Add medial tire decomposition paper

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2026-06-08 15:34:53 -04:00
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commit 20fe6c24ca
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@@ -16,6 +16,7 @@ All papers are at `papers/<name>/paper.tex`. The current set:
| `iterated_reduction_in_reduced_dual` | An Iterated Reduction in the Reduced Dual | | `iterated_reduction_in_reduced_dual` | An Iterated Reduction in the Reduced Dual |
| `level_resolutions_of_maximal_planar_graphs` | Level Resolutions of Maximal Planar Graphs | | `level_resolutions_of_maximal_planar_graphs` | Level Resolutions of Maximal Planar Graphs |
| `level_switching` | Level Switching | | `level_switching` | Level Switching |
| `medial_tire_decompositions_of_plane_triangulations` | Medial Tire Decompositions of Plane Triangulations |
| `nested_tire_decompositions_of_plane_triangulations` | Nested Tire Decompositions of Plane Triangulations | | `nested_tire_decompositions_of_plane_triangulations` | Nested Tire Decompositions of Plane Triangulations |
| `plane_depth` | Plane Depth | | `plane_depth` | Plane Depth |
| `plane_depth_sequencing` | Plane Depth Sequencing | | `plane_depth_sequencing` | Plane Depth Sequencing |
@@ -0,0 +1,350 @@
"""Compare full and reduced medial tire graphs on generated tires.
The new medial decomposition paper defines:
* full medial tire graph: the subgraph of M(G) induced by medial
vertices corresponding to edges incident to tread triangles;
* reduced medial tire graph: delete same-boundary medial edges and
chord-only medial edges.
For a tire tread inside an ambient triangulation, the medial edges
visible in the tread come from annular triangular faces. This script
checks whether any same-boundary medial edges are actually present in
that model. It also compares against the older standalone drawing
model, which added artificial outer/inner boundary faces.
"""
from __future__ import annotations
import argparse
import itertools
import random
from collections import Counter
Edge = tuple[int, int]
MedialEdge = tuple[Edge, Edge]
def random_tire(m: int, k: int, n_chords: int = 0, seed: int | None = None) -> dict:
"""Generate the same labelled annular tires used in earlier experiments."""
rng = random.Random(seed)
outer = list(range(m))
inner = list(range(m, m + k))
edges: set[Edge] = set()
for i in range(m):
edges.add(edge_key(outer[i], outer[(i + 1) % m]))
for j in range(k):
edges.add(edge_key(inner[j], inner[(j + 1) % k]))
inner_chords = set()
candidates = []
for a in range(k):
for b in range(a + 2, k):
if not (a == 0 and b == k - 1):
candidates.append((a, b))
rng.shuffle(candidates)
for a, b in candidates:
if len(inner_chords) >= n_chords:
break
if any((a < a2 < b < b2) or (a2 < a < b2 < b) for a2, b2 in inner_chords):
continue
inner_chords.add((a, b))
edges.add(edge_key(inner[a], inner[b]))
edges.add(edge_key(outer[0], inner[0]))
moves = ["O"] * m + ["I"] * k
rng.shuffle(moves)
triangles = []
i, j = 0, 0
for move in moves:
if move == "O":
tri = (outer[i % m], inner[j % k], outer[(i + 1) % m])
triangles.append(tri)
edges.add(edge_key(inner[j % k], outer[(i + 1) % m]))
i += 1
else:
tri = (outer[i % m], inner[j % k], inner[(j + 1) % k])
triangles.append(tri)
edges.add(edge_key(outer[i % m], inner[(j + 1) % k]))
j += 1
return {
"m": m,
"k": k,
"n_chords": len(inner_chords),
"outer": outer,
"inner": inner,
"edges": sorted(edges),
"triangles": triangles,
"inner_chords": sorted(inner_chords),
"lattice_path": "".join(moves),
"seed": seed,
}
def tire_from_path(m: int, k: int, chords: tuple[tuple[int, int], ...], path: str) -> dict:
outer = list(range(m))
inner = list(range(m, m + k))
edges: set[Edge] = set()
for i in range(m):
edges.add(edge_key(outer[i], outer[(i + 1) % m]))
for j in range(k):
edges.add(edge_key(inner[j], inner[(j + 1) % k]))
for a, b in chords:
edges.add(edge_key(inner[a], inner[b]))
edges.add(edge_key(outer[0], inner[0]))
triangles = []
i, j = 0, 0
for move in path:
if move == "O":
tri = (outer[i % m], inner[j % k], outer[(i + 1) % m])
triangles.append(tri)
edges.add(edge_key(inner[j % k], outer[(i + 1) % m]))
i += 1
else:
tri = (outer[i % m], inner[j % k], inner[(j + 1) % k])
triangles.append(tri)
edges.add(edge_key(outer[i % m], inner[(j + 1) % k]))
j += 1
return {
"m": m,
"k": k,
"n_chords": len(chords),
"outer": outer,
"inner": inner,
"edges": sorted(edges),
"triangles": triangles,
"inner_chords": sorted(chords),
"lattice_path": path,
"seed": None,
}
def chord_crosses(c1: tuple[int, int], c2: tuple[int, int]) -> bool:
a, b = c1
