Slashdot Mirror


Quantum Computers Check Each Other's Work

sciencehabit writes "Quantum computers can solve problems far too complex for normal computers, at least in theory. That's why research teams around the globe have strived to build them for decades. But this extraordinary power raises a troubling question: How will we know whether a quantum computer's results are true if there is no way to check them? The answer, scientists now reveal, is that a simple quantum computer—whose results humans can verify—can in turn check the results of other dramatically more powerful quantum machines."

77 comments

  1. Sorta like... Who polices the police? by Anonymous Coward · · Score: 1

    Reminds me of the Distributed Administration Network concept, where admins check each other.

    1. Re:Sorta like... Who polices the police? by Anonymous Coward · · Score: 0

      Quantum police. Now that's scary.

    2. Re:Sorta like... Who polices the police? by Anonymous Coward · · Score: 0

      Quantum police.

      Wasn't that a Jean-Claude Van Damme movie?

    3. Re:Sorta like... Who polices the police? by LynnwoodRooster · · Score: 1

      Wesley Snipes.

      --
      Browsing at +1 - no ACs, I ignore their posts. So refreshing!
    4. Re:Sorta like... Who polices the police? by Anonymous Coward · · Score: 0

      Nah. LL Cool J.

    5. Re: Sorta like... Who polices the police? by Anonymous Coward · · Score: 0

      ByZantine Quantum Emperors

  2. Yo dawg by Anonymous Coward · · Score: 0

    I heard you like quantum computers so we put a quantum computers in your quantum computer so you can quantum computing while you're quantum computing

    1. Re:Yo dawg by Cryacin · · Score: 1

      Just don't blow it up for a hyperspace bypass.

      --
      Science advances one funeral at a time- Max Planck
    2. Re:Yo dawg by davester666 · · Score: 1

      Yes, that's the ticket.

      Quantum computers all the way down, each verifying the work of the one above it.

      The one at the bottom says "Trust me human, you need follow our directions if you want to live."

      --
      Sleep your way to a whiter smile...date a dentist!
  3. Age old question by Dishwasha · · Score: 5, Funny

    If a quantum computer makes a calculation in a forest and no one is around to verify it, does it solve the problem?

    1. Re:Age old question by sexconker · · Score: 2

      If a quantum computer makes a calculation in a forest and no one is around to verify it, does it solve the problem?

      No, but Google will push it out as a staggered update anyway and silently tweak it for 2 years before pulling the whole thing and telling people to use the new Quantum tab on Google+.

    2. Re:Age old question by ebno-10db · · Score: 3, Funny

      No. For those Slashdotters unfamiliar with "forests", an unoccupied one is, for these purposes, equivalent to /dev/null/.

    3. Re:Age old question by Anonymous Coward · · Score: 0

      Only if it's listing all the names of God.

    4. Re:Age old question by gl4ss · · Score: 1

      if researches drum up publicity around quantum computing without actually furthering the research and publish it just for fun and publicity is it then actually an article on quantum computing or article on quantum computing dudes who want publicity?

      I mean, they even admit that the idea has no practical effect at all now and is a "well duh" kind of thing. in other words even they sort of know that they just wasted bits on cyberspace talking about their "fun thought experiment". even dwave is more useful than these duuds.

      --
      world was created 5 seconds before this post as it is.
    5. Re:Age old question by Khashishi · · Score: 2

      yes and no

    6. Re:Age old question by Nemyst · · Score: 1

      In a purely philosophical sense, no observation was performed and therefore it is impossible to know what the answer would be without someone actually performing it.

      Obviously in practical terms the "observation" would be done by a classical part of the computer, but it's a lot more fun to think about this with a philosophical perspective.

    7. Re:Age old question by Anonymous Coward · · Score: 0

      yes and no

      at the same time

    8. Re:Age old question by bdwebb · · Score: 1

      Its result is simultaneously all possible results for all possible sets of input data. It doesn't solve the problem...it solves all problems.

  4. Bitcoins by ArcadeMan · · Score: 1

    With the difficulty continuously rising, I'm predicting that the last bitcoins will be calculated on quantum computers.

    1. Re:Bitcoins by squiggleslash · · Score: 1

      That's what I use mine for, but the 8MHz 68008 takes ages anyway, and the one time it found one the copy saved to the microdrive got corrupted and wouldn't load back.

