TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Private-coin SMP equality requires Ω(√n) bits

Statement

For a classical simultaneous-message Equality protocol on n-bit inputs, with independent private randomness, no shared resource, and worst-case error at most 1/3, let c be the maximum of the two senders’ worst-case message lengths. For the fixed odd t used in the proof, c+1≥sqrt(n/(60 ln 2))/t; consequently the total communication is Ω(√n).

Facts & Assumptions

Given: A classical private-coin SMP protocol for Equality on $n \geq 1$ bits, with no shared resource, worst-case error at most $1 / 3$ , and maximum sender message length $c$ bits.

[F1]

Constant private-coin error reduction by repetition: For every odd $t$ , $t$ independent repetitions and majority have error at most $e^{- t / 48}$ and each sender sends at most $t (c + 1)$ bits after fixed-length encoding.

[F2]

Packing bound for separated randomized message rows: If row-stochastic $N \times m$ matrices $A, B$ and an $m \times m$ matrix $M \in [0, 1]^{m \times m}$ have diagonal entries of $A M B^{T}$ at least $0.99$ and off-diagonal entries at most $0.01$ , then $N \leq e^{60 (\ln m)^{2}}$ for $m \geq 2$ .

Proof

technique · reduction to the finite matrix-packing lemma

1.1givenF1construct

Choose any fixed odd integer $t > 48 \ln 400$ . By [F1], repetition gives error less than $1 / 400$ . Each sender's repeated transcript takes at most $2^{t (c + 1)}$ possible values. Let $N = 2^{n}$ and $m = 2^{t (c + 1)}$ ; define row $x$ of $A$ and row $y$ of $B$ as the probability distributions of Alice's and Bob's transcript, padding unused message labels with zeroes. Independence of their private random tapes makes the joint transcript law the product $A_{x} (i) B_{y} (j)$ . Let $M_{i j}$ be the referee's acceptance probability on transcript pair $(i, j)$ ; then $M \in [0, 1]^{m \times m}$ and $(A M B^{T})_{x y}$ is the acceptance probability on $(x, y)$ .

2.1step 1.1F1F2algebra∎

Repeated-protocol correctness gives diagonal acceptance at least $399 / 400 > 0.99$ and off-diagonal acceptance at most $1 / 400 < 0.01$ . Apply [F2] to obtain $2^{n} \leq \exp (60 (t (c + 1) \ln 2)^{2})$ . Taking logarithms and rearranging yields $c + 1 \geq \frac{1}{t} \sqrt{\frac{n}{60 \ln 2}} .$ Since $t$ is a fixed constant, $c = Ω (\sqrt{n})$ as $n \to \infty$ , and the total communication is at least $c$ . The proof uses only finite message distributions and finite conditioning; it invokes no choice axiom.

Depends on

Used by

Private-coin SMP Equality lower bound and source recovery Remark

Dependency tree · two levels

14 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)