Violate the extended church turing thesis the idea

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Unformatted text preview: s. The idea behind proving a superpolynomial gap (which we will outline below) is to make each query itself be the answer to a Fourier sampling problem. Now each query itself is much easier for the quantum algorithm than for any probabilistic algorithm. Carrying this out recursively for log n levels leads to the superpolynomial speedup for quantum algorithms. 1/2n/2 4.6 Recursive Fourier Sampling Our goal is to give a superpolynomial separation between quantum computation and classical probabilistic computation. The idea is to define a recursive version of the fourier sampling problem, where each query to the function (on an input of length n) is itself the answer to a recursive fourier sampling problem (on an input of length n/2). Intuitively a classical algorithm would need to solve n subproblems to solve a problem on an input of length n (since it must make n queries). Thus its running time satisfies the recurrence T (n) ≥ nT (n/2) + O(n) which has solution T (n) = Ω(nlog n ). The quantum algorithm needs to make only two queries and thus its running time satisfies the recurrence T (n) = 2T (n/2) + O(n), which solves to T (n) = O(n log n). Here is how it works for two levels: we are given a black box computing a function f : {0, 1}3n/2 → {0, 1}, with the promise that for every n bit string x, the function fx : {0, 1}n/2 → {0, 1} defined by fx (y ) = f (xy ) (xy is denotes the concatenation of x and y ) satisfies fx (y ) = sx · y for some sx ∈ {0, 1}n/2 . We are also given a black box g : {0, 1}n/2 → {0, 1} which satisfies the condition that if we construct a boolean function h on n bits as h(x) = g (sx ), then h(x) = s · x for some n-bit string s. The challenge is to figure out s. 4.7. SIMON’S ALGORITHM 43 The proof that no classical probabilistic algorithm can reconstruct s is somewhat technical, and establishes that for a random g satisfying the promise, any algorithm (deterministic or probabilistic) that makes no(log n) queries to g must give the wrong answer on at least 1/2 − o(1) fraction of g ’s. This lemma continues to hold even if the actual queries are chosen by a helpful(but untrusted) genie who knows the answer. For those with a background in computational complexity theory – this establishes that relative to an oracle BQP ￿⊆ M A. M A is the probabilistic generalization of N P . It is conjectured that recursive fourier sampling does not lie in the polynomial hierarchy. In particular, it is an open question to show that, relative to an oracle, recusive fourier sampling does not lie AM or in BP P N P . 4.7 Simon’s Algorithm The Problem Suppose we are given a black box for computing a 2-to-1 function f : Zn → Zn (from n-bit strings to n-bit strings), with the promise that there is 2 2 a non-zero string s ∈ Zn \ {0} such that 2 for all x ￿= y , f (x) = f (y ) if and only if x ⊕ y = s. 01 1101 Here ⊕ is the bitwise direct sum modulo 2. For example, ⊕ 0 1 1 1 or ⊕ 0 1 . 1010 00 A quick example o...
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