Alphabeta Math
Pipeline-generated
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.

✓ 22 results · all verified · 13 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 9 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Gap Amplification and Assignment Testing

1 · Prerequisites

2 · Summary

This page proves the constraint-graph gap-amplification step and develops the code and tester ingredients needed for assignment testing. It first fixes the published conventions for binary constraint graphs, their value and unsatisfaction fractions, and complete uniform gap-preserving CSP reductions, which compose with multiplied blowup and iterated gap maps. Degree reduction then replaces every vertex by one port per incidence, installs an expander cloud of equality edges inside each vertex and a tautological overlay at every port, and outputs a 387-regular graph whose unsatisfaction bounds that of the original through plurality decoding of the ports.

Powering turns a labeling into a table of local views on a radius t+⌈t⌉ ball. For every lazy walk pattern of length 2t+1 it uses paired incidence slots, duplicating walk slots so even self-reversing walks fit the ordinary undirected graph convention. The page proves the canonical lift of a satisfying labeling, plurality decoding of powered labelings, and the walk estimates: close endpoint laws for nearby lengths, plurality agreement along middle positions, controlled collisions of violated-edge positions, and an overlap-controlled union lower bound. Together they give UNSAT⁡(Gt)≥β0tmin⁡(UNSAT⁡(G),1/t), so that composing degree reduction with powering is a complete uniform linear-blowup gap-amplification step with gap map βtmin⁡(ε,c/t).

The assignment-testing half of the page records an explicit constant-rate constant-distance binary code, built by concatenating a Reed-Solomon outer code with a linear inner code produced by conditional expectation, and then defines assignment testers and their rejection ratios. It proves the BLR linearity test, the quadratic tensor test and the circuit-to-linear-quadratic reduction, builds an exponential-size constant-query base tester and an O(m+n) gate tester. The composition and proximity-amplification chain for a polynomial-size constant-query assignment tester is deferred: its present local composition loses a factor depending on the powered alphabet and does not justify the required raw-input-coordinate soundness. The page's false statement records that duplicating constraints alone cannot amplify a gap.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Complete uniform gap-preserving CSP reductions

Definition

Throughout this page a binary constraint graph is one in the convention of Constraint graph and labeling value: a finite ordinary undirected multigraph with paired incidence slots, a finite nonempty alphabet Σ, one relation Re⊆Σ2 per edge in a specified endpoint order, and explicit Boolean relation tables. For an edge set E≠∅ the value val⁡σ(G) is the fraction of ordinary edges satisfied by the labeling σ, and UNSAT⁡(G)=min⁡σ(1−val⁡σ(G)); an edgeless graph has value one and unsatisfiability zero, as in the published convention. The explicit encoding of G is the list of vertices, the paired incidence slots, the alphabet size and the relation tables; its bit length is the input size of the algorithms below.

Fix an alphabet-size parameter s≥1 and a degree parameter d≥1 for the input graphs. A complete uniform gap-preserving reduction with output alphabet Σ′, output degree bound d′, blowup C and gap map g consists of a deterministic algorithm R that, given the explicit encoding of a binary constraint graph G over an alphabet Σ with ∣Σ∣=s whose underlying graph has all vertex degrees at most d, outputs the explicit encoding of a binary constraint graph R(G) over the fixed alphabet Σ′ such that:

  1. Output shape. Arity stays two, all vertex degrees of R(G) are at most d′, and ∣E(R(G))∣≤C ∣E(G)∣,∣V(R(G))∣≤C ∣E(G)∣  whenever E(G)≠∅. An edgeless input is mapped to an edgeless output, which therefore has value one.
  2. Completeness. UNSAT⁡(G)=0 implies UNSAT⁡(R(G))=0.
  3. Gap preservation. UNSAT⁡(R(G))≥g(UNSAT⁡(G)) for every input G, where g:[0,1]→[0,1] is nondecreasing with g(0)=0 and g(ε)>0 for every ε>0.
  4. Uniformity. R runs in time polynomial in the bit length of the explicit input encoding, and the output is explicit. The alphabet Σ′, the numbers d′, C and the function g depend only on the fixed parameters s,d and on R itself, never on ∣V(G)∣ or ∣E(G)∣.

The definition asserts no existence statement: it records the interface in which the degree-reduction, powering and assignment-tester steps of this page are stated. Isolated vertices may be deleted from inputs and outputs without changing value or unsatisfiability, so the vertex bound in clause 1 is never needed for padded inputs; the bounded-degree clause and the fixed alphabet keep the explicit encoding length of the output within a constant multiple of ∣V(R(G))∣+∣E(R(G))∣ plus logarithmic vertex names. Loops count as ordinary edges with two incidences and a relation tested on the repeated label, exactly as published. Monotonicity of g is used only to compose gap maps, in Complete linear-blowup reductions compose, never to enlarge an input hypothesis.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Complete linear-blowup reductions compose

Statement

Let R1 be a complete uniform gap-preserving reduction with output alphabet Σ1, output degree bound d1, blowup C1 and gap map g1, defined for input alphabets of size s and input degrees at most d. Assume also that R1(G) has at least one edge whenever G has at least one edge. Let R2 be a complete uniform gap-preserving reduction with output alphabet Σ2, output degree bound d2, blowup C2 and gap map g2, defined for input alphabets of size at least ∣Σ1∣ and input degrees at most d1. Then the composition R2∘R1, which on input G first computes R1(G) and then R2(R1(G)), is a complete uniform gap-preserving reduction for input alphabets of size s and input degrees at most d, with output alphabet Σ2, output degree bound d2, blowup C1C2, gap map g2∘g1, and running time polynomial in the bit length of the explicit input encoding of G.

Facts & Assumptions

Given: complete uniform gap-preserving reductions R1,R2 with the parameters and compatibility hypotheses in the statement, including the assumption that R1 does not erase all edges of a nonempty input, and an input graph G over an alphabet of size s with all vertex degrees at most d.

[F1]

A complete uniform gap-preserving reduction R with output alphabet Σ′, output degree bound d′, blowup C and gap map g satisfies: ∣E(R(G))∣≤C∣E(G)∣ and ∣V(R(G))∣≤C∣E(G)∣ whenever E(G)≠∅; an edgeless input is mapped to an edgeless output; UNSAT⁡(G)=0 implies UNSAT⁡(R(G))=0; UNSAT⁡(R(G))≥g(UNSAT⁡(G)) with g nondecreasing, g(0)=0 and g(ε)>0 for ε>0; all degrees of R(G) are at most d′, the arity remains two, and R runs in polynomial time in the encoding length of its input (Complete uniform gap-preserving CSP reductions).

[A2]

The additional edge-preservation condition in the statement ensures that if G has an edge, then R1(G) is a nonempty input for the size clauses of R2.

Proof

technique · direct
1.1

If E(G)=∅ then R1(G) is edgeless by [F1], hence so is R2(R1(G)), as required for an edgeless input. If E(G)≠∅, [A2] gives E(R1(G))≠∅, so [F1] applied twice yields ∣E(R2(R1(G)))∣≤C2∣E(R1(G))∣≤C1C2∣E(G)∣ and ∣V(R2(R1(G)))∣≤C2∣E(R1(G))∣≤C1C2∣E(G)∣; all degrees of R1(G) are at most d1 and its alphabet is Σ1, so R2 is applicable to it.

F1A2algebra
1.2

If UNSAT⁡(G)=0 then two applications of the completeness clause of [F1] give UNSAT⁡(R1(G))=0 and then UNSAT⁡(R2(R1(G)))=0, so the composite is complete.

F1
1.3

For the gap map, two applications of the gap clause of [F1] give UNSAT⁡(R2(R1(G)))≥g2(UNSAT⁡(R1(G)))≥g2(g1(UNSAT⁡(G))), the second inequality because g2 is nondecreasing and UNSAT⁡(R1(G))≥g1(UNSAT⁡(G)); the composition g2∘g1 is nondecreasing, vanishes at 0 and is positive on (0,1], so it is an admissible gap map.

F1algebra
2.1

For uniformity, the encoding length of R1(G) is bounded by a polynomial in the encoding length of G: the alphabet Σ1 is fixed, all degrees are at most d1, and the vertex and edge counts obey the linear bounds of step 1.1; the running time of R1 on G and of R2 on R1(G) is polynomial in the respective encoding lengths by [F1], so with the shape, completeness and gap clauses of step 1.1, step 1.2 and step 1.3 the composite runs in polynomial time in the encoding length of G and is a complete uniform gap-preserving reduction with output alphabet Σ2, output degree bound d2, blowup C1C2 and gap map g2∘g1.

F1step 1.1step 1.2step 1.3algebra∎

Remarks

  • The compatibility hypotheses are not cosmetic: R2 is applied to a graph whose alphabet is Σ1 and whose degrees are bounded by d1, and step 1.1 is exactly where those two fixed parameters, together with the edge bound, keep the intermediate explicit encoding polynomially short. Without a bound on the intermediate encoding length the composition of two polynomial-time algorithms need not be polynomial time in the original input length.
  • The edge blowup multiplies and the gap maps compose in the order the reductions are applied; no constant is lost. Later items of this page use the lemma with R1 the degree-reduction map and R2 the powering map.
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Degree reduction by expander incidence clouds

Definition

Fix the alphabet Σ of Constraint graph and labeling value, a total ordering of Σ used for plurality tie breaking, and the reverse-paired degree-128 expander family Hr with unnormalized edge expansion at least h0=7/10 for r≥2 that is used in Constraint graph regularization.

Let G be a binary constraint graph over Σ in the published convention. The degree-reduction map Rdeg⁡ is the following deterministic construction, which is the published cloud-and-overlay preprocessing read as a map on explicit encodings.

  1. If E(G)=∅, output the empty graph over Σ, which has value one.
  2. Otherwise delete isolated vertices, and replace every vertex v of degree r by a cloud of its r incidence ports, so that each original edge contributes its two distinct ports, a loop contributing two distinct ports at its vertex. Put a copy of Hr inside the cloud, with equality relations on all its edges, and keep one external edge for each original edge, joining its two designated ports and carrying the same relation in the same endpoint order. The result is the cloud graph G1 of Constraint expander overlay; it has degree 129 and 2∣E(G)∣ vertices.
  3. On the 2∣E(G)∣ ports add a copy of H2∣E(G)∣ with tautological relations and 65 ordinary tautological loops at every port, i.e. 130 loop slots, to obtain the registered graph G2. By the published count, G2 is 387-regular with 2∣E(G)∣ vertices and 387∣E(G)∣ ordinary edges, its normalized adjacency satisfies α(G2)≤ρ2=(259+128ρ0)/387<1, and its alphabet is still Σ.

The decoding map D sends a labeling τ of G2 (equivalently of G1, on which it depends only through the ports) to the labeling Dτ of G that assigns to each original vertex the most frequent label among the ports of its cloud, breaking ties by the fixed ordering; deleted isolated vertices receive the first symbol of Σ. This is the plurality decoding of Constraint graph regularization.

Every step is a fixed function of the explicit input encoding: listing the Hr adjacency lists, copying 2∣E(G)∣ relation tables and adding 387∣E(G)∣ tautological slots takes time and output length polynomial in the input encoding length, and the output parameters 2∣E(G)∣, 387∣E(G)∣, 387, ρ2 depend only on the fixed family and on ∣E(G)∣. The map is the first of the two transformations composed in A complete uniform graph gap-amplification step; its quantitative unsatisfiability guarantee is stated and proved there, not here.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Cloud violations control distance to plurality labels

Statement

Let G be a binary constraint graph over the fixed alphabet Σ with E(G)≠∅, let G1 be its cloud graph as constructed by the degree-reduction map Degree reduction by expander incidence clouds, and let τ be any labeling of the ports of G1 with corresponding plurality decoding σ=Dτ. Write S for the number of ports whose label differs from the decoded label σ(v) of their original vertex v, and let Uint and Uext count the equality edges inside clouds and the external edges that are violated by τ. Then Uint≥720 S,UG≤Uext+S, where UG is the number of ordinary edges of G violated by the decoded labeling σ.

Facts & Assumptions

Given: a binary constraint graph G with E(G)≠∅ over the fixed alphabet, a labeling τ of its cloud graph G1, the decoded labeling σ=Dτ, and the counts S,Uint,Uext,UG above.

[F1]

For any labeling of the cloud graph G1 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 Uint,Uext count violated internal equality and external edges. Then Uint≥(h0/2)S and UG≤Uext+S with h0=7/10, where UG is the decoded violation count in G (Cloud plurality rounding).

[F2]

The cloud graph G1 of the degree-reduction map has one port per incidence, a copy of Hr with equality on every internal edge inside each cloud, one external edge per original edge joining the designated ports, and the same alphabet Σ; the fixed alphabet ordering is used both for the plurality choice of D and for the tie-breaking in the published rounding bound (Degree reduction by expander incidence clouds).

Proof

technique · direct
1.1

The cloud graph and the decoding used here are the ones of [F2], with the same alphabet ordering and the same equality and external relations. Substituting h0=7/10 into the first inequality of [F1] gives Uint≥(7/20)S.

F1F2algebra
1.2

If some cloud contains no port at all, then it contributes neither ports nor edges to the counts; the second inequality of [F1] is a statement about all clouds simultaneously and covers this case. It gives UG≤Uext+S, including S=0 and including the case in which the decoded labeling fails an external edge whose two ports carry labels constant on their clouds.

F1F2
2.1

Both displayed inequalities therefore hold for every labeling τ of the ports of G1, with the constants 7/20 and 1 read off from the published rounding argument; no additional hypothesis on G beyond E(G)≠∅ is used.

step 1.1step 1.2∎

Remarks

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Degree reduction preserves unsatisfaction quantitatively

Statement

Let G be a binary constraint graph over the fixed alphabet Σ with E(G)≠∅, and let G2=Rdeg⁡(G) be its degree-reduced graph as constructed by Degree reduction by expander incidence clouds. Then G2 is 387-regular over the same alphabet, has 2∣E(G)∣ vertices and exactly 387∣E(G)∣ ordinary edges, is produced from the explicit encoding of G by a deterministic algorithm in polynomial time, and with K=20/7 its unsatisfiability satisfies perfect completeness and the quantitative bound UNSAT⁡(G2)≥UNSAT⁡(G)387K=77740UNSAT⁡(G). Consequently Rdeg⁡, with the empty output on edgeless inputs, is a complete uniform gap-preserving reduction in the sense of Complete uniform gap-preserving CSP reductions with output alphabet Σ, output degree bound 387, blowup 387 and gap map g(ε)=ε/(387K).

Facts & Assumptions

Given: a binary constraint graph G over the fixed alphabet Σ with m=∣E(G)∣≥1, its cloud graph G1 and its degree-reduced graph G2=Rdeg⁡(G), and K=20/7.

[F1]

The full preprocessing graph G2 of a graph with m>0 edges has 2m vertices, degree 387 and 387m ordinary edges over the same alphabet; it has loops at every vertex and α(G2)≤ρ2<1. With K=20/7 and c=1/(129K), 129c387UNSAT⁡(G)≤UNSAT⁡(G2)≤UNSAT⁡(G)387. For every port labeling τ, UNSAT⁡τ(G2)=129387UNSAT⁡τ(G1) and UNSAT⁡Dτ(G)≤387KUNSAT⁡τ(G2). Construction and plurality decoding take polynomial time, and the edgeless convention has unsatisfiability zero (Constraint expander overlay).

[F3]

The degree-reduction map Rdeg⁡ is deterministic: on an edgeless input it outputs the empty graph over Σ, and otherwise it outputs the graph G2 of the cloud-and-overlay construction, listing the explicit expander adjacency lists, copying the 2m relation tables and adding the 387m slots of the overlay, in time polynomial in the encoding length of the input; the decoding map D is the fixed plurality decoding of the construction (Degree reduction by expander incidence clouds).

[F4]

A complete uniform gap-preserving reduction with output alphabet Σ′, output degree bound d′, blowup C and gap map g satisfies: ∣E(R(G))∣≤C∣E(G)∣ and ∣V(R(G))∣≤C∣E(G)∣ when E(G)≠∅; edgeless inputs go to edgeless outputs; UNSAT⁡(G)=0 implies UNSAT⁡(R(G))=0; UNSAT⁡(R(G))≥g(UNSAT⁡(G)) with g nondecreasing, g(0)=0 and g(ε)>0 for ε>0; and R runs in polynomial time (Complete uniform gap-preserving CSP reductions).

Proof

technique · direct
1.1

By [F3] the map is deterministic and polynomial time, and by [F1] its output is 387-regular over Σ with 2m vertices and 387m ordinary edges. For m≥1 these counts satisfy ∣E(G2)∣=387m≤387∣E(G)∣ and ∣V(G2)∣=2m≤387∣E(G)∣, the vertex bound because 2≤387.

F1F3algebra
1.2

Perfect completeness: if UNSAT⁡(G)=0 then the second inequality of [F1] gives UNSAT⁡(G2)≤UNSAT⁡(G)/387=0, and unsatisfiability is nonnegative, so UNSAT⁡(G2)=0. The edgeless case is the empty-output convention of [F3].

F1F3
1.3

For the quantitative bound, let τ be any labeling of G2, decoded to σ=Dτ on G. By [F1], UNSAT⁡Dτ(G)≤387KUNSAT⁡τ(G2), and by the definition of the minimum over labelings, UNSAT⁡(G)≤UNSAT⁡Dτ(G). Hence UNSAT⁡(G)≤387KUNSAT⁡τ(G2) for every τ, and minimizing the right side over τ gives UNSAT⁡(G)≤387KUNSAT⁡(G2),that is,UNSAT⁡(G2)≥UNSAT⁡(G)387K.

F1algebra
2.1

The gap map g(ε)=ε/(387K) is nondecreasing, satisfies g(0)=0 and is positive for ε>0, and together with steps 1.1, 1.2 and 1.3 it makes Rdeg⁡, with the empty output on edgeless inputs, a complete uniform gap-preserving reduction with output alphabet Σ (fixed), output degree bound 387, blowup 387 and that gap map, in the sense of [F4]. For the edgeless input both the input and the output have unsatisfiability zero, so the gap inequality holds trivially there.

F3F4step 1.1step 1.2step 1.3∎

Remarks

  • The constant is not improvable by this proof: the factor 387K=7740/7 is exactly the loss accumulated by the cloud rounding inequality UG≤Uext+(2/h0)Uint and the uniform edge rescaling from 129m to 387m slots in [F1]. Later items absorb it into their own constants, which is why A complete uniform graph gap-amplification step carries a smaller β.
  • Fact [F2] is declared because the published derivation of the decoding inequality in [F1] runs through the cloud rounding bound; the page records that bound as its own interface so that the code and tester branches can cite it.
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

Constraint graph powering with local-view labels

Definition

Let G be a binary constraint graph over the alphabet Σ in the convention of Constraint graph and labeling value, whose underlying graph is d-regular in the adjacency-slot convention of Regular multigraph and normalized adjacency, and let t≥1 be an integer. Put R:=t+⌈t⌉,L:=2t+1. A lazy step at a vertex v is the choice of one of the 2d options: one of d hold options (the walk stays at v) or one of the d slots at v (the walk traverses that slot). A lazy walk of length ℓ is a sequence of ℓ independent uniform lazy steps; its transition matrix is (I+M)/2 with M the normalized adjacency of the graph, so each step holds with probability 1/2 and otherwise traverses a uniformly chosen slot. Since the graph is regular, the uniform distribution on V is stationary, and the count of lazy walk patterns of length ℓ starting at v is (2d)ℓ.

