ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

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.

Uniform coordinate projections can hide a poor joint projection

Example

Use the four-node rank-one lattice rule in dimension two with $N = 4$ and $z = (1, 1)$ . Both one-coordinate projections are the equally spaced set ${0, 1 / 4, 1 / 2, 3 / 4}$ , but the two-coordinate set lies on the diagonal. It is not a $(0, 2, 2)$ -net in base $2$ , and its star discrepancy is at least $1 / 4$ .

Facts & Assumptions

Given: Dimension $d = 2$ , node count $N = 4$ , and rank-one lattice generator $z = (1, 1)$ .

[F1]

The rank-one lattice rule uses nodes $x_{n, j} = {n z_{j} / N}$ for $n = 0, \dots, N - 1$ , with ${a} = a - ⌊ a ⌋$ . (Digital point-set and rank-one lattice cube quadrature)

[F2]

A $(t, m, d)$ -net in base $b$ has $b^{m}$ points and each elementary interval with exponent sum $m - t$ contains exactly $b^{t}$ points; $D_{N}^{*}$ is the supremum of anchored-box count error. (Star discrepancy, digital nets, and digital sequences)

[F3]

The rank-one input is admissible: $1$ is a common divisor of $1$ and $4$ , and every common divisor $d$ satisfies $d \leq ∣ 1 ∣ = 1$ by If $d ∣ a$ and $a \neq 0$ then $d \neq 0$ and $∣ d ∣ \leq ∣ a ∣$ ; hence the set of divisors of a nonzero integer is bounded above by $∣ a ∣$ ; hence $\gcd (1, 4) = 1$ by Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ .

Proof

1.1F1F3givenalgebra

By [F1] and [F3], the nodes are $P = {(0, 0), (1 / 4, 1 / 4), (1 / 2, 1 / 2), (3 / 4, 3 / 4)}$ . Each coordinate projection has one point in each base- $2$ interval of length $1 / 4$ , so each is a $(0, 2, 1)$ -net.

2.1F2step 1.1

The elementary interval $E = [0, 1 / 2) \times [0, 1 / 2)$ has volume $1 / 4$ and contains $(0, 0)$ and $(1 / 4, 1 / 4)$ , hence two points. A $(0, 2, 2)$ -net would require exactly $2^{0} = 1$ point in every such interval, so $P$ is not one.

3.1F2step 1.1algebra∎

At $u = (1 / 2, 1 / 2)$ , the anchored box contains $2$ of the $4$ nodes, so its empirical proportion is $1 / 2$ while its volume is $1 / 4$ . Thus $D_{4}^{*} (P) \geq ∣ 1 / 2 - 1 / 4 ∣ = 1 / 4$ .

Depends on

Used by

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Dependency tree · two levels

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Sources

Art B. Owen, Monte Carlo Book: The Quasi-Monte Carlo Parts, Chapter 15 §15.7 (standard reference, not scraped)
Harald Niederreiter, Random Number Generation and Quasi-Monte Carlo Methods, Chapter 4 §4.1 (standard reference, not scraped)