Forbidden Directed Minors and Directed Pathwidth

Shiva Kintali ∗ Qiuyi Zhang †

March 15, 2015

Abstract Undirected graphs of pathwidth at most one are characterized by two forbidden minors i.e., (i) K3 the complete graph on three vertices and (ii) S2,2,2 the spider graph with three legs of length two each [BFKL87]. Directed pathwidth is a natural generalization of pathwidth to digraphs. In this paper, we prove that digraphs of directed pathwidth at most one are characterized by a ﬁnite number of forbidden directed minors. To achieve our goal, we present a new decomposition theorem for digraphs of directed pathwidth ≥ 2 and prove several properties of the forbidden directed minors, which are of independent interest.

Keywords: directed pathwidth, forbidden minors, graph minors, pathwidth.

1 Introduction

Pathwidth measures how close an undirected graph is to being a path. Directed pathwidth, introduced by Reed, Seymour and Thomas (see [Bar06]), measures how close a digraph is to being a directed path. For an undirected graph (resp. digraph) G, let pw(G) (resp. dpw(G)) denote its pathwidth (resp. directed pathwidth). Directed pathwidth is a generalization of pathwidth to ↔ digraphs i.e., for an undirected graph G, let G be the digraph obtained by replacing each edge {u, v} ↔ of G by two directed edges (u, v) and (v, u), then, dpw(G) = pw(G)[Bar06, Lemma 1]. Pathwidth has several equivalent characterizations in terms of vertex separation number, node searching number, interval thickness (i.e., one less than the maximum clique size in an interval supergraph) and cops and (invisible and eager) robber games. Similarly, directed pathwidth is characterized by directed vertex separation number and cops and (invisible and eager) robber games on digraphs [YC08]. Both pathwidth and directed pathwidth have associated decomposition structures called path decomposition and directed path decomposition respectively. The graph minor theorem [RS04] (i.e., undirected graphs are well-quasi-ordered under the minor relation) implies that every minor-closed family of undirected graphs has a ﬁnite set of minimal forbidden minors. In particular, it implies that for all k ≥ 0, graphs of treewidth (or pathwidth) ≤ k are characterized by a ﬁnite set of forbidden minors. The complete sets of forbidden minors are known for small values of treewidth and pathwidth. A graph has treewidth at most one (resp. two) if and only if it is K3-free (resp. K4-free). Graphs of treewidth at most three are characterized by four forbidden minors (K5, the graph of the octahedron, the graph of the pentagonal prism,

∗Department of Computer Science, Princeton University, Princeton, NJ 08540. Email : [email protected] †Mathematics Department, Princeton University, Princeton, NJ 08540. Email : [email protected]

1 Figure 1: The forbidden minors of undirected graphs of pathwidth at most one: K3 and S2,2,2 and the Wagner graph) [APC90, ST90]. Graphs of pathwidth at most one are characterized by two forbidden minors [BFKL87] i.e., (i) K3 the complete graph on three vertices and (ii) S2,2,2 the spider graph with three legs of length two each (see Figure1). Graphs of pathwidth at most two are characterized by 110 forbidden minors [KL94]. A natural question is “are digraphs of bounded directed pathwidth characterized by a finite set of forbidden directed minors ?”. Unfortunately there is no generalization of the graph minor theorem for digraphs yet. Existing notions of directed minors (eg. directed topological minors [Hun07], butterfly minors [JRST01], strong contractions [KS12], directed immersions [CS11]) do not imply well-quasi-ordering of all digraphs (see [Kin13] for more details). Recently, the first author [Kin13] introduced a notion of directed minors based on several operations of contracting special subsets of directed edges, conjectured that digraphs are well-quasi- ordered under the proposed directed minor relation and proved the conjecture for some special classes of digraphs. This conjecture implies that every family of digraphs closed under the directed minor operations (see Section2) are characterized by a finite set of minimal forbidden directed minors. In particular, it implies that for all k ≥ 0, digraphs of Kelly-width (or DAG-width, or directed pathwidth) ≤ k are characterized by a finite set of forbidden directed minors (see [Kin13] for more details). Using this directed minor relation, the authors studied the forbidden directed minors for digraphs with Kelly-width one and two (i.e., partial 0-DAGs and partial 1-DAGs) and proved that partial 0-DAGs (simply called DAGs) are characterized by one forbidden directed minor (i.e., K2) and partial 1-DAGs are characterized by three forbidden directed minors (see [KZ13] for more details). Kelly-width is a generalization of treewidth to digraphs. A digraph has directed pathwidth zero if and only if it is a DAG and hence it is characterized by one minimal forbidden minor i.e., K2 (see [KZ13]). In this paper, we show the finiteness of the set of minimal forbidden directed minors of digraphs with directed pathwidth one.

1.1 Notation and Preliminaries We use standard graph theory notation and terminology (see [Die05]). For a directed graph (digraph) G, we write V (G) for its vertex set (or node set) and E(G) for its edge set. All digraphs in this paper are ﬁnite and simple (i.e., no self loops and no multiple edges) unless otherwise stated. For X ⊆ V (G), let |X| denote the number of vertices in X. For an edge e = (u, v), we say that e is an edge from u to v. We say that u is the tail of e and v is the head of e. We also say that u is an in-neighbor of v and v is an out-neighbor of u. The out-neighbors of a vertex u is given by

2 Nout(u) = {v :(u, v) ∈ E} and the in-neighbors of a vertex v is given by Nin(v) = {u :(u, v) ∈ E}. Let dout(u) = |Nout(u)| (resp. din(v) = |Nin(v)|) denote the out-degree of u (resp. in-degree of v). For S ⊆ V (G) we write G[S] for the subgraph induced by S, and G\S for the subgraph induced by V (G) \ S. For F ⊆ E(G), we write G[F ] for the subgraph with vertex set equal to the set of endpoints of F , and edge set equal to F . For a digraph G, let G be the undirected graph, where V (G) = V (G) and E(G) = {{u, v} : (u, v) ∈ E(G)}. We say that G is the underlying undirected graph of G. For an undirected graph ↔ G, let G be the digraph obtained by replacing each edge {u, v} of G by two directed edges (u, v) ↔ and (v, u). We say that G is the bidirected graph of G. A directed (simple) path in G is a sequence of vertices v1, v2, . . . , vl such that for all 1 ≤ i ≤ l−1, (vi, vi+1) ∈ E(G). For a subset X ⊆ V (G), the set of vertices reachable from X is defined as: ReachG(X) := {v ∈ V (G) : there is a directed path to v from some u ∈ X}. We say that G is weakly connected if G is connected. We say that G is strongly connected if, for every pair of vertices u, v ∈ V (G), there is a directed path from u to v and a directed path from v to u. We use the term DAG when referring to directed acyclic graphs. Let T be a DAG. For two distinct nodes i and j of T , we write i ≺T j if there is a directed walk in T with first node i and last node j. For convenience, we write i ≺ j whenever T is clear from the context. For nodes i and j of T , we write i j if either i = j or i ≺ j. For an edge e = (i, j) and a node k of T , we write e ≺ k if either j = k or j ≺ k. We write e ∼ i (resp. e ∼ j) to mean that e is incident with i (resp. j). Let W = (Wi)i∈V (T ) be a family of finite sets called node bags, which associates each node i of [ T to a node bag Wi. We write Wi to denote Wj. ji The following notion of guarding is useful : Definition 1. [Guarding] Let G be a digraph and W, X ⊆ V (G). We say X guards W if W ∩X = ∅, and for all (u, v) ∈ E(G), if u ∈ W then v ∈ W ∪ X. In other words, X guards W means that there is no directed path in G \ X that starts from W and leaves W . Definition 2. [Edge contraction] Let G be an undirected graph, and e = {u, v} ∈ E(G). The vertices and edges of the graph G0 obtained from G by contracting e are: • V (G0) = V (G) \ u • E(G0) = (E(G) ∪ {{x, v} : {x, u} ∈ E(G)}) \ {{x, u} : x ∈ V (G)} Definition 3. [Minor] Let G and H be undirected graphs. We say that H is a minor of G, (denoted by H ≤ G), if H is isomorphic to a graph obtained from G by a sequence of vertex deletions, edge deletions and edge contractions. These operations may be applied to G in any order, to obtain its directed minor H. Theorem 4. (Robertson-Seymour theorem [RS04]) Undirected graphs are well-quasi-ordered by the minor relation ≤. Corollary 5. Every minor-closed family of undirected graphs has a finite set of minimal forbidden minors. In particular, for all k ≥ 0, graphs of treewidth (or pathwidth) ≤ k are characterized by a finite set of forbidden minors.

3 2 Directed Minors

In this section, we present a subset of the directed minor operations from [Kin13]. The directed minor relation in [Kin13] has more operations that are not necessary for our results in the current paper. When we perform the following operations on a digraph G to obtain a digraph G0, we remove any resulting self-loops and multi-edges from G0. Cycle contraction is a generalization of the edge contraction from Deﬁnition2.

Deﬁnition 6. [Cycle contraction] Let G be a graph, and C = {v1, v2, . . . , vl} ⊆ V (G) be a directed cycle in G i.e., (vi, vi+1) ∈ E(G) for 1 ≤ i ≤ l − 1 and (vl, v1) ∈ E(G). The vertices and edges of the graph G0 obtained from G by contracting C are: • V (G0) = {V (G) \ C} ∪ {w}, where w is a new vertex.

0 • E(G ) = {E(G) \ {{(x, vi) : 1 ≤ i ≤ l, x ∈ V (G)} ∪ {(vi, x) : 1 ≤ i ≤ l, x ∈ V (G)}}} ∪ {{(x, w):(x, vi) ∈ E(G) for some x ∈ V (G) \ C, 1 ≤ i ≤ l} ∪ {(w, x):(vi, x) ∈ E(G) for some x ∈ V (G) \ C, 1 ≤ i ≤ l}} Butterfly contractions (defined by Johnson et al [JRST01]) allow us to contract a directed edge e = (u, v) if either e is the only edge with head v, or it is the only edge with tail u, or both. The following operations, out-contraction and in-contraction, are slightly general and allows any edge to be out-contracted or in-contracted after removing certain incident edges. Out-contracting an edge (u, v) is equivalent to removing all the out-going edges of u and identifying u and v. Similarly, in-contracting an edge (u, v) is equivalent to removing all the in-coming edges of v and identifying u and v. Hence, we can out-contract or in-contract any edge of G without creating new paths in G. Definition 7. [Out contraction] Let G be a graph, and e = (u, v) ∈ E(G). The vertices and edges of the graph G0 obtained from G by out-contracting e are: • V (G0) = V (G) \ u • E(G0) = (E(G) ∪ {(x, v):(x, u) ∈ E(G)}) \{(x, u), (u, x): x ∈ V (G)} Note that we delete the vertex u, the tail of the edge e = (u, v). The vertex v exists in both G and G0. Definition 8. [In contraction] Let G be a graph, and e = (u, v) ∈ E(G). The vertices and edges of the graph G0 obtained from G by in-contracting e are: • V (G0) = V (G) \ v • E(G0) = (E(G) ∪ {(u, x):(v, x) ∈ E(G)}) \{(x, v), (v, x): x ∈ V (G)} Note that we delete the vertex v, the head of the edge e = (u, v). The vertex u exists in both G and G0. Definition 9. [Directed minor] Let G and H be digraphs. We say that H is a directed minor of G, (denoted by H G), if H is isomorphic to a graph obtained from G by a sequence of vertex deletions, edge deletions, cycle contractions, out/in contractions. These operations may be applied to G in any order, to obtain its directed minor H.

4 3 Directed Pathwidth

3.1 Deﬁnition Deﬁnition 10. [Directed path decomposition and Directed pathwidth [Bar06]] A Directed path decomposition of a digraph G is a sequence X1,X2,...,Xr of subsets (node bags) of V (G), such that: S • 1≤i≤r Xi = V (G). (DPW-1)

• For all i, j, k ∈ [r], if i ≤ j ≤ k, then Xi ∩ Xk ⊆ Xj. (DPW-2) • For all arcs (u, v) ∈ E(G), there exist i, j with 1 ≤ i ≤ j ≤ r such that u ∈ Xi and v ∈ Xj. (DPW-3)

The width of a directed path decomposition X = (Xi)i∈[r] is deﬁned as max{|Xi| : 1 ≤ i ≤ r} − 1. The directed pathwidth of G, denoted by dpw(G), is the minimum width over all possible directed path decompositions of G.

(DPW-2) can be replaced by the following equivalent statement:

• For any v ∈ V (G), {i : Xi ∩ v 6= ∅, 1 ≤ i ≤ r} is an integer interval. (DPW-2’) (DPW-3) can be replaced by the following equivalent statement: [ [ • For any i with 1 < i < r, there is no edge from Xj to Xj in G \ Xi. (DPW-3’) i+1≤j≤r 1≤j≤i−1 A directed path decomposition is called incremental if it satisﬁes the following condition:

• For any i ∈ [r − 1], |Xi∆Xi+1| = 1. (DPW-4)

3.2 Basic properties

A digraph D is a directed union of digraphs D1 and D2 if D1 and D2 are induced subgraphs of D, V (D1) ∪ V (D2) = V (D), and no edge of D has head in V (D1) and tail in V (D2). Directed pathwidth is closed under directed unions (see [YC08]). The following theorem is immediate.

Theorem 11. ([YC08]) Directed pathwidth of a digraph G is equal to the maximum directed pathwidth taken over the strongly-connected components of G.

Hence, all digraphs in the rest of this paper are strongly connected, unless otherwise stated. Kim and Seymour [KS12, Kim13] proved that directed pathwidth is closed under strong minors and butterﬂy minors. The following lemma is immediate.

Lemma 12. Let H G. Then, dpw(H) ≤ dpw(G). In other words, directed pathwidth is monotone under the directed minor operations mentioned in Section2.

5 4 Basic Forbidden Minors

Let F be the set of all minor-minimal (w.r.t ) digraphs of directed pathwidth two. Our goal is to prove that F is ﬁnite. By Theorem 11 we may assume that all digraphs in F are strongly-connected. For a digraph G, let G˜ be the digraph obtained by reversing the directions of all the edges in G. Lemma 13. ([YC08, Tam11]) For any digraph G, pw(G) = pw(G˜). Lemma 14. For digraphs G and H, H G if and only if H˜ G˜. Theorem 15. (Kintali, Zhang [KZ13]) Let G be a simple digraph such that every vertex in V (G) has out-degree ≥ 2. Then G contains K3,N4 or M5 (see Figure2) as a directed minor. Theorem 16. Let G be a simple digraph such that every vertex in V (G) has in-degree ≥ 2. Then G contains K3, N˜4 or M˜5 as a directed minor.

Corollary 17. K3,N4,M5, N˜4, M˜5 ∈ F. We say that these ﬁve digraphs are the “basic forbidden minors”.

Figure 2: K3, N4 and M5

0 Let F = F\{K3,N4,M5, N˜4, M˜5}. Since dpw(S2,2,2) = 2, (and S2,2,2 is a forbidden minor for 0 0 undirected graphs of pathwidth at most one) we have S2,2,2 ∈ F . So F is non-empty. The rest of this paper is aimed at proving that F 0 is ﬁnite. Let G ∈ F 0. The following three properties are crucial. • G is strongly-connected (PROP-1) • G has at least one vertex of out-degree equal to one (PROP-2) • G has at least one vertex of in-degree equal to one (PROP-3)

5 A New Decomposition Theorem

In this section, we present a new structural characterization of digraphs with directed pathwidth ≥ 2. For X ⊆ V (G), the complement of X is deﬁned as Xc := V (G) \ X. For X,Y ⊆ V (G) we say that X hits Y if Y \ X guards X. X has an out-neighbor in Y if there exists u ∈ X and v ∈ Y such that (u, v) ∈ E(G). A vertex u has k out-neighbors in Y if there exist distinct v1, .., vk ∈ Y such that (u, vi) ∈ E(G) for all i ∈ [k].

