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LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-27
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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.

Depends on

Used by

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Sources