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Add medial tire graph definition and color bound

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didericis
2026-06-08 14:44:35 -04:00
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papers/coloring_nested_tire_graphs/experiments/draw_medial_tire_graph.py
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"""Draw the medial tire graph for the tire example in Figure 2.
The example uses the same random tire parameters as fig_tire_example:
m=6, k=4, one chord of O, seed=3. The medial tire graph is built from
the plane graph obtained by omitting the chord edges of O, and then
deleting medial edges between two outer-boundary edges or two
inner-boundary edges. Its vertices are placed at the midpoints of the
retained tire edges.
"""
import os
import sys
HERE = os.path.dirname(os.path.abspath(__file__))
DUAL_EXP = os.path.abspath(
os.path.join(HERE, '..', '..', 'coloring_nested_tire_dual_graphs', 'experiments')
)
sys.path.insert(0, DUAL_EXP)
import matplotlib.pyplot as plt
import matplotlib.patches as patches
from tire_graph import random_tire, planar_positions
def edge_key(u, v):
return tuple(sorted((u, v)))
def is_inner_chord(edge, m, k):
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 is_outer_boundary_edge(edge, outer):
outer_set = set(outer)
if not set(edge) <= outer_set:
return False
m = len(outer)
idx = {v: i for i, v in enumerate(outer)}
a, b = idx[edge[0]], idx[edge[1]]
return (a - b) % m in (1, m - 1)
def is_inner_boundary_edge(edge, inner):
inner_set = set(inner)
if not set(edge) <= inner_set:
return False
k = len(inner)
idx = {v: i for i, v in enumerate(inner)}
a, b = idx[edge[0]], idx[edge[1]]
return (a - b) % k in (1, k - 1)
def suppress_boundary_medial_edge(e1, e2, outer, inner):
return (
is_outer_boundary_edge(e1, outer) and is_outer_boundary_edge(e2, outer)
) or (
is_inner_boundary_edge(e1, inner) and is_inner_boundary_edge(e2, inner)
)
def face_edges(face):
return [edge_key(face[i], face[(i + 1) % len(face)]) for i in range(len(face))]
def build_medial_tire(tire):
m, k = tire['m'], tire['k']
outer = tire['outer']
inner = tire['inner']
omitted = {edge_key(m + a, m + b) for a, b in tire['inner_chords']}
retained_edges = sorted(
edge_key(u, v)
for u, v in tire['edges']
if edge_key(u, v) not in omitted
)
retained = set(retained_edges)
faces = [tuple(tri) for tri in tire['triangles']]
faces.append(tuple(outer))
faces.append(tuple(reversed(inner)))
medial_edges = set()
for face in faces:
boundary = [e for e in face_edges(face) if e in retained]
for i, e in enumerate(boundary):
nxt = boundary[(i + 1) % len(boundary)]
if suppress_boundary_medial_edge(e, nxt, outer, inner):
continue
medial_edges.add(tuple(sorted((e, nxt))))
return retained_edges, sorted(medial_edges), omitted
def draw_medial_tire_graph(tire, filename):
m, k = tire['m'], tire['k']
outer_set = set(tire['outer'])
inner_set = set(tire['inner'])
pos = planar_positions(tire, R_out=1.0, R_in=0.45)