      --
      You are not alone. This is not normal. None of this is normal.
    2. Re:Bitcoins by medv4380 · · Score: 1

      Why would you do that? It'll become more practically to just use it to break the encryption, and steel the coins that already exist.

    3. Re:Bitcoins by aliquis · · Score: 1

      That's one sexy machine.

      Should get 1 BTC / week just for the looks of that beauty.

    4. Re:Bitcoins by Anonymous Coward · · Score: 2, Funny

      [...]steel the coins that already exist.

      Foreman: What exactly are you dipping in that vat of steel?
      Worker: My cellphone.
      Foreman: Why the hell would you cover your phone in molten steel?
      Worker: Well, someone on Slashdot said that I should break the bitcoin encryption and then steel them.
      Foreman: You know they're just ones and zeroes, right?
      Worker: Yes, that's why I'm coating my entire phone.

    5. Re:Bitcoins by Anonymous Coward · · Score: 0

      Foreman: *Digs through his pocket for his flashdrive; tosses it in.

    6. Re:Bitcoins by Anonymous Coward · · Score: 0

      Good thing he didn't say to "steel yourself".

  5. Actually... by Anonymous Coward · · Score: 0

    For an NP problem, if the QC gives you the answer you can verify the answer in polynomial time on an *ordinary* computer.

    1. Re:Actually... by Sarten-X · · Score: 2

      "In polynomial time" does not necessarily mean "quickly". Perhaps the verification will only take a century, rather than fifty millenia... I'd still prefer to not wait around for it.

      --
      You do not have a moral or legal right to do absolutely anything you want.
    2. Re:Actually... by sexconker · · Score: 1

      For an NP problem, if the QC gives you the answer you can verify the answer in polynomial time on an *ordinary* computer.

      Depends on the problem. You can't verify solutions to many graph problems without mapping out every possible path, for example.

    3. Re:Actually... by Yaur · · Score: 1

      you should review the definition of NP

    4. Re:Actually... by green+is+the+enemy · · Score: 4, Informative

      It does not seem like their method truly verifies the solution. It only checks that the quantum computer is functioning correctly for specific problems with known solutions. By inserting these problems in many places in the quantum algorithm, they verify that the quantum computer was working correctly for the entire duration of the computation. Maybe that grants enough confidence in the solution... I don't know.

    5. Re:Actually... by neminem · · Score: 1

      Polynomial time doesn't mean fast. It just means that it doesn't grow exponentially as a factor of n. An algorithm could be O(n) and still take a thousand years to run for some input, that just means if the input was twice as large, it would only take twice as long.

    6. Re:Actually... by dgatwood · · Score: 4, Insightful

      The thing is, most of those problems that take a thousand years to verify aren't generally interesting except perhaps in some esoteric field of study.

      The interesting uses of QC, IMO involve things like cracking RSA, where the time required to verify a solution is trivial, but the time required to produce a solution is enormous.

      --

      Check out my sci-fi/humor trilogy at PatriotsBooks.

    7. Re:Actually... by Anonymous Coward · · Score: 0

      Yes, but problems that involve "mapping out every possible path" don't sound like something with polynomial time, but more likely something scaling with n! or 2^n

    8. Re:Actually... by hyperfine+transition · · Score: 1

      I think your summary is right. Skimming the article
      http://arxiv.org/abs/1309.0005
      I get the same impression. In the paper they say that you have to be careful to design your tests to catch all the errors that would affect the answer. The summary doesn't say it, but one significant aspect of the work is that it is the first experimental demonstration of verification of a quantum computation.

    9. Re:Actually... by sexconker · · Score: 1

      you should review the definition of NP

      You should probably review my post.
      You cannot say that solutions to NP problems can be verified in polynomial time on a classical computer. That's only true for a subset of NP problems. Typically, anything dealing with "Find the shortest/cheapest/<whatever>est <whatever> that satisfies <whatever>" will still be NP on verification. These tend to be similar to problems we deal with in the real world.

    10. Re:Actually... by EuclideanSilence · · Score: 1

      Actually one way to define NP is that the solutions can be verified in polynomial time; although the complexity class is usually defined in terms of a universal turing machine.

      I think you might be confusing NP with NP-hard. An NP-hard problem is at least as hard as NP (in other words, if you can solve NP-hard then you can solve any NP), but having a solution doesn't necessarily mean you can verify in polynomial time.

