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Strong Sensitivity Conjecture (bs(f) ≤ s(f)^2)

This file formalizes the strong sensitivity conjecture, asserting:

For every Boolean function f : {0,1}^n → {0,1}, bs(f) ≤ s(f)^2, where bs(f) denotes block sensitivity and s(f) denotes sensitivity.

Huang's theorem proves a quartic upper bound, bs(f) ≤ s(f)^4, thereby resolving the most widely known form of the sensitivity conjecture.

We now ask whether a stronger upper bound holds. Interestingly, the original paper of Nisan and Szegedy, where the sensitivity conjecture first appeared, already speculated that a quadratic upper bound might be the correct relation. On the lower bound side, Rubinstein (https://link.springer.com/article/10.1007/BF01200762) constructed Boolean functions exhibiting the first quadratic separation. The best currently known gap, due to Ambainis and Sun (https://arxiv.org/abs/1108.3494), is bs(f) ≥ (2/3)⋅s(f)^2.

References:

namespace StrongSensitivityConjecture open Finset Function Classical section Sensitivity variable {n : }

Flip operator, flip x B returns input x with bits in block B inverted.

def flip (x : Fin n Bool) (B : Finset (Fin n)) : Fin n Bool := fun i => if i B then !(x i) else x i

Local sensitivity s(f,x), number of indices where flipping one bit changes the value of f.

def sensitivityAt (f : (Fin n Bool) Bool) (x : Fin n Bool) : := #{i | f (flip x {i}) f x}

Global sensitivity s(f), maximum sensitivity of f over all inputs.

def sensitivity (f : (Fin n Bool) Bool) : := univ.sup (sensitivityAt f)

Check validity of block collection (disjoint and sensitive), A collection of blocks cB is valid for f at x if the blocks are disjoint and flipping any block changes f(x).

def IsValidBlockConfig (f : (Fin n Bool) Bool) (x : Fin n Bool) (cB : Finset (Finset (Fin n))) : Prop := (cB : Set (Finset (Fin n))).PairwiseDisjoint id B cB, f (flip x B) f x

Local block sensitivity bs(f,x), maximum size of a collection of sensitive, disjoint blocks for f at x.

noncomputable def blockSensitivityAt (f : (Fin n Bool) Bool) (x : Fin n Bool) : := Finset.sup {cB | IsValidBlockConfig f x cB} card

Global block sensitivity of f, maximum block sensitivity of f over all inputs.

noncomputable def blockSensitivity (f : (Fin n Bool) Bool) : := univ.sup (blockSensitivityAt f)

Strong Sensitivity Conjecture, for every Boolean function f : {0,1}^n → {0,1}, bs(f) ≤ s(f)^2.

We call this the strong sensitivity conjecture because the original sensitivity conjecture only asked for a polynomial bound in terms of s(f). Huang's celebrated result (often called the sensitivity theorem) gives a quartic bound, bs(f) ≤ s(f)^4, thereby settling the original conjecture.

@[category research open, AMS 68] theorem declaration uses 'sorry'strong_sensitivity_conjecture {n : } (f : (Fin n Bool) Bool) : blockSensitivity f sensitivity f ^ 2 := n:f:(Fin n Bool) BoolblockSensitivity f sensitivity f ^ 2 All goals completed! 🐙

Simple test example, A Boolean function whose block sensitivity is strictly greater than its sensitivity. Source: Nisan1989.

nisanExample(x) = 1 iff the Hamming weight of x is either n/2 or n/2 + 1. We assume n is a multiple of 4. The function is symmetric, so its value only depends on the Hamming weight of the input.

@[category test, AMS 68] def nisanExample (n : ) (x : Fin n Bool) : Bool := let w := #{i | x i} decide ((w : ) ({(n / 2 : ), (n / 2 : ) + 1} : Finset ))

Assuming n is a multiple of 4, the sensitivity of nisanExample is n/2, achieved by any x with Hamming weight n/2.

@[category test, AMS 68] lemma declaration uses 'sorry'nisanExample_sensitivity (n : ) (hn : 4 n) : sensitivity (nisanExample n) = n / 2 := n:hn:4 nsensitivity (nisanExample n) = n / 2 All goals completed! 🐙

Assuming n is a multiple of 4, the block sensitivity of nisanExample is 3n/4, achieved by any x with Hamming weight n/2. An optimal block configuration uses all n/2 1-bits as singleton blocks and forms n/4 disjoint size-2 blocks from the 0-bits.

@[category test, AMS 68] lemma declaration uses 'sorry'nisanExample_blockSensitivity (n : ) (hn : 4 n) : blockSensitivity (nisanExample n) = 3 * n / 4 := n:hn:4 nblockSensitivity (nisanExample n) = 3 * n / 4 All goals completed! 🐙 end Sensitivityend StrongSensitivityConjecture