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K6MINORS IN LARGE 6-CONNECTED GRAPHS
Ken-ichi Kawarabayashi
National Institute of Informatics
2-1-2 Hitotsubashi, Chiyoda-ku, Tokyo 101-8430, Japan
Serguei Norine1
Department of Mathematics
Princeton University
Princeton, NJ 08544, USA
Robin Thomas2
School of Mathematics
Georgia Institute of Technology
Atlanta, Georgia 30332-0160, USA
and
Paul Wollan
Mathematisches Seminar der Universit¨ at Hamburg
Bundesstrasse 55
D-20146 Hamburg, Germany
ABSTRACT
Jørgensen conjectured that every 6-connected graph G with no K6minor
has a vertex whose deletion makes the graph planar. We prove the conjecture
for all sufficiently large graphs.
8 April 2005, revised 22 May 2009.
1Partially supported by NSF under Grant No. DMS-0200595.
2Partially supported by NSF under Grants No. DMS-0200595 and. DMS-0354742.
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1 Introduction
Graphs in this paper are allowed to have loops and multiple edges. A graph is a minor of
another if the first can be obtained from a subgraph of the second by contracting edges. An
H minor is a minor isomorphic to H. A graph G is apex if it has a vertex v such that G\v
is planar. (We use \ for deletion.) Jørgensen [4] made the following beautiful conjecture.
Conjecture 1.1 Every 6-connected graph with no K6minor is apex.
This is related to Hadwiger’s conjecture [3], the following.
Conjecture 1.2 For every integer t ≥ 1, if a loopless graph has no Kt minor, then it is
(t − 1)-colorable.
Hadwiger’s conjecture is known for t ≤ 6. For t = 6 it has been proven in [12] by show-
ing that a minimal counterexample to Hadwiger’s conjecture for t = 6 is apex. The proof
uses an earlier result of Mader [6] that every minimal counterexample to Conjecture 1.2 is
6-connected. Thus Conjecture 1.1, if true, would give more structural information. Further-
more, the structure of all graphs with no K6minor is not known, and appears complicated
and difficult. On the other hand, Conjecture 1.1 provides a nice and clean statement for
6-connected graphs. Unfortunately, it, too, appears to be a difficult problem. In this paper
we prove Conjecture 1.1 for all sufficiently large graphs, as follows.
Theorem 1.3 There exists an absolute constant N such that every 6-connected graph on at
least N vertices with no K6minor is apex.
We use a number of results from the Graph Minor series of Robertson and Seymour,
and also three results of our own that will be proved in other papers. The first of those is
a version of Theorem 1.3 for graphs of bounded tree-width. We will not define tree-width
here, because it is sufficiently well-known, and because we do not need the concept per se,
only several theorems that use it.
Theorem 1.4 For every integer w there exists an integer N such that every 6-connected
graph of tree-width at most w on at least N vertices and with no K6minor is apex.
Theorem 1.4 reduces the proof of Theorem 1.3 to graphs of large tree-width. By a result of
Robertson and Seymour [8] those graphs have a large grid minor. However, for our purposes
it is more convenient to work with walls instead. Let h ≥ 2 be even. An elementary wall of
height h has vertex-set
{(x,y) : 0 ≤ x ≤ 2h + 1,0 ≤ y ≤ h} − {(0,0),(2h + 1,h)}
and an edge between any vertices (x,y) and (x′,y′) if either
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Figure 1: An elementary wall of height 4.
• |x − x′| = 1 and y = y′, or
• x = x′, |y − y′| = 1 and x and max{y,y′} have the same parity.
Figure 1 shows an elementary wall of height 4. A wall of height h is a subdivision of an
elementary wall of height h. The result of [8] (see also [2, 7, 13]) can be restated as follows.
Theorem 1.5 For every even integer h ≥ 2 there exists an integer w such that every graph
of tree-width at least w has a subgraph isomorphic to a wall of height h.
The perimeter of a wall is the cycle that bounds the infinite face when the wall is drawn
as in Figure 1. Now let C be the perimeter of a wall H in a graph G. The compass of H
in G is the restriction of G to X, where X is the union of V (C) and the vertex-set of the
unique component of G\V (C) that contains a vertex of H. Thus H is a subgraph of its
compass, and the compass is connected. A wall H with perimeter C in a graph G is planar
if its compass can be drawn in the plane with C bounding the infinite face. In Section 2 we
prove the following.
Theorem 1.6 For every even integer t ≥ 2 there exists an even integer h ≥ 2 such that if a
5-connected graph G with no K6minor has a wall of height at least h, then either it is apex,
or has a planar wall of height t.
Actually, in the proof of Theorem 1.6 we need Lemma 2.4 that will be proved elsewhere.
The lemma says that if a 5-connected graph with no K6minor has a subgraph isomorphic
to subdivision of a pinwheel with sufficiently many vanes (see Figure 2), then it is apex.
By Theorem 1.6 we may assume that our graph G has an arbitrarily large planar wall H.
Let C be the perimeter of H, and let K be the compass of H. Then C separates G into K
and another graph, say J, such that K∪J = G, V (K)∩V (J) = V (C) and E(K)∩E(J) = ∅.
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Next we study the graph J. Since the order of the vertices on C is important, we are lead
to the notion of a “society”, introduced by Robertson and Seymour in [9].
Let Ω be a cyclic permutation of the elements of some set; we denote this set by V (Ω). A
society is a pair (G,Ω), where G is a graph, and Ω is a cyclic permutation with V (Ω) ⊆ V (G).
Now let J be as above, and let Ω be one of the cyclic permutations of V (C) determined by
the order of vertices on C. Then (J,Ω) is a society that is of primary interest to us. We call
it the anticompass society of H in G.
We say that (G,Ω,Ω0) is a neighborhood if G is a graph and Ω,Ω0are cyclic permutations,
where both V (Ω) and V (Ω0) are subsets of V (G). Let Σ be a plane, with some orientation
called “clockwise.” We say that a neighborhood (G,Ω,Ω0) is rural if G has a drawing Γ in
Σ without crossings (so G is planar) and there are closed discs ∆0⊆ ∆ ⊆ Σ, such that
(i) the drawing Γ uses no point of Σ outside ∆, and none in the interior of ∆0, and
(ii) for v ∈ V (G), the point of Σ representing v in the drawing Γ lies in bd(∆) (respectively,
bd(∆0)) if and only if v ∈ V (Ω) (respectively, v ∈ V (Ω0)), and the cyclic permutation of
V (Ω) (respectively, V (Ω0)) obtained from the clockwise orientation of bd(∆) (respectively,
bd(∆0)) coincides (in the natural sense) with Ω (respectively, Ω0).
We call (Σ,Γ,∆,∆0) a presentation of (G,Ω,Ω0).
Let (G1,Ω,Ω0) be a neighborhood, let (G0,Ω0) be a society with V (G0)∩V (G1) = V (Ω0),
and let G = G0∪ G1. Then (G,Ω) is a society, and we say that (G,Ω) is the composition
of the society (G0,Ω0) with the neighborhood (G1,Ω,Ω0). If the neighborhood (G1,Ω,Ω0)
is rural, then we say that (G0,Ω0) is a planar truncation of (G,Ω). We say that a society
(G,Ω) is k-cosmopolitan, where k ≥ 0 is an integer, if for every planar truncation (G0,Ω0)
of (G,Ω) at least k vertices in V (Ω0) have at least two neighbors in V (G0). At the end of
Section 2 we deduce
Theorem 1.7 For every integer k ≥ 1 there exists an even integer t ≥ 2 such that if G is
a simple graph of minimum degree at least six and H is a planar wall of height t in G, then
the anticompass society of H in G is k-cosmopolitan.
For a fixed presentation (Σ,Γ,∆,∆0) of a neighborhood (G,Ω,Ω0) and an integer s ≥ 0
we define an s-nest for (Σ,Γ,∆,∆0) to be a sequence (C1,C2,...,Cs) of pairwise disjoint
cycles of G such that ∆0⊆ ∆1 ⊆ ··· ⊆ ∆s⊆ ∆, where ∆idenotes the closed disk in Σ
bounded by the image under Γ of Ci. We say that a society (G,Ω) is s-nested if it is the
composition of a society (G1,Ω0) with a rural neighborhood (G2,Ω,Ω0) that has an s-nest
for some presentation of (G2,Ω,Ω0).
Let Ω be a cyclic permutation. For x ∈ V (Ω) we denote the image of x under Ω by
Ω(x). If X ⊆ V (Ω), then we denote by Ω|X the restriction of Ω to X. That is, Ω|X is the
permutation Ω′defined by saying that V (Ω′) = X and Ω′(x) is the first term of the sequence
Ω(x),Ω(Ω(x)),... which belongs to X. Let v1,v2,...,vk∈ V (Ω) be distinct. We say that
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(v1,v2,...,vk) is clockwise in Ω (or simply clockwise when Ω is understood from context)
if Ω′(vi−1) = vifor all i = 1,2,...,k, where v0means vkand Ω′= Ω|{v1,v2,...,vk}. For
u,v ∈ V (Ω) we define uΩv as the set of all x ∈ V (Ω) such that either x = u or x = v or
(u,x,v) is clockwise in Ω.
A separation of a graph is a pair (A,B) such that A ∪ B = V (G) and there is no edge
with one end in A − B and the other end in B − A. The order of (A,B) is |A ∩ B|. We say
that a society (G,Ω) is k-connected if there is no separation (A,B) of G of order at most
k − 1 with V (Ω) ⊆ A and B − A ?= ∅. A bump in (G,Ω) is a path in G with at least one
edge, both ends in V (Ω) and otherwise disjoint from V (Ω).
Let (G,Ω) be a society and let (u1,u2,v1,v2,u3,v3) be clockwise in Ω. For i = 1,2 let Pi
be a bump in G with ends uiand vi, and let L be either a bump with ends u3and v3, or
the union of two internally disjoint bumps, one with ends u3and x ∈ u3Ωv3and the other
with ends v3and y ∈ u3Ωv3. In the former case let Z = ∅, and in the latter case let Z be
the subinterval of u3Ωv3with ends x and y, including its ends. Assume that P1,P2,L are
pairwise disjoint. Let q1,q2∈ V (P1∪ V2∪ v3Ωu3) − {u3,v3} be distinct such that neither of
the sets V (P1) ∪ v3Ωu1, V (P2) ∪ v2Ωu3includes both q1and q2. Let Q1and Q2be two not
necessarily disjoint paths with one end in u3Ωv3−Z −{u3,v3} and the other end q1and q2,
respectively, both internally disjoint from V (P1∪P2∪L)∪V (Ω). In those circumstances we
say that P1∪P2∪L∪Q1∪Q2is a turtle in (G,Ω). We say that P1,P2are the legs, L is the
neck, and Q1∪ Q2is the body of the turtle.
Let (G,Ω) be a society, let (u1,u2,u3,v1,v2,v3) be clockwise in Ω, and let P1,P2,P3be
disjoint bumps such that Pihas ends uiand vi. In those circumstances we say that P1,P2,P3
are three crossed paths in (G,Ω).
Let (G,Ω) be a society, and let u1,u2,u3,u4,v1,v2,v3,v4 ∈ V (Ω) be such that either
(u1,u2,u3,v2,u4,v1,v4,v3) or (u1,u2,u3,u4,v2,v1,v4,v3) or (u1,u2,u3,v2 = u4,v1,v4,v3) is
clockwise. For i = 1,2,3,4 let Pibe a bump with ends uiand visuch that these bumps
are pairwise disjoint, except possibly for v2 = u4.
P1,P2,P3,P4is a gridlet.
Let (G,Ω) be a society and let (u1,u2,v1,v2,u3,u4,v3,v4) be be clockwise in Ω. For
i = 1,2,3,4 let Pi be a bump with ends ui and vi such that these bumps are pairwise
disjoint, and let P5be a path with one end in V (P1) ∪ v4Ωu2− {u2,v1,v4}, the other end
in V (P3) ∪ v2Ωu4− {v2,v3,u4}, and otherwise disjoint from P1∪ P2∪ P3∪ P4. In those
circumstances we say that P1,P2,...,P5is a separated doublecross.
A society (G,Ω) is rural if G can be drawn in a disk with V (Ω) drawn on the boundary
In those circumstances we say that
of the disk in the order given by Ω. A society (G,Ω) is nearly rural if there exists a vertex
v ∈ V (G) such that the society (G\v,Ω\v) obtained from (G,Ω) by deleting v is rural.
In Sections 4–9 we prove the following. The proof strategy is explained in Section 5. It
uses a couple of theorems from [9] and Theorem 4.1 that we prove in Section 4.
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Theorem 1.8 There exists an integer k ≥ 1 such that for every integer s ≥ 0 and every
6-connected s-nested k-cosmopolitan society (G,Ω) either (G,Ω) is nearly rural, or G has a
triangle C such that (G\E(C),Ω) is rural, or (G,Ω) has an s-nested planar truncation that
has a turtle, three crossed paths, a gridlet, or a separated doublecross.
Finally, we need to convert a turtle, three crossed paths, gridlet and a separated double-
cross into a K6minor. Let G be a 6-connected graph, let H be a sufficiently high planar wall
in G, and let (J,Ω) be the anticompass society of H in G. We wish to apply to Theorem 1.8
to (J,Ω). We can, in fact, assume that H is a subgraph of a larger planar wall H′that
includes s concentric cycles C1,C2,...,Cs surrounding H and disjoint from H, for some
suitable integer s, and hence (J,Ω) is s-nested. Theorem 1.8 guarantees a turtle or paths
in (J,Ω) forming three crossed paths, a gridlet, or a separated double-cross, but it does not
say how the turtle or paths might intersect the cycles Ci. In Section 10 we prove a theorem
that says that the cycles and the turtle (or paths) can be changed such that after possibly
sacrificing a lot of the cycles, the remaining cycles and the new turtle (or paths) intersect
nicely. Using that information it is then easy to find a K6minor in G. We complete the
proof of Theorem 1.3 in Section 11.
2 Finding a planar wall
Let a pinwheel with four vanes be the graph pictured in Figure 2. We define a pinwheel with
k vanes analogously. A graph G is internally 4-connected if it is simple, 3-connected, has at
least five vertices, and for every separation (A,B) of G of order three, one of A,B induces
a graph with at most three edges.
Figure 2: A pinwheel with four vanes.
We assume the following terminology from [10]: distance function, perimeter, (l,m)-star
over H, external (l,m)-star over H, subwall, dividing subwall, flat subwall, society of a wall.
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The objective of this section is to prove the following theorem.
Theorem 2.1 For every even integer t ≥ 2 there exists an even integer h such that if H is
a wall of height at least h in an internally 4-connected graph G, then either
(1) G has a K6minor, or
(2) G has a subgraph isomorphic to a subdivision of a pinwheel with t vanes, or
(3) G has a planar wall of height t.
We begin with the following easy lemma. We leave the proof to the reader.
Lemma 2.2 For every integer t there exist integers l and m such that if a graph G has a
wall H with an external (l,m)-star, then it has a subgraph isomorphic to a pinwheel with t
vanes.
We need one more lemma, which follows immediately from [10, Theorem 8.6].
Lemma 2.3 Every flat wall in an internally 4-connected graph is planar.
Figure 3: A K6minor in a grid with two crosses.
Proof of Theorem 2.1. Let t ≥ 1 be given, let l,m be as in Lemma 2.2, let p = 6, and let
k,r be as in [10, Theorem 9.2]. If h is sufficiently large, then H has k +1 subwalls of height
at least t, pairwise at distance at least r. If at least k of these subwalls are non-dividing,
then by [10, Theorem 9.2] G either has a K6minor, or an (l,m)-star over H, in which case
it has a subgraph isomorphic to a pinwheel with t vanes by Lemma 2.2. In either case the
theorem holds, and so we may assume that at least two of the subwalls, say H1and H2, are
dividing. We may assume that H1and H2are not planar, for otherwise the theorem holds.
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Let i ∈ {1,2}. By Lemma 2.3 the wall Hiis not flat, and hence its perimeter has a cross
Pi∪ Qi. Since the subwalls H1and H2are dividing, it follows that the paths P1,Q1,P2,Q2
are pairwise disjoint. Thus G has a minor isomorphic to the graph shown in Figure 3, but
that graph has a minor isomorphic to a minor of K6, as indicated by the numbers in the
figure. Thus G has a K6minor, and the theorem holds. ?
