Steps our limit says 4 by n runs of the algorithm

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Unformatted text preview: is summed up by the following circuit. | 0￿ H H Uf | 0￿ | f ( x) ￿ Figure 4.3: Circuit for Simon’s Algorithm In summary, the above circuit for Simon’s algorithm corresponds to the 4.7. SIMON’S ALGORITHM 47 following sequence of transformations. 1￿ H ⊗n |0￿ |0￿ −→ √ |x￿ |0￿ 2n x Uf 1￿ −→ √ | x ￿ |f ( x ) ￿ 2n x measure 1 −→ √ (|x0 ￿ + |x0 ⊕ s￿) ⊗ |f (x0 )￿ 2 1￿ H ⊗n −→ √ α y | y ￿ |f ( x0 ) ￿ 2n y for some numbers αy . As above, √ each y , if s · y = 1, then αy = 0, whereas if s · y = 0, then for x0 ·y 2. αy = (−1) When we observe the first register, we get a uniformly random y such that s · y = s1 y1 + · · · + sn yn = 0. We repeat to collect more and more equations, and recover s from n − 1 linearly independent equations. Example ￿ 00 if x = 00 or 10 Let n = 2 and f (x) = so that s = 10. 01 if x = 01 or 11 First, apply the Hadamard transform to prepare |x￿, then Uf to prepare | f ( x) ￿ H ⊗2 |0￿ |0￿ −→ 3 3 x=0 x=0 Uf 1 ￿ 1￿ |x￿ |0￿ −→ | x￿ |f ( x) ￿ 2 2 Then measure the second register to finalize the first step of the process. For the purpose of argument, lets suppose we measure f (x) = 01, so that x0 = 01 and x0 ⊕ s = 11: 3 1￿ measure 1 |x￿ |f (x)￿ −→ √ (|01￿ + |11￿) ⊗ |01￿ 2 2 x=0 We then impose the Hadamard transform to achieve: 0 11 1 1 1 1 −1 1 −1 1 −1 11 1 = √ √ 2 2 1 1 −1 −1 0 2 0 1 1 −1 −1 1 0 We expect it will take 2 runs through the above process to measure y = 01. Then the only nonzero solution for s is s = 10....
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