Counterexample · D-UNMERGED-DUPLICATES-FAIL

Splitting duplicates breaks strict geometry

Counterexamplepublication: internal

Meaning

Coincident score rows assigned separately can defeat strict Voronoi closure, explaining the merged-atom hypothesis.

Scientific context

The exchange-implies-Voronoi theorem assumes coincident score rows have been merged into one weighted atom. Without that assumption it is false: identical score rows sitting in different singleton cells can make a partition vacuously stable — there is no move that changes anything — while strict nearest-centre assignment and deterministic label reproduction both fail. The witness is one-dimensional with three singleton cells and two rows equidistant from their competing centres (CE-D-UNMERGED-DUPLICATES-001).

Merging duplicates is therefore part of the contract, not an optimization step. The solver refuses a terminal state whose own Mahalanobis rule would relabel a row for more than the gain tolerance, citing the same fixture, and the same code appears if a compile attempt hits the same condition.

Statement

If coincident positive-weight score atoms may occupy different singleton cells, a positive-definite labeling can be vacuously one-point exchange stable while strict nearest-centroid assignment and deterministic score-only label reproduction fail.

boundary condition for the finite D theorem and compiler

Where it stops

Refuted by

CE-D-UNMERGED-DUPLICATES-001
One-point exchange stability implies strict self-consistent nearest-centroid assignment even when coincident positive-weight score atoms may be assigned to different cells.

Relationships

Open next door

In the library

Enforced by the refusal CE-D-UNMERGED-DUPLICATES-001: the library declines the operation and names the fixture.

Every object and refusal

Local graph

Prerequisites to the left, what it enables or raises to the right, boundary cases below. Open in the map.

The example

CE-D-UNMERGED-DUPLICATES-001 — split duplicate atoms block strict D-Voronoi closure

Status: exact rational boundary counterexample.

Claim falsified:

One-point exchange stability implies strict self-consistent nearest-centroid assignment even when coincident positive-weight score atoms may be assigned to different cells.

Take scalar scores $(1,1,-1)$ , weights $(1/4,1/4,1/2)$ , and put each row in one of $K=3$ singleton cells. Then $I_q=1$ , all cells are nonempty, and no nonempty-preserving relocation exists. The labeling is therefore vacuously exchange stable. The first two centroids coincide, however, so strict assignment fails and no deterministic score-only rule can reproduce the split labels.

Boundary resolution: merge coincident score atoms before optimization, or require labels to be constant on each duplicate class. The audited theorem then forces distinct centroids and strict assignment.

Fixture: CE-D-UNMERGED-DUPLICATES-001.json.

From COUNTEREXAMPLES/README.md

Evidence: COUNTEREXAMPLES/CE-D-UNMERGED-DUPLICATES-001.json. Measured, never theorem authority.