To deduce Theorem 1.6 we need the following lemma, proved in [5].
Lemma 2.4 If a 5-connected graph G with no K6minor has a subdivision isomorphic to a
pinwheel with 20 vanes, then G is apex.
Proof of Theorem 1.6. Let t ≥ 2 be an even integer. We may assume that t ≥ t0, where t0
is as in Lemma 2.4. Let h be as in Theorem 2.1, and let G be a 5-connected graph with no K6
minor. From Theorem 2.1 we deduce that either G satisfies the conclusion of Theorem 1.6,
or has a subdivision isomorphic to a pinwheel with t0vanes. In the latter case the theorem
follows from Lemma 2.4. ?
We need the following theorem of DeVos and Seymour [1].
Theorem 2.5 Let (G,Ω) be a rural society such that G is a simple graph and every vertex
of G not in V (Ω) has degree at least six. Then |V (G)| ≤ |V (Ω)|2/12 + |V (Ω)|/2 + 1.
Proof of Theorem 1.7. Let k ≥ 1 be an integer, and let t be an even integer such that if
W is the elementary wall of height t and |V (W)| ≤ ℓ2/12+ℓ/2+1, then ℓ > 6k −6. Let K
be the compass of H in G, let (J,Ω) be the anticompass society of H in G, let (G0,Ω0) be a
planar truncation of (J,Ω), and let ℓ = |V (Ω0)|. Thus (J,Ω) is the composition of (G0,Ω0)
with a rural neighborhood (G′,Ω,Ω0). Then |V (H)| ≤ ℓ2/12 + ℓ/2 + 1 by Theorem 2.5
applied to the society (K ∪G′,Ω0), and hence ℓ > 6k−6. Let L be the graph obtained from
K ∪ G′by adding a new vertex v and joining it to every vertex of V (Ω0) and by adding an
edge joining every pair of nonadjacent vertices of V (Ω0) that are consecutive in Ω0. Then L
is planar. Let s be the number of vertices of V (Ω0) with at least two neighbors in G0. Then
all but s vertices of K ∪ G′have degree in L at least six. Thus the sum of the degrees of
vertices of L is at least 6|V (K ∪ G′)|−6s +ℓ. On the other hand, the sum of the degrees is
at most 6|V (L)| − 12, because L is planar, and hence s ≥ k, as desired. ?
3 Rural societies
If P is a path and x,y ∈ V (P), we denote by xPy the unique subpath of P with ends x and y.
Let (G,Ω) be a society. An orderly transaction in (G,Ω) is a sequence of k pairwise disjoint
bumps T = (P1,...,Pk) such that Pihas ends uiand viand u1,u2,...,uk,vk,vk−1,...,v1
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is clockwise. Let M be the graph obtained from P1∪ P2∪ ··· ∪ Pkby adding the vertices
of V (Ω) as isolated vertices. We say that M is the frame of T . We say that a path Q in G
is T -coterminal if Q has both ends in V (Ω) and is otherwise disjoint from it and for every
i = 1,2,...,k the following holds: if Q intersects Pi, then their intersection is a path whose
one end is a common end of Q and Pi.
Let (G,Ω) be a society, and let M and T be as in the previous paragraph. Let i ∈
{1,2,...,k} and let Q be a T -coterminal path in G\V (Pi) with one end in viΩuiand the
other end in uiΩvi. In those circumstances we say that Q is a T -jump over Pi, or simply a
T -jump.
Now let i ∈ {0,1,...,k} and let Q1,Q2be two disjoint T -coterminal paths such that Qj
has ends xj,yjand (ui,x1,x2,ui+1,vi+1,y1,y2,vi) is clockwise in Ω, where possibly ui= x1,
x2= ui+1, vi+1= y1, or y2= vi, and u0means x1, uk+1means x2, vk+1means y1, and v0
means y2. In those circumstances we say that (Q1,Q2) is a T -cross in region i, or simply a
T -cross.
Finally, let i ∈ {1,2,...,k} and let Q0, Q1, Q2be three paths such that Qj has ends
xj,yjand is otherwise disjoint from all members of T , x0,y0∈ V (Pi), the vertices x1,x2are
internal vertices of x0Piy0, y1,y2?∈ V (Pi), y1∈ ui−1Ωui∪viΩvi−1, y2∈ uiΩui+1∪vi+1Ωvi, and
the paths Q0, Q1, Q2are pairwise disjoint, except possibly x1= x2. In those circumstances
we say that (Q0,Q1,Q2) is a T -tunnel under Pi, or simply a T -tunnel.
Intuitively, if we think of the paths in T as dividing the society into “regions”, then
a T -jump arises from a T -path whose ends do not belong to the same region. A T -cross
arises from two T -paths with ends in the same region that cross inside that region, and
furthermore, each path in T includes at most two ends of those crossing paths. Finally,
a T -tunnel can be converted into a T -jump by rerouting Pialong Q0. However, in some
applications such rerouting will be undesirable, and therefore we need to list T -tunnels as
outcomes.
Let M be a subgraph of a graph G. An M-bridge in G is a connected subgraph B of G
such that E(B)∩E(M) = ∅ and either E(B) consists of a unique edge with both ends in M,
or for some component C of G\V (M) the set E(B) consists of all edges of G with at least
one end in V (C). The vertices in V (B)∩V (M) are called the attachments of B. Now let M
be such that no block of M is a cycle. By a segment of M we mean a maximal subpath P of
M such that every internal vertex of P has degree two in M. It follows that the segments of
M are uniquely determined. Now if B is an M-bridge of G, then we say that B is unstable if
some segment of M includes all the attachments of B, and otherwise we say that B is stable.
A society (G,Ω) is rurally 4-connected if for every separation (A,B) of order at most
three with V (Ω) ⊆ A the graph G[B] can be drawn in a disk with the vertices of A ∩ B
drawn on the boundary of the disk. A society is cross-free if it has no cross. The following,
a close relative of Lemma 2.3, follows from [9, Theorem 2.4].
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Theorem 3.1 Every cross-free rurally 4-connected society is rural.
Lemma 3.2 Let (G,Ω) be a rurally 4-connected society, let T = (P1,...,Pk) be an orderly
transaction in (G,Ω), and let M be the frame of T . If every M-bridge of G is stable and
(G,Ω) is not rural, then (G,Ω) has a T -jump, a T -cross, or a T -tunnel.
Proof. For i = 1,2,...,k let ui and vi be the ends of Pi numbered as in the defintion
of orderly transaction, and for convenience let P0 and Pk+1 be null graphs.
k + 1 cyclic permutations Ω0,Ω1,...,Ωk as follows. For i = 1,2,...,k − 1 let V (Ωi) :=
V (Pi) ∪ V (Pi+1) ∪ uiΩui+1∪ vi+1Ωviwith the cyclic order defined by saying that uiΩui+1
is followed by V (Pi+1) in order from ui+1to vi+1, followed by vi+1Ωvifollowed by V (Pi) in
order from vito ui. The cyclic permutation Ω0is defined by letting v1Ωu1be followed by
V (P1) in order from u1to v1, and Ωkis defined by letting ukΩvkbe followed by V (Pk) in
order from vkto uk.
Now if for some M-bridge B of G there is no index i ∈ {0,1,...,k} such that all
attachments of B belong to V (Ωi), then (G,Ω) has a T -jump. Thus we may assume that
such index exists for every M-bridge B, and since B is stable that index is unique. Let us
denote it by i(B). For i = 0,1,...,k let Gibe the subgraph of G consisting of Pi∪ Pi+1,
the vertex-set V (Ωi) and all M-bridges B of G with i(B) = i. The society (Gi,Ωi) is rurally
4-connected. If each (Gi,Ωi) is cross-free, then each of them is rural by Theorem 3.1 and it
follows that (G,Ω) is rural. Thus we may assume that for some i = 0,1,...,k the society
(Gi,Ωi) has a cross (Q1,Q2). If neither Pinor Pi+1includes three or four ends of the paths
Q1and Q2, then (G,Ω) has a T -cross. Thus we may assume that Piincludes both ends
of Q1and at least one end of Q2. Let xj,yjbe the ends of Qj. Since the M-bridge of G
containing Q2is stable, it has an attachment outside Pi, and so if needed, we may replace
Q2by a path with an end outside Pi(or conclude that (G,Ω) has a T -jump). Thus we may
assume that ui,x1,x2,y1,vioccur on Piin the order listed, and y2?∈ V (Pi).
The M-bridge of G containing Q1has an attachment outside Pi. If it does not include
Q2and has an attachment outside V (Pi) ∪ {y2}, then (G,Ω) has a T -jump or T -cross, and
so we may assume not. Thus there exists a path Q3with one end x3in the interior of Q1
and the other end y3∈ V (Q2) − {x2} with no internal vertex in M ∪ Q1∪ Q2. We call the
triple (Q1,Q2,Q3) a tripod, and the path y3Q2y2the leg of the tripod. If v is an internal
vertex of x1Piy1, then we say that v is sheltered by the tripod (Q1,Q2,Q3). Let L be a path
that is the leg of some tripod, and subject to that L is minimal. From now on we fix L and
will consider different tripods with leg L; thus the vertices x1,y1,x2,x3may change, but y2
and y3will remain fixed as the ends of L.
Let x′
internal vertex of x′
of x′
We define
1,y′
1∈ V (Pi) be such that they are sheltered by no tripod with leg L, but every
1Piy′
1and all tripods with leg L that shelter some internal vertex of x′
1is sheltered by some tripod with leg L. Let X′be the union
1Piy′
1Piy′
1, let X :=
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X′\V (L)\{x′
we deduce that the set {x′
exists a path P in G\{x′
no internal vertex in X ∪ Y . Let (Q1,Q2,Q3) be a tripod with leg L such that either x is
sheltered by it, or x ∈ V (Q1∪ Q2∪ Q3). If y ?∈ V (L ∪ Pi), then by considering the paths
P,Q1,Q2,Q3it follows that either (G,Ω) has a T -jump or T -tunnel. If y ∈ V (L), then there
is a tripod whose leg is a proper subpath of L, contrary to the choice of L. Thus we may
assume that y ∈ V (Pi), and that y ∈ V (Pi) for every choice of the path P as above. If
x ∈ V (Q1∪Q2∪Q3) then there is a tripod with leg L that shelters x′
Thus x ∈ V (Pi). Let B be the M-bridge containing P. Since y ∈ V (Pi) for all choices of
P it follows that the attachments of B are a subset of V (Pi) ∪ {y2}. But B is stable, and
hence y2is an attachment of B. The minimality of L implies that B includes a path from y
to y3, internally disjoint from L. Using that path and the paths P,Q1,Q2,Q3it is now easy
to construct a tripod that shelters either x′
1,y′
1} and let Y := V (M∪L)−x′
1,y′
1,y′
1Piy′
1−{y3}. Since (G,Ω) is rurally 4-connected
1,y3} does not separate X from Y in G. It follows that there
1,y3} with ends x ∈ X and y ∈ Y . We may assume that P has
1or y′
1, a contradiction.
1or y′
1, a contradiction. ?
4 Leap of length five
A leap of length k in a society (G,Ω) is a sequence of k + 1 pairwise disjoint bumps
P0,P1,...,Pk such that Pi has ends ui and vi and u0,u1,u2,...,uk,v0,vk,vk−1,...,v1, is
clockwise. In this section we prove the following.
Theorem 4.1 Let (G,Ω) be a 6-connected society with a leap of length five. Then (G,Ω)
is nearly rural, or G has a triangle C such that (G\E(C),Ω) is rural, or (G,Ω) has three
crossed paths, a gridlet, a separated doublecross, or a turtle.
The following is a hypothesis that will be common to several lemmas of this section, and
so we state it separately to avoid repetition.
Hypothesis 4.2 Let (G,Ω) be a society with no three crossed paths, a gridlet, a separated
doublecross, or a turtle, let k ≥ 1 be an integer, let
(u0,u1,u2,...,uk,v0,vk,vk−1,...,v1)
be clockwise, and let P0,P1,...,Pkbe pairwise disjoint bumps such that Pihas ends uiand
vi. Let T be the orderly transaction (P1,P2,...,Pk), let M be the frame of T and let
Z = u1Ωuk∪ vkΩv1∪ V (P2) ∪ V (P3) ∪ ··· ∪ V (Pk−1) − {u1,uk,v1,vk}.
Let Z1= v1Ωu1− {u0,u1,v1} and Z2= ukΩvk− {v0,uk,vk}.
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If H is a subgraph of G, then an H-path is a path of length at least one with both ends
in V (H) and otherwise disjoint from H. We say that a vertex v of P0is exposed if there
exists an (M ∪ P0)-path P with one end v and the other in Z.
Lemma 4.3 Assume Hypothesis 4.2 and let k ≥ 3. Let R1,R2be two disjoint (M ∪ P0)-
paths in G such that Rihas ends xi∈ V (P0) and yi∈ V (M) − {u0,v0}, and assume that
u0,x1,x2,v0 occur on P0 in the order listed, where possibly u0= x1, or v0= x2, or both.
Then either y1∈ V (P1)∪v1Ωu1, or y2∈ V (Pk)∪ukΩvk, or both. In particular, there do not
exist two disjoint (M ∪ P0)-paths from V (P0) to Z.
Proof. The second statement follows immediately from the first, and so it suffices to prove
the first statement. Suppose for a contradiction that there exist paths R1,R2satisfying the
hypotheses but not the conclusion of the lemma. By using the paths P2,P3,...,Pk−1we
conclude that there exist two disjoint paths Q1,Q2in G such that Qihas ends xi∈ V (P0)
and zi∈ V (Ω), and is otherwise disjoint from V (P0)∪V (Ω), and if Qiintersects some Pjfor
j ∈ {1,2,...,k}, then j ∈ {2,...,k−1} and Qi∩Pjis a path one of whose ends is a common
end of Qiand Pj. Furthermore, z1 ∈ u1Ωv1− {u1,v1} and z2∈ vkΩuk− {uk,vk}. From
the symmetry we may assume that either (u0,v0,z2,z1), or (u0,z1,v0,z2) or (u0,v0,z1,z2) is
clockwise. In the first two cases (G,Ω) has a separated doublecross (the two pairs of crossing
bumps are P1and Q1∪ u0P0x1, and Pkand Q2∪ v0P0x2, and the fifth path is a subpath
of P2), unless the second case holds and z1∈ ukΩv0or z2∈ v1Ωu0, or both. By symmetry
we may assume that z1 ∈ ukΩv0. Then, if z2 ∈ vk−2Ωu0, (G,Ω) has a gridlet formed by
the paths Pk,Pk−1,u0P0x1∪ Q1and v0P0x2∪ Q2. Otherwise, z2∈ vkΩvk−2− {vk,vk−2} and
(G,Ω) has a turtle with legs Pkand v0P0x2∪ Q2, neck P1and body u0P0x2∪ Q1.
Finally, in the third case (G,Ω) has a turtle or three crossed paths. More precisely, if
z2∈ v0Ωv1− {v1}, then (G,Ω) has a turtle described in the paragraph above. Otherwise,
by symmetry, we may assume that z2∈ v1Ωu0and z1∈ v0Ωvk, in which case v0P0x2∪ Q2,
u0P0x1∪ Q1and P2are the three crossed paths. ?
Lemma 4.4 Assume Hypothesis 4.2 and let k ≥ 2. Then (G\V (P0),Ω\V (P0)) has no T -
jump.
Proof. Suppose for a contradiction that (G\V (P0),Ω\V (P0)) has a T -jump. Thus there is
an index i ∈ {1,2,...,k} and a T -coterminal path P in G\V (P0∪ Pi) with ends x ∈ viΩui
and y ∈ uiΩvi. Let j ∈ {1,2,...,k}−{i}. Then using the paths P0,Pi,Pjand P we deduce
that (G,Ω) has either three crossed paths or a gridlet, in either case a contradiction. ?