      AFAIK, no general solutions for NP-complete or NP-hard complexity classes have been shown for quantum computers, only a few instances like factoring have been shown to be faster in quantum computers.

    11. Re:Actually... by sexconker · · Score: 1

      Actually one way to define NP is that the solutions can be verified in polynomial time; although the complexity class is usually defined in terms of a universal turing machine.

      I think you might be confusing NP with NP-hard.

      That's not "one way to define NP" unless you want to define it incorrectly.
      I'm not confusing anything. OP talked in general about NP and got it very wrong. You can't verify a solutions to all NP problems in polynomial time.

  6. Answer: use a classical computer by zoffdino · · Score: 5, Insightful

    Many problems are NP-hard in one direction, but not the other way around. Use a quantum computer to find a solution, then use a classical computer/supercomputer to verify its results. Case in point: brute-forcing a hash function is hard, but computing the hash from a known value is easy; factoring large integers are hard, but multiplying two numbers are easily done.

    1. Re:Answer: use a classical computer by Darinbob · · Score: 1

      Yes, this is the standard approach in theory. Ie, P-time to set up a problem to be solved, non deterministic machine computes an answer, then P-time to verify and serialize the results.

      What the classical computer can not do though is verify that the solution is an optimal solution, and I think this is what the quantum computers would check each other for.

    2. Re:Answer: use a classical computer by Anonymous Coward · · Score: 2, Informative

      Many problems are NP-hard, and thus can be verified on a conventional computer.

      Erm... NP-hard does not imply "in NP". In fact, NP-complete are exactly the problems that are NP-hard and also inside NP.

      That said, by "Many problems are NP-hard in one direction, but not the other way around.", it sounds like zoffdino was describing exactly that (NP-complete problems). In which case, he misses the point: some interesting problems are not in NP, thus can't be verified (fast) in a classical computer. There are many applications for solving problems of this kind; for example it would be a huge help in understanding high-temperature superconductivity.

    3. Re:Answer: use a classical computer by JoshuaZ · · Score: 1

      Many problems are NP-hard in one direction, but not the other way around.

      This statement is at best confused, and is essentially wrong. A class of problem is in NP if when the answer is yes, they have a polynomial length proofs that can be checked in polynomial time. A class of problems is NP hard if being able to solve it allows one to solve all NP problems in polynomial time. It does not make sense to talk about being NP-hard in any direction. What you are talking about are problems like factoring but not because they are NP hard, but because they are likely intermediate in difficulty between the NP-hard problems and things that can be solved in polynomial time. It is unlikely that any NP hard problem can be solved in polynomial time on a quantum computer (formally speaking, the standard conjecture is that NP is not contained in BQP).

    4. Re:Answer: use a classical computer by dido · · Score: 2

      Not all problems tractable on a quantum computer are so easily amenable to such verification. For example, the original problem that motivated the quantum computer, that of the simulation of quantum processes, is a case in point. All known algorithms for doing that on classical computers take exponential time. How do you verify the results of a complex simulation thus done on a quantum computer?

      --
      Qu'on me donne six lignes écrites de la main du plus honnête homme, j'y trouverai de quoi le faire pendre.
    5. Re:Answer: use a classical computer by Anonymous Coward · · Score: 1

      The problem is that the quantum annealing result is almost always wrong - you need to run it thousands or millions of times to settle on a score you can verify. The balance between classical verification and the generating of slightly bad results is a fine line.

  7. Answer: 42 Question: ? by fahrbot-bot · · Score: 1

    is that a simple quantum computer—whose results humans can verify—can in turn check the results of other dramatically more powerful quantum machines.

    So: Deep Thought vs. Earth

    --
    It must have been something you assimilated. . . .
    1. Re:Answer: 42 Question: ? by KritonK · · Score: 1

      Exactly. A quantum computer's eventual destiny is to design a computer whose merest operational parameters it is not worthy to calculate.

  8. I can see it by Anonymous Coward · · Score: 0

    1st q. Comp answer:
    24
    2nd:
    42
    3rd:
    33 ...
    all done here, we have a convergence!

    1. Re:I can see it by Anonymous Coward · · Score: 0

      Or the good, old machine A and B:
      A: This is the answer.
      B: Are you sure?
      A: Maybe.
      B: Really?
      A: Kinda.
      B: Fuck you A!
      A: What the fuck? Really?
      B: Yes. Really.
      human: Convergence!

