The Case Against Quantum Computing

Discussion in 'Computing and Networks' started by nsaspook, Dec 3, 2018.

  1. cmartinez

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  2. nsaspook

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  3. cmartinez

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    Yes... I noticed that the article says that 50 qubits will be the real breakthrough
     
  4. nsaspook

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    Even the 50 qubit quantum supremacy level is suspect with modern computing advances.
     
  5. cmartinez

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    But you do agree that this sort of reaserch should be pursued nevertheless?
     
  6. cmartinez

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    And here's yet another relevant development:

    https://www.sciencedaily.com/releases/2019/01/190107112955.htm

     
  7. nsaspook

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    Yes, but I'm pessimistic that 'Quantun' properties will result in 'Classical' computer capabilities greater that 'Classical devices beyond some very narrow problem constraints.
     
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  8. cmartinez

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    ... glad to hear that you're not an unreasonable skeptic ... ;)
     
  9. nsaspook

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    https://spectrum.ieee.org/tech-talk/computing/networks/the-practical-limits-of-quantum-cryptography

    https://blog.trailofbits.com/2018/10/22/a-guide-to-post-quantum-cryptography/
     
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  10. WBahn

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    My understanding is that for many categories of classical cryptography the effect of full-blown quantum computing enabled attacks will be an effective cutting in half of the key length. Thus, for those algorithms, all that needs to be done is to double the key length and the algorithms will have the same computational protection against quantum computers as they currently have against classical attack computers.

    Many (most?) of the main public-key algorithms will be rendered ineffective by quantum computing (of sufficient capacity), but there are other public-key algorithms that fall into the same category as most of the symmetric algorithms, requiring a doubling or so of the key lengths.
     
  11. MrAl

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    To bring QC down to earth i think an analogy would be calculating in the time domain vs in the frequency domain.
    In the time domain we solve for one value at one point in time, but in the frequency domain in theory we can solve for all time.
    Thus we get many results in the frequency domain at the same time but only one in the time domain so the advantage is obvious.

    Also, i had thought about the applicability of QC also and it seems that there is another parallel analogy. The difference between parallel computing and non parallel computing, or to put it another way, some problems lend themselves to parallel computing and others simply do not.

    A real test would be a chess program in QC that could beat all other high rated chess programs running on reasonably fast equipment.
    It would be interesting, very interesting, to see how deep the QC program could go as to ply depth. Would be really amazing if it could see right though to the endgame, or to the end itself, although reading out the resutls could probably take longer than anything we could really do i bet.
    Chess has been a subject of study for expert systems so if it does well then that means chess is dead and nobody has a job anymore in any field because they would all be solved :)
     
  12. bogosort

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    Can you give an example of such a calculation? As time and frequency are Fourier duals, one side doesn't have any more "information" than the other. While it's certainly true that certain operations are easier to perform after a Fourier transform -- e.g., convolution in one domain is equivalent to multiplication in the other -- this fact applies to both domains equally.

    If you mean that QC is like parallel computing strictly in the sense that only a small subset of problems are capable of quantum speedup -- just as only a small subset of problems are parallelizable -- then you're 100% correct. To be clear, however, we should note that QC is not in any other way like parallel computing. A common misconception is that quantum speedup happens because quantum computers try every possible solution in parallel, which is definitely not the case.

    It's an interesting idea, and if possible would reveal that chess positions have some deep structure beyond the obvious stuff. The challenge would be figuring out how to prepare the system to represent chess positions. I suspect that's a harder problem than figuring out optimal chess play using classical computing.
     
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  13. MrAl

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    As to the first paragraph, if we look at a simple time function:
    y=A*sin(w*t)

    when we input a value for 't' we get only one solution:
    0=A*sin(w*0)

    but in the frequency domain it is taken to be a continuous wave that goes on forever and given simply by:
    y=A

    or in the Laplace domain:
    y=(w*A)/(w^2+s^2)

    So it is like we are getting the entire solution in one shot rather than having to go through every single time value. That's because we can look at it as being the entire solution for all time (y=A) because wherever we look we see the same thing, A.
    It's like all the solutions are computed already because A is a constant..
    I could explain more if needed.

    As to the last paragraph, yes it would be interesting to see any structure if one turned up.
    Yes i agree, setting up the solution even in a regular language like C is not easy because we want it to be fast yet still go through every possible combination if possible. This leads to a brute force:
    for K=1 to 8 do
    for Q=1 to Qn do
    for R=1 to Rn do
    etc., etc.
    which would require one loop for every piece leading to a ton of combinations after just a few moves.
    Add to that the complexity of the fact that not every piece can move into all the empty board positions that's what really makes it hard.
    That's where stored opening books and stored endgame solutions and heuristics come into play.
     
  14. bogosort

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    You have it backwards. As a function of time, sin(ωt) has infinite support -- it is, as you say, a continuous wave that goes on forever. But in the frequency domain, it has only two nonzero values, precisely at ±ω. Regardless, both representations contain exactly the same amount of information.

    As in the time domain representation, y(t) = sin(ωt), the ω in the Laplace representation is a constant. And just as plugging in a value of t to y(t) gives you a single result, plugging in a value of s to Y(s) also gives you a single result. Neither representation contains more or less information than the other.

    In linear algebraic terms, the mathematical object described by a sinusoid is independent of the choice of basis, just as the mathematical object described by a number is independent of the chosen base.

    Please do, as I don't see the connection to QC.
     
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