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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.

✓ 11 results · all verified · 2 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 9 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Algorithm Composition, Precision Budgets, and Dequantization

1 · Prerequisites

2 · Summary

This page defines end-to-end channel and output contracts and proves product-state and tensor-channel error bounds, including final-measurement variation for arbitrary reference inputs. The finite hybrid argument proves the composed-channel bound using the current channel-norm interface. The page allocates precision and failure budgets, composes declared cost coordinates, and separates classical output, quantum-state output, input-access costs, and heralded success. Its vector sampling interface uses squared moduli for complex vectors; the ModFKV and Tang matrix algorithms specialize to real matrices. The ModFKV approximation is proved from exponential Gram concentration and a deterministic threshold-band argument. Tang's recommendation sampler uses a computable adaptive row-refinement rule, an explicit finite accepted branch, and a countable probability model for its almost-sure termination claim.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Threshold-band singular-value approximation

Definition

Let A:Rn→Rm have a singular value decomposition A=UΣVT as in Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition, let σ>0, and let 0<η≤1. Write P> for the orthogonal projection onto the span of right singular vectors whose singular values are at least σ(1+η), and E for the span of right singular vectors whose singular values lie in [σ(1−η),σ(1+η)). For any orthogonal projection PE with range contained in E, a threshold-band target is Aσ,η:=A(P>+PE).

The choice inside the threshold band is allowed to vary; all singular directions above the band are retained and all directions below it are discarded. The definition is independent of the choice of orthonormal bases within singular subspaces because it is stated using their subspaces. If a singular value equals the upper endpoint it belongs to P>; equality at the lower endpoint belongs to E. A range contained in ker⁡A contributes zero to Aσ,η. However, an allowed PE can mix nullspace directions with nonzero singular directions when both lie in E, so changing that projection can change the target.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Truncated singular values give the best Frobenius rank-k approximation

Statement

Let A∈Rm×n have singular values σ1≥σ2≥⋯≥0, padded by zeros to min⁡(m,n). For an integer 0≤k≤min⁡(m,n) let Ak be the truncation to the first k singular terms of an SVD of A. Then rank⁡(Ak)≤k, and for every matrix B with rank⁡(B)≤k, ∥A−Ak∥F2=∑j>kσj2≤∥A−B∥F2.

Facts & Assumptions

Given: The finite real matrix and integer k in the Statement.

[F1]

Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: the full SVD supplies orthonormal bases v1,…,vn of Rn and u1,…,um of Rm, with Avj=σjuj and Avj=0 for zero singular values; in particular, the right singular vectors for positive singular values are orthogonal to ker⁡A and can be completed within ker⁡A.

[F4]

The orthogonal projection PWv is the W-component in V=W⊕W⊥: every finite-dimensional subspace has an orthogonal projection onto it.

Proof

1.1givenF1F2F3algebra

By [F1], write A=∑j=1rσjujvjT for its r positive singular values, and set Ak=∑j=1min⁡(k,r)σjujvjT. Its image lies in the span of at most k left singular vectors, so its rank is at most k. The residual is the sum of the positive singular terms with index greater than k; padding the singular-value list by zeros, [F2]–[F3] give ∥A−Ak∥F2=∑j>kσj2.

1.2givenF2F4algebra

Let P be the orthogonal projection onto the row space of B, which exists by [F4]. Its rank is at most k, every row of B lies in ran⁡P, and hence B=BP. Rowwise, A−B=A(I−P)+(AP−B), where the first summand has each row in ran⁡(I−P) and the second has each row in ran⁡P. These subspaces are orthogonal, so summing the rowwise Pythagorean identities gives ∥A−B∥F2≥∥A(I−P)∥F2=∥A∥F2−∥AP∥F2.

1.3givenF1F2algebra

Let v1,…,vr be the right singular vectors for the positive singular values, and complete them to an orthonormal basis v1,…,vn of Rn using vectors in ker⁡A; set σj=0 for j>r. The SVD gives AP=∑j=1rσjuj(Pvj)T, so orthonormality of u1,…,ur and [F2] give ∥AP∥F2=∑j=1nσj2bj, where bj=∥Pvj∥22∈[0,1]. If w1,…,ws is an orthonormal basis of ran⁡P, then Parseval in the basis v1,…,vn gives ∑jbj=∑ℓ=1s∑j∣⟨vj,wℓ⟩∣2=s=rank⁡P≤k. For k=0, every bj=0. For k≥1, put λj=σj2 and use λj≥λk for j≤k and λj≤λk for j>k to obtain ∑j=1kλj−∑j=1nλjbj=∑j=1kλj(1−bj)−∑j=k+1nλjbj≥λk(k−∑j=1nbj)≥0. Thus ∥AP∥F2≤∑j≤kσj2 in either case.

2.1step 1.2step 1.3F3algebra∎

Substitute the bound from step 1.3 into step 1.2 and use ∥A∥F2=∑jσj2 from [F3]. This gives ∥A−B∥F2≥∑j>kσj2=∥A−Ak∥F2, proving the claim.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Tensor-product state errors add under trace distance

Statement

For a positive integer L and density operators ρj,σj on finite-dimensional Hj, j=1,…,L, D(⨂j=1Lρj,⨂j=1Lσj)≤∑j=1LD(ρj,σj). Consequently every finite POVM on the product output changes its outcome law by at most this sum.

Facts & Assumptions

Given: The finite density-operator pairs in the Statement.

[F1]

Density operator: each state is nonnegative of trace one.

[F2]

Trace Distance: D(ρ,σ)=12∥ρ−σ∥1.

[F3]

Trace norm of a tensor product: ∥A⊗B∥1=∥A∥1∥B∥1.

[F4]

Measurement Variation At Most Trace Distance: a finite POVM changes total variation by at most the input trace distance.

[F5]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: every finite-dimensional nonnegative self-adjoint operator has an orthonormal eigenbasis with nonnegative eigenvalues.

[F6]

Tensor products of finitely many linear maps: tensor maps act factorwise on product vectors; Ordered tensor registers, their inner product, and wire permutations supplies the finite product-basis convention.

[F7]

Trace Norm Duality And Norm Properties: the trace norm satisfies the triangle inequality.

[F8]

Trace Norm: the trace norm is the sum of singular values.

[F9]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: the trace is the sum of diagonal entries in any orthonormal basis.

Proof

1.1givenF1F5F6F9algebra

By [F5], write each ρj in an orthonormal eigenbasis with nonnegative eigenvalues. By [F9] their sum is tr⁡ρj=1. The tensor-product basis from [F6] diagonalizes ⨂jρj with eigenvalues all products of those nonnegative eigenvalues. Their sum is ∏jtr⁡ρj=1, so the product is a density operator; the same argument applies to ⨂jσj.

1.2givenconstructalgebra

Add and subtract the products with their first r factors changed, for r=0,…,L, to obtain ⨂j=1Lρj−⨂j=1Lσj=∑r=1L(⨂j<rρj)⊗(ρr−σr)⊗(⨂j>rσj). This is a finite telescoping identity.

2.1step 1.2F1F2F3F5F7F8F9algebra

Apply [F7] to the trace-norm triangle inequality for the finite sum in step 1.2. By [F1], [F5], [F8], and [F9], each unchanged state factor has trace norm one; by [F3], the rth summand has trace norm exactly ∥ρr−σr∥1. Divide by 2 and use [F2] to obtain the asserted sum.

3.1step 1.1step 2.1F4∎

Apply [F4] to the product output states verified in step 1.1 and the declared finite POVM. The resulting outcome total variation is at most their trace distance, hence at most the sum from step 2.1.

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

End-to-end channel and output error contract

Definition

A finite L-stage composition contract consists of nonzero finite-dimensional complex Hilbert spaces H0,…,HL and, for each j=1,…,L, an ideal channel Φj:L(Hj−1)→L(Hj) and an implemented channel Φ~j with the same domain and codomain. Channels are completely positive and trace-preserving on the full operator space, as in Finite-dimensional quantum channel. Products act right to left, so ΦL⋯Φ1 applies stage 1 first, following Denotational semantics of a finite layered circuit.

For channels on the same nonzero finite-dimensional complex Hilbert input and output spaces define the normalized diamond distance d⋄(Ψ,Φ):=12∥Ψ−Φ∥⋄, where ∥⋅∥⋄ is the unhalved supremum of induced trace norms over all finite reference spaces, as in Diamond Norm. For density operators ρ,σ define D(ρ,σ)=12∥ρ−σ∥1, as in Trace Distance. For probability distributions p,q on a common finite output alphabet define dTV(p,q)=12∑a∣p(a)−q(a)∣.

