Talk:BQP

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Is the number of qubits of the quantum computer held constant or may it vary with input size? --AxelBoldt

I think it is assumed that there's always enough of them, just as we do with Turing tapes. --Seb

What does this 1/4-clause mean in long run ? That chance of failure in many runs is (1/4)^N, 1/2*(1/2)^N or what ? --Taw

The probability that the algorithm fails N times in a row is (1/4)N. Actually I think 1/4 is more or less arbitrary; choosing any other (rational?) number in ]0,1/2[ would not change the class. --Seb
Right. It works for irrational numbers too. But not complex, quaturnion, or surreal. :-) --LC

i removed the paragraph "Because randomness is inherent in quantum computers, there is no notion of deterministic algorithms, such as those implemented by ordinary Turing machines. This makes BQP the primary class of practical quantum algorithms that is studied." it is possible to have quantum algorithms that end in an eigenstate of the measurement basis. these algorithms give the same output every time, so they are not "random". neither is the process of running the algorithm, since quantum dynamics is deterministic in the absence of measurement. cheers. -- dave kielpinski

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