LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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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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Cloud plurality rounding

Statement

For any labeling of the cloud graph $G_{1}$ of a nonempty-edge constraint graph $G$ , decode each original vertex by its cloud's plurality label, using fixed tie breaking. Let $S$ count the ports disagreeing with that label, and let $U_{i n t}, U_{e x t}$ count violated internal equality and external edges. Then $U_{i n t} \geq \frac{h_{0}}{2} S, U_{G} \leq U_{e x t} + S, h_{0} = 7 / 10,$ where $U_{G}$ is the decoded violation count in $G$ .

Facts & Assumptions

Given: the objects and hypotheses in the statement above.

[F1]

Use the degree- $128$ graphs $H_{r}$ of the stated convention, whose unnormalized edge expansion is at least $h_{0} = 7 / 10$ when $r \geq 2$ . For a constraint graph $G$ as in the stated convention with $E \neq \emptyset$ , remove isolated vertices and replace each vertex of degree $r$ by a cloud of its $r$ incidence ports. Put $H_{r}$ inside that cloud, with equality on every edge. Keep one external edge for every original edge, joining its two designated ports and carrying its original relation. A loop's two ports are distinct. Call the resulting graph $G_{1}$ . On the $2 ∣ E ∣$ ports, add a copy of $H_{2 ∣ E ∣}$ with tautological relations, and at each port add $65$ ordinary tautological loops, i.e. $130$ loop slots. Call this $G_{2}$ . Its degree is $129 + 128 + 130 = 387$ ; $G_{1}$ has degree $129$ . The alphabet is unchanged. For an edgeless input, output the empty graph with value one; positive-degree and nonempty-size claims about $G_{1}, G_{2}$ are restricted to $E \neq \emptyset$ . Fix an alphabet ordering for plurality tie breaking and for decoding removed isolated vertices. (Constraint graph regularization).

Proof

1.1

In a cloud, every label class other than the chosen largest class has size at most half the cloud: a class larger than half would be the unique largest. Its outgoing boundary therefore has at least $h_{0}$ times its size in edges. Each such edge violates equality, and summing over these classes counts any edge at most twice. Sum also over clouds to obtain $2 U_{i n t} \geq h_{0} S$ . Empty classes contribute nothing; a singleton cloud has no disagreeing port.

2.1

Compare the labeling with the labeling constant at its decoded label on each cloud. Every originally violated constraint whose external copy was satisfied must have a changed port at one endpoint. Each changed port is incident to exactly one external edge, so at most $S$ external constraints can newly fail. The constant labeling's external violations equal $U_{G}$ , including original loops whose two incidence ports now carry the same label. This proves $U_{G} \leq U_{e x t} + S$ , also when $S = 0$ .

step 1.1algebra∎

Depends on

Constraint graph regularization

Used by

Regularization preserves value quantitatively Lemma

Dependency tree · two levels

3 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

Irit Dinur, The PCP theorem by gap amplification; §4 proof of Lemma4.1, PDF pages13–14. (standard reference, not scraped)