The local channel error at stage j is d⋄(Φ~j,Φj). The final channel error is d⋄(Φ~L⋯Φ~1,ΦL⋯Φ1). A final output-distribution error is total variation distance between the finite laws obtained by applying a declared final POVM to the two output states. A failure probability is the probability of a declared failure event in the finite execution history. These four quantities have different definitions and must be reported separately.

When an input is entangled with a finite reference register, include that reference in the density operator of Density operator and retain it through the channel call; the diamond distance already takes the supremum over such references. Channel, state, and classical output contracts are distinct. Cost coordinates for input description and preprocessing, state preparation, gates, queries including controlled and inverse access, workspace, measurement, postprocessing, and classical output are specified separately by Classical output, quantum-state output, and access costs and Circuit-family and computation cost model.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Hybrid bound for a composed channel circuit

Statement

Let L≥1, let H0,…,HL be nonzero finite-dimensional complex Hilbert spaces, and let Φj,Φ~j:L(Hj−1)→L(Hj) be channels for j=1,…,L. Then d⋄(Φ~L⋯Φ~1,ΦL⋯Φ1)≤∑j=1Ld⋄(Φ~j,Φj), where d⋄ is the normalized distance in End-to-end channel and output error contract. For any finite reference space R, input density operator ρ on H0⊗R, and finite final POVM on HL⊗R, the total variation distance between the two final outcome laws is at most the same sum. The claim is uniform over the reference dimension; no product-input assumption is made.

Facts & Assumptions

Given: The finite channel sequence and the finite input/reference/measurement specified in the Statement.

[F1]

End-to-end channel and output error contract: d⋄(Ψ,Φ)=12∥Ψ−Φ∥⋄, where the diamond norm takes the supremum over every finite reference space.

[F2]

Channel Composition Hybrid Bound: for two matching pairs of channels between nonzero finite-dimensional complex Hilbert spaces, replacing both stages changes the raw diamond norm by at most the sum of the two local raw diamond errors.

[F3]

Finite-dimensional channels compose and tensor: the composition of finite-dimensional channels is a channel.

[F4]

Measurement Variation At Most Trace Distance: the output total variation of a finite POVM is at most the trace distance between its input density operators.

[F5]

Trace Norm Duality And Norm Properties: the trace norm satisfies the triangle inequality.

[F6]

Density operator: a density operator is positive semidefinite and has trace one.

[F7]

Trace Norm: the trace norm is the sum of singular values.

[F8]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: positive semidefinite operators have orthonormal eigenbases with nonnegative eigenvalues; those eigenvalues are their singular values.

[F9]

The basis-independent trace of an endomorphism of a finite-dimensional vector space: the trace is the sum of diagonal entries in any orthonormal basis.

[F10]

For every nonzero finite-dimensional space R, id⁡R is a channel: each block amplification fixes every operator, so it is completely positive, and it preserves trace because it is the identity (Completely positive maps, Finite-dimensional quantum channel).

Proof

1.1F1F5algebra

For any two maps A,B with matching domains, the triangle inequality for the induced trace norm holds at each finite reference space by [F5]. Taking the supremum over reference spaces in [F1] gives ∥A+B∥⋄≤∥A∥⋄+∥B∥⋄.

1.2givenF1F2F3F10algebra

For j=1,…,L, let Pj=Φj−1⋯Φ1 and Qj=Φ~L⋯Φ~j+1. If either product is empty, it is the identity on the corresponding nonzero finite-dimensional space and is a channel by [F10]; otherwise its factors compose to a channel by [F3]. Apply [F2] first to the identical pair Pj,Pj followed by Φ~j,Φj, and then to the pair Φ~j∘Pj,Φj∘Pj followed by the identical pair Qj,Qj. After dividing by 2, the two applications give d⋄(QjΦ~jPj,QjΦjPj)≤d⋄(Φ~j,Φj). The finite telescoping identity Φ~L⋯Φ~1−ΦL⋯Φ1=∑j=1L(QjΦ~jPj−QjΦjPj) follows by changing stages one at a time.

2.1step 1.1step 1.2algebra

Apply the diamond-norm triangle inequality from step 1.1 to the finite sum in step 1.2. Each summand has normalized distance at most its local value, so the final normalized diamond distance is at most ∑jd⋄(Φ~j,Φj).

3.1givenF1F3F6F7F8F9F10step 2.1algebra

The identity id⁡R is a channel by [F10], so tensor-channel closure [F3] makes (Φ~L⋯Φ~1)⊗id⁡R and (ΦL⋯Φ1)⊗id⁡R channels. Their complete positivity at ancilla dimension one sends the nonnegative ρ to nonnegative operators, and trace preservation keeps trace one, so the outputs ρ~L,ρL are density operators by [F6] and the channel definition. By the supremum in [F1], [F7]–[F9] give D(ρ~L,ρL)≤d⋄(Φ~L⋯Φ~1,ΦL⋯Φ1) because ∥ρ∥1=1.

4.1step 2.1step 3.1F4algebra∎

Apply [F4] to ρ~L,ρL and the declared final POVM. Its outcome-law total variation is at most D(ρ~L,ρL), hence at most the sum from step 2.1.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Tensor-product channel error bound

Statement

For a positive integer L and a finite family of matching channel pairs Φj,Ψj:L(Hj)→L(Kj), j=1,…,L, on nonzero finite-dimensional complex Hilbert spaces, d⋄(⨂j=1LΦj,⨂j=1LΨj)≤∑j=1Ld⋄(Φj,Ψj). For every nonzero finite-dimensional reference space R, every density operator ρ on (⨂j=1LHj)⊗R, and every finite POVM on (⨂j=1LKj)⊗R, the total variation distance between the outcome laws after applying (⨂jΦj)⊗id⁡R and (⨂jΨj)⊗id⁡R to ρ is at most the same sum. No product-input assumption is made; ρ may be entangled across the component inputs and reference.

Facts & Assumptions

Given: The finite channel pairs, product spaces, nonzero finite reference space R, input density operator ρ, and final finite POVM specified in the Statement.

[F1]

End-to-end channel and output error contract: d⋄ and D each include a factor 1/2; the raw diamond norm does not.

[F2]

Channel Tensor Error Bound: for two matching channel pairs, the raw diamond distance between tensor products is at most the sum of the raw local diamond errors; the diamond norm is multiplicative under tensor products.

[F3]

Finite-dimensional channels compose and tensor: tensor products and compositions of finite-dimensional channels are channels.

[F4]

Hybrid bound for a composed channel circuit: for matching finite-dimensional channels, every finite reference input and final POVM has outcome total variation at most the normalized diamond distance between the channels.

Proof

1.1basegivenF1algebra

For L=1 the channel inequality is equality by [F1].

1.2ihF1F2F3algebra

Assume the bound for L factors. By [F3], the first L channel factors in each tensor product form channels. Apply [F2] to this prefix channel pair and the final channel pair. Dividing the raw diamond inequality by 2 gives the normalized distance of the full tensor products at most the prefix normalized distance plus d⋄(ΦL+1,ΨL+1). The induction hypothesis bounds the prefix term by the first L summands.

2.1step 1.1step 1.2discharge-induction: step 1.2

Steps 1.1 and 1.2 prove the channel inequality for every finite L by induction.

3.1givenF3F4step 2.1discharge-induction: step 2.1∎

By [F3], the product maps Φ:=⨂jΦj and Ψ:=⨂jΨj are channels. Apply [F4] to the one-stage pair Φ,Ψ, with the declared reference R, input ρ, and final POVM. Its outcome total variation is at most d⋄(Φ,Ψ), which step 2.1 bounds by ∑jd⋄(Φj,Ψj).

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Allocate local error budgets through a pipeline

Definition

Inputs: a finite list of L stage compilers C1,…,CL, each accepting a rational local precision and, when it uses randomized verification, a rational local failure budget; rational targets 0<ε,δ<1; and declared stage input/output interfaces T0,…,TL, including encodings or register spaces. Stage j must accept Tj−1 and return Tj at every admissible budget, so adjacent interfaces match. For a channel pipeline these are the matching spaces and channel interfaces of End-to-end channel and output error contract. Each compiler is assumed to terminate with an implemented stage, its output type, and its actual cost record under Classical output, quantum-state output, and access costs. The sequential pipeline is run under the same declared cost-accounting model.

Output: a pipeline from T0 to TL, together with the budget lists and cost records below. If L=0, a single interface T0 is supplied; return its identity procedure, empty budget lists, no stage records, and an auxiliary record charging any identity-interface and allocator work. If L≥1, set εj=ε/L and δj=δ/L for each j. For j=1,…,L in order, invoke Cj(εj,δj) once when stage j uses randomized verification, and otherwise invoke Cj(εj) once and mark its failure coordinate not applicable. Retain each returned implementation, output type, and actual stage cost record, then compose the stages chronologically. Record shared input/preprocessing, allocation arithmetic, and any interstage wiring, control, postprocessing, or final-output costs in a separate auxiliary record, counting each operation once. Return the composed implementation, local budget lists, per-stage records, and auxiliary record. The correctness theorem gives an additive resource bound from these records under its explicit accounting assumptions; a requested precision alone does not determine an implementation or its cost.