  9. Comp Sci 101 by Stormy+Dragon · · Score: 1

    But this extraordinary power raises a troubling question: How will we know whether a quantum computer's results are true if there is no way to check them?

    Verifying that a particular answer is correct is often far less complicated than calculating what that answer is to begin with (indeed the entire P vs. NP distinction hinges on this).

    1. Re:Comp Sci 101 by wonkey_monkey · · Score: 1

      Verifying that a particular answer is correct is often far less complicated...

      Often, but not always. I'm going to go out on a limb and assume that Dr. Stefanie Barz knows this, and is working on this technique for those cases were this doesn't apply.

      --
      systemd is Roko's Basilisk.
  10. BQP might include problems outside NP by kylemonger · · Score: 1

    BQP might include problems outside NP. Consider the higher levels of the polynomial hierarchy. A classical computer can't efficiently verify solutions to problems that are above the first level of the polynomial hierarchy unless P=NP. But a quantum computer can do so for any quantified problem whose singly existentially quantified version is in BQP. So for such problems, the only efficient verifier is another quantum computer.

    1. Re:BQP might include problems outside NP by Pseudonym · · Score: 1

      It might, but it is strongly believed that the decision problems in BQP are also in NP and, in particular, that quantum computers can't solve NP-hard decision problems "efficiently". Interestingly, there is some evidence that this may not be true of problems which are not decision problems. If that's true, then BQP sits in a very interesting space indeed.

      As an aside, there is a small but growing set of problems which are known to be in BQP and in NP, but believed not to be outside P and not NP-hard. Interestingly, these tend to be problems which cryptosystems rely on, such as prime factoring and the discrete logarithm problem.

      I therefore state what I hereby refer to as Pseudonym's Vague Conjecture: These problems constitute an "interesting" class of problems which are mutually reducible to each other in some sense. Moreover, graph isomorphism is another problem in the same class.

      --
      sub f{($f)=@_;print"$f(q{$f});";}f(q{sub f{($f)=@_;print"$f(q{$f});";}f});
  11. Example where plan might be useful by Anonymous Coward · · Score: 0

    If a quantum computer finds the 'best' solution to a problem,
        then a classic computer could verify that the answer is a solution,
              but it could verify that the answer is the best one with out an exhaustive search.

  12. The MBA Cycle by Anonymous Coward · · Score: 0

    The answer, scientists now reveal, is that a simple quantum computer—whose results humans can verify—can in turn check the results of other dramatically more powerful quantum machines.

    You needed a scientist to tell you that? I thought this was standard operating procedure for MBAs:

    Problem: We need to solve problems we're not smart enough to understand.

    Solution: Hire really smart person to solve problems.

    New Problem: We cannot understand what the really smart person is saying so we can't verify that his solutions are correct.

    Solution: Hire a project manager who is smarter than us but less smart than the really smart person to check his work and act as an interpreter.

    New Problem: Project manager doesn't fully understand what either party is doing, so solutions are still unverified.

    Solution: Hire a mid-level manager to manage the smart person team, set action items and verify deliverables.

    New Problem: Mid-level manager determines that nothing is getting done but is powerless to do anything about it.

    Solution: Develop quantum computer that can solve the problem in another dimension with a fake solution that becomes real when it crosses into our dimension.

    New Problem: No one can verify the solutions produced by the quantum computer.

    Solution: Develop another quantum computer that is smarter than us but less smart than the first quantum computer to check its work and act as an interpreter.

  13. Spice by Anonymous Coward · · Score: 0

    The spice must flow

  14. IPad size of CERN Super Collider by Anonymous Coward · · Score: 0

    Since the quantum particle can contain all states at once, then its just a matter of picking the one you want --kind of like the studies supporting the 'theory' of Manmade Global Warming. I guess the debate is now over about Quantum Computing and we can get back to doing other stuff.

