Metadata
slug: hausmann-2026-question-5-4-refutation
bibkey: hausmann2026weightedmaxcut
doi: null
url: https://arxiv.org/html/2611.05178v1
triage: theorem
motivation_gids:
- D5/S3/Combinatorics/Graph/ComponentPacking/HausmannQuestion54Refutation.result
A counterexample to Hausmann Question 5.4
Problem
Will Hausmann, Weighted Max Cut in Subcubic Graphs, arXiv:2611.05178v1, Question 5.4, asks:
Does every subcubic graph of odd girth exactly five admit a partition such that is independent and every component of has order at most three?
The quotation uses the primary HTML’s words and mathematical alttext, with spacing normalized. Section 1 fixes finite simple graphs and defines subcubic as degree at most three. The full affirmative assertion is: for every finite vertex set and every simple undirected graph on , if every vertex has degree at most three, has a simple cycle of length five, and every odd simple cycle of has length at least five, then there exist such that
Here is the actual induced graph, its components include singleton components, and order means number of vertices. Even cycles are allowed. The quantification has no vertex-count cutoff and imposes no connectedness, planarity, regularity, sparsity, maximality of , edge-weight, or special construction hypothesis. Finite decidable presentations in the Lean claim do not restrict this class.
The answer is no: the explicitly labelled generalized Petersen graph below meets every input condition, and every independent leaves a component of at least four vertices in its complement. The settling statement is the negation of this entire universal assertion.
Motivation
The source proposes a small-component independent-side partition as a sufficient structural route to a weighted maximum-cut bound. An obstruction to that partition distinguishes the structural proposal from the weighted bound it was intended to establish. The counterexample excludes the whole partition assertion on its stated class; it is not a failure of a particular partition algorithm or a restricted positive search.
The source supplies the question and its relation to weighted Max Cut. The explicit counterexample and residual obstruction are repository-derived results, not results attributed to Hausmann’s paper.
Gap
Hausmann’s Lemma 4.7 states that triangle-free subcubic graphs admit an independent side with residual components of order at most six. It does not supply order three. Theorem 2.2 assumes a component-union bipartiteness structure and gives a weighted-cut inequality; it does not construct the partition in Question 5.4. These are literature comparisons, not suppliers on this Lean module’s proof path.
The accepted bounded literature intake also identifies an order-three forest-partition result under an additional maximum-average-degree bound in Cranston–Yancey, arXiv:2006.11445v2. That extra hypothesis is absent from Question 5.4. This comparison is attributed to the accepted intake; it is not a fresh reading or a compiled dependency of this module.
The bounded Lean-library comparison found repository
Brooks/Subcubic.brooks_cubic and its private induced-colouring helper,
which supply three-colourability, and the
ComponentPacking/AhanjidehEkimYildizZ matching/extremal constructions.
Neither supplies the required partition or its refutation. The inspected
third-party Vilin97/lean-pool revision
91c154506e3d08a1a25e4966c22c99212bf9df54 likewise supplies three-colourability.
The proof directly reuses Mathlib’s three-clique/three-cycle equivalence,
cycle-length lower bound, adjacency-to-component membership, and
Nat.card_le_card_of_injective; these interfaces support the new finite
obstruction rather than already proving it. This is a bounded supplier
comparison, not an exhaustive ecosystem or novelty claim.
The missing target fact is settled negatively by the connected-four-vertex obstruction for one legal graph. The inspected source and accepted bounded prior conclusions establish no claim of global openness, novelty, priority, or exhaustive third-party-library clearance.
Route
The labelled graph
For modulo seven, let and . Use exactly the edges
Equivalently, on vertices , the unordered edge set is
The explicit edge set fixes the labels; no graph-encoding relabelling is needed. There are fourteen vertices and twenty-one edges.
Legal source input
Both layers are seven-cycles: adding two modulo seven traverses the entire inner layer. The spokes form a matching between the layers. Each vertex has two distinct layer neighbours and one spoke neighbour, so the graph is simple and cubic, hence subcubic.