The correctness claim is separate: Allocated local errors meet the global budget proves the global channel and failure bounds and composes the recorded cost coordinates under explicit stage and pipeline accounting assumptions. This algorithm does not create implementations or amplify a compiler's error; each compiler's correctness method and cost remain part of its own guarantee.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Classical output, quantum-state output, and access costs

Definition

An algorithm result contract declares exactly one output type:

  • Classical string: a finite bit string in a stated encoding. Its cost includes every output bit written and any required formatting.
  • Classical sample: an element of a stated finite set, with the output law and the cost of writing the sample encoding.
  • Quantum state: a density operator, as in Density operator, on an explicitly named finite register, produced by a declared state-preparation procedure. This does not mean that its amplitude list or a classical description of the state is output.

The accompanying cost record keeps separate: input encoding length and preprocessing; each data-structure or oracle operation; state preparation and its inverse or controlled forms; ordinary gates, controlled gates, and inverse gates; register width and clean/dirty/other workspace; measurement and classical postprocessing; finite-gate approximation precision; and classical output bits. A coordinate that the model does not use is marked not applicable. No state-preparation oracle, QRAM, controlled access, inverse access, or full classical state description is assigned zero cost without an explicit interface.

For the exact continuous-rotation table model, QC24 State Preparation Cost Model and QC24 State Preparation specify the table input and circuit procedure; QC24 Explicit State-Preparation Cost Guarantee gives the corresponding D=2n gate/ancilla upper bound. That bound uses exact continuous rotations and gives no finite-gate or uniform bit-time claim. For a fixed finite gate set, Solovay–Kitaev Unitary Compiler for SU(2) and Solovay–Kitaev Error and Resource Bound supply a separate approximation and resource contract. The circuit-family coordinates are those of Circuit-family and computation cost model.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Heralded success and conditional verification contract

Definition

A heralded procedure has an explicit finite outcome set containing success and failure, a success probability ps, and a declared output type. For an input density operator ρ as in Density operator, its success branch is the subnormalized state Is(ρ) of the instrument Finite instruments and conditional states, with ps=tr⁡(Is(ρ)). If ps>0, its conditional output state is ρs=Is(ρ)/ps; if ps=0, no normalized conditional state is assigned and the branch is denoted ⊥. The outcome law is specified by the instrument and the POVM conventions of Finite POVMs and their outcome law.

The contract states separately (i) the success probability, (ii) the guarantee conditional on success, (iii) any unconditional guarantee including failure, and (iv) the cost of verification and of outputting or recording the success flag. An acceptance probability from Quantum acceptance probability for a finite promise computation is not itself a conditional-state guarantee. A claim about a quantum state output uses the type and cost distinctions in Classical output, quantum-state output, and access costs.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

ℓ2 sample/query access to vectors and matrices

Definition

For a finite complex vector v=(v1,…,vn)∈Cn, put ∥v∥2=(∑i=1n∣vi∣2)1/2, where ∣⋅∣ is the complex modulus of Real and imaginary parts, complex conjugation, and modulus. Its squared-entry distribution is Dv(i)=∣vi∣2/∥v∥22 on [n] when ∥v∥2>0. A vector has ℓ2 sample/query access when the interface supplies: (a) an entry query returning vi; (b) its norm ∥v∥2; and (c) a sample distributed according to Dv. For any finite adaptive sequence of sample calls, conditioned on the preceding history and the current requested sampler, each returned index has the declared law and uses fresh randomness independent of that history; in particular, repeated calls to a fixed sampler are mutually independent in the sense of Pairwise and mutual independence of finite-valued random variables. Every draw is charged. The interface states the cost and precision of every query, norm computation, and sample, and any preprocessing or storage cost. If v=0, Dv is undefined and the sampler returns the explicit status ZERO; it does not return a fictitious normalized sample.

For a finite real matrix A∈Rm×n, put ∥A∥F=(∑i=1m∑j=1nAij2)1/2, the Frobenius norm of The Frobenius norm ∥A∥F=(∑i,j∣aij∣2)1/2 on real or complex matrices. Row and column ℓ2 access supplies entry queries, row and column norms, and samplers for each nonzero row and column according to its squared-entry distribution. These real rows and columns are instances of the vector interface above. It also declares preprocessing, storage, and whether their costs are amortized. The common row sampler i∼DA has law P(i)=∥Ai,∗∥22/∥A∥F2 for A≠0, and the column sampler j∼DAT has law P(j)=∥A∗,j∥22/∥A∥F2; zero rows and columns have probability zero. An all-zero matrix returns ZERO for a request for a norm-normalized sample. All row-vector, column-vector, matrix-level row, and matrix-level column samplers obey the same conditional fresh-randomness rule; every draw is charged. Access to entries alone does not imply either sampler. The access and output costs are recorded separately under Classical output, quantum-state output, and access costs, and the distributions are finite probability laws as in Finite probability spaces, outcome weights, events, and event probabilities.

This is an input-data structure assumption. It is not a quantum circuit, state-preparation oracle, QRAM, or an assertion that the preprocessing needed to implement it is sublinear. The Tang matrix algorithms on this page specialize to real matrices; no complex-matrix extension of those algorithms is asserted.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Writing an N-bit output costs at least N bit operations

Statement

In an output-interface model with N designated output positions, where one bit operation can write at most one position and every output position must be written, emitting an N-bit classical string costs at least N bit operations. This lower bound concerns the output interface; it is not a lower bound on preparing a quantum state or on producing a succinct circuit description.

Facts & Assumptions

Given: An integer N≥0 and the output-interface convention in the Statement.

[F1]

Classical output, quantum-state output, and access costs: a classical string output is a finite bit string in a declared encoding, and its cost includes every output bit written.

Proof

1.1givenF1cases

If N=0, the empty string requires no output positions and the lower bound is 0. If N>0, the output has N distinct designated positions by [F1] and the stated interface.

2.1givenstep 1.1algebra∎

Each bit operation writes at most one position, while all N positions must be written. Hence any execution that emits the string uses at least N bit operations.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

ModFKV low-rank description procedure

Definition

Inputs: a nonzero real matrix A∈Rm×n with the exact row/column ℓ2 sample/query interface of ℓ2 sample/query access to vectors and matrices, a threshold 0<σ≤∥A∥F, tolerances 0<ϵ≤1 and 0<η≤1, and an integer sample count q≥1. The procedure returns a compact description; it does not write all mn entries of the output matrix.

  1. Sample i1,…,iq independently from the row distribution DA. Form the conceptual matrix S∈Rq×n whose rth row is Air,∗/qDA(ir). The entries of S can be queried through the stored row indices; no zero-probability row is selected.
  2. Define a column distribution F by choosing r uniformly in [q] and then sampling a column from DSr,∗. Draw j1,…,jq independently from F. Form W∈Rq×q with Wr,c=Sr,jc/qF(jc). A sampled column has F(jc)>0; otherwise return INVALID.
  3. Compute a singular value decomposition of W. Retain the left singular vectors u(1),…,u(k) whose singular values σ^1,…,σ^k are strictly greater than σ, and set U^=[u(1) ⋯ u(k)] and Σ^=diag⁡(σ^1,…,σ^k). If k=0, return the zero matrix description. Otherwise return the row indices and the matrices needed to evaluate D:=ASTU^Σ^−2U^TS.

The sample count q, matrix-entry and norm queries, sample operations, arithmetic, SVD, storage, and any finite-precision implementation are charged in the declared access model. Correctness, success probability, approximation error, and the parameter-dependent choice of q are separate claims of ModFKV low-rank approximation supplier.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Finite rejection sampler for a linear combination

Definition

Inputs: an integer k≥1, vectors v1,…,vk∈Rn with ℓ2 sample/query access, real coefficients w1,…,wk, and a positive integer cap R. Set x=∑j=1kwjvj and S=∑j=1k∥wjvj∥22. Output: an index i∈[n] or FAIL.

If S=0, return FAIL. Otherwise, for each of at most R trials:

  1. Using fresh randomness, choose j with probability ∥wjvj∥22/S, then sample i from Dvj. A zero-weight term is never chosen, so every requested vector distribution is defined.
  2. Query all vh,i and set ai:={(∑h=1kwhvh,i)2k∑h=1k(whvh,i)2,∑h(whvh,i)2>0,0,∑h(whvh,i)2=0. By Cauchy--Schwarz, 0≤ai≤1, so this is a valid acceptance probability.
  3. With a fresh independent acceptance coin, accept index i with probability ai and return i upon acceptance.

