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Extra info for 4-folds with numerically effective tangent bundles and second Betti numbers greater than one

Example text

We are given > 0, We want to estimate the mean M up to a precision . Again, classically, this will take O( 12 ), assuming that N is extremely large. Grover suggested a quantum algorithm to solve this problem in O( 1 ) steps[111]. Instead of showing Grover’s version, I will show a simple classical reduction[199] which allows solving the mean estimation problem given the median algorithm. The idea is that for Boolean functions the mean and median problems coincide. f1 (i)f2(i)f3(i)..... up to log( 2 ) digits, where fj (i) is the j th bit of f (i).

Now that we understand some of the limitations and advantages of the quantum model, let us go on to the subject of quantum noise. 11 Worries about Decoherence, Precision and Inaccuracies Learning about the possibilities which lie in quantum computation gave rise to a lot of enthusiasm, but many physicist[135, 189, 57, 19] were at the same time very sceptic about the entire field. The reason was that all quantum algorithms achieve their advantage over classical algorithms when assuming that the gates and wires operate without any inaccuracies or errors.

Why are such k s ”good”? Given an integer k which satisfies the criterion 26, we can find r with reasonably high probability. Note that for “good” k’s, there exists an integer m such that: m 1 k . | − |≤ Q r 2Q Remember that Q is chosen to be much larger than N , say Q ≥ N 2. This means k , a fraction with denominator ≥ N 2, can be approximated by m that Q r , a fraction with 1 denominator smaller than N , to within N 2 There is only one fraction with such a small denominator that approximates a fraction so well with such large denominator.

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