CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27

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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.
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Duplicating constraints does not change UNSAT

Statement refuted

Duplicating every constraint of a CSP instance the same number of times strictly increases its unsatisfaction fraction (Repeating constraints amplifies the gap).

Facts & Assumptions

Given: the alphabet $Σ : = {0, 1}$ and the constraint graph $G$ on the single variable $x_{1}$ whose two edges are loops, the first carrying the relation $R_{1} : = {(0, 0), (1, 1)}$ and the second the empty relation $R_{2} : = \emptyset$ .

[F1]

A loop is a constraint on a repeated variable, it is satisfied by a labeling $σ$ exactly when $(σ (x_{1}), σ (x_{1})) \in R$ , and the value of a system is the fraction of listed constraints that are satisfied, duplicated constraints counting with multiplicity (Constraint graph and labeling value).

[F2]

The unsatisfaction fraction of a system is the minimum over labelings of the fraction of the listed constraints that the labeling violates; a system all of whose constraints are unsatisfiable by every labeling has $UNSAT = 1$ , and a system all of whose constraints are universally satisfied has $UNSAT = 0$ (Repeating constraints amplifies the gap, Constraint graph and labeling value).

Counterexample

technique · direct

1.1F1F2algebra

The first loop carries $R_{1}$ , and $(0, 0), (1, 1) \in R_{1}$ , so every labeling of $x_{1}$ satisfies it; the second loop carries $R_{2} = \emptyset$ , so no labeling satisfies it. Hence the two-constraint list $G$ has ${UNSAT}_{σ} (G) = 1 / 2$ for every labeling $σ$ , and therefore $UNSAT (G) = 1 / 2$ .

1.2F1algebra

For $r \geq 1$ let $G^{(r)}$ list each of the two loops $r$ times. Every labeling satisfies exactly the $r$ copies of the first loop and violates exactly the $r$ copies of the second, so ${UNSAT}_{σ} (G^{(r)}) = r / (2 r) = 1 / 2$ for every $σ$ , and the list has $2 r$ constraints.

2.1step 1.1step 1.2F2∎

Thus $UNSAT (G^{(r)}) = 1 / 2 = UNSAT (G)$ for every $r \geq 1$ : the repetition changes neither the value nor the unsatisfaction fraction, and in particular it never strictly increases it, so the statement refuted above fails at this witness.

Depends on

Used by

Nothing in the library uses this result yet.

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

Irit Dinur, The PCP theorem by gap amplification, §1.5 and Definition 1.1, printed pp. 3-4. (standard reference, not scraped)
Arora and Barak, Computational Complexity: A Modern Approach, §18.5, printed p. 371. (standard reference, not scraped)