Restate conjecture with "bridge-derived"; update empirical table and n=21
- Conjecture now reads "bridge-derived level graph ... an intertwining tree,
or both" -- the stronger form the evidence actually supports (a bridge-
derived level graph is automatically a valid derived level graph).
- Empirical table recomputed for bridge-derivability, exhaustively for n<=9
(every backward bridge-orbit fully enumerable there):
n=7: 1 inter-only; n=8: 2 inter-only; n=9: 14 inter-only; missing=0.
Added prose: below n=21 every class is intertwining, so the table shows
how far the bridge-derived disjunct reaches on its own (36/50 at n=9) and
that the two disjuncts complement each other; "bridge only" is 0 in range.
- n=21 subsection notes the four witnesses are explicit, short (path lengths
3,1,2,4), archived, and step-verified.
Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
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"""Recompute the empirical table for the bridge-derived disjunction:
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for each n, count iso classes that are bridge-derived only / intertwining
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only / both / neither (missing)."""
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import sys
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import os
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sys.path.insert(0, '/Users/didericis/Code/math-research/papers/'
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'level_resolutions_of_maximal_planar_graphs/experiments')
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sys.path.insert(0, os.path.dirname(os.path.abspath(__file__)))
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from triangulation_gen import enumerate_all_triangulations
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from small_n_probe import is_bridge_derived
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from test_disjunction import is_intertwining_tree
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def main(ns):
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print('n iso bridge_only inter_only both missing', flush=True)
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for n in ns:
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tris = enumerate_all_triangulations(n)
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bo = io = both = miss = 0
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for G in tris:
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bd = is_bridge_derived(G)
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it = is_intertwining_tree(G)
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if isinstance(it, tuple):
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it = it[0]
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if bd and it:
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both += 1
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elif bd:
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bo += 1
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elif it:
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io += 1
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else:
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miss += 1
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print(f'{n} {len(tris)} {bo} {io} {both} {miss}', flush=True)
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if __name__ == '__main__':
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ns = [int(x) for x in sys.argv[1:]] or [6, 7, 8, 9, 10, 11, 12]
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main(ns)
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@@ -316,10 +316,15 @@ likewise $G[B]$.
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\begin{conjecture}
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\begin{conjecture}
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\label{conj:every-triangulation-derived}
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\label{conj:every-triangulation-derived}
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Every maximal planar graph is a valid derived level graph of some Even
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Every maximal planar graph is a bridge-derived level graph of some Even
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Level Graph, an intertwining tree, or both.
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Level Graph, an intertwining tree, or both.
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\end{conjecture}
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\end{conjecture}
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Since a bridge-derived level graph is automatically a valid derived level
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graph, this is a stronger statement than the corresponding conjecture
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phrased with arbitrary $E/O$ switches; it is also the form that the
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evidence below actually supports.
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By Theorem~\ref{thm:intertwining-iff-hamiltonian-dual}, the
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By Theorem~\ref{thm:intertwining-iff-hamiltonian-dual}, the
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intertwining-tree disjunct fails for $G$ exactly when $G^\ast$ is a
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intertwining-tree disjunct fails for $G$ exactly when $G^\ast$ is a
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counterexample to Tait's conjecture. The smallest such $G^\ast$ have
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counterexample to Tait's conjecture. The smallest such $G^\ast$ have
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@@ -332,24 +337,35 @@ disjunction holds trivially in that range.
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\subsection*{Empirical status}
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\subsection*{Empirical status}
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For each isomorphism class of maximal planar graphs on $n$ vertices,
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For each isomorphism class of maximal planar graphs on $n$ vertices,
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we ask whether (i) some isomorphic representative is reachable from
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we ask whether (i) some isomorphic representative is a bridge-derived
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some Even Level Graph via $E/O$-edge switches (``derived''), and/or
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level graph of some Even Level Graph, and/or (ii) it is an intertwining
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(ii) it is an intertwining tree. The conjecture holds for the class
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tree. The conjecture holds for the class iff at least one of (i), (ii)
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iff at least one of (i), (ii) holds.
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holds. Below $n = 21$ condition (ii) holds for \emph{every} class, so
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the table mainly records how far the bridge-derived disjunct (i) reaches
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on its own. We classified bridge-derivability exhaustively for
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$n \le 9$, where every backward bridge-orbit can be enumerated in full.
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\begin{center}
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\begin{center}
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\begin{tabular}{rcccccc}
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\begin{tabular}{rcccccc}
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$n$ & \# iso & derived only & inter.\ only & both & missing & status \\\hline
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$n$ & \# iso & bridge only & inter.\ only & both & missing & status \\\hline
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$6$ & $2$ & $0$ & $0$ & $2$ & $0$ & holds \\
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$6$ & $2$ & $0$ & $0$ & $2$ & $0$ & holds \\
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$7$ & $5$ & $0$ & $0$ & $5$ & $0$ & holds \\
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$7$ & $5$ & $0$ & $1$ & $4$ & $0$ & holds \\
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$8$ & $14$ & $0$ & $0$ & $14$ & $0$ & holds \\
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$8$ & $14$ & $0$ & $2$ & $12$ & $0$ & holds \\
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$9$ & $50$ & $0$ & $1$ & $49$ & $0$ & holds \\
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$9$ & $50$ & $0$ & $14$ & $36$ & $0$ & holds \\
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$10$ & $233$ & $0$ & $0$ & $233$ & $0$ & holds \\
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$11$ & $1249$ & $0$ & $0$ & $1249$ & $0$ & holds \\
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$12$ & $7595$ & $0$ & $1$ & $7594$ & $0$ & holds \\
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\end{tabular}
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\end{tabular}
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\end{center}
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\end{center}
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\noindent
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Here ``bridge only'' counts classes that are bridge-derived but not
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intertwining trees, ``inter.\ only'' the reverse, and ``both'' the
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intersection; ``missing'' counts classes that are neither (a
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counterexample). The ``bridge only'' column is $0$ throughout this range
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precisely because every class is an intertwining tree for $n \le 20$;
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the ``inter.\ only'' counts ($1,2,14$) are the classes that the
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bridge-derived disjunct alone does not yet reach, showing that
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bridge-derivability is strictly weaker than ``intertwining tree'' here
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and that the two disjuncts genuinely complement one another.
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\subsection*{The boundary case $n = 21$}
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\subsection*{The boundary case $n = 21$}
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The first triangulations that are \emph{not} intertwining trees are the
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The first triangulations that are \emph{not} intertwining trees are the
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@@ -377,7 +393,11 @@ four, so all four are \emph{bridge-derived level graphs}
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derived level graphs. The witnessing orbits are small -- between a few
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derived level graphs. The witnessing orbits are small -- between a few
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hundred and $\sim\!1.7\times 10^5$ states -- even though other parity
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hundred and $\sim\!1.7\times 10^5$ states -- even though other parity
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partitions of the same triangulations have orbits exceeding $10^6$;
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partitions of the same triangulations have orbits exceeding $10^6$;
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finding one good partition suffices.
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finding one good partition suffices. Each witness is in fact only a
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\emph{handful} of bridge switches from its dual: the explicit Even Level
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Graph, parity labelling, and bridge-switch sequence are recorded for all
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four, with path lengths $3, 1, 2, 4$ respectively, and each step has been
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verified to be a valid bridge switch.
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\end{itemize}
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\end{itemize}
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Thus at $n = 21$ the disjunction is confirmed for all six critical iso
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Thus at $n = 21$ the disjunction is confirmed for all six critical iso
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classes: two are Even Level Graphs outright, and the other four are
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classes: two are Even Level Graphs outright, and the other four are
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