How to test a 'float' or 'double' zero numerically?

Discussion in 'C++' started by Peng Yu, Sep 13, 2008.

  1. Peng Yu

    Peng Yu Guest

    Hi,

    Suppose T is 'float' or 'double'.

    T x;

    x < 10 * std::numeric_limits<T>::epsilon();

    I can use the above comparison to test if 'x' is numerically zero. But
    I'm wondering what should be a good multiplicative constant before
    epsilon?

    Thanks,
    Peng
     
    Peng Yu, Sep 13, 2008
    #1
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  2. I can use the above comparison to test if 'x' is numerically zero.

    No, as x can also be negative.
    Epsilon is the smallest value such such that 1.0 + epsilon != 1.0. You
    need to scale it with the numbers to compare with. Comparing against
    zero is always hard. You are probably best of with using abs(x) <
    your_own_epsilon. Set your_own_epsilon to what ever you want, such as
    0.00000001 perhaps.

    Regards,
    Anders Dalvander
     
    Anders Dalvander, Sep 13, 2008
    #2
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  3. Peng Yu

    Peng Yu Guest

    Right, I meant std::abs(x).
    Therefore, there is no general accept such epsilon?

    Thanks,
    Peng
     
    Peng Yu, Sep 13, 2008
    #3
  4. No you can't. A value of x distinct from zero might also test as
    "zero" with that.
     
    Juha Nieminen, Sep 13, 2008
    #4
  5. Peng Yu

    Ron AF Greve Guest

    Hi,

    Consider a machine where the smallest number that can be represented is
    0.0001

    Lets asume I have the following calculation (lets assume the 0.00005 would
    be the result of some calculaton).
    0.0001 -0.00005 - 0.00005
    Now it is obvious that this should result in zero. However the last two
    results would be zero since the machine can only have up to four digits
    behind the dot. So what should be zero is actually 0.0001 so a correct value
    for a multiplier for epsilon would be 0.0002. Reasoning 0.0001 < 0.0002
    therefore it is zero?

    Consider then the following

    The same formula only we also divide by 0.0001 afterwards
    ( 0.0001 -0.00005 - 0.00005 ) / 0.0001 = 1 However the one actually should
    be a zero therefore our first conclusion was incorrect. A correct multiplier
    for epsilon should be 10001

    Of course one could go on, epsilons multiplier could be anything.

    Conclusion there is not a correct multiplier for epsilon. There can be one
    per formula but that is probably not very practical.


    Regards, Ron AF Greve

    http://www.InformationSuperHighway.eu
     
    Ron AF Greve, Sep 13, 2008
    #5
  6. Peng Yu

    Peng Yu Guest

    I see. Then the problem is how to derive it for a particular formula.

    Probably, I need to write down the formula and take the derivatives of
    all its arguments, check how much errors there could be for each
    arguments. Then I would end up with a bound of the rounding error
    (epsilon is equivalent to it). Right?

    Thanks,
    Peng
     
    Peng Yu, Sep 14, 2008
    #6
  7. Peng Yu

    Rune Allnor Guest

    Numerical analysis is an art in itself. There are departments
    in universities which deal almost exclusively with the analysis
    of numerics, which essentially boils down to error analysis.

    In my field of work certain analytical solutions were formulated
    in the early '50s, but a stable numerical solution wasn't found
    until the early/mid '90s.

    You might want to check with the math department at your local
    university on how to approach whatever problem you work with.

    Rune
     
    Rune Allnor, Sep 14, 2008
    #7
  8. Peng Yu

    Peng Yu Guest

    In my field of work certain analytical solutions were formulated
    Would you please give some example references on this?

    Thanks,
    Peng
     
    Peng Yu, Sep 14, 2008
    #8
  9. No, different applications requires different precision, some would
    consider a variable equal to zero if it was 0.0001 from zero while
    others might require 0.0000001. You have to analyse your problem to find
    a value that suites you.
     
    Erik Wikström, Sep 14, 2008
    #9
  10. Peng Yu

    Rune Allnor Guest

    At the risk of becoming inaccurate, as I haven't reviewed
    the material in 5 years and write off the top of my head:

    Around 1953-55 Tompson and Haskell proposed a method to
    compute the propagation of seismic waves through layered
    media. The method used terms on the form

    x = (exp(y)+1)/(exp(z)+1)

    where y and z were of large magnitude and 'almost equal'.
    In a perfect formulation x would be very close to 1.

    Since y and z are large an one uses an imperfect numerical
    representation, the computation errors in the exponents
    become important. So basically the terms that should
    cancel didn't, and one was left with a numerically unstable
    solution.

    There were made several attempts to handle this (Ng and Reid
    in the '70s, Henrik Schmidt in the '80), with varoius
    degrees of success. And complexity. As far as I am concerned,
    the problem wasn't solved until around 1993 when Sven Ivansson
    came up with a numerically stable scheme.

    What all these attempts had in common was that they took
    the original analytical formulation and organized the terms
    in various ways to avoid the complicated, large-magnitude
    internal terms.

    I am sure there are simuilar examples in other areas.

    As for an example on error analysis, you could check out the
    analysis of Horner's rule for evaluating polynomials, which
    is tretaed in most intro books on numerical analysis.

    Rune
     
    Rune Allnor, Sep 14, 2008
    #10
  11. Peng Yu

    Rune Allnor Guest

    Typo correction: The problematics terms were on the form

    x = exp(y)-exp(z)

    where y and z are large and x is small.

    Rune
     
    Rune Allnor, Sep 14, 2008
    #11
  12. Peng Yu

    Ron AF Greve Guest

    Hi,

    You could indeed do an analysis that way. Actually that kind of thing is
    also done when measuring something and one has to know the error in the
    measurement. Taking in account the accuracy of measuring equipment and the
    kind of operation you (multiplying, addition etc) you can the tell what the
    error range is (like I measured 5V +/- 0.5V.

    It is a lot of work though.

    Regards, Ron AF Greve

    http://www.InformationSuperHighway.eu
     
    Ron AF Greve, Sep 14, 2008
    #12
  13. Peng Yu

    James Kanze Guest

    If you want to test whether x is numerically zero, "x == 0.0" is
    the only correct way.
    There isn't one, since the idiom is broken (in general---there
    are specific cases where it might be appropriate).
     
    James Kanze, Sep 15, 2008
    #13
  14. Peng Yu

    James Kanze Guest

    It depends on the computation. There are a lot of contexts
    where you get 0.0 exactly, and that's what you want to test for.
    There are less contexts where this is true for other values (0.0
    is a bit special), but they also exist.
     
    James Kanze, Sep 15, 2008
    #14
  15. Peng Yu

    James Kanze Guest

    No. The problem is how to implement the formula so that it
    gives the correct results.
    Not necessarily. You need to better understand how machine
    floating point works, and the mathematics which underlies it.

    Think of it for a minute. If I had a system in which sin(0.0)
    returned anything but 0.0 (exactly), I'd consider it defective.
    For other values, this is somewhat less obvious, but 0.0 (and in
    some contexts, 1.0 and -1.0) are a bit special.
     
    James Kanze, Sep 15, 2008
    #15
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