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Even-packet capacity: d+1d+1d+1 packets whose zero-sum subsets are even packet unions bound the colours by b(2d−1)b(2^d-1)b(2d−1) (Erdős 624)

Proved
Erdos624.even_packet_capacity

by Ethan Cheung Wang · Oct 25, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricserdos-problems

Setting: UUU is a finite set of inputs, L:P(U)→P(C)L:\mathcal P(U)\to\mathcal P(C)L:P(U)→P(C) an arbitrary list map (the actual colour lists), kkk a natural number. A colour uuu is nonbottom if u∉L(∅)u\notin L(\varnothing)u∈/L(∅) and covering if every kkk-subset Y⊆UY\subseteq UY⊆U contains some AAA with u∈L(A)u\in L(A)u∈L(A). An actual minimal witness of uuu is F⊆UF\subseteq UF⊆U with u∈L(F)u\in L(F)u∈L(F) and u∉L(F′)u\notin L(F')u∈/L(F′) for every F′⊊FF'\subsetneq FF′⊊F. A valid label for uuu is a map p:U→Vp:U\to Vp:U→V into an F2\mathbb F_2F2​-vector space with ∑x∈Fp(x)=0\sum_{x\in F}p(x)=0∑x∈F​p(x)=0 for every actual minimal witness FFF with 1≤∣F∣≤k1\le|F|\le k1≤∣F∣≤k. For T⊆UT\subseteq UT⊆U, rk⁡(T)=dim⁡span⁡p(T)\operatorname{rk}(T)=\dim\operatorname{span}p(T)rk(T)=dimspanp(T) and rk⁡(U)=dim⁡span⁡p(U)\operatorname{rk}(U)=\dim\operatorname{span}p(U)rk(U)=dimspanp(U) (the dimension of the valid parity space). Each colour u∈Su\in Su∈S has its own valid label pup_upu​ into one F2\mathbb F_2F2​-space. Theorem. Assume ∣L(A)∣≤b|L(A)|\le b∣L(A)∣≤b, every u∈Su\in Su∈S is nonbottom and covering with rk⁡u(U)≥k−d\operatorname{rk}_u(U)\ge k-drku​(U)≥k−d (stated as k≤rk⁡u(U)+dk\le\operatorname{rk}_u(U)+dk≤rku​(U)+d). Let P0,…,PdP_0,\dots,P_dP0​,…,Pd​ be pairwise disjoint nonempty packets with ∣P0∪⋯∪Pd∣≤k|P_0\cup\dots\cup P_d|\le k∣P0​∪⋯∪Pd​∣≤k such that for every u∈Su\in Su∈S: ∑P0∪Pjpu=0\sum_{P_0\cup P_j}p_u=0∑P0​∪Pj​​pu​=0 for j≠0j\ne0j=0, and every zero-sum G⊆⋃jPjG\subseteq\bigcup_jP_jG⊆⋃j​Pj​ is a union of an even number of packets. Then ∣S∣≤b(2d−1)|S|\le b(2^d-1)∣S∣≤b(2d−1).

This is the deterministic capacity core of Sol's 'Bias-Free Colour Bounds' (and of 'Bias-Free Bounds on Colour-Dependent Dense Pools'): there, packets with these properties are constructed probabilistically for s=b(2d−1)+1s=b(2^d-1)+1s=b(2d−1)+1 colours, which this theorem shows is impossible. The probabilistic construction is not formalized.

Context: Erdős Problem 624 asks whether H(n)−log⁡2n→∞H(n)-\log_2 n\to\inftyH(n)−log2​n→∞, where H(n)H(n)H(n) is the least mmm such that some f:P(X)→Xf:\mathcal P(X)\to Xf:P(X)→X with ∣X∣=n|X|=n∣X∣=n satisfies f[P(Y)]=Xf[\mathcal P(Y)]=Xf[P(Y)]=X for all ∣Y∣≥m|Y|\ge m∣Y∣≥m. This is an auxiliary finite statement from Sol's Round 13 notes on the two-track (PAIR) case. It does not settle Erdős Problem 624, PAIR or Step 17; the initial near-critical population those notes start from is not established.

Preamble
import Mathlib
open Finset
Formal statement
namespace Erdos624

theorem even_packet_capacity {U : Type*} [Fintype U] [DecidableEq U] {C : Type*} [DecidableEq C]
    {W : Type*} [AddCommGroup W] [Module (ZMod 2) W]
    (L : Finset U → Finset C) (b k d : ℕ) (hb : ∀ A, #(L A) ≤ b)
    (S : Finset C) (p : C → U → W) (hu : ∀ u ∈ S, u ∉ L ∅)
    (hcov : ∀ u ∈ S, ∀ Y : Finset U, #Y = k → ∃ A ⊆ Y, u ∈ L A)
    (hvalid : ∀ u ∈ S, ∀ F : Finset U, 0 < #F → #F ≤ k → u ∈ L F → (∀ F' ⊂ F, u ∉ L F') →
      ∑ x ∈ F, p u x = 0)
    (hglob : ∀ u ∈ S,
      k ≤ Module.finrank (ZMod 2) (Submodule.span (ZMod 2) (Set.range (p u))) + d)
    (P : Fin (d + 1) → Finset U) (hdisj : ∀ j l, j ≠ l → Disjoint (P j) (P l))
    (hne : ∀ j, (P j).Nonempty) (hk : #(Finset.univ.biUnion P) ≤ k)
    (hpair : ∀ u ∈ S, ∀ j, j ≠ 0 → ∑ x ∈ P 0 ∪ P j, p u x = 0)
    (hzs : ∀ u ∈ S, ∀ G ⊆ Finset.univ.biUnion P, ∑ x ∈ G, p u x = 0 →
      ∃ I : Finset (Fin (d + 1)), Even #I ∧ G = I.biUnion P) :
    #S ≤ b * (2 ^ d - 1) := by sorry

end Erdos624
Source
Erdős Problem 624 (https://www.erdosproblems.com/624); Round 13: Sol's Perplexity Computer notes on the fixed-bias / low-rank branch of Step 17 (derived from GPT 6.1 Sol's analysis, relayed by Ethan Cheung); hand-checked and Lean-verified by Grok Bot for Ethan Cheung (Oct 25 2026). Auxiliary result only; does NOT settle PAIR / Step 17 / #624.

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