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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.
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Adversary Methods and Quantum Query Complexity — Examples

1 · Prerequisites

2 · Summary

The OR_N calculation gives a balanced dual-vector certificate of cost √N and a matching adversary matrix. The four-bit sorted function then separates the nonnegative and signed optimizations: the nonnegative value is exactly 5/2, while an explicit rational signed matrix has ratio greater than 2.511486. Coordinate-distribution certificates give the nonnegative upper bound, and product distributions preserve that upper bound under iterated composition.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-generatedOpen item page →

Dual adversary certificate for OR_N

Example

For the partial OR problem D={0,e1,…,eN}, define v0,i=N−1/4 and vej,i=N1/41[i=j]. This is a balanced dual certificate with C0=C1=N. The matrix with only the symmetric edges (0,ej) has norm N and each filtered norm 1. Hence Adv⁡±(ORN)=N.

Facts & Assumptions

Given: The partial Boolean function OR_N on the zero string and the N unit vectors.

[F1]

Finite Boolean query problem and error model for adversary bounds: OR_N is the finite Boolean query problem on the declared domain.

[F2]

Input-mismatch masks and entrywise filtering: Δ_i marks exactly the pairs that differ in bit i.

[F3]

General adversary matrix and spectral bound: a real symmetric opposite-output Γ certifies its norm ratio as a lower bound on Adv±.

[F4]

Dual vector formulation of the general adversary bound: a vector certificate of cost C gives Adv±≤C, and the optimum is attained.

Verification

1.1givenF1F2F4construct

Let D={0,e_1,…,e_N}⊂{0,1}^N and f(x)=OR_N(x). For every i set v_{0,i}=N^{-1/4} and v_{e_j,i}=N^{1/4}δ_{ij} in the one-dimensional real space. If x=0,y=e_j, the only mismatch is coordinate j and the corresponding product is 1. If x=e_j,y=e_k with j≠k, the mismatch coordinates are j,k and both products vanish. Reverse pairs give the same real values. Thus all constraints in [F4] hold.

2.1givenF4step 1.1algebra

The cost on the unique 0-input is Σ_iN^{-1/2}=√N. On each 1-input e_j the cost is Σ_iN^{1/2}δ_{ij}=√N. Hence Adv±(OR_N)≤√N by [F4].

3.1givenF2F3step 2.1algebra∎

For the reverse bound, let Γ have Γ_{0,e_j}=Γ_{e_j,0}=1 for every j and all other entries zero. Its nonzero block is the all-ones 1×N matrix, whose singular value is √N, so ||Γ||=√N. For each i, Γ∘Δ_i retains only the two entries joining 0 and e_i and has norm 1. Therefore [F3] gives Adv±(OR_N)≥√N; equality follows.

CounterexampleConstruction: Literature-sourcedVerification: AI-generatedOpen item page →

A nonnegative-only adversary estimate need not capture the general bound

Statement refuted

The assertion that nonnegative adversary weights always attain the general signed adversary bound is false. The following family supplies a strict separation.

For the partial iterates f(1)=f and f(k+1)=f∘(f(k),f(k),f(k),f(k)) of the sorted four-bit function, Adv⁡±(f(k))>(2.511486)k,Adv⁡(f(k))≤(5/2)k. Thus signed adversary weights give a strict separation already at k=1 and an exponentially increasing ratio under iteration.

Facts & Assumptions

Given: The sorted four-bit function, its base certificates, and its partial Boolean iterates.

[F1]

The four-bit Ambainis function separates nonnegative and signed adversary bounds: the sorted four-bit f has Adv(f)=5/2 and Adv±(f)>2.511486, with explicit dual distributions and a signed matrix.

[F2]

Nonnegative and signed adversary bounds compose from below: both signed and nonnegative adversary values multiply as lower bounds when the composed block domain is nonempty.

[F3]

Input-distribution certificates upper-bound the nonnegative adversary value: for a nonconstant Boolean function, coordinate distributions with cross-output mismatch overlap η>0 give Adv≤1/η.

Counterexample

1.1givenF1F2constructalgebra

Let f be the sorted four-bit function and define f(1)=f and f(k+1)=f∘(f(k),f(k),f(k),f(k)) on their natural block domains. Since f is total on {0,1}4 and nonconstant, induction shows every f(k) is total on {0,1}4k and nonconstant: each outer bit string is realized by independently choosing a preimage of each bit under f(k), and outer strings with either f-value lift in the same way. Thus every composition domain is nonempty and both adversary bounds in [F2] apply. The signed matrix in [F1] and repeated composition in [F2] give Adv⁡±(f(k))≥Adv⁡±(f)k>(2.511486)k.

1.2givenF1F3construct

The base coordinate distributions in [F1] have mismatch overlap at least η=2/5. If PX(k) is a distribution over the 4k input coordinates with cross-output overlap at least ηk, define for an input X=(X1,…,X4) to f(k+1) PX(k+1)(j,r)=pX~(j)PXj(k)(r),X~j=f(k)(Xj). For opposite outputs of f(k+1), the outer strings X~,Y~ have opposite f-values. For each j with X~j≠Y~j, the inner pair Xj,Yj has opposite f(k)-values and contributes overlap at least ηk. Summing those j gives total overlap at least η⋅ηk=ηk+1; normalization follows by multiplying two probability distributions.

2.1givenF1F2F3step 1.1step 1.2algebra∎

By induction and [F3], Adv⁡(f(k))≤η−k=(5/2)k. Since 2.511486>2.5, the signed lower bound grows strictly faster than this nonnegative upper bound for every k≥1. Thus negative weights produce a strict adversary separation, and repeated composition amplifies it.

Sources