6 5.1 Elimination Orderings

Definition 18. Let G be a digraph and u ∈ V (G) with dout(u) = 1, then R(G, u) is defined as the digraph obtained by out-contracting u. We say that R(G, u) is obtained by eliminating u or reducing u in G. If dout(u) 6= 1, then R(G, u) is not defined. In other words, eliminating or reducing a vertex u in a digraph G yields digraph G0 := R(G, u) that is obtained from G by deleting vertex u and adding arcs from all in-neighbors of u to the unique out-neighbor of u, without introducing loops or multiple edges. In the rest of this paper whenever we use the notation R(G, u), we assume that dout(u) = 1 in G. Observation 19. If G is strongly-connected then R(G, u) is also strongly-connected.

Deﬁnition 20. We say that u1, ..., um is an elimination ordering in G, if G0 := G and Gi := R(Gi−1, ui) is deﬁned for i ∈ [m] i.e., dout(ui) = 1 in Gi−1. The corresponding out-sequence for this elimination ordering is a sequence of vertices v1, ..., vm, where vi is the unique out-neighbor of ui in Gi−1 for i ∈ [m]. We also denote Gi as R(G, [u1, ..., ui]).

Observation 21. Let u1, ..., um be an elimination ordering in G and v1, ..., vm be its out-sequence, G then Nout(ui) ⊆ {u1, ..., ui−1, v1, ..., vi−1, vi}.

Deﬁnition 22. Let u1, ..., um be an elimination ordering in G and v1, ..., vm be its out-sequence. We say that this elimination ordering is linear if either ui+1 = vi or vi+1 = vi holds for 1 ≤ i < m.

Observation 23. Let u1, ..., um be a linear elimination ordering and v1, ..., vm be its corresponding out-sequence, then vi ∈ {u1, ..., um, vm} for i ∈ [m].

Lemma 24. Let u1, ..., um be a linear elimination ordering and v1, ..., vm be its corresponding out-sequence, then {u1, ..., ui} hits {vi} for all 1 ≤ i ≤ m

Proof. We proceed with induction on i for 1 ≤ i ≤ m. Note that {u1} hits {v1} since u1 has only one out-neighbor. Assume that {u1, ..., ui−1} hits {vi−1}. Since ui has out-degree 1 at time of contraction, we have Nout(ui) ⊆ {u1, ..., ui−1, vi}. By linearity, either vi−1 = ui or vi−1 = vi. Case 1: vi−1 = ui. Now, consider any w ∈ {u1, ..., ui}. If w 6= ui, then by the inductive hypothesis, Nout(w) ⊆ {u1, ..., ui−1, ui}. If w = ui, then Nout(w) ⊆ {u1, ..., ui−1, vi}. We conclude that {u1, ..., ui} hits {vi}. Case 2: vi−1 = vi. Now, consider any w ∈ {u1, ..., ui}. If w 6= ui, then by the inductive hypothesis, Nout(w) ⊆ {u1, ..., ui−1, vi}. If w = ui, then Nout(w) ⊆ {u1, ..., ui−1, vi}. We conclude that {u1, ..., ui} hits {vi}. In the rest of this paper, all elimination orderings are linear. We say a digraph G admits an elimination ordering if there exists an elimination ordering u1, ..., um such that R(G, [u1, ..., um]) is a single vertex.

Lemma 25. A strongly connected digraph G has directed pathwidth at most 1 if and only if it admits an elimination ordering. In other words, digraphs in F do not admit an elimination ordering.

Deﬁnition 26. An elimination ordering u1, ..., um in G is maximal if there does not exist um+1 ∈ V (G) such that u1, ..., um+1 is also an elimination ordering in G.

7 Observation 27. Let G ∈ F. Let u1, ..., um be an elimination ordering in G and v1, ..., vm be its corresponding out-sequence. It is maximal if and only if in R(G, [u1, ..., um]), vm has outdegree ≥ 2 and all in-neighbors of vm have outdegree ≥ 2. 0 0 Lemma 28. Let u1, ..., um be a maximal elimination ordering and let u1, ..., ul be an elimination 0 0 ordering. If u1 ∈ {u1, ..., um}, then ui ∈ {u1, ..., um} for all 1 ≤ i ≤ l. 0 0 0 0 Proof. Let u1, ..., ul be an elimination ordering and v1, ..., vl be its corresponding out-sequence. We use strong induction on i. For the base case of i = 1, it holds by assumption. Now, assume that 0 uj ∈ {u1, ..., um} for 1 ≤ j < i. 0 0 0 0 For the sake of contradiction, assume ui 6∈ {u1, ..., um}. Since ui can be contracted in R(G, [u1, ..., ui−1]), 0 0 0 0 0 0 we see that in G, Nout(ui) ⊆ u1, ..., ui−1, vi . Also, by Lemma 24, we see that u1, ..., ui−1 hits 0 0 0 vi−1 . Thus, since u1, ..., ui−1 were reduced in the elimination ordering u1, ..., um, we see that in 0 0 0 R(G, [u1, ..., um]), the set u1, ..., ui−1, vi−1 must have been contracted to at most 1 vertex. Now, 0 0 0 0 by linearity, either vi = vi−1 or ui = vi−1. 0 0 0 0 0 0 0 0 0 Case 1: vi = vi−1. Then, Nout(ui) ⊆ u1, ..., ui−1, vi−1 . In R(G, [u1, ..., um]), since u1, ..., ui−1, vi−1 0 0 0 is now at most 1 vertex, dout(ui) = 1 and in fact, Nout(ui) = {vm}. Therefore, we may let um+1 = ui and vm+1 = vm, contradicting maximality. 0 0 0 0 0 Case 2: ui = vi−1. Then, in R(G, [u1, ..., um]), u1, ..., ui−1, ui is now at most 1 vertex, and 0 0 0 the vertex is ui. In fact, we can ﬁnd k such that vk = ui and by linearity, we must have ui = vm. 0 0 0 0 0 0 Since Nout(ui) ⊆ u1, ..., ui−1, vi , we see that dout(ui) = 1 and letting um+1 = ui = vm contradicts maximality.

0 0 Lemma 29. Let u1, ..., um be a maximal elimination ordering and u1, ..., ul be another maximal 0 0 0 elimination ordering with u1 = u1. Then, l = m and u1, ..., ul is a permutation of u1, ..., um. 0 Furthermore, if v1, ..., vm and v1, ..., vl are the corresponding out-sequences respectively, then vm = 0 vl. 0 0 0 Proof. Since u1 ∈ {u1, ...ul} and u1 ∈ {u1, ..., um}, by Lemma 28, we see that {u1, ..., um} and 0 0 0 0 {u1, ..., ul} must contain each other and thus have the same elements. Hence u1, ..., ul is a permutation of u1, ..., um and l = m. 0 0 0 0 Let X = {u1, ...um} = {u1, ..., ul}. By Lemma 24, X hits vm and X hits vl. Also, vm, vl 6∈ X. 0 Hence, vm = vl.

Lemma 30. If u1, ..., um and w1, ..., wm are two elimination orderings that are permutations of each other, then R(G, [u1, ..., um]) = R(G, [w1, ..., wm]).

5.2 Elimination Sets

Deﬁnition 31. Let u1, ..., um be the unique maximal (up to permutation) elimination ordering Sm starting at u1 = u and let v1, ..., vm be its out-sequence. We say that Su = i=1 {ui, vi} is the elimination set of u. The unique “end vertex” vm is called the bad vertex of Su. We denote Sm this bad vertex by bu := vm. Note that Su = {vm} ∪ i=1 {ui}.

Deﬁnition 32. Let Su be the elimination set of u, then R(G, Su) is deﬁned to be the unique digraph (uniqueness is implied by Lemma 30) obtained by eliminating vertices according to a maximal elimination ordering starting at u. In the rest of this paper, all elimination orderings are maximal.

8 c Lemma 33. If x ∈ Su, y ∈ Su and (x, y) ∈ E(G), then x = bu.

Proof. Lemma 24 implies that Su \{bu} hits {bu}. The lemma follows.

c Lemma 34. Let Su be the elimination set of u. Then, bu has at least 2 out-neighbors in Su. c c Furthermore, if v ∈ Su has out-neighbors in Su, then v has out-neighbors in Su.

Proof. By the maximality of the elimination ordering, bu must have at least 2 out-neighbors in c c Su. If v ∈ Su has out-neighbors in Su, then v is an in-neighbor of bu in R(G, Su). Again by the c maximality of the elimination ordering v must have out-neighbors in Su. Lemma 35. Let u be a vertex of out-degree 1 and v be its unique out-neighbor. Let w be another vertex. If Sw contains u or v, then Su ⊆ Sw.

Proof. By maximality, we see that if Sw contains either u or v, then u must have been eliminated in the elimination ordering starting at w. Let u1, ..., um be the elimination ordering starting at u1 = u. Therefore, by Lemma 28, we see that {u1, ..., um} ⊆ Sw. By Lemma 24, {u1, ..., um} hits {vm}, so we see that vm ∈ Sw. We conclude that Su ⊆ Sw.

5.3 Elimination Decomposition

Let G be a strongly-connected digraph. Let |V (G)| = n and V (G) = {w1, . . . , wn}. Let {Sw1 , ..., Swn } be the set of all elimination sets in G. Each elimination set Swi is a subset of V (G). If there are sets

Swi = Swj then we remove one of them arbitrarily. Let S = {Sv1 , ..., Svl } be the set of elimination sets without duplicates. The subset relation deﬁnes a natural partial order on the sets in S. Let

S∗ = {Su1 , ..., Sus } be the largest anti-chain of the sets in S.

Deﬁnition 36. We say that S∗ = {Su1 , ..., Sus } is an elimination decomposition of G. We Ss deﬁne S0 := V (G) \ i=1 Sui

As an example, we now exhibit an elimination decomposition of G = S2,2,2, the spider graph with three legs of length two each. S2,2,2 has three legs connected by a vertex (say w). Let ui, vi be the vertices in the i-th branch such that (ui, vi), (vi, ui), (vi, w), (w, vi) ∈ E(G). Then, the elimination sets are Sui = {ui, vi, w} for 1 ≤ i ≤ 3. S∗ = {Su1 ,Su2 ,Su3 } is the elimination decomposition of G. Theorem 37 (Characterization Theorem). Let G be a strongly connected digraph. Then, G has directed pathwidth ≥ 2 if and only if V (G) can be decomposed into sets V0,V1, ..., Vs, with V (G) = Ss i=0 Vi and satisfying the following properties:

• Property 1: for 1 ≤ i ≤ s, Vi is a proper subset of V (G)

• Property 2: V0 is a set of vertices of out-degree ≥ 2

c • Property 3: for 1 ≤ i ≤ s, if vi ∈ Vi has out-neighbors in Vi , then vi has at least 2 such out-neighbors

c c • Property 4: for 1 ≤ i ≤ s, if u ∈ Vi has out-neighbors in Vi, then u has out-neighbors in Vi .

9 Proof. If G is of directed pathwidth ≥ 2, then consider its elimination decomposition Su1 , ..., Sus Ss and let Vi = Sui and V0 = S0. By definition of S0, we see that i=1 Vi = V (G). Note that if there exists i such that Svi = V (G), then G would have directed pathwidth ≤ 1, so Svi must be a strict subset. If there exists v ∈ S0 that has outdegree ≤ 1, then Sv is an elimination set, but it is not necessarily maximal or in the elimination decomposition. However, we see that Sv ⊆ Sw for some Sw in the elimination decomposition. But v ∈ Sw contradicts v ∈ S0. Properties 3 and 4 follow from Lemma 33 and Lemma 34. If we can find such vertex sets V0, ...Vn, then assume for contradiction that G is of pathwidth ≤ 1. Thus, since G is strongly connected, we can find a linear elimination ordering u1, ..., um and its outsequence v1, ..., vm such that G is reduced to a single vertex. Since u1 has outdegree 1, we see that u1 ∈ Vi for some i ≥ 1. Since Vi is only a strict subset of V (G), there must exists l minimal such that either ul 6∈ Vi or vl 6∈ Vi. Case 1: ul 6∈ Vi. Since ul can be contracted and the minimality of l, we know that Nout(ul) ⊆ {u1, ...ul−1, vl−1} ⊆ Vi. Note that ul must have at least one out-neighbor in Vi, so this contradicts property 4. Case 2: vl 6∈ Vi. Since ul ∈ Vi and vl 6∈ Vi, by property 3, ul has at least 2 out-neighbors in c c Vi . Since {u1, ..., ul} ∈ Vi, we see that the two out-neighbors of ul in Vi were not eliminated and ul has outdegree ≥ 2 at time of elimination, a contradiction.

6 Intersection Properties

6.1 Lonely Sets

Deﬁnition 38. Let S∗ = {Su1 , ..., Sus } be an elimination decomposition of G. Then, X ⊆ V (G) is S lonely if there exists i such that j6=i Suj ∩ X = ∅

Lemma 39. Let Su be an elimination set in S∗ and let v be the unique out-neighbor of u. Then, {u, v} is lonely.

6.2 Double Intersection

Lemma 40. Let Sv1 , ..., Svt be a subset of elimination sets in S∗ such that Svi hits Svi+1 for St 1 ≤ i ≤ t − 1, and Svt hits Sv1 . Then, V (G) = i=1 Svi and there are no other elimination sets in S∗. St Proof. Let H = i=1 Svi . For any vertex v ∈ Svi , all its out-neighbors are in Svi ∪ Svi+1 . Hence, H has no out-neighbors in Hc. By strong connectivity of G, Hc = ∅.

Lemma 41. If Su ∩ Sv 6= ∅, then {bu, bv} ∩ Su ∩ Sv 6= ∅.

c c c Proof. By strong connectivity, Su ∩ Sv must have out-neighbors in (Su ∩ Sv) = Su ∪ Sv. Without c loss of generality, assume it has out-neighbors in Su, but by Lemma 33, the only vertex in Su that c has out-neighbors in Su is bu. Therefore, bu ∈ Su ∩ Sv.

Corollary 42. For an elimination set Su, if {bu} is lonely, then Su is lonely.

Lemma 43. If bu ∈ Sv and bu 6= bv, then Su hits Sv.

10 Proof. Consider the linear elimination order of Sv. Since bu 6= bv, we must have reduced bu during the elimination process starting at v. Hence, Nout(bu) ⊆ Sv. By Lemma 33, Su hits Sv.

Corollary 44. If Su ∩ Sv 6= ∅ and bv 6∈ Su, then Su hits Sv.

Corollary 45. If bu, bv ∈ Su ∩ Sv and bu 6= bv, then Su and Sv hit each other and V (G) = Su ∪ Sv.

6.3 Triple Intersection

Now we consider the case when there are three distinct elimination sets Su,Sv,Sw in S∗ that pairwise intersect. We say that Su ∩ Sv ∩ Sw is the “triple intersection”.

Lemma 46. If Su,Sv,Sw pairwise intersect and Su ∩ Sv ∩ Sw = ∅, then Su,Sv and Sw hit each other in a cycle and V (G) = Su ∪ Sv ∪ Sw.

