TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Exhaustive empirical search attains a global quantizer

Statement

For every finite indexed data set $x_{1}, \dots, x_{N} \in R^{d}$ with $N, d, k \geq 1$ and positive normalized weights, the algorithm of Exhaustive global empirical squared-error quantizer terminates after checking exactly $k^{N}$ assignments and returns a nearest-center quantizer whose empirical squared-error distortion is no greater than that of any $k$ -center tuple. Its exact arithmetic operation count is $O (k^{N} (N + k) d)$ ; this is an exponential exhaustive procedure, not a polynomial-time claim.

Facts & Assumptions

Given: $N, d, k \geq 1$ , points $x_{i} \in R^{d}$ , weights $w_{i} > 0$ summing to one, and exact arithmetic and comparisons.

[F1]

Exhaustive global empirical squared-error quantizer: the procedure checks every ordered assignment, uses a weighted centroid for each nonempty cluster and $x_{1}$ for each empty cluster, retains a minimum-cost assignment, then reassigns points to least-index nearest centers.

[F2]

The weighted centroid minimizes finite-cluster squared error: replacing any cluster center by its weighted centroid cannot increase its assigned squared-error cost.

[F3]

Put $X_{N} : = {x_{i} : 1 \leq i \leq N}$ , a finite Borel subset of $R^{d}$ , and let $μ_{N} : = \sum_{i = 1}^{N} w_{i} δ_{x_{i}}$ on $X_{N}$ , where $δ_{x_{i}}$ is unit point mass at $x_{i}$ . This is a probability measure on the relative Borel sigma algebra of $X_{N}$ with finite second moment $\int_{X_{N}} ∥ x ∥_{2}^{2} d μ_{N} (x) = \sum_{i = 1}^{N} w_{i} ∥ x_{i} ∥_{2}^{2} < \infty$ . For an ordered center tuple $c$ , let $Q_{c} : X_{N} \to {1, \dots, k}$ be its least-index nearest-center rule. Each fiber is Borel because $X_{N}$ is finite, so Finite measurable squared-error quantizer and nearest-center rule gives

$D_{μ_{N}} (Q_{c}; c_{1}, \dots, c_{k}) = \sum_{i = 1}^{N} w_{i} ∥ x_{i} - c_{Q_{c} (x_{i})} ∥_{2}^{2} = \sum_{i = 1}^{N} w_{i} \min_{1 \leq j \leq k} ∥ x_{i} - c_{j} ∥_{2}^{2} = : J (c) .$

Proof

1.1givenF1construct

The assignment set ${1, \dots, k}^{N}$ has exactly $k^{N}$ elements, so the finite ordered loop in [F1] terminates and retains an assignment $a^{*}$ of minimum centroid-adjusted assigned cost $L_{*}$ .

1.2givenF2F3algebra

Fix any center tuple $c$ and assign each point to its least-index nearest center, obtaining $a (c)$ . By [F2], the centroid-adjusted cost of this assignment is at most its cost at $c$ , so $L_{*} \leq L (a (c)) \leq J (c)$ by [F3].

2.1step 1.1step 1.2F1F3algebra

Let $c^{*}$ be the tuple retained for $a^{*}$ . Reassigning $a^{*}$ to nearest centers cannot increase its cost, hence $J (c^{*}) \leq L_{*}$ . Step 1.2 gives $L_{*} \leq J (c)$ for every $c$ , in particular $L_{*} \leq J (c^{*})$ ; therefore $J (c^{*}) = L_{*}$ and $J (c^{*}) \leq J (c)$ for every $k$ -center tuple. The returned tuple is globally minimizing.

3.1givenF1step 1.1algebra∎

Each of the $k^{N}$ assignments can be grouped, centered and costed in $O ((N + k) d)$ exact operations; the final nearest reassignment costs $O (N k d)$ , which is bounded by $O (k^{N} (N + k) d)$ for $N, k \geq 1$ . This proves the stated operation count without a bit-complexity claim.

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Sources

Robert M. Gray, Source Coding Theory, first edition (standard reference, not scraped)
Robert Gallager, 6.450 Principles of Digital Communications I, Chapter 3: Quantization (standard reference, not scraped)
Robert M. Gray and David L. Neuhoff, Quantization, IEEE Transactions on Information Theory 44(6) (1998) (standard reference, not scraped)