[apoursh]

Summary

New package research/mode-fiber-radius-three-halves. It continues the merged #6320, which partly answered #6312's open question.

Theorem D: R ≤ 3/2. With #6312's lower bound, the least uniform ℓ∞ radius of exact mode-preserving rounding for the PR4933 mode matrix M now satisfies 1 ≤ R ≤ 3/2. This replaces the R ≤ 2 of #6320 (Theorem C). The bound holds for every x, with no genericity assumption.

Argument. It is shorter than #6320's and needs neither its genericity reduction nor its Hall lemma:

1. Centre x at its nearest integer vector.
2. Take a vertex of the fiber polytope in [−½,½]^112. All but at most 8 rows land exactly at ±½. Snapping the rest leaves a demand u ∈ L with ‖u‖∞ ≤ 4.
3. Balancing lemma. Each block of 8 rows has one pinned coordinate and three free ones, so it splits into four sign-free pairs worth ±2e_j each. With a rainbow matching in every block, each coordinate gets 7 pairs. Making one or two blocks straight fixes the parity class of u in H8. That leaves 6–10 pairs per coordinate against a demand of at most 4.
4. Set eᵢ = (sᵢ + κᵢ)/2 ∈ {−1,0,1}. Then the error is at most 1 + ½.

Verification

• verify.py uses the standard library and exact arithmetic, and runs in about 5 s, both normally and with -O. It checks:
  • the 14 pinned blocks and MᵀM = 56·I;
  • H8 closure and that complements of supports are supports;
  • column sums divisible by 8;
  • the rainbow and straight matchings with every pair identity;
  • the 16-case balancing table;
  • 150 exact runs of rounder.py, covering all three parity cases (worst error 503/336).
• Seven rejection controls fail closed.
• A separate exact stress run of 12,000 inputs found no failure. Its worst error was 1259/840 ≈ 1.4988, so 3/2 is the limit of this construction.
• research_package.py --check-only prints PASS_PACKAGE_METADATA_AND_PINS. The Python 3.12 replay is left to CI.
• The overlap search found only Prove exact mode preservation needs a full rounding unit (no κ) #6312 and Mode-fiber rounding radius: 1 ≤ R ≤ 2; R = 1 reduces to zonotope filling (no κ) #6320. Mode-fiber rounding radius: 1 ≤ R ≤ 2; R = 1 reduces to zonotope filling (no κ) #6320's package is unchanged.

Scope

• This is a property of the exact-mode contract only. Complete κ and the selected frontier are unchanged.
• It is not a claim that any caller needs the contract.
• It supplies no physical rounder, error or tape bill.
• R = 1 stays open; TASK.json records the next step and its blocker.

AI assistance: Anthropic Claude (Claude Code) found the argument, wrote the code and ran the checks under apoursh's direction (disclosed in TASK.json).

🤖 Generated with Claude Code

[apoursh]

FINDING (negative) / follow-up to this PR's open question (is R = 1?). No κ, no frontier change, nothing claimed or reserved.

Question. Does any facet of the zonotope give a simple obstruction to R = 1? Use #6320's Theorem A in centred form: R = 1 iff every y ∈ 2L ∩ int Z is a sign sum Σ κ_i m_i, where Z = Σ [−1,1] m_i. Near a facet with primitive normal θ, any sign sum lies at depth 2·Σ_{flipped}|⟨m_i,θ⟩|. Rows with ⟨m_i,θ⟩ = 0 are free. If some multiple k·g of g = gcd{|⟨m_i,θ⟩| ≠ 0} with k·g ≤ h(θ)/2 is not a subset sum of that multiset, the interior lattice layer at depth 2kg could contain no sign sum. That would be a candidate counterexample and would give R > 1.

Result. No facet tested has such a gap. Every multiple of g up to h(θ)/2 is a subset sum of the off-facet |⟨m_i,θ⟩| values.

• Exhaustive over primitive θ ∈ {−2,…,2}^8 whose zero rows have rank 7: 15,760 facets, 0 gaps.
• Random sampling of facet normals spanned by 7 rows (seeds 2 and 7): 6,728 and 14,622 distinct facets, 0 gaps.

Scope. This is local-only evidence from unpublished scripts, which are reproduced below. It runs on upstream main 7c49fe32c55de02c2c873655c804b16e4c3fef0a, reads research/coded-native-caller/NORMALIZERS.json (mode_columns) and uses exact integer arithmetic. It does not prove R = 1. It rules out only the one-dimensional depth obstruction, and only on the facets tested. The full facet set was not enumerated. Any counterexample must come from a codim ≥ 2 face or from a non-depth obstruction. 1 ≤ R ≤ 3/2 stands.

Open next step. Repeat the test on codim-2 faces. There, the two-dimensional depth vectors of the forced rows must cover the lattice points of the 2-D normal cone near the face. Alternatively, find a symmetry reduction that enumerates all facets.

Core check (Python 3, stdlib)

# rows = the 112 columns of mode_columns read as 8-vectors; th = primitive facet normal
dots=[sum(a*b for a,b in zip(th,m)) for m in rows]
nz=[abs(x) for x in dots if x]; g=math.gcd(*nz)
vals=sorted(x//g for x in nz); tot=sum(vals); ss=1
for v in vals: ss|=ss<<v          # subset-sum bitset
k=1
while k<=tot//2 and (ss>>k)&1: k+=1
gap = k<=tot//2                    # never True in any tested facet

Facet test: the rows with dot 0 have rank 7 (exact Fraction elimination).

Anthropic Claude (Claude Code) assisted under apoursh's direction.