Proof. Consider Su,Sv and note that we cannot have bu, bv ∈ Su ∩Sv or else due to the empty triple intersection, we have bu, bv 6∈ Sw. But since Sw intersects both Su,Sv, we must have bw ∈ Su,Sv. But, bw is in the triple intersection, a contradiction. Since one of bu, bv must be in the intersection, so without loss of generality, let bu ∈ Sv and bv 6∈ Su. By Corollary 44, Su hits Sv. Simiarly, Sw intersects Su, but since bu ∈ Su ∩ Sv and the triple intersection is empty, we conclude that bu 6∈ Sw. By Corollary 44, we conclude that Sw hits Su and by Lemma 41, bw ∈ Su. Lastly note that bw 6∈ Sv since the triple intersection is empty. We conclude that Sv hits Sw. Thus, we have 3 elimination sets that hit each other and by Lemma 40, V (G) = Su ∪ Sv ∪ Sw.

Lemma 47. If Su ∩ Sv ∩ Sw 6= ∅, then one of bu, bv, bw must be in the triple intersection.

c Proof. Let H = Su ∩ Sv ∩ Sw. By strong connectivity of G, H must have an out-neighbor in H . c c c c Let u ∈ H, v ∈ H such that (u, v) ∈ E(G). Note that either v ∈ Su or v ∈ Sv or v ∈ Sw. By Lemma 33, we conclude that u must be either bu or bv or bw respectively.

Lemma 48. If we have Su,Sv,Sw such that bu, bv, bw ∈ Su ∩ Sv ∩ Sw, then bu = bv = bw. Proof. By Corollary 45, if any of the bad vertices are not equal, then we have only two elimination sets, contradicting the existence of Su,Sv and Sw.

Lemma 49. If we have Su,Sv,Sw such that Su ∩ Sv ∩ Sw 6= ∅ and bu is in the triple intersection and bv is not in the triple intersection, then Su hits Sv and 1) If bu = bw, then Sw hits Sv 2) If bw = bv, then Su hits Sw 3) If all vertices are distinct, then Su hits Sw. Also, Sw hits Sv or Sv hits Sw.

Proof. Assume bu is in the triple intersection set H = Su ∩ Sv ∩ Sw and bv is not. Note ﬁrst that bv 6∈ Su, or else by Lemma 43, bu 6= bv implies we only have two elimination sets, contradicting existence of Sw. By Corollary 44, Su hits Sv. If bu = bw, then bw ∈ H and bv 6= bw. As previous, we conclude that Sw hits Sv. If bw = bv, then we note that bw 6= bu, so Su hits Sw. Otherwise, we see that bu, bv, bw are all distinct vertices and note also by Lemma 43, bv, bw 6∈ Su. bw 6= bu still gives Su hits Sw. Then, one of bv, bw must be in Sv ∩ Sw, so either Sv hits Sw or vice versa.

11 Lemma 50. If we have Su,Sv,Sw such that Su ∩ Sv ∩ Sw 6= ∅ and Su hits Sv, then one of the following holds: 1) bu is in the triple intersection and bv is not 2) bw is in the triple intersection and bu, bw are not and Sw hits Su,Sv.

Proof. Note that if bu = bv, then since Su hits Sv, all the out-neighbors of bu = bv are contained in Su ∪ Sv, which means Su,Sv hit each other and by Lemma 40, this contradicts existence of Sw. Also, if bv ∈ Su, then bu 6= bv implies Sv hits Su, which again contradicts existence of Sw. Therefore, bv 6∈ Su and so bv is not in the triple intersection. If bu is in the triple intersection, we are done. Otherwise, bu, bv are both not in the triple intersection and by Lemma 47, bw is in the triple intersection. By Lemma 49, Sw hits both Su,Sv.

6.4 More Triple Intersections c In this section, we study intersections of the form Su ∩ Sv ∩ Sw, under the assumption that the triple intersection is not empty.

Lemma 51. Let Su ∩ Sv 6= ∅ and let u1, ..., um be the elimination ordering of Su and v1, .., vm be the corresponding out-sequence. If ul = bv and there exists t ≥ l such that vt ∈ Sv, then Su,Sv hit each other and V (G) = Su ∪ Sv.

Proof. We see that by Lemma 24, {u1, ..., ut} hits vt in G. Since ul = bv and vt ∈ Sv, we see that by Lemma 33, {u1, ..., ut} ∪ Sv has no out-neighbors to its complement. So, by strong connectivity, G = {u1, ..., ut} ∪ Sv. Therefore, V (G) = Su ∪ Sv and there are no other vertices. Since there are only two elimination sets, Su,Sv hit each other trivially.

Lemma 52. Let Su ∩ Sv 6= ∅ and let u1, ..., um be the elimination ordering of Su and v1, .., vm be the corresponding out-sequence. Assume there exists t minimal such that ut ∈ Sv and there exists c minimal p ≥ t such that vp 6∈ Sv. If (Su ∪ Sv) 6= ∅, then up = bv and bu 6∈ Sv.

Proof. We see that by deﬁnition, vl ∈ Sv for t ≤ l < p. Now, by linearity, we see that either vp−1 = vp or vp−1 = up. Since vp−1 ∈ Sv, vp−1 = up must hold. Realize bv cannot be eliminated before up, since if p = t, then minimality of t is contradicted and if p > t, since vp−1 ∈ Sv, by c Lemma 51, we see that (Su ∪ Sv) = ∅, a contradiction.

Since up ∈ Svj and vp 6∈ Svj and bv was not eliminated, we conclude that up = bv. Also, we claim bu 6∈ Sv. If bu ∈ Sv and since bv ∈ Su, then if bu 6= bv, we have Su,Sv hit each other. By c Lemma 40, this contradicts (Su ∪ Sv) = ∅. Otherwise, bu = bv, but that is impossible since bu cannot be reduced in the elimination ordering of Su.

Lemma 53. If bu ∈ Su ∩Sv ∩Sw and let u1, ..., um be the elimination ordering of Sv and v1, .., vm be c the corresponding out-elimination ordering. Assume there exists t minimal such that ut ∈ Su ∩ Sw. c Then, vt 6∈ Sw and ut has out-neighbors in Su ∩ Sv ∩ Sw.

Proof. Consider Nout(ut) and note that Nout(ut) ⊆ {u1, ..., um, vt}, so we know that Nout(ut) ⊆ Sv. By Lemma 33, since ut 6= bu, we also know Nout(ut) ⊆ Su. Next, note that ut cannot only have c out-neighbors in Sw, or else ut ∈ Sw. So, ut must have out-neighbors in Su ∩ Sv ∩ Sw. Now, if vt ∈ Sw, then note that bw could not have been previously reduced by Lemma 51. Therefore, if vt has any out-neighbors in Sw, they cannot have been reduced to its out-neighbors c c in Sw. Since ut has out-neighbors in Su ∩ Sv ∩ Sw, they must have been reduced previously in u1, ..., ut−1, contradicting minimality of t.

12 c c Lemma 54. Let G ∈ F. If bu ∈ Su ∩ Sv ∩ Sw, then either Su ∩ Sv ∩ Sw = ∅ or Su ∩ Sv ∩ Sw = ∅.

Proof. Assume otherwise that both sets are non-empty. Consider the elimination order of Su, u1, ..., um and its corresponding out-sequence v1, .., vm, where u1 = u and vm = bu. Note that c c c u1 ∈ (Sv ∪ Sw) and we know that since bu 6∈ Su ∩ Sv ∩ Sw,Su ∩ Sv ∩ Sw, there must exist minimal l > 1 such that ul ∈ Sv ∪ Sw. Without loss of generality, we may assume that ul ∈ Sv. Assume that there exists a minimal c r ≥ l such that vr 6∈ Sv. But (Su ∪ Sv) 6= ∅ and bu ∈ Sv, which contradicts Lemma 52. c So, no such r exists. Now, since bu 6∈ Su ∩ Sw ∩ Sv and it is a non-empty set, we can ﬁnd a c minimal q > l such that uq ∈ Su ∩ Sw ∩ Sv. Note vq ∈ Sv, which contradicts Lemma 53.

Lemma 55. Let G be a minimal forbidden minor. If bu ∈ Su ∩ Sv ∩ Sw and bv, bw are not, then c c either Su ∩ Sv ∩ Sw = ∅ or Su ∩ Sv ∩ Sw = ∅. c Proof. Since bu 6= bv, bw, ﬁrst note that the out-neighbors of bu in Su are contained in Sv ∩ Sw and there are at least 2 such out-neighbors. Assume that both sets are non-empty. Consider the elimination order of Sw, u1, ...um, and its corresponding out-sequence v1, ..., vm, where u1 = c w, vm = bw. Note that u1 ∈ (Su ∪ Sv) , bu ∈ Su ∪ Sv, so we can ﬁnd minimal l > 1 such that ul ∈ Su ∪ Sv. Case 1: ul ∈ Su. Note that bw 6∈ Su, or else Su,Sw will hit one another. So, there must exist minimal t ≥ l such that vt 6∈ Su. By Lemma 52, ut = bu. But bu has at least two out-neighbors in c Su ∩ Sv ∩ Sw and so one of its out-neighbors must have been previously reduced and there exists a c minimal q < t such that uq ∈ Su ∩ Sv ∩ Sw. We see that q > l. c So, uq 6∈ Su, vq ∈ Su, thus in G uq has an out-neighbor x ∈ Su in G that must have been previously reduced. We see that if x ∈ Svj , then the minimality of y is contradicted. We must have x 6∈ Sv and since uq ∈ Sv, uq = bv by Lemma 33 and bv ∈ Sw. Note that vt−1 = ut = bu is forced since vt−1 = vt contradicts the minimality of t. But, vt−1 = bu ∈ Sv and t − 1 ≥ q, where uq = bv, contradicting Lemma 51. c c Case 2: ul ∈ Sv ∩ Su. Since Su ∩ Sv ∩ Sw 6= ∅, we see that there exists a minimal t ≥ l such c c that ut ∈ Sv ∩ Su or vt ∈ Sv ∩ Su. c c Case 2.1: ut ∈ Sv ∩ Su. Then by Lemma 53, ut has an out-neighbor x in G in Sv. Note either x was previously reduced or exists k < t such that vk = x. Either case, x ∈ Su contradicts the minimality of t. So x 6∈ Su, but ut 6= bu contradicts Lemma 33. c c Case 2.2: vt ∈ Sv ∩ Su. Let q ≤ t be minimal such that vq ∈ Sv. By Lemma 52, we see that uq = bv. If bu was reduced on ut or before, then vt ∈ Svi contradicts Lemma 51. Since bu 6= bw, there must exist a minimal r > t such that ur = bu. Note that if vr−1 = bu ∈ Sv, then since bv is already contracted, this contradicts Lemma 51. Therefore, vr−1 = vr. However, we claim that vr−1 ∈ Su. Assume not and consider the minimal t < p < r such that vp 6∈ Su. Then, we see that vp−1 = up and so, up ∈ Su. However, since bu is not yet reduced, this forces up = bu, a contradiction to the minimality of r. So, vr = vr−1 ∈ Su. But ur = bu and vr ∈ Su contradicts Lemma 51.

c c Lemma 56. If bu, bv ∈ Su ∩ Sv ∩ Sw, then Su ∩ Sv ∩ Sw = ∅ or Su ∩ Sv ∩ Sw = ∅.

Proof. Note that if bw is in the triple intersection, then this follows from Lemma 54. So, we may assume bw is in the triple intersection. Assume that both sets are nonempty. Consider the elimination ordering of Sw as u1, ..., um and its corresponding out-sequence v1, .., vm. Now, since bu ∈ Su ∩ Sv ∩ Sw, we can ﬁnd a minimal l

13 c such that ul ∈ Su ∪ Sv. Without loss of generality, let ul ∈ Su. Now, note that bw ∈ (Su ∪ Sv) , or c else we get two sets that hit each other by Corollary 45. So, since Su ∩ Sv ∩ Sw 6= ∅, we can ﬁnd c c minimal q ≥ l such that uq ∈ Su ∩ Sv or vq ∈ Su ∩ Sv. c c Case 1: uq ∈ Su ∩ Sv. Then by Lemma 53, uq has an out-neighbor x in Su. Note that the out-neighbor cannot be in Sv since x must be previously reduced or reduced to, contradicting minimality. Thus, since uq 6= bu, we have a contradiction. c Case 2: vq ∈ Su ∩ Sv. By Lemma 52, uq = bv = bj and vq ∈ Sv, contradicting Lemma 51. c c Lemma 57 (Non-Intersection Theorem). If Su∩Sv ∩Sw 6= ∅, then Su∩Sv ∩Sw = ∅ or Su∩Sv ∩Sw = ∅.

Proof. If bu is in the triple intersection, then it follows by Lemma 54. If bu is not in the triple intersection, but both bv, bw are, then it follows by Lemma 56. If both bu, bv are not in the triple intersection, or if bu, bw are not in the triple intersection, then it follows by Lemma 55.

6.5 Applications of Intersection Theorems

Deﬁnition 58. Su is an A-type (elimination set) if there exists an elimination set Sv such that Su hits Sv. If Su is not an A-type, we call it a non A-type.

Lemma 59. If Su,Sv are non A-types and Su ∩ Sv 6= ∅, then bu, bv ∈ Su ∩ Sv and bu = bv.

Proof. Note that if bu 6∈ Su ∩ Sv, then by Corollary 44, Sv must hit Su, contradicting that Su are non A-types. Therefore, bu, bv ∈ Su ∩ Sv and by Lemma 43, bu = bv.

Lemma 60. If Su hits Sv and Sw hits Sx and bu = bw, then either Su hits Sx or Sw hits Sv. Also, Sv ∩ Sx 6= ∅.

Proof. Assume that Sw does not hit Sv, then there is an out-neighbor of bw = bu, v1, such that c v1 6∈ Sw ∪ Sv. But Su hits Sv, so v1 ∈ Su ∪ Sv and therefore we conclude that v1 ∈ Su ∩ Sv. c Symmetrically, we see that v1 ∈ Sx ∩ Sw. Therefore, v1 ∈ Su ∩ Sx. Now, we split into cases. Case 1: bu 6∈ Sx. By Corollary 44, Sx hits Su and bx ∈ Su. If bx 6∈ Sv, then again, Sv hits Sx. But now, Su,Sv,Sx hit each other in a cycle, contradicting that existence of Sw. Lemma 40 So, bx ∈ Sv. Now, we see bx ∈ Su ∩ Sv and so by Corollary 44, we must have bu ∈ Sv or else Sv would hit Su, a contradiction. In this case, we have that bu ∈ Su ∩ Sv ∩ Sw and Su hits Sv. Note that if bu = bv, then Su,Sv hits each other, contradicting the existence of Sw. Thus bu 6= bv, and so Nout(bu) ⊆ Sv. Therefore, Sw hits Sv, a contradiction. Case 2: bu ∈ Sx. Then, bu = bw ∈ Su ∩ Sw ∩ Sx and since Sw hits Sx, we can repeat the same argument above to show that Nout(bw) ⊆ Sx. In this case, Su hits Sx. So, in either case there exists an elimination set S that hits both Sv,Sx. Therefore, we conclude that Sv ∩ Sx 6= ∅.

Lemma 61. If Su hits Sv and Sw is a non A-type and bu = bw, then Su ∩ Sv = ∅.

Proof. Assume that Su ∩ Sv 6= ∅, then Sv cannot hit Su, or else Su,Sv hit each other. So, bu ∈ Sv. Therefore, bu ∈ Su ∩ Sv ∩ Sw and Su hits Sv and Sw is a non A-type, so by Lemma 50, we conclude that bu is in the triple intersection and bv is not. However, since bu = bw, by Lemma 49, Sw must be an A-type that hits Sv, contradicting the fact that Sw is a non A-type.

14 7 Minimal Forbidden Minor Properties

Let G ∈ F. As noted before, G is strongly-connected and dpw(G) ≥ 2. We say that Su shares its bad vertex with Sv if bu = bv.