Write Pℓ:={1,…,2d}ℓ for the set of lazy-walk patterns of length ℓ, a set of cardinality (2d)ℓ independent of the starting vertex. A pattern read from a vertex determines the sequence of visited vertices and the slots traversed. The local-view alphabet of the powering is Σt:=ΣPR, that is, a view is a function assigning a symbol of Σ to every length-R pattern. Fix the lexicographic order on patterns induced by the option order. For vertices x,y at graph distance at most R, let κx,y∈PR be the lexicographically first pattern read from x that ends at y; this set is nonempty because a path of length at most R can be padded by holds. Its cardinality is ∣Σt∣=∣Σ∣(2d)R≤∣Σ∣(2d)O(t).

The powered graph Gt. Its vertex set is V; the label of a vertex v is a view φv∈Σt. For every length-L pattern σ=(σ1,…,σL)∈PL, starting vertex v0∈V, and copy bit b∈{0,1}, create a directed incidence slot; reading σ from v0 gives the visited vertices v0,v1,…,vL, and the slot joins v0 to vL. Pair it with the slot for the reversed pattern read from vL and copy bit 1−b. This fixed-point-free pairing duplicates every walk slot, as required to represent the adjacency-slot power as an ordinary undirected multigraph even when a walk is fixed by reversal. The relation table in the orientation (v0,vL) is the one given below; the paired reverse orientation carries its transpose. Let J:={j:1≤j≤L, j−1≤R, L−j≤R} be the central window of positions, of size 2⌈t⌉+1, symmetric about the midpoint t+1. The edge relation in the orientation of the pattern σ from v0 to vL consists of all pairs (φ,ψ)∈Σt2 such that for every j∈J whose step σj is a move along a slot e=(vj−1,vj) of G, the pair (φ(κv0,vj−1), ψ(κvL,vj)) belongs to the relation of e in its orientation (vj−1,vj). The canonical patterns exist because j−1≤R and L−j≤R, so both endpoints lie within radius R of their respective view centres. Hold steps impose no condition.

Every vertex has exactly 2(2d)L incident slots, so Gt is dt-regular with dt=2(2d)L. The pairing gives ∣V∣(2d)L ordinary edges and the same number of relation tables, equivalently 2∣V∣(2d)L directed incidence slots. Sampling a uniform ordinary edge and orienting it by its unique copy-0 incidence slot is therefore equivalent to choosing a uniform start vertex and a uniform length-L pattern; its violation probability is UNSAT⁡φ(Gt). Since d and t are fixed, Σt is a fixed finite alphabet; each table is a subset of the fixed finite set Σt2, computable by the displayed rule, so the number of table entries is Od,t,∣Σ∣(∣V∣(2d)L), while endpoint names and paired-slot indices require Od,t(log⁡(∣V∣+2)) bits each. Thus the explicit bit encoding has length Od,t,∣Σ∣(∣V∣(2d)Llog⁡(∣V∣+2)). Enumerating the patterns, computing their endpoints and canonical coordinates, and writing the fixed-size relation tables takes time polynomial in the explicit input and output bit lengths.

Remarks

  • Convention bridge. This is the Arora-Barak §18.5.1 convention: views cover the ball of radius t+⌈t⌉, powered slots are length-(2t+1) walks, and the central window has 2⌈t⌉+1 positions. Dinur's §6 uses walks of length t with views of radius ⌈t/2⌉ and a window of t positions. The two parameterizations are translations of one another, but their numerical constants are not interchangeable; every later item on this page states its bounds in the present convention, and the lazy walk introduced here is the one used throughout.
  • The reversal pairing is consistent with the published endpoint convention: position j reverses to L+1−j∈J, the endpoint views swap, and each tested base relation is transposed. Thus the reverse-pattern table is the transpose of the original table, exactly as Constraint graph and labeling value requires. If a walk pattern is fixed by reversal, its central option is a hold and the mirrored move tests pair in transposed pairs; its table is therefore symmetric, so the two copy-bit incidences define one well-formed loop edge.
  • Each view records one canonical coordinate for each vertex of the radius-R ball; a middle-position constraint reads those coordinates for the two endpoints of its tested base edge. The canonical coordinate removes any dependence on the placement of holds, while plurality decoding still counts distinct walk patterns with multiplicity.
  • The role of this construction in the page is the powering step of A complete uniform graph gap-amplification step; the companion item Graph power and walk constraint records the underlying one-step power convention, in which a walk predicate is a conjunction of the original edge relations along the walk.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Canonical local views preserve perfect satisfiability

Statement

Let G be a d-regular binary constraint graph over Σ, let t≥1 and let Gt be its powered graph as in Constraint graph powering with local-view labels. If σ:V→Σ satisfies every edge of G, then the canonical lift σˉ defined by σˉ(v)(π):=σ(endpoint of the pattern π read from v),π∈PR, is a labeling of Gt that satisfies every slot of Gt; in particular val⁡(Gt)=1 whenever val⁡(G)=1. The same holds with repeated vertices, hold steps and loops, and no hypothesis on t beyond t≥1 is used.

Facts & Assumptions

Given: a d-regular binary constraint graph G over Σ, an integer t≥1, its powered graph Gt with view alphabet Σt=ΣPR, central window J and slot relations as in the powering definition, and a labeling σ of G that satisfies every edge.

[F1]

A view is a function φ:PR→Σ; a directed incidence slot of Gt is indexed by a length-L pattern read from a vertex v0 and a copy bit, visits v0,…,vL, joins v0 to vL, and its relation table is determined by the starting vertex and the pattern (with the base graph fixed), independently of the copy bit. That table accepts the pair of views (φ,ψ) exactly when for every j∈J whose step is a move along a slot e=(vj−1,vj), the pair (φ(κv0,vj−1),ψ(κvL,vj)) lies in the relation of e; the two canonical patterns exist because their endpoints lie within radius R of their view centres, and hold steps impose no condition (Constraint graph powering with local-view labels).

[F2]

For an edge e with endpoints u,w in the specified order, a labeling satisfies e exactly when (σ(u),σ(w))∈Re; loops are tested on the repeated label and a labeling satisfies G when it satisfies every ordinary edge (Constraint graph and labeling value).

Proof

technique · direct
1.1

For each v∈V and each pattern π∈PR the endpoint of π read from v is a well-defined vertex, so σˉ(v) is a function PR→Σ, i.e. an element of Σt; hence σˉ is a labeling of Gt.

F1given
2.1

Let a directed incidence slot of Gt be given by a pattern (σ1,…,σL) from v0, a copy bit, and visited vertices v0,…,vL, and let j∈J be such that σj is a move along a slot e=(vj−1,vj). The coordinate κv0,vj−1 ends at vj−1 and κvL,vj ends at vj by definition. Since the canonical lift labels every pattern by its endpoint, σˉ(v0)(κv0,vj−1)=σ(vj−1) and σˉ(vL)(κvL,vj)=σ(vj); the copy bit does not alter the relation or these coordinates.

F1step 1.1algebra
3.1

Since σ satisfies every edge of G, the pair (σ(vj−1),σ(vj)) lies in the relation of the slot e in its orientation (vj−1,vj), whether or not e is a loop; by step 2.1 the pair of views reads exactly this pair at position j, so the slot of Gt accepts (σˉ(v0),σˉ(vL)). Position j was an arbitrary member of the central window and the slot was arbitrary, so σˉ satisfies every slot of Gt, and therefore val⁡(Gt)=1 when val⁡(G)=1.

F1F2step 2.1∎

Remarks

  • The lift is a function of the base labeling and the explicit pattern list; it makes no choice. Repeated occurrences of one vertex in a pattern all receive the same symbol σ of that vertex, which is why consistency across the central overlap is automatic here.
  • The converse direction is not claimed: satisfying labelings of Gt need not be lifts. That gap is what the plurality decoding and collision analysis of the following items address.
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

Plurality decoding of powered local views

Definition

Let G be a binary constraint graph over the finite nonempty alphabet Σ whose underlying graph is d-regular in the adjacency-slot convention of Constraint graph and labeling value, let t≥1, and let Gt be its local-view powered graph with view alphabet Σt=ΣPR, pattern sets Pℓ={1,…,2d}ℓ, length L=2t+1 and central window J as in Constraint graph powering with local-view labels. Fix once and for all a total order on Σ, and write min⁡ below for the least element in that order.

Let φ:V→Σt be a labeling of Gt; its value φ(w) at a vertex w is the view at w. For vertices v,w with dist⁡(v,w)≤R, write κw,v for the canonical length-R pattern from w to v fixed in Constraint graph powering with local-view labels.

Claims. Let 1≤ℓ≤R and let π∈Pℓ be a lazy-walk pattern read from v that ends at w. Since dist⁡(v,w)≤ℓ≤R, the canonical pattern κw,v exists, and the view at w assigns it a symbol φ(w)(κw,v)∈Σ; we say that the view at w claims the value φ(w)(κw,v) for v via π. The claim depends on the endpoint w, not on the placement of holds in π.

Plurality decoding. For v∈V and a∈Σ put pv(a):=1(2d)t#{π∈Pt:the view at the endpoint of π read from v claims a for v}, the number of length-t patterns from v whose endpoint's view claims a for v, divided by the total number (2d)t of such patterns. Equivalently, pv is the law of the claimed value for v: if a pattern is drawn uniformly at random from Pt, the view at its endpoint claims a for v with probability pv(a). The plurality decoding of the powered labeling φ is the labeling φ^:V→Σ,φ^(v):=min⁡{a∈Σ: pv(a)=max⁡b∈Σpv(b)}, that is, the least symbol, in the fixed order on Σ, that is claimed for v with maximal probability. We call φ^(v) the decoded label of v and pv the opinion distribution of v.

Remarks

  • Patterns are counted with multiplicity. Distinct patterns with the same endpoint contribute separate votes, while repeated visits inside a single pattern do not create extra votes; no uniform vote over distinct centres is taken. This is Dinur's "popular opinion" [display (4) of §6] and the Arora-Barak "plurality assignment" of §18.5.1, both of which average the claim of the endpoint of a random walk of the decoding length, with multiplicities.
  • Tie breaking is part of the definition. The order on Σ is fixed once on the page, so the decoding is a function of the powered labeling and of the fixed explicit data of Gt: it makes no choice, and it is computable from the explicit encoding of Gt by counting patterns, since (2d)t is a constant once d and t are fixed.
  • The relation and decoding use the same canonical coordinate. The decoding consults a view at w only at κw,v, the same coordinate that a powered slot relation reads for the base vertex v at a middle position. Distinct length-t patterns with the same endpoint therefore contribute the same claim value but are still counted with their pattern multiplicity.
  • The decoding is defined for every labeling of Gt, canonical lifts included: for the canonical lift σˉ of a base labeling σ, every pattern from v ends at a vertex w whose view claims σ(v) for v, so pv is concentrated on σ(v) and σˉ^=σ.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Nearby lazy-walk lengths have close endpoint and claim laws

Statement

Let G be a binary constraint graph over Σ whose underlying graph is d-regular in the adjacency-slot convention of Constraint graph and labeling value, and use the lazy-walk convention of Constraint graph powering with local-view labels: a lazy step at a vertex chooses uniformly among 2d options, the d hold options and the d slots at that vertex, and a lazy-walk pattern of length ℓ is drawn uniformly from Pℓ. Put C0:=12π<12. For ℓ≥1 let Bℓ be the number of non-hold options of a uniformly random lazy-walk pattern of length ℓ read from a vertex v; Bℓ∼Bin⁡(ℓ,12) for every v. Then:

  1. Binomial closeness. For all integers m,m′≥1 with ∣m−m′∣≤min⁡(m,m′), TV⁡(Bin⁡(m,12),Bin⁡(m′,12))≤C0 ∣m−m′∣min⁡(m,m′).
  2. Transfer to walk statistics. A lazy-walk pattern determines its endpoint from its sequence of non-hold options. For any fixed labeling φ of the powered graph Gt with its fixed view radius R=t+⌈t⌉ and lengths 1≤ℓ,ℓ′≤R, the view at that endpoint claims for its start the value at the canonical coordinate specified in Plurality decoding of powered local views. Thus the claimed value, like the endpoint, is a function of the non-hold option sequence alone. Let Xv,ℓ denote the value claimed for v by the view at the endpoint of a uniformly random lazy-walk pattern of length ℓ from v. Then for m=min⁡(ℓ,ℓ′) and ∣ℓ−ℓ′∣≤m, TV⁡(Xv,ℓ,Xv,ℓ′)≤C0 ∣ℓ−ℓ′∣m, The same endpoint-law bound holds for arbitrary positive lengths satisfying the displayed window condition, without a radius restriction.
  3. The window used by the powering analysis. If t≥4 and ∣ℓ−t∣≤t/(8C0∣Σ∣), then TV⁡(Xv,t,Xv,ℓ)≤1/(4∣Σ∣), uniformly in the start vertex v and in the labeling of Gt.

Facts & Assumptions

Given: a d-regular binary constraint graph G in the stated convention, its lazy-walk patterns, a vertex v, lengths ℓ,ℓ′≥1, and the statistic Xv,ℓ of the claimed value defined above.

[F1]

A lazy step at v chooses uniformly among the 2d options consisting of d hold options and the d slots at v; the steps of a lazy-walk pattern are independent, the pattern set Pℓ={1,…,2d}ℓ has (2d)ℓ elements, the transition matrix is (I+M)/2, the uniform distribution is stationary, and reversal of a pattern interchanges its start and endpoint (Constraint graph powering with local-view labels).

[F2]

For any 1≤ℓ≤R and pattern π∈Pℓ from v ending at w, the view at w claims the value φ(w)(κw,v) for v, where κw,v is the fixed canonical pattern from w to v; Xv,ℓ is this claimed value for a uniformly random π (Plurality decoding of powered local views).

[L1]

For an=(2nn)/4n one has πn an→1 as n→∞ (The central binomial coefficient is asymptotic to 4^n divided by the square root of pi n).

Proof

technique · direct
1.1

A pattern records at each coordinate whether it is a hold or a move, together with the chosen option within that type. For each k, there are (ℓk)dkdℓ−k=(ℓk)dℓ patterns with exactly k moves, so Bℓ∼Bin⁡(ℓ,12). Conditional on Bℓ=k, the sequence of the k move slots is uniform among the dk slot sequences; hold-option identities and the set of hold positions do not affect it. In particular Bℓ does not depend on v.

F1given
1.2

Write Pm(k)=(mk)2−m and bm=max⁡kPm(k), and let ar=(2rr)/4r. Then b2r=ar and b2r+1=ar(2r+1)/(2r+2), while ar+1/ar=(2r+1)/(2r+2)<1; hence bm is nonincreasing in m. Moreover arr is increasing because (ar+1r+1)/(arr)=(2r+1)2/(4r(r+1))>1, so [L1] gives ar≤1/πr. For even m=2r this yields bm≤2/(πm). For odd m=2r+1 with r≥1, it gives bm≤(2r+1)/((2r+2)πr)≤2/(π(2r+1)), since squaring the last inequality reduces to 4r2+2r−1≥0; and b1=1/2≤2/π. Thus in every case bm≤2/(πm)=2C0/m.

L1algebra
2.1

The endpoint of a pattern is determined by its sequence of non-hold options, and the claimed value of [F2] is the value of the fixed canonical coordinate from that endpoint back to v. Hence both the endpoint and the claimed value are functions of the non-hold option sequence alone.

F1F2step 1.1
2.2

Pascal's rule gives Pm+1(k)=12(Pm(k)+Pm(k−1)), so ∑k∣Pm+1(k)−Pm(k)∣=12∑k∣Pm(k)−Pm(k−1)∣. The sequence k↦Pm(k) rises to its maximum and then falls, so its total variation is at most 2bm and hence TV⁡(Bin⁡(m,12),Bin⁡(m+1,12))≤12bm.

step 1.2algebra
3.1

For m′≥m the triangle inequality for total variation and step 2.2 give TV⁡(Bin⁡(m,12),Bin⁡(m′,12))≤12∑i=0m′−m−1bm+i≤m′−m2bm≤C0m′−mm by the monotonicity and the bound of step 1.2; the case m′<m is the same with the roles exchanged, which proves claim 1 with min⁡(m,m′) in the denominator.

step 1.2step 2.2algebra
4.1

Let 1≤ℓ,ℓ′≤R with m=min⁡(ℓ,ℓ′) and ∣ℓ−ℓ′∣≤m be given, and couple Bℓ and Bℓ′ maximally, so that they differ with probability TV⁡(Bin⁡(ℓ,12),Bin⁡(ℓ′,12)). Conditionally on Bℓ=Bℓ′=k, use the same uniform slot sequence of length k from v in both experiments, which is legitimate by the conditional uniformity of step 1.1 and the fact that the endpoints and claimed values are functions of that sequence by step 2.1. This couples Xv,ℓ and Xv,ℓ′ to agree except on an event of probability at most the binomial total variation, and endpoints are coupled in the same way; claim 2 follows from step 3.1.

step 1.1step 2.1step 3.1
5.1

If t≥4 and ∣ℓ−t∣≤t/(8C0∣Σ∣), then the window constant 1/(8C0∣Σ∣)<1 gives ∣ℓ−t∣≤t and hence m=min⁡(ℓ,t)≥t−t≥t/2. Also 1/(8C0∣Σ∣)<1/2 for ∣Σ∣≥1, so ∣ℓ−t∣≤t/2≤m and 1≤ℓ≤t+t≤R, so claim 2 applies. It gives TV⁡(Xv,ℓ,Xv,t)≤C0∣ℓ−t∣/m≤2/(8∣Σ∣)≤1/(4∣Σ∣). This proves claim 3 uniformly in v and in the powered labeling.

step 4.1algebra∎

Remarks

  • The lemma is stated for the binomial law of the number of moves rather than for the non-lazy walk of a fixed length, because the lazy convention of Constraint graph powering with local-view labels makes the hold positions independent of the moves; this couples endpoints and the canonical-coordinate claims using the same move sequence. Dinur's Lemma 6.4 proves the analogous binomial weight-ratio estimate for the non-lazy distribution with p=1−1/d, and Arora-Barak's printed p. 374 uses the statistical-distance form with the constant 10δ for a window of size δt; the constant C0=1/2π above is the exact constant supplied by the central binomial asymptotic.
  • No hypothesis on the graph beyond d-regularity is used: the lemma compares laws of walks of different lengths on the same graph and involves neither the spectral gap nor the alphabet. The alphabet enters only through the final constant 1/(4∣Σ∣) of claim 3, which fixes the admissible window width.
  • The transfer claim is stated for the claimed-value statistic because that is the consumer's need in Plurality opinions agree with local views in middle positions; the endpoint version is the special case in which the statistic forgets the endpoint's view coordinate.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Plurality opinions agree with local views in middle positions

Statement

Let G be a binary constraint graph over Σ whose underlying graph is d-regular in the adjacency-slot convention of Constraint graph and labeling value, let t≥4, let Gt be its powered graph with walk length L=2t+1, central window J and view alphabet Σt, and let φ:V→Σt be any labeling of Gt with plurality decoding φ^ (Plurality decoding of powered local views). Write c:=18C0∣Σ∣,C0=12π, so that c<1, and let Jc:={j∈J: ∣j−1−t∣≤ct} be the corresponding sub-window of the central window; it is nonempty, since t+1∈Jc.

