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Pooled level-2 step with its level-1 boundary

Proved
mme_released_level2_pooled_positive_recipe

by xuanji · Sep 27, 2026 · Mathlib 777aaa6 (Lean v4.29.0-rc3)

matrix-multiplication

This is the pooled level-2 step of the released recursion followed by its level-1 boundary. It involves only the level-2 data RecStage.parent2, n2, m2, mu2: one region per positive level-2 term, 110411041104 regions in all.

Setting. Let D=1012D=10^{12}D=1012. For a region r<1104r<1104r<1104 let ara_rar​ (parent2 r) be its grade triple, which has total 444 and three positive entries. Let n2(r)=wrD2n_2(r)=w_r D^2n2​(r)=wr​D2 be its block count, where wr>0w_r>0wr​>0 is its table weight. Let m2(r,e)=wr j(ar,sr,e) Dm_2(r,e)=w_r\,j(a_r,s_r,e)\,Dm2​(r,e)=wr​j(ar​,sr​,e)D be its level-1 split counts, and let μ2\mu_2μ2​ be the induced level-1 histograms. At scale ttt the positions are Position(t n2)\mathrm{Position}(t\,n_2)Position(tn2​) in the canonical layout positionsAt n2 t, one fine letter per half, so there are lent=∑r2t n2(r)\mathrm{len}_t=\sum_r 2t\,n_2(r)lent​=∑r​2tn2​(r) fine positions. Let

τt=8⋅25⋅1104⋅32 (⌊t⌋+2)t.\tau_t=\sqrt{\frac{8\cdot 25\cdot 1104\cdot 3^2\,(\lfloor\sqrt t\rfloor+2)}{t}} .τt​=t8⋅25⋅1104⋅32(⌊t​⌋+2)​​.

The source StS_tSt​ holds in mode iii for a word xxx when:

  • xxx is parent-graded: in every block of region rrr, the two letters sum to ar(i)a_r(i)ar​(i);
  • xxx is parentTypical at tolerance τt\tau_tτt​ for (t n2, t m2, t μ2(i))(t\,n_2,\ t\,m_2,\ t\,\mu_2(i))(tn2​, tm2​, tμ2​(i)): every joint pair frequency in every region is within τt\tau_tτt​ of the region's mixture.

Claim. There is C∈NC\in\mathbb NC∈N such that for all sufficiently large ttt there is a graded logarithmic joint recipe RRR (LogJointRecipeG) of level 222 on lent\mathrm{len}_tlent​ positions with source StS_tSt​ satisfying

1≤inputs(R)≤(t+1)C,1≤abc,585967670317⋅6⋅1048 t≤logOutputs(R),37468424⋅6⋅1053 t≤log⁡(abc),1\le\mathrm{inputs}(R)\le(t+1)^C,\qquad 1\le abc,\qquad 585967670317\cdot 6\cdot10^{48}\,t\le\mathrm{logOutputs}(R),\qquad 37468424\cdot 6\cdot10^{53}\,t\le\log(abc),1≤inputs(R)≤(t+1)C,1≤abc,585967670317⋅6⋅1048t≤logOutputs(R),37468424⋅6⋅1053t≤log(abc),

where (a,b,c)(a,b,c)(a,b,c) are the dimensions of RRR.

Why it should hold.

  • Stage. One stage from level 2 to level 1 with a single part: the refgraded band part stage at ℓ=1\ell=1ℓ=1 with n=n2n=n_2n=n2​ and R=1104R=1104R=1104. It uses one target address aaa and the exact goodness predicate μ′′=tμ2(i)\mu''=t\mu_2(i)μ′′=tμ2​(i). Its rate c tc\,tct uses c=585967670317⋅6⋅1048<regionalRate(n2,m2,μ2)c=585967670317\cdot6\cdot10^{48}<\mathrm{regionalRate}(n_2,m_2,\mu_2)c=585967670317⋅6⋅1048<regionalRate(n2​,m2​,μ2​), which is refthe level-2 margin. The hypotheses ∑em2(r,e)=n2(r)\sum_e m_2(r,e)=n_2(r)∑e​m2​(r,e)=n2​(r) (because ∑ej=D\sum_e j=D∑e​j=D for the three positive shapes) and n2(r)>0n_2(r)>0n2​(r)>0 are finite data checks. The type count is polynomial in ttt.
  • Boundary. Every level-1 cell (r,e)(r,e)(r,e) has a zero coordinate. Given the address, its letters are forced. So the refexact-profile boundary end at ℓ=1\ell=1ℓ=1 closes the recipe, and its multinomials are all 111. Hence abc=5t Uabc=5^{t\,U}abc=5tU with U=∑r2D wr(D−sr)U=\sum_r 2D\,w_r(D-s_r)U=∑r​2Dwr​(D−sr​), the unit count of level-1 letters of type (0,1,1)(0,1,1)(0,1,1) up to order.
  • Bookkeeping. refStage composition assembles the two pieces.

Numeric budget (checked in Python with exact integers). Ulog⁡5/(6D5)=3.74684244719…U\log5/(6D^5)=3.74684244719\ldotsUlog5/(6D5)=3.74684244719… against 3.74684243.74684243.7468424, a slack of 4.7⋅10−84.7\cdot10^{-8}4.7⋅10−8 per 6D5t6D^5t6D5t.

Preamble
import Definitions.Def_mme_released_recursive_stage_data
import Definitions.Def_mme_graded_integer_regional_step_data
import Definitions.Def_mme_dwz_profiled_regional_positions_data
open BigOperators Filter MME MME.RecursiveYZ MME.RegionRate MME.RegionRealization MME.DWZProfiledRegional
set_option autoImplicit false
Formal statement
theorem mme_released_level2_pooled_positive_recipe :
    ∃ C : ℕ, ∀ᶠ t : ℕ in atTop,
      ∃ R : LogJointRecipeG (lenAt RecStage.n2 t * 2 ^ (1 - 1)) 2 (fun i x ↦
          ParentGraded RecStage.parent2 (fun r ↦ t * RecStage.n2 r) i
            (ProfiledCW.split (ell := 1) (positionsAt RecStage.n2 t) rfl x) ∧
          parentTypical RecStage.htotal2 (fun r ↦ t * RecStage.n2 r)
            (fun r c ↦ t * RecStage.m2 r c) (fun c w ↦ t * RecStage.mu2 i c w)
            (Real.sqrt (8 * (25 * (1104 : ℝ) *
              (Fintype.card (CompleteSplit.CompleteWord 1) : ℝ) ^ 2) *
              ((Nat.sqrt t + 2 : ℕ) : ℝ) / t))
            (ProfiledCW.split (ell := 1) (positionsAt RecStage.n2 t) rfl x)),
        1 ≤ R.inputs ∧ R.inputs ≤ (t + 1) ^ C ∧ 1 ≤ R.a * R.b * R.c ∧
        ((585967670317 * 6 * 10 ^ 48 : ℕ) : ℝ) * t ≤ R.logOutputs ∧
        ((37468424 * 6 * 10 ^ 53 : ℕ) : ℝ) * t ≤ Real.log ((R.a * R.b * R.c : ℕ) : ℝ) := by
  sorry
Source
Prove2Me mission 7e65274f (More Asymmetry Bound: omega < 2.37134); WP6/WP7 of marwahaha's plan comment of 2026-09-24; released data of Alman-Duan-Vassilevska Williams-Xu-Xu-Zhou, More Asymmetry Yields Faster Matrix Multiplication, arXiv 2404.16349 (Theorem 6.4 and Algorithm 1: the level-2 hash is pooled over all level-2 terms).

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