Lemma 4.5 Assume Hypothesis 4.2 and let k ≥ 2. Let v ∈ V (P0) be such that there is no
(M ∪ P0)-path in G\v from vP0v0 to vP0u0∪ V (P1∪ P2∪ ··· ∪ Pk−1) ∪ vkΩuk− {vk,uk}
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and none from vP0u0to V (P2∪ P3∪ ··· ∪ Pk) ∪ u1Ωv1− {u1,v1}. Then (G\v,Ω\v) has no
T -jump.
Proof. The hypotheses of the lemma imply that every T -jump in (G\v,Ω\v) is disjoint
from P0. Thus the lemma follows from Lemma 4.4. ?
Lemma 4.6 Assume Hypothesis 4.2, let k ≥ 3, and let v ∈ V (P0) be such that no vertex in
V (P0)−{v} is exposed. Let i ∈ {0,1,...,k} be such that (G\v,Ω\v) has a T -cross (Q1,Q2)
in region i. Then i ∈ {0,k} and v is not exposed. Furthermore, assume that i = 0, and that
there exists an (M ∪P0)-path Q with one end v and the other end in P1∪v1Ωu1−{u0}, and
that v0P0v is disjoint from Q1∪Q2. Then for some j ∈ {1,2} there exist p ∈ V (Qj∩u0P0v)
and q ∈ V (Qj∩ Q) such that pP0v and qQv are internally disjoint from Q1∪ Q2.
Proof. If i ?∈ {0,k}, then the T -cross is disjoint from P0by the choice of v, and hence the
T -cross and P0give rise to three crossed paths. To complete the proof of the first assertion
we may assume that i = 0 and that v is exposed. Thus there exists a T -coterminal path
Q′from v to Z ∩ V (Ω) disjoint from P0∪ P1∪ Pk\v. If (Q′∪ vP0v0) ∩ (Q1∪ Q2) = ∅ then
(G,Ω) has a separated doublecross, where one pair of crossed paths is obtained from the
T -cross, the other pair is Pkand Q′∪ vP0v0, and the fifth path is a subpath of P2. Thus
we may assume that there exists x ∈ (V (Q′′)) ∩ V (Q1) and that x is chosen so that xQ′′y is
internally disjoint from Q1∪Q2, where Q′′= Q′∪vP0v0and y is the end of Q′in Z ∩V (Ω).
Let x′∈ (V (P0) ∩ (V (Q1) ∪ V (Q2)) ∪ {u0} be chosen so that x′P0v0is internally disjoint
from Q1∪ Q2. Let z1∈ v1Ωu1− {v1,u1} be an end of Q1. If x ∈ V (Q′), then Q1is disjoint
from P0, because v is the only exposed vertex. Thus z1Q1x ∪ xQ′y is a T -jump disjoint
from P0, contrary to Lemma 4.4. It follows that x ∈ V (v0P0v), and Q′is disjoint from
Q1∪ Q2. Let j ∈ {1,2} be such that x′∈ V (Qj), let zj∈ v1Ωu1− {v1,u1} be an end of Qj
and let P′
jump, disjoint from P′
Lemma 4.3 applied to T and the path P′
of u0P0v ∪ Q′. This proves the first assertion of the lemma.
To prove the second statement of the lemma we assume that i = 0 and that Q is a path
from v to v′∈ v1Ωu1− {u0}, disjoint from M ∪ P0\v, except that P1∩ Q may be a path
with one end v′. Let the ends of Q1,Q2be labeled as in the definition of T -cross. If P0
is disjoint from Q1∪ Q2, then (G,Ω) has three crossed paths (if (y2,u0,x1) is clockwise) or
a gridlet with paths Q1,Q2,P0,P2(if (x1,u0,x2) or (y1,u0,y2) is clockwise), or a separated
doublecross with paths Q1,Q2,P0,P2,Pk(if (v1,u0,y1) or (x2,u0,u1) is clockwise). Thus we
may assume that P0intersects Q1∪ Q2. (Please note that v0P0v is disjoint from Q1∪ Q2
by hypothesis.) Similarly we may assume that Q intersects Q1∪Q2, for otherwise we apply
the previous argument with P0replaced by Q ∪ vP0v0. Let p ∈ V (Q1∪ Q2) ∩ u0P0v and
0:= v0P0x′∪ x′Qjzj. If x′Qjzjdoes not intersect u0P0v, then u0P0v ∪ Q′is a T -
0, contrary to Lemma 4.4; otherwise there exist two paths contradicting
0: one is a subpath of Qjand the other is a subpath
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q ∈ V (Q1∪ Q2) ∩V (Q) be chosen to minimize pP0v and qQv. If p and q belong to different
paths Q1,Q2, then (G,Ω) has a turtle with legs Q1,Q2, neck Pkand body pP0v0∪qQv. Thus
p and q belong to the same Qjand the lemma holds. ?
In the proof of the following lemma we will be applying Lemma 3.2. To guarantee that
the conditions of Lemma 3.2 are satisfied, we will need a result from [5]. We need to precede
the statement of this result by a few definitions.
Let M be a subgraph of a graph G, such that no block of M is a cycle. Let P be a
segment of M of length at least two, and let Q be a path in G with ends x,y ∈ V (P) and
otherwise disjoint from M. Let M′be obtained from M by replacing the path xPy by Q;
then we say that M′was obtained from M by rerouting P along Q, or simply that M′was
obtained from M by rerouting. Please note that P is required to have length at least two,
and hence this relation is not symmetric. We say that the rerouting is proper if all the
attachments of the M-bridge that contains Q belong to P. The following is proved in [5,
Lemma 2.1].
Lemma 4.7 Let G be a graph, and let M be a subgraph of G such that no block of M is
a cycle. Then there exists a subgraph M′of G obtained from M by a sequence of proper
reroutings such that if an M′-bridge B of G is unstable, say all its attachments belong to
a segment P of M′, then there exist vertices x,y ∈ V (P) such that some component of
G\{x,y} includes a vertex of B and is disjoint from M\V (P).
Lemma 4.8 Assume Hypothesis 4.2, and let k ≥ 4. If every leap of length k−1 has at most
one exposed vertex, (G,Ω) is 4-connected and (G\v,Ω\v) is rurally 4-connected for every
v ∈ V (P0), then (G,Ω) is nearly rural.
Proof. Since (G,Ω) has no separated doublecross it follows that it does not have a T -cross
both in region 0 and region k. Thus we may assume that it has no T -cross in region k.
Similarly, it follows that it does not have a T -tunnel under both P1and Pk, or a T -cross in
region 0 and a T -tunnel under Pk. Thus we may also assume that (G,Ω) has no T -tunnel
under Pk. If some leap of length k in (G,Ω) has an exposed vertex, then we may assume
that v is an exposed vertex. Otherwise, let the leap (P0,P1,...,Pk) and v ∈ V (P0) be chosen
such that either v = u0or there exists an (M ∪ P0)-path with one end v and the other end
in P1∪ v1Ωu1− {u0}, and, subject to that, vP0v0is as short as possible.
By Lemma 4.7 we may assume, by properly rerouting M if necessary, that every M-bridge
of G\v is stable. Since the reroutings are proper the new paths Piwill still be disjoint from
P0, and the property that defines v will continue to hold. Similarly, the facts that there is no
T -cross in region k and no T -tunnel under Pkremain unaffected. We claim that v satisfies
the lemma.
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We apply Lemma 3.2 to the society (G\v,Ω\v) and orderly transaction T . We may
assume that (G\v,Ω\v) is not rural, and hence by Lemma 3.2 the society (G\v,Ω\v) has
a T -jump, a T -cross or a T -tunnel. By the choice of v there exists a path Q from v to
v′∈ vkΩuk−{vk,uk} such that Q does not intersect Pk∪P0\v and intersects at most one of
P1,P2,...,Pk−1. Furthermore, if it intersects Pifor some i ∈ {1,2,...,k −1} then Pi∩Q is
a path with one end a common end of both.
We claim that v satisfies the hypotheses of Lemma 4.5. To prove this claim suppose for
a contradiction that P is an (M ∪ P0)-path violating that hypothesis. Suppose first that P
and Q are disjoint. Then P joins different components of P0\v by Lemma 4.3. But then
changing P0to the unique path in P0∪P that does not use v either produces a leap with at
least two exposed vertices, or contradicts the minimality of vP0v0. Thus P and Q intersect.
Since no leap of length k has two or more exposed vertices, it follows that v is not exposed.
Thus P has one end in u0P0v by the minimality of vP0v0, and the other end in Pk∪ ukΩvk,
because v is not exposed. But then P ∪ Q includes a T -jump disjoint from P0, contrary
to Lemma 4.4. This proves our claim that v satisfies the hypotheses of Lemma 4.5. We
conclude that (G\v,Ω\v) has no T -jump.
Assume now that (G\v,Ω\v) has a T -cross (Q1,Q2) in region i. Then by the first part of
Lemma 4.6 and the assumption made earlier it follows that i = 0 and v is not exposed. But
the existence of Q and the second statement of Lemma 4.6 imply that some leap of length k
has at least two exposed vertices, a contradiction. (To see that let j,p,q be as in Lemma 4.6.
Replace P1by Q3−jand replace P0by a suitable subpath of Qj∪ pP0v0∪ qQv.)
We may therefore assume that (G\v,Ω\v) has a T -tunnel (Q0,Q1,Q2) under Pifor some
i ∈ {1,2,...,k}. Then the leap L′= (P0,P1,...,Pi−1,Pi+1,...,Pk) of length k − 1 ≥ 3 has
a T′-cross, where T′is the corresponding orderly society, and the result follows in the same
way as above. ?
Lemma 4.9 Assume Hypothesis 4.2 and let k ≥ 3. If there exist at least two exposed
vertices, then there exists a cycle C and three disjoint (M ∪ C)-paths R1,R2,R3such that
Rihas ends xi∈ V (C) and yi∈ V (M), C\{x1,x2,x3} is disjoint from M, y1= u0, y2= v0
and y3∈ Z.
Proof. Let x1be the closest exposed vertex to u0on P0, and let x2be the closest exposed
vertex to v0. Let R1= P0[x1,u0] and let R2= P0[x2,v0]. For i = 1,2 let Sibe an (M ∪ P0)-
path with one end xiand the other end in Z. By Lemma 4.3 S1and S2intersect, and so
we may assume that S1∩ S2is a path R3containing an end of both S1and S2, say y3. Let
x3be the other end of R3. Then P0∪ S1∪ S2includes a unique cycle C. The cycle C and
paths R1,R2,R3are as desired for the lemma. ?
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If the cycle C in Lemma 4.9 can be chosen to have at least four vertices, then we say
that the leap (P0,P1,...,Pk) is diverse.
Lemma 4.10 Assume Hypothesis 4.2, let k ≥ 4, and let there be no diverse leap of length
k. If C is as in Lemma 4.9 and (G\E(C),Ω) is rurally 4-connected, then (G\E(C),Ω) is
rural.
Proof. Since the leap (P0,P1,...,Pk) is not diverse, it follows that C is a triangle. Let
R1,R2,R3and their ends be numbered as in Lemma 4.9. We may assume that P0= R1∪
R2+ x1x2. Since there is no diverse leap, Lemma 4.3 implies that there is no path in
G\E(C)\V (Pk) from x2to vkΩuk, and none in G\E(C)\V (P1) from x1to u1Ωv1. It also
implies that no vertex on P0is exposed in G\x1x3\x2x3.
As in Lemma 4.8, we can apply Lemma 4.7 and assume, by properly rerouting M if neces-
sary, that the conditions of Lemma 3.2 are satisfied. We assume that the society (G\E(C),Ω)
has a T -jump, a T -cross, or a T -tunnel, as otherwise by Lemma 3.2 (G\E(C),Ω) is rural.
By the observation at the end of the previous paragraph this T -jump, T -cross, or T -tunnel
cannot use both x1and x2; say it does not use x2. But that contradicts Lemma 4.5 or the
first part of Lemma 4.6, applied to v = x2and the graph G\x1x3, in case of a T -jump or a
T -cross.
Thus we may assume that (G\E(C)\x2,Ω\x2) has a T -tunnel (Q0,Q1,Q2) under Pifor
some i ∈ {1,2,...,k}. But then the leap L′= (P0,P1,...,Pi−1,Pi+1,...,Pk) of length
k − 1 ≥ 3 has a T′-cross (Q′
obtained from Piby rerouting along Q0and Q′
of Pijoining the ends of Q1and Q2. By the first half of Lemma 4.6 applied to the graph
G\x1x3, the leap L′, v := x2 and the T′-cross (Q′
that y3∈ v2Ωu2− {u0}. By the second half of Lemma 4.6 applied to the same entities and
Q := R3+ x3x2there exist j ∈ {1,2}, p ∈ V (Q′
and qQx2are internally disjoint from Q′
and the leap (P0,P1,...,Pk) is diverse, as a subpath of Q0joins a vertex of R1to a vertex
of Q in G\x1x3. If j = 2 then we obtain a diverse leap from (P0,P1,...,Pk) by replacing P1
by Q′
1,Q′
2), where T′is the corresponding orderly transaction, Q′
2is the union of Q1∪ Q2with the subpath
1is
1,Q′
2) we may assume that i = 1 and
j∩ R1) and q ∈ V (Q′
j∩ Q) such that pP0x2
1∪Q′
2. If j = 1, then p,q belong to the interior of Q0,
1and replacing P0by a suitable subpath of Q ∪ v0P0p ∪ Q′
2. ?
Lemma 4.11 Assume Hypothesis 4.2, let k ≥ 3, let (G,Ω) be 4-connected, let C,R1,R2,R3
be as in Lemma 4.9, and assume that C is not a triangle. Then there exist four disjoint
(M ∪ C)-paths, each with one end in V (C) and the other end respectively in the sets {u0},
{v0}, Z and V (P1∪ Pk).
Proof. By an application of the proof of the max-flow min-cut theorem there exist four
disjoint (M ∪C)-paths, each with one end in V (C) and the other end respectively in the sets
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{u0}, {v0}, Z and V (M). By Lemma 4.3 the fourth path does not end in V (M) − V (P1) −
V (Pk). The result follows. ?
Lemma 4.12 Assume Hypothesis 4.2, let k ≥ 3, let C,R1,R2,R3be as in Lemma 4.9, let
D := M ∪ C ∪ R1∪ R2∪ R3, and let R4be a D-path with ends x4∈ V (C) − {x1,x2,x3}
and y4∈ V (P1). Then x1,x2,x3,x4occur on C in the order listed. Furthermore, if R is a
D-path with ends x ∈ V (C) − {x1,x2,x3} and y ∈ V (M), then x1,x2,x3,x occur on C in
the order listed and y ∈ V (P1).
Proof. The vertices x1,x2,x3,x4occur on C in the order listed by Lemma 4.3. Now let R
be as stated. By Lemma 4.3 we have y ∈ V (P1∪ Pk), and so by the first part of the lemma
we may assume that y ∈ V (Pk). By the symmetric statement to the first half of the lemma
it follows that x1,x2,x,x3occur on C in the order listed. We may assume that P0is the
unique path from u0to v0in R1∪ R2∪ C\x3. Then R4∪ R ∪ C\V (P0) includes a T -jump
disjoint from P0, contrary to Lemma 4.4. ?
We need to further upgrade the assumptions of Hypothesis 4.2, as follows.
Hypothesis 4.13 Assume Hypothesis 4.2. Let C be a cycle with distinct vertices x1,x2,x3
such that C\{x1,x2,x3} is disjoint from M. Let R1,R2,R3be pairwise disjoint (M ∪ C)-
paths such that Rihas ends xiand yi, where y1= u0, y2= v0, and y3∈ Z. By a ray we mean
an (M ∪C)-path from C to M, disjoint from R1∪R2∪R3. We say that a vertex v ∈ V (P1)
is illuminated if there is a ray with end v. Let x4,x5∈ V (P1) be illuminated vertices such
that either x4= x5, or u1,x4,x5,v1occur on P1in the order listed, and x4P1x5includes all
illuminated vertices. Let R4:= u1P1x4and R5:= v1P1x5, and let y4:= u1and y5:= v1.
Let S4and S5be rays with ends x4and x5, respectively, and let A0:= V (M) − V (P1) and
B0:= V (C ∪ S4∪ S5∪ x4P1x5).
Lemma 4.14 Assume Hypothesis 4.13, let k ≥ 3, and let (G,Ω) be 6-connected. Then
x4?= x5, and the path x4P1x5has at least one internal vertex.