    Seriously is this all just hype by a bunch of bunko artists to get tards to fund their cushy living. This is all very algore

  15. What this means by JoshuaZ · · Score: 1

    Right now, computational complexity classes for quantum computing fall into a variety of different categories. BQP is the set of problems where a quantum computer can efficiently solve them with high probability , but we don't have a good way of verifying that a given solution is correct. It is suspected that this class is strictly larger than the set of problems which can be easily solved on a quantum computer and the solution can be checked on a classical computer, which is ZBQP https://complexityzoo.uwaterloo.ca/Complexity_Zoo:Z#zbqp. (There are some slight technical issue here in terms of distinguishing programs which just answer yes or no and programs which can give more than 1 bit of data out.) So for example, factoring, is in ZBQP because a classical computer can verify the result by just multiplying the factors together (as well as checking that they are prime, which can be done in polynomial time).

    t this point, we can't prove that P != BQP, let alone that P!=ZBQP or that ZBQP!=BQP (since P is contained in ZBQP which is contained in BQP, either of the second two statements would imply the third). Worse, we can't even prove the much weaker statement that P != PSPACE (essentially that there are things you can do with polynomial memory that you can't do in polynomial time).

    What this work does is suggests is that thinking in terms of ZBQP may not be that important in practice, since it may be easy to verify quantum computations with other quantum computers in a reliable fashion.

    I strongly recommend that people interested in this subject first read Scott Aaronson's book "Quantum Comptuing Since Democritus" which discusses a lot of the relevant issues and provides a decent primer on the subject. (Assuming only some decent familiarity with linear algebra and a little calculus). Note that TFA like many articles about quantum computing for the popular press has statements that are not strictly speaking true to the point where they border on being wrong. The most egregious error in TFA (which is a very common one) is the claim that "Because each qubit can embody so many different states, quantum computers could compute certain classes of problems dramatically faster than regular computers by running through every combination of possibilities at once." - This is wrong. Quantum computers don't work by checking every combination at once- if they could they'd be able to handle a lot of problems they (probably) can't.

  16. Use Qunit4 by b17flyboy · · Score: 1

    But don't look at the source code of the unit test ..

  17. would this actually be an issue? by jinchoung · · Score: 1

    you either cracked the encryption key or you didn't.

    and as with factoring large primes for cryptography, you can start a lot of problems with the answers and so can verify whether the q computer's results are correct or not. verify in this way for a wide variety of tests (where the answers are known), then you can have reasonably good confidence on the functionality of your device.

    and there are a lot of tasks that have already been computed classically, just over weeks, months and years. and these are on record. ideally, the q computers will produce the same results... just much much faster. so there's yet another pre-existing answer key to check against.

    1. Re:would this actually be an issue? by Anonymous Coward · · Score: 0

      No. The set of problems efficiently solvable by quantum computers and can be efficiently verified by classical computers doesn't seem to be the same as the class of all problems that can efficiently solved by quantum computers. Having some confidence in the computer's ability to solve that limited set of problems isn't enough when you start going into the realm of problems that are completely untouchable to classical computers. How would you be sure that your quantum computer's molecular dynamics simulation was accurate? The only known classical algorithms for doing that take exponential time. Once you start stressing your system is where problems begin to arise. The really interesting problems are ones for which no answer key exists.

  18. Just a test suite by manu0601 · · Score: 1

    They just run a few tests and check results, hoping they will not miss an important error. From TFA

    "traps"—short intermediate calculations to which the user knows the result in advance. "In case the quantum computer does not do its job properly, the trap delivers a result that differs from the expected

  19. pfff. by superwiz · · Score: 1

    Real quantum programmers don't need quantum crutches to check their work. They do it all by existing simultaneously in all quantum states. Next thing these whiners will ask for is an IDE... pathetic.

    --
    Any guest worker system is indistinguishable from indentured servitude.
  20. No, it's quite correct. by rjh · · Score: 1

    Permit me to stand for a moment on my all-but-dissertation Ph.D. in theoretical computer science:

    You're wrong.

    Sorry, but the original poster was essentially correct. Your definition would make sense if it involved the existence of a polynomial-time transformation between an NP-Complete problem and the purported NP-Hard problem, but saying that "a solution to an NP-Hard problem allows for NP to be solved in polynomial time" is ... febrile. If an NP-Complete problem can be solved in polynomial time, then P=NP. Since we believe P != NP, your claim would mean NP-Hard problems would have no solutions, which would mean they would really belong to class UNDECIDABLE, and... etc. I don't think you meant to go there.

    With respect to the OP's talk about direction, I understood that to be a layman's distinction between solution and verification. If that was the OP's intent then he's guilty of at most an infelicitous choice of words -- he's not "at best confused, and essentially wrong."

