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One fixed transformation has constant-factor growth

Statement

Let t be the integer fixed in One fixed-alphabet transformation doubles small gaps and let T:=Tt be the transformation of One fixed-alphabet Dinur transformation at this t. Then there are constants CE,CV,CD≥1, depending only on Σ⋆ and on this fixed t (and therefore fixed before any input graph is given), such that for every finite binary constraint graph G over Σ⋆ with m=∣E(G)∣ edge records ∣E(T(G))∣≤CE m,∣V(T(G))∣≤CV m, every vertex of T(G) has degree at most CD, and T is deterministic and computable in time polynomial in the bit length of the explicit encoding of G; explicitly one may take CE=6MΣtCt, CV=(2ℓt+qmax⁡,t+MΣt)Ct and CD=6MΣtdt, where MΣt,ℓt,qmax⁡,t are the constants attached to the input alphabet Σt by Alphabet reduction controls explicit size and degree. In particular the growth factor is bounded by constants independent of m, and T maps edgeless inputs to the empty graph.

Facts & Assumptions

Given: Fix the alphabet Σ⋆ and the integer t of One fixed-alphabet transformation doubles small gaps, with T:=Tt.

[F1]

For every integer t≥t0 and every finite binary Σ⋆-graph G one has Tt(G)=AΣt(Rt(G)), and Tt is a deterministic map from finite Σ⋆-graphs to finite Σ⋆-graphs. (One fixed-alphabet Dinur transformation)

[F2]

The alphabet Σt is finite of size ∣Σt∣=∣Σ⋆∣(2D)R≥2 with D=387 and R=t+⌈t⌉. (One fixed-alphabet Dinur transformation)

[F3]

The intermediate graph Rt(G) is a binary constraint graph over Σt with at most Ct∣E(G)∣ ordinary edges, and it is edgeless whenever G is edgeless. (One fixed-alphabet Dinur transformation)

[F4]

The gap-amplification step at Σ⋆ has output degree bound dt:=2(2D)2t+1=DO(t) and blowup Ct:=D⋅(2D)2t+1=DO(t). (A complete uniform graph gap-amplification step)

[F5]

The map Rt is deterministic and runs in time polynomial in the bit length of the explicit encoding of G, and the parameters Σt,dt,Ct depend only on ∣Σ∣ and t, never on ∣V(G)∣ or ∣E(G)∣. (A complete uniform graph gap-amplification step)

[F6]

For every finite alphabet Σ with ∣Σ∣≥2 and every finite Σ-graph H with m′ edge records, ∣E(AΣ(H))∣≤6MΣm′, where MΣ=lcm⁡(1,…,QΣ) depends only on Σ. (Alphabet reduction controls explicit size and degree)

[F7]

For the same input H, ∣V(AΣ(H))∣≤(2ℓ+qmax⁡+MΣ)m′, where ℓ=2⌈log⁡2W⌉<2W and qmax⁡≤QΣ are the length and largest gadget size attached to Σ. (Alphabet reduction controls explicit size and degree)

[F8]

Every vertex of AΣ(H) has degree at most 6MΣd when the input has maximum degree at most d, and AΣ is computable by a deterministic algorithm in time polynomial in the bit length of the explicit encoding of H. (Alphabet reduction controls explicit size and degree)

[F9]

The integer t satisfies t≥t0 and depends only on the absolute constants κ,β,c,t0, never on an input graph. (One fixed-alphabet transformation doubles small gaps)

Proof

Given: Use the fixed alphabet Σ⋆, the fixed integer t of [F9] and the map T=Tt.

1.1F1F2F3F4F9given

By [F9] the integer t≥t0 lies in the transformation domain, so by [F1] and [F2] the map T sends finite Σ⋆-graphs to finite Σ⋆-graphs by G↦AΣt(Rt(G)), with Σt a finite alphabet of size at least two; by [F3] the graph Rt(G) has at most Ct∣E(G)∣ edge records and is edgeless when G is edgeless, and by [F4] its degrees are at most dt.

1.2F4F5F6F7F8algebra

Let MΣt,ℓt,qmax⁡,t be the constants attached to the alphabet Σt by [F6]–[F8], and set CE:=6MΣtCt, CV:=(2ℓt+qmax⁡,t+MΣt)Ct and CD:=6MΣtdt. These are constants depending only on Σ⋆ and t, because Σt,dt,Ct do by [F4] and [F5], while MΣt,ℓt,qmax⁡,t do by their definition at the input alphabet Σt.

2.1F1F2F3F6step 1.1step 1.2algebra

Let G be an arbitrary finite Σ⋆-graph with m=∣E(G)∣ edge records. Applying [F6] with Σ:=Σt and H:=Rt(G), which is a finite Σt-graph by [F1]–[F3], gives ∣E(T(G))∣=∣E(AΣt(Rt(G)))∣≤6MΣt∣E(Rt(G))∣≤6MΣtCt m=CE m, the last inequality using [F3].

2.2F1F2F3F7step 1.1step 1.2algebra

Applying [F7] to the same input H=Rt(G) gives ∣V(T(G))∣≤(2ℓt+qmax⁡,t+MΣt)∣E(Rt(G))∣≤(2ℓt+qmax⁡,t+MΣt)Ct m=CV m, again using the edge bound of [F3].

2.3F1F2F4F8step 1.1step 1.2algebra

Applying [F8] to H=Rt(G) with degree parameter d=dt, which bounds the degrees of Rt(G) by [F4], shows that every vertex of T(G)=AΣt(Rt(G)) has degree at most 6MΣtdt=CD.

2.4F1F5F8step 1.1

By [F5] the first stage computes Rt(G) deterministically in time polynomial in the bit length of the explicit encoding of G, so the explicit encoding of Rt(G) has polynomially bounded length; by [F8] the second stage computes AΣt(Rt(G)) deterministically in time polynomial in the bit length of that encoding. Composing the two deterministic polynomial-time algorithms exhibits a deterministic algorithm computing T(G) in time polynomial in the bit length of the explicit encoding of G.

3.1F1F3F6F7step 2.1step 2.2step 2.3step 2.4∎

The graph G was arbitrary, the constants CE,CV,CD depend only on Σ⋆ and t by step 1.2, and all three output bounds and the uniformity clause were established in steps 2.1–2.4; hence T has the claimed constant-factor growth. If G is edgeless, then Rt(G) is edgeless by [F3], and [F6] and [F7] applied with m′=0 give ∣E(T(G))∣=0 and ∣V(T(G))∣=0: the output is the empty graph and all bounds hold as 0≤0.

Remarks

The constants are enormous — CE and CV are built from the least common multiple MΣt of the local gadget sizes — but they are fixed before any input is read, which is all the later iteration needs. Degree reduction, powering and alphabet reduction each blow up the graph by a constant factor once t and the alphabet are frozen, and the composition of the three maps stays deterministic polynomial time because each stage's explicit output encoding has polynomial length. No choice principle is used.

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