medial_vertices, medial_edges, omitted = build_medial_tire(tire)
medial_pos = {
e: ((pos[e[0]][0] + pos[e[1]][0]) / 2, (pos[e[0]][1] + pos[e[1]][1]) / 2)
for e in medial_vertices
}
C = {
'outer_cycle': '#1f77b4',
'inner_cycle': '#d62728',
'inner_chord': '#ff7f0e',
'annular': '#c7c7c7',
'medial_edge': '#0f766e',
'medial_vertex': '#134e4a',
}
fig, ax = plt.subplots(figsize=(7.2, 7.2))
for r in (1.04, 0.41):
ax.add_patch(patches.Circle((0, 0), r, fill=False,
edgecolor='lightgray',
linewidth=0.5, linestyle='--'))
for u, v in sorted(edge_key(u, v) for u, v in tire['edges']):
x1, y1 = pos[u]
x2, y2 = pos[v]
e = edge_key(u, v)
if e in omitted:
color, lw, ls, alpha = C['inner_chord'], 1.6, (0, (4, 3)), 0.45
elif u in outer_set and v in outer_set:
color, lw, ls, alpha = C['outer_cycle'], 2.0, '-', 0.35
elif u in inner_set and v in inner_set:
color, lw, ls, alpha = C['inner_cycle'], 2.0, '-', 0.35
else:
color, lw, ls, alpha = C['annular'], 1.0, '-', 0.55
ax.plot([x1, x2], [y1, y2], color=color, linewidth=lw,
linestyle=ls, alpha=alpha, zorder=1)
for e1, e2 in medial_edges:
x1, y1 = medial_pos[e1]
x2, y2 = medial_pos[e2]
ax.plot([x1, x2], [y1, y2], color=C['medial_edge'],
linewidth=1.8, alpha=0.92, zorder=3)
for idx, e in enumerate(medial_vertices):
x, y = medial_pos[e]
ax.scatter([x], [y], s=58, color=C['medial_vertex'],
edgecolors='white', linewidths=0.8, zorder=4)
ax.annotate(f"$m_{{{idx}}}$", (x, y), xytext=(3, 3),
textcoords='offset points', color=C['medial_vertex'],
fontsize=6.5, zorder=5)
for v in tire['outer']:
x, y = pos[v]
ax.scatter([x], [y], s=150, color=C['outer_cycle'],
edgecolors='white', linewidths=0.8, alpha=0.6, zorder=2)
for v in tire['inner']:
x, y = pos[v]
ax.scatter([x], [y], s=135, color=C['inner_cycle'],
edgecolors='white', linewidths=0.8, alpha=0.6, zorder=2)
legend_items = [
plt.Line2D([], [], color=C['medial_edge'], linewidth=1.8,
label=r'medial tire graph edge'),
plt.Line2D([], [], marker='o', color='w',
markerfacecolor=C['medial_vertex'], markeredgecolor='white',
markersize=7, label=r'medial vertex at an edge midpoint'),
plt.Line2D([], [], color=C['inner_chord'], linewidth=1.6,
linestyle=(0, (4, 3)), alpha=0.55,
label=r'chord of $O$ omitted from medial construction'),
plt.Line2D([], [], color=C['annular'], linewidth=1.0,
label=r'underlying tire graph, faint'),
]
ax.legend(handles=legend_items, loc='upper left',
bbox_to_anchor=(1.0, 1.0), fontsize=9, frameon=False)
ax.set_xlim(-1.24, 1.24)
ax.set_ylim(-1.24, 1.24)
ax.set_aspect('equal')
ax.axis('off')
ax.set_title(
r"Medial tire graph $M_{\mathrm{tire}}(T)$ for Figure 2's tire"
"\n"
r"midpoint vertices; no medial edges between consecutive boundary edges",
fontsize=11,
)
fig.savefig(filename, dpi=180, bbox_inches='tight')
plt.close(fig)
print(f"wrote {filename}")
print(f"retained tire edges: {len(medial_vertices)}")
print(f"medial edges: {len(medial_edges)}")
print(f"omitted O chords: {sorted(omitted)}")
def main():
paper_dir = os.path.abspath(os.path.join(HERE, '..'))