Proof. If x4= x5or x4P1x5has no internal vertex, then by Lemma 4.12 the set {x1,x2,...,x5}
is a cutset separating C from M\V (P1), contrary to the 6-connectivity of (G,Ω). Note that
V (C) − {x1,x2,...,x5} is non-empty as it includes an end of a ray. ?
Assume Hypothesis 4.13. By Lemma 4.14 the paths R1,R2,...,R5 are disjoint paths
from A0to B0. The following lemma follows by a standard “augmenting path” argument.
Lemma 4.15 Assume Hypothesis 4.13, and let k ≥ 2. If there is no separation (A,B) of
order at most five with A0⊆ A and B0⊆ B, then there exist an integer n and internally
disjoint paths Q1,Q2,...,Qnin G, where Qihas distinct ends aiand bisuch that
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(i) a1∈ A0− {y1,y2,...,y5} and bn∈ B0− {x1,x2,...,x5},
(ii) for all i = 1,2,...,n−1, ai+1,bi∈ V (Rt) for some t ∈ {1,2,...,5}, and yt,ai+1,bi,xt
are pairwise distinct and occur on Rtin the order listed,
(iii) if ai,bj∈ V (Rt) for some t ∈ {1,2,...,5} and i,j ∈ {1,2,...,5} with i > j +1, then
either ai= bj, or yt,bj,ai,xtoccur on Rtin the order listed, and
(iv) for i = 1,2,...,n, if a vertex of Qibelongs to A0∪ B0∪ V (R1∪ R2∪ ··· ∪ R5), then
it is an end of Qi.
The sequence of paths (Q1,Q2,...,Qn) as in Lemma 4.15 will be called an augmenting
sequence.
Lemma 4.16 Assume Hypothesis 4.13, and let k ≥ 3. Then there is no augmenting sequence
(Q1,Q2,...,Qn), where Q1is disjoint from P2.
Proof. Suppose for a contradiction that there is an augmenting sequence (Q1,Q2,...,Qn),
where Q1is disjoint from P2, and let us assume that the leap (P0,P1,...,Pk), cycle C, paths
R1,R2,R3,S4,S5 and augmenting sequence (Q1,Q2,...,Qn) are chosen with n minimum.
Let the ends of the paths Qibe labeled as in Lemma 4.15. We may assume that P0is the
unique path from u0to v0in R1∪ R2∪ C\x3. We proceed in a series of claims.
(1)
The vertex bnbelongs to the interior of x4P1x5.
To prove (1) suppose for a contradiction that bn∈ V (C ∪ S4∪ S5). By Lemma 4.12, the
choice of x4,x5and the fact that an?= x4,x5by Lemma 4.15(ii) we deduce that an∈ V (Ri)
for some i ∈ {1,2,3}. Then we can use Qn to modify C to include anRixi (and modify
R1,R2,R3accordingly), in which case (Q1,Q2,...,Qn−1) is an augmentation contradicting
the choice of n. This proves (1).
(2)ai,bi∈ V (Rj) for no i ∈ {1,2,...,n} and no j ∈ {1,2,...,5}.
To prove (2) suppose to the contrary that ai,bi ∈ V (Rj).
rerouting Rj along Qi we obtain an augmentation (Q1,Q2,...,Qi−2,Qi−1∪ bi−1Rjai+1∪
Qi+1,Qi+2,...,Qn), contrary to the minimality of n. This proves (2).
Then 1 < i < n and by
(3)ai,bi∈ V (R1∪ R2∪ R3) for no i ∈ {1,2,...,n}.
Using (2) the proof of (3) is analogous to the argument at the end of the proof of Claim (1).
(4)ai,bi∈ V (R4∪ R5) for no i ∈ {1,2,...,n}.
By (2) one of ai,bibelongs to R4and the other to R5. We can reroute P1along Qi, and then
(Q1,Q2,...,Qi−1) becomes an augmentation, contrary to the minimality of n.
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(5)
For i = 1,2,...,n − 1, the graph Qi∪ R1∪ R2∪ R3includes no T -jump.
This claim follows from (3), Lemma 4.3 and Lemma 4.4 applied to P0.
(6)a1?∈ v1Ωu1.
To prove (6) suppose for a contradiction that a1∈ v1Ωu1. Since a1?= y1, we may assume
from the symmetry that a1∈ v1Ωy1−{y1}. Then b1∈ V (P1∪R1) by (5). But if b1∈ V (Ri),
where i = 1 or i = 5, then by rerouting Rialong Q1we obtain an augmenting sequence
(Q2∪ x1Ria2,Q3,Q4,...,Qn), contrary to the choice of n. Thus b1∈ u1P1x5. By replacing
P1by the path Q1∪u1P1b1and considering the paths R3and S5∪R5we obtain contradiction
to Lemma 4.3. This proves (6).
(7)a1?∈ ukΩvk.
Similarly as in the proof of (6), if a1∈ ukΩvk, then b1∈ V (R2) by (5), and we reroute
R2along Q1to obtain a contradiction to the minimality of n. This proves (7).
(8)a1∈ V (Pk).
To prove (8) we may assume by (6) and (7) that a1∈ Z. Then b1∈ V (R3∪ P1) by (5).
If b1∈ V (R3), then we reroute R3along Q1as before. Thus b1∈ V (P1). It follows from (5)
and the hypothesis V (P2) ∩ V (Q1) = ∅ that a1∈ u1Ωu2− {u1,u2} or a1∈ v2Ωv1− {v1,v2},
and so from the symmetry we may assume the latter.
Let us assume for a moment that y3∈ a1Ωv1. We reroute P1along Q1∪b1P1v1. The union
of R3, R2and a path in C between x2and x3, avoiding x1,x4,x5, will play the role of P0
after rerouting. If b1∈ x4P1v1−{x4}, then R1∪C ∪S4∪R4includes two disjoint paths that
contradict Lemma 4.3 applied to the new frame and new path P0. Therefore b1∈ V (R4),
and hence (u1P1a2∪Q2,Q3,...,Qn) is an augmenting sequence after the rerouting, contrary
to the choice of n.
It follows that y3?∈ a1Ωv1. If b1∈ V (R5), we replace P1by Q1∪ u1P1b1; then (v1P1a2∪
Q2,Q3,...,Qn) is an augmenting sequence that contradicts the choice of n. So it follows
that b1∈ u1P1x5. But now (G,Ω) has a gridlet using the paths P0,Pk, Q1∪ u1P1b1and a
subpath of R5∪ S5∪ R3∪ C\V (P0). This proves (8).
(9)n > 1.
To prove (9) suppose for a contradiction that n = 1. Thus b1belongs to the interior of x4Px5
by (1), and a1∈ V (Pk) by (8). But then Q1is a T -jump, contrary to (5).
(10) b1∈ V (R3).
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To prove (10) we first notice that b1 ∈ V (R2∪ R3) by (5), (9) and (1). Suppose for a
contradiction that b1∈ V (R2). Then a2∈ V (R2), but b2?∈ V (R1∪ R2∪ R3) by (3) and
b2?∈ V (P1) by (5), a contradiction. This proves (10).
Let P12and P34be two disjoint subpaths of C, where the first has ends x1,x2, and the
second has ends x3,x4. By (8) and (10) the path Q1∪b1R3x3∪P34∪S4is a T -jump disjoint
from R1∪ P12∪ R2, contrary to Lemma 4.4. ?
We are now ready to prove Theorem 4.1.
Proof of Theorem 4.1. Let (G,Ω) be a 6-connected society with a leap of length five. Thus
we may assume that Hypothesis 4.2 holds for k = 5. By Lemma 4.8 either (G,Ω) is nearly
rural, in which case the theorem holds, or there exists a leap of length at least four with at
least two exposed vertices. Thus we may assume that there exists a leap of length four with
at least two exposed vertices. Let C be a cycle as in Lemma 4.9. If there is no diverse leap,
then C is a triangle, (G\E(C),Ω) is rurally 4-connected and hence rural by Lemma 4.10,
and the theorem holds. Thus we may assume that the cycle C is not a triangle, and so by
Lemma 4.11 we may assume that Hypothesis 4.13 for k = 4 holds. By Lemma 4.14 and the
6-connectivity of G there is no separation (A,B) as described in Lemma 4.15, and hence by
that lemma there exists an augmenting sequence (Q1,Q2,...,Qn). By Lemma 4.16 the path
Q1intersects P2, and hence Q1is disjoint from P3, contrary to Lemma 4.16 applied to the
leap (P0,P1,P3,P4) of length three and an augmenting sequence (Q′
is the union of Q1and a1P2u2or a1P2v2. ?
1,Q2,...,Qn), where Q′
1
5 Societies of bounded depth
Let (G,Ω) be a society. A linear decomposition of (G,Ω) is an enumeration {t1,...,tn} of
V (Ω) where (t1,...,tn) is clockwise, together with a family (Xi: 1 ≤ i ≤ n) of subsets of
V (G), with the following properties:
(i)?(Xi: 1 ≤ i ≤ n) = V (G),
(ii) for 1 ≤ i ≤ n, ti∈ Xi, and
(iii) for 1 ≤ i ≤ i′≤ i′′≤ n, Xi∩ Xi′′ ⊆ Xi′.
The depth of such a linear decomposition is
max(|Xi∩ Xi′| : 1 ≤ i < i′≤ n),
and the depth of (G,Ω) is the minimum depth of a linear decomposition of (G,Ω). Theo-
rems (6.1), (7.1) and (8.1) of [9] imply the following.
Theorem 5.1 There exists an integer d such that every 4-connected society (G,Ω) either
has a separated doublecross, three crossed paths or a leap of length five, or some planar
truncation of (G,Ω) has depth at most d.
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In light of Theorems 4.1 and 5.1, in the remainder of the paper we concentrate on societies
of bounded depth. We need a few definitions. Let (G,Ω) be a society, let u1,u2,...,u4tbe
clockwise in Ω, and let P1,P2,...,P2tbe disjoint bumps in G such that for i = 1,2,...,2t
the path P2i−1has ends u4i−3and u4i−1, and the path P2ihas ends u4i−2and u4i. In those
circumstances we say that (G,Ω) has t disjoint consecutive crosses.
Now let u1,v1,w1,u2,v2,w2,...,ut,vt,wt be clockwise in Ω, let x ∈ V (G) − {u1,v1,
w1,...,ut,vt,wt}, for i = 1,2,...,t let Pi be a path in G\x with ends ui and wi and
otherwise disjoint from V (Ω), let Qibe a path with ends x and viand otherwise disjoint
from V (Ω), and assume that the paths Piand Qiare pairwise disjoint, except that the paths
Qimeet at x. Let W be the union of all the paths Piand Qi. We say that W is a windmill
with t vanes, and that the graph Pi∪ Qiis a vane of the windmill.
Finally, let u1,u2,...,utand v1,v2,...,vtbe vertices of V (Ω) such that for all xi∈ {ui,vi}
the sequence x1,x2,...,xtis clockwise in Ω. Let z1,z2∈ V (G)−{u1,v1,...,ut,vt} be distinct,
for i = 1,2,...,t let Pibe a path in G\z2with ends z1and uiand otherwise disjoint from
V (Ω), and let Qibe a path in G\z1with ends z2and viand otherwise disjoint from V (Ω).
Assume that the paths Piand Qjare disjoint, except that the Pishare z1, the Qishare z2
and Piand Qiare allowed to intersect. Let F be the union of all the paths Piand Qi. Then
we say that F is a fan with t blades, and we say that Pi∪Qiis a blade of the fan. The vertices
z1and z2will be called the hubs of the fan. In Section 8 we prove the following theorem.
Theorem 5.2 For every two integers d and t there exists an integer k such that every 6-
connected k-cosmopolitan society (G,Ω) of depth at most d contains one of the following:
(1) t disjoint consecutive crosses, or
(2) a windmill with t vanes, or
(3) a fan with t blades.
Unfortunately, windmills and fans are nearly rural, and so for our application we need to
improve Theorem 5.2. We need more definitions.
Let x,ui,vi,wi,Pi,Qibe as in the definition of a windmill W with t vanes, let a,b,c,d ∈
V (G) be such that u1,v1,w1,...,ut,vt,wt,a,b,c,d is clockwise in Ω, and let (P,Q) be a cross
disjoint from W whose paths have ends in {a,b,c,d}. In those circumstances we say that
W ∪ P ∪ Q is a windmill with t vanes and a cross.
Now let ui,vi,Pi,Qibe as in the definition of a fan F with t blades, and let a,b,c,d ∈ V (Ω)
be such that all xi∈ {ui,vi} the sequence x1,x2,...,xt,a,b,c,d is clockwise in Ω. Let (P,Q)
be a cross disjoint from F whose paths have ends in {a,b,c,d}. In those circumstances we
say that W ∪ P ∪ Q is a fan with t blades and a cross.
Let z1,z2,ui,vi,Pi,Qibe as in the definition of a fan F with t blades, and let a1,b1,c1,a2,
b2,c2∈ V (G) be such that all xi∈ {ui,vi} the sequence x1,x2,...,xt,a1,b1,c1,a2,b2,c2is
clockwise in Ω, except that we permit c1= a2. For i = 1,2 let Libe a path in G\V (F) with
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ends aiand ciand otherwise disjoint from V (Ω), and let Sibe a path with ends ziand bi
and otherwise disjoint from V (F) ∪ V (Ω). If the paths L1,L2,S1,S2are pairwise disjoint,
except possibly for L1intersecting L2at c1= a2, then we say that F ∪ L1∪ L2∪ S1∪ S2is
a fan with t blades and two jumps.
Now let ui,vi,Pi,Qibe as in the definition of a fan F with t+1 blades, and let a,b ∈ V (Ω)
be such that all xi∈ {ui,vi} the sequence x1,x2,...,xt,a,xt+1,b is clockwise in Ω. Let P be
a path in G\V (F) with ends a and b, and otherwise disjoint from V (F). We say that F ∪P
is a fan with t blades and a jump. In Section 9 we improve Theorem 5.2 as follows.
Theorem 5.3 For every two integers d and t there exists an integer k such that every 6-
connected k-cosmopolitan society (G,Ω) of depth at most d is either nearly rural, or contains
one of the following:
(1) t disjoint consecutive crosses, or
(2) a windmill with t vanes and a cross, or
(3) a fan with t blades and a cross, or
(4) a fan with t blades and a jump, or
(5) a fan with t blades and two jumps.
For t = 4 each of the above outcomes gives a turtle, and hence we have the following
immediate corollary.
Corollary 5.4 For every integer d there exists an integer k such that every 6-connected
k-cosmopolitan society (G,Ω) of depth at most d is either nearly rural, or has a turtle.
The next three sections are devoted to proofs of Theorems 5.2 and 5.3. The proof of
Theorem 5.2 will be completed in Section 8 and the proof of Theorem 5.3 will be completed
in Section 9. At that time we will be able to deduce Theorem 1.8.
6 Crosses and goose bumps
In this section we prove that a society (G,Ω) either satisfies Theorem 5.2, or it has many
disjoint bumps. If X is a set and Ω is a cyclic permutation, we define Ω\X to be Ω|(V (Ω)−
X). Let P1,P2,...,Pkbe a set of pairwise disjoint bumps in (G,Ω), where Pihas ends ui
and vi and u1,v1,u2,v2,...,uk,vk is clockwise in Ω. In those circumstances we say that
P1,P2,...,Pkis a goose bump in (G,Ω) of strength k.
Lemma 6.1 Let b,d and t be positive integers, and let (G,Ω) be a society of depth at most
d. Then either (G,Ω) has a goose bump of strength b, or there is a set X ⊆ V (G) of size at
most (b − 1)d such that the society (G\X,Ω\X) has no bump.
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Proof. Let (t1,t2,...,tn) and (X1,X2,...,Xn) be a linear decomposition of (G,Ω) of depth
at most d, and for i = 1,2,...,n − 1 let Yi= Xi∩ Xi+1. If P is a bump in (G,Ω), then the
axioms of a linear decomposition imply that
IP:= {i ∈ {1,2,...,n − 1} : Yi∩ V (P) ?= ∅}
is a nonempty subinterval of {1,2,...,n − 1}. It follows that either there exist bumps
P1,P2,...,Pb such that IP1,IP2,...,IPbare pairwise disjoint, or there exists a set I ⊆
{1,2,...,n−1} of size at most b− 1 such that I ∩IP?= ∅ for every bump P. In the former
case P1,P2,...,Pbis a desired goose bump, and in the latter case the set X :=?
desired. ?
i∈IYiis as
The proof of the following lemma is similar and is omitted.