If all R trials reject, return FAIL. When x=0 but S>0, every defined acceptance probability is zero and the cap still forces a finite FAIL output. No normalized target distribution is assigned when x=0. All samples, entry queries, random bits, cap, and output writing are charged. Correctness and cost of the linear-combination sampler ↗ proves the target law on the accepted branch and the failure and query bounds under its separate hypotheses on x, the cancellation ratio, and R.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Two-stage ℓ2 sampling controls Gram matrices

Statement

Let M∈Ra×b be nonzero, let q1,…,qa be a probability distribution satisfying qi≥α∥Mi,∗∥22/∥M∥F2 for some 0<α≤1, and let p be a positive integer. Draw p independent row indices I1,…,Ip from this distribution and form N∈Rp×b with row t equal to MIt,∗/pqIt; a zero row with qi=0 is never sampled. Then E∥MTM−NTN∥F2≤∥M∥F4αp, and for every θ>0, P(∥MTM−NTN∥F≥θ∥M∥F2)≤1αpθ2. For column sampling, assume the analogous probability lower bound on columns and apply the result to MT.

In particular, let A∈Rm×n be nonzero with F=∥A∥F, and use p samples at each of two stages. First draw independent Ir with probabilities di=∥Ai,∗∥22/F2 and set Sr,∗=AIr,∗/pdIr. Conditional on S, draw independent Jc with probabilities fj=∥S∗,j∥22/F2 and set W∗,c=S∗,Jc/pfJc. These are the exact normalized two-stage sketches; zero-probability indices are never selected. Then ∥S∥F=∥W∥F=F, both Gram estimators are unbiased (the second conditional on S), and E∥ATA−STS∥F2≤F4/p,E[∥SST−WWT∥F2∣S]≤F4/p. For every θ>0, the probability that either Gram error is at least θF2 is at most 2/(pθ2); the two errors need not be independent.

Facts & Assumptions

Given: The finite matrices, sampling probabilities, positive sample count, and independent samples within each stage in the Statement; second-stage sampling is conditional on the realized first stage.

[F1]

Finite probability spaces, outcome weights, events, and event probabilities: a finite probability law assigns nonnegative weights summing to one; here the assumed row law assigns weight qi to index i.

[F2]

Pairwise and mutual independence of finite-valued random variables: independent sampled indices make their entrywise matrix-valued contributions independent.

[F4]

Markov's inequality on a finite probability space: for nonnegative Y and c>0, P(Y≥c)≤E[Y]/c.

Proof

1.1givenF1algebra

The assumed lower bound on qi forces Mi,∗=0 whenever qi=0. For one sampled row I, define the random matrix X=MI,∗TMI,∗/qI when qI>0, and X=0 on zero-probability zero rows. For each matrix entry (r,s), EXrs=∑i:qi>0qiMirMisqi=∑iMirMis=(MTM)rs. Thus NTN=p−1∑t=1pXt is unbiased.

1.2givenF1F3algebra

The squared Frobenius norm of one contribution is ∥X∥F2=∥MI,∗∥24/qI2. Therefore E∥X∥F2=∑i:qi>0∥Mi,∗∥24qi≤∥M∥F2α∑i∥Mi,∗∥22=∥M∥F4α, using the lower bound on each qi.

2.1step 1.1step 1.2F2algebra

Expand ∥p−1∑t(Xt−EXt)∥F2 entrywise. For distinct s,t, independence [F2] and zero means make each cross expectation zero. Hence its expectation is p−1E∥X−EX∥F2≤p−1E∥X∥F2≤∥M∥F4/(αp) by step 1.2.

3.1step 2.1F3F4algebra

Apply [F4] to the nonnegative random variable Y=∥MTM−NTN∥F2 with threshold θ2∥M∥F4>0. Step 2.1 yields the probability bound. Transposing M interchanges rows and columns, proving the column assertion.

4.1givenF1F3step 1.1step 2.1step 3.1algebra

In the exact two-stage construction every sampled row of S has squared norm F2/p, so ∥S∥F2=F2 and ∑jfj=1. Every sampled column of W likewise has squared norm F2/p, giving ∥W∥F=F. Apply steps 1.1–2.1 with M=A, α=1, and row law d for the first Gram estimator. For each fixed S, apply the transposed result with M=ST, α=1, and row law f for the second estimator. This proves both unbiasedness assertions and both second-moment bounds, with the second conditional on S.

5.1F1step 3.1step 4.1algebra∎

Step 3.1 bounds the first-stage tail by 1/(pθ2) and bounds the second-stage tail by the same number conditional on every S. Averaging the latter bound over the finite first-stage histories gives the same unconditional bound. Pointwise the indicator of either failure is at most the sum of the two failure indicators, so summing their finite probability weights gives 2/(pθ2). No independence between the stages' errors is used.

LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Squared-vector distributions are close in total variation

Statement

Let x,y∈Rn be nonzero, and define Dx(i)=xi2/∥x∥22 and Dy(i)=yi2/∥y∥22. Then dTV(Dx,Dy)≤2∥x−y∥2∥x∥2. The bound is valid even when its right side exceeds 1.

Facts & Assumptions

Given: Nonzero real vectors x,y∈Rn.

[F1]

ℓ2 sample/query access to vectors and matrices: Dx(i)=xi2/∥x∥22 for nonzero x, and likewise for y.

[F2]

End-to-end channel and output error contract: dTV(p,q)=12∑i∣pi−qi∣ for distributions on a common finite alphabet.

Proof

1.1givenF1construct

Put a=∥x∥2>0, b=∥y∥2>0, u=x/a, and v=y/b. Then ∥u∥2=∥v∥2=1 by their definitions.

2.1step 1.1F2F3algebra

Using ∣ui2−vi2∣=∣ui−vi∣∣ui+vi∣ and [F3], dTV(Dx,Dy)=12∑i∣ui−vi∣∣ui+vi∣≤12∥u−v∥2∥u+v∥2≤∥u−v∥2.

3.1step 1.1step 2.1F3algebra∎

Since u−v=(x−y)/a+y(1/a−1/b), [F3] gives ∥u−v∥2≤∥x−y∥2a+b∣1a−1b∣=∥x−y∥2+∣b−a∣a≤2∥x−y∥2a, where ∣b−a∣≤∥x−y∥2 is the reverse triangle inequality. Combining this with step 2.1 proves the result.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Estimate an inner product from ℓ2-sampling access

Statement

Let x∈Rn be nonzero with exact norm, ℓ2 sample access, and entry-query access, and let y∈Rn have entry-query access and known norm. For 0<ϵ≤1, a positive integer m≥16/ϵ2, and a positive integer r with 2−r≤δ<1, there is a finite estimator using the lower median of r block means, with mr samples from x and at most mr entry queries to each of x and y, such that P(∣μ^−⟨x,y⟩∣>ϵ∥x∥2∥y∥2)≤δ. If y=0, return 0 without sampling. All sample and entry-query costs are charged.

Facts & Assumptions

Given: The access interfaces, ϵ,m,r,δ and vectors in the Statement.

[F1]

ℓ2 sample/query access to vectors and matrices: sampling index i from nonzero x gives probability xi2/∥x∥22.

[F2]

Chebyshev's inequality on a finite probability space: for a finite real variable Z and t>0, P(∣Z−EZ∣≥t)≤Var⁡(Z)/t2.

[F3]

Pairwise and mutual independence of finite-valued random variables: independent samples give independent copies of any deterministic function of the sampled index.

[F4]

The finite union bound: the probability of a finite union is at most the sum of the event probabilities.

Proof

1.1givenF1algebracases

If y=0, the prescribed output 0 equals ⟨x,y⟩ and has zero error. Suppose henceforth y≠0. For one sample I∼Dx, query xI and yI and define Z=∥x∥22yI/xI when xI≠0 and Z=0 when xI=0. Since a zero coordinate has sampling probability zero, [F1] gives EZ=∑i:xi≠0xiyi=⟨x,y⟩.

2.1step 1.1F1algebra

Directly from the sampling law [F1], E[Z2]=∥x∥22∑i:xi≠0yi2≤∥x∥22∥y∥22. Hence Var⁡(Z)≤E[Z2]≤∥x∥22∥y∥22.

3.1givenstep 1.1step 2.1F2F3constructalgebra

Partition mr independent samples into r disjoint blocks of size m. Let Mb be the mean of Z in block b and define μ^ to be the ⌈r/2⌉-th order statistic of M1,…,Mr (the lower median). Independence gives each block mean variance at most ∥x∥22∥y∥22/m. Apply [F2] with the positive threshold t=ϵ∥x∥2∥y∥2; each block mean is outside the target interval with probability at most 1/(mϵ2)≤1/16.