c, d = c2
return (a < c < b < d) or (c < a < d < b)
def chord_sets(k: int, max_chords: int) -> list[tuple[tuple[int, int], ...]]:
candidates = []
for a in range(k):
for b in range(a + 2, k):
if not (a == 0 and b == k - 1):
candidates.append((a, b))
out = [()]
def rec(start: int, chosen: tuple[tuple[int, int], ...]) -> None:
if len(chosen) >= max_chords:
return
for idx in range(start, len(candidates)):
chord = candidates[idx]
if any(chord_crosses(chord, old) for old in chosen):
continue
nxt = chosen + (chord,)
out.append(nxt)
rec(idx + 1, nxt)
rec(0, ())
return out
def lattice_paths(m: int, k: int):
for o_positions in itertools.combinations(range(m + k), m):
o_set = set(o_positions)
yield "".join("O" if idx in o_set else "I" for idx in range(m + k))
def edge_key(u: int, v: int) -> Edge:
return tuple(sorted((u, v)))
def face_edges(face: tuple[int, ...]) -> list[Edge]:
return [edge_key(face[i], face[(i + 1) % len(face)]) for i in range(len(face))]
def is_cycle_edge(edge: Edge, cycle: list[int]) -> bool:
cycle_set = set(cycle)
if not set(edge) <= cycle_set:
return False
n = len(cycle)
idx = {v: i for i, v in enumerate(cycle)}
a, b = idx[edge[0]], idx[edge[1]]
return (a - b) % n in (1, n - 1)
def is_inner_chord(edge: Edge, m: int, k: int) -> bool:
u, v = edge
if not (m <= u < m + k and m <= v < m + k):
return False
a, b = u - m, v - m
d = abs(a - b)
return min(d, k - d) != 1
def suppress_reason(e1: Edge, e2: Edge, tire: dict) -> str | None:
outer = tire["outer"]
inner = tire["inner"]
if is_cycle_edge(e1, outer) and is_cycle_edge(e2, outer):
return "outer_boundary"
if is_cycle_edge(e1, inner) and is_cycle_edge(e2, inner):
return "inner_boundary"
m, k = tire["m"], tire["k"]
if is_inner_chord(e1, m, k) or is_inner_chord(e2, m, k):
return "inner_chord"
return None
def medial_from_faces(faces: list[tuple[int, ...]], retained: set[Edge]) -> set[MedialEdge]:
medial_edges: set[MedialEdge] = set()
for face in faces:
boundary = [e for e in face_edges(face) if e in retained]
if len(boundary) < 2:
continue
for i, e in enumerate(boundary):
nxt = boundary[(i + 1) % len(boundary)]
if e != nxt:
medial_edges.add(tuple(sorted((e, nxt))))
return medial_edges
def compare_tire(tire: dict, *, standalone_boundary_faces: bool) -> dict:
annular_faces = [tuple(tri) for tri in tire["triangles"]]
faces = list(annular_faces)
if standalone_boundary_faces:
faces.append(tuple(tire["outer"]))
faces.append(tuple(reversed(tire["inner"])))
# Definition 3.1 includes edges incident to at least one tread triangle.
retained = {e for face in annular_faces for e in face_edges(face)}
full_edges = medial_from_faces(faces, retained)
removed = {me for me in full_edges if suppress_reason(me[0], me[1], tire)}
reduced_edges = full_edges - removed
reasons = Counter(suppress_reason(me[0], me[1], tire) for me in removed)
reasons.pop(None, None)
return {
"vertices": len(retained),
"full_edges": len(full_edges),
"reduced_edges": len(reduced_edges),
"removed": len(removed),
"reasons": reasons,
"examples": sorted(removed)[:5],
}
def run_sweep(args: argparse.Namespace) -> None:
ambient_cases = 0
ambient_differ = []
standalone_cases = 0
standalone_differ = []
ambient_reasons: Counter[str] = Counter()
standalone_reasons: Counter[str] = Counter()
max_chords = args.max_chords
for m in range(args.min_cycle, args.max_cycle + 1):
for k in range(args.min_cycle, args.max_cycle + 1):
for chords in range(max_chords + 1):
for seed in range(args.seeds):
tire = random_tire(m=m, k=k, n_chords=chords, seed=seed)
ambient = compare_tire(tire, standalone_boundary_faces=False)
ambient_cases += 1
ambient_reasons.update(ambient["reasons"])
if ambient["removed"]:
ambient_differ.append((m, k, chords, seed, tire, ambient))
standalone = compare_tire(tire, standalone_boundary_faces=True)
standalone_cases += 1
standalone_reasons.update(standalone["reasons"])
if standalone["removed"]:
standalone_differ.append((m, k, chords, seed, tire, standalone))
print("ambient tread-face model")
print(f" cases checked: {ambient_cases}")
print(f" cases where full != reduced: {len(ambient_differ)}")
print(f" removed-edge reasons: {dict(sorted(ambient_reasons.items()))}")
if ambient_differ:
m, k, chords, seed, tire, result = ambient_differ[0]
print(" first difference:")
print(f" m={m} k={k} requested_chords={chords} seed={seed}")
print(f" path={tire['lattice_path']} chords={tire['inner_chords']}")
print(f" removed examples={result['examples']}")
print()
print("standalone tire-with-boundary-faces model")
print(f" cases checked: {standalone_cases}")
print(f" cases where full != reduced: {len(standalone_differ)}")
