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OR has a square-root approximate-degree lower bound and corresponding query lower bounds

Statement

For N≥1, define ORN(x)=0 at x=0N and 1 at every other Boolean vector. For 0≤ϵ<1/2 and d=deg⁡~ϵ(ORN;{0,1}N), d2≥1−2ϵ2N. Consequently every local bit-query algorithm computing OR with error at most ϵ has at least ⌈d/2⌉ queries, and Qϵbit(ORN;{0,1}N)≥⌈121−2ϵ2N⌉.

Facts & Assumptions

Given: An integer N≥1, an error 0≤ϵ<1/2, and the full-cube OR function.

[F1]

Acceptance polynomials and approximate degree on Boolean domains: approximate degree is attained by a polynomial that approximates on the promise domain.

[F2]

Symmetrizing a Boolean-cube polynomial gives a Hamming-weight polynomial: symmetrization preserves approximation and yields a polynomial in Hamming weight with no larger degree.

[F3]

Finite-grid Markov derivative inequality: if a degree-at-most-d polynomial has values in [0,1] on 0,1,…,N, then D:=max⁡[0,N]∣q′∣≤(d2/N)h, where h is its continuous range height; with Bounded integer-grid values control excursions by the derivative, h≤1+D.

[F5]

A T-query bit algorithm has an acceptance polynomial of degree at most 2T: for the local oracle-only model, Qϵbit(f;D)≥⌈deg⁡~ϵ(f;D)/2⌉.

[F6]

The reals form a totally ordered field: R has field arithmetic compatible with a total order, so positive denominators can be multiplied or divided in inequalities.

Proof

1.1givenF1F2constructalgebra

Choose an ϵ-approximating polynomial of degree at most d from [F1]. Apply [F2] to obtain a univariate polynomial q(t) of degree at most d whose values at integer k approximate ORN on the weight-k layer. Define r(t):=(q(t)+ϵ)/(1+2ϵ) and c:=(1−2ϵ)/(1+2ϵ). At k=0, q(0)∈[−ϵ,ϵ], so r(0)∈[0,2ϵ/(1+2ϵ)]; at each 1≤k≤N, q(k)∈[1−ϵ,1+ϵ], so r(k)∈[1/(1+2ϵ),1]. Thus all grid values lie in [0,1] and r(1)−r(0)≥c>0.

2.1step 1.1F3F4F6algebra

Let h=max⁡[0,N]r−min⁡[0,N]r and D=max⁡[0,N]∣r′∣. By [F4] on [0,1], D≥r(1)−r(0)≥c. By [F3], D≤ah with a:=d2/N, and h≤1+D. If a≥1, then a≥c/(1+c) since 0<c≤1. If a<1, then D≤a(1+D), so (1−a)D≤a and D≤a/(1−a). Combining c≤D≤a/(1−a) gives a≥c/(1+c). In both cases a≥c/(1+c)=(1−2ϵ)/2, proving d2≥(1−2ϵ)N/2 by ordered-field algebra [F6].

3.1step 2.1F5algebra∎

The local query lower bound follows from [F5]. Since d≥(1−2ϵ)N/2 by step 2.1, monotonicity of the ceiling gives the displayed explicit query estimate.

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