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different thing. How can we compare the number of
Boolean gates to the number of unitary quantum gates? (like comparing oranges and bananas ) and put it in the same world diagram of P and NP ( which is about different Turing machine computation step (head movement )? Boolean gates delay time is much different thing than Unitary quantum gates delay time ( such a thing even exist ? )
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So imagine an algorithm that, on input x of length n, calls polynomially many algorithms A_1, A_2, ..., A_n as subroutines, each A_i running in poly-time on "its input" as follows: the input on A_i, i <= 2 <= n is the output of A_{i-1} where each A_i is the previous A (on input x output xx).
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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
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Is your objection to the sentence "If a polynomial time algorithm calls as a subroutine polynomially many polynomial time algorithms, the resulting algorithm is still polynomial time."? That seems right to me too. The first polynomial time machine, A, is allowed to call as a subroutine polynomially
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This is the informal argument used in the text of why BQP is low for itself: "Informally, this is true because polynomial time algorithms are closed under composition. If a polynomial time algorithm calls as a subroutine polynomially many polynomial time algorithms, the resulting algorithm is still
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It should be stated that quantum computer time complexity doesn't measure actual time but "computational steps". as far as i know the "computational steps" in quantum computer is the number of unitary quantum gates. However the link attached is for
Boolean gates complexity class which is a totally
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Your example is not the correct interpretation of BQP with a BQP oracle. Consider two poly-time algorithms, A and B, such that A can call B as a subroutine at unit cost. In this scenario, you can claim that there is another algorithm C that runs in polynomial time and simulates algorithm A.
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It seems that this argument is wrong: consider an algorithm A that on input a string x outputs xx (i.e. A(x) = xx) and consider the result of an application of A to itself n times, n is the initial length of x i.e. n = |x|. In other words compute A(A(A(...A(x)...))), the result is clearly
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Sure! I'm just commending on the sentence, not on the result. If the cost is unit, as we assume in the oracle scenario, then it's fine. My comment is only about this sentence which seems apparently false in the general setting (again: not in the oracle setting).
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Surely most of the examples given are not decision problems but function problems? Like computing the Jones polynomial at a root of unity, you want a number; or simulating a quantum system, you want a set of probability amplitudes?
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The probability that the algorithm fails N times in a row is (1/4). Actually I think 1/4 is more or less arbitrary; choosing any other (rational?) number in ]0,1/2[ would not change the class. --
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many (say p(n) many) algorithms B_1, B_2, ..., B_{p(n)}. These algorithms are not allowed to call other algorithms as subroutines. This is different from your example. --
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by the second claim, which is absurd. Given that I doubt I have suddenly solved the P = NP problem, something is clearly wrong here. Can anybody explain this better?
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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 ? --
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related articles on
Knowledge. If you would like to participate, please visit the project page, where you can join
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I don't speak
Spanish or German but when I go to those other pages I do see the fraction 1/4 used instead of 1/3.
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Is the number of qubits of the quantum computer held constant or may it vary with input size? --AxelBoldt
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problems. Paired with the fact that many practical BQP problems are suspected to exist outside of
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Knowledge. If you would like to participate, please visit the project page, where you can join
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This article was the subject of a Wiki
Education Foundation-supported course assignment, between
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Right. It works for irrational numbers too. But not complex, quaturnion, or surreal. :-) --
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Clearly the resulting algorithm is exponential on the input x (the output has length 2^n).
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I think it is assumed that there's always enough of them, just as we do with Turing tapes. --
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which is to say that quantum computing is more powerful than classical. However, it says "
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Requested articles/Applied arts and sciences/Computer science, computing, and
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dynamics is deterministic in the absence of measurement. cheers. -- dave kielpinski
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Find pictures for the biographies of computer scientists (see
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in the banner shell. Please resolve this conflict if possible.
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if I understand right, it should say "There is no proof that
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Unclear section on the relationship between BQP, P and NP?
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is inherent in quantum computers, there is no notion of
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841:{\displaystyle NP\not \subseteq BQP}
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1004:{\displaystyle P\subseteq BGP}
873:{\displaystyle P\subseteq BQP}
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909:. Proof by contradiction: If
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109:and see a list of open tasks.
1116:{\displaystyle P\subset BQP}
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112:Mathematics
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59:Mathematics
30:Start-class
1165:Categories
1144:Homo logos
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367:Infobox
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1047:P NP
938:then
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709:Robin
431:Stubs
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564:talk
525:and
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630:BQP
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993:B
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967:P
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958:⊈
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926:P
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920:=
917:P
897:P
894:N
888:P
868:P
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