print(f" removed-edge reasons: {dict(sorted(standalone_reasons.items()))}")
if standalone_differ:
m, k, chords, seed, tire, result = standalone_differ[0]
print(" first difference:")
print(f" m={m} k={k} requested_chords={chords} seed={seed}")
print(f" path={tire['lattice_path']} chords={tire['inner_chords']}")
print(f" full_edges={result['full_edges']} reduced_edges={result['reduced_edges']}")
print(f" removed examples={result['examples']}")
def run_exhaustive(args: argparse.Namespace) -> None:
ambient_cases = 0
ambient_differ = []
standalone_cases = 0
standalone_differ = []
for m in range(args.min_cycle, args.max_cycle + 1):
for k in range(args.min_cycle, args.max_cycle + 1):
for chords in chord_sets(k, args.max_chords):
for path in lattice_paths(m, k):
tire = tire_from_path(m, k, chords, path)
ambient = compare_tire(tire, standalone_boundary_faces=False)
ambient_cases += 1
if ambient["removed"]:
ambient_differ.append((m, k, chords, path, ambient))
standalone = compare_tire(tire, standalone_boundary_faces=True)
standalone_cases += 1
if standalone["removed"]:
standalone_differ.append((m, k, chords, path, standalone))
print("exhaustive ambient tread-face model")
print(f" cases checked: {ambient_cases}")
print(f" cases where full != reduced: {len(ambient_differ)}")
if ambient_differ:
m, k, chords, path, result = ambient_differ[0]
print(" first difference:")
print(f" m={m} k={k} chords={chords} path={path}")
print(f" removed examples={result['examples']}")
print()
print("exhaustive standalone tire-with-boundary-faces model")
print(f" cases checked: {standalone_cases}")
print(f" cases where full != reduced: {len(standalone_differ)}")
if standalone_differ:
m, k, chords, path, result = standalone_differ[0]
print(" first difference:")
print(f" m={m} k={k} chords={chords} path={path}")
print(f" full_edges={result['full_edges']} reduced_edges={result['reduced_edges']}")
print(f" removed examples={result['examples']}")
def main() -> None:
parser = argparse.ArgumentParser()
parser.add_argument("--min-cycle", type=int, default=3)
parser.add_argument("--max-cycle", type=int, default=8)
parser.add_argument("--max-chords", type=int, default=3)
parser.add_argument("--seeds", type=int, default=50)
parser.add_argument("--exhaustive", action="store_true")
args = parser.parse_args()
if args.exhaustive:
run_exhaustive(args)
else:
run_sweep(args)
if __name__ == "__main__":
main()
@@ -0,0 +1,82 @@
# Full vs Reduced Medial Tire Findings
Question: do Definition 3.1 (full medial tire graph) and Definition 3.2
(reduced medial tire graph) differ?
## Experiment
Script:
```bash
python3 papers/medial_tire_decompositions_of_plane_triangulations/experiments/compare_full_reduced_medial_tires.py
```
The script compares two models.
- Ambient tread-face model: medial edges are contributed by annular
triangular faces of the tire tread inside the ambient triangulation.
- Standalone tire-with-boundary-faces model: the outer and inner
boundary walks are also treated as faces, as in the older drawing
script.
## Random Sweep
Command:
```bash
python3 papers/medial_tire_decompositions_of_plane_triangulations/experiments/compare_full_reduced_medial_tires.py
```
Result:
```text
ambient tread-face model
cases checked: 7200
cases where full != reduced: 0
removed-edge reasons: {}
standalone tire-with-boundary-faces model
cases checked: 7200
cases where full != reduced: 7200
removed-edge reasons: {'inner_boundary': 39600, 'outer_boundary': 39600}
first difference:
m=3 k=3 requested_chords=0 seed=0
path=IOOOII chords=[]
full_edges=24 reduced_edges=18
removed examples=[((0, 1), (0, 2)), ((0, 1), (1, 2)), ((0, 2), (1, 2)), ((3, 4), (3, 5)), ((3, 4), (4, 5))]
```
## Exhaustive Small Sweep
Command:
```bash
python3 papers/medial_tire_decompositions_of_plane_triangulations/experiments/compare_full_reduced_medial_tires.py --exhaustive --max-cycle 5 --max-chords 2
```
Result:
```text
exhaustive ambient tread-face model
cases checked: 5578
cases where full != reduced: 0
exhaustive standalone tire-with-boundary-faces model
cases checked: 5578
cases where full != reduced: 5578
first difference:
m=3 k=3 chords=() path=OOOIII
full_edges=24 reduced_edges=18
removed examples=[((0, 1), (0, 2)), ((0, 1), (1, 2)), ((0, 2), (1, 2)), ((3, 4), (3, 5)), ((3, 4), (4, 5))]
```
## Interpretation
For the intended ambient-triangulation definition, the experiments
support the suspicion that Definition 3.1 and Definition 3.2 coincide:
same-boundary medial edges do not arise from annular triangular tread
faces, and inner chords are not incident to tread triangles.
