K_6 minors in large 6-connected graphs

Abstract

Jorgensen conjectured that every 6-connected graph with no K_6 minor has a
vertex whose deletion makes the graph planar. We prove the conjecture for all
sufficiently large graphs.

Full text

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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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1Introduction
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.
2Finding 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. ?
3Rural 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. ?
15