A triangle cannot lie in either seven-cycle. A triangle meeting both layers
would use two spokes, which is impossible because distinct spokes have no
common endpoint. Thus there is no triangle. The sequence
is a simple five-cycle. Simplicity and triangle-freeness exclude
shorter odd simple cycles, so the odd girth is exactly five. The private
Lean helper legal discharges the graph’s actual degree and odd-girth
premises at the call to the universal claim.
A connected residual set for every independent side
For every independent , the private theorem
four_obstruction supplies four distinct vertices with
Its exhaustive membership split uses independence to exclude neighbours of vertices placed in . Every completed branch supplies the four outside vertices, their distinctness, and the displayed attachment edges. The argument covers arbitrary independent sets, including empty and nonmaximal sets; it has no cardinality restriction on . The attachment edges connect the four vertices. They need not form an entire component, and no assertion that their induced graph is exactly a tree is needed.
If , all four vertices belong to . Their attachment edges lie
in and put them in the same actual connected component. Distinctness
gives an injection from a four-element set into that component’s vertex
support. Its order is therefore at least four, contradicting the required
bound three. This is the private theorem residual_large, which does not
need the disjointness assumption. The public result instantiates the
universal claim on this legal graph and applies that obstruction to the
asserted partition. A single legal negative instance refutes the universal
statement over all finite graphs.
Falsifier
A defect in this refutation would require a failure of source correspondence, a missing source input premise for the specified graph, an uncovered independent set in the four-vertex obstruction, or an invalid passage from the attachment edges to the order of an actual induced component. An independent in this exact graph whose complementary components all have order at most three would directly contradict the stated obstruction. The accepted mathematical evidence settles these checks for this graph. A different graph’s successful partition would not refute this counterexample.
Evidence
Exact formal result and scope
The formal module
D5/S3/Combinatorics/Graph/ComponentPacking/HausmannQuestion54Refutation.lean
defines claim : Prop and proves
D5/S3/Combinatorics/Graph/ComponentPacking/HausmannQuestion54Refutation.result : Not claim.
Its canonical frozen membership and module identity are maintained in
the owning state record.
The corresponding Blueprint gives the full claim and proof mechanism.
The proof is self-contained in Lean with Mathlib imports; the axiom
closure is contained in {propext, Classical.choice, Quot.sound}.
The finite evaluator’s output is not an assumption of that proof.
The declaration-level classification is: four_obstruction,
residual_large, and result are content; legal is a consumed
bind-only helper whose consumer is result. Admission is by
escape-witness, with four_obstruction on the live component-injection
path. There are no frozen D5 suppliers. The unique designated public
result is exempt from four-slot Reg registration; the other authored
theorems are private, consumed helpers.
Independent exhaustive computation
The completed independent evaluator reconstructed this same labelled graph and examined all subsets. Exactly subsets were independent, and zero gave a permissible complementary partition. For each independent side, a connected residual set of at least four vertices was checked; a separate union-find evaluation also rejected every complement. These counts concern this single graph, not all graphs of a bounded order or a minimal-counterexample search.
The accepted evaluator identifies the program as
hausmann26496-independent-evaluation-1104m/verify.py in the private
the-omega-institute/trureturing-experiments repository, with SHA-256
cf57e9ab6361dd60489fef9ad2a4f407af94b812d10f0cec8e4a7485194def99.
Its reported command, from the main repository root with CPython 3.13.12
selected on PATH, is
python3 ../trureturing-experiments/hausmann26496-independent-evaluation-1104m/verify.py
The evaluator reports exit code zero. These are attributed completed evaluation results; no rerun is asserted here. The experiment’s immutable repository revision is ASSUMED-UNVERIFIED. The program hash identifies the reported procedure, while the Lean obstruction supplies the deductive refutation independently of its output.
Triage
theorem; a full refutation of the affirmative assertion in Question 5.4.
Issue 26496
registers the exact finite-simple, subcubic, odd-girth-exactly-five target.
