Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: 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.

Complete-graph search matrix and gap

Example

For the complete graph search instance with any finite N≥2, let ∣w⟩ be the marked vertex and let ∣r⟩ be the uniform vector on the N−1 unmarked vertices. Compute the invariant 2×2 generator matrix in the ordered basis (∣w⟩,∣r⟩) and its eigenvalue gap. Include the boundary case N=2. Also verify the supplied uniform preparation for N=2 and N=4 against the one- and two-qubit Hadamard circuits.

Facts & Assumptions

Given: N≥2, the encoded complete-graph adjacency, marked projector, and Hamiltonian H=−A/N−Pw of the algorithm.

[F1]

Complete-graph continuous-time search: on the active label basis, A∣x⟩=∑y≠x∣y⟩, Pw=∣w⟩⟨w∣, and H=−A/N−Pw.

[F2]

Graph adjacency and Laplacian generator conventions: symmetric graph adjacency is Hermitian and the ordered vertex basis is fixed.

[F3]

Dirac Notation Conversion: ∣x⟩⟨y∣ acts by z↦⟨z,y⟩x.

[F4]
[F5]

Exact and approximate success of complete-graph continuous-time search: with the supplied uniform preparation, exact evolution under H=−A/N−Pw at t=πN/2 gives marked-label probability exactly 1.

[F6]

Hadamard preparation of a uniform binary state: for N=2n, the standard-binary uniform state is H⊗n∣0n⟩, with n Hadamard gates when primitive.

Verification

technique · direct
1.1givenF4choosealgebra

The N−1 unmarked encoded basis vectors are pairwise orthonormal by [F4], so ∣r⟩=(N−1)−1/2∑x≠w∣x⟩ is unit and orthogonal to the unit vector ∣w⟩. The uniform vector is ∣s⟩=N−1/2∣w⟩+(N−1)/N∣r⟩, whose squared coefficient norms sum to 1.

2.1givenF1step 1.1algebra

By [F1], A∣w⟩=∑x≠w∣x⟩=N−1∣r⟩. For ∣r⟩, each summand A∣x⟩ contains ∣w⟩ once, so its ∣w⟩ coefficient is N−1; for each unmarked y, the sum contains ∣y⟩ once for each x≠y,w, hence N−2 times. Thus A∣r⟩=N−1∣w⟩+(N−2)∣r⟩.

3.1F1F2F3step 2.1algebra

The two vectors span an invariant subspace for A by step 2.1. By [F3], Pw∣w⟩=∣w⟩ and Pw∣r⟩=0, so Pw also preserves this span. The symmetric adjacency is Hermitian by [F2], and the generator matrix is H=(−1−N−1/N−N−1/N−(N−2)/N).

4.1step 3.1algebra

Put K=H+(N−1)I/N. Direct multiplication gives K2=I/N, and tr⁡K=0. Equivalently, its characteristic polynomial is λ2−1/N, so its eigenvalues are ±1/N. Therefore the eigenvalues of H are −(N−1)/N±1/N, and their gap is 2/N.

5.1step 1.1step 3.1step 4.1algebra

At N=2, ∣r⟩ is the single unmarked basis vector and the formula in step 3.1 becomes H=(−1−1/2−1/20). The shifted matrix has eigenvalues ±1/2, so the gap is 2; all normalizations and denominators used above remain valid.

5.2F5step 1.1step 3.1step 4.1

For the uniform input ∣s⟩ in step 1.1, [F5] gives marked-label probability 1 after exact evolution at t=πN/2. The invariant matrix and its gap were derived locally in steps 3.1–4.1.

6.1F1F2F6algebra∎

Specialize to the standard binary encoding for N=2 and N=4. For N=2, the preparation is H∣0⟩=(∣0⟩+∣1⟩)/2, which is the required uniform state with one Hadamard gate. For N=4, applying the second factor gives H⊗2∣00⟩=(∣00⟩+∣01⟩+∣10⟩+∣11⟩)/2, the required uniform state with two Hadamard gates. These are the n=1,2 cases of [F6]; no preparation or gate cost is inferred from the search matrix.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources