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A fixed Grover iteration count can fail when a is unknown

Statement refuted

For every pair of initial success weights a1,a2∈(0,1), an iteration count maximizing success for a1 over 0≤k≤⌈1/a1⌉ gives success probability at least 1/2 for a2 when reused unchanged.

Facts & Assumptions

Given: Two prepared success weights and a common fixed iteration count.

[F1]

Exact success probability after amplitude amplification: after k iterations, success is sin⁡2((2k+1)θ) when sin⁡2θ=a.

[F2]

Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions gives sin⁡(π/6)=cos⁡(π/3) and cos⁡(2π/3)=−cos⁡(π/3); Double-angle and quadratic power-reduction identities gives cos⁡(2x)=2cos⁡2x−1; Pi is the first positive zero of sine gives sin⁡x>0 for 0<x<π; Parity and the Pythagorean identity for sine and cosine gives sin⁡2x+cos⁡2x=1; Quarter-turn values and shifts by pi/2 and pi gives sin⁡(π/2)=1 and sin⁡π=0.

Counterexample

1.1givenF1F2algebra

Put c=cos⁡(π/3)=sin⁡(π/6) by [F2]. Since 0<π/6<π, [F2] gives c>0. Also [F2] gives cos⁡(2π/3)=−c, while [F2] gives cos⁡(2π/3)=2c2−1. Hence (2c−1)(c+1)=0, and positivity gives c=1/2. By [F2], sin⁡2(π/3)=1−cos⁡2(π/3)=3/4. For a1=1/4, take θ1=π/6; the allowed range is 0≤k≤⌈1/a1⌉=2. The probabilities for k=0,1,2 are 1/4,1,1/4, respectively, so k=1 is the unique maximizer.

2.1givenF1F2step 1.1algebra∎

For a2=3/4, take θ2=π/3. Reusing k=1 gives sin⁡2(3θ2)=sin⁡2π=0<1/2, while direct sampling at k=0 succeeds with probability 3/4. Thus the first weight's maximizing count violates the claimed lower bound at the second weight.

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