The accepted conservative research-line route and escape-witness admission
are retained; no automatic Tier 1/2 or open-problem-resolution admission
exception is asserted.
| Mathematical item | Result and evidence | Boundary |
|---|---|---|
| Legality of the explicit | Proved in the frozen Lean module through the private consumed legal helper: degree at most three and odd girth exactly five. | A legal instance of the source class, without an added source hypothesis. |
| Uniform residual obstruction | Proved in the frozen Lean module by private four_obstruction, consumed by residual_large. | Every independent subset of this fixed labelled graph leaves four distinct connected residual vertices. |
| Failure of every required partition | Proved in the frozen Lean module by residual_large and public result : Not claim. | Full universal Question 5.4 is refuted; the existing result Describe binds this single theorem as Refuted. |
| Exhaustive finite counts | Computed in the accepted independent evaluation: 16,384 subsets, 435 independent sides, zero admissible partitions; program, hash and attributed execution are given in Evidence. | One fixed graph; no rerun asserted; immutable experiment revision ASSUMED-UNVERIFIED. |
| Surviving order-six result | Proved in the cited literature: Hausmann Lemma 4.7, attributed there to Berke–Szabó. | A source-paper result, not an additional theorem or proof dependency of this module. It gives six, not three. |
| Surviving weighted bounds | Proved in the source paper: Theorem 1.4 gives for triangle-free subcubic graphs; Theorem 1.5 gives at odd girth at least seven; Theorem 1.6 gives at odd girth at least nine, for nonnegative edge weights. | Literature results, not additional compiled conclusions here. Refuting the proposed partition does not refute these bounds. |
| Weighted conjecture on the wider triangle-free class | Open in this dossier; no proof or counterexample is supplied. | The refuted sufficient structural route does not decide the weighted inequality. |
| Question 5.3 | Open in this dossier; its exact source condition is bipartiteness of for every residual component and every , with independent in a triangle-free subcubic graph. | Its component-union condition is distinct from the order-three condition; no settlement is claimed. |
| Question 5.5 | Open; the exact external next candidate is separately registered as issue 26528. | Odd girth at least seven and residual component order at most five; no result is transferred from Question 5.4. |
| Size minimality, a classification of obstructions, and global novelty or priority | Unsettled here; novelty and priority are ASSUMED-UNVERIFIED. | The accepted result requires only this explicit counterexample. |
The exact next external source question is Hausmann Question 5.5:
Does every subcubic graph of odd girth at least seven admit a partition such that is independent and every component of has order at most five?
Thus it quantifies over every finite simple graph of maximum degree at most three with no odd simple cycle of length less than seven, and asks for disjoint exhaustive , independent , and at most five vertices in every actual component of . The source’s odd-girth-at-least classes include bipartite graphs. This candidate belongs to 26528 and remains unresolved under the supplied research premise. The present witness has a five-cycle and supplies no counterexample to that different input class.
ASSUMED-UNVERIFIED
The primary HTML at the exact frontmatter URL has SHA-256
49bc3e301c8bea370cc9e9114b211a64ef529a70d268016c6e41d83254abe2d3.
The supplied plaintext has SHA-256
35663c84f57daddde3971e6957bb4da43eb8fdafe066016cf99a1166de52bda0;
it is a transcription aid. Source correspondence uses Section 1,
Theorem 2.2, Lemma 4.7/Section 4.3, and Section 5 Questions 5.3–5.5.
Worldwide literature coverage, ownership clearance, novelty and priority
are ASSUMED-UNVERIFIED beyond these bounded readings and the attributed
library comparison. No such claim is needed for the explicit refutation.
The independent computation is attributed to the completed evaluator; its immutable experiment-repository revision is ASSUMED-UNVERIFIED. Its files remain in the private experiments repository. The Lean proof does not depend on replaying that program. Minimality, classification of all obstructions, the wider weighted conjecture, and Questions 5.3 and 5.5 are not settled by this result.