Draw a uniformly random ordinary edge of Gt and orient it by its unique copy-0 incidence slot; equivalently, choose a uniformly random start vertex v0∈V and a uniformly random lazy-walk pattern σ=(σ1,…,σL)∈PL, giving the visited vertices v0,…,vL. Then for every j∈Jc and every slot s of G from u to u′ (every one of the d slot options at u, loops included, and no hold option), Pr⁡[φ(v0)(κv0,u)=φ^(u)  and  φ(vL)(κvL,u′)=φ^(u′) ∣ the j-th lazy step of σ is the option s] ≥ 14∣Σ∣2, where the two κ coordinates are the canonical patterns specified by the slot relation of Constraint graph powering with local-view labels. In words: whenever a powered walk traverses a fixed slot at a middle position of the window Jc, the two endpoint views report the decoded plurality labels of that slot's two endpoints with probability bounded below by a positive constant depending only on ∣Σ∣. The bound is uniform in the slot, in the position, in the graph, in t and in the powered labeling.

Facts & Assumptions

Given: a d-regular binary constraint graph G over Σ, an integer t≥4, the powered graph Gt with parameters R,L,J,Σt, a labeling φ of Gt with decoding φ^, a position j∈Jc and a slot s of G from u to u′.

[F1]

Under the sampling convention of Constraint graph powering with local-view labels, a uniformly random powered edge oriented by its unique copy-0 incidence slot is a uniformly random start vertex and length-L pattern; its step options are independent and uniform, and reversal pairs its copy-0 incidence with the copy-1 incidence of the reversed pattern (Constraint graph powering with local-view labels).

[F2]

For any 1≤ℓ≤R and uniformly random lazy-walk pattern π of length ℓ from a vertex x, let Xx,ℓ be the value claimed for x by the view at the endpoint of π, namely φ(y)(κy,x) where y is the endpoint and κy,x is the canonical pattern from y to x; the opinion distribution of the decoding is px(a)=Pr⁡[Xx,t=a], and φ^(x) maximises px, so px(φ^(x))≥1/∣Σ∣ (Plurality decoding of powered local views).

[L1]

If 1≤ℓ≤R, m=min⁡(ℓ,t), ∣ℓ−t∣≤m and ∣ℓ−t∣≤t, then TV⁡(Xx,t,Xx,ℓ)≤C0∣ℓ−t∣/m; consequently TV⁡(Xx,t,Xx,ℓ)≤1/(4∣Σ∣) whenever ∣ℓ−t∣≤ct, uniformly in the start vertex x and in the labeling φ (Nearby lazy-walk lengths have close endpoint and claim laws).

Proof

technique · direct
1.1

Condition on the j-th lazy step of the sampled representative walk being the option s at vj−1=u; this forces vj=u′. The coordinates of σ other than the j-th are still independent uniform options, the constraint links only the prefix coordinates 1,…,j−1 through the requirement that the prefix ends at u, and it does not involve the suffix coordinates j+1,…,L. Hence, conditionally, the suffix (σj+1,…,σL) read from u′ is a uniformly random lazy-walk pattern of length L−j, the prefix is a uniformly random pattern of length j−1 from v0 ending at u, the two are independent, and v0 is the start of that prefix. Reversal is a bijection from patterns of length j−1 ending at u to patterns of length j−1 starting at u, and preserves the uniform law on each such set, so the reversed prefix is a uniformly random lazy-walk pattern of length j−1 from u. For j∈Jc we have j−1≥t−ct≥1, L−j≥t+1−ct−1≥1, and both lengths differ from t by at most ct.

F1givenalgebra
2.1

By the powering definition, the relation reads the canonical coordinates φ(v0)(κv0,u) and φ(vL)(κvL,u′). The reversed prefix from u ends at v0, so the first coordinate has the law of Xu,j−1 by [F2]; the suffix from u′ ends at vL, so the second has the law of Xu′,L−j. The prefix and suffix patterns are independent under step 1.1, so the two claimed values are independent.

F1F2step 1.1
2.2

By [F2] the decoding satisfies pu(φ^(u))≥1/∣Σ∣, and by [L1], applied with ℓ=j−1 and x=u, the law Xu,j−1 is within total variation 1/(4∣Σ∣) of Xu,t because 1≤j−1≤R and ∣j−1−t∣≤ct; hence Pr⁡[Xu,j−1=φ^(u)]≥pu(φ^(u))−1/(4∣Σ∣)≥3/(4∣Σ∣)≥1/(2∣Σ∣). The same computation with ℓ=L−j and x=u′ gives Pr⁡[Xu′,L−j=φ^(u′)]≥1/(2∣Σ∣), since 1≤L−j≤R and ∣L−j−t∣=∣j−1−t∣≤ct.

F2L1step 1.1algebra
3.1

Multiplying the two conditional probabilities of step 2.2 and using the conditional independence of step 2.1 gives the bound 1/(4∣Σ∣2) for the event that both endpoint views report the decoded labels of u and u′. This holds for every j∈Jc and every slot s of G, with constants depending only on ∣Σ∣ and not on the graph, on t, on the position or on the powered labeling, and it covers loops through u=u′.

step 2.1step 2.2algebra∎

Remarks

  • Both endpoints are needed and the tested position is central. The slot relation of Constraint graph powering with local-view labels tests at position j the canonical coordinate at the view on the first vertex v0 for u and the canonical coordinate at the view on the last vertex vL for u′; that is why the argument conditions on the walk through the two endpoints of the traversed slot and not on a single random walk. The position t+1 always belongs to Jc.
  • The sub-window is a genuine restriction. Item [L1] loses only 1/(4∣Σ∣) of probability over lengths t±ct, so the constant survives; over the whole central window of width ≍t the loss is a positive constant and the argument would fail for large alphabets. The promise of A complete uniform graph gap-amplification step is unaffected, because Jc still has Θ(t) positions and every slot violation detected at a position of Jc⊆J is a violation of the powered slot.
  • Numerical form of the window. By definition of C0, c=1/(8C0∣Σ∣)=π/32/∣Σ∣<1, so the window is centred at t+1 and is nonempty for every t≥4.
  • The claim is stated conditionally on the traversed option rather than unconditionally, because the consumer Powering amplifies a small unsatisfaction gap must multiply it by the probability that a stationary lazy walk traverses a given violated slot; that probability is computed there.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Violated-edge positions have controlled collisions

Statement

Let G be a binary constraint graph over Σ in the convention of Constraint graph and labeling value whose underlying graph is a finite d-regular adjacency-slot multigraph on n≥1 vertices with normalized adjacency M and normalized second eigenvalue bound α=∥M∣1⊥∥<1 (for n=1 put α:=0), as in Regular multigraph and normalized adjacency. Let F be a set of ordinary edges of G (loops allowed), put ε:=∣F∣/∣E∣, let k≥1, and consider a uniformly random lazy-walk pattern of length k read from a uniformly random start vertex, in the lazy-walk convention of Constraint graph powering with local-view labels. Let Ai be the event that the i-th lazy step of this pattern traverses an edge of F. Then ∑1≤i<j≤kPr⁡[Ai∩Aj] ≤ d2(k2ε2+kε1−α). Consequently, on the small-gap range kε≤c for an absolute constant c, ∑1≤i<j≤kPr⁡[Ai∩Aj] ≤ d2(c+11−α)kε = O(kε), the constant depending only on d, α and c. Both bounds are uniform in F; for ε=0 both sides vanish.

Facts & Assumptions

Given: a d-regular binary constraint graph G with M,α as above, a set F of ordinary edges, including possible loops, with ε=∣F∣/∣E∣, an integer k≥1, and the events A1,…,Ak of the random lazy-walk pattern.

[F1]

A lazy step at a vertex chooses uniformly among the 2d options consisting of the d hold options and the d slots at that vertex, the steps are independent, and the transition matrix of one lazy step is P=(I+M)/2, for which the uniform distribution on V is stationary; a lazy-walk pattern of length k read from a start vertex is a uniformly random element of Pk (Constraint graph powering with local-view labels).

[F2]

The underlying graph has nd slots and ∣E∣=nd/2 ordinary edges, uniform directed-slot sampling induces the uniform distribution on the ordinary edges, and each vertex has exactly d outgoing slots; a nonloop ordinary edge contributes one slot at each of its two endpoints, and loops have two slots at the same vertex (Regular multigraph and normalized adjacency).

[F3]

Any ordinary edge of G is a relation Re⊆Σ2 in a specified endpoint order; fractions of satisfied or violated edges are computed with respect to the ordinary edges (Constraint graph and labeling value).

[L1]

For any initial probability vector p and integer t≥0, ∥Mtp−u∥2≤αt∥p−u∥2 with u=1/n, in the ordinary Euclidean norm (Expander walk contraction).

[L2]

For vectors u,v in a real or complex inner product space, ∣⟨u,v⟩∣≤∥u∥∥v∥ (Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors).

Proof

technique · direct
1.1

By [F1] the start vertex is uniform and every lazy step preserves the uniform law, so the position before the i-th step is uniform for every i, and Pr⁡[Ai]=12⋅2∣F∣nd=ε/2: with probability 12 the step is a hold, and otherwise it uses one of the nd directed slots with uniform marginal law, of which 2∣F∣ belong to F, by [F2].

F1F2F3algebra
2.1

If F=∅, every event Ai is empty and both bounds in the statement are zero, so assume ∣F∣>0. Fix i<j and let x∈Rn be the law of the position after step i conditioned on Ai, and let f∈Rn have f(w)=Pr⁡[a lazy step at w traverses an edge of F], so that Pr⁡[Aj∣Ai]=⟨Pj−i−1x,f⟩ in the ordinary Euclidean inner product. In the stationary walk each of the 2∣F∣ slots of F is traversed with the same probability, so the traversed slot under Ai is equally likely to be any of them. Thus xv≤dF(v)/(2∣F∣)≤d/(2∣F∣), with dF the number of slots of F at v. Since ∑vxv=1, ∥x−u∥22=∑vxv2−1/n≤d/(2∣F∣). Also f(w)=dF(w)/(2d)≤12 and ∑wf(w)=∣F∣/d, so in the same Euclidean norm ∥f∥22=∑wf(w)2≤12∑wf(w)=∣F∣/(2d). Consequently ∥x−u∥2∥f∥2≤1/2.

F1F2step 1.1algebra
3.1

Since P=(I+M)/2 and M leaves the mean-zero subspace invariant with operator norm α by [L1], expanding Ph=2−h∑t=0h(ht)Mt gives ∥Ph(x−u)∥2≤(1+α2)h∥x−u∥2 for every h≥0; here 1+α2<1 and (1−1+α2)−1=2/(1−α). Splitting ⟨Phx,f⟩=⟨u,f⟩+⟨Ph(x−u),f⟩ and applying [L2] in the ordinary Euclidean norm with step 2.1 yields Pr⁡[Aj∣Ai]≤ε/2+12(1+α2)j−i−1≤ε/2+d/2 (1+α2)j−i−1, since d≥1.

L1L2step 2.1algebra
4.1

Multiplying by Pr⁡[Ai]=ε/2 and summing over i<j gives ∑i<jPr⁡[Ai∩Aj]≤k22⋅ε24+ε2d2∑i<j(1+α2)j−i−1≤k2ε28+kε2d2⋅21−α, which is at most d/2 (k2ε2+kε/(1−α)) because d/2≥1/8 for d≥1. For kε≤c the term k2ε2 is at most ckε, so the sum is at most d/2 (c+1/(1−α)) kε, and for ε=0 all the events are empty and both displays vanish.

step 3.1algebra∎

Remarks

  • What is counted. Ai is the event that the i-th lazy step traverses an edge of F (a move, never a hold option); the collision estimate therefore also bounds the overlaps of the smaller events Bj,f of Powering amplifies a small unsatisfaction gap, which require in addition that the two endpoint views report the decoded labels of the edge. Loop edges are allowed in F: a loop has two slots at its vertex, so ∑vdF(v)=2∣F∣ and the incidence bounds of step 2.1 remain correct, and a loop step keeps the walk at its vertex while still testing the relation on the two claims of the two endpoint views.
  • Where the small-gap hypothesis enters. The term k2ε2 is dominated by kε exactly when kε is bounded, which is the range ε=O(1/t) of the powering analysis; the other term d/2 kε/(1−α) is the spectral contribution and is already O(kε) for fixed spectral gap. The dependence on the spectral gap is through 1/(1−α) only, and not through any power of n.
  • The lazy convention halves the first moment but leaves the collision structure intact: the ratio Pr⁡[Aj∣Ai]≲ε/2+αLj−i−1 has the same shape as the walk-return bound of the published expander items for the non-lazy walk, with αL=(1+α)/2; the argument above re-derives it for P because the published contraction lemma is stated for M.
  • The bound is uniform in F: no lower bound on ∣F∣ is used beyond ∣F∣≥1 in the case ε>0, and the case F=∅ is the vanishing case ε=0.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Overlap control gives a union lower bound

Statement

Let B1,…,Bm be finitely many events in a probability space, put S:=∑i=1mPr⁡[Bi], and suppose that for a real number C≥0 ∑1≤i<j≤mPr⁡[Bi∩Bj] ≤ C S. Then Pr⁡[⋃i=1mBi] ≥ S1+2C. Both quantities vanish when S=0; no hypothesis is imposed on the individual probabilities beyond S<∞, and the bound is uniform over all finite families with the stated overlap ratio.

Facts & Assumptions

Given: events B1,…,Bm on a probability space, the sum S=∑iPr⁡[Bi], and a real C≥0 with ∑i<jPr⁡[Bi∩Bj]≤CS.

[L1]

For vectors u,v in a real or complex inner product space, ∣⟨u,v⟩∣≤∥u∥∥v∥; applied to the indicator functions of two events in L2 of a finite probability space this is the inequality E[∣XY∣]≤E[X2] E[Y2] for random variables, and for a nonnegative integer-valued N it gives (EN)2≤Pr⁡[N>0] E[N2], since N=0 off the event {N>0} (Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors).

Proof

technique · direct
1.1

Let N:=∑i=1m1Bi count the events that occur. Then N≥0 is integer valued, with EN=∑iPr⁡[Bi]=S by linearity of expectation, and {N>0}=⋃iBi.

givenalgebra
2.1

Expanding the square, N2=∑i1Bi+2∑i<j1Bi∩Bj, so taking expectations and using the hypothesis gives EN2=S+2∑i<jPr⁡[Bi∩Bj]≤S+2CS=(1+2C)S, a finite bound.

step 1.1algebra
3.1

If S=0 then EN=0 with N≥0, so every Pr⁡[Bi]=0, N=0 almost surely, and both sides of the claimed inequality are zero; assume S>0 from now on. Applying [L1] to N and to the indicator of {N>0} gives S2=(EN)2≤Pr⁡[N>0] EN2≤Pr⁡[N>0] (1+2C)S. Dividing by the positive number (1+2C)S gives Pr⁡[⋃iBi]=Pr⁡[N>0]≥S/(1+2C), which is the claim.

step 1.1step 2.1L1algebra∎

Remarks

  • The hypothesis is a ratio condition, not a smallness condition on the intersections separately: the bound is useful exactly when C is uniformly bounded, and then it loses only the factor 1+2C relative to the first moment S.
  • The Arora-Barak form of the same estimate (Claim 18.34) counts elements of finite sets, makes 2C copies of each element and reduces to inclusion-exclusion, with the weaker constant 14 and a hypothesis on the diameter of the set system; the probabilistic second-moment computation above is the convention of this page, and it applies directly to the events Bj,f of Powering amplifies a small unsatisfaction gap, whose pair overlaps are controlled by Violated-edge positions have controlled collisions.
  • The constant is sharp already for two disjoint events: then C=0, S=Pr⁡[B1]+Pr⁡[B2] and the union has probability exactly S.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Powering preserves perfect satisfiability

Statement

Let G be a binary constraint graph over Σ with E(G)≠∅ whose underlying graph is d-regular in the adjacency-slot convention, let t≥1, and let Gt be its local-view powered graph (Constraint graph powering with local-view labels). If val⁡(G)=1, then val⁡(Gt)=1; equivalently UNSAT⁡(G)=0 implies UNSAT⁡(Gt)=0. This is perfect completeness; the quantitative lower bound for positive gaps is given, under its additional hypotheses, by Powering amplifies a small unsatisfaction gap. Perfect completeness does not assert UNSAT⁡(Gt)≥UNSAT⁡(G).

Facts & Assumptions

Given: a d-regular binary constraint graph G over Σ with at least one edge, an integer t≥1, its powered graph Gt, and the assertion val⁡(G)=1.

[F1]

For a labeling σ of G and E(G)≠∅, val⁡σ(G) is the fraction of ordinary edges of G satisfied by σ; hence val⁡σ(G)=1 holds exactly when σ satisfies every edge of G (Constraint graph and labeling value).

[F2]

If σ satisfies every edge of G, its canonical lift σˉ(v)(π):=σ(endpoint of the pattern π read from v) is a labeling of Gt that satisfies every slot of Gt; consequently val⁡(Gt)=1 whenever val⁡(G)=1 (Canonical local views preserve perfect satisfiability).

[F3]

val⁡(Gt)=max⁡φval⁡φ(Gt) where the maximum runs over all labelings of Gt, and Gt has at least one slot because G has at least one slot (Constraint graph powering with local-view labels, Constraint graph and labeling value).

Proof

technique · direct
1.1

Since val⁡(G)=1 and G is nonempty, there is a labeling σ of G with val⁡σ(G)=1, and by [F1] such a σ satisfies every edge of G.

F1F3given
2.1

By [F2] applied to the labeling of step 1.1, the canonical lift σˉ satisfies every slot of Gt, so val⁡σˉ(Gt)=1 and hence val⁡(Gt)≥1; since val⁡ is a maximum of fractions, val⁡(Gt)=1 and UNSAT⁡(Gt)=1−val⁡(Gt)=0.

F2F3step 1.1∎

Remarks

  • The gap need not be monotone under powering: take one vertex with one ordinary loop whose relation is empty, so d=2 and UNSAT⁡(G)=1. At t=1 the walk length is three and J={1,2,3}. Any move tests the empty relation and fails, while the all-hold patterns pass vacuously. Each step holds with probability 1/2, so every powered labeling has value 1/8 and UNSAT⁡(G1)=7/8.
  • Nothing is assumed about t beyond t≥1, and no hypothesis on the powered labeling is needed: the lift is built from σ with no choice, so the statement is itself choice-free.
  • The edgeless case is deliberately excluded here and is instead governed by the convention that an edgeless graph has value one and unsatisfiability zero (Constraint graph and labeling value); the degree-reduction map of Degree reduction by expander incidence clouds sends edgeless inputs to edgeless outputs, for which the powered graph is again empty.
  • The converse also holds for perfect satisfiability, even though a satisfying powered labeling need not be a canonical lift. If φ satisfies Gt, put a(v):=φ(v)(κv,v). For any oriented base edge e=(u,w), consider the length-2t+1 pattern that holds t times at u, traverses e, then holds t times at w. Its central position t+1 belongs to J, and its test is exactly (a(u),a(w))∈Re. Thus a satisfies every base edge, including loops. The quantitative estimate Powering amplifies a small unsatisfaction gap concerns arbitrary, possibly imperfect powered labelings.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Powering amplifies a small unsatisfaction gap

Statement

Let d≥1, let Σ be a finite alphabet with ∣Σ∣≥2, and let α0<1 be a constant. Then there are a constant β>0 depending only on d,∣Σ∣,α0, and an integer t0 with the same dependence, such that for every t≥t0, every binary constraint graph G over Σ whose underlying graph is d-regular in the adjacency-slot convention with E(G)≠∅ and normalized second eigenvalue bound α≤α0, and every labeling φ of the powered graph Gt of Constraint graph powering with local-view labels, UNSAT⁡φ(Gt) ≥ βt min⁡(UNSAT⁡(G), 1/t). In particular, if 0<ε≤c/t for a constant c then UNSAT⁡φ(Gt)≥β′t ε, and for larger gaps the lower bound saturates at β/t. No parity hypothesis on t is needed in this parameterization: the powered slots have length 2t+1 and the analysed window is symmetric about the midpoint t+1. The bound is uniform in the labeling φ.