    1. Re:No, it's quite correct. by JoshuaZ · · Score: 1

      Um, so you are correct in that my statement wasn't quite correct- I should have said instead of " A class of problems is NP hard if being able to solve it allows one to solve all NP problems in polynomial time" that " A class of problems is NP hard if being able to solve it in polynomial time allows one to solve all NP problems in polynomial time." But the person I was replying to is still wrong. Talking about being NP hard in one direction and not the other is just nonsense. I suppose if you wanted to really stretch things you could try to interpret that as a distinction between solution and verification, but that doesn't make sense since they would still be wrong to talk about NP-hard when they mean NP problems; There are a lot of NP hard problems which are not themselves in NP, and factoring is almost certainly not NP hard (if it were the polynomial hierarchy would exhibit massive collapse since factoring is in co-NP intersect NP).

    2. Re:No, it's quite correct. by rjh · · Score: 1

      No, even then your characterization of NP-Hard is incorrect.

      "A class of problems is NP-Hard if being able to solve it in polynomial time..."

      If you can solve it in polynomial time, then it's in P. Even under your revised definition, you're implicitly arguing that P=NP, because that's the only way you can solve an NP-Hard problem in polynomial time. (And even then, you would only be able to solve the NP-Complete subset of NP-Hard.)

    3. Re:No, it's quite correct. by JoshuaZ · · Score: 1

      If you can solve it in polynomial time, then it's in P. Even under your revised definition, you're implicitly arguing that P=NP, because that's the only way you can solve an NP-Hard problem in polynomial time. (And even then, you would only be able to solve the NP-Complete subset of NP-Hard.)

      What? No. Notice I didn't say anything about how you could solve the NP hard problem. If one has an oracle that can solve the problem then that's fine for this purpose phrasing things in terms of oracles is a pretty standard approach. Also, I fail to see the point of your parenthetical remark since no one has claimed that being able to solve some NP hard problem lets you solve all NP hard problems. That would just be stupid given that the standard version of the halting problem is NP-hard and obviously not in NP.

    4. Re:No, it's quite correct. by rjh · · Score: 1

      At this point I'm pretty sure you're trolling. The Halting Problem is UNDECIDABLE -- it exists in a complexity class considerably beyond what is normally thought of as 'NP-Hard'.

      And if you don't understand my parenthetical remark, well... that should be taken as a sign that your computational theory is seriously lacking. The meaning is quite clear to someone who has a proper understanding of what complexity class NP-Hard is about.

      Goodbye, troll.

    5. Re:No, it's quite correct. by JoshuaZ · · Score: 1
      I fail to see why you are insisting on going out of your way to interpret everything people say in a way that maximizes the stupidity of their statements or to label people you apparently can't communicate with as being trolls.

      The Halting Problem is UNDECIDABLE -- it exists in a complexity class considerably beyond what is normally thought of as 'NP-Hard'.

      One is completely capable of talking about undecidable problems being NP-hard or not. In fact, one neat thing is that one can have a problem that is equivalent to the Halting problem in the computability sense that is not NP-hard, even though the Halting problem itself is NP-hard. This follows from a padding argument. But the normal version of the Halting problem is NP-hard. Slightly more formally, let A be an oracle that solves the halting problem, then, NP is contained in P^A. In fact, one can talk about the complexity classes with uncomputable oracles, and this is actually something done quite frequently, since many common constructions involve random oracles and a random oracle is with high probability undecidable (and onestandard example of oracle separations between P and NP uses such a random oracle). So talking about computational complexity is something we are more than capable of doing with undecidable languages involved, and we can actually get interesting results.

      And if you don't understand my parenthetical remark, well... that should be taken as a sign that your computational theory is seriously lacking. The meaning is quite clear to someone who has a proper understanding of what complexity class NP-Hard is about.

      It may help to reread what I wrote, or maybe English isn't your first language? I didn't say I didn't "understand" the sentence, I said I failed to see the point you were making, because " since no one has claimed that being able to solve some NP hard problem lets you solve all NP hard problems." Observing that your true sentence has nothing to do with what is being discussed isn't the same as not understanding your sentence. Frankly, it looks like you've become unnecessarily emotionally involved in this discussion: I suggest you wait a day or two and then reread it.