tire = random_tire(m=6, k=4, n_chords=1, seed=3)
out = os.path.join(paper_dir, 'fig_medial_tire_example.png')
draw_medial_tire_graph(tire, out)
if __name__ == '__main__':
main()
papers/coloring_nested_tire_graphs/fig_medial_tire_example.png
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papers/coloring_nested_tire_graphs/paper.aux
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@@ -19,60 +19,64 @@
\newlabel{fig:dual-depth}{{1}{3}}
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\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces The medial tire graph for the tire in Figure\nonbreakingspace 2\hbox {}. The chord of $O$ is drawn faintly and omitted before taking the medial graph; medial edges between consecutive outer-boundary edges or consecutive inner-boundary edges are also omitted. Each medial vertex is placed at the midpoint of its corresponding retained tire edge.}}{5}{}\protected@file@percent }
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\citation{bauerfeld-nested-tire-duals}
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\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces Case 2 ($R$ = annulus) with $O$ a barbell. $B_{\mathrm {out}}$ is the outer hexagon (red); $O$ has two triangles $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ joined by the bridge $a_3\text {--}b_1$ (all light red). The annulus is triangulated by $14$ annular triangles: $6$ ``outer-cap'' triangles (one per outer edge), $6$ ``inner-cap'' triangles (one per non-bridge edge of $O$), and $2$ ``bridge-cap'' triangles $\{u_0, a_3, b_1\}$ and $\{u_3, a_3, b_1\}$ adjacent to the bridge. Each blue dot sits at the centroid of an annular triangle; blue edges connect dual vertices whose triangles share an interior annular edge (spoke or bridge). The two bridge-cap vertices have $\Gamma $-degree $3$ (their triangles have no boundary edge) and are joined by the dashed blue \emph {chord} corresponding to the bridge; the remaining $13$ edges form the Hamilton cycle that wraps around the annulus. All $14$ vertices lie on the outer face of the cycle-with-chord embedding, so $\Gamma \cong \Theta (1, 7, 7)$ is outerplanar.}}{11}{}\protected@file@percent }
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\@writefile{lof}{\contentsline {figure}{\numberline {4}{\ignorespaces Case 2 ($R$ = annulus) with $O$ a barbell. $B_{\mathrm {out}}$ is the outer hexagon (red); $O$ has two triangles $\{a_1, a_2, a_3\}$ and $\{b_1, b_2, b_3\}$ joined by the bridge $a_3\text {--}b_1$ (all light red). The annulus is triangulated by $14$ annular triangles: $6$ ``outer-cap'' triangles (one per outer edge), $6$ ``inner-cap'' triangles (one per non-bridge edge of $O$), and $2$ ``bridge-cap'' triangles $\{u_0, a_3, b_1\}$ and $\{u_3, a_3, b_1\}$ adjacent to the bridge. Each blue dot sits at the centroid of an annular triangle; blue edges connect dual vertices whose triangles share an interior annular edge (spoke or bridge). The two bridge-cap vertices have $\Gamma $-degree $3$ (their triangles have no boundary edge) and are joined by the dashed blue \emph {chord} corresponding to the bridge; the remaining $13$ edges form the Hamilton cycle that wraps around the annulus. All $14$ vertices lie on the outer face of the cycle-with-chord embedding, so $\Gamma \cong \Theta (1, 7, 7)$ is outerplanar.}}{10}{}\protected@file@percent }
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\@writefile{lof}{\contentsline {figure}{\numberline {5}{\ignorespaces Tire-tree decomposition (Theorem\nonbreakingspace 1.23\hbox {}) on a $13$-vertex maximal planar example $G$ with five BFS levels. $(a)$ $G$ with vertex source $v_0$ and $\ell _G \in \{0,1,2,3,4\}$; four nested seams are highlighted, $C_{T_R} = \{a,b,c\}$ (orange), $C_{T_L} = \{a,c,d\}$ (red, including the chord $a$-$c$ shared with $C_{T_R}$), $C_{T_{LL}} = \{f_1, f_2, f_3\}$ (purple), $C_{T_{LLL}} = \{g_1, g_2, g_3\}$ (teal). Inset: the rooted tree of tire treads $\mathcal {T}(G, \{v_0\})$ branches at $T_0$ into the leaf $T_R$ (containing $e$) and a chain $T_L \to T_{LL} \to T_{LLL}$ (the highlighted sub-tree). $(b)$ The disk $G_{T_L}$ inside the seam $C_{T_L}$, drawn standalone with $C_{T_L}$ as cycle source and vertex labels rotated to match the new (cycle-source) role of the boundary triangle. $\ell _{G_{T_L}}(\cdot ) = \ell _G(\cdot ) - 1$ on $V(G_{T_L})$ (verified by the generator script), and $\mathcal {T}(G_{T_L}, C_{T_L})$ is the chain $T_L \to T_{LL} \to T_{LLL}$, iso to the highlighted sub-tree of $(a)$.}}{16}{}\protected@file@percent }
