Floating point multiplication in python

Discussion in 'Python' started by xyz, Sep 6, 2011.

  1. xyz

    xyz Guest

    hi all:

    As we know , 1.1 * 1.1 is 1.21 .
    But in python ,I got following :

    >>> 1.1 * 1.1

    1.2100000000000002

    why python get wrong result? Who can tell me where's the 0.0000000000000002 from?
     
    xyz, Sep 6, 2011
    #1
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  2. On Tue, 6 Sep 2011 03:57 pm xyz wrote:

    > hi all:
    >
    > As we know , 1.1 * 1.1 is 1.21 .
    > But in python ,I got following :
    >
    >>>> 1.1 * 1.1

    > 1.2100000000000002



    The problem is that 1.1 is a *decimal* number, but computers use *binary*,
    and it is impossible to express 1.1 exactly as a binary number. So when you
    type 1.1 into nearly all computer languages, what you are actually getting
    is a tiny bit more than 1.1:

    >>> repr(1.1)

    '1.1000000000000001'


    which is the closest number to 1.1 that is possible using a C double
    floating point number.

    Normally you don't see it, because Python truncates the result when
    printing:

    >>> str(1.1)

    '1.1'

    but the difference is there.


    --
    Steven
     
    Steven D'Aprano, Sep 6, 2011
    #2
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  3. Am 06.09.2011 07:57 schrieb xyz:
    > hi all:
    >
    > As we know , 1.1 * 1.1 is 1.21 .
    > But in python ,I got following :
    >
    >>>> 1.1 * 1.1

    > 1.2100000000000002
    >
    > why python get wrong result? Who can tell me where's the 0.0000000000000002 from?


    1.1 does not fit in a binary floating point number. It is approximated -
    in binary! - as 1.000110011001100110011 ... (periodically).

    Note that, while in the decimal system we normally use, only numbers
    which have other components in the denominator than 2 or 5 are
    periodically, in the binary systems only components with 2 are allowed
    in order not to be periodically.

    Example: 3.453 is not periodically, because it is 3453/100 and 100 has
    only the factors 2 and 5, each twice.

    1/3 = .3333333... is periodically, because it has the factor 3. The same
    applies to 1/6, which has 2 and 3 as factors. The latter destroys the
    non-periodical behaviour.

    As said, in the dual system, only the 2 is allowed.

    ..5 (10) = 2** -1 = .1 (2).
    ..25 (10) = 2 ** -2 = .01 (2).
    ..75 (10) = their sum = .11 (2).

    But .1 (1/10) is more complicated, -2 would be as well.

    As the IEEE floating point representation is limited, there is a slight
    error value which makes the stored value differ from the intended one.

    Look here:



    >>> x=(1,0,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1,0,0,1,1)
    >>> a=0
    >>> for n,i in enumerate(x): a += i*2**-n; print a, a-1.1, i*2**-n, a-olda