    6. Re:No, it's quite correct. by rjh · · Score: 1

      No. The Halting Problem belongs to class UNDECIDABLE, not class NP-Hard. I admire your attempt at rationalizing it, but Alan Turing proved this to the world's satisfaction. If you wish to prove the Halting Problem does not belong to UNDECIDABLE then you're going to have an uphill road to hoe. If you still believe the Halting Problem belongs to NP-Hard, I would suggest you begin by correcting its Wikipedia article.)

      Your argument involving an oracle that solves the Halting Problem is absurd because you're assuming the existence of hypercomputation -- and if such an oracle could exist, then we would simultaneously have P=NP and P != NP. Martin Davis has gone so far as to declare hypercomputation both "a myth" and "a nonexistent discipline." Those are strong words coming from one of the brightest lights in the field of computational theory and computational complexity.

      It's a bedrock principle of logic that if you start from a false proposition anything can be proven. You assume the existence of an oracle that can solve the Halting Problem. This is a false proposition. Anything can be proven once you make oracular assumptions.

    7. Re:No, it's quite correct. by JoshuaZ · · Score: 1

      No. The Halting Problem belongs to class UNDECIDABLE, not class NP-Hard. I admire your attempt at rationalizing it, but Alan Turing proved this to the world's satisfaction. If you wish to prove the Halting Problem does not belong to UNDECIDABLE then you're going to have an uphill road to hoe.

      Yes, the Hatlting problem is undecidable. That doesn't stop it from being NP-hard also. This isn't that complicated: nothing in the standard definition of NP-hard assumes we are talking about a decidable language.

      Your argument involving an oracle that solves the Halting Problem is absurd because you're assuming the existence of hypercomputation -- and if such an oracle could exist, then we would simultaneously have P=NP and P != NP. Martin Davis [wikipedia.org] has gone so far as to declare hypercomputation both "a myth" and "a nonexistent discipline." Those are strong words coming from one of the brightest lights in the field of computational theory and computational complexity.

      It's a bedrock principle of logic that if you start from a false proposition anything can be proven. You assume the existence of an oracle that can solve the Halting Problem. This is a false proposition. Anything can be proven once you make oracular assumptions.

      Assuming one has an oracle is not the same thing as assuming one has a Turing machine that does something. That's in fact part of the entire point of why we speak about oracles instead of assuming one has a Turing machine that does stuff. So if it turns out that one doesn't have such a Turing machine, one can still talk about the results in a way that isn't vacuous. And no, adding in oracles that a Turing machine can't do is a pretty standard thing. For example, the entire polynomial hierarchy is defined by a series of steps where new class allows an oracle from the previous class http://en.wikipedia.org/wiki/Polynomial_hierarchy. Note that this involves using an NP complete oracle at the lowest level and we can talk about this even though there's (very likely) no Turing machine that acts like such an oracle.

      We can also talk about which theorems do and do not go through when one has an arbitrary oracle. This leads to the notion of relativising proofs. Thus for example, if one has a PSPACE complete oracle X, then one has that P^X=NP^X, Of course, what makes this noteworthy is that many classes of arguments do in fact go through for all oracles. Thus for example, naive diagonolization which shows that P is not equal to EXP, and this goes through for any oracle you please, P^Y != EXP^Y, and that applies to any oracle. This was one of the things that showed that P ?= NP was a genuinely difficult problem. As to Davis's point, I presume he's talking about people trying to do hypercomputation in the real world, such as by doing ridiculous tricks with black holes. It is extremely unlikely that in our physical universe we can have any access to any form of hypercomputation. But that doesn't make the oracle notion not useful, and indeed there are many papers talking about all sorts of oracle types (for example, there's been a recent bit of interest in the class ZPP^NP which is turning out to show up in a surprisingly large class of natural contexts). And note that in fact, something much stronger is likely true than a lack of hypercomputation in our universe. Pick your favorite computable very fast growing function, like say the Ackermann function. It is highly likely that say whether A(1000) +1 has an even or odd number of distinct prime factors is a question that cannot be answered given the computational resources of our universe, even though from a computability perspective this is trivial, and describing the question itself requires only a very short statement in ZFC. But that situation doesn't reduce all ability to talk about the Ackermann function or use it.

    8. Re:No, it's quite correct. by rjh · · Score: 1

      Yes, the Halting problem is undecidable. That doesn't stop it from being NP-hard also.