7.1 A-Types In A Cycle When there are A-types that hit each other in a cycle, we ﬁrstly want to be able to bound the length of the cycle or the number of A-types in the cycle. Using minimality, we show that the number of A-types can be at most 4.

Theorem 62. Let G ∈ F. If there exists a number of A-types that hit each other in a cycle, then there are at most 4 such A-types.

Proof. Let Sv1 , ..., Svn be A-types such that Svi hits Svi+1 , where vn+1 = v1, for 1 ≤ i ≤ n and n > 4. By Lemma 40, we know that no other vertices or maximal elimination sets exist. Furthermore, we may assume that n is minimal. Therefore, for any i, Svi can only hit Svi+1 , or else we can ﬁnd a smaller cycle of A-types that hit each other, a contradiction to minimality. Similarly, Svi+1 can only be hit by Svi .

Assume that we have more than 4 A-types. Then, our ﬁrst claim is for any i, Svm cannot intersect both Svi ,Svi+1 for m 6= i, i + 1. Assume, for contradiction, that Svm ∩ Svi 6= ∅,Svm ∩ Svi+1 6= ∅.

Case 1: Svi ∩ Svi+1 = ∅. By Lemma 44, since Svm cannot hit Svi+1 , bvi+1 ∈ Svm . Similarly, Svi cannot hit Svm , so bvm ∈ Svi . Since Svi ∩ Svi+1 = ∅, we deduce bvm 6∈ Svi+1 . So, Svi+1 must hit Svm .

If bvi 6∈ Svm , then Svm must hit Svi , so there are at most 3 A-types, a contradiction. So, bvi ∈ Svm .

By Lemma 45, bvi = bvm and Nout(bvm ) = Nout(bvi ) ⊆ Svi ∪ Svi+1 . Then, H = Svi ∪ Svi+1 ∪ Svm has no out-neighbors in Hc, so G = H only has 3 elimination sets, a contradiction.

Case 2: Svi ∩ Svi+1 6= ∅. Then all Svi ,Svi+1 ,Svm pairwise intersect and note that if Svm ∩

Svi ∩ Svi+1 = ∅, then by Lemma 46, we have 3 A-types that hit each other, a contradiction. So,

Svm ∩ Svi ∩ Svi+1 6= ∅, and Svi hits Svi+1 , so by Lemma 50, either bvi is in the triple intersection and bvi+1 is not, or Svm hits both Svi ,Svi+1 . As noted before, the second case cannot happen, so we must have bvi is in the triple intersection and bvi+1 is not. By Lemma 49, we see that we must have either Svm hits Svi+1 or Svi hits Svm , both contradictions. So, our claim holds.

Our second claim is if Svi intersects Svj , then j = i, i − 1, i + 1, where v0 = vn, v1 = vn+1.

Assume that Svi and Svj intersects and j 6= i, i − 1, i + 1. Then, we cannot have Svi hits Svj or

Svj hits Svi since j 6= i, i − 1, i + 1. Therefore, by Lemma 44, bvi , bvj ∈ Svi ∩ Svj and since we have more than 2 elimination sets, bvi = bvj must hold. So, by Lemma 60, we know that Svi hits Svj+1 , or Svj hits Svi+1 , both of which are impossible since j 6= i, i − 1, i + 1.

Finally, we proceed with our proof. First, assume that we can ﬁnd an A-type Svi , such that vi is not an in-neighbor of bvi . Without loss of generality, let Sv2 be our A-type that satisﬁes this. 0 0 Then, note that we may reduce v2 in Sv2 to get Sv2 and consider the graph G = R(G, v2). It is strongly connected and so we claim that we can appeal to the characterization theorem with 0 0 V0 = S0 = ∅,V1 = Sv1 ∪ Sv2 ,V2 = Sv2 ∪ Sv3 and Vi = Svi+1 for 3 ≤ i ≤ n − 1. Note that all Vi are strict subsets of G for 3 ≤ i ≤ n − 1, trivially. Furthermore, V1,V2 are also strict subsets since they do not contain the lonely vertices in Sv4 . Sn−1 0 Clearly, we still have i=0 Vi = V (G ). Now, note that v2 and its out-neighbor w2 are lonely vertices in Sv2 and are not in any other elimination sets. So, for i ≥ 3, note that Vi = Svi+1 does not have v2, w2 as out-neighbors. This is because if Svi+1 does, then since Svi+1 hits Svi+2 , we see

15 that either v2, w2 ∈ Svi+1 ∪ Svi+2 . But, by the loneliness of v2, w2, one of Svi+1 ,Svi+2 must be Sv2 , which is impossible. 0 c So, in G , bvi+1 ∈ Vi is still the unique vertex with at least 2 edges from vi to Vi . For V1, we 0 claim that b2 also satisﬁes this constraint. Note that in G , there are two out-neighbors of b2 that c are in Sv2 , however, the problem is that they could be in Sv1 . Assume w is an out-neighbor of b2 in

Sv1 , then note that w must be in Sv3 since Sv2 hits Sv3 . But, Sv1 cannot intersect Sv3 when there are more than 3 A-types, by our second claim. Similarly, this holds for V2. c Lastly, we claim that property 4 holds. If u ∈ V1 and there exists v ∈ V1 such that (u, v) ∈ 0 E(G ), then in G, we know that v ∈ Sv1 or v ∈ Sv2 , and u 6∈ Sv1 ,Sv2 . c Case 1: v ∈ Sv1 . Then, we can ﬁnd w ∈ Sv1 such that (u, w) ∈ E(G). If w 6∈ Sv2 , then note 0 that w 6∈ Sv1 ∪ Sv2 and (u, w) ∈ E(G ), so we are done.

If w ∈ Sv2 , then we see that since S0 = ∅, we claim that u must be a bad vertex of another

A-type, Svk and clearly Svk 6= Sv1 ,Sv2 . If u is not the bad vertex, then there must be an elimination set Svm that contains u and its out-neighbors. But, Svm intersects Sv1 ,Sv2 , a contradiction.

So, there must exist an A-type Svk such that u = bvk . Note that Svk hits Svk+1 and v, w ∈

Svk ∪ Svk+1 . If both v, w ∈ Svk , then note that Svk intersects both Sv1 ,Sv2 , a contradiction to our ﬁrst claim.

If v ∈ Svk , then Svk must be Sv0 = Svn , by our second claim. Then, we know that Svk+1 = Sv1 .

But, w ∈ Svk+1 = Sv1 , which contradicts w 6∈ Sv1 .

So we must have w ∈ Svk and by claim 2, we must have Svk = Sv3 . Since v ∈ Svk+1 , we must have that Svk+1 = Sv0 = Svn . So, we see that n = 4, contradiction.

Case 2: v ∈ Sv2 . This case is symmetric to case 1. Therefore, V1 satisfies property 4. Similarly, we may repeat the same argument for V2. c 0 For i ≥ 3, if u ∈ Vi and there exists v ∈ Vi such that (u, v) ∈ E(G ), then note that (u, v) ∈ E(G) c c 0 0 so we can find w ∈ Vi such that (u, w) ∈ E(G). We see that w ∈ Vi in G and (u, w) ∈ E(G ), unless w = v2 and u = w2. Note that by choice, u = w2 is not the bad vertex, so we must have v ∈ Sv2 ∩ Vi in G, contradicting claim 2. So, all properties are satisfied and we may use the characterization theorem and get a contradiction to minimality. Therefore, G has at most 4 A-types that hit each other in a a cycle.

Note that if we cannot ﬁnd an A-type Svi such that vi is not an in-neighbor of bvi , then all bvi is lonely. By Lemma 42, all Svi are lonely. Now, we may repeat the argument above, except during the last paragraph: c 0 For i ≥ 3, if u ∈ Vi and there exists v ∈ Vi such that (u, v) ∈ E(G ), then note that (u, v) ∈ E(G) c c 0 0 so we can ﬁnd w ∈ Vi such that (u, w) ∈ E(G). We see that w ∈ Vi in G and (u, w) ∈ E(G ), unless w = v2 and u = w2 = bv2 . However, Nout(bv2 ) ⊆ Sv2 ∪ Sv3 , we have that Vi either intersects

Sv2 or Sv3 , contradicing that Vi is lonely.

7.2 Connected Sets

Recall that Su shares its bad vertex with Sv if bu = bv. Note that Su shares its bad vertex with itself.

Deﬁnition 63. An elimination set Su is connected to a non A-type Sv with a connection of length L if there exists elimination sets Sv1 , ...., SvL+1 such that Su = Sv1 and Sv shares its bad vertex with SvL+1 and for 1 ≤ i ≤ L, Svi shares its bad vertex with an elimination set Swi such that Swi hits Svi+1 .

16 It is important to note that an elimination set can only be connected to a non A-type elimination set. Connection to an A-type elimination set is not deﬁned and is omitted for simplicity. Note that an elimination set Su could have multiple connections to multiple non A-types. If Su is a non A-type, then Su is connected to itself with length 0. Also, any A-type Su is connected to some non A-type Sv because there are no cycle of A-types that hit each other. Indeed, if Su hits a non A-type Sv, then Su is connected to Sv. If not, Su hits another A-type, which must hit another elimination set. Eventually an A-type must hit a non A-type, Sv and so Su is connected to Sv. A non A-type elimination set is connected to itself and possibly others if it shares its bad vertex with another elimination set.

Observation 64. For any elimination set Su, there exists Sv such that Su is connected to Sv.

Deﬁnition 65. Let Su be a non A-type in G and let Sv1 , ..., Svk be all elimination sets connected Sk to Su. The connected set of Su is Cu = i=1 Svi

If Su is connected to Sv and Sv is connected to Sw, then Su is connected to Sw by simply concatenating the connections. Similarly, if Su hits Sv and Sv is connected to Sw with length L, then Su is connected to Sw with length L + 1. So these “connections” deﬁne a quasi-order, i.e., they are reﬂexive and transitive.

c Lemma 66. If (u, v) ∈ E(G) and u ∈ Cw, v ∈ Cw, then u = bw. c Proof. Assume that (u, v) ∈ E(G) and u ∈ Cw, v ∈ Cw. There exists some Sx such that u ∈ Sx ⊆ Cw. By Lemma 33, we see that u = bx is forced. If Sx is connected to Sw with length 0, then bx = bw. Therefore, u = bx = bw.

Otherwise, Sx is connected to Sw with length ≥ 1, so there exists Sv1 , ...., SvL+1 such that

Sx = Sv1 and Sv shares its bad vertex with SvL+1 and for 1 ≤ i ≤ L, Svi shares its bad vertex with an elimination set Swi such that Swi hits Svi+1 . By deﬁnition, it is clear that Sw1 ,Sv2 also is connected to Sw and so Sw1 ,Sv2 ⊆ Cw. Since bw1 = bx and Sw1 hits Sv2 , we conclude that S c Nout(bx) ⊆ Sw1 Sv2 ⊆ Cw, contradicting v ∈ Cw.

Lemma 67. If an A-type Su is connected to Sv with length at least L ≥ 1 and Su hits Sw, then Sw ⊆ Cv and Sw is connected to Sv with length ≤ L.

Proof. Since Su is connected to Sv with length L ≥ 1, we see that there are maximal elimination sets Sv1 , ...SvL+1 such that Su = Sv1 and Sv shares its bad vertex with SvL+1 and for 1 ≤ i ≤ L, Svi shares its bad vertex with an elimination set Swi such that Swi hits Svi+1 .

First note that Su shares its bad vertex with Sw1 that hits Sv2 , where Sv2 shares its bad vertex with Sw2 . Since Su hits Sw and bu = bw1 , by Lemma 60, Sw ∩ Sv2 6= ∅. Then, if Sw hits Sv2 , then

Sw is connected to Sv with length L. Sw ⊆ Cv. So, we conclude that bv2 ∈ Sw.

Now, we will proceed by induction on L. For the base case L = 1, note bv2 = bv, so Sv ∩Sw 6= ∅. Since Sv is a non A-type, by Corollary 44, we must have bw ∈ Sv. Now, by Corollary 45, we conclude that bw = bv and Sw ⊆ Cv with length L − 1. Now, we induct on L and assume that our result holds for L = i ≥ 1. Then, let L = i + 1.

Note that if Sv2 hits Sw, then we are done by the inductive hypothesis since there is a connection from Sv2 to Sv with length i. Otherwise, Sv2 does not hit Sw, so bw ∈ Sv2 . By Corollary 45, bw = bv2 = bw2 and so, Sw ⊆ Cv and Sv is connected to Sw with length ≤ L

Lemma 68. Let Su be any elimination set and Su ∩ Cv 6= ∅, then Su ⊆ Cv.

17 Proof. If Su ∩ Cv 6= ∅, then there exists some elimination set Sw ⊆ Cv such that Sw ∩ Su 6= ∅. If Su hits Sw, then Su ⊆ Cv. Otherwise, bw ∈ Su. Also, if bu = bw, then Su ⊆ Cv. Therefore, we conclude that bu 6∈ Sw and Sw hits Su. Now, we do casework based on whether Sw shares its bad vertex as Sv. Case 1: bw = bv. Since Sv is a non A-type and Sw hits Su, by Lemma 61, Sw ∩ Su = ∅, a contradiction. Case 2: bw 6= bv. Then, Sw is connected to Sv with length at least 1 and Sw hits Su, so Su ⊆ Cv by Lemma 67.

Lemma 69. Let Cu,Cv be connected sets and Cu ∩ Cv 6= ∅, then either Cu ⊆ Cv or Cv ⊆ Cu.

Proof. Let w ∈ Cu ∩ Cv, then by Lemma 68, note that any elimination set that contains w will be in Cu,Cv. So, we can ﬁnd some elimination set Sx ⊆ Cu,Cv. As before, note that either bx = bu or bx 6= bu. Case 1: bx = bu. Then Su ∩ Cv 6= ∅ and so, Su ⊆ Cv. Since Su is connected to Sv, we conclude that any set in Su is also connected to Sv, by transitivity. So, Cu ⊆ Cv. Case 2: bx 6= bu. Note that Sx must share its bad vertex with some Sy that is an A-type connected to Su. But then, Su ∩ Cv 6= ∅ and Su ⊆ Cv. If Su shares a bad vertex with Sv, then repeating the argument in case 1 gives Cv ⊆ Cu. 0 Otherwise, Su is an A-type of length at least 1 in both Cu,Cv, so it hits Su and by Lemma 67, 0 Su ⊆ Cu,Cv. Then, repeat the argument until we see either Su,Sv and deduce that either Cu ⊆ Cv or vice versa.

7.3 Minimally Connected Sets

Deﬁnition 70. Let Cu be the connected set of Su and let Cv1 , ..., Cvk be all the properly contained Tk c connected sets of Cu. The minimally connected set of Su is Mu = Cu ∩ i=1 Cvi .

Note that by deﬁnition, if Mu1 , ..., Muk are all the minimally connected sets and Cu1 , ..., Cuk are all the connected sets, then k k [ [ c Mui = Cui = S0 i=1 i=1 The last equality holds because every elimination set is connected to some elimination set.

Lemma 71. If Mu ∩ Mv 6= ∅, then Mu = Mv.

Proof. Note that Cu ∩ Cv 6= ∅. By Lemma 69, either Cu ⊆ Cv or Cv ⊆ Cu. Without loss of generality, Cu ⊆ Cv. If Cu ( Cv, then Mv ∩ Cu = ∅, a contradiction. Therefore, Cu = Cv and by deﬁnition, Mu = Mv. Minimally connected sets that are not equal do not intersect at all, allowing us to isolate the eﬀects of minor operations in one minimally connected set.

