LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck pass

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.

Conditioning on a private message separates average acceptance

Statement

Let A,B be N×m row-stochastic matrices, M∈[0,1]^{m×m}, and let C=AMBᵀ satisfy C_xx≥0.99 and C_xy≤0.01 for x≠y. For every probability row vector D with ||D||₂²≤0.01 there are weights q_j and probability rows D_j with D=Σ_jq_jD_j such that, for rows X_j=(M_ij)_{i=1}^m (the transposes of the columns of M), Σ_j q_j(DA)X_jᵀ≤0.02 and Σ_jq_j(D_jA)X_jᵀ≥0.99.

Facts & Assumptions

Given: $N, m \geq 1$ , row-stochastic $A, B \in [0, 1]^{N \times m}$ , $M \in [0, 1]^{m \times m}$ , and $C = A M B^{T}$ with $C_{x x} \geq 0.99$ and $C_{x y} \leq 0.01$ when $x \neq y$ . Let $D$ be a probability row vector with $∥ D ∥_{2}^{2} \leq 0.01$ .

[F1]

Finite distributions, entropy, divergence, and stochastic matrices: A matrix $A \in [0, 1]^{N \times m}$ is row-stochastic when each row sums to one.

Proof

technique · direct

1.1givenF1algebraconstruct

For each message index $j$ , set $q_{j} = \sum_{x} D_{x} B_{x j}$ . If $q_{j} > 0$ , put $D_{x}^{(j)} = D_{x} B_{x j} / q_{j}$ ; if $q_{j} = 0$ , put $D^{(j)} = D$ . Then each $D^{(j)}$ is a probability vector, $\sum_{j} q_{j} = 1$ , and $\sum_{j} q_{j} D^{(j)} = D$ , because $\sum_{j} B_{x j} = 1$ ; zero-weight terms contribute nothing.

2.1givenstep 1.1F1algebra∎

Let $X_{j} = (M_{i j})_{i = 1}^{m}$ be the transpose of column $j$ of $M$ . Reordering finite sums gives $\sum_{j} q_{j} (D^{(j)} A) X_{j}^{T} = \sum_{x} D_{x} (A M B^{T})_{x x} \geq 0.99.$ Also $\sum_{j} q_{j} (D A) X_{j}^{T} = D A M B^{T} D^{T}$ . Write $C = I + E$ . Since all entries of $C$ are probabilities, $∣ E_{x y} ∣ \leq 0.01$ ; hence this last quantity is at most $\sum_{x} D_{x}^{2} + 0.01 (\sum_{x} D_{x})^{2} \leq 0.02$ . These are the required decomposition and acceptance bounds.

Depends on

Finite distributions, entropy, divergence, and stochastic matrices

Used by

Dependency tree · two levels

2 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

Ilan Newman and Mario Szegedy, Public vs. Private Coin Flips in One Round Communication Games (standard reference, not scraped)
Ronald de Wolf, Quantum Computing: Lecture Notes, arXiv:1907.09415v6 (standard reference, not scraped)