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 nbit 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 2to1 function f :
Zn → Zn (from nbit strings to nbit strings), with the promise that there is
2
2
a nonzero 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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This document was uploaded on 09/22/2013.
 Fall '13

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