Record level-cycle coloring conjectures

papers/coloring_nested_tire_graphs/experiments/check_level_cycle_three_color.py

@@ -0,0 +1,298 @@
+"""Empirical test for the level-cycle three-colour conjecture.
+
+The weakened conjecture says: for every maximal planar graph G, there
+is some level source S and some proper 4-vertex-colouring c such that
+every simple cycle contained in a single level G[L_d] uses at most
+three colours.
+
+Important: this checks the cycle-by-cycle version.  Two cycles in the
+same level or the same inner outerplanar component may omit different
+colours.
+
+Run examples:
+    sage -python experiments/check_level_cycle_three_color.py 4 9
+    sage -python experiments/check_level_cycle_three_color.py 4 8 --quantifier all-sources
+    sage -python experiments/check_level_cycle_three_color.py 4 8 --sources all
+"""
+
+from __future__ import annotations
+
+import argparse
+from collections import deque
+from itertools import combinations
+from typing import Any, Iterable, Iterator, Sequence, cast
+
+from sage.all import Graph, graphs  # type: ignore[attr-defined]  # pylint: disable=no-name-in-module
+from sage.graphs.graph_coloring import all_graph_colorings  # type: ignore[attr-defined]  # pylint: disable=no-name-in-module
+
+
+Coloring = dict[Any, int]
+Source = tuple[Any, ...]
+
+
+def vertex_key(v: Any) -> str:
+    """Stable ordering key for Sage vertex labels."""
+    return repr(v)
+
+
+def is_induced_cycle(g: Graph, vertices: Sequence[Any]) -> bool:
+    """Return True iff G[vertices] is a simple cycle."""
+    if len(vertices) < 3:
+        return False
+    h = cast(Graph, g.subgraph(list(vertices)))
+    return h.is_connected() and h.num_edges() == len(vertices) and all(
+        h.degree(v) == 2 for v in h.vertices()
+    )
+
+
+def induced_cycle_sources(g: Graph, max_size: int | None = None) -> Iterator[Source]:
+    """Yield every vertex set inducing a simple cycle in G."""
+    vertices = sorted(g.vertices(), key=vertex_key)
+    upper = len(vertices) if max_size is None else min(max_size, len(vertices))
+    for k in range(3, upper + 1):
+        for subset in combinations(vertices, k):
+            if is_induced_cycle(g, subset):
+                yield tuple(subset)
+
+
+def level_sources(g: Graph, mode: str, max_cycle_source_size: int | None) -> Iterator[Source]:
+    """Yield level sources according to the requested mode."""
+    if mode in ("vertex", "all"):
+        for v in sorted(g.vertices(), key=vertex_key):
+            yield (v,)
+    if mode in ("cycle", "all"):
+        yield from induced_cycle_sources(g, max_cycle_source_size)
+
+
+def distances_from_source(g: Graph, source: Source) -> dict[Any, int]:
+    """Shortest-path distance from a vertex or cycle source."""
+    if len(source) == 1:
+        return dict(g.shortest_path_lengths(source[0]))
+    distances = {v: 0 for v in source}
+    queue: deque[Any] = deque(source)
+    while queue:
+        v = queue.popleft()
+        for w in g.neighbor_iterator(v):
+            if w in distances:
+                continue
+            distances[w] = distances[v] + 1
+            queue.append(w)
+    return distances
+
+
+def simple_cycle_vertex_sets(g: Graph) -> set[frozenset[Any]]:
+    """Enumerate vertex sets of simple cycles in an undirected graph.
+
+    The DFS only starts a cycle at its least vertex under vertex_key, which
+    avoids most duplicates; the final frozenset de-duplicates cycles with the
+    same vertex set.
+    """
+    vertices = sorted(g.vertices(), key=vertex_key)
+    index = {v: i for i, v in enumerate(vertices)}
+    cycles: set[frozenset[Any]] = set()
+
+    def dfs(start: Any, current: Any, path: list[Any], seen: set[Any]) -> None:
+        for nxt in g.neighbor_iterator(current):
+            if nxt == start:
+                if len(path) >= 3:
+                    cycles.add(frozenset(path))
+                continue
+            if nxt in seen:
+                continue
+            if index[nxt] <= index[start]:
+                continue
+            seen.add(nxt)
+            path.append(nxt)
+            dfs(start, nxt, path, seen)
+            path.pop()
+            seen.remove(nxt)
+
+    for start in vertices:
+        dfs(start, start, [start], {start})
+    return cycles
+
+
+def level_cycle_violation(
+    g: Graph, distances: dict[Any, int], coloring: Coloring
+) -> tuple[int, frozenset[Any], set[int]] | None:
+    """Return the first level cycle using all four colours, if any."""
+    by_level: dict[int, list[Any]] = {}
+    for v, d in distances.items():
+        by_level.setdefault(d, []).append(v)
+
+    for d in sorted(by_level):
+        if len(by_level[d]) < 3:
+            continue
+        h = cast(Graph, g.subgraph(by_level[d]))
+        for cycle in simple_cycle_vertex_sets(h):
+            used = {coloring[v] for v in cycle}
+            if len(used) > 3:
+                return d, cycle, used
+    return None
+
+
+def coloring_witness(
+    g: Graph,
+    source: Source,
+    max_colorings: int | None,
+) -> tuple[Coloring | None, int, bool]:
+    """Find a proper 4-colouring satisfying the conjectured restriction.
+
+    Returns (witness, checked_count, exhausted).  If witness is None and
+    exhausted is False, max_colorings was reached before a decision.
+    """
+    distances = distances_from_source(g, source)
+    checked = 0
+    for raw in all_graph_colorings(g, 4, vertex_color_dict=True):
