On Parity Decision Trees for Fourier-Sparse Boolean Functions

Abstract

We study parity decision trees for Boolean functions. The motivation of our study is the log-rank conjecture for XOR functions and its connection to Fourier analysis and parity decision tree complexity. Our contributions are as follows. Let f : 𝔽₂ⁿ → {-1, 1} be a Boolean function with Fourier support 𝒮 and Fourier sparsity k.  
- We prove via the probabilistic method that there exists a parity decision tree of depth O(√k) that computes f. This matches the best known upper bound on the parity decision tree complexity of Boolean functions (Tsang, Wong, Xie, and Zhang, FOCS 2013). Moreover, while previous constructions (Tsang et al., FOCS 2013, Shpilka, Tal, and Volk, Comput. Complex. 2017) build the trees by carefully choosing the parities to be queried in each step, our proof shows that a naive sampling of the parities suffices. 
- We generalize the above result by showing that if the Fourier spectra of Boolean functions satisfy a natural "folding property", then the above proof can be adapted to establish existence of a tree of complexity polynomially smaller than O(√ k). More concretely, the folding property we consider is that for most distinct γ, δ in 𝒮, there are at least a polynomial (in k) number of pairs (α, β) of parities in 𝒮 such that α+β = γ+δ. We make a conjecture in this regard which, if true, implies that the communication complexity of an XOR function is bounded above by the fourth root of the rank of its communication matrix, improving upon the previously known upper bound of square root of rank (Tsang et al., FOCS 2013, Lovett, J. ACM. 2016). 
- Motivated by the above, we present some structural results about the Fourier spectra of Boolean functions. It can be shown by elementary techniques that for any Boolean function f and all (α, β) in binom(𝒮,2), there exists another pair (γ, δ) in binom(𝒮,2) such that α + β = γ + δ. One can view this as a "trivial" folding property that all Boolean functions satisfy. Prior to our work, it was conceivable that for all (α, β) ∈ binom(𝒮,2), there exists exactly one other pair (γ, δ) ∈ binom(𝒮,2) with α + β = γ + δ. We show, among other results, that there must exist several γ ∈ 𝔽₂ⁿ such that there are at least three pairs of parities (α₁, α₂) ∈ binom(𝒮,2) with α₁+α₂ = γ. This, in particular, rules out the possibility stated earlier.