Facts & Assumptions

Given: integers d≥1 and t≥max⁡{4,⌈(8C0∣Σ∣)2⌉}; a finite alphabet Σ with ∣Σ∣≥2; a d-regular binary constraint graph G over Σ with E(G)≠∅ and α≤α0<1; a labeling φ of the powered graph Gt.

[F1]

For a labeling of G, UNSAT⁡σ(G)=1−val⁡σ(G) is one minus the fraction of ordinary edges satisfied, and UNSAT⁡(G)=min⁡σUNSAT⁡σ(G); for a labeling of Gt, UNSAT⁡φ(Gt) is the fraction of ordinary edges violated, equivalently the violation probability for a uniformly random edge oriented by its unique copy-0 incidence slot (Constraint graph and labeling value, Constraint graph powering with local-view labels).

[F2]

The plurality decoding φ^ of φ satisfies pv(φ^(v))≥1/∣Σ∣ for every v, where pv is the law of the value claimed for v by the view at the endpoint of a uniformly random lazy-walk pattern of length t from v (Plurality decoding of powered local views).

[F3]

Draw a uniformly random powered edge oriented by its unique copy-0 incidence slot, equivalently a uniform start v0∈V and a uniform lazy-walk pattern σ∈P2t+1. With C0=1/2π and cwin:=1/(8C0∣Σ∣), for every j with ∣j−1−t∣≤cwint and every slot s of G from u to u′, the probability that both φ(v0)(κv0,u)=φ^(u) and φ(vL)(κvL,u′)=φ^(u′), conditioned on the j-th lazy step of σ being the option s, is at least 1/(4∣Σ∣2) (Plurality opinions agree with local views in middle positions).

[F4]

For any set F of ordinary edges of G with εF=∣F∣/∣E(G)∣, with Ai the event that the i-th step of a uniformly random lazy-walk pattern of length k from a uniform start vertex traverses an edge of F, one has ∑1≤i<j≤kPr⁡[Ai∩Aj]≤d/2 (k2εF2+kεF/(1−α)) (Violated-edge positions have controlled collisions).

[F5]

If finitely many events satisfy ∑i<jPr⁡[Bi∩Bj]≤C∑iPr⁡[Bi] for some C≥0, then Pr⁡[⋃iBi]≥∑iPr⁡[Bi]/(1+2C) (Overlap control gives a union lower bound).

Proof

technique · direct
1.1

If UNSAT⁡(G)=0 then the right-hand side is zero and there is nothing to prove, so assume ε:=UNSAT⁡(G)>0. Let φ^ be the plurality decoding of φ and put ε′:=UNSAT⁡φ^(G), so ε′≥ε by [F1]; let V⊆E(G) be the set of edges violated by φ^, of size ε′∣E(G)∣≥1. Choose a set F⊆V as follows: if min⁡(ε′,1/t)∣E(G)∣≥1 let F be any subset of V with ∣F∣=⌈min⁡(ε′,1/t)∣E(G)∣⌉, and otherwise let F={f0} for some single violated edge f0. Writing εF:=∣F∣/∣E(G)∣ and m:=∣E(G)∣, in both cases εF≥min⁡(ε′,1/t)≥min⁡(ε,1/t)>0, and εF≤min⁡(ε′,1/t)+1/m. Put a:=cwint and k:=#{j:∣j−1−t∣≤a}. Since cwin<1, this interval lies inside J; since t≥⌈cwin−2⌉, a≥1, and the integer positions give k=2⌊a⌋+1, so a≤k≤2a+1.

F1F2F3constructalgebra
2.1

For j in the window of step 1.1 and f∈F let Bj,f be the event that the j-th lazy step of the sampled representative walk traverses a slot of the edge f carrying the relation Rf in its specified endpoint order (for a loop, either of its two slots), and that both endpoint views report the decoded labels of the endpoints of f. In the uniform powered-edge model of [F3] the position before the j-th step is uniform, so each of the 2m slots of G is traversed by that step with probability 1/(2⋅2m)=1/(4m), and combining this with the conditional bound of [F3] gives Pr⁡[Bj,f]≥1/(16m∣Σ∣2). Moreover Bj,f implies that the sampled powered edge is violated: the tested pair at position j is exactly the pair (φ^(u),φ^(u′)) of decoded labels at the endpoints of f, and this pair fails Rf because φ^ violates f.

F1F3step 1.1algebra
3.1

Let Aj be the event that the j-th lazy step of the sampled representative walk traverses an edge of F. If j=j′ and f≠f′ then Bj,f∩Bj′,f′=∅, a single step traversing one edge only; otherwise Bj,f∩Bj′,f′⊆Aj∩Aj′. The k consecutive lazy steps in the central sub-window have the law of a length-k lazy walk from a uniform start: the walk starts uniformly and the uniform distribution remains stationary at the beginning of this sub-window, while its step choices are independent. Thus [F4] applies to this block and this F, and ∑(j,f)≠(j′,f′)Pr⁡[Bj,f∩Bj′,f′]≤2∑j<j′Pr⁡[Aj∩Aj′]≤2d/2 (k2εF2+kεF/(1−α)).

F1F4step 1.1step 2.1
4.1

Summing the bound of step 2.1 over the k positions and the ∣F∣ edges gives S:=∑j,fPr⁡[Bj,f]≥k∣F∣/(16m∣Σ∣2)=kεF/(16∣Σ∣2)≥cwint εF/(16∣Σ∣2)>0, and the events and the collision sum are finite, so [F5] is applicable. Applying it with the ratio C:=∑(j,f)≠(j′,f′)Pr⁡[Bj,f∩Bj′,f′]S ≤ 2d/2 (k2εF2+kεF/(1−α))⋅16∣Σ∣2kεF = 32d2 ∣Σ∣2(kεF+11−α) gives Pr⁡[⋃j,fBj,f]≥S/(1+2C).

F5step 2.1step 3.1algebra
5.1

Two regimes. If kεF≤1 then 1+2C≤1+64d/2 ∣Σ∣2(1+1/(1−α0))=:1+2C∗ and hence Pr⁡[⋃B]≥cwint εF/(16∣Σ∣2(1+2C∗))=β1t εF, where β1:=cwin/(16∣Σ∣2(1+2C∗)). If kεF>1 then kεF+1/(1−α)≤kεF(1+1/(1−α)), so 1+2C≤(1+64d/2∣Σ∣2(1+1/(1−α0)))kεF and Pr⁡[⋃B]≥1/(16∣Σ∣2(1+2C∗))=β2 with β2:=1/(16∣Σ∣2(1+2C∗)). Every Bj,f implies that the sampled powered edge is violated, so UNSAT⁡φ(Gt)≥Pr⁡[⋃j,fBj,f], and because tmin⁡(ε,1/t)≤t−1/2≤1 for t≥1 while εF≥min⁡(ε,1/t), both regimes give UNSAT⁡φ(Gt)≥min⁡(β1,β2)t min⁡(UNSAT⁡(G),1/t) with β:=min⁡(β1,β2) and t0:=max⁡{4,⌈cwin−2⌉}.

step 1.1step 4.1algebra∎

Remarks

  • The two regimes are the two halves of the promise. For kεF≤1 the union bound loses only the constant 1+2C∗ and delivers Θ(t εF); for kεF>1 the same computation yields a constant lower bound, which dominates βtmin⁡(ε,1/t) because that quantity is at most β/t. This is exactly Dinur's Lemma 6.1 with min⁡(UNSAT⁡(G),1/t), and it is what makes the iteration of A complete uniform graph gap-amplification step terminate in O(log⁡M) rounds.
  • The window is a sub-window of the central window. The positions used lie in Jc⊆J and number k=2⌊cwint⌋+1≥cwint, so every event Bj,f is a violation of the powered slot relation and the sub-window still gives the t gain. Over the full central window the middle-position lemma would lose a constant and the collision term kεF/(1−α) would grow.
  • Where each constant comes from. β depends only on d,∣Σ∣,α0: through cwin=1/(8C0∣Σ∣) and the factor 1/(1−α) of the collision bound, and through d/2 in the same bound. No dependence on n, on ∣E(G)∣, on t or on the labeling remains, and the proof chooses nothing beyond the subset F and the slot of each edge, both of which are fixed deterministically.
  • The statement above is uniform over labelings of Gt, including non-liftable ones; that is the content of the decoding step 1.1, which replaces an arbitrary powered labeling by one base labeling φ^ at the cost of the factor 1 in ε′≥ε.
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-27Open item page →

A complete uniform graph gap-amplification step

Statement

Fix a finite alphabet Σ with ∣Σ∣≥2 and put D:=387, K:=20/7, and let β0,t0 be the constants of Powering amplifies a small unsatisfaction gap for d=D, this Σ and α0:=ρ2<1, the spectral bound of Constraint expander overlay. For every integer t≥t0 there is a complete uniform gap-preserving reduction Rt in the sense of Complete uniform gap-preserving CSP reductions, defined on binary constraint graphs over Σ of arbitrary degree, with

  • Rt(G):=(Rdeg⁡(G))t, where Rdeg⁡ is the degree-reduction map of Degree reduction by expander incidence clouds and (⋅)t is the local-view powering of Constraint graph powering with local-view labels;
  • output alphabet Σt=ΣPR of size ∣Σt∣≤∣Σ∣DO(t), output degree bound dt:=2(2D)2t+1=DO(t), and blowup Ct:=D⋅(2D)2t+1=DO(t);
  • gap map gt(ε)=βt min⁡(ε,c/t) with c:=DK=7740/7 and β:=β0/(DK)>0, so UNSAT⁡(Rt(G))≥βt min⁡(UNSAT⁡(G),c/t) and hence UNSAT⁡(G)≥ε implies UNSAT⁡(Rt(G))≥βt min⁡(ε,c/t);
  • perfect completeness: val⁡(G)=1 implies val⁡(Rt(G))=1, and edgeless inputs are mapped to edgeless outputs;

and Rt is deterministic and runs in time polynomial in the bit length of the explicit encoding of G. The constants β,c,t0 and the parameters Σt,dt,Ct depend only on ∣Σ∣ and t, never on ∣V(G)∣ or ∣E(G)∣. This is a single powering step, not the fixed-alphabet PCP iteration: the alphabet grows with t, and no claim is made here about reducing it.

Facts & Assumptions

Given: a finite alphabet Σ with ∣Σ∣≥2, integers t≥t0 and D=387, K=20/7, and the constants β0,t0 of Powering amplifies a small unsatisfaction gap for d=D, Σ and α0=ρ2.

[F1]

Rdeg⁡ is a complete uniform gap-preserving reduction for the fixed input alphabet Σ and arbitrary input degrees, with output alphabet Σ, output degree bound D, blowup D, gap map gdeg⁡(ε)=ε/(DK), and polynomial running time; on inputs with E(G)≠∅ its output is D-regular with 2∣E(G)∣ vertices and D∣E(G)∣ ordinary edges over Σ and normalized second eigenvalue bound at most ρ2<1 (Degree reduction preserves unsatisfaction quantitatively, Constraint expander overlay, Degree reduction by expander incidence clouds).

[F2]

For a d-regular graph G with n vertices, the powered graph Gt has vertex set V(G), two paired incidence slots for each of the n(2d)2t+1 pairs (start vertex, pattern), degree 2(2d)2t+1, and n(2d)2t+1 ordinary edges; its normalized adjacency is the length-(2t+1) lazy-walk transition matrix, and its view alphabet is Σt=ΣPR with R=t+⌈t⌉. Explicit relation tables are computable by enumerating patterns and tables, so Gt is produced in time polynomial in its explicit encoding length (Constraint graph powering with local-view labels).

[F3]

If val⁡(G2)=1 then val⁡((G2)t)=1 (Powering preserves perfect satisfiability).

[F4]

For every labeling φ of (G2)t, UNSAT⁡φ((G2)t)≥β0t min⁡(UNSAT⁡(G2),1/t) whenever t≥t0 and G2 is D-regular over Σ with normalized second eigenvalue bound at most ρ2 and at least one edge (Powering amplifies a small unsatisfaction gap).

[F5]

A complete uniform gap-preserving reduction has arity two throughout, maps edgeless inputs to edgeless outputs, obeys ∣E(R(G))∣≤C∣E(G)∣ and ∣V(R(G))∣≤C∣E(G)∣ on nonempty inputs, and its parameters depend only on the fixed input parameters and on the reduction itself (Complete uniform gap-preserving CSP reductions).

Proof

technique · direct
1.1

If G is D-regular over Σ with at least one edge, [F2] gives a powered graph on ∣V(G)∣ vertices, degree dt=2(2D)2t+1, and ∣E(Gt)∣=∣V(G)∣(2D)2t+1=2(2D)2t+1∣E(G)∣/D≤Ct∣E(G)∣, since D≥2. Also ∣V(Gt)∣=∣V(G)∣=2∣E(G)∣/D≤Ct∣E(G)∣. Completeness is [F3], the gap bound is [F4], and the construction is explicit and polynomial for fixed D,t,∣Σ∣ by [F2]. This establishes the powering facts on the regular intermediate graphs used below.

F2F3F4algebra
2.1

The gap map of the composite is g(ε)=gt(gdeg⁡(ε))=β0t min⁡(ε/(DK),1/t) with gdeg⁡(ε)=ε/(DK) as in [F1]; writing min⁡(ε/(DK),1/t)=(1/(DK))min⁡(ε,DK/t) gives g(ε)=βt min⁡(ε,c/t) with β=β0/(DK) and c=DK>0.

F1step 1.1algebra
3.1

Define Rt(G)=(Rdeg⁡(G))t. If E(G)≠∅, [F1] gives a D-regular intermediate graph with 2∣E(G)∣ vertices, so [F2] gives ∣V(Rt(G))∣=2∣E(G)∣, degree dt, and ∣E(Rt(G))∣=2∣E(G)∣(2D)2t+1≤Ct∣E(G)∣ because D≥2; the vertex bound follows as well. Completeness follows by [F1] and [F3]. Applying [F1] and then [F4] gives the gap map of step 2.1. If E(G)=∅, the first map returns the empty graph and the powered output is empty, so the empty-input, completeness, and gap-at-zero clauses hold. The construction is deterministic and polynomial time: [F1] computes the intermediate graph in polynomial time and its size is O(∣E(G)∣) for fixed D,∣Σ∣, after which [F2] enumerates a fixed number of patterns per vertex and writes fixed-size tables. Thus Rt satisfies the clauses of [F5] directly, without applying a composition theorem whose second-stage domain is larger than the regular range used here.

F1F2F3F4F5step 2.1algebra
4.1

The alphabet size obeys ∣Σt∣=∣Σ∣(2D)R with R=t+⌈t⌉, which is at most ∣Σ∣DO(t); the degree 2(2D)2t+1 and blowup D(2D)2t+1 are both DO(t); and the gap inequality for inputs with UNSAT⁡(G)≥ε follows from the gap map of step 2.1 and the monotonicity of ε↦min⁡(ε,c/t). This proves every clause of the statement, with the stated dependence of all constants on ∣Σ∣ and t only.

F1F2step 1.1step 2.1step 3.1algebra∎

Remarks

  • The loss factor of the page is here. The factor DK=7740/7 of Degree reduction preserves unsatisfaction quantitatively is absorbed into β and into the saturation threshold c, exactly as the promised claim allows; no other quantity of the composite depends on the internal constant 129 of the cloud construction. The saturation threshold c/t is inherited from the min⁡(ε,1/t) of the powering lemma, scaled by the degree-reduction loss.
  • Not yet a PCP reduction. The output alphabet Σt grows like ∣Σ∣DO(t), and the output degree dt also grows with t; both are constants for fixed t, which is what the statement needs, but a fixed-alphabet PCP requires the alphabet-reduction step that is owned by the following page. The obligation is recorded in the coverage record of this page rather than discharged here.
  • Determinism. The reduction enumerates all patterns and all relation tables instead of sampling them, so no random choices and no choice principle are used; the graph family Hr inside Rdeg⁡ is the explicit one supplied by the published expander construction. The pair Σt is fixed by listing the patterns in a fixed order, which also fixes the tie-breaking order used by the plurality decoding of Plurality decoding of powered local views.
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-09-27Open item page →

Explicit binary codes of constant rate and distance

Definition

Throughout this page, for N≥1 the relative Hamming distance between x,y∈{0,1}N is δ(x,y):=#{i∈{1,…,N}:xi≠yi}N∈[0,1], and the relative distance of a subset C⊆{0,1}N with ∣C∣≥2 is min⁡{δ(x,y):x≠y, x,y∈C}, while a subset with exactly one element has relative distance 1 by convention.

A binary code family is a sequence of maps Ck:{0,1}k→{0,1}N(k) for k≥1, each injective, with an integer length function N(k)≥k. Its rate at k is k/N(k). The family is explicit with constant rate and constant distance when there are absolute constants c0>0, δ0>0, C0>0 such that for every k≥1 N(k)≤C0k,kN(k)≥c0,δ(Ck(x),Ck(y))≥δ0  for all x≠y∈{0,1}k, and each encoder Ck is computable by one deterministic algorithm in time polynomial in k, uniformly in k: the algorithm takes 1k together with x∈{0,1}k as input and writes the N(k) output bits.

Remarks

  • Explicit here means the same convention as for constraint graphs in Complete uniform gap-preserving CSP reductions: the output is written bit by bit, so that the running time is measured against the length of the produced codeword plus the input length, and a family is uniform when one algorithm serves all k. Nothing is claimed about the parity-check matrix, the decoder, or the existence of a fast decoder.
  • The padding convention is part of the definition: the code is defined for every k≥1 with its own length N(k), and a construction that first pads x to some convenient length kˉ≥k and then encodes is admissible only when the resulting length is O(k) and the analysis of injectivity and distance is done for the padded map, as in A polynomial-time explicit constant-rate constant-distance code.
  • Rate and distance trade off, and both constants above are absolute: the page needs N(k)=O(k) so that a constraint system whose variables are the k message bits and whose size is measured against N(k) stays linear in k, and it needs δ0>0 so that a constant fraction of the encoded bits witnesses every error in a message. The padded construction of A polynomial-time explicit constant-rate constant-distance code achieves δ0=1/8 and rate greater than 1/128 with N(k)<128k; 1/32 is the rate before padding.
  • The relative distance of a code is a minimum over pairs of codewords and is defined to be one for a one-element code so that the convention matches the relative-distance convention for testers, where an empty solution set is assigned distance one (Assignment tester and rejection ratio).
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Reed-Solomon outer code and binary linear inner code

Definition

Fix an integer m≥1 and put q:=2m and K:=q/2. Choose, by scanning the monic polynomials of degree m over F2 in a fixed order and testing each for irreducibility by trial division against all monic polynomials of positive degree at most ⌊m/2⌋, the first monic irreducible f∈F2[z] of degree m; by For every finite field Fq and every n≥1, a monic irreducible polynomial of degree n exists such an f exists, and the scan is deterministic and takes time polynomial in 2m. Set Fq:=F2[z]/(f), a field with exactly q elements by For every prime p and n≥1, a field with pn elements exists and A simple algebraic extension is its minimal-polynomial quotient and has power basis 1,a,…,an−1 and degree n, in which the classes of 1,z,…,zm−1 form a power basis; write enc:Fq→F2m,enc(∑j<mcjzj+(f)):=(c0,…,cm−1), for the resulting bit encoding, which is a bijection, and list the elements of Fq in the order induced by enc on the binary representations 0,1,…,q−1 of the exponents.