+        coloring = cast(Coloring, raw)
+        checked += 1
+        if level_cycle_violation(g, distances, coloring) is None:
+            return coloring, checked, True
+        if max_colorings is not None and checked >= max_colorings:
+            return None, checked, False
+    return None, checked, True
+
+
+def source_label(source: Source) -> str:
+    if len(source) == 1:
+        return f"vertex:{source[0]}"
+    return "cycle:{" + ",".join(str(v) for v in source) + "}"
+
+
+def test_graph(
+    g: Graph,
+    sources: Iterable[Source],
+    max_colorings: int | None,
+    stop_first: bool,
+    quantifier: str,
+) -> tuple[bool, bool, int]:
+    """Test a graph over selected sources.
+
+    Returns (passed, complete, sources_checked).
+    """
+    checked_sources = 0
+    complete = True
+    found_any_source = False
+    for source in sources:
+        checked_sources += 1
+        witness, n_checked, exhausted = coloring_witness(g, source, max_colorings)
+        if witness is None:
+            if not exhausted:
+                complete = False
+            status = "FAIL" if exhausted else "UNKNOWN"
+            print(
+                f"    {status}: source={source_label(source)}, "
+                f"colorings_checked={n_checked}"
+            )
+            if exhausted and quantifier == "all-sources":
+                distances = distances_from_source(g, source)
+                first = next(all_graph_colorings(g, 4, vertex_color_dict=True), None)
+                if first is not None:
+                    violation = level_cycle_violation(g, distances, cast(Coloring, first))
+                    print(f"      first_coloring_violation={violation}")
+                return False, complete, checked_sources
+            if stop_first:
+                if quantifier == "all-sources":
+                    return True, False, checked_sources
+                if not exhausted:
+                    return True, False, checked_sources
+        else:
+            found_any_source = True
+            print(
+                f"    pass: source={source_label(source)}, "
+                f"colorings_checked={n_checked}"
+            )
+            if quantifier == "exists-source":
+                return True, complete, checked_sources
+
+    if quantifier == "exists-source" and not found_any_source:
+        return False, complete, checked_sources
+    return True, complete, checked_sources
+
+
+def parse_args() -> argparse.Namespace:
+    parser = argparse.ArgumentParser(description=__doc__)
+    parser.add_argument("n_min", type=int, nargs="?", default=4)
+    parser.add_argument("n_max", type=int, nargs="?", default=9)
+    parser.add_argument(
+        "--sources",
+        choices=("vertex", "cycle", "all"),
+        default="vertex",
+        help=(
+            "which candidate level sources to search; cycle means induced "
+            "cycle sources"
+        ),
+    )
+    parser.add_argument(
+        "--quantifier",
+        choices=("exists-source", "all-sources"),
+        default="exists-source",
+        help=(
+            "exists-source tests the weakened conjecture; all-sources tests "
+            "the stronger earlier version"
+        ),
+    )
+    parser.add_argument(
+        "--max-cycle-source-size",
+        type=int,
+        default=None,
+        help="optional cap on induced cycle source size",
+    )
+    parser.add_argument(
+        "--max-colorings",
+        type=int,
+        default=None,
+        help="optional cap per graph/source; capped searches report UNKNOWN",
+    )
+    parser.add_argument(
+        "--full",
+        action="store_true",
+        help="continue after failures/unknowns instead of stopping at first",
+    )
+    return parser.parse_args()
+
+
+def main() -> int:
+    args = parse_args()
+    stop_first = not args.full
+    total_graphs = 0
+    total_sources = 0
+    unknown = 0
+
+    for n in range(args.n_min, args.n_max + 1):
+        print(f"=== n={n} ===")
+        for idx, g in enumerate(graphs.triangulations(n), start=1):
+            total_graphs += 1
+            source_list = list(
+                level_sources(g, args.sources, args.max_cycle_source_size)
+            )
+            print(f"  graph #{idx}: sources={len(source_list)}")
+            passed, complete, checked_sources = test_graph(
+                g, source_list, args.max_colorings, stop_first, args.quantifier
+            )
+            total_sources += checked_sources
+            if not complete:
+                unknown += 1
+            if not passed:
+                print(
+                    f"COUNTEREXAMPLE candidate: n={n}, graph_index={idx}, "
+                    f"source_mode={args.sources}, quantifier={args.quantifier}"
+                )
+                print(f"  edges={sorted(tuple(sorted(e)) for e in g.edges(labels=False))}")
+                return 1
+            if not complete and stop_first:
+                print(
+                    f"UNKNOWN: n={n}, graph_index={idx}, "
+                    f"source_mode={args.sources}, quantifier={args.quantifier}"
+                )
+                return 2
+
+    print(
+        f"PASS: checked {total_graphs} triangulations and {total_sources} "
+        f"sources; unknown_graphs={unknown}"
+    )
+    return 0 if unknown == 0 else 2
+
+
+if __name__ == "__main__":
+    raise SystemExit(main())

