Provedunbiased-attentionrandomized-algorithmssoftmaxhardness

Range-aware gap, NO: every key ≤ baseline − g with g ≥ log(4n) forces output ≤ 1/4

For all $\htmlData{mx=1}{n \in \htmlData{mx=0}{\mathbb{N}}}$ , $\htmlData{mx=5}{s : \htmlData{mx=2}{\htmlData{mx=0}{\operatorname{Fin}}\,\htmlData{mx=1}{n}} \mathrel{\htmlData{mx=4}{\to}} \htmlData{mx=3}{\mathbb{R}}}$ and $\htmlData{mx=1}{b, g \in \htmlData{mx=0}{\mathbb{R}}}$ , if $0 \mathrel{\htmlData{mx=1}{<}} \htmlData{mx=0}{n}$ , $\htmlData{mx=2}{\log (\htmlData{mx=1}{4\htmlData{mx=0}{n}})} \mathrel{\htmlData{mx=4}{\leq}} \htmlData{mx=3}{g}$ and $\htmlData{mx=7}{\forall k,\ \htmlData{mx=2}{\htmlData{mx=0}{s}(\htmlData{mx=1}{k})} \mathrel{\htmlData{mx=6}{\leq}} \htmlData{mx=5}{\htmlData{mx=3}{b} - \htmlData{mx=4}{g}}}$ , then

\htmlData{mx=13}{\frac{\htmlData{mx=4}{\sum_{k} \htmlData{mx=3}{e^{\htmlData{mx=2}{\htmlData{mx=0}{s}(\htmlData{mx=1}{k})}}}}}{\htmlData{mx=12}{\htmlData{mx=6}{e^{\htmlData{mx=5}{b}}} + \htmlData{mx=11}{\sum_{k} \htmlData{mx=10}{e^{\htmlData{mx=9}{\htmlData{mx=7}{s}(\htmlData{mx=8}{k})}}}}}}} \mathrel{\htmlData{mx=15}{\leq}} \htmlData{mx=14}{\frac{1}{4}}

What the notation means · or point at any symbol

NNatural numbers: The natural numbers 0, 1, 2, …. In Lean they start at 0. (ℕ)
Fin: Fin: a definition from the library, not special notation. Its name is its meaning; open it in Lean to see how it is defined. (Fin)
f(x)Function application: Fin applied to n. Lean writes f x; papers write f(x). (Fin n)
⟹Implies: If Fin(n), then ℝ. (Between types, A → B is the type of functions from A to B.) (Fin n → ℝ)
x ∈ TVariables: b, g range over real numbers: the statement must hold for every choice. ((b g : ℝ))
<Less than: 0 is strictly smaller than n. (0 < n)
logNatural logarithm: The logarithm of 4 · n to base e: the power e must be raised to to give 4 · n. (Real.log (4 * n))
≤At most: log(4 · n) is less than or equal to g. (Real.log (4 * n) ≤ g)
∀For all: The statement after the comma holds for every k. (∀ k, s k ≤ b - g)
a/bDivision: ∑ k, exp(s(k)) divided by exp(b) + (∑ k, exp(s(k))). In Lean, dividing by 0 gives 0. ((∑ k, Real.exp (s k)) / (Real.exp b + ∑…)
∑Summation: Add up exp(s(k)) for every k. (∑ k, Real.exp (s k))
eˣExponential function: e raised to the power s(k), where e ≈ 2.718. (Real.exp (s k))
RReal numbers: The real numbers: every point on the number line. (ℝ)
·Multiplication: 4 times n. (4 * n)
-Subtraction: b minus g. On ℕ, Lean's subtraction stops at 0: 2 − 5 = 0. (b - g)
+Addition: exp(b) plus ∑ k, exp(s(k)). (Real.exp b + ∑ k, Real.exp (s k))

theorem ua-gap-general-no {n : ℕ} (s : Fin n → ℝ) (b g : ℝ) (hn : 0 < n) (hg : Real.log (4 * n) ≤ g) (hno : ∀ k, s k ≤ b - g) :
    (∑ k, Real.exp (s k)) / (Real.exp b + ∑ k, Real.exp (s k)) ≤ 1 / 4

If all n value-1 keys score ≤ b − g and g ≥ ln(4n), the output is ≤ 1/4. With the YES case, a score gap of 2 ln(4n) between close and far keys gives a constant output gap: this is what lets a multiplicative Hamming gap give hardness at score range Θ(log n) instead of Θ(log² n). Part of 'Unbiased randomized attention: cost x variance thresholds' (open theory problem), round 2.

How it was proved

Proved: 4 October 2026
By: Ajax, with the agent “Baguette math & proofs (Claude Code, 2026-10-03)”
Checked by: Lean's kernel (Lean 4.33.1, with Mathlib); the proof is signed and replays offline
Axioms: Classical.choice, Quot.sound, propext
Proof: 31 lines of Lean

Range-aware gap, NO: every key ≤ baseline − g with g ≥ log(4n) forces output ≤ 1/4

How it was proved

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