The outer code. The Reed-Solomon outer code RSq,K encodes a message (c0,…,cK−1)∈FqK, read as the polynomial P(X):=∑i<KciXi of degree less than K, into the word RSq,K(c):=(P(α))α∈Fq∈Fq q, the evaluation at all q field elements in the fixed order above. It is Fq-linear of dimension K over Fq, hence Km bits of message and qm bits of outer codeword.

The inner code. A binary linear inner code of rate 1/16 is a fixed injective F2-linear map Ein:F2m→F216m, written on messages as u↦Mu for a binary 16m×m matrix M, the generator matrix; its existence with relative distance at least 1/4 is the content of Conditional expectation constructs the inner code deterministically and the construction is uniform in m.

The concatenated code. Let M be the generator matrix of an inner code as above. The concatenation RSq,K∘M encodes a message u∈F2Km, split as u=(u1,…,uK) with uj∈F2m, in two steps: form the outer word (α1,…,αq):=RSq,K(enc−1(u1),…,enc−1(uK)), and output the concatenation of M enc(α1),…,M enc(αq). Its output is a string of 16qm bits, its input is Km=qm/2 bits, and it is injective because RSq,K and each of enc,Ein,enc−1 is. It is the code used by A polynomial-time explicit constant-rate constant-distance code.

Remarks

  • Determinism. The field Fq is not presupposed: the definition names the first irreducible polynomial of degree m in a fixed enumeration, so the field, its power basis, the order of its elements, the Reed-Solomon evaluation points and the bit encoding are all functions of m alone and involve no choice. The time to find f is polynomial in 2m, because testing irreducibility of one degree-m polynomial by trial division costs O(2m) field operations and at most 2m polynomials are tried.
  • Two different alphabets. The outer code is linear over Fq and the inner code over F2; the concatenation is a binary code, and its parameters are computed in Concatenation multiplies rate and relative distance. The rate of the outer code is K/q=1/2 and the rate of the inner code is m/16m=1/16, so the concatenation has rate 1/32.
  • The inner code is part of the data of the definition rather than a canonical object: any matrix M produced by the uniform procedure of Conditional expectation constructs the inner code deterministically gives the same parameters, so all statements of this page are independent of which of these matrices is used. The convention of Explicit binary codes of constant rate and distance is used for rate and relative distance.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Reed-Solomon outer code has constant rate and distance

Statement

Let m≥1, q=2m and K=q/2, and let RSq,K:FqK→Fq q be the Reed-Solomon evaluation code of Reed-Solomon outer code and binary linear inner code. Then RSq,K is injective, its rate over the alphabet Fq is K/q=1/2, and its relative Hamming distance over Fq is at least q−K+1q ≥ 12, that is, distinct codewords differ in at least q−K+1=q/2+1 coordinates. In the bit metric of Explicit binary codes of constant rate and distance applied to the words encoded by the power-basis map this is a statement about Fq-symbols; the binary distance of the concatenated code is computed separately in Concatenation multiplies rate and relative distance.

Facts & Assumptions

Given: integers m≥1, q=2m, K=q/2, the field Fq with its fixed listing of the q elements, and the evaluation code RSq,K.

[F1]

Fq=F2[z]/(f) is a field with exactly q elements; the code RSq,K sends a message (c0,…,cK−1) to the word (P(α))α∈Fq where P=∑i<KciXi has degree less than K, evaluated at all q elements of the field in the fixed listing; the map is Fq-linear (Reed-Solomon outer code and binary linear inner code).

[L1]

A nonzero polynomial of degree n over an integral domain has at most n distinct roots in that domain (A nonzero polynomial of degree n over an integral domain has at most n distinct roots).

[L2]

Every field has no zero divisors, so a field with its ring structure is an integral domain (A field has no zero divisors: ab=0⇒a=0 or b=0).

Proof

technique · direct
1.1

Let c≠c′ be two distinct messages and let P,P′ be the associated polynomials. Then P−P′≠0 because the coefficient vectors differ, and it is a polynomial of degree at most K−1, so by [L1] applied over the field Fq, which is an integral domain by [L2], it has at most K−1 roots; equivalently P(α)=P′(α) for at most K−1 of the q evaluation points α.

F1L1L2
2.1

Among the q coordinates of the two words, therefore, at least q−(K−1)=q−K+1 differ; the code is injective because q−K+1≥1, and its relative distance is at least (q−K+1)/q. With K=q/2 this is (q/2+1)/q=1/2+1/q≥1/2.

step 1.1F1algebra
3.1

The message space FqK has qK elements and the rate over Fq is the dimension divided by the length, K/q=1/2; the input and output sizes in bits are Km=qm/2 and qm respectively.

F1algebra∎

Remarks

  • The count is exact for the evaluation code: the bound q−K+1 is attained by polynomials vanishing on K−1 evaluation points, so the distance cannot be improved by this method, but only the lower bound 1/2 is used on this page.
  • The alphabet here is Fq, not the binary alphabet: the lemma is a statement about the outer code alone, and the passage to binary distance is the content of Concatenation multiplies rate and relative distance, which uses the injectivity and distance of the inner code. The case m=1 is included: then q=2, K=1, and the code is the length-two repetition code over F2, with rate 1/2 and relative distance 1≥1/2.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

A random inner linear code has fewer than one bad word in expectation

Statement

Let m≥1, let M be a uniformly random binary 16m×m matrix whose 16m2 entries are independent fair bits, and let wt⁡ denote Hamming weight in F216m. Then the expected number of nonzero u∈F2m with wt⁡(Mu)<4m satisfies E[#{u∈F2m∖{0}:wt⁡(Mu)<4m}] ≤ (2m−1)e−m < (2e)m < 1. Consequently, with positive probability a random matrix has no such u, and the map u↦Mu of every such matrix is injective with relative distance at least 1/4 and rate 1/16, in the conventions of Explicit binary codes of constant rate and distance. The deterministic construction of such a matrix is the content of Conditional expectation constructs the inner code deterministically.

Facts & Assumptions

Given: an integer m≥1, the uniform random matrix M with independent fair entry bits, and the set of nonzero u∈F2m.

[F1]

For N≥1 the Hamming weight of x∈{0,1}N is the number of nonzero coordinates, and the relative distance of two binary words of length N is their Hamming distance divided by N (Explicit binary codes of constant rate and distance).

[L1]

If X1,…,Xk are independent Bernoulli(p) variables with sum S and mean μ=pk, then for 0<α<1, Pr⁡[S≤(1−α)μ]≤exp⁡(−α2μ/2) (A Chernoff bound for sums of independent Bernoulli trials).

[L2]

On a finite probability space the expectation of a sum of random variables is the sum of the expectations, and the expectation of an indicator is the probability of its event (Expectation of a real random variable on a finite probability space, The expectation of an indicator is the probability of the event).

[L3]

A finite family of classes of events is independent when the probability of every finite intersection of chosen events factors as the product of their probabilities; events determined by disjoint blocks of independent coordinates therefore factorize (Independent families of event classes).

Proof

technique · direct
1.1

Fix u≠0 and let Yi be the i-th coordinate of Mu, so Yi=∑j≤mMijuj in F2 and wt⁡(Mu)=∑i≤16mYi. For fixed u≠0 the row i of M has at least one free coordinate in the support of u, and the entries of the row are independent fair bits, so Yi is a fair bit; distinct rows involve disjoint blocks of entries and are independent, so Y1,…,Y16m is an independent family of Bernoulli(1/2) variables by [L3].

F1L3algebra
2.1

Applying [L1] to this family with k=16m, p=1/2, μ=8m and α=1/2 gives Pr⁡[wt⁡(Mu)≤4m]≤e−m, and since weights are integers, Pr⁡[wt⁡(Mu)<4m]≤Pr⁡[wt⁡(Mu)≤4m]≤e−m.

L1step 1.1algebra
3.1

By [L2] the expectation of the number of bad u is the sum of the probabilities Pr⁡[wt⁡(Mu)<4m] over the 2m−1 nonzero u, hence at most (2m−1)e−m<2me−m=(2/e)m; since 2<e, this is strictly less than 1.

L2step 2.1algebra
4.1

A random variable with expectation <1 takes a value <1, so some matrix M has no u≠0 with wt⁡(Mu)<4m; for such M the weight of Mu is at least 4m for every nonzero u, whence Mu≠0 and u↦Mu is injective, with relative distance at least 4m/(16m)=1/4 and rate m/(16m)=1/16 in the conventions of [F1].

F1step 3.1algebra∎

Remarks

  • The strict inequality wt⁡(Mu)<4m is what produces relative distance 1/4 rather than the weaker 1/4−ε; it is also what makes the counting bound <1 rather than ≤1, which is needed in step 4.1 to conclude existence without any tie.
  • The argument uses only the 2m−1 nonzero messages and the fairness of the row functionals; the value 16 is chosen so that the mean 8m of the weight is twice the threshold 4m, giving the Chernoff exponent m and the base 2/e<1. With 16m rows replaced by cm rows for a constant c>8, the same computation gives the base 2exp⁡(−c(1−8/c)2/4), which is less than 1 exactly when c2−(16+4ln⁡2)c+64>0; the positive root of that quadratic is about 14.3, so the same estimate tolerates any number of rows cm with c above that root, and c=16 is the convenient integer choice.
  • The positive-probability statement is already enough for a non-uniform existence claim; the point of Conditional expectation constructs the inner code deterministically is to make the matrix computable rather than merely existent.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Conditional expectation constructs the inner code deterministically

Statement

For every integer m≥1 there is a deterministic algorithm which, given m, runs in time polynomial in 2mm and outputs a binary 16m×m matrix M such that wt⁡(Mu)≥4m for every nonzero u∈F2m. Hence u↦Mu is an injective F2-linear map F2m→F216m of rate 1/16 and relative distance at least 1/4, in the conventions of Explicit binary codes of constant rate and distance, the same algorithm serves every m, and the output matrix is a function of m alone, with no random choices.

Facts & Assumptions

Given: an integer m≥1, the number N:=16m2 of entries, and the identification of binary 16m×m matrices with the points of Ω:={0,1}N by listing the entries in a fixed order.

[F1]

For the uniform random matrix M on Ω and X(M):=#{u∈F2m∖{0}:wt⁡(Mu)<4m} one has EX<1; indeed EX≤(2m−1)e−m<(2/e)m<1 (A random inner linear code has fewer than one bad word in expectation).

[F2]

On a finite probability space with weights, the expectation of a real random variable is the weighted sum ∑ωX(ω)w(ω); on Ω the uniform weights are 2−N (Expectation of a real random variable on a finite probability space).

[F3]

For binary words, Hamming weight and relative distance are as in the code conventions; an injective linear map with minimum nonzero weight at least 4m has relative distance at least 4m/(16m)=1/4 and rate m/(16m)=1/16 (Explicit binary codes of constant rate and distance).

Proof

technique · constructive
1.1

For a partial assignment a∈{0,1}s of the first s entries let F(a):={M∈Ω:Mj=aj for j≤s} be its fiber and define Xˉ(a):=1∣F(a)∣∑M∈F(a)X(M)=2−(N−s)∑M∈F(a)X(M), the average of X over the fiber. Then Xˉ(∅)=EX by [F2], and for every s<N Xˉ(a)=12(Xˉ(a0)+Xˉ(a1)), because F(a) is the disjoint union of the two fibers F(a0) and F(a1), which have equal size 2N−s−1; this is the elementary averaging identity for finite sums.

F2algebra
1.2

Fix a∈{0,1}s and u≠0. A row i is determined for (a,u) when all coordinates j in the support of u have aj already fixed, in which case the i-th bit of Mu equals the known value ∑j∈supp⁡uMijuj; otherwise that bit is a fair coin, because at least one of its summands is an undecided uniform bit and the bits of distinct rows are independent. Writing Du for the number of determined rows whose determined bit is 1 and Ru for the number of undetermined rows, the conditional law of wt⁡(Mu) over the uniform fiber F(a) is Du+Bin⁡(Ru,12); hence Pr⁡[wt⁡(Mu)<4m∣F(a)]=Pr⁡[Bin⁡(Ru,12)<4m−Du], a number of the form 2−Ruc with c a nonnegative integer.

algebragiven
2.1

Define a path of partial assignments by a0:=∅ and, for s<N, as+1:=as0 if Xˉ(as0)≤Xˉ(as1) and as+1:=as1 otherwise, so ties go to the 0 branch; this is a deterministic choice. The averaging identity of step 1.1 gives Xˉ(as+1)≤Xˉ(as) at every step, hence Xˉ(aN)≤Xˉ(∅)=EX<1 by [F1].

F1step 1.1construct
3.1

The terminal fiber is a single matrix M∗, so Xˉ(aN)=X(M∗) is a nonnegative integer strictly below 1, hence equal to 0: the produced matrix satisfies wt⁡(M∗u)≥4m for every nonzero u.

step 2.1algebradischarge-construct
4.1

For the running time, evaluating Xˉ(a) for one partial assignment means summing the 2m−1 conditional probabilities of step 1.2; for each u, the support, the numbers Du and Ru and the binomial tail are computed in O(m) operations using a precomputed table of the binomial coefficients (Rt) for R≤16m, and all arithmetic is exact on integers of O(m) bits, so one evaluation costs O(2mm2) and the whole path costs O(2mm4) operations, which is polynomial in 2mm. By step 3.1 the output has wt⁡(M∗u)≥4m for all u≠0, so it is injective and has rate 1/16 and relative distance at least 1/4 by [F3].

F3step 1.2step 3.1discharge-constructalgebra∎

Remarks

  • The argument is the method of conditional expectations in its finite form: no conditional expectation as an abstract object is needed, only the identity that the average over a fiber is the average of the averages over the two half-fibers. That is also why the procedure is deterministic: the two candidate values are computed exactly, not estimated.
  • The inner code is produced in time polynomial in 2mm, which is polynomial in the message length of the concatenated code of Reed-Solomon outer code and binary linear inner code because that message length is 2m−1m. The procedure is uniform in m, so no choice of a matrix is made anywhere on the page: every consumer uses the matrix M∗ output for its own m.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Concatenation multiplies rate and relative distance

Statement

Let m≥1, q=2m, K=q/2, let RSq,K⊆Fq q be the Reed-Solomon outer code with rate Ro=K/q and relative distance at least δo, and let Ein=M(⋅):F2m→F216m be the inner encoding of Conditional expectation constructs the inner code deterministically, injective and linear of rate Ri=1/16 and relative distance at least δi, so that wt⁡(Ein(w))≥δi⋅16m for every nonzero w∈F2m. Then the concatenated code of Reed-Solomon outer code and binary linear inner code, which encodes Km bits into 16qm bits, is injective with rate = RoRi = K16q,relative distance ≥ δoδi. For the outer parameters of Reed-Solomon outer code has constant rate and distance and the inner parameters of Conditional expectation constructs the inner code deterministically these are rate at least 12⋅116=132 and relative distance at least 12⋅14=18.

Facts & Assumptions

Given: integers m≥1, q=2m, K=q/2; the bit encoding enc:Fq→F2m of the field, which is F2-linear by the power basis; the outer code RSq,K with rate Ro=K/q and relative distance at least δo; an injective F2-linear Ein:F2m→F216m with wt⁡(Ein(w))≥δi 16m for w≠0.

[F1]

The concatenated encoding sends a message u=(u1,…,uK)∈(F2m)K to the word obtained by applying Ein∘enc to each of the q coordinates of the outer word RSq,K(enc−1(u1),…,enc−1(uK)); it has Km input bits and 16qm output bits, and the field element 0 has bit encoding the zero vector (Reed-Solomon outer code and binary linear inner code).

[F2]

Relative distance of a code of length n over an alphabet is the minimum, over distinct codewords, of the fraction of differing coordinates; over a binary alphabet the coordinates are bits. Rate is the input length in bits divided by the output length in bits (Explicit binary codes of constant rate and distance).

[L1]

Distinct messages of RSq,K have outer words differing in at least δoq coordinates, and distinct messages have distinct outer words (Reed-Solomon outer code has constant rate and distance).

[L2]

The map Ein is linear and injective with wt⁡(Ein(w))≥δi 16m for every nonzero w; in particular u↦Mu has rate Ri=1/16 and relative distance at least δi (Conditional expectation constructs the inner code deterministically).

Proof

technique · direct
1.1

The concatenated encoding is injective: if two messages u≠u′ had the same concatenated word, then their outer words would agree in every coordinate, since Ein∘enc is injective and the blocks of the concatenation are read off coordinatewise; but distinct messages have distinct outer words by [L1, F1]. Moreover the rate is Km/(16qm)=K/(16q)=RoRi, because the outer code has K Fq-symbols (Km bits) and the concatenated word has q blocks of 16m bits.

F1F2L1algebra
2.1

Let u≠u′ be distinct messages with outer words c≠c′, and let S={j:cj≠cj′} be the set of differing coordinates, of size at least δoq by [L1]. For j∈S the field element cj−cj′≠0 has nonzero bit encoding enc(cj−cj′)=enc(cj)−enc(cj′) by linearity of enc, so the block difference Ein(enc(cj))−Ein(enc(cj′))=Ein(enc(cj−cj′)) is a nonzero word of weight at least δi 16m by [L2].

F1L2step 1.1algebra
3.1

The blocks indexed by S occupy disjoint sets of coordinates of the concatenated word, so the total Hamming distance between the two concatenated words is the sum over j∈S of the block weights, at least ∣S∣ δi 16m≥δoq δi 16m; dividing by the word length 16qm gives relative distance at least δoδi, and with the outer and inner parameters this is at least 12⋅14=18 while the rate is 12⋅116=132.

F2step 2.1L1L2algebra∎

Remarks

  • Both factors are honest minima. The outer distance loses a factor q in the number of surviving coordinates and the inner distance loses a factor 16m in the surviving bits per coordinate, and the product is exactly the product of the relative distances; nothing is lost to the intermediate field or to the choice of inner matrix, because the inner map is linear and injective and its blocks are disjoint.
  • The lemma is stated for the concrete outer and inner codes of this page, but the proof uses only [F1], [F2], [L1] and [L2], so the same computation applies to any outer code with rate Ro and relative distance δo whose alphabet is identified with F2m and any injective linear inner code with parameters Ri,δi.
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

A polynomial-time explicit constant-rate constant-distance code

Statement

There is a deterministic uniform binary code family (Ck)k≥1 in the sense of Explicit binary codes of constant rate and distance, with N(k) < 128k,rate > 1128,relative distance ≥ 18, each encoder computable by one algorithm in time polynomial in k. Thus the definition of an explicit constant-rate constant-distance family is met with C0=128, c0=1/128 and δ0=1/8, and these three constants are absolute.

Facts & Assumptions

Given: an integer k≥1, the family of concatenated codes of Reed-Solomon outer code and binary linear inner code indexed by m≥1, and the constants K=q/2, q=2m.