papers/coloring_nested_tire_graphs/paper.aux

@@ -38,12 +38,15 @@
 \newlabel{rem:tree-coloring-factorisation}{{1.24}{15}}
 \@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 }
 \newlabel{fig:tire-tree-decomposition}{{5}{16}}
-\newlabel{def:level-cycle-three-colour-restriction}{{1.25}{16}}
-\newlabel{conj:level-cycle-three-colour}{{1.26}{16}}
-\newlabel{def:seam}{{1.27}{16}}
-\newlabel{def:partial-tire-tree}{{1.28}{17}}
-\newlabel{lem:seam-edge-shared}{{1.29}{17}}
-\newlabel{conj:seam-counterexample}{{1.30}{17}}
+\newlabel{rem:level-cycle-motivation}{{1.25}{16}}
+\newlabel{def:level-cycle-three-colour-restriction}{{1.26}{16}}
+\newlabel{conj:false-universal-level-cycle-three-colour}{{1.27}{17}}
+\newlabel{ex:universal-level-cycle-counterexample}{{1.28}{17}}
+\newlabel{conj:level-cycle-three-colour}{{1.29}{17}}
+\newlabel{def:seam}{{1.30}{17}}
+\newlabel{def:partial-tire-tree}{{1.31}{18}}
+\newlabel{lem:seam-edge-shared}{{1.32}{18}}
+\newlabel{conj:seam-counterexample}{{1.33}{18}}
 \bibcite{tait-original}{1}
 \bibcite{bauerfeld-depth}{2}
 \bibcite{bauerfeld-nested-tire-duals}{3}
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papers/coloring_nested_tire_graphs/paper.tex

