Y s 0 1 1 we now show that h 2 2 x0 2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: r all states |y ￿ such that y · s = 0. This means that Fourier sampling 1 1 ( √2 |x0 ￿ + √2 |x0 ⊕ s￿ results in a uniform superposition of y such that y · s = 0. Recall that ￿ 1￿ H ⊗n | x ￿ = αy |y ￿ where αy = (−1)x·y Ny ￿ 1 So H ⊗2 √2 (|x0 ￿ + |x0 ⊕ s￿) = 1 y αy |y ￿ where 2 1 αy = √ (−1)x0 ·y + (−1)(x0 ⊕s)·y 2 1 = √ (−1)x0 ·y (1 + (−1)s·y ) 2 1 Now it is easy to see that if s · y = 0, αy = ± √2 , but if s · y = 1, αy = 0. Therefore when we measure the first register, we will measure a y such that y · s = 0. Step 3: Repeat until there are enough such y ’s that we can classically solve for s. There are exactly n linearly independent values of y such that y · s = y1 s1 + y2 s2 + · · · yn sn = 0, and one of these is the trivial solution y = 0. Therefore, there are n−1 non-trivial, linearly independent solutions to y ·s = 0. But if y1 and y2 are linearly independent solutions, (y1 + y2 ) · s = y1 · s + y2 · s = 0 so linear combinations of solutions are also solutions. This gives us a total of 2n−1 y ’s such that y · s = 0. To solve for s, we need to find exactly n − 1 non-trivial, linearly independent y such that y · s = 0. For example, if s = 010, then y0 = 000, y1 = 001, and y2 = 100 are linearly independent solutions to y · s = 0. But the linear combination y1 + y2 = 101 is 46 CHAPTER 4. FOURIER SAMPLING & SIMON’S ALGORITHM also a solution. We need only find two of {y1 , y2 , y1 + y2 } in order to classically solve for s. How long should we expect this to take? The probability that we fail on the first run is the probability that we find y = 0, which is one value out of 2n−1 . So P1 = 1/2n−1 , where P1 denotes the probability of failing on the first run. Lets call the first nontrivial solution y1 . We fail when looking for y2 if we find 0 or y1 , so P2 = 2/2n−1 = 1/2n−2 . When looking for y3 , we fail if we find any of {0, y1 , y2 , y1 + y2 }, so P3 = 4/2n−1 = 1/2n−3 . Carrying on in this way, the probability of failing to find yi is Pi = 1/2n−i . The chance that we fail up to and including yi can be approximated by P < 1/2n−1 + 1/2n−2 + · · · + 1/2n−i . If we push this approximation all the way to i = n − 1, we see that we fail with probability less than 1 (compute the geometric sum). That’s not a strong enough approximation, so instead notice that our probability of failure up to and including i = n − 2 is less than 1/2. Then our probability of success up to the n − 2st run is greater than 1/2. We find the final linearly independent term on the last run with probability 1/2 (if you don’t believe this, notice that half of the solutions are linear combinations that include yn−1 ). Finally our total probability of success is P (success) > 1/2 ∗ 1/2 = 1/4. Therefore, we expect our process to take O(n) steps (our limit says 4 by n runs of the algorithm should be enough for success). Simon’s algorithm...
View Full Document

Ask a homework question - tutors are online