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evaluation. This is ±0i a0very1iinteresting = f (1)
not
p
if f (0) problem (since there are
<
2 it is a warmup for the DeutschJozsa
only four such functions), but
 3i =
(1.44) algorithm.
0i 1i
: ±1i p 2 if f (0) 6= f (1). ¶3. It could be expensive to decide on a classical computer. For example,
Realizing that f (0) f (1) is 0 if f (0) = f (1) and 1 otherwise, we can rewrite this result
suppose f (0) = the millionth digit of ⇡ and f (1) = the millionth digit
concisely as
of e. Then the problem is to decide if the millionth digits of ⇡ and e 0i 1i
are the same.
p
 3 i = ±f (0) f (1)i
,
(1.45)
2
It is mathematically simple, but computationally complex.
so by measuring the ﬁrst qubit we may determine f (0) f (1). This is very interesting indeed: the quantum Begin with the the ability i determine
¶4. Initial state: circuit has given us qubits  0to = 01i. a global property of f (x), namely f (0) f (1), using only one evaluation of f (x)! This is faster than is possible
5
This is the 1998 i...
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 Fall '13
 BruceMacLennan

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