4.1step 3.1F3F4algebra∎

If the median μ^ lies outside the target interval, at least ⌈r/2⌉ block means lie outside it. For each fixed subset of that size the probability all its blocks fail is at most (1/16)⌈r/2⌉ by [F3]; there are at most 2r such subsets. By [F4], the failure probability is at most 2r(1/16)⌈r/2⌉≤2−r≤δ.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passOpen item page →

Adaptive conditional failure budgets compose by a union bound

Statement

Consider a procedure with finitely many stages j=1,…,L and a finite probability space of execution histories. Let δj≥0 for every stage, and let Fj be the event that stage j is the first failing stage. If, after every successful history of positive probability that reaches stage j, the conditional probability of failure at stage j is at most δj, then P(some stage fails)≤∑j=1Lδj. No independence hypothesis is required. If r fresh trials are mutually independent and each succeeds with probability at least p∈[0,1], their miss probability is at most (1−p)r≤e−pr. For a common success probability p>0, the expected number of trials capped at r is [1−(1−p)r]/p, which tends to 1/p as r→∞.

Facts & Assumptions

Given: The finite staged procedure and its conditional failure bounds.

[F1]

Finite probability spaces, outcome weights, events, and event probabilities: event probabilities are sums of weights in a finite probability space.

[F2]

Normalization, nonnegativity, monotonicity, complements, and differences in a finite probability space: probabilities are nonnegative, and probabilities of disjoint events add.

[F3]

Pairwise and mutual independence of finite-valued random variables: events determined by distinct members of a mutually independent family factor.

[F4]

1+x≤exp⁡(x) for every real x, hence (1−p)m≤exp⁡(−mp): for 0≤p≤1 and natural r, (1−p)r≤e−pr.

Proof

1.1givenF1construct

The stage-failure events are part of the given execution model. The events F1,…,FL are pairwise disjoint: one execution cannot have two different first failing stages. Their union is exactly the event that some stage fails, because every failing finite history has a least index at which failure first occurs.

1.2givenF3F4algebra

For r mutually independent trials with failure events Ej, [F3] gives P(⋂j=1rEj)=∏j=1rP(Ej)≤(1−p)r. Apply [F4] to obtain the stated exponential bound, including p=0 and p=1.

2.1step 1.1F1F2algebra

For fixed j, partition Fj by the finite successful histories h that reach stage j. For each such history, the hypothesis gives P(Fj∣h)≤δj; histories of probability zero contribute zero. Summing the conditional probabilities weighted by P(h) gives P(Fj)≤δjP(stage j is reached successfully)≤δj. By [F2] and step 1.1, P(⋃jFj)=∑jP(Fj)≤∑jδj.

3.1givenF1step 1.2algebra∎

Let Tr be the number of trials used by the capped repeat-until-success procedure, with Tr=r if all r trials fail. For 1≤j≤r, {Tr≥j} is the event that the first j−1 trials fail, so P(Tr≥j)=(1−p)j−1. For a finite nonnegative integer-valued variable, E[Tr]=∑j=1rP(Tr≥j) by summing its finite outcome weights; hence for p>0 the finite geometric sum is [1−(1−p)r]/p. By step 1.2, (1−p)r→0 as r→∞, so this capped expectation tends to 1/p.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Allocated local errors meet the global budget

Statement

Suppose L≥1 stage compilers called by Allocate local error budgets through a pipeline return implemented channels with local normalized diamond errors at most ε/L and, for every successful history reaching stage j, conditional failure probability at most δ/L. Then the final channel error is at most ε and the probability of any stage failure is at most δ. Use the finite execution-history model of End-to-end channel and output error contract. No independence between stage failures is assumed.

For resource composition, let cj be each stage's actual cost record and let caux record, exactly once, shared input/preprocessing, allocator arithmetic, and every interstage wiring/control, postprocessing, and final-output operation not already included in a cj. For each additive count, length, or sequential-time coordinate r of Classical output, quantum-state output, and access costs and Circuit-family and computation cost model—including input/circuit-description length, preprocessing and sequential compiler time, each named data-structure or oracle operation, state-preparation and its controlled/inverse calls, gate occurrences by type, measurements, classical predicate/postprocessing, and output-bit writes—assume the pipeline performs precisely the operations in these records and has no unrecorded work. Then its total is ctotal,r=caux,r+∑j=1Lcj,r. If any records are upper bounds rather than exact counts, replace the equality by the corresponding upper bound. For serial circuit layers, write the actual gate word as B0C1B1⋯CLBL, with Cj the complete stage-j word and B0,…,BL its chronological auxiliary blocks, allowing empty blocks. Put dj=d(Cj) and daux:=∑ℓ=0Ld(Bℓ), retaining each auxiliary block at its position rather than minimizing the depth after deleting stages. Then dtotal≤daux+∑jdj. For each wire/workspace category s (total wires, clean ancillas, dirty ancillas, or other workspace), let wj,s be the stage allocation and waux,s the shared/interface allocation. Identify these allocations with physical wire sets covering all wires of the total circuit, with each category inherited from the total circuit's declared classification. For clean or dirty ancilla counts, require the declared data/ancilla partitions in the stage and auxiliary records to be restrictions of the corresponding total partition under these wire identifications, and require all corresponding exact return contracts; apply this condition separately to each ancilla coordinate. A coordinate without these compatible supplied contracts is not covered by the bound. If stage workspaces are allocated disjointly, then wtotal,s≤waux,s+∑j=1Lwj,s. The local precision entries εj and error/failure probabilities remain separate fields; they are not resource counts to add.

Facts & Assumptions

Given: The compilers and local guarantees in the Statement, with rational 0<ε,δ<1.

[F1]

Allocate local error budgets through a pipeline: for L≥1 it assigns εj=ε/L and δj=δ/L to every stage and calls each compiler once.

[F2]

Hybrid bound for a composed channel circuit: the final normalized channel error is at most the sum of local normalized channel errors.

[F3]

Adaptive conditional failure budgets compose by a union bound: conditional failure bounds δj after every successful history imply total failure at most ∑jδj, without independence.

[F4]

Classical output, quantum-state output, and access costs and Circuit-family and computation cost model: cost records separate input/preprocessing, named data access, state preparation, gate types, width/workspace, measurement, classical processing, output writing, and finite-gate precision; unprovided access or connecting operations are not free.

Proof

1.1givenF1algebra

By [F1], each stage error is at most ε/L, so the sum of the L local errors is at most L(ε/L)=ε.

1.2givenF1F3algebra

By [F1], each stage has conditional failure bound δ/L. Apply [F3] to the finite execution history to obtain total failure at most L(δ/L)=δ.

2.1step 1.1F2

Apply [F2] to those L stage channels and use step 1.1; the final normalized diamond error is at most ε.

3.1givenF4algebra∎

For each count-valued additive coordinate r, the stated pipeline accounting partitions its operations into the auxiliary operations in caux and the stage operations in c1,…,cL, each counted once. Counting this disjoint union gives ctotal,r=caux,r+∑jcj,r; replacing exact stage counts by upper bounds gives the corresponding inequality. Choose a minimum layering separately for each finite word B0,C1,B1,…,CL,BL, using no layers for an empty word. Concatenate these layerings in that order. Within each block every shared-wire precedence is respected, and between blocks every earlier gate is scheduled first. This gives an admissible layering of the actual full word with ∑ℓd(Bℓ)+∑jdj=daux+∑jdj layers, proving the minimum-depth upper bound. For a wire/workspace category s, let Ts be the total set counted and let As,W1,s,…,WL,s be the sets counted in the auxiliary and stage records. Coverage and the compatible category restrictions give Ts⊆As∪⋃jWj,s. Thus finite cardinality gives ∣Ts∣≤∣As∣+∑j∣Wj,s∣, as asserted. In particular a total ancilla cannot be recorded only as stage data; the clean and dirty bounds each use their own compatible partitions and supplied return contracts. Precision, error, and failure fields remain separate by [F4].

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passOpen item page →

Correctness and cost of the linear-combination sampler

Statement

Run Finite rejection sampler for a linear combination with k≥1, x=∑j=1kwjvj≠0, and positive integer cap R. Put C:=∑j∥wjvj∥22/∥x∥22, and suppose a proof-side bound C0≥C and 0<δ<1 satisfy R≥kC0ln⁡(1/δ). The routine itself receives R but does not test or require a certificate for C0; these are hypotheses for this theorem's guarantee. A proposal is accepted with probability 1/(kC), and conditional on acceptance its index has law Dx. The probability of returning FAIL is at most δ. The worst-case entry-query count is at most kR, with at most R mixture draws and R vector-sample calls in addition. If an entry query, a mixture draw, and a vector-sample call have uniform cost upper bounds Q, Bmix, and Bsamp in the declared model, their total access cost is at most Bsetup+R(kQ+Bmix+Bsamp), where Bsetup charges the initial norm queries and mixture setup. Arithmetic, acceptance randomness, and output writing are charged separately; no unit-cost sampling or bit-time bound is assumed. As R→∞, the capped expected number of trials tends to kC.