Lemma 6.2 Let t and d be positive integers, and let (G,Ω) be a society of depth at most d.
Then either (G,Ω) has t disjoint consecutive crosses, or there is a set X ⊆ V (G) of size at
most (t − 1)d such that the society (G\X,Ω\X) is cross-free.
Lemma 6.3 Let d,b,t be positive integers, let k ≥ (b − 1)d + (t − 1)?(b−1)d
(G,Ω) be a 3-connected society of depth at most d such that at least k vertices in V (Ω) have
2
?+ 1 and let
at least two neighbors in V (G). Then (G,Ω) has either a fan with t blades, or a goose bump
of strength b.
Proof. By Lemma 6.1 we may assume that there exists a set X ⊆ V (G) of size at most
(b − 1)d such that (G\X,Ω\X) has no bump. There are at least (t − 1)?(b−1)d
in V (Ω) − X with at least two neighbors in V (G). Let v be one such vertex, and let H
be the component of G\X containing v. Since (G\X,Ω\X) has no bumps it follows that
V (H) ∩ V (Ω) = {v}. By the fact that v has at least two neighbors in G (if V (H) = {v}) or
the 3-connectivity of (G,Ω) (if V (H) ?= {v}) it follows that H has at least two neighbors in
X. Thus there exist distinct vertices z1,z2such that for at least t vertices of v ∈ V (Ω) − X
the component of G\X containing v has z1and z2as neighbors. It follows that (G,Ω) has
a fan with t blades, as desired. ?
2
?+ 1 vertices
7 Intrusions, invasions and wars
Let Ω be a cyclic permutation. A base in Ω is a pair (X,Y ) of subsets of V (Ω) such that
|X ∩Y | = 2, X ∪Y = V (Ω) and for distinct elements x1,x2∈ X and y1,y2∈ Y the sequence
(x1,y1,x2,y2) is not clockwise. Now let (G,Ω) be a society. A separation (A,B) of G is
called an intrusion in (G,Ω) if there exists a base (X,Y ) in Ω such that X ⊆ A, Y ⊆ B and
there exist disjoint paths (Pv)v∈A∩B, each with one end in X, the other end in Y and with
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v ∈ V (Pv). The intrusion (A,B) is minimal if there is no intrusion (A′,B′) of order |A ∩ B|
with base (X,Y ) such that A′is a proper subset of A. The paths Pvwill be called longitudes
for the intrusion (A,B). We say that (A,B) is based at (X,Y ), and that (X,Y ) is a base
for (A,B). An intrusion (A,B) in (G,Ω) is an invasion if |A ∩ B ∩ V (Ω)| = 2.
Lemma 7.1 Let d be a positive integer, and let (G,Ω) be a society of depth at most d − 1.
Then for every base (X,Y ) in Ω there exists an intrusion of order at most 2d based at (X,Y ).
Proof. Let (t1,t2,...,tn) and (X1,X2,...,Xn) be a linear decomposition of (G,Ω) of depth
at most d − 1, and let X ∩ Y = {ti,tj}. Let i′,j′∈ {1,2,...,n} be such that |i − i′| =
|j − j′| = 1, and let Z := (Xi∩ Xi′) ∪ (Xj∩ Xj′) ∪ {ti,tj}. It follows from the axioms of a
linear decomposition that |Z| ≤ 2d and that Z separates X from Y in G. Thus there exists
a separation (A,B) of G of order at most 2d with X ⊆ A and Y ⊆ B. Any such separation
(A,B) with |A ∩ B| minimum is as desired by Menger’s theorem. ?
An intrusion (A,B) in a society (G,Ω) is t-separating if (G,Ω) has goose bumps P1,P2,...,Pt
and Q1,Q2,...,Qtsuch that V (Pi) ⊆ A − B and V (Qi) ⊆ B − A for all i = 1,2,...,t.
Lemma 7.2 Let d,s,t be positive integers, and let (G,Ω) be a society of depth at most d−1
with a goose bump of strength t(s + 2d). Then there exist s-separating minimal intrusions
(A1,B1),(A2,B2),...,(At,Bt) of order at most 2d such that Ai∩ Aj⊆ Bi∩ Bjfor all pairs
of distinct indices i,j = 1,2,...,t.
Proof. Let P be the set of paths comprising a goose bump of strength t(s+2d). Thus there
exist bases (X1,Y1),(X2,Y2),...(Xt,Yt) such that the sets Xiare pairwise disjoint and for
each i = 1,2,...,t exactly s+2d of the paths in P have both ends in Xi. By Lemma 7.1 there
exists, for each i = 1,2,...,t, an intrusion (Ai,Bi) of order at most 2d based at (Xi,Yi).
Let us choose, for each i = 1,2,...,t, an intrusion (Ai,Bi) of order at most 2d based at
(Xi,Yi) in such a way that
t
?
i=1
We claim that Ai∩ Aj ⊆ Bi∩ Bj. To prove the claim suppose to the contrary that say
x ∈ A1∩ A2− B1∩ B2. Let
|Ai| is minimum.(1)
A′
1= A1∩ B2,
B′
1= A2∪ B1,
A′
2= A2∩ B1,
B′
2= A1∪ B2.
Then (A′
Y2⊆ B′
1,B′
1) and (A′
2,B′
2) are separations of G with X1 ⊆ A′
1, Y1 ⊆ B′
1, X2 ⊆ A′
2and
2. We have
|A1∩ B1| + |A2∩ B2| = |A′
1∩ B′
1| + |A′
2∩ B′
2|.
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Page 25
Furthermore, since each longitude for (A1,B1) intersects A′
|A1∩B1|, and similarly |A′
and hence the longitudes for (A1,B1) are also longitudes for (A′
for (A2,B2) are longitudes for (A′
is an intrusion in (G,Ω) based at (Xi,Yi) of order |Ai∩ Bi|. Since A1∩ A2− (B1∩ B2) =
(A1∩ A2− B1) ∪ (A1∩ A2− B2) we may assume that x ∈ A1− B2. But then replacing
(A1,B1) by (A′
that Ai∩ Aj⊆ Bi∩ Bjfor all distinct integers i,j = 1,2,...,t.
Since at most 2d of the paths in P with ends in Xican intersect Ai∩Bi, we deduce that
each intrusion (Ai,Bi) is s-separating. Moreover, each (Ai,Bi) is clearly minimal by (1). ?
1∩B′
1we deduce that |A′
1∩B′
1| ≥
2∩B′
2| ≥ |A2∩B2|. Thus the last two inequalities hold with equality,
1,B′
1), and the longitudes
2,B′
2). It follows that for i = 1,2 the separation (A′
i,B′
i)
1,B′
1) produces a set of intrusions that contradict (1). This proves our claim
We need a lemma about subsets of a set.
Lemma 7.3 Let d and t be nonnegative integers, and let F be a family of 2(d
subsets of a set S, where each member of F has size at most d. Then there exist a set X ⊂ S
of size at most?d
2
?and a family F′⊆ F of size at least t such that F ∩ F′⊆ X for every
two sets F,F′∈ F′.
2)tddistinct
Proof. We proceed by induction on d + t. If d = 0 or t = 0, then the lemma clearly holds,
and so we may assume that d,t > 0. Let F0∈ F be minimal with respect to inclusion. If F
has a subfamily F1of at least 2(d
the induction hypothesis applied to F1and by adding F0to the family thus obtained. If the
family F2= {F − F0: F ∈ F,F ∩ F0?= ∅} includes at least 2(d−1
the result follows from the induction hypothesis applied to F2by adding F0to the set thus
obtained. Thus we may assume neither of the two cases holds. Thus
2)(t−1)dsets disjoint from F0, then the result follows from
2)td−1distinct sets, then
|F| ≤ 2(d
2)(t − 1)d− 1 + 2d2(d−1
2)td−1− 1 + 1 < 2(d
2)td,
a contradiction. ?
Lemma 7.4 Let d,s,t be positive integers, and let (G,Ω) be a society of depth at most d−1
with a goose bump of strength 2(2d
at most?2d
2
?and s-separating intrusions (A1,B1),(A2,B2),...,(At,Bt) in (G\X,Ω\X) such
that Ai∩ Aj= ∅ for all pairs of distinct indices i,j = 1,2,...,t.
2)t2d(s + 2d). Then there exist a set X ⊆ V (G) of size
Proof. Let T = 2(2d
(A2,B2),...,(AT,BT) of order at most 2d such that Ai∩ Aj ⊆ Bi∩ Bj for all pairs of
distinct indices i,j = 1,2,...,t. By Lemma 7.3 applied to the sets Ai∩ Bi there exist
a set X ⊆?T
2
(A1,B1),(A2,B2),...,(At,Bt), such that Ai∩ Bi∩ Aj∩ Bj ⊆ X for all distinct integers
i,j = 1,2,...,t. It follows that (Ai− X,Bi− X) are as required for (G\X,Ω\X). ?
2)t2d. By Lemma 7.2 there exist s-separating minimal intrusions (A1,B1),
i=1(Ai∩ Bi) of size at most
?2d
?
and a subset of t of those intrusions, say
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Our next objective is to prove, albeit with weaker bounds, that the conclusion of Lemma 7.4
can be strengthened to assert that the intrusions (Ai,Bi) therein are actually invasions.
Let (A,B) be an intrusion in a society (G,Ω) based at (X,Y ). A path P in G[A] is a
meridian for (A,B) if its ends are the two vertices of X ∩ Y . If P is a meridian for (A,B)
and (Lv)v∈A∩Bare longitudes for (A,B), then the graph (P ∪?
a frame for (A,B).
v∈A∩BLv)\(B − A) is called
Lemma 7.5 Let λ and s be positive integers, let s′= (s − 1)(λ − 1) + 1, let (G,Ω) be a
cross-free society, and let (A,B) be an s′-separating minimal intrusion in (G,Ω) of order
at most λ. Then there exists an s-separating minimal invasion (C,D) in (G,Ω) of order at
most λ with a frame F such that V (F) − V (Ω) ⊆ A.
Proof. We may assume that
(1)
there is no integer λ′≤ λ and an ((s − 1)(λ′− 1) + 1)-separating minimal intrusion
(A′,B′) in (G,Ω) of order at most λ′with A′a proper subset of A,
for if (A′,B′) exists, and it satisfies the conclusion of the lemma, then so does (A,B). We
first show that (A,B) has a meridian. Indeed, suppose not. Let (X,Y ) be a base of (A,B)
and let X ∩Y = {u,v}; then G[A] has no u-v path. Since (G,Ω) is cross-free it follows that
G[A] has a separation (A1,A2) of order zero such that both X1= X ∩A1and X2= X ∩A2
are intervals in Ω. It follows that there exist Y1,Y2such that (X1,Y1) and (X2,Y2) are bases.
Thus (A1,A2∪B ∪(X1∩Y1)) and (A2,A1∪B ∪(X2∩Y2)) are minimal intrusions, and one
of them violates (1). This proves that (A,B) has a meridian.
Let M be a meridian in (A,B), let (Lv)v∈A∩Bbe a collection of longitudes for (A,B) and
let F = M ∪?
that
v∈A∩B(Lv\(B −A)). By the same argument that justifies (1) we may assume
(2) there is no integer λ′< λ and an ((s − 1)(λ′− 1) + 1)-separating minimal intrusion
(A′,B′) in (G,Ω) of order at most λ′with frame F′such that F′\V (Ω) is a subgraph of F.
We claim that |A ∩ B ∩ V (Ω)| = 2. We first prove that A ∩ B ∩ X = {u,v}. To this
end suppose for a contradiction that w ∈ A ∩ B ∩ X − {u,v}; then w divides X into two
cyclic intervals X1 and X2 with ends u,w and w,v, respectively. Let Y1 and Y2 be the
complementary cyclic intervals so that (X1,Y1) and (X2,Y2) are bases.
For i = 1,2 let Aiconsist of w and all vertices a ∈ A such that there exists a path in
G[A]\w with one end a and the other end in Xi− {w}, and let A3 = A − A1− A2. It
follows that A1∩ A2= {w}, for if P is a path in G[A]\w with one end in X1and the other
end in X2, then (P,Pw) is a cross in (G,Ω), a contradiction. Thus (A1,A2∪ A3∪ B) and
(A2,A1∪ A3∪ B) are minimal intrusions based on (X1,Y1) and (X2,Y2), respectively, with
A1,A2⊆ A. Thus one of them violates (2).
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Next we show that |A ∩ B ∩ Y | = 2, and so we suppose for a contradiction that there
exists z ∈ A ∩ B ∩ Y − {u,v}. We define B1,B2,B3,X1,Y1,X2,Y2analogously as in the
previous paragraph, but with the roles of A and B reversed. Similarly we find that one of
(A∪B1∪B3,B2) and (A∪B2∪B3,B1) is an ((s−1)(λ′−1)+1)-separating minimal intrusion
in (G,Ω) of order at most λ′, for some λ′< λ, and so from the symmetry we may assume
that (A∪B1∪B3,B2) has this property. Since (M,Pz) is not a cross in (G,Ω) it follows that
M and Pzintersect. Thus M ∪ Pzincludes a meridian for (A ∪ B1∪ B3,B2). Finally, since
Z = B2∩ (A ∪ B1∪ B3) ⊆ A ∩ B, the paths (Lv)v∈Zform longitudes for (A ∪ B1∪ B3,B2),
contrary to (2).
Thus we have shown that A ∩ B ∩ V (Ω) = {u,v}. Let Z be the set of all vertices z ∈ A
such that there is no path in G[A] with one end z and the other end in X, let C = A − Z
and D = B∪Z. Then (C,D) is an intrusion with C∩D = A∩B and F is a frame for (C,D)
with V (F) − V (Ω) ⊆ C. Since the order of (C,D) is at least two, it satisfies the conclusion
of the lemma. ?
We are ready to deduce the main result of this section. By a war in a society (G,Ω)
we mean a set W of minimal invasions such that each invasion in W has a meridian, and
A ∩ A′= ∅ for every two distinct invasions (A,B),(A′,B′) ∈ W. We say that the war W
is s-separating if each invasion in W is s-separating, we say W has order at most λ if each
member of W has order at most λ, and we say that W is a war of intensity |W|.
Lemma 7.6 Let s, t and d be positive integers, and let b = 2(2d
if a cross-free society (G,Ω) of depth at most d − 1 has a goose bump of strength b, then it
has a set X of at most?2d
2
?vertices such that the society (G\X,Ω\X) has an s-separating
war of intensity t and order order at most 2d.
2)(2dt)2d(s(2d−1)+2). Then
Proof. Let s′= (2d−1)(s−1)+1. By Lemma 7.4 there exist a set X ⊆ V (G) with at most
?2d
2
?elements and s′-separating intrusions (A1,B1),(A2,B2),...,(A2dt,B2dt) in (G\X,Ω\X)
of order at most 2d such that Ai∩ Aj = ∅ for every pair i,j = 1,2,...,2dt of distinct
integers. By 2dt applications of Lemma 7.5 there exist, for each i = 1,2,...,2dt, and s-
separating minimal invasion (Ci,Di) in (G\X,Ω\X) of order at most 2d with a frame Fi
such that V (Fi) − V (Ω) ⊆ V (Ai). Let Mibe a meridian for (Ci,Di), and let (Xi,Yi) be the
base for (Ci,Di). Since (G,Ω) has depth at most d there exists a set I ⊆ {1,2,...,2dt} of
size t such that the sets {Xi}i∈I are pairwise disjoint. By symmetry we may assume that
I = {1,2,...,t}. We claim that (C1,D1),(C2,D2),...,(Ct,Dt) are as desired. To prove the
claim suppose for a contradiction that say x ∈ Ci∩ Cj. Since (Ci,Di) is an invasion there
exists a path in G[Ci] from x to Xi⊆ Yj; therefore this path intersects Cj∩ Dj. Thus there
exists a vertex v ∈ Cj∩Dj∩Ci; let L be the longitude of Fjthat includes v. But L connects
v ∈ Cito a vertex of Xj⊆ Yi⊆ Di, and hence intersects Ci∩ Di⊆ V (Fi). Thus Fiand Fj
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intersect. But V (Fi)∩V (Fj)−V (Ω) ⊆ Ai∩Aj= ∅ and V (Fi)∩V (Fj)∩V (Ω) ⊆ Xi∩Xj= ∅,
a contradiction. Thus (C1,D1),(C2,D2),...,(Ct,Dt) satisfy the conclusion of the lemma. ?