Deﬁnition 72. If Su is connected to a non A-type Sv, then the distance from Su to Sv is the minimum L ≥ 0 such that Su has a connection of length L to Sv. We denote this by d(Su,Sv).

Deﬁnition 73. Su is minimally connected to Sv if there does not exist Sw such that d(Su,Sw) < d(Su,Sv).

18 Lemma 74. If Su is minimally connected to Sv and d(Su,Sv) > 0, then there are no non A-types that share the same vertex as Su.

Proof. If Sw is a non A-type and bw = bu, then d(Su,Sw) = 0 is a contradiction of minimality.

Lemma 75. If Su,Sv are connected to Sw and d(Su,Sw) − d(Sv,Sw) ≥ 2, then Su ∩ Sv = ∅.

Proof. Assume otherwise that Su ∩ Sv 6= ∅. Then, by Lemma 43, either Su hits Sv or Sv hits Su or bu = bv. If Su hits Sv, then d(Su,Sw) ≤ d(Sv,Sw) + 1, a contradiction. If Sv hits Su, then by Lemma 67, d(Su,Sw) ≤ d(Sv,Sw). Note that if Sv has distance 0, then Su cannot intersect Sv, by Lemma 60. If bu = bv, then d(Su,Sw) = d(Sv,Sw), a contradiction.

Lemma 76. If Su ⊆ Cv ( Cw, then d(Su,Sv) < d(Su,Sw).

Proof. Note Cv ( Cw implies that d(Sv,Sw) > 0, or else Sv,Sw would share the same vertex and

Cv = Cw. So, Su has a connection of length L = d(Su,Sw) > 0 to Sw, and there exists Sv1 , ...., SvL+1 such that Su = Sv1 and Sw shares its bad vertex with SvL+1 and for 1 ≤ i ≤ L, Svi shares its bad vertex with an elimination set Swi such that Swi hits Svi+1 . But by Lemma 66, the only out-neighbors of Cv is from bv and since Sw 6⊆ Cv, there must exists k ≤ L such that bv ∈ Svk . So, Su is connected to Svk with smaller length and Svk is connected to Sv with length 0, so d(Su,Sv) < d(Su,Sw).

Lemma 77. Mu is the set of all minimally connected sets to Su.

Proof. Let Sv be minimally connected to Su. If it is in Cw ( Cu, then by Lemma 76, we have a contradiction of minimality. Therefore, all minimally connected sets to Su are in Mu. Let Su ⊆ Mu, then note that it must be connected to Su. If it is not minimally connected to Su and is minimally connected to another elimination set Sw, then Cw ( Cu. But, this contradicts Su ⊆ Mu, so Mu is in the set of all minimally sets to Su.

Therefore, we note that if Sw ⊆ Mu is a non A-type, then d(Sw,Su) = 0. And, we see that bw = bu is implied. We now consider an equivalence relation on all minimally connected sets

Mv1 , ..., Mvr using set equivalence avoiding duplicates. The choice of the representatives of these equivalence classes is arbitrary. Under the equivalence relation quotient, by Lemma 71 and Lemma 77, note that none of the minimally connected sets intersect. Furthermore, it is easy to see that if (u, v) ∈ E(G) and u ∈ c Mw, v ∈ Mw, then u = bw, as for connected sets.

8 Finiteness of Minimal Forbidden Minors

8.1 Finiteness of Minimally Connected Sets

Deﬁnition 78. Let Mu be a minimally connected set. Deﬁne the extra set of Mu, Tu ⊆ S0, to be c the smallest subset of S0 such that there does not exist v ∈ S0 ∩ Tu such that Nout(v) ⊆ Mu ∪ Tu.

Note that this is a recursive deﬁnition of Tu and Tu is not the maximal/maximum set in S0 that hits Mu. However, it is true that Tu hits Mu. If there are no vertices in S0 with all of its out-neighbors in Mu, then Tu = ∅. Since Tu hits Mu, we have that if (w, v) ∈ E(G) and c w ∈ Mu ∪ Tu, v ∈ (Mu ∪ Tu) , then w = bu, as before.

19 Lemma 79. (Mu ∪ Tu) ∩ (Mv ∪ Tv) = ∅ for u 6= v.

c Proof. Let H = (Mu ∪ Tu) ∩ (Mv ∪ Tv). If H is non-empty, then it must have out-neighbors in H . So, by our above observation, either bu ∈ Mv ∪ Tv or bv ∈ Mu ∪ Tu. But since Mu ∩ Mv = ∅, we either have bu ∈ Tv or bv ∈ Tu. However, Tu,Tv ⊆ S0, a contradiction.

Lemma 80. Let Su be an non A-type and H be the set of all elimination sets that share the same c bad vertex as Su. Then, Su has at most 2 out-neighbors in H .

c Proof. Assume otherwise and let w1, w2, w3 ∈ H be three out-neighbors of Su. Consider deleting 0 0 0 (bu, w1) to obtain G . Note that G might not be strongly connected. If so, we can ﬁnd A, B ⊆ V (G ) such that w1 ∈ A, bu ∈ B and all edges go from A to B. Note that if v ∈ B, then Nout(v) ⊆ B. 00 0 Now, consider G = G [B], which is strongly connected. If Su1 , ..., Sus is the elimination decomposition of G, then we claim that we can appeal to the characterization theorem with V0 = S0 ∩ B Ss 00 and Vj = Suj ∩ B. It is clear that i=0 Vi = V (G ). 00 00 Next, we claim that Vj 6= G , j ≥ 1. If Vj = G , then bu ∈ Vj, and so bu ∈ Suj in G. If buj 6= bu, then Su is an A-type, a contradiction, so we must have bu = buj . But, this means that c 0 there are at least two out-neighbors from buj to Suj in G and in G , we have only deleted one edge, c so there must still be an out-neighbor of buj in Suj . Note that the out-neighbor is in B, since it is 00 c 00 an out-neighbor of bu ∈ B. Therefore, it is in G ∩ Suj , contradicting Vj = G . We conclude that Vj are still strict subsets. 0 00 We see that if v ∈ S0, then v still has outdegree ≥ 2 in G . If v ∈ G , then its outdegree is not reduced, by our previous observation. Thus, V0 is a set of vertices with outdegree ≥ 2. 00 c Now, if Vj does not contain buj , then Vj = ∅ since otherwise Vj has no out-neighbors in G ∩Vj . c For property 3, note that if buj 6= bu and buj ∈ B, then Vj still has at least two out-neighbors to Vj as Nout(buj ) ⊆ B. If bu = buj , then u2, u3 6∈ Vj since u2, u3 6∈ H and Suj ⊆ H. So property 3 holds. c 00 Now, for property 4, if wj ∈ Vj and xj ∈ Vj such that (wj, xj) ∈ E(G ), then we see that c 0 in G, there existed yj ∈ Vj such that (wj, yj) ∈ E(G). Note that if (wj, yj) ∈ E(G ), then since 00 wj ∈ B, yj ∈ B, so (wj, yj) ∈ E(G ). This does not hold only if wj = bu and yj = u1. Note that if u2 or u3 is not in Vj, then property 4 still holds. We only run into a problem if we can ﬁnd an c elimination set Sv1 such that u1 ∈ Sv1 and u2, u3 ∈ Sv1 .

Now, consider deleting (bu, u2), (bu, u3), and we can ﬁnd Sv2 ,Sv3 so that u2 6∈ Sv2 , u1, u3 ∈

Sv2 and u3 6∈ Sv3 , u1, u2 ∈ Sv3 . It is clear that Sv1 ,Sv2 ,Sv3 mutually intersect. We see that

Sv1 ∩Sv2 ∩Sv3 6= ∅ or else, by Lemma 46, they would form 3 A-types that hit each other, contradicting c c the existence of Su. But u1 ∈ Sv1 ∩ Sv2 ∩ Sv3 and u2 ∈ Sv1 ∩ Sv2 ∩ Sv3 contradicts Lemma 57. So property 4 holds, and G00 is a minor of G with directed pathwidth ≥ 2, contradicting the minimality of G.

c Corollary 81. Let Mu be a minimally connected set. Mu has at most 2 out-neighbors to Mu. c Lemma 82. If Mu has at least 2 out-neighbors in (Mu ∪ Tu) , then there exists another Mv that has at least 2 out-neighbors in Mu ∪ Tu.

0 0 Proof. Consider reducing u in Mu to get Mu and we claim that our resulting graph G is still of directed pathwidth 2, unless we can ﬁnd a minimally connected set Mv with at least two out- neighbors in Mu ∪ Tu.

20 Assume for the sake of contradiction that such a Mv does not exist. Let Su1 , ..., Sus be the c elimination decomposition of G. We try to apply characterization theorem with V0 = S0 ∩ Tu and c 0 Ss+1 0 Vj = Suj ∩ Mu, for 1 ≤ j ≤ s. Lastly, we let Vs+1 = Mu ∪ Tu. Clearly, i=0 Vi = V (G ). Note that all Vj are strict subsets as before and Vs+1 is a strict subset since Mu ∪ Tu has out- c 0 neighbors in (Mu ∪ Tu) in G and thus, Vs+1 is a strict subset in G . Furthermore, note that S0 still have vertices of outdegree ≥ 2 in G0. c Let w be the out-neighbor of u. For property 3, for 1 ≤ j ≤ s, if Vj = Suj ∩Mu is non-empty and c has lost out-neighbors in Vj after the contraction of u, then Vj must have u, w as its out-neighbors c in G. We see that Vj ⊆ Mu and so there must exist some Mv such that Vj ⊆ Mv. Clearly, Mv has at least two out-neighbors in Mu ∪ Tu and this contradicts our assumption that no such Mv exists. So, we see that property 3 holds. For Vs+1, property 3 holds by assumption. c For property 4, for 1 ≤ j ≤ s, realize in G, if x ∈ Suj and there exists y ∈ Suj such that c c (x, y) ∈ E(G), then there exists z ∈ Suj such that (x, z) ∈ E(G). When Vj = Suj ∩ Mu is non- empty, note that unless we have x = w and z = u,(x, z) ∈ E(G0) holds or even if z = u, then 0 c c (x, w) ∈ E(G ) and w ∈ Suj holds. Since Vj ⊆ Mu, it follows that x = w = bu. 0 If bu has all of its out-neighbors in Suj in G , then we see that in G, Nout(bu) ⊆ Suj ∪ {u}. This implies that Su hits Suj , contradicting that Su is a non A-type. 0 c For Vs+1 = Mu ∪ Tu, and if x ∈ Vs+1 and there exists y ∈ Vs+1 such that (x, y) ∈ E(G), then we consider the following cases. c 0 Case 1: If x ∈ S0, then we can find z ∈ Vs+1 such that (x, z) ∈ E(G ) unless and Nout(x) ⊆ Vs+1. However, this implies that x ∈ Tu, a contradiction. Case 2: If x ∈ Mv, then x = bv. Note that in G, we can find z 6= x such that (x, z) ∈ E(G). If 0 z 6∈ Mu ∪ Tu, then z 6∈ Vs+1 and (x, z) ∈ E(G ), so property 4 holds. Otherwise, z ∈ Mu ∪ Tu and Mv has at least two out-neighbors y, z ∈ Vs+1, but this contradicts our assumption. Now, by the characterization theorem, G0 having pathwidth ≥ 2 contradicts the minimality of G. Therefore, we must be able to find Mv such that Mv has at least two out-neighbors in Mu ∪ Tu.

c Lemma 83. If Mu has at least 2 out-neighbors in (Mu ∪ Tu) , then there are exactly two minimal connected sets and its extra set and they hit each other.

Proof. By Lemma 82, we can ﬁnd Mu1 such that it has at least two out-neighbors in Mu ∪ Tu. By c Lemma 79, Mu ∪ Tu ⊆ (Mu1 ∪ Tu1 ) and by Lemma 80, Mu ∪ Tu has exactly 2 out-neighbors in

Mu ∪ Tu. In fact, the outdegree restriction implies Mu1 ∪ Tu1 hits Mu ∪ Tu. And appealing to 82 again, we can ﬁnd Mu2 such that it has at least two out-neighbors in Mu1 ∪ Tu1 .

Let Mu = Mu0 . We can keep doing this to find a sequence of Mu1 ,Mu2 , ... and since the graph is finite, there must exist 0 ≤ m < n such that Mum = Mun . We recall that Mui ∪ Tui hits Sn−1 c Mui+1 ∪ Tui+1 in this sequence for i ≥ 0. So, H = k=m(Muk ∪ Tuk ) has no out-neighbors in H . So, our graph G is made up t different minimally connected sets and their extra sets that hit each other in a cycle. Assume that t > 2 and relabel our minimally connected sets to be Mu1 , ..., Mut .

Let Sv1 , ..., Svs be the elimination decomposition of G. 0 Consider the contraction u2 in Mu2 to get Mu2 . Now, we appeal to characterization theorem in 0 c the resulting graph G with V0 = S0 ∩ Tu and for 1 ≤ j ≤ s ( 2 S ∩ (M 0 )c b 6= b V = vj u2 vj u1 j 0 Svj ∪ Mu2 ∪ Tu2 bvj = bu1

21 s [ 0 0 It is clear that Vi = V (G ). Note that these sets Vj are all strict subsets of V (G ) since each i=0

Svj is in one of Mu1 , ..., Mut and t > 2, so there must exists some Mv such that Vj 6⊆ Mv.

For property 3, for 1 ≤ j ≤ s, if bvj 6= bu1 , then Vj does not have out-neighbors in Mu2 ∪ Tu2 , 0 c c 0 so Vj = Svj ∩ Mu2 ) still has two out-neighbors in Vj . If bvj = bu1 , then Vj = Svj ∪ Mu2 ∪ Tu2 and note that bu2 has at least two out-neighbors in Mu3 ∪ Tu3 . Therefore, property 3 holds. 0 c For property 4, for 1 ≤ j ≤ s, if bvj 6= bu1 , then if (x, y) ∈ E(G ) with x ∈ Vj , y ∈ Vj, then c 0 we can find z ∈ Svj such that (x, z) ∈ E(G). Note that (x, z) ∈ E(G ) unless z = u2. Let w2 0 c c be the out-neighbor of u2. If x 6= w2, then (x, w2) ∈ E(G ) and w2 ∈ Vj since Svj ⊆ Mu2 . If x = w2, z = u2, then x = w2 = bu2 is forced. But, Nout(u2) ⊆ Svj ∪ {u2} implies that Su2 hits Svj , c 0 contradicting that Su2 is a non A-type. So, we can find z ∈ Vj with (x, z) ∈ E(G ). 0 If bvj = bu1 , then we claim that y ∈ Svj . If y 6∈ Svj , then y ∈ Mu2 ∪Tu2 . But since only Mu1 ∪Tu1 has out-neighbors in Mu2 ∪ Tu2 , we must have x ∈ Mu1 ∪ Tu1 , implying x = bu1 , a contradiction c since x ∈ Vj. Therefore, y ∈ Svj and we can find z ∈ Svj such that (x, z) ∈ E(G). Note that z 6∈ Mu2 ∪ Tu2 since x ∈ Mu1 forces x = bu1 , leading to the same contradiction. Therefore, z 6∈ Vj and (x, z) ∈ E(G0). So, G0 has directed pathwidth ≥ 2, contradicting the minimality of G. So, t ≤ 2, but we see that t ≥ 2 since by Lemma 82, there must exist another Mv ∪ Tv that hits Mu ∪ Tu. So, t = 2. Theorem 84. There are at most 2 minimally connected sets.

c Proof. If there exists Mu such that Mu ∪ Tu has at least 2 out-neighbors in (Mu ∪ Tu) , then by Lemma 83, we have exactly 2 minimally connected set. So, all minimally connected sets Mu has c at most 1 out-neighbor in (Mu ∪ Tu) . Assume that there are more than 2 minimally connected sets. Let Mu,Mv be two diﬀerent such sets. By strong connectivity, there is a directed path from bu to Mv, denoted as P (bu,Mv). We see c that except for the endpoints, the vertices of P (bu,Mv) are in (Mu ∪ Tu ∪ Mv ∪ Tv) . Similarly, we can ﬁnd a reverse path P (bv,Mu) with such properties. If x, y ∈ P (bv,Mu) ∩ P (bu,Mv), then note that either there is a directed cycle that includes x, y and we can cycle contract, or if not, then x 6= bu, bv, so we may just out-contract. By cycle contraction, it worth noting that bu is now reduced to bv. So, we may assume that the paths share at most one common vertex. c Otherwise, out-contract each vertex in (Mu ∪ Tu ∪ Mv ∪ Tv) until each path at most 2 edges (one from bu to the common vertex and the common vertex to Mv). Using the characterization theorem, we claim that this subgraph is still of pathwidth ≥ 2. Consider the new graph obtained and let H = Mu ∪ Tu ∪ Mv ∪ Tv ∪ {c}, where c is the possibly non-existent common vertex. Now, let Sv1 , ..., Svs be the elimination sets contained in Mu,Mv and Ss let V0 = Tu ∪ Tv ∪ {c}, Vj = Svj for 1 ≤ j ≤ s. Clearly, i=0 Vj = V (H) and all Vj are strict subsets of H. c Next, note that Tu,Tv has no out-neighbors in (Mu ∪ Mv) in G, so they are still of out-degree ≥ 2. If c exists, then note that is has one out-neighbor in Mu and one in Mv. So, V0 has vertices of out-degree ≥ 2. c It is important to observe that in G, bu has at most 1 out-neighbor in (Mu ∪ Tu) , meaning c that bu has an out-neighbor w ∈ Vj ∩ (Mu ∪ Tu). In H, note that bu also has an out-neighbor in c c (Mu ∪ Tu) and so, it has two distinct vertices in Vj .

