Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-30
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Alphabet reduction controls explicit size and degree

Statement

Fix a finite alphabet Σ with ∣Σ∣≥2 and let G be a finite binary constraint graph over Σ with m edge records and maximum degree at most d, where the degree of a vertex counts the incidence slots of its incident edge records and a loop therefore contributes two. Let AΣ be the alphabet reduction of Fixed-alphabet reduction with constant gap retention and let Σ^ be its fixed output alphabet of 66 symbols.

  1. Edges. ∣E(AΣ(G))∣≤6MΣm, where MΣ=lcm⁡(1,…,QΣ) with QΣ=27NΣ2+4 and NΣ=2⌈log⁡2W⌉+1+cW3 for the absolute constant c of Shared codeword blocks and edge acceptance circuits; both MΣ and QΣ depend only on Σ.
  2. Vertices. ∣V(AΣ(G))∣≤(2ℓ+qmax⁡+MΣ)m, where ℓ=2⌈log⁡2W⌉<2W and qmax⁡≤QΣ is the largest local gadget size; hence the vertex count is OΣ(m).
  3. Degree. Every vertex of AΣ(G) has degree at most 6MΣd, a constant depending only on Σ and d.
  4. Uniformity. All relation tables of AΣ(G) use the fixed alphabet Σ^, and AΣ is computable by a deterministic algorithm in time polynomial in the bit length of the explicit encoding of G.

Facts & Assumptions

Given: The fixed alphabet Σ, the input graph G with m edge records and maximum degree d, and the map AΣ of Fixed-alphabet reduction with constant gap retention, applied to G through the composition H=G∘P and the arity-six conversion.

[F1]

The map AΣ is deterministic, binary, runs in polynomial time in the explicit input encoding, uses the fixed alphabet Σ^={B(0),B(1)}⊔{0,1}6 of size 66, and satisfies ∣E(AΣ(G))∣≤CΣm and ∣V(AΣ(G))∣≤CΣm for a constant CΣ depending only on Σ. (Fixed-alphabet reduction with constant gap retention)

[F2]

If E(G)≠∅, every edge of G contributes exactly M constraints of the composition H=G∘P, where M=lcm⁡{qe} over the positive local gadget sizes, so H has ∣E(G)∣M constraints in total. If G is edgeless then H has no variables and no constraints. (Composition of an edge system with an assignment tester)

[F3]

Each active vertex v of G has one shared block Bv=((v,1),…,(v,ℓ)) of ℓ coordinates, shared by all incident edge gadgets, and every non-named gadget variable has a private copy (e,z) used by exactly one edge e. (Composition of an edge system with an assignment tester)

[F4]

The code length satisfies W≤ℓ<2W and each edge circuit has at most O(W2ℓ)=O(W3) gates besides its 2ℓ formal input bits. (Shared codeword blocks and edge acceptance circuits)

[F5]

A two-piece tester applied to a circuit with N wires has a finite constraint list of at most 27N2+4 constraints, each of arity at most six. (A two-piece constant-query PCP of proximity)

[F6]

The conversion keeps one shared vertex per input variable of the Boolean system and adds one private tuple vertex per listed constraint, joining the tuple vertex to the constraint's variables by at most q edge records in total for arities at most q. (Bounded-arity Boolean constraints become binary graph constraints)

[F7]

For each edge circuit, the local tester's total variable count is at most its constraint count qe; this follows from the explicit table-variable and nine-family counts in step 1.3 of Fixed-alphabet reduction with constant gap retention. Hence the private auxiliary variables of one edge gadget are at most qe≤qmax⁡.

Proof

Given: Fix Σ, W=∣Σ∣≥2, ℓ, the input graph G, and the constants NΣ, QΣ, MΣ of the statement.

1.1F2F4F5givenconstruct

Put H=G∘P and A=AΣ(G), the arity-six conversion of H. If E(G)≠∅, then H has Mm constraints by [F2], where each qe is the size of the local gadget of edge e. Every edge circuit has at most 2ℓ+O(W3) wires by [F4], so each qe≤QΣ by [F5], and therefore M divides MΣ=lcm⁡(1,…,QΣ). In particular M≤MΣ, a constant depending only on Σ.

1.2F1F2F6givencases

If E(G)=∅, then H has no variables and no constraints by [F2], and the conversion of an empty system is an edgeless graph, so all three bounds in clauses 1--3 hold with m=0 and every vertex degree zero.

2.1F2F6step 1.1algebra

Assume E(G)≠∅. Each of the Mm constraints of H has arity at most six, so the conversion creates at most six edge records per constraint by [F6]. Hence ∣E(A)∣≤6∣E(H)∣=6Mm≤6MΣm, as asserted in clause 1.

2.2F2F3F6F7step 1.1algebra

The vertex set of A consists of the vertices of H together with one private tuple vertex per constraint of H by [F6], so ∣V(A)∣=∣V(H)∣+Mm. The block coordinates account for at most 2ℓm coordinates, since each of the at most 2m nonisolated vertices contributes ℓ coordinates by [F3]. By [F7], the edge-private auxiliary variables of one gadget number at most qmax⁡, giving ∣V(H)∣≤(2ℓ+qmax⁡)m. Therefore ∣V(A)∣≤(2ℓ+qmax⁡+MΣ)m by step 1.1, proving clause 2.

2.3F2F3F5step 1.1algebra

Consider a vertex of A that comes from a coordinate x of a shared block Bv of H. By [F3] this coordinate appears in the gadgets of exactly the edge records incident to v, and v is incident to at most d edge records by hypothesis. In one copy of the gadget of such an edge, the coordinate occurs in at most qe constraints, each of arity at most six, so at most 6qe times; after the uniform duplication of [F2] it occurs at most 6qe⋅M/qe=6M≤6MΣ times in that gadget. Summing over the at most d incident records gives deg⁡A(x)≤6MΣd.

3.1F3F6step 2.3algebra

A vertex of A that comes from an edge-private auxiliary variable of H belongs to the gadget of exactly one edge, so the same occurrence count gives degree at most 6M≤6MΣ; a private tuple vertex created by the conversion has one edge record per variable occurrence of its constraint, hence degree at most six by [F6]. Since d≥1 whenever E(G)≠∅, all these degrees are at most 6MΣd. This proves clause 3.

3.2F1F6step 2.1step 2.2algebra

The determinism, polynomial running time and fixed output alphabet are inherited from the reduction AΣ by [F1]; the edge and vertex counts of clauses 1 and 2 are polynomial in m by steps 2.1 and 2.2, and all relation tables are the constant-size tables of the 66-symbol conversion.

4.1F1F2F4F5step 1.1step 1.2step 2.1step 2.2step 2.3step 3.1step 3.2algebradischarge-construct∎

Clauses 1, 2, 3 and 4 hold: the edgeless case is step 1.2, the edge and vertex bounds are steps 2.1 and 2.2, the degree bound is steps 2.3 and 3.1, and uniformity is step 3.2. Every constant produced is a function of Σ and d alone, namely 6MΣ, 2ℓ+qmax⁡+MΣ and 6MΣd.

Remarks

The degree bound is what makes the iterated transformation self-contained: after one application the output has constant degree depending only on the fixed input alphabet and the input degree bound, so the next round can use the same reduction with the same constants. The count MΣ is a constant for fixed Σ even though it is enormous, because each local tester has constant size once the alphabet is fixed. The proof is choice-free: the composition, the least common multiple and the conversion are deterministic finite constructions.

Depends on

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