8 Using wars
Lemma 8.1 Let l,t,r be positive integers such that r ≥ (t − 1)?l
nected society, and let Z ⊆ V (G) be a set of size at most l such that the society (G\Z,Ω\Z)
has a war W of intensity r such that for every (A,B) ∈ W at least two distinct members of
Z have at least one neighbor in A. Then (G,Ω) has a fan with t blades.
2
?+ 1, let (G,Ω) be a con-
Proof. There exist distinct vertices z1,z2∈ Z and a subset W′of W of size t such that
for every (A,B) ∈ W′both z1and z2have a neighbor in A. Furthermore, since (A,B) is a
minimal intrusion, it follows that for every vertex a ∈ A there exists a path in G[A] from a
to V (Ω). It follows that (G,Ω) has a fan with t blades, as desired. ?
Let (A,B) be an invasion in a cross-free society (G,Ω), based at (X,Y ), and let (Lv)v∈A∩B
be longitudes for (A,B). Let Ω′be a cyclic permutation in A defined as follows: for each
u ∈ Y , if u is an end of Lv, then we replace u by v, and otherwise we delete u. Then
(G[A],Ω′) is a society, and we will call it the society induced by (A,B). Since (G,Ω) is
cross-free the definition does not depend on the choice of longitudes for (A,B).
Assume now that (G[A],Ω′) is rural. A path P in G[A] is called a perimeter path in
(G[A],Ω′) if A ∩ B ⊆ V (P) and G[A] has a drawing in a disk with vertices of Ω′appearing
on the boundary of the disk in the order specified by Ω′and with every edge of P drawn in
the boundary of the disk.
The next lemma is easy and we omit its proof.
Lemma 8.2 Let (A,B) be an invasion with longitudes {Pv}v∈A∩B in a cross-free society
(G,Ω). Then the society induced by (A,B) is cross-free.
Lemma 8.3 Let (G,Ω) be a 5-connected society, let Z ⊆ V (G) be such that (G\Z,Ω\Z) is
cross-free, and let (A,B) be an invasion in (G\Z,Ω\Z). If at most one vertex of Z has a
neighbor in A, then the society induced in (G\Z,Ω\Z) by (A,B) is rural and has a perimeter
path.
Proof. Let (G[A],Ω′) be the society induced in (G\Z,Ω\Z) by (A,B). By Lemma 8.2 it
is cross-free and by Theorem 3.1 it is rural. Thus it has a drawing in a disk ∆ with V (Ω′)
drawn on the boundary of ∆ in the order specified by Ω′. When ∆ is regarded as a subset
of the plane, the unbounded face of G[A] is bounded by a walk W. Let P be a subwalk of
W containing A ∩ B. If P is not a path, then it has a repeated vertex, say x, and G[A] has
a separation (C,D) with C ∩D = {x} and A∩B ∩V (Ω) ⊆ C. Since (G[A],Ω′) is cross-free,
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the latter inclusion implies that D − C is disjoint from V (Ω) or from A ∩ B. However,
the latter is impossible, which can be seen by considering the drawing of G[A] in ∆. Thus
(D−C)∩V (Ω) = ∅, and since (A,B) has longitudes we deduce that |(D−C)∩A∩B| ≤ 1.
Let z ∈ Z be such that no vertex of Z−{z} has a neighbor in A. Since (G,Ω) is 4-connected,
the fact that ((D −C)∩A∩B)∪{x,z} does not separate G implies that D −C consists of
a unique vertex, say d, and d ∈ A ∩ B. Furthermore, the only neighbor of d in A is x. But
then (A−{d},B ∪{x}) contradicts the minimality of (A,B). This proves that P is a path,
and it follows that it is a perimeter path for (G[A],Ω′). ?
Let (G,Ω) be a society. A set T of bumps in (G,Ω) is called a transaction in (G,Ω) if
there exist elements u,v ∈ V (Ω) such that each member of T has one end in uΩv and the
other end in V (Ω) − uΩv. The first part of the next lemma is easy, and the second part is
proved in [9, Theorem (8.1)].
Lemma 8.4 Let (G,Ω) be a society, and let d ≥ 1 be an integer. If (G,Ω) has depth d, then
it has no transaction of cardinality exceeding 2d. Conversely, if (G,Ω) has no transaction of
cardinality exceeding d, then it has depth at most d.
Lemma 8.5 Let (G,Ω) be a society of depth d, and let X ⊆ V (G). Then the society
(G\X,Ω\X) has depth at most 2d.
Proof. By Lemma 8.4 the society (G,Ω) has no transaction of cardinality exceeding 2d.
Then clearly (G\X,Ω\X) has no transaction of cardinality exceeding 2d, and hence has
depth at most 2d by another application of Lemma 8.4. ?
We need one last lemma before we can prove Theorem 5.2. The lemma we need is
concerned with the situation when a society of bounded depth “almost” has a windmill with
t vanes, except that the paths Piare not necessarily disjoint and their ends do not necessarily
appear in the right order. We begin with a special case when the ends of the paths Pido
appear in the right order.
Lemma 8.6 Let t ≥ 1 be an integer, and let ρ = d(t−1)(t′−1)+1, where t′= d(t−1)2+t.
Let (G,Ω) be a society of depth d, let (u1,z1,v1,u2,z2,v2,...,uρ,zρ,vρ) be clockwise, let
z ∈ V (G), for i = 1,2,...,ρ let Pibe a bump with ends uiand vi, and let Qibe a path of
length at least one with ends z and zidisjoint from V (Ω) − {z,zi}. Assume that the paths
Qiare pairwise disjoint except for z, and that each is disjoint from every Pj. Then (G,Ω)
has either a windmill with t vanes, or a fan with t blades.
Proof. By the proof of Lemma 6.1 applied to the paths Pieither some t of those paths
are vertex-disjoint, in which case (G,Ω) has a windmill with t vanes, or there exists a set
X ⊆ V (G) of size at most (t−1)d such that each Piuses at least one vertex of X. We may
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therefore assume the latter. For i = 1,2,...,ρ the path Pihas a subpath P′
ui, the other end xi∈ X and no internal vertex in X. Thus there exist x ∈ X and a set
I ⊆ {1,2,...,ρ} of size t′such that x = xifor all i ∈ I. Let H be the union of all P′
i ∈ I. By an application of Lemma 6.1 to the graph H\x we deduce that either H\x has a
iwith one end
iover
goosebump of strength t, in which case (G,Ω) has a windmill with t vanes, or H has a set
Y of size at most (t − 1)d such that H\Y \x has no bumps. In the latter case for each i ∈ I
there is a path P′′
from Y ∪ {x}. Thus there is a vertex y ∈ Y ∪ {x} and a set J ⊆ I of size t such that yi= y
for every i ∈ J. Since H\Y \x has no bumps it follows that P′′
distinct j,j′∈ J. Thus (G,Ω) has a fan with t blades, as desired. ?
iin H with one end ui, the other end yi∈ Y ∪ {x} and otherwise disjoint
jand P′′
j′ share only y for
Now we are ready to prove the last lemma in full generality.
Lemma 8.7 Let t ≥ 1 be an integer, and let ξ = (d+1)ρ, where ρ is as in Lemma 8.6. Let
(G,Ω) be a society of depth d, let z ∈ V (G), for i = 1,2,...,ξ let (ui,zi,vi) be clockwise,
and let (u1,z1,u2,z2,...,uξ,zξ) be clockwise. Let Pibe a bump with ends uiand vi, and let
Qibe a path of length at least one with ends z and zidisjoint from V (Ω) − {z,zi}. Assume
that the paths Qiare pairwise disjoint except for z, and that each is disjoint from every Pj.
Then (G,Ω) has either a windmill with t vanes, or a fan with t blades.
Proof. Let (t1,t2,...,tn) be a clockwise enumeration of V (Ω), and let (X1,X2,...,Xn) be a
corresponding linear decomposition of (G,Ω) of depth d. Let us fix an integer i = 1,2,...,ρ,
and let I = {(i − 1)(d + 1) + 1,(i − 1)(d + 1) + 2,...,i(d + 1)}. For each such i we will
construct paths P∗
will make use of the paths Pjand Qjfor j ∈ I.
If (uj,zj,vj,ui(d+1)+1) is clockwise for some j ∈ I, then we put P∗
Otherwise, letting s be such that ts= ui(d+1), we deduce that Pjintersects Xts∩ Xts+1for
all j ∈ I. Since |I| > |Xts∩ Xts+1| it follows that there exist j < j′∈ I such that Pjand Pj′
intersect. Let P∗
This completes the construction. The lemma follows from Lemma 8.6. ?
iand Q∗
isatisfying the hypothesis of Lemma 8.6. In the construction we
i= Pjand Q∗
i= Qj.
ibe a subpath of Pj∪ Pj′ with ends ujand uj′, and let Q∗
i= Qj.
Proof of Theorem 5.2. Let the integers d and t be given, let ξ be as in Lemma 8.7, let
ℓ = 2(t − 1)d +?4d+2
22
Lemma 7.6 with s = 1, t = τ and d replaced by 4d+1, and let k be as in Lemma 6.3 applied
to b, t, and 4d. We will prove that k satisfies the conclusion of the theorem.
?, let τ = (t − 1)?ℓ
?+?2(t − 1)d +?8d+2
2
??(6ξ − 1) + 1, let b be as in
To that end let (G,Ω) be a k-cosmopolitan society of depth at most d, and let (G0,Ω0) be
a planar truncation of (G,Ω). Let S ⊆ V (Ω0). We say that S is sparse if whenever u1,u2∈ S
are such that there does not exist w ∈ S such that (u1,w,u2) is clockwise, then there exist
two disjoint bumps P1,P2in (G0,Ω0) such that uiis an ends of Pi. The reader should notice
that if H is one of the graphs listed as outcomes (1)-(3) of Theorem 5.2, then V (H)∩V (Ω0)
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is sparse. We say that (G0,Ω0) is weakly linked if for every sparse set S ⊆ V (Ω0) there exist
|S| disjoint paths from S to V (Ω) with no internal vertex in V (G0). Thus if the conclusion of
the theorem holds for some weakly linked truncation of (G0,Ω0), then it holds for (G,Ω) as
well. Thus we may assume that (G0,Ω0) is a weakly linked truncation of (G,Ω) with |V (G0)|
minimum. We will prove that (G0,Ω0) satisfies the conclusion of Theorem 5.2. Since (G0,Ω0)
is weakly linked, Lemma 8.4 implies that (G0,Ω0) has no transaction of cardinality exceeding
2d, and hence has depth at most 2d by Lemma 8.4.
By Lemma 6.2 there exists a set Z1⊆ V (G0) such that |Z1| ≤ 2(t − 1)d and the society
(G0\Z1,Ω1\Z1) is cross-free. By Lemma 8.5 the society (G0\Z1,Ω0\Z1) has depth at most
4d. By Lemma 6.3 we may assume that (G0\Z1,Ω0\Z1) has a goose bump of strength b. By
Lemma 7.6 there exists a set Z2⊆ V (G) − Z1such that |Z2| ≤?4d+2
(G0\Z,Ω0\Z) there exists a 1-separating war W of intensity τ and order at most 8d + 2,
where Z = Z1∪ Z2. If there exist at least (t − 1)?ℓ
least two distinct vertices of Z have a neighbor in A, then the theorem holds by Lemma 8.1.
We may therefore assume that this is not the case, and hence W has a subset W′of size at
least |Z|(6ξ−1)+1 such that for every (A,B) ∈ W′at most one vertex of Z has a neighbor
in A.
Let (A,B) ∈ W′and let z ∈ Z be such that no vertex in Z − {z} has a neighbor in
A. By Lemma 8.3 the society (G0[A],Ω′) induced in (G0\Z,Ω0\Z) by (A,B) is rural and
has a perimeter path P. It follows that (A ∪ {z},B ∪ {z}) is a separation of G0. Let
A ∩ B = {w0,w1,...,ws}, and let Libe the longitude containing wi. Let the ends of Libe
ui∈ A and vi∈ B. We may assume that (u0,u1,...,us) is clockwise. The vertices widivide
P into paths P0,P1,...,Ps, where Pihas ends wi−1and wi. We claim that no Piincludes
all neighbors of z. For suppose for a contradiction that say Pi does. Let (G,Ω) be the
composition of (G0,Ω0) with a rural neighborhood (G1,Ω,Ω0). Let G′
let G′
by z followed by wi−1,wi−2,...,w1. Since (G[A],Ω′) is rural and all neighbors of z belong
to Pi, it follows that (G′
(G′
that (G′
a minimal intrusion there exists a set P′of |S′| disjoint paths from S′to V (Ω0) with no
internal vertex in G′
it follows that S is sparse (G0,Ω0). Since (G0,Ω0) is weakly linked there exists a set P of
|S| disjoint paths in G from S to V (Ω) with no internal vertex in G0. By taking unions of
members of P and P′we obtain a set of paths proving that (G′
desired. Since W is 1-separating this contradicts the minimality of G0, proving our claim
that no Piincludes all neighbors of z. The same argument, but with G′
Ω′
2
?and in the society
2
?+ 1 invasions (A,B) ∈ W such that at
1= G1∪ G[A ∪ {z}],
0= G0\(A − B) and let Ω′
0consist of wsΩw0followed by ws−1,ws−2,...,wifollowed
1,Ω,Ω′
0) is a rural neighborhood and (G,Ω) is the composition of
0) with this neighborhood. Thus (G′
0,Ω′
0,Ω′
0,Ω′
0) is a planar truncation of (G,Ω). We claim
0) is weakly linked. To prove that let S′⊆ V (Ω′
0) be sparse. Since (A,B) is
0; let S be the set of their ends in V (Ω0). Since S′is sparse in (G′
0,Ω′
0),
0,Ω′
0) is weakly linked, as
1= G1∪ G[A] and
0not including z shows that z has a neighbor in A − B.
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We have shown, in particular, that exactly one vertex of Z has a neighbor in A−B. Thus
there exists a subset W′′of W′of size 6ξ and a vertex z ∈ Z such that for every (A,B) ∈ W′′
the vertex z has a neighbor in A − B. Now let w = (A,B) ∈ W′′, and let the notation be
as before. We will construct paths Pw, Qwsuch that the hypotheses of Lemma 8.7 will be
satisfied for at least half the members w ∈ W′′.
The facts that (A,B) is a minimal intrusion and that z has a neighbor in A − B imply
that there exists a path Qwin G[A∪{z}] from z to zw∈ V (Ω0)∩A and a choice of longitudes
(Lv: v ∈ A∩B) for (A,B) such that Qwis disjoint from all Lv. Referring to the subpaths Pi
of the perimeter path P defined above, since no Piincludes all neighbors of z it follows that
there exists v ∈ A ∩ B − V (Ω0). We define Pwto be a path obtained from Lvby suitably
modifying Lvinside B such that Pwintersects A′for at most one (A′,B′) ∈ W′′−{(A,B)}.
Such modification is easy to make, using the perimeter path of (A′,B′). Let uw∈ A and
vw∈ B be the ends of Pw.
The set W′′has a subset W′′′of size ξ such that, using to the notation of the previous
paragraph, either (uw,zw,vw) is clockwise for every w ∈ W′′′or (vw,zw,uw) is clockwise
for every w ∈ W′′′, and for every w ∈ W′′′the path Pw is disjoint from A′for every
(A′,B′) ∈ W′′′− {w}. The theorem now follows from Lemma 8.7. ?
9 Using lack of near-planarity
In this section we prove Theorems 5.3 and 1.8. The first follows immediately from Theo-
rem 5.2 and the two lemmas below.