Facts & Assumptions

Given: The access and coefficient hypotheses of the Statement, with x≠0.

[F1]

Finite rejection sampler for a linear combination: with S=∑j∥wjvj∥22>0, the proposal chooses term j proportionally to ∥wjvj∥22 and then samples from Dvj; its acceptance probability is the displayed ratio. The routine's finite cap does not depend on detecting the truth of C≤C0.

[F2]

ℓ2 sample/query access to vectors and matrices: each nonzero term sampler has law Dvj with fresh conditional randomness; entry queries, norm queries, and sample calls have separately declared costs.

[F5]

Adaptive conditional failure budgets compose by a union bound: for independent trials of success probability p>0, the expected capped number of trials tends to 1/p as the cap tends to infinity.

Proof

1.1givenF1F2algebra

Put S=∑j∥wjvj∥22>0. The proposal probability of coordinate i is P(i)=∑j(wjvj,i)2/S, because choosing j has weight ∥wjvj∥22/S and the conditional coordinate law is vj,i2/∥vj∥22. Its target probability is Q(i)=xi2/∥x∥22.

2.1step 1.1F3algebra

When ∑j(wjvj,i)2>0, [F3] shows the acceptance ratio Q(i)/(kCP(i))=xi2/[k∑j(wjvj,i)2] lies in [0,1]. If the denominator is zero, then every wjvj,i=0, so xi=0 and the prescribed acceptance is 0.

3.1step 1.1step 2.1F3algebra

Multiplying proposal and acceptance probabilities gives P(i)ai=Q(i)/(kC). Summing over i shows total acceptance probability is 1/(kC), and conditioning on acceptance yields Q(i). Also kC≥1 by [F3] applied to each coordinate and summed.

4.1givenF1F2F4F5algebra∎

Independent trials miss with probability (1−1/(kC))R. Since C≤C0, this is at most e−R/(kC0)≤δ by [F4] and the cap hypothesis. Each executed trial queries the k entries at its sampled coordinate, makes one mixture draw, and calls one vector sampler. At most R trials therefore use at most kR entry queries and R calls of each other type. Initial norms and mixture setup contribute Bsetup; multiplying the call counts by the declared uniform upper bounds gives Bsetup+R(kQ+Bmix+Bsamp). This count excludes arithmetic, acceptance randomness, and output writing, which retain their separate charges. With per-trial success probability 1/(kC), [F5] shows the capped expected trial count tends to kC as R→∞.

LemmaStatement: Literature-sourcedProof: AI-generatedOpen item page →

Exponential Frobenius concentration for normalized Gram sampling

Statement

Let A≠0, F=∥A∥F, and let S,W be its exact normalized two-stage sketches in Two-stage ℓ2 sampling controls Gram matrices, using q≥1 samples at each stage. For 0<δ<1 and t>0, if q≥t−2(1+2log⁡(2/δ))2, then, with probability at least 1−δ, both ∥ATA−STS∥F≤tF2,∥SST−WWT∥F≤tF2. Deterministically ∥S∥F=∥W∥F=F. The confidence bound has no matrix-dimension factor.

Facts & Assumptions

Given: The finite sampling laws, exact sketch normalization, and parameters in the Statement.

[F1]

Two-stage ℓ2 sampling controls Gram matrices proves the normalized Gram estimator is unbiased and has squared Frobenius mean error at most F4/q, at both sampling stages, with the second statement conditional on S.

Proof

1.1givenalgebra

We first prove a finite bounded-differences estimate. If a centered finite real variable Y lies in an interval of length c, let h(s)=log⁡EesY. Differentiating finite sums shows h′′(s) is the variance under the weights proportional to esY. Any variable in [a,b] has variance at most E[(Y−(a+b)/2)2]≤(b−a)2/4. Thus h′′(s)≤c2/4; since h(0)=h′(0)=0, integration gives EesY≤es2c2/8 for all real s.

2.1step 1.1F2algebraconstruct

Suppose a real function f of independent finite samples changes by at most cj if sample j is replaced. Reveal samples in order and let Mj be the finite conditional mean of f. For each fixed earlier history, the possible values of Mj lie in an interval of length at most cj, since averaging the future preserves the coordinate oscillation bound. Therefore Mj−Mj−1 is conditionally centered in an interval of that length. Applying step 1.1 conditionally and summing over successive histories gives Ees(f−Ef)≤es2∑jcj2/8. Markov applied to this exponential, followed by s=4u/∑jcj2, gives P(f−Ef≥u)≤e−2u2/∑jcj2 when u>0 and the denominator is positive. If all cj=0, f is constant on the product support and its upper tail is zero.

3.1givenF1F2step 2.1algebra

For a finite random vector X with ∥X∥2=L identically, set Z=q−1∑jXjXjT and H=EXXT. Independence cancels the mixed centered terms, and ∥XXT∥F2=L2, so E∥Z−H∥F2≤L2/q. By [F2], E∥Z−H∥F≤L/q. Replacing one Xj changes ∥Z−H∥F by at most 2L/q, by the triangle inequality. Step 2.1 now gives P(∥Z−H∥F>L(1+2log⁡(2/δ))/q)≤δ/2.

4.1givenF1step 3.1algebra∎

The row sampling is step 3.1 with X=Ai,∗T/DA(i) and L=F2. Every row of S has squared norm F2/q, so ∥S∥F=F. Its column mixture probability is ∥S∗,j∥2/F2; hence every normalized sampled column of W has squared norm F2/q, and ∥W∥F=F. Conditional on every realized S, step 3.1 applies to the column sampling with the same L=F2 and failure bound δ/2. Averaging that bound over the finite row histories and adding the first-stage failure bound proves the assertion. Zero-probability rows and columns are never selected and require no division by zero.

LemmaStatement: Literature-sourcedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

A deterministic threshold-band bound for two Gram sketches

Statement

Let A∈Rm×n be nonzero, F=∥A∥F, 0<σ≤F, K=F2/σ2, and 0<η≤1. Suppose S∈Rq×n and W∈Rq×q satisfy ∥W∥F=F and ∥ATA−STS∥F≤tF2,∥SST−WWT∥F≤tF2,0<tK≤12. Let U be the orthonormal left singular vectors of W with singular values strictly greater than σ, and let Σ contain those singular values. Write k for their number, V=STUΣ−1, and M=VVT, using empty matrices when k=0. Then k<K, the eigenvalues of VTV lie in [1−tK,1+tK], and some threshold-band target of Threshold-band singular-value approximation obeys ∥AM−Aσ,η∥F≤F(tK+77tK/η).

Facts & Assumptions

Given: The matrices and bounds in the Statement. Matrices are real and all projections below are orthogonal.

[F2]

Cauchy-Schwarz ∣⟨x,y⟩∣≤∥x∥2∥y∥2 with its equality case, the triangle inequality for ∥⋅∥2, the parallelogram law and polarisation applies to matrix entries, giving ∣tr⁡(XTY)∣≤∥X∥F∥Y∥F and ∥X∥op≤∥X∥F.

[F3]

A Threshold-band singular-value approximation is A(P1+Q), where P1 projects onto singular values at least σ(1+η) and Q is any projection supported in the band [σ(1−η),σ(1+η)).

Proof

1.1givenF1F2algebracases

Set H=ATA, G=STS, B=SST, C=WWT, and a=σ2. Since tr⁡C=F2 and each retained eigenvalue exceeds a, ka<F2, so k<K. For k>0, T=UTBU≥Σ2−tF2I≥(a−tF2)I>0. Therefore P=STUT−1UTS is the projection onto the range of V. Also VTV−I=Σ−1UT(B−C)UΣ−1 has Frobenius norm at most tK. By the SVD, the nonzero eigenvalues of M−P are those of VTV−I, giving ∥M−P∥F≤tK. For k=0, use P=M=0 and the same bounds with empty matrices.

1.2givenF1F2algebra

We record the spectral comparisons explicitly. For a symmetric J with decreasing eigenvalues, λj(J)=max⁡dim⁡E=jmin⁡x∈E,∥x∥=1xTJx: the span of its first j eigenvectors attains the value, and every j-dimensional E intersects the span of the eigenvectors numbered j onward. Thus ∣λj(J)−λj(J′)∣≤∥J−J′∥op. Moreover ∑j≤kλj(J)=max⁡rank⁡R=ktr⁡(RJ) over projections R: in an eigenbasis the diagonal entries of R are in [0,1] and sum to k, so moving their weight to the first k indices maximizes the sum. By [F2] the top-k sums for J,J′ differ by at most k∥J−J′∥F. Pad matrices by zero blocks when dimensions differ. Applying these comparisons to H,G and B,C, and using [F1] between G,B, gives ∣λj(H)−λj(C)∣≤2tF2 and a top-k sum difference at most 2ktF2.