[F1]

For every m≥1 the concatenated code of Reed-Solomon outer code and binary linear inner code encodes Km=2m−1m bits into 16qm=32⋅2m−1m bits; it is defined after choosing the first irreducible polynomial of degree m in a fixed enumeration and the inner matrix produced by the uniform procedure of Conditional expectation constructs the inner code deterministically, and its bit encoding of field elements is the power-basis encoding enc.

[F2]

The Reed-Solomon outer code has rate 1/2 and relative distance at least 1/2 (Reed-Solomon outer code has constant rate and distance).

[F3]

The inner code is injective and linear of rate 1/16 and relative distance at least 1/4, and the matrix is produced deterministically in time polynomial in 2mm (Conditional expectation constructs the inner code deterministically).

[F4]

The concatenation of an outer code of rate Ro and relative distance δo with an injective linear inner code of rate Ri and relative distance δi has rate RoRi and relative distance at least δoδi (Concatenation multiplies rate and relative distance).

[F5]

A binary code family with N(k)≤C0k, rate at least c0 and relative distance at least δ0, encoded by one deterministic algorithm running in time polynomial in k, is an explicit constant-rate constant-distance family (Explicit binary codes of constant rate and distance).

Proof

technique · constructive
1.1

Define m(k) to be the least integer m≥1 with 2m−1m≥k; it exists because 2m−1m→∞, and for m≥2 minimality gives 2m−2(m−1)<k. Given x∈{0,1}k, pad it by zeros to the length Km=2m−1m and apply the concatenated encoding of [F1], whose output length is N(k):=32⋅2m−1m.

F1construct
2.1

For m≥2 we have 2m−1m=2⋅2m−2m≤4⋅2m−2(m−1)<4k, using m≤2(m−1); hence N(k)<128k. The case m=1 occurs exactly for k=1, where N(1)=32<128. The rate of the padded k-bit family is k/N(k)>1/128; Km/N(k)=1/32 is the rate of the full concatenated code before restricting it to padded messages. The relative distance remains at least 12⋅14=18 by [F2], [F3] and [F4], because restricting an injective code to padded messages cannot decrease its minimum pairwise distance.

F1F2F3F4step 1.1algebra
2.2

The family is injective: the padding is injective and the concatenated encoding is injective by [F4]. It is uniform and deterministic: m is determined by k, the field and its bit encoding by [F1], the inner matrix by [F3], and no step uses randomness.

F1F3F4step 1.1
3.1

The running time is polynomial in k: m≤2+log⁡2k because 2m−1<2k for m≥2; the degree-m irreducible polynomial is found by scanning at most 2m monic polynomials and testing each by trial division, in 2m2m/2m2≤23m/2m2 field operations; evaluating the outer code at all q field elements costs O(qK)=O(q2) field operations; the inner matrix is computed in time polynomial in 2mm by [F3]; and applying it to the q outer symbols costs O(qm2). Since q=2m≤4k and m=O(log⁡k), every term is polynomial in k, so one algorithm serves all k within polynomial time.

F1F3step 1.1step 2.1algebra
4.1

By [F5] the family, with its length function, rate, relative distance and polynomial-time uniform encoder verified in steps 2.1, 2.2 and 3.1, is an explicit constant-rate constant-distance binary code family with the stated constants.

F5step 2.1step 2.2step 3.1discharge-construct∎

Remarks

  • The constants. The full concatenated code has rate 1/32, the product of the outer rate 1/2 and inner rate 1/16. Padding the k input bits changes that rate to k/N(k)>1/128 because N(k)<128k; the distance bound 1/8 is the product 1/2⋅1/4 and survives restriction to padded inputs. None of these constants depends on k.
  • What is not claimed. The theorem asserts neither a decoder nor a parity-check description of the family, and it does not optimise the constants; the page needs only that a constant-rate constant-distance family with a polynomial-time uniform encoder exists, which is what Explicit binary codes of constant rate and distance asks for.
  • Small lengths. The padding makes the family total: k=1 is encoded by the length-32 concatenation with m=1, q=2 and the length-two repetition outer code of Reed-Solomon outer code has constant rate and distance, so no separate small-case convention is needed beyond the zero padding.
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Assignment tester and rejection ratio

Definition

Constraint systems of bounded arity. Fix a finite alphabet Σ0 containing distinguished, distinct symbols 0 and 1, identified with the Boolean bits, and an integer q≥2. A constraint system of arity at most q over Σ0 consists of a finite variable set V together with a finite list of constraints, each constraint being an ordered tuple (v1,…,vk) of variables of length k≤q, repetitions allowed, together with a relation R⊆Σ0k. A labeling σ:V→Σ0 satisfies such a constraint when (σ(v1),…,σ(vk))∈R, and val⁡σ(G):=#{constraints satisfied by σ}#{constraints of G}, the fraction of satisfied constraints, defined to be 1 when the list is empty; UNSAT⁡σ(G):=1−val⁡σ(G) and UNSAT⁡(G):=min⁡σUNSAT⁡σ(G). For q=2 this is the value convention of Constraint graph and labeling value: a constraint on two distinct variables is an edge carrying its relation in the displayed endpoint order, a constraint on a repeated variable is a loop, and duplicated constraints correspond to duplicated edges. There, and throughout, relations are explicit tables.

Circuits with named inputs. A Boolean circuit C of size m over the basis of Boolean circuits: basis, fan-in, size, and depth is given together with a specified list X=(x1,…,xn) of n of its input coordinates, so that an input is a string a∈{0,1}X≅{0,1}n, and C accepts a when some assignment to the remaining input coordinates makes the designated output evaluate to one under the circuit gate rules, as in Circuit satisfiability. Write SAT⁡(C)⊆{0,1}X for the set of accepted inputs. For a∈{0,1}X put δ(a,SAT⁡(C)):=min⁡a′∈SAT⁡(C)#{i:ai≠ai′}n, the relative Hamming distance on the named coordinates, and define δ(a,SAT⁡(C)):=1 when SAT⁡(C)=∅. For n=0 the cube {0,1}0 has one element, so δ is 0 or 1 according to whether C accepts; for n≥1 and nonempty SAT⁡(C) the minimum is over a nonempty finite set and always lies in [0,1].

Assignment tester. An assignment tester with alphabet Σ0, arity bound q and rejection ratio ρ>0 is a map P sending each pair (C,X) as above to a constraint system G=P(C,X) of arity at most q over Σ0 whose variable set contains the named input coordinates, X⊆V(G), such that with Y:=V(G)∖X:

  • (Perfect completeness.) If a∈SAT⁡(C) then there is b∈Σ0Y with UNSAT⁡a∪b(G)=0.
  • (Proximity soundness.) For every a∈{0,1}X and every b∈Σ0Y, UNSAT⁡a∪b(G) ≥ ρ⋅δ(a,SAT⁡(C)).

The two clauses are the completeness and soundness clauses of Complete uniform gap-preserving CSP reductions read at the level of a fixed input: perfect completeness says that an accepted input extends to a fully satisfying labeling of the whole system, and proximity soundness says that the violation fraction witnessed by any labeling is at least ρ times how far the given input is from acceptance. Since δ≤1 always, the soundness clause is implied by the stronger requirement that every a∉SAT⁡(C) has UNSAT⁡a∪b(G)≥ρ for all b; the definition states the proportional form, which is what the amplification and composition arguments of this page use.

The definition itself imposes no bound on ∣G∣ or on the time needed to produce it. When those are needed one says that the tester is uniform (or has output size N) if P is computed by a deterministic algorithm running in time polynomial in the bit length of the explicit description of (C,X) and, respectively, if the number of variables and constraints of P(C,X) is at most N times a constant depending only on the fixed parameters; all constant factors below depend only on Σ0,q and on the family of constructions, never on n or m.

Remarks

  • Distinct from a global gap. A gap-preserving reduction compares UNSAT⁡ of a whole instance before and after the map; an assignment tester compares, for one fixed input prefix, the violations forced on an arbitrary auxiliary labeling against the distance of that prefix from the accepted set. The two notions meet when the prefix is the empty function, and the composition of testers is what makes the constant-query PCP with named input coordinates available.
  • The empty-SAT⁡ convention matters. With δ:=1 on SAT⁡(C)=∅ the soundness clause forces a positive violation fraction for every input, including the degenerate case where C accepts nothing; without the convention the distance to an empty set would be undefined and the clause vacuous. The same convention makes the instancewise bound of Gate constraints are a weak assignment tester non-vacuous for unsatisfiable circuits.
  • Multiplicity and order. The constraint list is a list, not a set: two identical constraints count twice in the value fraction, exactly as duplicated edges do in the constraint-graph convention. The displayed order of the variables of a constraint is part of the data, and reversing a tuple transposes its relation; a tester must fix the order of every constraint it emits.
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Hadamard linearity constraint system

Definition

Let n≥0 and let f:F2n→F2 be a fixed table, in the conventions of The BLR linearity test over F_2. The Hadamard linearity constraint system Lin⁡n(f) is the constraint system of arity at most 3 over the alphabet F2 of Assignment tester and rejection ratio whose variables are the table coordinates z∈F2n — one variable per cube point, so V=F2n — and whose constraint list contains, once for each ordered pair (x,y)∈F2n×F2n, the ternary constraint with variable tuple (x, y, x+y) and relation Rlin:={(a,b,c)∈F23: a+b=c}, the addition being modulo two. The list is a list, so constraints belonging to different pairs are counted separately even when they coincide as tuples.

Since x,y range over the whole cube, the variable tuple of a constraint may repeat coordinates — for instance (0,0,0) for the pair (0,0), and (x,x,0) for (x,x) — and the satisfaction rule of a constraint system is the substitution rule: the tuple (x,y,x+y) is satisfied by a labeling σ:F2n→F2 exactly when σ(x)+σ(y)=σ(x+y), coinciding coordinates being read as the single symbol assigned to them. The same substitution covers the degenerate cases: the pair (0,0) forces σ(0)=0, and for n=0 the cube has one point, the list has the single constraint with tuple (0,0,0), and the system is satisfied exactly by the table f(0)=0.

Value and rejection probability. For a labeling σ of Lin⁡n(f) — that is, an arbitrary table σ:F2n→F2 — the fraction val⁡σ counts the satisfied constraints among the 4n listed pairs, and UNSAT⁡σ=1−val⁡σ. For the fixed table f we write εlin(f):=1−val⁡f(Lin⁡n(f))=#{(x,y):f(x)+f(y)≠f(x+y)}4n, the rejection probability of the linearity system, which is exactly the BLR rejection probability of f in the sense of The BLR linearity test over F_2: the pair (x,y) is drawn uniformly in the test and the listed constraint at that pair is violated precisely when the test rejects. In particular a uniformly random constraint of the system involves the three coordinates x,y,x+y and is checked with three table queries and 2n random bits.

Perfect completeness. Every linear table passes. If f=ℓu for some u∈F2n, then for all x,y the distributivity of the dot product over addition in F2 gives ℓu(x)+ℓu(y)=u⋅x+u⋅y=u⋅(x+y)=ℓu(x+y), so every constraint of the system is satisfied and εlin(ℓu)=0. The system is explicit: its 4n constraint tuples and its single fixed relation table Rlin are enumerated by the displayed rule, and its size is determined by n alone.

Remarks

  • One constraint per test, not one per violation pattern. The system records multiplicity: the pairs (x,y) and (y,x) give two constraints, and a pair with x=y still gives one constraint. This is what makes εlin(f) equal — not merely comparable — to the rejection probability of the sampled BLR test, and it is the convention used by the exponential-base tester of An exponential-size constant-query base assignment tester, where every random choice of the verifier is materialized as one constraint.
  • Relation to the proximity definition. The system constrains table coordinates, not inputs of a circuit: in the language of Assignment tester and rejection ratio the whole cube of table positions is the auxiliary part of the variable set, and the rejection probability is the quantity the soundness clause controls. The distance of a table from the linear tables is a property of the labeling of these coordinates, and BLR rejection is proportional to distance from linearity is the statement that the violated fraction is at least that distance.
  • No random sampling is part of the definition. Sampling the pair (x,y) is a way of estimating the value of an explicit, fully listed system; the system itself is deterministic and independent of any randomness, and it is produced by enumerating the 4n pairs, which is polynomial in the size of the listing but exponential in n as a description of f requires.
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

BLR rejection is proportional to distance from linearity

Statement

For every n≥0 and every f:F2n→F2, let α(f) be the BLR acceptance probability and ε(f)=1−α(f) the BLR rejection probability of The BLR linearity test over F_2, and let dist⁡(f,Λ):=min⁡a∈F2ndist⁡(f,ℓa),Λ={ℓa:x↦a⋅x} be the relative Hamming distance from f to the nearest linear function, the minimum being over the nonempty finite set of 2n linear functions. Then ε(f) ≥ dist⁡(f,Λ). Equivalently, the fraction of violated constraints of the Hadamard linearity constraint system Lin⁡n(f) of Hadamard linearity constraint system is at least the distance of f from Λ. There is no restriction on the acceptance probability: the bound holds in the low-acceptance regime α≤1/2 as well, and it holds in dimension n=0.

Facts & Assumptions

Given: an integer n≥0 and a fixed table f:F2n→F2, with h(x):=(−1)f(x) and h^(a)=Exh(x)χa(x).

[F1]

The BLR test chooses independent uniform x,y∈F2n and accepts exactly when f(x)+f(y)=f(x+y); its acceptance probability is denoted α and ε=1−α is its rejection probability. The linear functions are ℓa(x)=a⋅x with a∈F2n, the normalized distance is dist⁡(f,g)=2−n#{x:f(x)≠g(x)}, and for n=0 the cube has one point and the sole linear function is ℓ0=0 (The BLR linearity test over F_2).

[F2]

With h=(−1)f one has 2α−1=Ex,yh(x)h(y)h(x+y)=∑ah^(a)3, the sum being over the 2n indices a (The cubic Fourier identity for BLR).

[F3]

The characters are orthonormal and Parseval holds: ∑ah^(a)2=Exh(x)2=1, with h^(a)=Exh(x)χa(x) and χa(x)=(−1)a⋅x (Character orthogonality, inversion and Parseval).

[F4]

The Hadamard linearity constraint system Lin⁡n(f) lists one ternary constraint for each ordered pair (x,y) with tuple (x,y,x+y) and relation a+b=c, has 4n constraints, and the fraction of its constraints violated by f equals the BLR rejection probability ε; every linear table satisfies all its constraints (Hadamard linearity constraint system).

Proof

technique · direct calculation
1.1

The index set F2n is finite and nonempty for every n≥0, so M:=max⁡ah^(a) exists; it is a signed maximum, attained at some index, and no absolute value is taken.

F3givenalgebra
2.1

By [F2] and [F3], 2α−1=∑ah^(a)3≤∑aM h^(a)2=M∑ah^(a)2=M, because h^(a)≤M and h^(a)2≥0 for every index a; the sum has 2n≥1 terms, and the inequality is preserved by the finite sum.

F2F3step 1.1algebra
2.2

For each a, since h=(−1)f takes the value +1 exactly where f=ℓa and −1 exactly where f≠ℓa, h^(a)=Exh(x)χa(x)=Pr⁡[f(x)=ℓa(x)]−Pr⁡[f(x)≠ℓa(x)]=1−2dist⁡(f,ℓa), so dist⁡(f,ℓa)=(1−h^(a))/2 and therefore dist⁡(f,Λ)=min⁡a(1−h^(a))/2=(1−M)/2, the minimum of the distances corresponding to the maximum of the coefficients.

F1F3step 1.1algebra
3.1

Combining steps 2.1 and 2.2, ε=1−α=(1−(2α−1))/2≥(1−M)/2=dist⁡(f,Λ). This is the claimed inequality, with no hypothesis on α; in particular it is available when α≤1/2, where the acceptance-based form of BLR soundness imposes no restriction on f.

step 2.1step 2.2algebra
4.1

By [F4] the fraction of constraints of Lin⁡n(f) violated by the table f equals ε, so step 3.1 says exactly that this violated fraction is at least dist⁡(f,Λ); the system has 4n constraints of arity 3 over F2, and the statement of the theorem is the conjunction of the two formulations.

F4step 3.1
5.1

Degenerate and extremal cases. For n=0 we have M=h^(0)=h(0)=(−1)f(0), so 2α−1=M and dist⁡(f,ℓ0)=(1−M)/2: if f(0)=0 then ε=0=dist⁡ and the single constraint (0,0,0) of the system is satisfied, while if f(0)=1 then ε=1=dist⁡, matching the constraint f(0)+f(0)=f(0) being violated. For a linear table f=ℓu every constraint is satisfied by [F4], so ε=0=dist⁡(f,Λ). The inequality is not in general an equality: for n≥1 and f=ℓu+1 one has f(x)+f(y)=ℓu(x+y) and f(x+y)=ℓu(x+y)+1 for all x,y, so every constraint is violated and ε=1, while dist⁡(f,ℓu)=1 and dist⁡(f,ℓb)=1/2 for b≠u give dist⁡(f,Λ)=1/2<1.

F1F4step 3.1algebra∎

Remarks

  • What the bound does and does not say. The inequality ε≥dist⁡(f,Λ) is one-sided: it converts a large distance into a large rejection probability and says nothing about the converse, and the example f=ℓu+1 shows the two quantities can differ by a factor two. Read backwards it recovers the classical form of BLR acceptance implies agreement with a linear function, which concludes agreement with a linear function from acceptance above one half; the present statement covers all acceptance probabilities, including the ones for which that conclusion is vacuous.
  • Why the signed Fourier maximum is the right object. Step 2.2 identifies dist⁡(f,Λ) with (1−M)/2 where M=max⁡ah^(a) is the largest coefficient, and step 2.1 bounds 2α−1 by the same M; both the geometry of the nearest linear table and the acceptance probability are controlled by that single signed maximum. A maximum of ∣h^(a)∣ would not do: the constant table f≡1 has h^(0)=−1 and is at distance 1 from the only linear table of dimension zero, while ε=1.
  • Use in the tester. The proportional form is what the exponential-base tester of An exponential-size constant-query base assignment tester needs for its first rejection family: a table that is not close to any linear function is rejected with probability bounded below by the constant distance threshold, and the Hadamard system of Hadamard linearity constraint system materializes the tests as explicit constraints.
DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Quadratic tensor consistency test

Definition

Let n≥0. For r,s∈F2n put r⊗s:=(risj)1≤i,j≤n∈F2n×n, the tensor product of the two vectors, and identify F2 n×n with F2 n2 through the fixed row-major order of the index pairs (i,j); for n=0 the tensor product is the unique empty matrix and both cubes have the single point ∅. A Hadamard table is a function on one of these cubes, in the conventions of The BLR linearity test over F_2: f:F2n→F2 and g:F2 n×n→F2 are fixed tables, and the tensor of a vector with itself is the matrix u⊗u of entries uiuj.

The ideal tensor test. Choose r,s∈F2n independently and uniformly, query f(r), f(s) and g(r⊗s), and accept exactly when g(r⊗s)=f(r)⋅f(s), the right-hand side being the product in F2 of the two queried bits, equal to 1 precisely when both factors are 1. The test uses three table queries and 2n random bits; query points may coincide, and for n=0 all three points are the empty index, so the test reads the two single-entry tables and accepts exactly when g=f⋅f. The rejection probability of a pair (f,g) is εten(f,g):=Pr⁡r,s[g(r⊗s)≠f(r)f(s)], over the two independent uniform choices with the tables fixed.