@@ -1257,6 +1257,21 @@ This is the structural setup underlying the chain-pigeonhole
 program for tire treads.
 \end{remark}
 
+\begin{remark}[Motivation for level-cycle restrictions]
+\label{rem:level-cycle-motivation}
+The tire-tree decomposition reduces global colouring questions to local
+choices on treads together with compatibility along nested boundary
+cycles.  Without further structure, the number of boundary colour states
+can grow quickly as one descends the tree: each seam or level cycle may
+in principle carry any proper restriction of a $4$-colouring.  The
+following restriction is meant to test whether this state space can be
+compressed.  If level cycles can always be made to omit one colour, then
+each such interface behaves like a three-colour boundary object, while
+still allowing different cycles to omit different colours.  This would
+not by itself solve the gluing problem, but it would give a simpler
+target class of boundary states for arguments about nested tire trees.
+\end{remark}
+
 \begin{definition}[Level-cycle three-colour restriction]
 \label{def:level-cycle-three-colour-restriction}
 Let $G$ be a maximal planar graph, let $S \subseteq V(G)$ be a level
@@ -1276,13 +1291,51 @@ tread, or the same inner outerplanar component are not required to
 omit the same colour.
 \end{definition}
 
-\begin{conjecture}[Level-cycle three-colour conjecture]
-\label{conj:level-cycle-three-colour}
+\begin{conjecture}[False universal-source form]
+\label{conj:false-universal-level-cycle-three-colour}
 Let $G$ be a maximal planar graph and let $S \subseteq V(G)$ be any
 level source.  Then $G$ admits a proper $4$-vertex-colouring with the
 level-cycle three-colour restriction with respect to $S$.
 \end{conjecture}
 
+\begin{example}[Counterexample to Conjecture~\ref{conj:false-universal-level-cycle-three-colour}]
+\label{ex:universal-level-cycle-counterexample}
+Let $G$ be the maximal planar graph on vertex set
+$\{1,2,3,4,5,6,7,8\}$ with edge set
+\[
+\begin{aligned}
+E(G)=\{&
+12,13,14,15,16,17,23,26,27,34,35,36,38,\\
+&45,56,58,67,68\}.
+\end{aligned}
+\]
+Here $ij$ denotes the edge $\{i,j\}$.
+Take the vertex source $S=\{7\}$.  The corresponding levels are
+\[
+L_0=\{7\},\qquad L_1=\{1,2,6\},\qquad
+L_2=\{3,4,5,8\}.
+\]
+Inside $G[L_2]$ the vertices $(3,4,5,8)$ form a simple cycle.  In
+every proper $4$-vertex-colouring of $G$, these four vertices receive
+four distinct colours.  The edges $34$, $45$, $58$, $38$, and $35$
+force all pairs among $\{3,4,5,8\}$ except possibly $\{4,8\}$ to have
+distinct colours.  If $4$ and $8$ had the same colour, then vertex $6$,
+which is adjacent to $3$, $5$, and $8$, would have to use the fourth
+colour; but vertex $1$ is adjacent to $3$, $4$, $5$, and $6$, and
+would then be adjacent to all four colours, impossible in a proper
+$4$-colouring.  Hence $4$ and $8$ also have distinct colours, so the
+level cycle $(3,4,5,8)$ uses all four colours in every proper
+$4$-colouring of $G$.  Therefore no proper $4$-colouring has the
+level-cycle three-colour restriction with respect to $S=\{7\}$.
+\end{example}
+
+\begin{conjecture}[Level-cycle three-colour conjecture]
+\label{conj:level-cycle-three-colour}
+Let $G$ be a maximal planar graph.  Then there exists a level source
+$S \subseteq V(G)$ such that $G$ admits a proper $4$-vertex-colouring
+with the level-cycle three-colour restriction with respect to $S$.
+\end{conjecture}
+
 \begin{definition}[Seam]
 \label{def:seam}
 A \emph{seam} of a maximal planar graph $G$ is a simple cycle