2.1step 1.1step 1.2F1F2algebracases

If k>0, UTB2U−T2=UTB(I−UUT)BU is positive semidefinite. Hence tr⁡(PG)=tr⁡[(UTB2U)T−1]≥tr⁡T≥∑j≤kλj(C)−ktF2. Also tr⁡(PH)≥tr⁡(PG)−ktF2. Combining these inequalities with the top-k comparisons in step 1.2 gives Δ:=∑j≤kλj(H)−tr⁡(PH)≤4ktF2. The variational formula gives Δ≥0. If k=0, Δ=0.

3.1step 1.1step 1.2step 2.1algebra

Let r count eigenvalues of H strictly greater than a. Between indices k and r, every crossing eigenvalue lies within 2tF2 of a, by step 1.2 and the strict threshold rule for C. There are at most K crossing indices: if r>k, then r<K since each of its first r eigenvalues exceeds a and tr⁡H=F2; if k>r, then k<K. Thus the difference between the sums of the largest k values of λj(H)−a and all positive values of λj(H)−a is at most 2KtF2 . Put pj=⟨ej,Pej⟩ in an eigenbasis of H. Since ∑jpj=k, the nonnegative threshold deficit is Δa=∑λj>a(λj−a)(1−pj)+∑λj≤a(a−λj)pj≤Δ+2KtF2≤6KtF2. For k=0 the same crossing argument applies, and zero eigenvalues can be padded without changing the sums.

4.1step 3.1algebra

Let P1,P2,P3 project onto eigenvalues respectively in [a(1+η)2,∞), [a(1−η)2,a(1+η)2), and [0,a(1−η)2). Write E for the part of Δa on the first and third sets, and L=tr⁡[P1(I−P)]+tr⁡(P3P). Both outside gaps from a are at least ηa, so L≤E/(ηa) and E≤6KtF2≤7KtF2. This uses no strict separation of eigenvalues within the band.

5.1F1step 4.1algebraconstruct

On the band space, R=P2PP2 is a positive contraction. Let Q be its spectral projection onto eigenvalues at least 1/2. In an eigenbasis of R, ∥P2P−Q∥F2=∑smin⁡(s,1−s)≤2∑ss(1−s)=2∥P2P(I−P2)∥F2≤2L. For the last inequality, the high cross block satisfies ∥P2PP1∥F=∥P2(I−P)P1∥F≤∥(I−P)P1∥F and the low block satisfies ∥P2PP3∥F≤∥PP3∥F. These squares sum to at most L.

6.1F2F3step 1.1step 4.1step 5.1algebra∎

Evaluate ∥A(P−P1−Q)∥F2 by summing its rows in an eigenbasis of H. Its high contribution is ∑P1λj(1−pj)≤2E/η, because λ/(λ−a)≤2/η there. Its low contribution is at most aL, and its band contribution is at most a(1+η)2∥P2P−Q∥F2≤8aL. Thus the sum is at most 11E/η≤77tKF2/η. By [F3], A(P1+Q) is a permitted target. Finally ∥A(M−P)∥F≤∥A∥op∥M−P∥F≤FtK; the triangle inequality gives the result. The same block formulas cover empty high, band or low spaces, k=0, repeated eigenvalues, η=1, and both endpoint conventions.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

ModFKV low-rank approximation supplier

Statement

Let A∈Rm×n be nonzero, F=∥A∥F, and suppose it has the exact row/column ℓ2 sample/query access of ℓ2 sample/query access to vectors and matrices. Let 0<σ≤F, 0<ϵ≤14σ/F, ϵ2≤η≤1, and 0<δ<1. Set K=F2/σ2 and ϵˉ=ηϵ2. For a sufficiently large universal constant in the sample count q=Θ ⁣(K4ϵˉ2log⁡2δ), the procedure ModFKV low-rank description procedure outputs a description of a rank-at-most-K matrix D=ASTU^Σ^−2U^TS such that, with probability at least 1−δ, there is a threshold-band target Aσ,η from Threshold-band singular-value approximation satisfying ∥D−Aσ,η∥F≤ϵF. The ModFKV description itself queries O(q2) matrix entries and computes an SVD of a q×q matrix at O(q3) arithmetic cost, before any later sampling from D's rows. These operation counts use the stated exact-access arithmetic model and do not include unprovided preprocessing for the input access structure.

Facts & Assumptions

Given: The matrix, exact access, and parameters of the Statement.

[F1]

Exponential Frobenius concentration for normalized Gram sampling controls both Gram errors with probability at least 1−δ, and proves ∥S∥F=∥W∥F=F exactly.

[F2]

A deterministic threshold-band bound for two Gram sketches gives the retained rank, near-isometry, and the threshold-band error from these two Gram errors.

Proof

1.1givenF1algebra

Set t=ϵˉ/(1000K2). Choose the integer q=⌈106K4ϵˉ−2(1+2log⁡(2/δ))2⌉. This has the asserted Θ order, uniformly for 0<δ<1, and [F1] gives both Gram errors at most tF2 except on an event of probability at most δ. The algorithm's mixture probability equals ∥S∗,j∥2/F2, so a zero-probability column is never drawn and its INVALID branch has probability zero.

2.1givenF2step 1.1algebra

Since K≥1, η≤1 and ϵ≤1/4, tK≤1/2. On the event in step 1.1, [F2] applies to the actual strictly retained singular values and to M=STUΣ−2UTS. It gives rank at most K and a threshold-band target with error at most F(tK+77tK/η)≤F(ϵ2/1000+ϵ77/1000)≤ϵF. Empty retention returns zero and is included in [F2]. No gap between actual singular values is assumed.

3.1givenstep 1.1step 2.1algebra∎

The q row samples, q mixture samples and the probabilities of the sampled columns can be obtained using at most O(q2) input queries: each mixture probability sums q known row probabilities and queried squared entries. Forming W uses q2 entries; its exact-arithmetic SVD and the stored matrices require O(q3) arithmetic. These costs give the stated counts for the description and make no claim to materialize A, S or D in full.

TheoremStatement: Literature-sourcedProof: AI-generatedOpen item page →

Tang's conditional recommendation sampler

Statement

Let A∈Rm×n have the row/column entry-query and ℓ2-sampling access of ℓ2 sample/query access to vectors and matrices, let F=∥A∥F>0, and fix a row i with ∥Di,∗∥2>0. Here D is the ModFKV matrix description from ModFKV low-rank description procedure and its success event is specified by ModFKV low-rank approximation supplier. Assume 0<σ≤F,0<ϵ≤14σ/F,ϵ2≤η≤1,0<δ<1. Assume countable choice and dependent choice, and model the finite sketch history together with the countably many potential fresh row-sample calls in the adaptive refinement by the joint product probability law supplied by Assuming countable and dependent choice, countable products of arbitrary probability spaces. After a finite row-refinement stop, the capped rejection sampler uses its finite conditional law; the joint execution law is the countable mixture over finite stopping histories together with any residual nontermination event, on which no rejection call occurs. There is a Tang recommendation procedure whose sketch succeeds unconditionally with probability at least 1−δ/3. For every fixed successful description having Di,∗≠0, its conditional probability of a successful row estimate and an accepted sample is at least 1−2δ/3. Thus the unconditional probability of a nonzero realized target row together with failure of these guarantees is at most δ. If the nonzero-row promise holds almost surely, this implies overall success probability at least 1−δ. For every fixed successful sketch and successful row estimate, the law of the accepted classical sample is within total variation ϵ (where dTV(p,q)=12∑a∣p(a)−q(a)∣ as in End-to-end channel and output error contract) of the normalized squared-entry distribution of Di,∗. A capped sampling failure returns FAIL; no conditional distribution is assigned to a zero target row. The nonzero-row hypothesis is a promise on the realized description, not an assumed ability to compute its full row norm. The failure bound is unconditional; it is not a claim that conditioning on a nonzero realized row preserves ModFKV's confidence. At constant precision its query and arithmetic-time bounds are respectively O~ ⁣(F18σ18ϵ8η4),O~ ⁣(F24σ24ϵ12η6). For the final ϵ total-variation guarantee, the finite row refinement below adds a query and arithmetic-time factor O~(∥Ai,∗∥22/∥Di,∗∥22). These bounds hold on the successful event; refinement terminates almost surely for every realized description satisfying the nonzero-row promise. Hidden logarithms, access-structure preprocessing, arithmetic precision, and the cost of writing the returned sample remain charged. The result is conditional on this exact input model and does not output D as a full matrix or a full classical amplitude list.