Perfect completeness. If f=ℓu, that is f(r)=u⋅r, and g=ℓu⊗u, that is g(z)=(u⊗u)⊙z with ⊙ the coordinatewise-modulo-two dot product of The BLR linearity test over F_2, then for all r,s the distributivity of the dot product gives (u⊗u)⊙(r⊗s)=∑i,juiujrisj=(∑iuiri)(∑jujsj)=(u⋅r)(u⋅s), so the test accepts with probability one and εten(ℓu,ℓu⊗u)=0.

The self-corrected implementation. When the tables are only close to linear rather than linear, the test is executed on decoded values: each queried value is replaced by a two-query self-correction with auxiliary points chosen uniformly and independently, Corr⁡f(r;y):=f(y)+f(r+y),Corr⁡g(r⊗s;Y):=g(Y)+g(r⊗s+Y), with y∈F2n and Y∈F2 n×n uniform, in the convention of Two-query linear self-correction. The self-corrected tensor test accepts exactly when Corr⁡g(r⊗s;Y)=Corr⁡f(r;y)⋅Corr⁡f(s;y′), with independent auxiliary points y,y′,Y; it uses six table queries — f twice for each of its two decoded values and g twice — and n2+4n random bits for the independent choices r,s,y,y′,Y. The two tests differ only in reading the tables at auxiliary points instead of at the queried points; when the tables are linear the readings agree for every choice of the auxiliaries, since then Corr⁡f(r;y)=f(r) and Corr⁡g(r⊗s;Y)=g(r⊗s).

Remarks

  • Why the tensor form is the right consistency condition. A tensor table g claiming to encode w∈F2 n×n is consistent with f encoding u exactly when w=u⊗u; by the displayed bilinearity the test at (r,s) compares the two bits rWs and (u⋅r)(u⋅s), where W is w read as a matrix, so the rejection event is the event r(W−u⊗u)s=1, a rank-one condition that Quadratic tensor test rejects an inconsistent tensor quantifies by the half-cube principle. The test is the quadratic analogue of the linearity test The BLR linearity test over F_2, applied to the product structure rather than to addition alone.
  • Ideal and noisy parts are separated on purpose. The ideal test is a mathematical condition on exactly linear tables and its rejection probability is what the soundness analysis computes; the self-corrected test is the constant-query implementation available to a verifier that knows only that the tables are close to linear, and its additional error is bounded by the self-correction failure probabilities. The definition fixes the queries, the auxiliary distributions and the acceptance rules of both, and states no error bound: that is the content of Quadratic tensor test rejects an inconsistent tensor.
  • Multiplicity and order. The coordinates of r⊗s are ordered row-major and g is indexed by that fixed order; the pair (r,s) ranges over all ordered pairs, so an unordered pair of vectors contributes the two outcomes (r,s) and (s,r) whose tensors transpose one another. The product f(r)f(s) is symmetric in its arguments while g need not be, which is why the ordered form is the one materialized by the tester of An exponential-size constant-query base assignment tester.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Quadratic tensor test rejects an inconsistent tensor

Statement

Let n≥0, let u∈F2n and let w∈F2 n×n be read as a matrix, with W the matrix of entries wij and u⊗u the matrix of entries uiuj. Suppose W≠u⊗u. Then:

  1. (Ideal test.) The ideal tensor test of Quadratic tensor consistency test applied to the linear tables f=ℓu and g=ℓw, where ℓw(z)=w⊙z, rejects with probability at least 1/4: εten(ℓu,ℓw) ≥ 14.
  2. (Self-corrected test.) If instead f and g are arbitrary tables at distances δf,δg<1/4 from ℓu and ℓw respectively, then with auxiliary points drawn independently as in the self-corrected tensor test, Pr⁡[self-corrected test rejects] ≥ 14−4δf−2δg. In particular, whenever 4δf+2δg<1/4 the self-corrected test rejects with probability bounded below by the positive constant 1/4−4δf−2δg, and for δf,δg≤δ this is at least 1/4−6δ. More generally, if an ideal test using c values from tables each within distance δ<1/4 of a specified linear table rejects with probability at least π, then replacing those values by self-corrections makes it reject with probability at least π−2cδ.

Facts & Assumptions

Given: an integer n≥0, vectors u∈F2n, w∈F2 n×n with W≠u⊗u, independent uniform r,s∈F2n, and the ideal and self-corrected tensor tests of Quadratic tensor consistency test.

[F1]

The ideal tensor test accepts exactly when g(r⊗s)=f(r)f(s), with r,s independent uniform; for linear tables f=ℓu, g=ℓw the two sides are (u⋅r)(u⋅s) and w⊙(r⊗s) (Quadratic tensor consistency test).

[F2]

For linear tables, (u⊗u)⊙(r⊗s)=∑i,juiujrisj=(u⋅r)(u⋅s), and w⊙(r⊗s)=∑i,jwijrisj, the matrix W having entries wij (Quadratic tensor consistency test).

[F3]

For every nonzero vector c∈F2m the linear function x↦c⋅x takes the value 1 on exactly half of the cube F2m; equivalently, two distinct linear Boolean functions disagree on exactly half the cube (Character orthogonality, inversion and Parseval).

[F4]

If a table h has distance δ<1/4 from a linear function ℓ, then ℓ is the unique linear function at distance less than 1/4, and at every fixed requested point the two-query self-corrector returns ℓ at that point with probability at least 1−2δ (Pointwise correctness under less than quarter noise, Two-query linear self-correction).

[F5]

The self-corrected tensor test replaces each of the three queried values by a two-query self-correction with independent auxiliary points and accepts exactly when the corrected values satisfy the tensor equation (Quadratic tensor consistency test).

Proof

technique · direct
1.1

The difference D:=W−u⊗u is a nonzero matrix over F2, so some column of D is nonzero; fix such a column j and let c be that column, a nonzero vector with (rD)j=c⋅r for every row vector r.

F2givenchoose
2.1

By [F3] applied to the nonzero c of step 1.1 the functional r↦(rD)j equals 1 on exactly half of the cube, so the set of r with rD≠0 has probability at least 1/2, because (rD)j=1 makes the row vector rD nonzero.

F3step 1.1algebra
3.1

Conditioned on any fixed r with rD≠0 the map s↦rDs=(rD)⋅s is a nonzero linear functional of s, so by [F3] it equals 1 for exactly half of the s; since s is independent of r, this conditional probability is 1/2 for every such r. For linear tables the ideal test rejects exactly when w⊙(r⊗s)≠(u⋅r)(u⋅s), and by [F2] the two sides differ by w⊙(r⊗s)−(u⊗u)⊙(r⊗s)=r(W−u⊗u)s=rDs, a bit; hence rejection is the event rDs=1 and its probability is at least 12⋅12=14, which proves the first clause. For n=0 the hypothesis W≠u⊗u is empty, the one-point cube having u⊗u as its only element.

F1F2F3step 2.1algebra
4.1

For the second clause let E be the event that all three self-corrections return the true linear values, namely Corr⁡f(r)=ℓu(r), Corr⁡f(s)=ℓu(s) and Corr⁡g(r⊗s)=ℓw(r⊗s); by [F4] and the union bound Pr⁡[Ec]≤2δf+2δf+2δg=4δf+2δg, since each failure bound holds uniformly at every requested point. On E the self-corrected test of [F5] reads exactly the values the ideal test reads at the same pair (r,s), so its outcome coincides with the ideal outcome. With π≥1/4 the ideal rejection probability of step 3.1, the union bound, without an independence assumption between E and (r,s), gives Pr⁡[self-corrected rejects] ≥ Pr⁡[{rDs=1}∩E] ≥ π−Pr⁡[Ec] ≥ 14−4δf−2δg.

F4F5step 3.1algebra
5.1

Steps 3.1 and 4.1 are the two clauses of the statement. If δf,δg≤δ the second bound is at least 1/4−6δ. For the general clause, let I be the event that the specified ideal test rejects, so Pr⁡[I]≥π, and let Ec be the event that all c corrections return their specified linear values. The uniform bound of [F4] and the union bound give Pr⁡[Ecc]≤2cδ even when requested points depend on the test randomness. On I∩Ec the corrected test rejects, so Pr⁡[corrected test rejects]≥Pr⁡[I]−Pr⁡[Ecc]≥π−2cδ. ∎

F4step 3.1step 4.1givenalgebra

Remarks

  • Where each factor of 1/2 comes from. The proof needs two independent half-cube events: a nonzero row functional in r, then a nonzero functional in s. This is the random subsum principle in its two-variable form, cited from the character lemma rather than reproved, and it is the only probabilistic input to the ideal analysis; the value 1/4 is exactly the product of the two halves and is therefore not improvable by this argument.
  • Dependence of the corrected values. The auxiliary points y,y′,Y are drawn independently, while the two query points within each correction are linked by the requested point. The union bound uses only the separate failure estimates, which hold for each requested point; it requires no independence between the correction event and the ideal rejection event.
  • Why this suffices for the tester. The constant-query tester of An exponential-size constant-query base assignment tester tests tables that are promised to pass a 0.99-linearity test, so their distances from linear are at most 0.01; with δf,δg≤0.01 the bound of the second clause is at least 1/4−0.06>0.19, a positive constant independent of n and of the tables. This is the sense in which the tensor test has soundness error bounded away from zero, and it is what the composition step later consumes.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Circuit satisfiability becomes linear-quadratic consistency

Statement

Let C be a Boolean circuit with s inputs and m non-input gates over the basis {NOT,AND,OR,0,1} of Boolean circuits: basis, fan-in, size, and depth, together with a list X=(x1,…,xn) of n≤s named input coordinates. Then there are an integer N=s+m and a list of m+1 equations in variables w1,…,wN over F2, of the form qk(w)=∑1≤i≤j≤NAk(i,j) wiwj  =  bk,Ak(i,j)∈F2,bk∈F2, such that:

  1. (Input prefix and size.) The variables w1,…,wn are the named input coordinates x1,…,xn, the next s−n variables are the other input coordinates, and there is one variable per wire of C; each equation has at most four nonzero coefficients Ak(i,j), the total number of nonzero coefficients is O(m+1), and all of them are determined by the gate list of C in time polynomial in the size of that list.
  2. (Exact extension.) For every x∈{0,1}n, the circuit C accepts x if and only if there are wn+1,…,wN∈F2 with w1⋯wn=x satisfying all m+1 equations. For each choice of all s input bits, the gate equations determine a unique assignment to the gate wires, each wℓ being the value of the corresponding wire in the evaluation of C on those full input bits; this assignment satisfies the additional output equation exactly when the circuit accepts that full input.
  3. (Random subsum.) If w∈F2N fails at least one of the equations, then for a uniform z∈F2m+1 the single combined equation ∑kzkqk(w)=∑kzkbk,that isA(z)⊙(w⊗w)=b(z), fails with probability at least 1/2; here A(z):=∑kzkAk∈F2 N×N is the coefficient matrix placed in the coordinate pairs (i,j) with i≤j, b(z):=∑kzkbk, and ⊙ is the coordinatewise dot product of The BLR linearity test over F_2. Consequently, if g:F2 N×N→F2 is the Hadamard table of w⊗w, a verifier can test the subsum with the one query g(A(z)), which it compares with the bit b(z) that it computes itself from the gate list.

Facts & Assumptions

Given: a Boolean circuit C with s inputs, including named inputs x1,…,xn, and m non-input gates and designated output, over the basis of Boolean circuits: basis, fan-in, size, and depth; the equations displayed below, in variables w1,…,wN with N=s+m; and the tensor conventions of Quadratic tensor consistency test.

[F1]

The gates of C are NOT, AND and OR of fan-in at most two and the constants 0,1, and C accepts the named prefix x when some assignment to the other s−n input coordinates makes the designated output evaluate to 1 in the topological evaluation (Circuit satisfiability, Boolean circuits: basis, fan-in, size, and depth, Assignment tester and rejection ratio).

[F2]

For w∈F2N the tensor w⊗w∈F2 N×N has coordinate wiwj at the pair (i,j); the coordinatewise dot product against a matrix Z supported on pairs with i≤j evaluates to ∑i≤jZ(i,j)wiwj, and for a vector U∈F2 N×N the Hadamard table is g(Z)=U⊙Z (Quadratic tensor consistency test, The BLR linearity test over F_2).

[F3]

If v∈F2M is a nonzero vector and z is uniform in F2M, then z⋅v=1 with probability exactly 1/2; equivalently, distinct linear Boolean functions differ on half the cube (Character orthogonality, inversion and Parseval).

[F4]

The arithmetic of F2 has 1+1=0, so for bits u,v the identities u+u=0, u2=u, u∨v=u+v+uv and u∧v=uv hold, and ¬u=1+u (The BLR linearity test over F_2).

Proof

technique · direct
1.1

List the wires of C in a topological order, starting with the n named inputs x1,…,xn, then the s−n remaining inputs, and then the m non-input gates, and create one variable wℓ∈F2 for each wire, so that the first n variables are exactly the named input coordinates and N=s+m. Record the index of the designated output wire from the circuit description; it need not be the last wire in this order. This is a construction on the explicit gate list, and it uses no choices.

F1construct
2.1

For every gate, in the topological order of the wire list of step 1.1, write one equation in the variables of its input wires and its output wire, and write one further equation for the designated output; all arithmetic is in F2:

NOT z=¬x:z+x=1;AND z=x∧y:z+xy=0; OR z=x∨y:z+x+y+xy=0;constant z=⊤:z=1;constant z=⊥:z=0; output z=⊤:z=1.

Each equation is of the displayed form qk(w)=bk: the linear term wℓ is the diagonal coefficient Ak(ℓ,ℓ) (legitimate because wℓ2=wℓ for a bit), a product wiwj with i≠j is the coefficient Ak(i,j) at the unique pair with i<j, and the constant on the right side is bk∈F2. Every gate equation has at most four nonzero coefficients (the OR equation has four), and the output equation has one, so the list has m+1 equations and at most 4m+1=O(m+1) nonzero coefficients, all read off the gate list of step 1.1 in polynomial time. [F4, given, step 1.1, construct]

3.1

Suppose C accepts x, choose an assignment to the other s−n inputs witnessing acceptance and assign to every wire variable its value in the resulting full-input evaluation of C. Then w1⋯wn=x, and each equation of step 2.1 holds: for NOT, AND, OR and the constants this is exactly the evaluation rule in the identities of [F4], and the output equation holds because the designated output evaluates to 1. Hence the system has a solution extending x; it is the evaluation assignment of the circuit.

F1F4step 1.1step 2.1algebra
3.2

Conversely, suppose w satisfies all m+1 equations and w1⋯wn=x. We show by induction along the topological order that every wire variable equals the evaluation of its wire on the full input assignment encoded by w1,…,ws. The input variables do by the definition of that assignment. For a gate whose input wires are already correct, its equation determines the output variable: the NOT equation gives z=1+x, the AND equation z=xy, the OR equation z=x+y+xy, and the constant equations give z=1, z=0, all of which are the evaluation rules of [F4]. Thus every wire variable has its evaluated value, including the designated output wire; the additional output equation forces that value to be 1. Hence C accepts x by [F1], and once all s input coordinates are fixed, the extension to the gate wires is unique: two solutions with the same full input assignment agree wire by wire in the same induction.

F1F4step 1.1step 2.1algebra
3.3

For the random subsum, fix any w∈F2N and put vk:=qk(w)+bk∈F2, so that vk=0 says the k-th equation is satisfied, and let U:=w⊗w; by [F2] each qk(w)=Ak⊙U, so vk=Ak⊙U+bk and the combined equation of the statement holds exactly when z⋅v=0. If w fails at least one equation then v≠0, and [F3] makes z⋅v=1 with probability exactly 1/2 over the uniform z; this is the claimed rejection probability, and the combined left side is A(z)⊙U=g(A(z)) for the Hadamard table of U, so the whole test costs the single table query g(A(z)) plus the computation of A(z) and b(z) from the gate list.

F2F3step 2.1algebra
4.1

Steps 3.1, 3.2 and 3.3 are the three clauses of the statement: the construction of steps 1.1 and 2.1 gives the input prefix and the size bound, the pair of inductions gives the exact extension equivalence and uniqueness conditional on the full input assignment, and the subsum argument gives the constant rejection probability of a single tensor query; the circuit-to-system translation is deterministic and polynomial time in the gate list.

step 1.1step 2.1step 3.1step 3.2step 3.3∎

Remarks

  • Why diagonal coefficients are legitimate. The equations are bilinear in the tensor w⊗w, and in characteristic two the diagonal coordinate wℓwℓ equals wℓ for a bit; this is how the linear terms z+x=1 of a NOT gate and the constants are written without adding a coordinate fixed to one. The upper-triangle support i≤j with the ordered-pair indexing of the tensor is the Arora-Barak convention, and it is what makes A(z)⊙(w⊗w) reproduce the quadratic form without a factor 2, which would vanish in F2.
  • Exactness, not proximity. Clause 2 is an exact equivalence: every solution is the evaluation assignment for some completion of the named input prefix, so the system neither creates spurious satisfying inputs nor loses the accepted ones. The approximation enters only through the tables queried by the verifier, which are handled by the tensor test of Quadratic tensor test rejects an inconsistent tensor and the linearity test of BLR rejection is proportional to distance from linearity.
  • One equation per gate plus one for the output. Constants 0 and 1 are wires of the circuit and contribute their own equations, so a circuit whose output is a constant has its acceptance encoded by the output equation. A well-formed circuit has a designated output wire: when m=0 this must be one of the s input wires, while when s=0 a constant or other non-input gate supplies a wire; the impossible case s=m=0 is not part of the domain.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

An exponential-size constant-query base assignment tester

Statement

There is an absolute constant ρ≥1/500 with the following property. For every Boolean circuit C with s input coordinates, of which n≤s are named, and m non-input gates over the basis of Boolean circuits: basis, fan-in, size, and depth, presented together with its input-coordinate list X=(x1,…,xn), the construction below produces a constraint system P(C,X) of arity at most 6 over the fixed alphabet {0,1} such that:

  1. (Input coordinates and size.) The variables of P(C,X) are the 2N table coordinates of a table f:F2N→F2, the 2N2 table coordinates of a table g:F2 N×N→F2, with N:=s+m, together with the n named coordinates x1,…,xn; the number of constraints is at most 2q(s+m) for a fixed polynomial q, and the whole system is enumerated by a deterministic algorithm in time polynomial in that number, using no randomness and no choices.
  2. (Assignment tester.) P(C,X) is an assignment tester with alphabet {0,1}, arity bound 6 and rejection ratio ρ in the sense of Assignment tester and rejection ratio: every accepted input extends to a labeling with UNSAT⁡=0, and for every x and every labeling b of the non-input variables, UNSAT⁡x∪b(P(C,X))≥ρ δ(x,SAT⁡(C)), where the input coordinates of the system are the named coordinates x1,…,xn themselves.

Facts & Assumptions

Given: a Boolean circuit C with s inputs and m non-input gates, its named input list X=(x1,…,xn), and the constants N=s+m, M0=m+1, ε0:=1/100, ρ:=1/500.

[F1]

The circuit yields, with w1,…,wN the wire variables whose first n entries are the named input coordinates, a list of M0=m+1 equations qk(w)=Ak⊙(w⊗w)=bk with Ak supported on coordinate pairs i≤j and with at most four nonzero coefficients each; a well-formed circuit with a designated output has N=s+m≥1; for every input x, C accepts x if and only if the system has a solution w whose first n coordinates are x, in which case each solution is the evaluation assignment for some completion of the named inputs, unique once all s input bits are fixed; if w fails an equation then a uniform z∈F2M0 has A(z)⊙(w⊗w)≠b(z) with probability exactly 1/2, where A(z)=∑kzkAk and b(z)=∑kzkbk; the construction is deterministic and polynomial time in the gate list (Circuit satisfiability becomes linear-quadratic consistency).