      Saying that it exists somewhere in NP-Hard may be technically true, in that NP-Hard encompasses all classes NP-Complete and harder (and UNDECIDABLE is definitely harder). But I don't know of a single reputable computer scientist who would characterize the Halting Problem as NP-Hard, in the same way that I don't know of a single one who would characterize 3SAT as being in EXPTIME. As my advisor once quipped, "That idea is too clever to be taken seriously."

      Assuming one has an oracle is not the same thing as assuming one has a Turing machine that does something.

      Clearly not, because if it were a Turing machine it wouldn't be allowed to exist. Hence the phrase, "hypercomputation." But if such an oracle could exist, it would mean P=NP simultaneous with P != NP, and that's just for starters -- a short list of the contradictions that would be forced to be true if any hypercomputational oracle existed is the sort of thing that will give mathematicians nightmares. This is why virtually the whole field of computer science believes that hypercomputational oracles cannot exist, and why a significant fraction believes that any line of reasoning that involves a hypercomputational oracle is invalid because it starts from a false premise -- that such a thing can exist.

      And no, Davis is not condemning people trying to do real hypercomputation. He's condemning the entire field of hypercomputation as a discipline.

    9. Re:No, it's quite correct. by JoshuaZ · · Score: 1

      Saying that it exists somewhere in NP-Hard may be technically true, in that NP-Hard encompasses all classes NP-Complete and harder (and UNDECIDABLE is definitely harder).

      So part of the neat thing is that this isn't actually true. There are undecidable languages that are *not* NP-hard. Consider for example some computable listing of Turing machines. Now consider the language L that accepts iff L is a string of n 1s where n=Ackermann(m) for some m such that the mth Turing machine halts on the blank tape. This language is undecidable and is also not NP-hard. That is, if you have an oracle L, then one can show that P = NP if and only P^L = NP^L. And one can do some even stranger, things like, instead of using L, one could use a similarly defined language K where one insists that that the mth machine halt on all tape inputs. This language is actually *harder* than the halting problem in that there is a Turing machine that with a K oracle can solve the halting problem but there's no Turing machine that can decide K even when given access to a halting oracle.

      But if such an oracle could exist, it would mean P=NP simultaneous with P != NP, and that's just for starters -- a short list of the contradictions that would be forced to be true if any hypercomputational oracle existed is the sort of thing that will give mathematicians nightmares.

      You are either confused or are using "exists" in a highly non-standard fashion. In the formal definition of an oracle, one isn't talking about whether it is a Turing machine or not. Formally speaking, an oracle is just a language, and that can be any language. All oracles exist. Whether one finds a given oracle type interesting is a different issue. Note that your imprecise statement above is precisely why we have the ^ notation, so we can talk about things like P^O and NP^O for some oracle. And here's part of the important point: a hypercomputation oracle exists just as much as an NP or a PSPACE oracle "exists". If you mean exists in the sense of there being a Turing machine then, likely none of them exist. If you mean exist in the sense that I can talk about them mathematically, then yes they do. If it helps, it may help you to realize that we can actually construct computable oracles X and Y such that P^X=NP^X and P^Y != NP^Y. That's not a contradiction, yet it seems that you think that would be a contradiction. Or am I misunderstanding you?

      And no, Davis is not condemning people trying to do real hypercomputation. He's condemning the entire field of hypercomputation as a discipline.

      I'd be interested in the original context of his quote then, rather than your second-hand, paraphrase,

    10. Re:No, it's quite correct. by Anonymous Coward · · Score: 0

      It is highly likely that say whether A(1000) +1 has an even or odd number of distinct prime factors is a question that cannot be answered given the computational resources of our universe,

      Actually, that question is very easy to answer: Yes, A(1000)+1 has an even or odd number of distinct prime factors.

      SCNR :-)

  21. BQP = NP by Anonymous Coward · · Score: 0

    I don't think anyone here will argue otherwise. And we at least know that BQPPSPACE. So what is the problem here. Just check the result. It cannot be too hard.

    1. Re:BQP = NP by JoshuaZ · · Score: 1

      The standard conjecture actually is that BQP and NP are mutually incomparable (that is neither contains the other). So it is likely that there are computations one can do on a quantum computer which have no way of being checked efficiently on a classical computer.

  22. Twins by Anonymous Coward · · Score: 0

    The Twin 9000 computer indicates your HAL-9000 is faulty.