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Lockheed Martin Purchases First Commercial Quantum Computer

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  • Re:Grammar (Score:5, Informative)

    by jd (1658) <imipak&yahoo,com> on Friday May 27, 2011 @04:53PM (#36267280) Homepage Journal

    Spellcheckers don't usually help with grammar.

  • by stevelinton (4044) <sal@dcs.st-and.ac.uk> on Friday May 27, 2011 @04:56PM (#36267336) Homepage

    An Adiabatic Quantum Computer is quite a different beast from a quantum computer in the usual sense, and even if it can solve the same class of problems in polynomial time (not at all obvious at this stage) it isn't at all clear that 1 qubit in this machine does the same work as 1 traditional qubit.

    They are, to be honest, being a little bit naughty calling this a quantum computer at all, although it does compute and has quanta, but so does my phone.

  • by Anonymous Coward on Friday May 27, 2011 @05:03PM (#36267442)

    Meh. Quantum computing, even at its *full* potential, is no threat to symmetric encryption. The recommended minimum key size will jump a moderate amount and you'll be all set again. The effect on asymmetric encryption depends on the type. Some could be severely compromised. BUT, seeing as operations are currently exceptionally fast for end users AND that asymmetric encryption is generally only used to *establish* symmetrically-protected channels over insecure networks, they could probably be jumped up by several orders of magnitude themselves without anything really bad happening. And if all else failed on the asymmetric side, an infrastructure for pre-shared keys isn't really all that difficult. It's just that we've never needed on before so it seems strange. But we already trust CA's to play their part in the asymmetric world - why wouldn't we trust them to act as a middle-man for symmetric key distribution?

  • by Anonymous Coward on Friday May 27, 2011 @05:17PM (#36267616)

    Adiabatic quantum computing != "classic" quantum computing.

    It does NOT runs the Shor algorithm.

    You can use SSL to download your porn safely tonight.

  • by retchdog (1319261) on Friday May 27, 2011 @05:35PM (#36267782) Journal

    think instead, that solving the hamiltonian is equivalent to (or potentially "harder than") solving the original problem, so that you can translate the original problem into a hamiltonian problem. it doesn't mean that you know the answer of either, but you do know that the solution of the hamiltonian will match up to a solution of the original problem. this is the spirit of it: http://en.wikipedia.org/wiki/Reduction_(complexity) [wikipedia.org]

    very, very roughly, think of it like rewriting java, for example, as c. you may not know what the particular code actually DOES in an overall sense, or what it will output, but you can nevertheless rewrite it sort of mechanically (like a compiler would) if you know both languages. furthermore, it's feasible that translating the code is easier than devising the algorithm from scratch. this is basically a reduction. if you can "easily" rewrite any java code as c code, that means java is "reducible" to c. the theory of computation essentially deals with reductions, not of code, but of entire problem classes, which is where P, NP and all that come from.

The trouble with the rat-race is that even if you win, you're still a rat. -- Lily Tomlin

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