Lemma 9.1 Let (G,Ω) be a rurally 5-connected society that is not nearly rural, and let t be
a positive integer. If (G,Ω) has a windmill with 4t + 1 vanes, then it has a windmill with t
vanes and a cross.
Proof. Let x,ui,vi,wi,Pi,Qibe as in the definition of a windmill W with 4t+1 vanes. Since
(G\x,Ω\{x}) is rurally 4-connected and not rural, it has a cross (P,Q) by Theorem 3.1. We
may choose the windmill W and cross (P,Q) in (G\x,Ω\{x}) such that W ∪P ∪Q is minimal
with respect to inclusion. If the cross does not intersect the windmill, then the lemma clearly
holds, and so we may assume that a vane Pi∪ Qiintersects P ∪ Q. Let v be a vertex that
belongs to both Pi∪Qiand P ∪Q such that some subpath R of Pi∪Qiwith one end v and
the other end in V (Ω) has no vertex in (P ∪Q)\v. If R has at least one edge, then P ∪Q∪R
has a proper subgraph that is a cross, contrary to the minimality of W ∪ P ∪ Q. Thus v is
an end of P or Q. Since P and Q have a total of four ends, it follows that P ∪ Q intersects
at most four vanes of W. By ignoring those vanes we obtain a windmill with 4(t − 1) + 1
vanes, and a cross (P,Q) disjoint from it. The lemma follows. ?
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Lemma 9.2 Let (G,Ω) be a rurally 6-connected society that is not nearly rural, and let t be
a positive integer. If (G,Ω) has a fan with 16t+5 blades, then it has a fan with t blades and
a cross, or a fan with t blades and a jump, or a fan with t blades and two jumps.
Proof. Let z1,z2be the hubs of a fan F2with 16t+5 blades. If (G\{z1,z2},Ω\{z1,z2}) has
a cross, then the lemma follows in the same way as Lemma 9.1, and so we may assume not.
Since (G\z1,Ω\{z1}) has a cross, an argument analogous to the proof of Lemma 9.1 shows
that there exists a subfan F1of F2with 4t + 1 blades (that is, F1is obtained by ignoring
a set of 12t + 4 blades), and two paths L2,S2with ends a2,c2and b2,z2, respectively, such
that x1,x2,...,x4t+1, a2,b2,c2is clockwise in Ω for every choice of x1,x2,...,x4t+1as in the
definition of a fan, and the graphs L2,S2\z2,F1are pairwise disjoint. By using the same
argument and the fact that (G\z2,Ω\{z2}) has a cross we arrive at a subfan F of F1with t
blades and paths L1,S1satisfying the same properties, but with the index 2 replaced by 1.
We may assume that F,L1,L2,S1,S2are chosen so that F ∪ L1∪ L2∪ S1∪ S2is minimal
with respect to inclusion. This will be referred to as “minimality.”
If the paths L1,L2,S1,S2are pairwise disjoint, except possibly for shared ends and pos-
sibly S1and S2intersecting, then it is easy to see that the lemma holds, and so we may
assume that an internal vertex of L1belongs to L2∪ S2. Let v be the first vertex on L1
(in either direction) that belongs to L2∪ S2, and suppose for a contradiction that v is not
an end of L1. Let L′
internal vertex in L2∪S2. Then by replacing a subpath of L2or S2by L′
contradiction to minimality, or a cross that is a subgraph of L1∪ L2∪ S1∪ S2\{z1,z2}, also
a contradiction. This proves that v is an end of L1, and hence both ends of L1are also ends
of L2or S2. In particular, L1and L2share at least one end.
Suppose first that one end of L1is an end of S2. Thus from the symmetry we may assume
that a1is an end of L2and c1= b2; thus a2= a1, because a2,b2,c2is clockwise. But now
c2is not an end of L1or S1, and so the argument of the previous paragraph implies that no
internal vertex of L2belongs to S1∪ L1. The paths S1,S2,L2now show that (G,Ω) has a
fan with t blades and a jump.
We may therefore assume that a1= a2and c1= c2. Let H be the union of L1,L2,S1\z1,
S2\z2, and V (Ω). Then the society (H,Ω) is rural, as otherwise (G\{z1,z2},Ω) has a cross.
Let Γ be a drawing of (H,Ω) in a disk ∆ such that the vertices of V (Ω) are drawn on the
boundary of ∆ in the clockwise order specified by Ω. Let ∆′⊆ ∆ be a disk such that ∆′
includes every path in Γ with ends a1and c1, and the boundary of ∆′includes a1Ωc1and
a path P of Γ from a1to c1. Then L1and L2lie in ∆′, and since Liis disjoint from Si\zi
it follows that S1\z1and S2\z2are inside ∆′and, in particular, are disjoint from P. By
considering P, S1and S2we obtain a fan with t blades and a jump. ?
1be a subpath of L1with one end v, the other end in V (Ω) and no
1we obtain either a
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Proof of Theorem 5.3. Let d and t be integers, let k be an integer such that Theorem 5.2
holds for d and 16t + 5, and let (G,Ω) be a 6-connected k-cosmopolitan society of depth at
most d. We may assume that (G,Ω) is not nearly rural, for otherwise the theorem holds. By
Theorem 5.2 the society (G,Ω) has t disjoint consecutive crosses, or a windmill with 4t + 1
vanes, or a fan with 16t + 5 blades. In the first case the theorem holds, and in the second
and third case the theorem follows from Lemma 9.1 and Lemma 9.2, respectively. ?
For the proof of Theorem 1.8 we need one more lemma.
Lemma 9.3 Let d and s be integers, let (G,Ω) be an s-nested society, and let (G′,Ω′) be a
planar truncation of (G,Ω) of depth at most d. Then (G,Ω) has an s-nested planar truncation
of depth at most 2(d + 2s).
Proof. By a vortical decomposition of a society (G,Ω) we mean a collection (Zv: v ∈ V (Ω))
of sets such that
(i)?(Zv: v ∈ V (Ω)) = V (G),
(ii) for v ∈ V (Ω), v ∈ Zv, and
(iii) if (v1,v2,v3,v4) is clockwise in Ω, then Zv1∩ Zv3⊆ Zv2∪ Zv4.
The depth of such a vortical decomposition is max|Zu∩ Zv|, taken over all pairs of distinct
vertices u,v ∈ V (Ω) that are consecutive in Ω, and the depth of (G,Ω) is the minimum
depth of a vortical decomposition of (G,Ω). Thus if (G,Ω) has depth at most d, then
the corresponding linear decomposition also serves as a vortical decomposition of depth at
most d.
Let (G,Ω) be an s-nested society, and let it be the composition of a society (G0,Ω0) with
a rural neighborhood (G1,Ω,Ω0), where the neighborhood has a presentation (Σ,Γ1,∆,∆0)
with an s-nest C1,C2,...,Cs. Let ∆0,∆1,...,∆s be as in the definition of s-nest. Let
(G′,Ω′) be a planar truncation of (G,Ω) of depth at most d. Then (G,Ω) is the composition
of (G′,Ω′) with a rural neighborhood (G2,Ω,Ω′), and we may assume that (G2,Ω,Ω′) has a
presentation (Σ,Γ2,∆,∆′), where ∆0⊆ ∆′. We may assume that the s-nest C1,C2,...,Cs
is chosen as follows: first we select C1such that ∆0⊆ ∆1and the disk ∆1is as small as
possible, subject to that we select C2such that ∆1 ⊆ ∆2 and the disk ∆2 is as small as
possible, subject to that we select C3, and so on.
Let ∆∗be a closed disk with ∆′⊆ ∆∗⊆ ∆. We say that ∆∗is normal if whenever an
interior point of an edge e ∈ E(Γ1) belongs to the boundary of ∆∗, then e is a subset of the
boundary of ∆∗. A normal disk ∆∗defines a planar truncation (G∗,Ω∗) in a natural way as
follows: G∗is consists of all vertices and edges that of G either belong to G′, or their image
under Γ1belongs to ∆∗, and Ω∗consists of vertices of G whose image under Γ1belongs to
the boundary ∆∗in the order determined by the boundary of ∆∗.
Given a normal disk ∆∗and two vertices u,v ∈ V (G) we define ξ∆∗(u,v), or simply
ξ(u,v) as follows. If u is adjacent to v, and the image e under Γ1of the edge uv is a subset of
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the boundary of ∆∗, and for every internal point x on e there exists an open neighborhood
U of x such that U ∩ ∆∗= U ∩ ∆i, then we let ξ(u,v) = i. Otherwise we define ξ(u,v) = 0.
A short explanation may be in order. If the image e of uv is a subset of the boundary of
∆∗, then this can happen in two ways: if we think of e as having two sides, either ∆∗and
∆iappear on the same side, or on opposite sides of e. In the definition of ξ it is only edges
with ∆∗and ∆ion the same side that count.
We may assume, by shrinking ∆′slightly, that the boundary of ∆′does not include
an interior point of any edge of Γ2. Then ∆′is normal, and the corresponding planar
truncation is (G′,Ω′). Since a linear decomposition of (G′,Ω′) of depth at most d may be
regarded as a vortical decomposition of (G′,Ω′) of depth at most d, we may select a normal
disk ∆∗that gives rise to a planar truncation (G∗,Ω∗) of (G,Ω), and we may select a vortical
decomposition (Zv: v ∈ V (Ω∗)) of (G∗,Ω∗) such that |Zu∩ Zv| ≤ d + 2ξ(u,v) for every pair
of consecutive vertices of Ω∗. Furthermore, subject to this, we may choose ∆∗such that the
number of unordered pairs u,v of distinct vertices of G with ξ(u,v) = s is maximum, subject
to that the number of unordered pairs u,v of distinct vertices of G with ξ(u,v) = s − 1 is
maximum, subject to that the number of unordered pairs u,v of distinct vertices of G with
ξ(u,v) = s − 2 is maximum, and so on.
We will show that (G∗,Ω∗) satisfies the conlusion of the theorem. Let (t1,t2,...,tn)
be an arbitrary clockwise enumeration of V (Ω∗), and let Xi := Zti∪ (Zt1∩ Ztn). Then
(X1,X2,...,Xn) is a linear decomposition of (G∗,Ω∗) of depth at most 2(d + 2s).
To complete the proof we must show that (G∗,Ω∗) is s-nested, and we will do that by
showing that each Ciis a subgraph of G∗. To this end we suppose for a contradiction that
it is not the case, and let i0∈ {1,2,...,s} be the minimum integer such that Ci0is not a
subgraph of G∗.
If Ci0has no edge in G∗, then we can construct a new society (G3,Ω3), where Ω3consists
of the vertices of Ci0in order, and obtain a contradiction to the choice of (G∗,Ω∗). Since
the construction is very similar but slightly easier than the one we are about to exhibit,
we omit the details. Instead, we assume that Ci0includes edges of both G∗and G\E(G∗).
Thus there exist vertices x,y ∈ V (Ci0) ∩ V (Ω∗) such that some subpath P of Ci0with ends
x and y has no internal vertex in V (Ω∗). Let B denote the boundary of ∆∗. There are three
closed disks with boundaries contained in B ∪ P. One of them is ∆∗; let D be the one that
is disjoint from ∆0. If the interior of D is a subset of ∆i0and includes no edge of Ci0, then
we say that P is a good segment. It follows by a standard elementary argument that there
is a good segment.
Thus we may assume that P is a good segment, and that the notation is as in the
previous paragraph. There are two cases: either D is a subset of ∆∗, or the interiors of D
and ∆∗are disjoint. Since the former case is handled by a similar, but easier construction,
we leave it to the reader and assume the latter case. Let (s0,s1,...,st+1) be clockwise in
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Ω∗such that s0,s1,...,st+1are all the vertices that belong to D ∩ ∆∗. Thus {s0,st+1} =
{x,y}. Let r0= s0,r1,...,rk,rk+1= st+1be all the vertices of P, in order, let H be the
subgraph of G∗consisting of all vertices and edges whose images under Γ1 belong to D,
and let X := {s0,s1,...,st+1,r0,r1,...,rk+1}. We can regard H as drawn in a disk with
the vertices s0,s1,...,st+1,rk,rk−1,...,r1drawn on the boundary of the disk in order. We
may assume that every component of H intersects X. The way we chose the cycles Ci0
implies that every path in H\{s1,s2,...,sk} that joins two vertices of P is a subpath of
P. We will refer to this property as the convexity of H. For i = 0,1,...,k + 1 let bi
be the maximum index j such that the vertex sjcan be reached from {r0,r1,...,ri} by a
path in H with no internal vertex in X. We define b−1 := −1, and let Ri be the set of
all vertices of H that can be reached from {ri,sbi−1+1,sbi−1+2,...,sbi} by a path with no
internal vertex in X. The convexity of H implies that for i < j the only possible member
of Ri∩ Rjis sbi. We now define a new society (G∗∗,Ω∗∗) as follows. The graph G∗∗will be
the union of G∗and H, and the cyclic permutation is defined by replacing the subsequence
s0,s1,...,st+1of Ω∗by the sequence r0,r1,...,rk,rk+1. We define the sets Z∗∗
For v ∈ V (Ω∗) − V (Ω∗∗) we let Z∗∗
v
:= Zv. If v = ri and bi > bi−1 we define Z∗∗
the union of Ri∪ {sbi,ri−1} and all Zsjfor j = bi−1+ 1,bi−1+ 2,...,bi. If v = ri and
bi= bi−1we define Z∗∗
that (G∗∗,Ω∗∗) is a planar truncation of (G,Ω) and that (Z∗∗
decomposition of (G∗∗,Ω∗∗). We claim that ξ∆∗(sj,sj+1) < i0for all j = 0,1,...,t. To prove
this we may assume that sjis adjacent to sj+1, and let e be the image under Γ1of the edge
sjsj+1. It follows that e is a subset of ∆i0, and hence if sjsj+1∈ E(Ck) for some k, then
k ≤ i0. Furthermore, if equality holds, then ∆i0and ∆∗lie on opposite sides of e, and hence
ξ∆∗(sj,sj+1) = 0. This proves our claim that ξ∆∗(sj,sj+1) < i0. Since for i = 0,1,...,k we
have Z∗∗
v as follows.
v
to be
v := Ri∪ {sbi,ri−1} ∪ (Zsbi∩ Zsbi+1). It is straightforward to verify
v
: v ∈ V (Ω∗∗)) is a vortical
ri∩ Z∗∗
ri+1⊆ (Zsbi∩ Zsbi+1) ∪ {ri,sbi}, and ξ∆∗∗(ri,ri+1) = i0, we deduce that
|Z∗∗
ri∩ Z∗∗
ri+1| ≤ |Zsbi∩ Zsbi+1| + 2 ≤ d + ξ∆∗(sbi,sbi+1) ≤ d + 2ξ∆∗∗(ri,ri+1).
Thus the existence of (G∗∗,Ω∗∗) contradicts the choice of (G∗,Ω∗). This completes our proof
that C1,C2,...,Csare subgraphs of G∗, and hence (G∗,Ω∗) is s-nested, as desired. ?
Proof of Theorem 1.8. Let d be as in Theorem 5.1, and let k be as in Corollary 5.4
applied to 2(d + 2s) in place of d. We claim that k satisfies Theorem 1.8. To prove that
let (G,Ω) be a 6-connected s-nested k-cosmopolitan society that is not nearly rural. Since
(G,Ω) is an s-nested planar truncation of itself, by Theorem 5.1 we may assume that (G,Ω)
has either a leap of length five, in which case it satisfies Theorem 1.8 by Theorem 4.1, or
it has a planar truncation of depth at most d. In the latter case it has an s-nested planar
truncation (G′,Ω′) of depth at most 2(d + 2s) by Lemma 9.3, and the theorem follows from
Corollary 5.4 applied to the society (G′,Ω′). ?
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10 Finding a planar nest
In this section we prove a technical result that applies in the following situation. We will
be able to guarantee that some societies (G,Ω) contain certain configurations consisting
of disjoint trees connecting specified vertices in V (Ω).
Theorem 10.3 below, states that if the society is sufficiently nested, then we can make sure
The main result of this section,
that the cycles in some reasonably big nest and the trees of the configuration intersect nicely.