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\@writefile{lof}{\contentsline {figure}{\numberline {6}{\ignorespaces Tire-tree decomposition (Theorem\nonbreakingspace 1.25\hbox {}) on a $13$-vertex maximal planar example $G$ with five BFS levels. $(a)$ $G$ with vertex source $v_0$ and $\ell _G \in \{0,1,2,3,4\}$; four nested seams are highlighted, $C_{T_R} = \{a,b,c\}$ (orange), $C_{T_L} = \{a,c,d\}$ (red, including the chord $a$-$c$ shared with $C_{T_R}$), $C_{T_{LL}} = \{f_1, f_2, f_3\}$ (purple), $C_{T_{LLL}} = \{g_1, g_2, g_3\}$ (teal). Inset: the rooted tree of tire treads $\mathcal {T}(G, \{v_0\})$ branches at $T_0$ into the leaf $T_R$ (containing $e$) and a chain $T_L \to T_{LL} \to T_{LLL}$ (the highlighted sub-tree). $(b)$ The disk $G_{T_L}$ inside the seam $C_{T_L}$, drawn standalone with $C_{T_L}$ as cycle source and vertex labels rotated to match the new (cycle-source) role of the boundary triangle. $\ell _{G_{T_L}}(\cdot ) = \ell _G(\cdot ) - 1$ on $V(G_{T_L})$ (verified by the generator script), and $\mathcal {T}(G_{T_L}, C_{T_L})$ is the chain $T_L \to T_{LL} \to T_{LLL}$, iso to the highlighted sub-tree of $(a)$.}}{18}{}\protected@file@percent }
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\@writefile{lot}{\contentsline {table}{\numberline {1}{\ignorespaces Exhaustive vertex-source search for the level-cycle three-colour conjecture on all triangulation isomorphism classes with $4 \leq n \leq 13$. Every triangulation in this range admits at least one vertex source witnessing the conjecture.}}{21}{}\protected@file@percent }
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\newlabel{conj:seam-counterexample}{{1.38}{22}}
\bibcite{tait-original}{1}
\bibcite{bauerfeld-depth}{2}
\bibcite{bauerfeld-nested-tire-duals}{3}
@@ -89,5 +93,5 @@
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\label{fig:tire-example}
\end{figure}
\begin{definition}[Medial tire graph]
\label{def:medial-tire-graph}
Let $T = (B_{\mathrm{out}}, O, E_{\mathrm{ann}})$ be a tire graph.
Let $T^{\circ}$ be the plane graph obtained from $T$ by deleting every
edge of $O$ that is not on the inner boundary walk $B_{\mathrm{in}}$.
Equivalently, $T^{\circ}$ keeps $B_{\mathrm{out}}$, $B_{\mathrm{in}}$,
and the annular edges, but omits the chords of the inner outerplanar
graph $O$.
The \emph{medial tire graph} of $T$, denoted $M_{\mathrm{tire}}(T)$,
is obtained from the medial graph $M(T^{\circ})$ by deleting every
medial edge whose two endpoint-vertices correspond either to two edges
of $B_{\mathrm{out}}$ or to two edges of $B_{\mathrm{in}}$. Thus
$V(M_{\mathrm{tire}}(T))$ is naturally indexed by the non-chord edges
of $T$, and two such vertices are adjacent exactly when the
corresponding edges are consecutive on the boundary of a face of
$T^{\circ}$, except for consecutive pairs lying wholly along one of
the two boundary walks. The medial vertices corresponding to edges of
$E_{\mathrm{ann}}$ are called the \emph{annular medial vertices}.
\end{definition}
\begin{figure}[h]
\centering
\includegraphics[width=0.78\textwidth]{fig_medial_tire_example.png}
\caption{The medial tire graph for the tire in
Figure~\ref{fig:tire-example}. The chord of $O$ is drawn faintly and
omitted before taking the medial graph; medial edges between consecutive
outer-boundary edges or consecutive inner-boundary edges are also
omitted. Each medial vertex is placed at the midpoint of its
corresponding retained tire edge.}
\label{fig:medial-tire-example}
\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}
\label{rem:tire-counts}
Let $\mu = |V(B_{\mathrm{out}})|$ and $\nu = |V(B_{\mathrm{in}})|$. By