This is 0i a0very1iinteresting f 1 not p if f 0

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Unformatted text preview: | evaluation. This is ±|0i a|0very1iinteresting = f (1) not p if f (0) problem (since there are < 2 it is a warmup for the Deutsch-Jozsa 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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