22 For property 3, if Vj = Svj does not satisfy bvj = bu or bvj = bv, then Nout(Svj ) is strictly c contained in Mu,Mv, so bvj still has two out-neighbors in Svj . If bvj = bu, then our observation above gives us property 3. Similarly if bvj = bv. c c For property 4, if (x, y) ∈ E(H) and x ∈ Vj, y ∈ Vj , then note that in G, there exists z ∈ Vj such that (x, z) ∈ E(G). If x 6= bu, bv, then z ∈ H and (x, z) ∈ E(H). Otherwise, x = bu and c if Vj ⊆ Mu ∪ Tu, then in H, bu has an out-neighbor in (Mu ∪ Tu) by the above observation. If c Vj ⊆ Mv ∪ Tv, then again, bu has an out-neighbor in Mu ∪ Tu ⊆ (Mv ∪ Tv) . Similarly if x = bv. So, by the characterization theorem, H contradicts the minimality of G and there are at most 2 minimally connected sets.

8.2 Finiteness of Elimination Sets In previous section we gave a bound on the number of minimally connected sets. Furthermore, we have bound the number of out-neighbors of a minimally connected set (and its extra set) to its complement to be at most 2. From this, we intuit that a minimally connected set cannot contain many elimination sets and these elimination sets cannot have large distance. In this section, we show that the number of elimination sets is indeed ﬁnite. We will show that for any minimally connected set Mu, the number of elimination sets in Mu is at most 4. To understand the behavior of Mu, we start by understanding the behavior of S0 and the extra sets.

Theorem 85. All vertices in S0 have outdegree exactly 2.

Proof. Let Su1 , ..., Sus be the elimination decomposition of G and assume that v ∈ S0 have more 0 0 than 2 out-neighbors, say w1, w2, w3. Consider deleting (v, w1) to get G . Note that G might not 0 be strongly connected. If so, we can ﬁnd A, B ⊆ V (G ) such that w1 ∈ A, v ∈ B and all edges go from A to B. Note that if v ∈ B, then Nout(v) ⊆ B. Now, consider G00 = G0[B], which is strongly connected. Consider appealing to the characteriza- Ss 00 tion theorem with V0 = S0 ∩ B and Vj = Suj ∩ B for 1 ≤ j ≤ s. Clearly, we have j=0 Vj = V (G ). 00 00 Furthermore, for 1 ≤ j ≤ s, Vj is a strict subset of G since v ∈ G and v 6∈ Vj. Also, V0 has 0 vertices of outdegree ≥ 2 since all vertices in V0, including v, still retained outdegree ≥ 2 in G . 00 Then, in G , those vertices in B did not lose outdegree, so V0 still has vertices of outdegree ≥ 2.

For property 3, if Vj 6= ∅, note that by Lemma 33, buj ∈ Vj. Now, buj is still the only vertex in c 00 Vj with out-neighbors in Vj and all its out-neighbors are in G . For property 4, we want to show 00 c c 00 that if (x, y) ∈ E(G ) and x ∈ Vj , y ∈ Vj, then there exists z ∈ Vj such that (x, z) ∈ E(G ). Note c that in G, we can ﬁnd z ∈ Svj such that (x, z) ∈ E(G). Unless x = v, z = w1, z ∈ B since x ∈ B and (x, z) ∈ E(G00).

So, we only have a problem when there exists Sv1 such that Nout(v) ⊆ Sv1 ∪ {w1} and w1 6∈ Sv1 .

Now, consider deleting (v, w2), (v, w3) and we can similarly ﬁnd Sv2 ,Sv3 such that Nout(v) ⊆ Sv2 ∪

{w2} and w2 6∈ Sv2 , Nout(v) ⊆ Sv3 ∪ {w3} and w3 6∈ Sv3 . c c c But, w1 ∈ Sv1 ∩ Sv2 ∩ Sv3 , w2 ∈ Sv1 ∩ Sv2 ∩ Sv3 , w3 ∈ Sv1 ∩ Sv2 ∩ Sv3 . Since Sv1 ,Sv2 ,Sv3 pairwise intersect, by Lemma 46, we must have Sv1 ∩ Sv2 ∩ Sv3 6= ∅, or else the existence of v would be contradicted. But, w1, w2 contradicts Lemma 57. We conclude that property 4 must hold for one of w1, w2, w3 and deleting the respective edge gives G0 with directed pathwidth ≥ 2, by the characterization theorem. This contradicts the minimality of G.

23 0 Lemma 86. Let u, v ∈ S0 and (u, v) ∈ E(G) and let G be graph obtained by out-contracting (u, v). 0 0 Then, if S0 = S0 \{u} has vertices of outdegree ≥ 2 in G and there does not exist w ∈ V (G) such that w has u, v as out-neighbors, then G0 has directed pathwidth ≥ 2.

0 Proof. Let Su1 , ..., Sus be the elimination decomposition of G and let G be defined as above. Note that G0 might not be strongly connected, in which case, we can find A, B ⊆ V (G0) such that v ∈ B. 00 0 0 Now, consider G = G [B] and V0 = S0 ∩B and Vj = Suj ∩B. Note that all the properties hold. The first assumption gives property 2 and the second assumption gives property 3,4. By the same argument in Theorem 85, the characterization theorem gives us that G00 has directed pathwidth ≥ 2. This also implies that G0 has directed pathwidth ≥ 2.

Theorem 87. For any Mu, |Tu| ≤ 3. There exists only one vertex v ∈ Tu such that Nout(v)∩Mu 6= ∅.

Proof. It is crucial to note that there are no directed cycles in G[Tu], which can be quickly seen from the definition. This is because if C ⊆ Tu has a directed cycle, then let H ⊆ Tu be all the vertices 0 0 in H with a directed path in G[Tu] to some vertex in C. Consider Tu = Tu \ H and notice that Tu 0 0 has out-neighbors only in Tu ∪ Mu since no vertex in Tu has out-neighbors in H. Furthermore, if 0 v ∈ H, then v has an out-neighbor in H. Therefore, Tu contradicts the minimality of Tu. c Case 1: Assume that we can find a Mu in G such Mu ∪ Tu has 2 out-neighbors in (Mu ∪ Tu) . In this case, by Lemma 83, the graph only contains another Mv ∪Tv such that Mu ∪Tu and Mv ∪Tv hit each other. By Lemma 80, Mu has no out-neighbors in Tu and similarly for Mv. If Mv does not have out-neighbors in Tu, then Tu = ∅ by strongly connectivity. Otherwise, Mv has out-neighbors in Tu and since G[Tu] is a directed acyclic graph, let w ∈ Tu be a vertex with no in-neighbors in Tu and some out-neighbor in Tu. Let x ∈ Tu be an out-neighbor of w in Tu. 0 Consider out-contracting (w, x) in Tu to get Tu. Note that Tu \{w} has no out-neighbors to 0 w, so only Mv (or bv) has w as an out-neighbor. Therefore, Tu still has vertices of outdegree ≥ 2. 0 Consider out-contracting w to its out-neighbor in Tu and since Tu still has vertices of outdegree ≥ 2. If bv does not have w, x as out-neighbors, then Lemma 86 gives a contradiction to the minimality of G. We immediately deduce that if Mv only has one out-neighbor in Tu, then |Tu| ≤ 1. Otherwise, Mv has two out-neighbors in Tu, then w, x must be its only out-neighbors in Tu. Note that x has a directed path Pxy to y ∈ Tu, where Nout(y) ⊆ Mv due to the DAG property. By Theorem 85, we see that w has one other out-neighbor z. If z is on the path, then consider out-contracting Pxz until Pxz becomes only an edge (x, z). Then, out-contract Pzy until z, y are identified. out-contraction preserves outdegree of w to be ≥ 2 since G[Tu] is a directed acyclic graph. Delete any other vertices in Tu. Note that w, x, z has outdegree ≥ 2 and by the characterization theorem, G still has directed pathwidth ≥ 2. Thus, |Tu| ≤ 3. Note that x could have out-neighbors in Mu. If x does, then let y be a be its out-neighbor in Mu. First, out-contract (x, a) and next out-contract (w, z) and note that z is the only remaining vertex in Tu and |Tu| ≤ 1. If z is not on the path and there is a path Pza, where a ∈ Pxy, then first out-contract Pza, and then follow the same argument above. Otherwise, there is a path Pza that does not intersect Pxy such that a ∈ Mu. Then fully out-contract Pxy until x, y are identified. Then, fully out-contract Pza until z and a are identified. Then, delete anything in Tu until only x = y is left in Tu and it has outdegree ≥ 2. By the characterization theorem, it can be checked that G still has directed pathwidth ≥ 2. Thus, |Tu| ≤ 1.

24 c Case 2: All Mu ∪ Tu only has at most 1 out-neighbor in (Mu ∪ Tu) . If there exists a Mu ∪ Tu c with exactly 1 out-neighbor in (Mu ∪ Tu) , then by Lemma 84, G has at most one other Mv ∪ Tv c with also exactly 1 out-neighbor in (Mu ∪ Tu) . Case 2.1: If Mv exists, note that Mu has at most 1 out-neighbor in Tu, by Lemma 80, and Mv also has at most 1 out-neighbor in Tu. We conclude that there does not exist w with two out-neighbors in Tu and so, the same argument in case 1 holds and we deduce that |Tu| ≤ 1. c Case 2.2: If Mv does not exist, then note that H = (Mu ∪ Tu) must be a subset of S0. But by Theorem 85, if w ∈ H, then w has outdegree ≥ 2 and since w 6∈ Tu, w has an out-neighbor in H. We conclude that w does not have two out-neighbors in Tu and so, the same argument in case 1 holds and we deduce that |Tu| ≤ 1. c Case 2.3: Mu ∪ Tu has no out-neighbors in (Mu ∪ Tu) . In this case, Mu has both its out- neighbors in Tu and the same argument in case 1 holds and we deduce that |Tu| ≤ 3.

c c c c Lemma 88. If bu, bv, bw ∈ Su ∩ Sv ∩ Sw and bu has out-neighbors in Su ∩ Sv ∩ Sw and Su ∩ Sv ∩ Sw, then Su,Sv,Sw are the only elimination sets in G.

c c Proof. Consider H = Sw ∩ Su ∩ Sv, which is not empty, since vk ∈ H. By strong connectivity, bu has a directed path to H. Note that if the path is completely contained in I = Su ∪ Sv ∪ Sw, then c c c there exists x ∈ I ∩H , y ∈ H such that (x, y) ∈ E(G). Since x ∈ Su ∪Sv and y ∈ Su ∩Sv, it follows that x = bu. Then, I is a minimal minor by using the characterization theorem on Su,Sv,Sw. c Otherwise, the simple path from bu to H is in I , except for the endpoints. We can out-contract all vertices in Ic along the path and deduce the resulting graph to have directed pathwidth ≥ 2.

Lemma 89. Let G ∈ F. If there are at least two bad vertices in G or S0 6= ∅, then there cannot exist Su such that bu has more than 1 out-neighbor in Su.

Proof. Let G be a minimal forbidden minor with at least 2 bad vertices or S0 6= ∅ and assume for contradiction there exists Su be a maximal elimination set such that bu has at least 2 out-neighbors in Su, say v, w. Let H be the vertex set that contains all Sx such that bx = bu. Since there are at least 2 bad vertices or S0 6= ∅, we know that H is a strict subset of G and by strong connectivity, c it must have an out-neighbor in H . So, by Lemma 33, it must be bu that has an out-neighbor in G/H, say h.

Let Sv1 , ..., Svs be the elimination decomposition of G. For now, consider deleting (bu, w) to 0 0 get G . Note that deleting (bu, w) might have removed strong connectivity, so in G , there might c be vertices in Su that cannot be reached from bu. Note that all other vertices in Su can still be 0 0 reached from bu in G since bu ∈ G . So, we can ﬁnd two subsets A, B such that all edges go from A to B and w ∈ A, A ⊆ Su, bu ∈ B. 00 0 Now, consider the graph G = G [B] and invoking the characterization theorem with V0 = S0 ∩ 0 00 B,Vi = Svi ∩ B for 1 ≤ i ≤ s. Note that removing A does not reduce outdegree from G to G . 00 And from G to G , only the outdegree of bu, which was reduced by 1. So, V0 still is a set of vertices of outdegree ≥ 2. And since A is a strict subset of Su, we see that we have V0,V1, ..., Vn are still 00 Ss 00 strict subsets of G with i=0 Vi = V (G ). Now, for property 3 to hold, we only run into a problem if there exists Sx with bx = bu that contains all out-neighbors of bu except w, h. Case 1: Assume such a Sx exists. Then, consider deleting (bu, v) instead and note that we run into a problem if there exists Sy with by = bu that contains all but v, h. However, we see that c c by ∈ Su ∩ Sx ∩ Sy, but v ∈ Su ∩ Sx ∩ Sy and w ∈ Su ∩ Sx ∩ Sy, a contradiction to Lemma 57

25 Now, consider property 4, and note that it holds unless there exists Sy that contains all out- neighbors of bu except v. Note that Su,Sv,Sw mutually intersect and therefore, by Lemma 46, they must have a triple intersection. However, this violates Lemma 57 as before. So property 4 must hold. Case 2: Such a Sx does not exist. By symmetry, we may assume there does not exist Sy with by = bu that contains all out-neighbors except v, h. So, property 3 holds if we delete (bu, w) or (bu, v). So, consider deleting (bu, w) again, and note that property 4 holds unless we can ﬁnd some Sy that contains all out-neighbors of bu except w. Then, delete (bu, v) and that fails only if Sz contains all out-neighbors of bu except v. But again, Su,Sy,Sz mutually intersect, and as before, we derive a contradiction. 0 Therefore, deleting one of (bu, v), (bu, w) gives G , which is still of directed pathwidth ≥ 2, a contradiction.

