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A nonnegative-only adversary estimate need not capture the general bound

Statement refuted

The assertion that nonnegative adversary weights always attain the general signed adversary bound is false. The following family supplies a strict separation.

For the partial iterates f(1)=f and f(k+1)=f∘(f(k),f(k),f(k),f(k)) of the sorted four-bit function, Adv⁡±(f(k))>(2.511486)k,Adv⁡(f(k))≤(5/2)k. Thus signed adversary weights give a strict separation already at k=1 and an exponentially increasing ratio under iteration.

Facts & Assumptions

Given: The sorted four-bit function, its base certificates, and its partial Boolean iterates.

[F1]

The four-bit Ambainis function separates nonnegative and signed adversary bounds: the sorted four-bit f has Adv(f)=5/2 and Adv±(f)>2.511486, with explicit dual distributions and a signed matrix.

[F2]

Nonnegative and signed adversary bounds compose from below: both signed and nonnegative adversary values multiply as lower bounds when the composed block domain is nonempty.

[F3]

Input-distribution certificates upper-bound the nonnegative adversary value: for a nonconstant Boolean function, coordinate distributions with cross-output mismatch overlap η>0 give Adv≤1/η.

Counterexample

1.1givenF1F2constructalgebra

Let f be the sorted four-bit function and define f(1)=f and f(k+1)=f∘(f(k),f(k),f(k),f(k)) on their natural block domains. Since f is total on {0,1}4 and nonconstant, induction shows every f(k) is total on {0,1}4k and nonconstant: each outer bit string is realized by independently choosing a preimage of each bit under f(k), and outer strings with either f-value lift in the same way. Thus every composition domain is nonempty and both adversary bounds in [F2] apply. The signed matrix in [F1] and repeated composition in [F2] give Adv⁡±(f(k))≥Adv⁡±(f)k>(2.511486)k.

1.2givenF1F3construct

The base coordinate distributions in [F1] have mismatch overlap at least η=2/5. If PX(k) is a distribution over the 4k input coordinates with cross-output overlap at least ηk, define for an input X=(X1,…,X4) to f(k+1) PX(k+1)(j,r)=pX~(j)PXj(k)(r),X~j=f(k)(Xj). For opposite outputs of f(k+1), the outer strings X~,Y~ have opposite f-values. For each j with X~j≠Y~j, the inner pair Xj,Yj has opposite f(k)-values and contributes overlap at least ηk. Summing those j gives total overlap at least η⋅ηk=ηk+1; normalization follows by multiplying two probability distributions.

2.1givenF1F2F3step 1.1step 1.2algebra∎

By induction and [F3], Adv⁡(f(k))≤η−k=(5/2)k. Since 2.511486>2.5, the signed lower bound grows strictly faster than this nonnegative upper bound for every k≥1. Thus negative weights produce a strict adversary separation, and repeated composition amplifies it.

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