LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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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.
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.

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Discriminant of the marked-state absorbing walk

Statement

Let $P$ be a finite symmetric row-stochastic matrix on $X$ , and let $M \subseteq X$ , $U = X ∖ M$ . Define the row-stochastic absorbing matrix

$P_{x y}^{'} = {\begin{cases} 1, & x = y \in M, \\ 0, & x \in M, y \neq x, \\ P_{x y}, & x \in U . \end{cases}$

For the general row-stochastic discriminant $D (P^{'})_{x y} = \sqrt{P_{x y}^{'} P_{y x}^{'}}$ of Szegedy row-preparation map and discriminant,

$D (P^{'}) = P_{U U} \oplus I_{M},$

where $P_{U U}$ is the principal submatrix on $U$ . In general $P^{'}$ need not be symmetric or reversible.

Facts & Assumptions

Given: A finite symmetric row-stochastic matrix $P$ , a marked set $M$ , and the stated row convention.

[F1]

Finite row-stochastic kernel: each row of $P$ is nonnegative and sums to $1$ .

[F2]

Szegedy row-preparation map and discriminant: the discriminant of a row-stochastic matrix has entries $D_{x y} = \sqrt{P_{x y} P_{y x}}$ .

[F3]

The Szegedy discriminant is self-adjoint and contractive: the discriminant entry formula holds for every finite row-stochastic kernel, without reversibility.

Proof

1.1givenF1construct

If $x \in U$ , row $x$ of $P^{'}$ equals row $x$ of $P$ and sums to one by [F1]. If $x \in M$ , row $x$ has a single $1$ at $y = x$ and zeros elsewhere. Thus $P^{'}$ is row-stochastic.

2.1F2F3step 1.1algebra∎

For $x, y \in U$ , $P_{x y}^{'} = P_{x y}$ and $P_{y x}^{'} = P_{y x} = P_{x y}$ , so [F2] gives $D (P^{'})_{x y} = P_{x y}$ . If $x, y \in M$ , the product $P_{x y}^{'} P_{y x}^{'}$ is $1$ exactly when $x = y$ and $0$ otherwise, giving the identity block. If exactly one of $x, y$ is marked, the transition from the marked state to the unmarked state is zero, so $D (P^{'})_{x y} = 0$ . These three disjoint cases give the displayed direct sum.

Depends on

Used by

Szegedy-walk marked-set detection Algorithm
Walk detection correctness and cost Theorem

Dependency tree · two levels

22 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

Andrew M. Childs, Lecture Notes on Quantum Algorithms, Chapter 18, §18.4 (standard reference, not scraped)
Frédéric Magniez, Ashwin Nayak, Jérémie Roland, and Miklos Santha, Search via Quantum Walk, §2 (standard reference, not scraped)