They differ only in the standalone tire-with-boundary-faces model,
where the artificial outer and inner boundary faces create medial edges
between consecutive boundary edges.
@@ -0,0 +1,34 @@
\relax
\citation{bauerfeld-nested-tire-decompositions}
\citation{bauerfeld-nested-tire-decompositions}
\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Introduction}}{1}{}\protected@file@percent }
\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Background}}{1}{}\protected@file@percent }
\citation{bauerfeld-nested-tire-decompositions}
\citation{bauerfeld-nested-tire-decompositions}
\newlabel{def:medial-graph}{{2.1}{2}}
\newlabel{prop:medial-dual-invariance}{{2.3}{2}}
\newlabel{cor:tait-medial}{{2.4}{2}}
\@writefile{toc}{\contentsline {section}{\tocsection {}{3}{Medial tire pieces}}{2}{}\protected@file@percent }
\newlabel{def:full-medial-tire}{{3.1}{2}}
\newlabel{thm:annular-medial-colour-bound}{{3.3}{3}}
\newlabel{def:boundary-medial-vertices}{{3.4}{3}}
\newlabel{def:medial-restriction-relation}{{3.5}{3}}
\citation{bauerfeld-nested-tire-decompositions}
\citation{bauerfeld-nested-tire-decompositions}
\@writefile{toc}{\contentsline {section}{\tocsection {}{4}{Decomposition}}{4}{}\protected@file@percent }
\newlabel{cor:medial-tire-decomposition}{{4.1}{4}}
\newlabel{def:compatible-family}{{4.2}{4}}
\newlabel{prop:gluing-criterion}{{4.3}{4}}
\@writefile{toc}{\contentsline {section}{\tocsection {}{5}{A medial pigeonhole programme}}{4}{}\protected@file@percent }
\bibcite{bauerfeld-nested-tire-decompositions}{1}
\bibcite{tait-original}{2}
\newlabel{tocindent-1}{0pt}
\newlabel{tocindent0}{12.7778pt}
\newlabel{tocindent1}{17.77782pt}
\newlabel{tocindent2}{0pt}
\newlabel{tocindent3}{0pt}
\newlabel{def:medial-boundary-state}{{5.1}{5}}
\newlabel{conj:medial-chain-pigeonhole}{{5.2}{5}}
\newlabel{conj:medial-route-fct}{{5.3}{5}}
\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{5}{}\protected@file@percent }
\gdef \@abspage@last{5}
@@ -0,0 +1,138 @@
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%% filename: amsart-template.tex
%% American Mathematical Society
%% AMS-LaTeX v.2 template for use with amsart
%% ====================================================================
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\begin{document}
\title{Medial Tire Decompositions of Plane Triangulations}
% author one information
\author{Eric Bauerfeld}
\address{}
\curraddr{}
\email{}
\thanks{}
\subjclass[2010]{Primary }
\keywords{plane graph, triangulation, medial graph, tire graph, Tait coloring, Four Colour Theorem}
\date{}
\dedicatory{}
\begin{abstract}
We use the nested tire decomposition of a plane triangulation to induce
a decomposition of its full medial graph into medial tire subgraphs.
For a plane triangulation $G$, the medial graph $M(G)$ is naturally
isomorphic to the medial graph of the planar dual $G^*$, and proper
$3$-vertex-colourings of $M(G)$ are equivalent to proper
$3$-edge-colourings of the cubic dual. Thus Tait's reformulation of
the Four Colour Theorem may be studied through proper vertex
$3$-colourings of medial subgraphs. We define medial tire pieces,
their boundary-state restriction relations, and a chain-pigeonhole
conjecture for compatible medial boundary states across the tire tree.
\end{abstract}
\maketitle
\section{Introduction}
A classical theorem of Tait recasts the Four Colour Theorem in dual,
edge-colouring terms: a plane triangulation $G$ is properly
$4$-vertex-colourable if and only if its dual cubic graph $G^*$ is
properly $3$-edge-colourable. The present paper records a medial
version of this viewpoint. The vertices of the medial graph $M(G)$
correspond to edges of $G$, and adjacency in $M(G)$ records
consecutiveness of edges around vertices and faces of $G$. Since
planar duality interchanges vertices and faces while preserving the
edge set, $M(G)$ is naturally isomorphic to $M(G^*)$.
Consequently a proper vertex $3$-colouring of $M(G)$ is the same
object as a proper edge $3$-colouring of $G^*$. This suggests another
route toward the Four Colour Theorem: rather than colouring the dual
cubic graph directly, decompose the full medial graph into local
annular pieces and try to prove that their proper vertex
$3$-colouring boundary restrictions always compose.
The structural input is the nested tire decomposition of
\cite{bauerfeld-nested-tire-decompositions}. A level source in a plane
triangulation determines a rooted tree of tire treads. Each tread is
an annular triangulated region with an outer boundary, an inner
outerplanar graph, and annular triangular faces. We show that this
decomposition induces a decomposition of $M(G)$ into medial tire
subgraphs. The boundary data of a medial tire are proper
$3$-colourings of the medial vertices corresponding to boundary edges
in the associated dual tire graph.
\section{Background}
Throughout, $G$ is a simple plane maximal planar graph with fixed
embedding, and $G^*$ denotes its full planar dual. We use the level
source, dual depth, tire graph, tire tread, and tire-tree terminology
of~\cite{bauerfeld-nested-tire-decompositions}. In particular, a level
source $S$ determines a rooted tire tree $\mathcal{T}(G,S)$ whose
vertices are tire treads and whose parent-child relation records
nested containment across level-cycle interfaces.
\begin{definition}[Medial graph]
\label{def:medial-graph}
Let $H$ be a plane graph. The \emph{medial graph} $M(H)$ has one
vertex $m_e$ for each edge $e \in E(H)$. Two medial vertices
$m_e,m_f$ are adjacent whenever $e$ and $f$ are consecutive in the
cyclic order of edges around a vertex of $H$ or around a face of $H$.
The embedding is the standard one obtained by placing $m_e$ at the
midpoint of $e$ and drawing medial edges through the vertex- and
face-corners of $H$.
\end{definition}
\begin{remark}
If $H$ has bridges or vertices of degree $1$, the usual medial
construction may create parallel edges or loops depending on the
chosen convention. In this paper the main application is to plane
triangulations and their cubic planar duals, where the medial graph is
a loopless $4$-regular plane graph.
\end{remark}
\begin{proposition}[Medial dual invariance]
\label{prop:medial-dual-invariance}
Let $H$ be a connected plane graph and let $H^*$ be its planar dual.
Then there is a natural plane-graph isomorphism
\[
M(H) \cong M(H^*).
\]
\end{proposition}
\begin{proof}
Each edge $e \in E(H)$ corresponds to a unique dual edge $e^* \in
E(H^*)$, giving a bijection $m_e \mapsto m_{e^*}$ between the vertices
of $M(H)$ and $M(H^*)$. In $M(H)$ two vertices $m_e,m_f$ are adjacent
exactly when $e$ and $f$ are consecutive around either a vertex or a
face of $H$. Under duality, vertices and faces are interchanged, and
the cyclic order of the corresponding dual edges around the dual face
or dual vertex is the same up to reversal. Thus the same pairs are
medial-adjacent in $M(H^*)$, and the midpoint construction identifies
the two embedded medial graphs.
\end{proof}
\begin{corollary}[Tait colourings as medial vertex colourings]
\label{cor:tait-medial}
Let $G$ be a simple plane triangulation. Proper vertex
$3$-colourings of $M(G)$ are in natural bijection with proper
$3$-edge-colourings of the cubic planar dual $G^*$.
\end{corollary}
\begin{proof}
By Proposition~\ref{prop:medial-dual-invariance}, $M(G) \cong
M(G^*)$. Vertices of $M(G^*)$ correspond to edges of $G^*$, and two
such vertices are adjacent exactly when the corresponding dual edges
are incident and consecutive around a vertex or face of $G^*$. Since
$G^*$ is cubic, proper vertex $3$-colouring of $M(G^*)$ is therefore
equivalent to assigning three colours to the edges of $G^*$ so that the
three edges incident to each dual vertex receive pairwise distinct
colours.
\end{proof}
\section{Medial tire pieces}
\begin{definition}[Full medial tire graph]
\label{def:full-medial-tire}
Let $T$ be a tire tread in the tire tree $\mathcal{T}(G,S)$ supplied
by~\cite{bauerfeld-nested-tire-decompositions}. The \emph{full medial
tire graph} of $T$, denoted $\mathsf{M}(T)$, is the subgraph of
$M(G)$ induced by the medial vertices $m_e$ with $e$ an edge of $G$
incident to at least one triangular face in the tread $T$. The medial
vertices corresponding to annular edges of $T$ are called
\emph{annular medial vertices}.