Facts & Assumptions

Given: The exact input access and parameters, and a realized nonzero target row as promised in the Statement.

[F1]

ModFKV low-rank approximation supplier and its proof give a successful sketch with failure probability at most its assigned budget; A deterministic threshold-band bound for two Gram sketches gives k<K and eigenvalues of VTV in [1−tK,1+tK] for V=STUΣ−1, where t=ηϵ2/(1000K2) and tK≤1/2. The retained singular values in Σ lie in [σ,F].

[F2]

With exact norm, fresh ℓ2 samples and entry queries to x, and entry queries and a known norm for y, Estimate an inner product from ℓ2-sampling access estimates ⟨x,y⟩ to error α∥x∥∥y∥ with failure at most β using O(α−2log⁡(2/β)) samples and queries.

[F3]

Apply Squared-vector distributions are close in total variation to the ordered pair (y,x) and use symmetry of total variation: for nonzero vectors, dTV(Dx,Dy)=dTV(Dy,Dx)≤2∥x−y∥/∥y∥.

[F4]

Correctness and cost of the linear-combination sampler proves exact sampling on acceptance and a miss probability at most the assigned budget under a proof-side cancellation bound and a finite cap.

[F5]

Under countable choice and dependent choice, the finite sketch history and the countable stream of repeated fresh samples from the fixed row distribution admit the joint product probability law stated in the Statement (Assuming countable and dependent choice, countable products of arbitrary probability spaces). Every finite stopping history has a finite conditional sampler law by [F4], so the stopped branches form a countable mixture; paths that never stop remain the residual nontermination event in the product row-sample space.

[F6]

A countable union of events has probability at most the sum of their probabilities (Finite and countable subadditivity of measures).

[F7]

If a sequence of events has summable probabilities, only finitely many occur almost surely (First Borel-Cantelli lemma for events).

Proof

1.1givenF1algebracases

Put b=∥Ai,∗∥. If b=0, return FAIL: then Di,∗=0 and the promised input is absent. Otherwise b>0 is available from exact access. Run ModFKV with failure budget δ/3, retaining its rank k, sample count, and V of [F1]. If the retained rank is k=0, return FAIL because again Di,∗=0. The sketch event of [F1] has probability at least 1−δ/3; all subsequent estimates use fresh randomness conditional on the actual sketch. Write h=Ai,∗ST and z=hUΣ−1, so Di,∗=zVT.

2.1F1F2step 1.1constructalgebra

At stages l=0,1,…, set ρl=2−l, βl=δ 2−(l+1)/3, and αl=ρl/K. Put ml=⌈16/αl2⌉ and choose the least positive integer rl with 2−rl≤βl/q. For each of the q entries of h, use [F2]'s lower-median estimator with ml samples per block, rl blocks, tolerance αl, and failure budget βl/q; use mutually independent fresh sample batches for these estimates. Each row of S has norm F/q. Thus on that stage's simultaneous success, ∥h^l−h∥≤αlbF, and the explicit vector z^l=h^lUΣ−1 obeys ∥z^l−z∥≤ρlb=:el. Stop at the first stage with ∥z^l∥≥(1+32/ϵ)el. All quantities in this rule are computable from stored coefficients and exact input row norms; no query of the unknown full output-row norm occurs.

3.1givenF2F5F6F7step 2.1algebra

Fix a realized sketch and use [F5] to predefine the disjoint finite sample batches for every potential stage l. Let El be the cylinder event that at least one of that stage's q inner-product estimates fails its tolerance. Fresh samples, [F2], and [F6] give P(El)≤βl. Since ∑l≥0βl=δ/3, [F6] gives P(⋃lEl)≤δ/3, and [F7] gives P(El infinitely often)=0. Thus with probability at least 1−δ/3 every stage estimate succeeds, and almost surely all sufficiently late stage estimates succeed. On those late stages ∥z^l−z∥≤el→0, so z^l→z. The nonzero-row promise implies z≠0; hence the stopping threshold (1+32/ϵ)el tends to zero while ∥z^l∥→∥z∥>0, and the procedure terminates almost surely, including for a sketch outside the successful sketch event whenever its realized target row is nonzero. On the event that every stage estimate succeeds, it stops no later than a stage with el≤ϵ∥z∥/64, because ∥z^l∥≥∥z∥−el≥(1+32/ϵ)el. Hence the stopped stage satisfies 2l≤max⁡{1,128b/(ϵ∥z∥)}.

4.1F1F3step 2.1step 3.1algebra

On the successful sketch and estimation event, at the stopping stage ∥z∥≥∥z^∥−e≥32e/ϵ. With x=z^VT and y=Di,∗=zVT, [F1] gives ∥x−y∥/∥y∥≤3e/∥z∥≤3ϵ/32. In particular x≠0, and [F3] yields dTV(Dx,Dy)≤3ϵ/16<ϵ. This holds for every fixed successful sketch and estimate, so any mixture of their accepted output laws has the same bound when compared with the corresponding mixture of target laws.

5.1F1F4step 4.1algebraconstruct

Form the coefficient row w=z^Σ−1UT, so x=wS. Its numerator in the cancellation ratio is ∑j∥wjSj,∗∥2=(F2/q)∥w∥2≤(F2/(qσ2))∥z^∥2. On the successful sketch, ∥x∥2≥12∥z^∥2, so the ratio is at most 2K/q. Take C0=2K/q as a proof-side bound and integer cap R=⌈2Klog⁡(3/δ)⌉, then invoke the finite sampler [F4] with q terms and cap R. The bound C≤C0 is established only on the successful sketch event; the procedure does not test or require a certificate for it. On every other history the sampler is still executable: if its component-square sum is zero it returns FAIL immediately, and otherwise its finite proposal/acceptance loop returns an accepted index or FAIL by the cap. Conditional on any fixed good coefficients, [F4] gives miss probability at most δ/3 and accepted law exactly Dx.

6.1F4F5F6step 1.1step 3.1step 4.1step 5.1algebra

Sketch failure has probability at most δ/3. Conditional on every fixed sketch, any row-estimation failure has probability at most δ/3 by step 3.1. Conditional on a successful sketch and row estimate with nonzero target row, capped rejection failure has probability at most δ/3 by [F4]. By [F5], the finite conditional rejection laws are attached to their countably many finite stopping histories, so the failure event is measurable in the joint execution model; [F6] bounds the union of the sketch, estimation, and applicable rejection failures by δ. Conditioning on rejection success does not manufacture confidence: for each fixed successful estimate its accepted law is already Dx by [F4]. Step 4.1 supplies the asserted total-variation conclusion on that branch. Outside it the routine may return FAIL or an uncontrolled sample; no assertion treats a failed event as successful.

6.2F1F2F4step 3.1step 5.1algebra

Stage l uses O(qK4llog⁡(2q/βl)) queries and arithmetic for inner products, plus O(qk) coefficient arithmetic. Summing geometrically to the stopping stage gives O~(qKϵ−2b2/∥z∥2+qK) on simultaneous success. Since ∥Di,∗∥≤3/2∥z∥, this is O~(qKϵ−2(1+b2/∥Di,∗∥2)). Rejection uses O(Klog⁡(3/δ)) trials with O(q) queries per trial; forming its weights costs O(qk). Even the looser O~(Kq2) bound suffices.

7.1F1F8step 1.1step 6.2algebra∎

Substituting q=O~(K4/(η2ϵ4)), the base query bound O~(Kq2) is O~(K9/(η4ϵ8)), and the SVD time O~(q3) is O~(K12/(η6ϵ12)). Since q≥ϵ−2, the refined estimate cost in step 6.2 is at most these base bounds times O~(1+b2/∥Di,∗∥2). On the successful sketch, Di,∗=Ai,∗VVT and [F1], [F8] give ∥VVT∥op=λmax⁡(VTV)≤1+tK≤3/2. Hence ∥Di,∗∥≤(3/2)b, so this last factor is O~(b2/∥Di,∗∥2). Replacing K by F2/σ2 gives the stated powers, with all access preprocessing, precision and writing costs charged separately.

RemarkRemark: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Dequantization is relative to the input access and output task

Remarks

Under the nonzero-row promise and countable-product sampling model stated in the theorem, Tang's real-matrix result matches a particular input structure: entry queries together with efficient row/column ℓ2-norm sampling, including the preprocessing and storage that provide those samplers. Its task is to sample from a specified recommendation row derived from a thresholded low-rank approximation, and its output is a sample rather than a full matrix or a classical list of amplitudes. It therefore establishes a classical counterpart for that access/task/output contract. It does not assert efficient classical simulation of arbitrary quantum circuits, arbitrary quantum-state output, or algorithms whose input model lacks the required ℓ2 samplers. Compare algorithms only after matching input access, output type, approximation metric, and charged preprocessing.

5 · Examples, counterexamples and false statements

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