ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A connected LLE neighbor graph can have more than one fixed vector

Example

Take $x_{1} = x_{3} = - 1$ , $x_{2} = x_{4} = 1$ , and $x_{5} = 0$ in $R$ , with $k = 2$ and deterministic distance ties broken by index. The undirected union neighbor graph is connected. The minimum-norm local weights give

$W = (\begin{array}{ccccc} 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 / 2 & 1 / 2 & 0 & 0 & 0 \end{array}),$

for which $\ker (I_{5} - W)$ has dimension two rather than being $span {1_{5}}$ . Thus graph connectivity alone does not satisfy the simple constant-kernel condition required by LLE.

The diagram shows the undirected union neighbor graph. Its edges record selected neighbor relations, not whether the corresponding directed LLE weight is nonzero.

diagram failed to render:
\begin{tikzpicture}[vertex/.style={draw,circle,minimum size=7mm}]
  \node[vertex] (x1) at (0,1) {$1:-1$};
  \node[vertex] (x3) at (0,-1) {$3:-1$};
  \node[vertex] (x5) at (2,0) {$5:0$};
  \node[vertex] (x2) at (4,1) {$2:1$};
  \node[vertex] (x4) at (4,-1) {$4:1$};
  \draw (x1)--(x3) (x1)--(x5) (x3)--(x5);
  \draw (x2)--(x4) (x2)--(x5) (x4)--(x5);
\end{tikzpicture}

Example references

Hastie, Tibshirani, and Friedman, ESL II, §14.9, printed pp. 573–575

Cameron, Linear and Nonlinear Dimensionality Reduction, §7, PDF pp. 13–16

Facts & Assumptions

Given: The five real points, $k = 2$ , deterministic tie-breaking by sample index, union symmetrization, and the LLE local minimum-norm rule.

[F1]

Neighbor lists sort by distance and then sample index; the union graph uses an edge if either endpoint selects the other (Deterministic finite nearest-neighbor lists and graph symmetrizations).

[F2]

Local weights minimize reconstruction error subject to summing to one, with a minimum-Euclidean-norm tie rule (Minimum-norm affine reconstruction weights for a finite LLE neighborhood).

[F3]

The finite minimum-norm rule defines a unique weight vector even for singular local covariance (Finite LLE affine least-squares weights have a unique minimum-norm minimizer).

[F4]

LLE requires $\ker (I - W) = span {1_{n}}$ before using the isolated constant zero eigenvector (Finite locally linear embedding with minimum-norm weights and a simple constant kernel).

Verification

1.1givenF1algebra

For $x_{1} = x_{3} = - 1$ , the two nearest indices to each are the duplicate at $- 1$ and $x_{5} = 0$ , giving $N_{1} = (3, 5)$ and $N_{3} = (1, 5)$ . For $x_{2} = x_{4} = 1$ , the lists are $N_{2} = (4, 5)$ and $N_{4} = (2, 5)$ . From $x_{5} = 0$ all four distances are $1$ , so $N_{5} = (1, 2)$ by the index tie rule.

2.1step 1.1F2F3algebra

For rows 1 and 3, the exact reconstruction weights are 1 on the duplicate point and 0 on $x_{5}$ ; for rows 2 and 4, they are 1 on the duplicate point and 0 on $x_{5}$ . For row 5, weights $(1 / 2, 1 / 2)$ on points $x_{1} = - 1, x_{2} = 1$ reconstruct zero exactly. Each is the unique minimum-norm choice by [F2] and [F3], yielding the displayed $W$ .

2.2step 1.1F1

The union edges are ${1, 3}$ , ${1, 5}$ , ${3, 5}$ , ${2, 4}$ , ${2, 5}$ , and ${4, 5}$ , so every vertex lies in one connected graph.

3.1step 2.1step 2.2F4algebra∎

The equation $W y = y$ gives $y_{1} = y_{3}$ , $y_{3} = y_{1}$ , $y_{2} = y_{4}$ , $y_{4} = y_{2}$ , and $y_{5} = (y_{1} + y_{2}) / 2$ . The parameters $y_{1}, y_{2}$ are free, so $\dim \ker (I_{5} - W) = 2$ , whereas the constant vectors form a one-dimensional subspace. The condition in [F4] fails despite connectivity, and LLE does not specify an embedding for this input.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

The Elements of Statistical Learning, second edition, corrected 12th printing (standard reference, not scraped)
Maria Cameron, Linear and Nonlinear Dimensionality Reduction, lecture notes (standard reference, not scraped)