Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Expander walk hits dense bad sets

Statement

Fix a finite d-regular adjacency-slot multigraph on n≥1 vertices, with normalized adjacency M, and put α=∥M∣1⊥∥.

Let α<1 and let B be a fixed vertex set of density δ∈[0,1]. For a walk begun from the uniform distribution and taking t≥0 steps (thus sampling t+1 vertices), Pr⁡[no visit to B]≤(1−δ)[1−(1−α)δ]t≤(1−δ)e−(1−α)δt. A zeroth power is interpreted as one even when its base is zero.

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

Let S have density β=∣S∣/n in a finite regular graph and let PS project onto functions supported in S. Then ∥PSMPS∥≤α+(1−α)β. For a stationary length-t walk, t≥0, its confinement probability is n−1⟨1S,(PSMPS)t1S⟩0, where the inner product is unnormalized. (Expander walk restricted operator).

Proof

1.1F1

Apply the confinement identity to S=V∖B. The restricted norm is at most q=α+(1−α)(1−δ)=1−(1−α)δ. Bounding the matrix power in the unnormalized inner product and using ∥1S∥02=n(1−δ) gives the first estimate. If S is empty the probability is zero directly.

2.1step 1.1algebra∎

For x≥0, 1−x≤e−x: the difference has value zero at zero and derivative 1−e−x≥0. Here x=(1−α)δ∈[0,1], so raising this inequality to the nonnegative integer t gives the second bound. At t=0 the probability is 1−δ; at δ=0 it is one and at δ=1 zero.

Depends on

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Sources