A target in a society (G,Ω) is a subgraph F of G such that
(i) F is a forest and every leaf of F belongs to V (Ω), and
(ii) if u,v ∈ V (Ω) belong to a component T of F, then there exists a component T′?= T
of F and w ∈ V (T′) ∩ V (Ω) such that (u,w,v) is clockwise.
We say that a vertex v ∈ V (G) is F-special if either v has degree at least three in F, or v
has degree at least two in F and v ∈ V (Ω).
Now let F be a target in (G,Ω) and let T be a component of F. Let P be a path in
G\V (Ω) with ends u,v such that u,v ∈ V (T) and P is otherwise disjoint from F. Let C be
the unique cycle in T ∪ P, and assume that C has at most one F-special vertex. If C\u\v
has no F-special vertex, then let P′be the subpath of C that is complementary to P, and if
C\u\v has an F-special vertex, say w, then let P′be either the subpath of C\u with ends
v and w, or the subpath of C\v with ends u and w. Finally, let F′be obtained from F ∪ P
by deleting all edges and internal vertices of P′. In those circumstances we say that F′was
obtained from F by rerouting.
A subgraph F of a rural neighborhood (G,Ω,Ω0) is perpendicular to an s-nest (C1,C2,...,Cs)
if for every component P of F
(i) P is a path with one end in V (Ω) and the other in V (Ω0), and
(ii) P ∩ Ciis a path for all i = 1,2,...,s.
The complexity of a forest F in a society (G,Ω) is
?
(degF(v) − 2)++
?
v∈V (Ω)
(degF(v) − 1)+,
where the first summation is over all v ∈ V (G) − V (Ω) and x+denotes max(x,0).
The following is a preliminary version of the main result of this section.
Theorem 10.1 Let w,s,k be positive integers, and let s′= 2w(k + 1) + s. Then for every
s′-nested society (G,Ω) such that G has tree-width at most w and for every target F0 in
(G,Ω) of complexity at most k there exists a target F in (G,Ω) obtained from F0by repeated
rerouting such that (G,Ω) can be expressed as a composition of some society with a rural
neighborhood (G′,Ω,Ω′) that has a presentation with an s-nest (C1,C2,...,Cs) such that
G′∩ F is perpendicular to (C1,C2,...,Cs).
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Proof. Suppose that the theorem is false for some integers w,s,k, a society (G,Ω) and target
F0, and choose these entities with |V (G)|+|E(G)| minimum. Let (G,Ω) be the composition
of a society (G0,Ω0) with a rural neighborhood (G1,Ω,Ω0). Let κ be the complexity of F∩G1
in the society (G1,Ω), and let s′′= 2w(κ + 1) + s. Since (G,Ω) is s′-nested and s′′≤ s′we
may choose a presentation (Σ,Γ,∆,∆0) of (G1,Ω,Ω0) and an s′′-nest (C1,C2,...,Cs′′) for
it. We may assume that G0,Ω0,G1,F,Σ,Γ,∆,∆0,C1,C2,...,Cs′′ are chosen to minimize κ.
The minimality of G implies that G = C1∪C2∪···∪Cs′ ∪F. Likewise, C1∪C2∪···∪Cs′ is
edge-disjoint from F, for otherwise contracting an edge belonging to the intersection of the
two graphs contradicts the minimality of G.
By a dive we mean a subpath of F ∩ G1with both ends in V (Ω0) and otherwise disjoint
from V (Ω0). Let P be a dive with ends u,v, and let P′be the corresponding path in Γ. Then
∆0∪P′separates Σ; let ∆(P′) denote the component of Σ−∆0−P′that is contained in ∆,
and let H(P) denote the subgraph of G1consisting of all vertices and edges that correspond
to vertices or edges of Γ that belong to the closure of ∆(P′). Thus P is a subgraph of
H(P). We say that a dive P is clean if H(P)\V (Ω0) includes at most one F-special vertex,
and if it includes one, say v, then v ∈ V (P), and no edge of E(F) − E(P) incident with
v belongs to H(P). The depth of a dive P is the maximum integer d ∈ {1,2,...,s′} such
that V (P) ∩ V (Cd) ?= ∅, or 0 if no such integer exists.
|V (P) ∩ V (Ci)| ≥ 2 for all i = 1,2,...,d − 1.
It follows from planarity that
(1) Every clean dive has depth at most 2w.
To prove the claim suppose for a contradiction that P1is a clean dive of depth d ≥ 2w+1.
Thus V (P1) ∩ V (Cd) ?= ∅. Assume that we have already constructed dives P1,P2,...,Pt
for some t ≤ w such that V (Pi) ∩ V (Cd−i+1) ?= ∅ for all i = 1,2,...,t and H(Pt) ⊆
H(Pt−1) ⊆ ··· ⊆ H(P1). Since V (Pt) ∩ V (Cd−t+1) ?= ∅, there exist distinct vertices x,y ∈
V (Pt)∩V (Cd−t). Furthermore, it is possible to select x,y such that one of subpaths of Cd−t
with ends x,y, say Q, is a subgraph of H(Pt) and no internal vertex of Q belongs to Pt.
We claim that some internal vertex of Q belongs to F. Indeed, if not, then we can reroute
xPty along Q to produce a target F′and delete an edge of xPty; since P1is clean and H(Pt)
is a subgraph of H(P1) this is indeed a valid rerouting as defined above. But this contradicts
the minimality of G, and hence some internal vertex of Q, say q, belongs to F. Since P1is
clean and H(Pt) is a subgraph of H(P1) it follows that q belongs to a dive Pt+1that is a
subgraph of H(Pt)\V (Pt). It follows that H(Pt+1) is a subgraph of H(Pt), thus completing
the construction.
The dives P1,P2,...,Pw+1just constructed are pairwise disjoint and all intersect Cd−w.
Since d ≥ 2w+1 this implies that P1,P2,...,Pw+1all intersect each of C1,C2,...,Cw+1, and
hence C1∪P1,C2∪P2,...,Cw+1∪Pw+1is a “screen” in G of “thickness” at least w +1. By
[14, Theorem (1.4)] the graph G has tree-width at least w, a contradiction. This proves (1).
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Our next objective is to prove that κ = 0. That will take several steps. To that end
let us define a dive P to be special if P\V (Ω0) contains exactly one F-special vertex. By a
bridge we mean a subgraph B of G1∩F consisting of a component C of G1\V (Ω0) together
with all edges from V (C) to V (Ω0) and all ends of these edges.
(2) If a bridge B includes an F-special vertex not in V (Ω0), then B includes a special dive.
To prove Claim (2) let B be a bridge containing an F-special vertex not in V (Ω0). For
an F-special vertex b ∈ V (B) − V (Ω0) and an edge e ∈ E(B) incident with b let Pe be
the maximal subpath of B containing e such that one end of Peis b and no internal vertex
of Peis F-special or belongs to V (Ω0). Let uebe the other end of Pe. The second axiom
in the definition of target implies that at most one vertex of F belongs to V (Ω). Since
every F-special vertex in V (G1)−V (Ω) has degree at least three, it follows that there exists
an F-special vertex b ∈ V (B) − V (Ω0) such that ue1,ue2∈ V (Ω0) for two distinct edges
e1,e2∈ E(B) incident with b. Then Pe1∪ Pe2is as desired. This proves (2).
By (2) we may select a special dive P with H(P) minimal. We claim that P is clean. For
let v ∈ V (P) − V (Ω0) be F-special. If some edge e ∈ E(F) − E(P) incident with v belongs
to H(P), then there exists a subpath P′of F containing e with one end v and the other end
in V (Ω0) ∪ V (Ω). But P′is a subgraph of H(P), and hence the other end of P′belongs to
V (Ω0) by planarity. It follows that P ∪ P′includes a dive that contradicts the minimality
of H(P). This proves that the edge e as above does not exist.
It remains to show that no vertex of H(P)\V (Ω0) except v is F-special. So suppose for
a contradiction that such vertex, say v′, exists. Then v′?∈ V (P), because P is special, and
hence v′belongs to a bridge B′?= B. But B′includes a special dive by (2), contrary to the
choice of P. This proves our claim that P is clean.
By (1) P has depth at most 2w. In particular, the image under Γ of some F-special vertex
belongs to the open disk ∆2w+1bounded by the image under Γ of C2w+1. Let G′
and all vertices and edges of G whose images under Γ belong to the closure of ∆2w+1, let G′
consist of all vertices and edges whose images under Γ belong to the complement of ∆2w+1,
and let Ω′
the order of V (C2w+1). Then (G,Ω) can be regarded as a composition of (G′
rural neighborhood (G′
where σ = 2wκ+s. On the other hand, the complexity of F ∩G′
0consist of G0
1
0be defined by V (Ω′
0) = V (C2w+1) and let the cyclic order of Ω′
0be determined by
0,Ω′
0) with the
1,Ω,Ω′
0). This rural neighborhood has a presentation with a σ-nest,
1is at most κ−1, contrary
to the minimality of κ. This proves our claim that κ = 0.
By repeating the argument of the previous paragraph and sacrificing 2w of the cycles Ci
we may assume that (G1,Ω,Ω0) has a presentation with an s-nest C1,C2,...,Csand that
there are no dives. It follows that every component P of F ∩ G1is a path with one end in
V (Ω) and the other in V (Ω0). To complete the proof of the theorem we must show that
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P ∩ Ciis a path for all i = 1,2,...,s. Suppose for a contradiction that that is not the case.
Thus for some i ∈ {1,2,...,s} and some component P of F ∩ G1the intersection P ∩ Ciis
not a path. Thus there exist distinct vertices x,y ∈ V (P ∩Ci) such that xPy is a path with
no edge or internal vertex in Ci. Let us choose P,i,x,y such that, subject to the conditions
stated, i is maximum. If i < s and xPy intersects Ci+1, then P ∩Ci+1is not a path, contrary
to the choice of i. If i = 1 or xPy does not intersect Ci−1, then by rerouting one of the
subpaths of Ciwith ends x,y along xPy we obtain contradiction to the minimality of G.
Thus we may assume that i > 1 and that xPy intersects Ci−1.
Exactly one of the subpaths of Ciwith ends x,y, say Q, has the property that the image
under Γ of xPy ∪Q bounds a disk contained in ∆ and disjoint from ∆0. If no component of
F ∩ G1other than P intersects Q, then by rerouting F along Q we obtain a contradiction
to the minimality of G. Thus there exists a component P′of F ∩ G1 other that P that
intersects Q, say in a vertex u. The vertex u divides P′into two subpaths P′
both P′
that say P′
otherwise disjoint from C1∪ C2∪ ··· ∪ Cs, and hence by rerouting Cialong P′′we obtain a
contradiction to the minimality of G. This completes the proof of the theorem. ?
1and P′
2. If
1and P′
1does not intersect Ci+1. But P′
2intersect Ci+1, then P′contradicts the choice of i. Thus we may assume
1includes a subpath P′′with both ends on Ciand
Before we state the main result of this section we need the following deep result from
[11]. A linkage in a graph G is a subgraph of G, every component of which is a path. A
linkage L in a graph G is vital if V (L) = V (G) and there is no linkage L′?= L in G such that
for every two vertices u,v ∈ V (G), the vertices u,v are the ends of a component of L if and
only if they are the ends of a component of L′.
Theorem 10.2 For every integer p ≥ 0 there exists an integer w such that every graph that
has a vital linkage with p components has tree-width at most w.
Now we are ready to state and prove the main theorem of this section. If F is a target
in a society (G,Ω) we say that a vertex v ∈ V (G) is critical for F if v is either F-special
or a leaf of F. We say that two targets F,F′are hypomorphic if they have the same set of
critical vertices, say X, and u,v ∈ X are joined by a path in F with no internal vertices in
X if and only if they are so joined in F′.
Theorem 10.3 For every two positive integers s,k there exists an integer s′such that for
every s′-nested society (G,Ω) and for every target F in (G,Ω) of complexity at most k there
exists a target F in (G,Ω) obtained from a target hypomorphic to F0by repeated rerouting
such that (G,Ω) can be expressed as a composition of some society with a rural neighbor-
hood (G′,Ω,Ω′) that has a presentation with an s-nest (C1,C2,...,Cs) such that G′∩ F is
perpendicular to (C1,C2,...,Cs).
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Proof. We proceed by induction on |V (G)|+|E(G)|. Let p = k+2, and let w be the bound
guaranteed by Theorem 10.2. By hypothesis (G,Ω) is the composition of a society (G0,Ω0)
with a rural neighborhood (G1,Ω,Ω0), where (G1,Ω,Ω0) has a presentation (Σ,Γ,∆,∆0) and
an s′-nest (C1,C2,...,Cs′). Let X be the set of all vertices critical for F, and let L = F\X.
Then L is a linkage in G\X. If it is vital, then G has tree-width at most |X|+w ≤ 2k+1+w,
and hence the theorem follows from Theorem 10.1.
Thus we may assume that L is not vital. Assume first that there exists a vertex v ∈
V (G)−V (L). If v ∈ V (Ci) for some i ∈ {1,2,...,s′}, then the theorem follows by induction
applied to the graph obtained from G by contracting one of the edges of Ciincident with v;
otherwise, the theorem follows by induction applied to the graph G\v.
Thus we may assume that V (L) = V (G), and hence there exists a linkage L′?= L linking
the same pairs of terminals. Thus there exists an edge e ∈ E(L) − E(L′). If e ∈ E(Ci)
for some i ∈ {1,2,...,s′}, then the theorem follows by induction by contracting the edge e;
otherwise it follows by induction by deleting e, because the linkage L′guarantees that G\e
has a target hypomorphic to F. ?
11Chasing a turtle
In this section we prove Theorem 1.3, but first we need the following two theorems.
Theorem 11.1 There is an integer s such that if an s-nested society (G,Ω) has a turtle,
then G has a K6minor.
Proof.
Let k be the maximum complexity of a turtle, let s = 3, and let s′′be as in
Theorem 10.3. We claim that s′′satisfies the theorem. Indeed, let (G,Ω) be an s′′-nested
society that has a turtle. Since every turtle is a target, and every target obtained from a
target hypomorphic to a turtle is again a turtle, we deduce from Theorem 10.3 that (G,Ω)
has a turtle F and can be expressed as a composition of a society with a rural neighborhood
(G′,Ω,Ω′) that has a presentation with a 3-nest (C1,C2,C3) such that G′∩F is perpendicular
to (C1,C2,C3). It is now fairly straightforward to deduce that G has a K6 minor. The
argument is illustrated in Figure 4. ?
Theorem 11.2 There is an integer s such that if an s-nested society (G,Ω) has three crossed
paths, a separated doublecross or a gridlet, then G has a K6minor.
Proof. The argument is analogous to the proof of the previous theorem, using Figures 5, 6
and 7 instead. We omit the details. ?
Proof of Theorem 1.3. Let s be an integer large enough that both Theorem 11.1 and
Theorem 11.2 hold for s. Let k be an integer such that Theorem 1.8 holds for this integer.
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Figure 4: A turtle giving rise to a K6minor.
Figure 5: Three crossed paths giving rise to a K6minor.
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5
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Figure 6: A gridlet giving rise to a K6minor.
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Figure 7: A separated doublecross giving rise to a K6minor.
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Let t be such that Theorem 1.7 holds for t and the integer k just defined. Let h be an
integer such that Theorem 1.6 holds with t replaced by t+2s. Let w be an integer such that
Theorem 1.5 holds for the integer h just defined. Finally, let N be as in Theorem 1.4.
Suppose for a contradiction that G is a 6-connected graph on at least N vertices that is
not apex. By Theorem 1.4 G has tree-width exceeding w. By Theorem 1.5 G has a wall of
height h. By Theorem 1.6 G has a planar wall H0of height t+2s. By considering a subwall
H of H0of height t and s cycles of H0\V (H) we find, by Theorem 1.7, that the anticompass
society (K,Ω) of H in G is s-nested and k-cosmopolitan. By Theorem 1.8 the society (K,Ω)
has a turtle, three crossed paths, a separated doublecross, or a gridlet. By Theorems 11.1
and 11.2 the graph G has a K6minor, a contradiction. ?
Acknowledgment
We would like to acknowledge the contributions of Matthew DeVos and Rajneesh Hegde,
who worked with us in March 2005 and contributed to this paper, but did not want to be
included as a coauthors.
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