Lemma 90. Let Mu be a minimally connected set with more than 4 elimination sets. For any Sv ⊆ Mu, there exists Sw such that either Sv hits Sw or Sw hits Sv or bv = bw. Furthermore, either c c 1) There exists a non A-type Sx such that bx has out-neighbors in Sv ∩ Sw and Sv ∩ Sw and Nout(bx) ⊆ Sv ∪ Sw c c 2) There exists x ∈ S0 such that such that x has out-neighbors in Sv ∩ Sw and Sv ∩ Sw and Nout(x) ⊆ Sv ∪ Sw c c 3) bv = bu and bu has an out-neighbor in Sw ∩ Sv and bu has only 1 out-neighbor in (Sw ∪ Sv) c c 4) bw = bu and bu has an out-neighbor in Sv ∩ Sw and bu has only 1 out-neighbor in (Sw ∪ Sv) Proof. By Lemma 84, we can have at most two minimally connected sets. Furthermore, it is important to note that by Lemma 82, if Sv is any elimination set in Mu, then bv has at most 2 c out-neighbors in Mu. 0 Now, consider any Sv ⊆ Mu and reducing v to get Sv. By minimality, there must exists an elimination ordering u1, ..., um and its corresponding out-sequence such that R(G, [u1, ..., um]) is only 1 vertex. Realize u1 6∈ S0 or else Nout(u1) ⊆ Sv follows. Also, u1 6∈ Sv, or else it is clear 0c that we cannot eliminate any vertex in Sv . Thus, u1 is in some elimination set Sw 6= Sv. We may assume that u1 = w. We claim that either Sv hits Sw or Sw hits Sv or bv = bw. If Sv ∩ Sw 6= ∅, then we are done by Lemma 41 and Lemma 43. So, we may assume that Sv ∩ Sw = ∅. Let p be minimal such that up 6∈ Sw or vp 6∈ Sw. 0 0 Case 1: up 6∈ Sw. We claim that up ∈ Sv. Since up can be contracted in G , we see that in G , Nout(up) ⊆ Sw. However, Nout(up) ( Sw, so it must be that up has v as an out-neighbor. Let x 0 be the out-neighbor of v. However, if up 6∈ Sv, then up would still have x as an out-neighbor in G and x 6∈ Sw since x is lonely. We conclude that up ∈ Sv. Note that since Sv,Sw do not intersect, p is minimal such that up ∈ Sv and since up has out- c neighbors in Sv, up = bv is forced. This implies that Nout(up) ⊆ Sv ∪ Sw, implies that Sv hits Sw. Case 2: vp 6∈ Sw. Similarly, we claim that vp ∈ Sv. Note that outdegree of up must have been 0 0 reduced in G , so up must have x as an out-neighbor in G . Since x 6∈ Sw, it follows that vp = x ∈ Sv. Then, since bw has not been reduced, we conclude that up = bw. So, Nout(up) ⊆ Sv ∪ Sw, implying

26 that Sw hits Sv.

Now, consider the minimal n such that either un 6∈ Sv ∪ Sw or vn 6∈ Sv ∪ Sw. Case 1: un 6∈ Sv ∪ Sw. In G, we see that Nout(un) ⊆ Sv ∪ Sw and by Lemma 34, un has an c c out-neighbor in w1 ∈ Sv ∩ Sw and w2 ∈ Sv ∩ Sw. We claim that un ∈ S0 or that it is a bad vertex of a non A-type. If not, then either un is a bad vertex of an A-type or un is contained in some elimination set Sy but is not the bad vertex of Sy. Case 1.1: un is a bad vertex of an A-type Sx and let Sx hits Sy. Since Nout(un) ⊆ Sx ∪ Sy, we c c see that w1, w2 ∈ Sx∪Sy. If w1, w2 ∈ Sx, then w1 ∈ Sv ∩Sw ∩Sx and w2 ∈ Sv ∩Sw ∩Sx, contradicting c c Lemma 57. Similarly for Sy. Without loss of generality, let w1 ∈ Sx ∩ Sy and w2 ∈ Sy ∩ Sx. Note that un = bx 6∈ Sv, but w1 ∈ Sx ∩ Sv, so Sv hits Sx and bv ∈ Sx. From above, we now split into cases: Sv hits Sw, Sw hits Sv, or bv = bw. Case 1.1.1: Sv hits Sw. Then, Sw ∩ Sx 6= ∅ and since bx 6∈ Sw, we conclude that Sw hits Sx and bw ∈ Sx. Note that Sy cannot hit Sw, or else Sx,Sy,Sw hits each other in a cycle. So, bw ∈ Sy. And, notice bw ∈ Sx ∩ Sy ∩ Sw. Since Sy cannot hit Sx, it also follows that bx ∈ Sy. But, c c bx ∈ Sx ∩ Sy ∩ Sw and w2 ∈ Sx ∩ Sy ∩ Sw, contradicting Lemma 57. Case 1.1.2: Sw hits Sv. If Sy hits Sw, then we have a cycle of A-types that hit each other in a cycle, so bw ∈ Sy. Now, either bw = by or bw 6= by. Case 1.1.2.1: bw = by. Then, since Sw hits Sv, Nout(by) ⊆ Sw ∪ Sv. Therefore, H = c Sv ∪Sw ∪Sx ∪Sy has no out-neighbors to H , contradicting that G = H has more than 4 elimination sets. Case 1.1.2.2: bw 6= by. Then, we know that by 6∈ Sw and Sw hits Sy. Since Sw hits Sv,Sy, we see that Sv ∩ Sy 6= ∅. and Sy cannot hit Sv, bv ∈ Sy. Notice bv ∈ Sx ∩ Sy ∩ Sv. Since Sy cannot hit c c Sx, bx ∈ Sy. But, bx ∈ Sx ∩ Sy ∩ Sv and w1 ∈ Sx ∩ Sy ∩ Sv contradicts Lemma 57.

Case 1.1.3: bv = bw. Then, since bv ∈ Sx, we see that bv ∈ Sx ∩ Sv ∩ Sw. Note that if Sy hits c Sw, then H = Sv ∪ Sw ∪ Sx ∪ Sy has no out-neighbors to H , contradicting that G = H has more than 4 elimination sets. Therefore, bw ∈ Sy and bw ∈ Sw ∩ Sx ∩ Sy. Since Sy cannot hit Sx, it also c c follows that bx ∈ Sy. But, bx ∈ Sx ∩ Sy ∩ Sw and w2 ∈ Sx ∩ Sy ∩ Sw, contradicting Lemma 57.

Case 1.2: un is contained in Sy and un is not the bad vertex. Then, Nout(un) ⊆ Sy, but c c w1 ∈ Sv ∩ Sw ∩ Sy and w2 ∈ Sv ∩ Sw ∩ Sy contradicts Lemma 57. We conclude that un ∈ S0 or that it is a bad vertex of a non A-type and they have out-neighbors c c in Sv ∩ Sw and Sv ∩ Sw. Note that Nout(un) ⊆ Sv ∪ Sw. Thus, either 1) or 2) holds.

Case 2: vn 6∈ Sv ∪Sw. Note that since vn 6∈ Sv ∪Sw and n is minimal, we conclude that un = bv or un = bw. We claim that un = bu. Assume for contradiction that un 6= bu, then un must be the bad vertex of an A-type in Mu. First, we realize that Nout(un) ⊆ Sv ∪ Sw ∪ {vn}. Now, we split into cases. Case 2.1: un = bv. Since Nout(un) ⊆ Sv ∪ Sw ∪ {vn}, we know that un must have an out- c neighbor w1 in Sw ∩Sv. Since Sv is an A-type, it hits Sx and vn ∈ Sx. Note that if Sx hits Sw, then c H = Sx ∪ Sw ∪ Sv has no out-neighbors in H , contradicting our assumption. Therefore, bw ∈ Sx. Also, note that Sv hits Sw is impossible since bv would have been reduced. So, either bv = bw or Sw hits Sv. Case 2.1.1: un = bv = bw. In this case, Sv,Sw are both A-types that hit Sx. However since

27 c c un = bw, un has at least two out-neighbors in Sw. Note that if both are in Sv, then bv cannot be c c reduced. So, un has an out-neighbor w2 ∈ Sw ∩ Sv. But, bw ∈ Sx ∩ Sv ∩ Sw and w1 ∈ Sw ∩ Sv ∩ Sx c and w2 ∈ Sw ∩ Sv ∩ Sx, contradicting Lemma 57. Case 2.1.2: Sw hits Sv. Note that Sv cannot hit Sw, so bw ∈ Sv. Since w1 ∈ Sx ∩ Sw and Sx cannot hit Sw, we conclude that bw ∈ Sx. Therefore, bw ∈ Sv ∩ Sw ∩ Sx. So, either Sw hits Sx or bw = bx. Case 2.1.2.1: Sw hits Sx. Then, Sw hits Sx,Sv. So, Sw must have out-neighbors w2 ∈ c c Sw ∩ Sx ∩ Sv. But, w1 ∈ Sw ∩ Sx ∩ Sv, contradicting Lemma 57. Case 2.1.2.2: bw = bx. Then, Nout(bx) ⊆ Sv ∪ Sw, so H = Sv ∪ Sw ∪ Sx has no out-neighbors in Hc, contradicting that G has more than 4 elimination sets.

Case 2.2: un = bw. Note that the same argument for Case 2.1 holds for Case 2.2.

c We conclude that un = bu and bu has at most one out-neighbor outside (Su ∪ Sw) .

Theorem 91. Let Mu be a minimally connected set with more than 4 elimination sets. Then, Mu c does not have 2 out-neighbors in (Mu ∪ Tu) .

c Proof. Assume that Mu has 2 out-neighbors in (Mu ∪ Tu) . By Lemma 83, we have Mu,Mv such c that Mu ∪ Tu and Mv ∪ Tv hit each other. Furthermore, Mv has 2 out-neighbors in (Mv ∪ Tv) . If Mv has 2 out-neighbors in Mu and ﬁrst assume they are contained in (possibly non-distinct) elimination sets Sx,Sy ⊆ Mu such that either bx = by, or Sx hits Sy or Sy hits Sx. Let us call c this the containment property. Now, if the two out-neighbors of Mu in Mu are contained in some elimination set Sz, then there exists a ∈ Mu such that (u, a) ∈ E(G). Otherwise, a is just an empty vertex. Also, note that Tu = ∅ since Mu,Mv cannot hit Tu by Lemma 80. c c c Now, out-contract vertices in Mu ∩Sw ∩Sx ∩{a, bu} . If there are 3 elimination sets in Mu, then the previous set was certain non-empty due to the existence of lonely vertices. Now, we appeal to c the characterization theorem with V0 = S0, Vj = Svj ∩ Mu for 1 ≤ j ≤ s, and Vs+1 = Sw ∪ {a, bu} and Vs+2 = Sx ∪ {a, bu}. Checking the properties in the characterization theorem, we see that in G0, we still have directed pathwidth≥ 2, a contradiction.

So, the containment property does not hold. Then, assume there exists a Sw ⊆ Mu such that d(Sw,Su) ≥ 2. By Lemma 90, we can ﬁnd Sx such that either Sx hits Sw or Sw hits Sx or bw = bx. Note that Sx ⊆ Mu or else Sw would contain all the out-neighbors of Mu and the containment c property holds. And, we claim that there is a vertex y ∈ Tu that has out-neighbors in Sx ∩ Sw and c Sx ∩ Sw. By Lemma 90, we have the following cases: c c Case 1: There exists a non A-type Sy such that by has out-neighbors in Sx ∩ Sw and Sx ∩ Sw and Nout(by) ⊆ Sx ∪ Sw. It suﬃces to show that Sy shares a bad vertex with Sv. Since there are only minimally connected sets, we see that Sy must share a bad vertex with Su or Sv. Assume that the former holds. Let H be the set of elimination sets that have bad vertex bu. Since Mu already has 2 out- c c neighbors in (Mu ∪ Tu) , by Lemma 80, Mu has no out-neighbor in H ∩ Mu. Therefore, if the c former does hold, then bu has an out-neighbor w1 ∈ Sx ∩ Sw and w1 ∈ H is forced. But, H only contain sets of distance 0 and by Lemma 75, Sw cannot have distance ≥ 2. Then, Mu has both out-neighbors in Sx,Sw, and so the containment property holds.

28 c c Case 2: There exists y ∈ S0 such that such that y has out-neighbors in Sx ∩ Sw and Sx ∩ Sw and Nout(y) ⊆ Sx ∪ Sw. Since Sx ⊆ Mu, we conclude that y ∈ Tu. c Case 3: bx = bu and bu has an out-neighbor in Sw ∩ Sx and bu has only 1 out-neighbor in c (Sw ∪ Sx) . Since bu has an out-neighbor in Sw, which is impossible by similar argumentation in Case 1. c Case 4: bw = bu and bu has an out-neighbor in Sx ∩ Sw and bu has only 1 out-neighbor in c (Sw ∪ Sx) . This is not possible because if bw = bu, then Sw has distance 0, a contradiction.

So, our claim holds and Mv must have an out-neighbor in Tu. If Mv has 1 out-neighbor in Mu, then note that the out-neighbor must be in Sx ∪ Sw. Otherwise, we can reduce Tu and we would c still have a graph of directed pathwidth ≥ 2, by the characterization theorem with V0 = S0 ∩ Tu and Vj = Svj . c c c However, we can follow the same procedure as described above to contract Mu∩Sw ∩Sx∩{a, bu} and derive a contradiction. If all Sw ⊆ Mu has d(Sw,Su) ≤ 1, then let Sx,Sy be the two elimination sets that contain all out-neighbors of Mv and Tu. Then, let Sx hit Sw1 or Sx = Sw1 , where bw1 = bu. And similarly for c c c c c Sy, deﬁne a Sw2 . Then out-contract any vertex in Mu ∩ Sx ∩ Sy ∩ Sw1 ∩ Sw2 ∩ {a, bu} and derive a contradiction.

Theorem 92. Let Mu be a minimally connected set with more than 4 elimination sets. Then, Mu c does not have any out-neighbors in (Mu ∪ Tu) .

c Proof. Assume that Mu has out-neighbors in (Mu ∪ Tu) . By Theorem 91, we know that Mu has c at most 1 out-neighbor in (Mu ∪ Tu) . By Lemma 84, we know that there is at most one other minimally connected set in G. Case 1: There exists Mv,Tv such that Mu ∪ Tu and Mv ∪ Tv hits each other and Mv ∪ Tv also c c has 1 out-neighbor in (Mv ∪ Tv) . If Mu has at least 2 out-neighbors in Mu, then we can follow the argument in Lemma 91 to deduce our result. c Otherwise Mu only has 1 out-neighbor in Mu. This means that Tu = ∅. Even if Mv hits Tu, then we can still contract Tu and it would still have directed pathwidth ≥ 2. Now, assume there exists a set Sw ⊆ Mu such that d(Sw,Su) ≥ 1. Then by Lemma 90, we can find Sx such that Sw hits Sx or Sx hits Sw or bx = bw. Furthermore, we have the following cases c Case 1.1: If there exists a non A-type Sy such that by has out-neighbors w1 ∈ Sw ∩ Sx and c c w2 ∈ Sx ∩ Sw. Note that by 6= bv since it has only 1 out-neighbor in (Mv ∪ Tv) . Therefore, by = bu. Let H be all set of all distance 0 elimination sets in Mu. Note that if w1, w2 6∈ H, then Lemma 80 is contradicted. So, one of w1, w2 ∈ H and we note that d(Sw,Su) ≤ 2 or else Sw,Sx would have distance ≥ 2, contradicting Lemma 75. If exactly one of w1, w2 6∈ H then WLOG, let it be w1. If d(Sw,Su) = 2, then we can find Sy,Sz c c c such that Sw hits Sy, which hits Sz, where bz = bu. We can contract all else in Mu ∩ Sw ∩ Sy ∩ Sz and show by the characterization theorem that directed pathwidth ≥ 2. So, there are at most 3 elimination sets in Mu. We use a similar argument if d(Sw,Su) = 1. Otherwise, we have w1, w2 ∈ H, there exists Sz of distance 0 such that w1 ∈ Sz. By the same c reasoning, we can find Sa of distance 0 such that w2 ∈ Sa. By Lemma 89, w1 ∈ Sz ∩ Sa and c c c w2 ∈ Sa ∩ Sz. Now, we may contract all else in Mu ∩ Sz ∩ Sa and show by equivalence that directed pathwidth ≥ 2. So, there are at most two elimination sets in Mu.

