False statementConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07

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Polynomial-size circuits imply a uniform polynomial-time generator

Statement

Every polynomial-size circuit family has a polynomial-time algorithm that, given $1^{n}$ , outputs its length- $n$ circuit.

Facts & Assumptions

Given: the nonuniform family convention.

[L1]

$P / p o l y$ requires only a separate polynomial-size circuit at each length and no generator, by Circuit families and P/poly.

[L2]

There is an undecidable language recognized by constant-size circuits, by Some undecidable languages have polynomial-size circuits.

Refutation

technique · direct

1.1

Let $(C_{n})$ be the constant-size family for the undecidable length language from [L2]. If a polynomial-time generator $G$ output $C_{n}$ on input $1^{n}$ , then on any word $x$ one could compute $G (1^{∣ x ∣})$ and evaluate that constant- size circuit on $x$ in polynomial time.

L2givenconstruct

2.1

The resulting algorithm would decide the language, contradicting its undecidability. Hence this polynomial-size family has no such generator.

step 1.1contradiction

3.1

The size condition in [L1] is therefore strictly nonuniform; effective generation is an additional hypothesis.

L1step 2.1∎

Depends on

Used by

Unary circuits for an undecidable language are not uniformly generated Counterexample

Dependency tree · two levels

4 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

Arora and Barak, Computational Complexity: A Modern Approach (standard reference, not scraped)
Lance Fortnow, Counting Complexity (standard reference, not scraped)