\end{definition}
\begin{remark}
In the ambient-triangulation setting, the full medial tire graph
$\mathsf{M}(T)$ coincides with the omitted-edge medial tire graph
studied in~\cite{bauerfeld-nested-tire-decompositions}. Indeed, the
medial edges of $\mathsf{M}(T)$ are contributed by corners of annular
triangular tread faces. Such a face contains at most one outer-boundary
edge and at most one inner-boundary edge, so it does not contribute a
medial edge between two outer-boundary edges or between two
inner-boundary edges. Similarly, chords of the inner outerplanar graph
lie outside the annular tread and are not incident to annular tread
faces. Thus the deletion rule used for the earlier reduced medial tire
graph removes no edges from the ambient object $\mathsf{M}(T)$.
The distinction only appears in the standalone drawing convention where
the outer and inner boundary walks are added as artificial faces before
forming a medial graph. Those artificial faces create same-boundary
medial edges, and the reduced construction deletes them.
\end{remark}
\begin{theorem}[Annular medial colour bound]
\label{thm:annular-medial-colour-bound}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire tread with
non-degenerate boundaries and simple inner boundary $B_{\mathrm{in}}$.
Let $A(T)$ be the subgraph of $\mathsf{M}(T)$ induced by the annular
medial vertices. For a graph $H$, write $\operatorname{Col}_3(H)$ for
the set of proper $3$-vertex-colourings of $H$. Then $A(T)$ is a cycle
and
\[
|\operatorname{Col}_3(\mathsf{M}(T))|
\;\leq\; |\operatorname{Col}_3(A(T))|.
\]
\end{theorem}
\begin{proof}
Since the tread is a triangulated annulus with no vertices in its
interior, each annular face has exactly one boundary edge, lying either
on $B_{\mathrm{out}}$ or on $B_{\mathrm{in}}$, and exactly two annular
edges. As the annular faces are traversed cyclically around the tread,
consecutive faces share one annular edge. Equivalently, the annular
edges occur in a cyclic order in which each annular face contains two
consecutive annular edges. Hence the subgraph of $\mathsf{M}(T)$
induced by the annular medial vertices is a cycle.
Consider the restriction map from proper $3$-colourings of
$\mathsf{M}(T)$ to colourings of this annular medial cycle $A(T)$. We
claim that this map is injective. Let $x$ be a non-annular medial
vertex. Then $x$ corresponds to an edge of $B_{\mathrm{out}}$ or
$B_{\mathrm{in}}$: chords of $O$ are not incident to annular tread
faces, and hence do not contribute vertices of $\mathsf{M}(T)$. This
boundary edge is incident to a unique annular face of the tread, and
the other two edges of that face are annular edges. Therefore $x$ is
adjacent in $\mathsf{M}(T)$ to the two annular medial vertices
corresponding to those two annular edges.
Those two annular medial vertices are adjacent to each other, because
their annular edges are consecutive on the same triangular annular
face. In any proper $3$-colouring they therefore receive two distinct
colours, and $x$ is forced to receive the remaining third colour. Thus
every non-annular medial vertex has its colour uniquely determined by
the colouring of $A(T)$. Two colourings of $\mathsf{M}(T)$ with the
same restriction to $A(T)$ are identical, so the restriction map is
injective. The stated inequality follows.
\end{proof}
\begin{definition}[Boundary medial vertices]
\label{def:boundary-medial-vertices}
Let $T$ be a tire tread and let $\Gamma_T$ be the corresponding dual
tire subgraph in $G^*$. A vertex $m_e \in V(\mathsf{M}(T))$ is an
\emph{outer boundary medial vertex} if the corresponding dual edge
$e^* \in E(G^*)$ lies on the outer boundary of $\Gamma_T$. It is an
\emph{inner boundary medial vertex} if $e^*$ lies on the inner boundary
of $\Gamma_T$. We write
\[
\partial_{\mathrm{out}}\mathsf{M}(T)
\quad\text{and}\quad
\partial_{\mathrm{in}}\mathsf{M}(T)
\]
for the two boundary sets.
\end{definition}
\begin{definition}[Medial tire restriction relation]
\label{def:medial-restriction-relation}
Let $\mathrm{Col}_3(X)$ denote the set of proper vertex
$3$-colourings of the induced subgraph on a vertex set $X$. The
\emph{medial tire restriction relation} of $T$ is
\[
R_T \subseteq
\mathrm{Col}_3(\partial_{\mathrm{out}}\mathsf{M}(T))
\times
\mathrm{Col}_3(\partial_{\mathrm{in}}\mathsf{M}(T)),
\]
where $(\alpha,\beta) \in R_T$ exactly when $\alpha \cup \beta$
extends to a proper vertex $3$-colouring of $\mathsf{M}(T)$.
\end{definition}
\begin{remark}
The definition deliberately records boundary colourings on medial
vertices corresponding to boundary edges in the dual tire graph. Under
Corollary~\ref{cor:tait-medial}, these are precisely edge-colouring
states on the boundary edges through which a dual tire piece meets its
parent and children.