29 Case 1.2: There exists y ∈ S0 with out-neighbors in Sw ∪ Sx. However, note that Sx ⊆ Mu c since Mv only has 1 out-neighbors in (Mv ∪ Tv) . So, y ∈ Tu but Tu = ∅, contradiction. c Case 1.3: bw = bu and and bu has an out-neighbor in Sx ∩ Sw. This contradicts the distance assumption of Sw. c Case 1.4: bx = bu and and bu has 1 out-neighbor in (Sw ∪ Sx) . has an out-neighbor in c w1 ∈ Sw ∩ Sx. If w1 6∈ H, then follow the argument in Case 1.1 to deduce that there are at most 2 elimination sets. Otherwise, w1 ∈ H. So, we can find Sy that contains w1 such that by = bu. Note c that bu contains two out-neighbors in Sy and it already contains an out-neighbor in Mv ∪ Tv. But, c c bu has only 1 out-neighbor in (Sw ∪ Sx) , so bu has an out-neighbor w2 ∈ Sx ∩ Sy. We conclude as in Case 1.1 If we cannot find a Sw with distance ≥ 1, we see that all elimination sets are of distance 0. c Since Mu only has 1 out-neighbor in Mu, there must exist Sx of distance 0 such that bu has an c out-neighbor in Sx. However, since bu must have at least 2 out-neighbors in Sx, we must be able to find Sy of distance 0 such that bu has an out-neighbor in Sy. By Lemma 89, these out-neighbors c c c c are in Sx ∩ Sy and Sx ∩ Sy. Then, we contract all else in Mu ∩ Sx ∩ Sy and deduce that there are at most two elimination sets in Mu. c c Case 2: There does not exist Mv,Tv. Therefore, (Mu ∪ Tu) ⊆ S0. However, if v ∈ S0 ∩ Tu, then v has at most 1 out-neighbor in Mu ∪ Tu since if v has both out-neighbors in Mu ∪ Tu, then by 85, v ∈ Tu must follow. Thus, we may conclude by similar argument in Case 1.

Theorem 93. Mu has at most 4 elimination sets.

Proof. Assume that Mu has more than 4 elimination sets. Then, by Theorem 91, Mu has no out-neighbors outside Mu ∪ Tu. So, G = Mu ∪ Tu. c Let H denote the set of all elimination sets of distance 0 in Mu. If Mu has out-neighbors in Mu, then by Lemma 87, there exists only one v ∈ Tu such that v has out-neighbors in Mu. Let Sx,Sy be the elimination sets that contain the 2 out-neighbors of v. If one of Sx,Sy has distance ≥ 2, c then if Mu has an out-neighbor w ∈ Mu ∩ H , then w ∈ Sx ∪ Sy, or else we may just out-contract all vertices in Tu into the elimination set of distance ≥ 2. So, we can follow the cases in Lemma 91. c So, we may assume that Mu has no out-neighbors in Mu and Tu = ∅. By Lemma 80, H has at c most 2 out-neighbors in H . First note that for any Sv of distance ≥ 2 in Mu, by Lemma 90, there c c must exists Sw such that bu has out-neighbors in Sw ∩ Sv and Sv ∩ Sw. We claim that Mu has no elimination sets of distance ≥ 4. Assume otherwise and note that we can ﬁnd Sv ⊆ Mu be an elimination set in Mu such that Sv is not hit by another elimination set and its distance is ≥ 4. Then, by Lemma 90, there must exist Sw such that either Sv hits Sw or c c bw = bv and bu has out-neighbors in w1 ∈ Sw ∩ Sv and w2 ∈ Sw ∩ Sv and Nout(bu) ⊆ Sv ∪ Sw. Note that Sv has distance ≥ 3. Note that Sw hits another elimination set Sx, which has distance ≥ 2, c so there exists Sy such that bu has out-neighbors in w3 ∈ Sy ∩ Sx and w4 ∈ Sx ∩ Sy. Note that c c w3 6= w4 and w1 6= w2. Since bu has at most 2 out-neighbors in H and w1, w2, w3 ∈ H by Lemma 75, we must have w2 = w3 and w4 ∈ H. But w4 ∈ H contradicts Nout(bu) ⊆ Sv ∪ Sw. If Sv ⊆ Mu is an elimination set of distance ≥ 2, then the same argument above shows that there must exists Sw such that Nout(bu) ⊆ Sv ∪ Sw. If Sw hits Sv, then since Sv is of distance 2, there exists Su1 ,Su2 such that Sv hits Su1 and Su1 hits Su2 with bu2 = bu. By strong connectivity, note that Sw,Sv,Su1 ,Su2 contains all its out-neighbors. So, we conclude that there are at most 4 elimination sets.

30 We can similarly conclude if Sv hits Sw or if bw = bv. So, we may assume that all elimination sets are of distance ≤ 1. If there exists Sv ⊆ Mu a elimination set of distance 1, then by 90, we have Sw with the given properties and the following cases: c c Case 1: Note that bu has out-neighbors in Sv ∩ Sw and Sv ∩ Sw and Nout(bu) ⊆ Sv ∪ Sw. Let

Sv hit Su1 and bu1 = bu, then clearly, Sv,Sw,Su1 contains all its out-neighbors and we conclude there are at most 4 elimination sets Case 2: This is impossible as Tu = ∅ Case 3: bv = bw is impossible as the distance of Sv is 1. c Case 4: bw = bu and bu has an out-neighbor in Sv ∩ Sw and bu has 1 out-neighbor in w1 ∈ c (Sw ∪ Sv) . Let Su1 contain w1 and since it is of distance ≤ 1, there exists Su2 , where Su1 = Su2 or

Su1 hits Su2 and bu2 = bu. Then, we conclude since Sv,Sw,Su1 ,Su2 contains all of its out-neighbors. So we may assume that G = H. Assume we can find Sv,Sw such that there exists out-neighbors c c of bu, u1, u2, such that u1 ∈ Sv ∩Sw and u2 ∈ Sv1 ∩Sv2 . If G = Sv1 ∪Sv2 , then it at most 4 elimination c sets. Otherwise, note that there must be an out-neighbor of bu in (Sv ∪ Sw) and say u3 ∈ Sx. Note that if both u1, u2 ∈ Sx, then since bu ∈ Sv ∪ Sw ∪ Sx, we contradict Lemma 57. Without c c loss of generality, let u1 ∈ Sv ∩ Sw ∩ Sx. By Lemma 88, we have at most 3 elimination sets since c c u3 ∈ Sv ∩ Sw ∩ Sx. If we cannot find such Sv,Sw, then we claim that we can find one elimination set Sx that contain Nout(bu). Assume otherwise, then we can find Sy such that |Sy ∩ Nout(bu)| is maximized. By our assumption, we can find u1 ∈ Nout(bu) such that u1 6∈ Sy. However, we can also find Sz c such that u1 ∈ Sz. Note that if there exists ui ∈ Sy ∩ Sz, then we are done. So, we conclude that |Sy ∩ Nout(bu)| < |Sz ∩ Nout(bu)|, contradicting our assumption. Therefore, Sx contains all the out-neighbors of bu and we have only 1 elimination set in G.

8.3 Finiteness of Forbidden Minors

Theorem 94. Let Su be an elimination set, then |Su| ≤ 160000.

Proof. We know that there exists a simple directed path from u to bu by following the elimination ordering (and disregarding in-contractions), let it be Pubu . We call this the main path. We first claim that this path is bounded. Let v1, ..., vk be vertices on this path, where v1 = u, vk = bu. 0 Then, for 1 ≤ i < k, consider the graph obtained by out-contracting (vi, vi+1), say G . Note that if G0 is not strongly connected, then we can consider strongly connected components with as long as all bad vertices are still in the strongly connected component. 0 We want to appeal to the characterization theorem with V0 = S0,Vj = Swj ∩ G for 1 ≤ j ≤ s. The first few properties hold trivially. Note that S0 has vertices of outdegree ≥ 2 since if x ∈ S0 has lost outdegree, then Nout(x) ⊆ Su, which would imply x ∈ Sv, a contradiction. So, by the characterization theorem, there must exists Vj = Svj such that either property 3 fails or property 4 fails. c If property 3 fails, then we see that either bwj had vi, vi+1 ∈ Vj as out-neighbors, bwj = vi c or bwj = vi+1 and it has vi ∈ Vj as an out-neighbor. If property 4 fails, then it must be that c vi ∈ Vj , vi+1 ∈ Vj. But, note that the number of elimination sets and bad vertices are bounded to c 8. Note that by if vi ∈ Vj , vi+1 ∈ Vj, then vk ∈ Vj unless there exists l such that vl = bwj . Therefore, we conclude that for each bad vertex bwj can only prevent the contraction of at most 16 vertices. Furthermore, the second condition can only prevent the contraction of at most 8 vertices. We also have to make sure that we do not disconnect an elimination set from its bad vertex. It suffices to

31 show that for any elimination set Sx, Pxbx is still a directed path. This is only a problem when vi is an intersection vertex for Pxbx and Pubu and vi+1 is not, since if both vi, vi+1 are intersection points, then we may either cycle contract or out-contract without problems. However, consider contracting vi+1 and note that we run into a problem only if property 3 fails and vi+1, vi+2 is a bad vertex, so there are at most 8 more possible vertices when this can be a problem. Since there are 8 different paths Pxbx , gives an additional 8*8 = 64 vertices. Therefore, Pubu has at most 100 vertices. Consider the maximal elimination ordering of Su, say u1, ..., um and it outsequence w1, ..., wm. If there exists a vertex x such that wi, ..., wi+k = x, then we claim that k ≤ 25. Now, consider ui+1, ..., ui+k and find a ui+l such that it does not have ui+1, ..., ui+k as its in-neighbor. Consider contracting directed path from ui+l to x in G. Note that property 3 holds, we can find Vj = Svj such that unless a bad vertex is on that path or that the bad vertex is w and one of its two out- c neighbors in Vj is on that path. Note that property 4 holds since ui+l does not have ui+1, ..., ui+k as its in-neighbor. So, there are at most 16 such paths. By the same argument before, there are at most 100 vertices on these paths.

So, our most conservative bound is that for each 100 vertices on Pubu , there could be these 16 paths that are reduced at each vertex along Pubu , each of which has 100 vertices. Hence Su has at most 100*16*100 = 160000 vertices.

Theorem 95. Let G be a forbidden minor, then |S0| ≤ 7

Proof. Note that if there exists two minimally connected sets, Mu,Mv, then by the argument in c c Lemma 83, Lemma 84, we see that |S0 ∩ Tu ∩ Tv | ≤ 1. Since |Tu| , |Tv| ≤ 3 by Lemma 87, we conclude that |S0| ≤ 7. Otherwise, note that if no minimally connected sets exist, then we must have A-types that hit each other in a cycle. This implies that S0 = ∅. This leaves us the case when we only have one c minimally connected set Mu. Note that if (Mu ∪ Tu) = ∅, then S0 = Tu, so |S0| ≤ 3. c c Otherwise, Mu ∪ Tu only has 1 out-neighbor x ∈ (Mu ∪ Tu) = S0 ∩ Tu. Let us consider the case c when (bu, x) 6∈ E(G) and (bu, x) ∈ E(G). Now, by Lemma 86, note that if we can ﬁnd u, v ∈ S0 ∩Tu, 0 c such that out-contracting (u, v) still preserves the outdegree of the vertices in S0 ∩Tu, then we derive a contradiction. Since |Tu| ≤ 3, we conclude that |S0| ≤ 7. Theorem 96. F 0 is ﬁnite.

Proof. It suﬃces to show that F 0 only contains graphs with a bounded vertex count. Let G ⊆ F 0. By Lemma 84 and Theorem 93, we conclude that there are at most 8 elimination sets in G. So, by Ps Theorem 94 and Theorem 95, we conclude that |V (G)| ≤ |S0| + i=1 |Svi | is bounded.

References

[APC90] Stefan Arnborg, Andrzej Proskurowski, and Derek G. Corneil. Forbidden minors characterization of partial 3-trees. Discrete Mathematics, 80(1):1–19, 1990.2

[Bar06] Janos Barat. Directed path-width and monotonicity in digraph searching. Graphs and Combinatorics, 22(2):161–172, 2006.1,5

32 [BFKL87] R.L. Bryant, M.R. Fellows, N.G. Kinnersley, and M.A. Langston. On ﬁnding obstruction sets and polynomial-time algorithms for gate matrix layout. Proceedings of the 25th Allerton Conference on Communication, Control, and Computing, 1987.1,2

[CS11] Maria Chudnovsky and Paul D. Seymour. A well-quasi-order for tournaments. J. Comb. Theory, Ser. B, 101(1):47–53, 2011.2

[Die05] Reinhard Diestel. Graph Theory. Springer, 3 edition, 2005.2

[Hun07] Paul William Hunter. Complexity and inﬁnite games on ﬁnite graphs. PhD thesis, University of Cambridge, November 2007.2

[JRST01] Thor Johnson, Neil Robertson, P. D. Seymour, and Robin Thomas. Directed tree-width. J. Comb. Theory Ser. B, 82(1):138–154, May 2001.2,4

[Kim13] Ilhee Kim. On Containment Relations in Directed Graphs. PhD Thesis, 2013.5

[Kin13] Shiva Kintali. Directed Minors I. Preprint, 2013.2,4

[KL94] Nancy G. Kinnersley and Michael A. Langston. Obstruction set isolation for the gate matrix layout problem. Discrete Applied Mathematics, 54(2-3):169–213, 1994.2

[KS12] Ilhee Kim and Paul D. Seymour. Tournament minors. CoRR, abs/1206.3135, 2012.2, 5

[KZ13] Shiva Kintali and Qiuyi Zhang. Forbidden Directed Minors and Kelly-width. Submitted. http://arxiv.org/abs/1308.5170, 2013.2,6

[MTV10] Daniel Meister, Jan Arne Telle, and Martin Vatshelle. Recognizing digraphs of kelly- width 2. Discrete Applied Mathematics, 158(7):741–746, 2010.

[RS04] Neil Robertson and Paul D. Seymour. Graph Minors. XX. Wagner’s conjecture. J. Comb. Theory, Ser. B, 92(2):325–357, 2004.1,3

[ST90] A. Satyanarayana and L. Tung. Forbidden minors characterization of partial 3-trees. Networks, 20(3):299–322, 1990.2

[Tam11] Hisao Tamaki. A polynomial time algorithm for bounded directed pathwidth. In WG, pages 331–342, 2011.6

[YC08] Boting Yang and Yi Cao. Digraph searching, directed vertex separation and directed pathwidth. Discrete Applied Mathematics, 156(10):1822–1837, 2008.1,5,6