Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Simon samples uniformly from s-perp

Statement

For every Simon hidden nonzero string s∈{0,1}n, the one-sample algorithm returns a string uniformly distributed on

s⊥:={y∈{0,1}n:s⋅y=0}.

The set s⊥ has 2n−1 elements. Conditional on the measured output value f(x0), the amplitude of y is 2−(n−1)/2(−1)x0⋅y when s⋅y=0 and is zero otherwise. Thus the outcome law is independent of which output fiber was observed.

Facts & Assumptions

Given: n≥1, the Simon promise, and a fresh one-sample execution.

[F1]

Simon query and output measurement produce a coset pair: conditional on output f(x0), the address state is (∣x0⟩+∣x0⊕s⟩)/2.

[F2]

Hadamard character sum on F_2^n: H⊗n∣x⟩=2−n/2∑y(−1)x⋅y∣y⟩.

[F3]

Binary dot product as parity: x⋅y is the parity ⨁i<n(xi∧yi).

[F4]

One Simon Fourier-sampling sample: after the target branch, the algorithm applies H⊗n to the address and measures in the computational basis.

[F5]

Computational-basis measurement weights are squared amplitudes: for a unit vector measured in the computational basis, the outcome weight is the squared modulus of its basis amplitude.

[F6]

The Simon annihilator is an (n−1)-dimensional binary subspace: s⊥ is a subspace of dimension d=n−1 and has exactly 2d elements.

[F7]

The computational basis is orthonormal: the address register's computational basis is orthonormal.

Proof

1.1givenF1F2F3algebra

Apply [F2] to the two terms in [F1]. The amplitude at y is 2−(n+1)/2((−1)x0⋅y+(−1)(x0⊕s)⋅y). By [F3], (x0⊕s)⋅y=(x0⋅y)⊕(s⋅y).

2.1step 1.1F3algebra

If s⋅y=1, the two signs in step 1.1 are opposites and the amplitude is zero. If s⋅y=0, they agree and the amplitude is 2−(n−1)/2(−1)x0⋅y. Thus its squared modulus is 2−(n−1) on every allowed y and zero otherwise.

3.1givenF2F6F7step 2.1algebra

There are 2n−1 allowed strings by [F6]. The vector after the final Hadamards has squared norm 2n−12−(n−1)=1 by step 2.1 and orthonormality [F7]. This also covers n=1, when the only allowed string is 0.

4.1F4F5step 2.1step 3.1algebra∎

The measured vector is unit by step 3.1, so [F5] gives probability 2−(n−1) for each allowed string and zero for every disallowed string. These weights sum to one; the phase depends on x0 but the probabilities do not. Thus every fresh execution has the same uniform sample law regardless of the observed coset.

Depends on

Used by

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Sources