\end{remark}
\section{Decomposition}
\begin{corollary}[Medial tire decomposition]
\label{cor:medial-tire-decomposition}
Let $G$ be a plane triangulation with level source $S$. The tire-tree
decomposition $\mathcal{T}(G,S)$ of
\cite{bauerfeld-nested-tire-decompositions} induces a rooted
decomposition of the full medial graph $M(G)$ into full medial tire
graphs $\{\mathsf{M}(T): T \in V(\mathcal{T}(G,S))\}$, glued along
their boundary medial vertex sets.
\end{corollary}
\begin{proof}
By the tire-tread partition theorem of
\cite{bauerfeld-nested-tire-decompositions}, the bounded triangular
faces of $G$ are partitioned into nested tire treads, with intersections
between parent and child treads occurring only along their level-cycle
interface data. Every edge of $G$ that is incident to a bounded face
therefore belongs to the closure of at least one tire tread, and an
edge lying in two closures lies on the interface between adjacent
treads in the tire tree. Passing to $M(G)$ sends edges of $G$ to
medial vertices. Thus each tread determines the induced subgraph
$\mathsf{M}(T)$ on its incident edge set, and overlaps between two such
subgraphs are exactly the medial vertices corresponding to interface
edges, namely the appropriate boundary medial vertex sets.
\end{proof}
\begin{definition}[Compatible family of medial tire colourings]
\label{def:compatible-family}
A \emph{compatible family of medial tire colourings} on
$\mathcal{T}(G,S)$ is a choice, for each tread $T$, of a proper
vertex $3$-colouring $\varphi_T$ of $\mathsf{M}(T)$ such that whenever
$T'$ is a child tread of $T$, the two colourings agree on
$
V(\mathsf{M}(T)) \cap V(\mathsf{M}(T')).
$
\end{definition}
\begin{proposition}[Gluing criterion]
\label{prop:gluing-criterion}
The full medial graph $M(G)$ has a proper vertex $3$-colouring if and
only if the tire tree $\mathcal{T}(G,S)$ admits a compatible family of
medial tire colourings.
\end{proposition}
\begin{proof}
A proper vertex $3$-colouring of $M(G)$ restricts to a proper vertex
$3$-colouring of every induced subgraph $\mathsf{M}(T)$, and these
restrictions agree on overlaps.
Conversely, suppose a compatible family is given. Define a colour on
each vertex $m_e$ of $M(G)$ by choosing any tread $T$ with
$m_e \in V(\mathsf{M}(T))$ and setting
$\varphi(m_e)=\varphi_T(m_e)$. Compatibility makes this independent of
the choice of $T$. Every medial edge of $M(G)$ is drawn in a corner of
some bounded triangular face of $G$ or along the outer boundary
interface. The relevant incident primal edges lie together in the
closure of a single tire tread or in a shared boundary interface, where
properness is already enforced by one of the local colourings. Hence
$\varphi$ is a proper vertex $3$-colouring of $M(G)$.
\end{proof}
\section{A medial pigeonhole programme}
The restriction relation $R_T$ records exactly the local information
needed to pass a medial $3$-colouring through a tire. In a nested
chain
\[
T_0 \supset T_1 \supset \cdots \supset T_k,
\]
the outer boundary state of $T_{i+1}$ must match an inner boundary
state allowed by $R_{T_i}$. Thus a proof of the Four Colour Theorem in
this framework would follow from a structural reason that these
restriction sets cannot remain mutually disjoint along every branch of
the tire tree.
\begin{definition}[Medial boundary state]
\label{def:medial-boundary-state}
A \emph{medial boundary state} on a boundary set
$\partial\mathsf{M}(T)$ is a proper vertex $3$-colouring of the
subgraph induced by that boundary set, considered up to permutation of
the three colours and the dihedral symmetries of the boundary walk
when that boundary is a cycle.
\end{definition}
\begin{conjecture}[Medial chain-pigeonhole principle]
\label{conj:medial-chain-pigeonhole}
There is a function $N(k)$ such that the following holds. Let
$T_0 \supset T_1 \supset \cdots \supset T_{N(k)}$ be a nested chain of
tire treads whose relevant boundary medial walks have length at most
$k$. Then two adjacent restriction relations in the chain have
compatible medial boundary states after colour permutation and boundary
symmetry. Equivalently, the chain contains a local gluing step that
cannot be obstructed by disjoint proper vertex $3$-colouring
restrictions.
\end{conjecture}
\begin{conjecture}[Medial tire route to the Four Colour Theorem]
\label{conj:medial-route-fct}
For every plane triangulation $G$ and every level source $S$, the
restriction relations $\{R_T : T \in V(\mathcal{T}(G,S))\}$ admit a
compatible selection of boundary states across the tire tree. Hence
$M(G)$ is properly vertex $3$-colourable, $G^*$ is properly
$3$-edge-colourable, and $G$ is properly $4$-vertex-colourable.
