note that for each z ∈ Z, 00 (s+2) L (zi) is a constant 1-element subset of L (z) for all i, so c is well-deﬁned. Therefore c is an L(s+2)-coloring of G and hence an L-coloring of G. (s+2) Let M be a c-monochromatic component. Since c is an L -coloring of G, if V (M)∩(Z∪Y1) 6= (s+1) ∗ ∅, then M is contained in G[Y1 ] by Claim 22.1, which contains at most hs+1(k + ξ) = η1 6 η 00 00 vertices. If V (M) ∩ (Z ∪ Y1) = ∅, then M is a c -monochromatic component contained in G , so ∗ ∗ M contains at most η2 6 η vertices. This proves that c has clustering at most η .

~~8 Open Problems~~

~~The 4-Color Theorem [41] is best possible, even in the setting of clustered coloring. That is, for all c there are planar graphs for which every 3-coloring has a monochromatic component of~~

86 size greater than c; see [46]. These examples have unbounded maximum degree. This is necessary since Esperet and Joret [15] proved that every planar graph with bounded maximum degree is 3-colorable with bounded clustering. The examples mentioned in Section1, in fact, contain large K2,t subgraphs. The following question naturally arises: does every planar graph with no K2,t subgraph have a 3-coloring with clustering f(t), for some function f? More generally, is every graph with layered treewidth k and with no K2,t subgraph 3-colorable with clustering f(k, t), for some function f? For s > 2, is every graph with layered treewidth k and with no Ks,t subgraph (s + 1)-colorable with clustering f(k, s, t), for some function f? In our companion paper [33] we prove an aﬃrmative answer to the weakening of this question with “layered treewidth” replaced by “treewidth” for s > 1.

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