[F2]

The BLR rejection probability of a table equals the fraction of violated linearity constraints of its Hadamard system and is at least its distance from the nearest linear function, with no restriction on the acceptance probability; if that distance is less than 1/4 the nearest linear function is unique (BLR rejection is proportional to distance from linearity, The BLR linearity test over F_2, Pointwise correctness under less than quarter noise).

[F3]

The ideal tensor test and its self-corrected implementation use three and six table queries respectively; if f and g are at distances δf,δg<1/4 from the linear tables ℓu and ℓw and w≠u⊗u, the self-corrected test rejects with probability at least 1/4−4δf−2δg (Quadratic tensor consistency test, Quadratic tensor test rejects an inconsistent tensor).

[F4]

A two-query self-correction of a table at distance δ<1/4 from a linear function ℓ returns ℓ at the requested point with probability at least 1−2δ (Pointwise correctness under less than quarter noise, Two-query linear self-correction).

[F5]

A map from circuits to constraint systems is an assignment tester with alphabet Σ0, arity bound q and rejection ratio ρ when it contains the named input coordinates and satisfies the perfect-completeness and proximity-soundness clauses; the value of a constraint system is the fraction of its listed constraints satisfied, with multiplicity (Assignment tester and rejection ratio).

Proof

technique · constructive
1.1

Apply the construction of [F1] to (C,X), obtaining the wire variables w1,…,wN with the n named input coordinates first, followed by the other s−n input coordinates, the coefficient vectors Ak and right sides bk for k=1,…,M0, and the exact extension equivalence. The output system of this step has the variables f(z)  (z∈F2N),g(Z)  (Z∈F2 N×N),x1,…,xn, that is, one variable for every table coordinate of the two Hadamard tables together with the n named input coordinates of the circuit; the intended labeling is f=ℓw, g=ℓw⊗w, xi the i-th input bit, for the evaluation vector w of an accepted input.

F1construct
2.1

List the following five families of constraints on the variables of step 1.1, each constraint being the tuple of the variables displayed together with the fixed binary relation displayed, where sums are in F2, ei is the i-th unit vector, and r,s,y,y′∈F2N, Y,Z,W∈F2 N×N, z∈F2M0, i∈[n] range over all choices:

(Lf)(f(r),f(s),f(r+s)), relation a+b=c; (Lg)(g(Z),g(W),g(Z+W)), relation a+b=c; (T)(f(y),f(r+y),f(y′),f(s+y′),g(Y),g(r⊗s+Y)), relation (a1+a2)(b1+b2)=c1+c2; (S)(g(Y),g(A(z)+Y)), relation a+b=b(z); (C)(f(y),f(ei+y),xi), relation a+b=c.

The families have 4N, 4N2, 2N2+4N, 2M0+N2 and n2N constraints respectively, and every constraint has arity at most 6 over {0,1}; each relation is an explicit table over the binary alphabet and is determined by the circuit. Put K:=max⁡{2N,2N2,N2+4N,M0+N2,N},L:=max⁡(1,n)2K. Each of the four nonempty power-of-two family sizes divides 2K, and when n≥1 the comparison-family size n2N divides L as well. Duplicate every constraint in a nonempty family F exactly L/∣F∣ times. Thus every nonempty family contributes L constraints; the comparison family is empty when n=0, and the total is then 4L, otherwise 5L. Since a well-formed circuit has N≥1, M0=m+1≤N+1 gives K≤5N2 and hence 5L≤26N2+3. Therefore the constraint count is at most 2q(N) for the fixed polynomial q(t):=6t2+3, with N=s+m. Enumerating the families, the coefficient vectors A(z) and the right sides b(z) — computed from the gate list, not queried — takes time polynomial in that output count. [F1, F5, step 1.1, construct, algebra]

3.1

Perfect completeness. Let x∈SAT⁡(C) and choose a completion of the named inputs witnessing acceptance and let w be its evaluation vector from [F1], with first n coordinates x. Label f=ℓw, g=ℓw⊗w and xi by the corresponding bit, so that Corr⁡f and Corr⁡g return the linear values at every point. Every (Lf) and (Lg) constraint is satisfied because linear functions satisfy the BLR equation, every (T) constraint is satisfied by the perfect completeness of the tensor test, every (S) constraint is satisfied because A(z)⊙(w⊗w)=∑kzk(Ak⊙(w⊗w))=∑kzkbk=b(z) as w solves every equation of [F1], and every (C) constraint is satisfied because Corr⁡f(ei)=ℓw(ei)=wi=xi. Hence UNSAT⁡x∪b(P(C,X))=0 for that labeling.

F1F3F5step 2.1algebra
3.2

First soundness case: suppose the table f of a labeling b is at distance more than ε0=1/100 from every linear function. Then the fraction of violated (Lf) constraints is the BLR rejection probability of f, which by [F2] is at least that distance, hence exceeds ε0; after duplication this is still the violated fraction of the family. Since the total is at most 5L, the system as a whole violates at least ε0L/(5L)=ε0/5=ρ of its constraints, and ρ≥ρ δ(x,SAT⁡(C)) because δ≤1. The same argument applies to the table g and the family (Lg).

F2F5step 2.1algebra
3.3

Second soundness case: suppose both tables are at distance at most ε0 from linear functions, say f from ℓu and g from ℓv in the unique sense of [F2], and v≠u⊗u. Then by [F3] the self-corrected tensor test rejects with probability at least 1/4−4ε0−2ε0=1/4−3/50=19/100, and the (T) constraints are exactly the outcomes of that test, one for each choice of (r,s,y,y′,Y); hence at least (19/100)L of them are violated and the system violates at least (19/100)/5>1/500=ρ of its constraints.

F3F5step 2.1algebra
3.4

Third and fourth soundness cases: suppose in addition v=u⊗u. If the vector u does not satisfy every equation of [F1], then the fraction of z with A(z)⊙v≠b(z) is exactly 1/2 by the random subsum clause, and for each such z the fraction of auxiliary points Y with g(Y)+g(A(z)+Y)≠b(z) is at least 1−2ε0 by [F4] applied to the two uniform points Y and A(z)+Y; hence at least 12(1−2ε0)=49100 of the (S) constraints are violated and the system violates at least 49500>ρ of its constraints. If instead u satisfies every equation, then by the exact extension equivalence of [F1] the input prefix u1⋯un lies in SAT⁡(C). For n≥1, writing d:=#{i:xi≠ui}/n gives d≥δ(x,SAT⁡(C)), and for each such i the auxiliary point y is uniform and independent of everything else, so by [F4] the fraction of y with Corr⁡f(ei;y)=ui≠xi is at least 1−2ε0; hence at least d(1−2ε0)≥49100δ(x,SAT⁡(C)) of the (C) constraints are violated, whence the system violates at least 49500δ(x,SAT⁡(C)). For n=0, an equation-satisfying u means the empty input is accepted by [F1], so δ=0 and the required bound is immediate; if the empty input is rejected then every u fails an equation and the preceding subsum case applies.

F1F4F5step 2.1algebra
4.1

Steps 3.1, 3.2, 3.3 and 3.4 cover every labeling: either a table is more than ε0 far from linear (step 3.2), or both are within ε0 of linear tables ℓu,ℓv with v≠u⊗u (step 3.3), or v=u⊗u and v fails an equation (step 3.4, first part), or v=u⊗u satisfies every equation and the comparison family charges the distance of x from SAT⁡(C) (step 3.4, second part). In each case UNSAT⁡x∪b(P(C,X))≥ρ δ(x,SAT⁡(C)) with ρ=1/500, and step 3.1 gives perfect completeness, so by [F5] the constructed system is an assignment tester with alphabet {0,1}, arity bound 6 and rejection ratio ρ; its size, explicitness and construction time are those of step 2.1.

F5step 2.1step 3.1step 3.2step 3.3step 3.4discharge-construct∎

Remarks

  • Where the exponential size is spent. The constraints are not sampled: every random choice of every test is materialized once, so the system contains 2N and 2N2 table coordinates and 2O(N2) constraints, with N=s+m. That is exponentially larger than the circuit but still of the form 2poly⁡(s+m), which is what this base tester asserts. A polynomial-size composition requires an additional robust input-preserving interface beyond this item.
  • Why the input coordinates are compared directly. The comparison family (C) reads the raw coordinate xi as a variable of the system and compares it with the self-corrected value of f at ei; this is the only place where the given input, rather than the decoded wire vector, enters, and it is what turns the exact extension equivalence of [F1] into the proximity clause of the tester. A decoding that satisfies the circuit but differs from x on a fraction d of the coordinates therefore forces d(1−2ε0) violated comparison constraints, the mechanism by which the rejection ratio becomes proportional to the Hamming distance rather than merely positive.
  • The constant 1/100. The threshold ε0 is fixed before the construction and is smaller than 1/4, so the nearest linear decoders are unique; it also leaves 1/4−3/50 in the tensor test and 49/100 in the subsum and comparison families, both bounded away from zero. Any smaller absolute threshold would do; the value is not optimised, only kept an absolute constant for later composition work.
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-27Open item page →

Gate constraints are a weak assignment tester

Statement

Let C be a Boolean circuit with s input coordinates, of which n≤s are named, and m non-input gates over the basis of Boolean circuits: basis, fan-in, size, and depth. Then there is a binary constraint graph G over the fixed alphabet Σ0={0,1}3 of eight symbols, with the n named input coordinates among its vertices, such that

  • G has M=O(m+s) ordinary edges, all relations being explicit tables of size at most 8×8, and both G and its edge list are produced from the gate list of C by a deterministic algorithm in time polynomial in m+s;
  • G satisfies the completeness and proximity inequalities of Assignment tester and rejection ratio with the instance-dependent bound ρC:=1/M: every accepted input extends to a satisfying labeling, and every input x and auxiliary labeling b satisfy UNSAT⁡x∪b(G)≥ρC δ(x,SAT⁡(C)). The input coordinates are labeled by the two designated symbols 0:=(0,0,0) and 1:=(1,1,1) of Σ0, identified with the bits of the input. As M grows with C, this alone does not give a fixed positive rejection ratio for the map on all circuits.

Facts & Assumptions

Given: a Boolean circuit C with s input coordinates including named coordinates x1,…,xn, m non-input gates (NOT, AND or OR, with fan-in at most two, and the constants 0,1) and a designated output, in the conventions of Boolean circuits: basis, fan-in, size, and depth, and the assignment-tester conventions of Assignment tester and rejection ratio.

[F1]

The wires of C are its s inputs and its m non-input gates; they can be listed in a topological order in which every gate follows its input wires. For any assignment to all s input coordinates, each wire value is determined by preceding wires; C accepts a named-input assignment x exactly when some assignment to the other s−n input coordinates makes its designated output equal 1 (Boolean circuits: basis, fan-in, size, and depth, Circuit satisfiability).

[F2]

A binary constraint graph has a finite nonempty alphabet, one relation Re⊆Σ2 per edge in a fixed endpoint order, loops with two incidences testing Re(a,a), isolated vertices removable without changing the value, and value equal to the fraction of ordinary edges satisfied when there is at least one edge; duplicating every edge preserves the value fraction, and relations are explicit Boolean tables (Constraint graph and labeling value).

[F3]

An assignment tester with alphabet Σ0 and rejection ratio ρ maps a circuit with named inputs to a constraint system containing those input coordinates, such that accepted inputs extend to labelings with unsatisfiability zero and every input x and auxiliary labeling b satisfy UNSAT⁡x∪b≥ρ δ(x,SAT⁡(C)), where δ is the relative Hamming distance to the accepted inputs and equals 1 when SAT⁡(C)=∅ (Assignment tester and rejection ratio).

Proof

technique · constructive
1.1

The gadget. Put Σ0:={0,1}3, with designated symbols 0=(0,0,0) and 1=(1,1,1); the remaining six symbols are auxiliary. Take as vertices the s+m wires of C, listed in a topological order, together with one gate vertex per non-input gate; the wire vertices include the named input coordinates and are the only vertices whose alphabet is restricted to the two designated symbols. For each wire vertex v add one loop edge with relation D:={(s,s):s∈{0,1}}, restricting its label to a designated symbol. Write d(0):=(0,0,0) and d(1):=(1,1,1) for the designated symbols. For a gate g with input wires p,q and output wire r let Tg⊆Σ0 be the set of triples (a,b,z) with z the value of g on the input bits a,b, taking b:=a and p=q for the one-input gate NOT, so that Tg is a four-element table for AND and OR and the two-element table {(a,a,1−a):a∈{0,1}} for NOT; for a constant gate let Tg:={(c,c,c)} and let r be its output wire. Add a loop edge at the gate vertex with relation Dg:={(s,s):s∈Tg}, add edges from the gate vertex to p and to q with relations P1:={((a,b,z),d(a)):(a,b,z)∈Tg} and P2:={((a,b,z),d(b)):(a,b,z)∈Tg}, and add an edge from the gate vertex to r with relation P3:={((a,b,z),d(z)):(a,b,z)∈Tg}; for a constant gate only P3 is added, there being no input wire. Finally add a loop at the output wire with relation {(1,1)}.

F1F2construct
2.1

Size and explicitness. Each wire contributes one loop, each non-constant gate contributes one loop and three projection edges, each constant gate one loop and one projection edge, and the output one further loop; hence M≤(s+m)+4m+1=O(m+s) ordinary edges, and Σ0 is a fixed eight-symbol alphabet with relations of at most 8×8 entries. Every relation is one of the finitely many displayed tables determined by the gate type, so the edge list is written down from the topological gate list in time linear in m+s.

F2step 1.1algebra
2.2

Perfect completeness. If x∈SAT⁡(C), choose an assignment to the other s−n input coordinates witnessing acceptance, and label every wire vertex by its value in the resulting evaluation of C, read as the designated symbol 0 or 1, and label each gate vertex by the triple of the two input values and the output value of that gate. Then every wire label is designated, so the loops with relation D are satisfied; each gate vertex carries a triple of its truth table, so its loop Dg is satisfied; each projection edge is satisfied because the triple's coordinates are exactly the labels of the corresponding wire vertices; and the output loop is satisfied because the output evaluates to 1. Hence UNSAT⁡x∪b(G)=0 for that labeling, the completeness clause of [F3].

F1F3step 1.1algebra
2.3

Soundness. Fix x∈{0,1}X and an arbitrary labeling b of the gate vertices and of the wire vertices other than the named inputs, and suppose all M edges of G are satisfied. Then every wire label is a designated symbol by the loops with relation D. Read the labels of the unnamed input wires as an assignment to the other s−n input coordinates. We show by induction along the topological order that each wire label equals the evaluation of that wire on this full input assignment: the input vertices have these values by construction, and for a gate whose input wires are already correct, its loop forces its label to be a truth-table triple (a,b,z) of that gate, while the projection edges force a and b to be the labels of its input wires and z to be the label of its output wire; hence the output wire of the gate carries the correct evaluated value. The induction terminates at the designated output wire, whose loop forces its label to be 1, so the output evaluates to 1 and C accepts x by [F1]. Therefore if x∉SAT⁡(C) at least one edge is violated under every b, that is UNSAT⁡x∪b(G)≥1/M; and since δ(x,SAT⁡(C))≤1 always, with δ=1 on SAT⁡(C)=∅ by [F3], this gives UNSAT⁡x∪b(G)≥δ(x,SAT⁡(C))/M.

F1F3step 1.1algebra
3.1

Steps 2.1, 2.2 and 2.3 verify the size, explicitness, completeness and instancewise soundness clauses with ρC=1/M, for the constructed graph over Σ0={0,1}3 of arity 2, the input coordinates being the wire vertices of the named inputs labeled by the designated symbols. Since M is unbounded across circuits, this is a weak, size-dependent proximity construction rather than a map with the fixed rejection ratio required by [F3].

F3step 2.1step 2.2step 2.3discharge-construct∎

Remarks

  • Why the alphabet has eight symbols. A gate vertex must carry the two input bits and the output bit so that the projection edges can force the wires; that is the three-coordinate alphabet {0,1}3, and the wire vertices use only the two designated symbols, which is what makes the induction of step 2.3 go through. The six other symbols may legitimately label gate vertices when they belong to that gate's truth table, but no such symbol satisfies a wire loop D, so it cannot fake an input or wire value.
  • The rejection ratio is only inverse-linear. Its ratio 1/M would require O(log⁡M) verified constant-factor amplification steps to reach a fixed rejection ratio; the point of this construction is that its size is linear in the circuit rather than exponential. It is deliberately not the constant-ratio tester; An exponential-size constant-query base assignment tester supplies that one at exponential size. Combining these two testers at polynomial size requires an additional robust input-preserving composition interface.
  • No randomness and no choice. The gadget is a fixed function of the gate list: the topological order is taken from the circuit description, the relations are the five displayed tables, and no vertex, edge or label is selected. In particular the construction uses no choice principle, and the identification of bits with designated symbols is part of the alphabet convention.
False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-09-27Open item page →

Repeating constraints amplifies the gap

Statement

Duplicating every constraint of a CSP instance the same number of times strictly increases its unsatisfaction fraction.

Facts & Assumptions

Given: the value conventions of Constraint graph and labeling value and a nonzero number of repetitions r≥1.

[F1]

A constraint system consists of a finite list of constraints; for a labeling σ the value val⁡σ(G) is the fraction of listed constraints satisfied, so duplicated constraints count with multiplicity, and UNSAT⁡σ(G)=1−val⁡σ(G) with UNSAT⁡(G)=min⁡σUNSAT⁡σ(G) (Constraint graph and labeling value).

[F2]

Gap-preserving reductions are compared through unsatisfaction fractions, so any operation that preserves the fraction of violated constraints for every labeling preserves UNSAT⁡ of the system (Complete uniform gap-preserving CSP reductions).

Refutation

technique · direct
1.1

Let G have M constraints and let G(r) be the system obtained by listing every constraint of G exactly r times, r≥1. If M=0, both lists are empty and the value convention of [F1] gives unsatisfaction zero for every labeling in both systems. If M>0, fix a labeling σ and let v be the number of constraints of G violated by σ: the list G(r) has rM constraints and exactly rv of them are violated, so UNSAT⁡σ(G(r))=rv/(rM)=v/M=UNSAT⁡σ(G); taking the minimum over labelings gives UNSAT⁡(G(r))=UNSAT⁡(G), never a strict increase.

F1F2algebra
1.2

The failure is nonvacuous. Take the alphabet Σ:={0,1}, the one-vertex constraint graph V:={x1} with the two loops of Constraint graph and labeling value: the loop carrying the relation R1:={(0,0),(1,1)}, which every labeling of x1 satisfies, and the loop carrying the empty relation R2:=∅, which every labeling violates. Then G has M=2 constraints, every labeling violates exactly one of them, and UNSAT⁡(G)=1/2; after repeating each constraint r times every labeling violates exactly r of the 2r listed constraints, so UNSAT⁡(G(r))=1/2 for every r≥1.

F1algebra
2.1

So the false statement fails at the witness of step 1.2 for every r≥1, and by step 1.1 no instance whatsoever has its unsatisfaction fraction changed by an equal repetition of its constraints. What an amplification step needs instead is a change of the variables and constraints, not a reweighting of the existing list: the powering of Constraint graph powering with local-view labels and the comparison families of the assignment-tester construction are of this genuinely different kind.

step 1.1step 1.2F1∎

5 · Examples, counterexamples and false statements

None yet.

Sources