\end{conjecture}
\begin{remark}
Conjecture~\ref{conj:medial-route-fct} is equivalent in strength to
the Four Colour Theorem when combined with Tait's correspondence. The
point of the formulation is not to weaken the target theorem, but to
move the obstruction into finite boundary-state restrictions carried by
annular medial tire pieces.
\end{remark}
\begin{thebibliography}{9}
\bibitem{bauerfeld-nested-tire-decompositions}
E.~Bauerfeld,
\emph{Nested Tire Decompositions of Plane Triangulations},
manuscript (math-research repository), 2026.
\bibitem{tait-original}
P.~G. Tait,
\emph{Remarks on the colourings of maps},
Proceedings of the Royal Society of Edinburgh \textbf{10} (1880),
729--729.
\end{thebibliography}
\end{document}
@@ -8,50 +8,48 @@
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@@ -1,5 +1,5 @@
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@@ -52,9 +52,8 @@ $G$ induces a BFS layering of $G$ and endows the inner planar dual
$G'$ with a \emph{dual depth} grading. The basic object of study is $G'$ with a \emph{dual depth} grading. The basic object of study is
the \emph{tire graph} $T$ --- a plane graph whose outer and inner the \emph{tire graph} $T$ --- a plane graph whose outer and inner
boundaries bound a closed planar region, the \emph{tire tread} $R$, boundaries bound a closed planar region, the \emph{tire tread} $R$,
triangulated by the \emph{annular edges} $E_{\mathrm{ann}}$. We define triangulated by the \emph{annular edges} $E_{\mathrm{ann}}$. Our main
medial tire graphs and prove a basic colour-count bound for their structural results are the
annular medial cycle. Our main structural results are the
\emph{tire-component lemma}, the \emph{tire-tread partition theorem}, \emph{tire-component lemma}, the \emph{tire-tread partition theorem},
and the rooted \emph{tire-tree decomposition}, which together organize and the rooted \emph{tire-tree decomposition}, which together organize
the bounded faces of $G$ into nested tire treads. the bounded faces of $G$ into nested tire treads.
@@ -283,52 +282,6 @@ corresponding retained tire edge.}
\label{fig:medial-tire-example} \label{fig:medial-tire-example}
\end{figure} \end{figure}
\begin{theorem}[Annular medial colour bound]
\label{thm:annular-medial-colour-bound}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph with
non-degenerate boundaries and simple inner boundary $B_{\mathrm{in}}$.
Let $A(T)$ be the subgraph of $M_{\mathrm{tire}}(T)$ induced by the
annular medial vertices. For a graph $H$, write
$\operatorname{Col}_3(H)$ for the set of proper $3$-vertex-colourings
of $H$. Then $A(T)$ is a cycle and
\[
|\operatorname{Col}_3(M_{\mathrm{tire}}(T))|
\;\leq\; |\operatorname{Col}_3(A(T))|.
\]
\end{theorem}
\begin{proof}
Since the tread is a triangulated annulus with no vertices in its
interior, each annular face has exactly one boundary edge, lying either
on $B_{\mathrm{out}}$ or on $B_{\mathrm{in}}$, and exactly two annular
edges. As the annular faces are traversed cyclically around the tread,
consecutive faces share one annular edge. Equivalently, the annular
edges occur in a cyclic order in which each annular face contains two
consecutive annular edges. Hence the subgraph of
$M_{\mathrm{tire}}(T)$ induced by the annular medial vertices is a
cycle.
Consider the restriction map from proper $3$-colourings of
$M_{\mathrm{tire}}(T)$ to colourings of this annular medial cycle
$A(T)$. We claim that this map is injective. Let $x$ be a
non-annular medial vertex. Then $x$ corresponds to an edge of
$B_{\mathrm{out}}$ or $B_{\mathrm{in}}$, since the chords of $O$ were
omitted before forming $M_{\mathrm{tire}}(T)$. This boundary edge is
incident to a unique annular face of $T^{\circ}$, and the other two
edges of that face are annular edges. Therefore $x$ is adjacent in
$M_{\mathrm{tire}}(T)$ to the two annular medial vertices corresponding
to those two annular edges.
Those two annular medial vertices are adjacent to each other, because
their annular edges are consecutive on the same triangular annular
face. In any proper $3$-colouring they therefore receive two distinct
colours, and $x$ is forced to receive the remaining third colour.
Thus every non-annular medial vertex has its colour uniquely determined
by the colouring of $A(T)$. Two colourings of $M_{\mathrm{tire}}(T)$
with the same restriction to $A(T)$ are identical, so the restriction
map is injective. The stated inequality follows.
\end{proof}
\begin{remark} \begin{remark}
\label{rem:tire-counts} \label{rem:tire-counts}
Let $\mu = |V(B_{\mathrm{out}})|$ and $\nu = |V(B_{\mathrm{in}})|$. By Let $\mu = |V(B_{\mathrm{out}})|$ and $\nu = |V(B_{\mathrm{in}})|$. By