    ....
    1 -0.1 1 1
    1.0 -0.1 0.0 0.0
    1.0 -0.1 0.0 0.0
    1.0 -0.1 0.0 0.0
    1.0625 -0.0375 0.0625 0.0625
    1.09375 -0.00625 0.03125 0.03125
    1.09375 -0.00625 0.0 0.0
    1.09375 -0.00625 0.0 0.0
    1.09765625 -0.00234375 0.00390625 0.00390625
    1.099609375 -0.000390625 0.001953125 0.001953125
    1.099609375 -0.000390625 0.0 0.0
    1.099609375 -0.000390625 0.0 0.0
    1.09985351562 -0.000146484375 0.000244140625 0.000244140625
    1.09997558594 -2.44140625001e-05 0.0001220703125 0.0001220703125
    1.09997558594 -2.44140625001e-05 0.0 0.0
    1.09997558594 -2.44140625001e-05 0.0 0.0
    1.09999084473 -9.15527343759e-06 1.52587890625e-05 1.52587890625e-05
    1.09999847412 -1.52587890634e-06 7.62939453125e-06 7.62939453125e-06
    1.09999847412 -1.52587890634e-06 0.0 0.0
    1.09999847412 -1.52587890634e-06 0.0 0.0
    1.0999994278 -5.72204589933e-07 9.53674316406e-07 9.53674316406e-07
    1.09999990463 -9.53674317294e-08 4.76837158203e-07 4.76837158203e-07
    1.09999990463 -9.53674317294e-08 0.0 0.0
    1.09999990463 -9.53674317294e-08 0.0 0.0
    1.09999996424 -3.57627869541e-08 5.96046447754e-08 5.96046447754e-08
    1.09999999404 -5.96046456636e-09 2.98023223877e-08 2.98023223877e-08
    1.09999999404 -5.96046456636e-09 0.0 0.0
    1.09999999404 -5.96046456636e-09 0.0 0.0
    1.09999999776 -2.23517426789e-09 3.72529029846e-09 3.72529029846e-09
    1.09999999963 -3.72529118664e-10 1.86264514923e-09 1.86264514923e-09
    1.09999999963 -3.72529118664e-10 0.0 0.0
    1.09999999963 -3.72529118664e-10 0.0 0.0
    1.09999999986 -1.3969847501e-10 2.32830643654e-10 2.32830643654e-10
    1.09999999998 -2.32831531832e-11 1.16415321827e-10 1.16415321827e-10
    1.09999999998 -2.32831531832e-11 0.0 0.0
    1.09999999998 -2.32831531832e-11 0.0 0.0
    1.09999999999 -8.73123795486e-12 1.45519152284e-11 1.45519152284e-11
    1.1 -1.45528034068e-12 7.27595761418e-12 7.27595761418e-12
    1.1 -1.45528034068e-12 0.0 0.0
    1.1 -1.45528034068e-12 0.0 0.0
    1.1 -5.45785638906e-13 9.09494701773e-13 9.09494701773e-13
    1.1 -9.10382880193e-14 4.54747350886e-13 4.54747350886e-13
    1.1 -9.10382880193e-14 0.0 0.0
    1.1 -9.10382880193e-14 0.0 0.0
    1.1 -3.41948691585e-14 5.68434188608e-14 5.68434188608e-14
    1.1 -5.77315972805e-15 2.84217094304e-14 2.84217094304e-14
    1.1 -5.77315972805e-15 0.0 0.0
    1.1 -5.77315972805e-15 0.0 0.0
    1.1 -2.22044604925e-15 3.5527136788e-15 3.5527136788e-15
    1.1 -4.4408920985e-16 1.7763568394e-15 1.7763568394e-15
    1.1 -4.4408920985e-16 0.0 0.0
    1.1 -4.4408920985e-16 0.0 0.0
    1.1 -2.22044604925e-16 2.22044604925e-16 2.22044604925e-16
    1.1 0.0 1.11022302463e-16 2.22044604925e-16
    1.1 0.0 0.0 0.0
    1.1 0.0 0.0 0.0
    1.1 0.0 1.38777878078e-17 0.0
    1.1 0.0 6.93889390391e-18 0.0

    So here we have reached the point where the maximum precision is reached
    - a doesn't change anymore, although it should. The error is about 1E-16.

    Now if you multiply two values with an error, the error also propagates
    into the result - PLUs the result can have its own error source - in the
    same order of magnitude.

    (a+e) * (a+e) = a*a + 2*a*e + e*e. So your new error term is 2*a*e + e*e
    or (2*a + e) * e.
     
    Thomas Rachel, Sep 6, 2011
    #3
  4. Thomas Rachel wrote:

    > Now if you multiply two values with an error, the error also propagates
    > into the result - PLUs the result can have its own error source - in the
    > same order of magnitude.
    >
    > (a+e) * (a+e) = a*a + 2*a*e + e*e. So your new error term is 2*a*e + e*e
    > or (2*a + e) * e.


    Your explanation about floating-point precision, which I already knew about
    but have only scanned here – so it might be flawed as well –,
    notwithstanding, it is not clear to me at all what you are trying to prove
    there.

    Computers (well, perhaps outside of mathematical software) do NOT compute an
    equation as humans would do, so the binomial theorem does NOT apply. In an
    algorithm of the real implementation,

    (a + e) * (a + e)

    would be computed as follows:

    b := a + e
    c := b * b
    or
    c := b + … + b

    [add the value of `b' to the value of `b' (b−1) times, since binary logic
    does not support multiplication]

    IOW, the error does propagate into the result indeed, but not as you
    described. Indeed, thanks to rounding on assignment and multiplication
    (i. e., setting register values or IEEE-754 floating-point mantissa and
    exponent), the error will be different, probably greater than you compute
    here.

    --
    PointedEars

    Bitte keine Kopien per E-Mail. / Please do not Cc: me.
     
    Thomas 'PointedEars' Lahn, Sep 6, 2011
    #4
  5. On Wed, 7 Sep 2011 02:07 am Thomas 'PointedEars' Lahn wrote:

    > Thomas Rachel wrote:
    >
    >> Now if you multiply two values with an error, the error also propagates
    >> into the result - PLUs the result can have its own error source - in the
    >> same order of magnitude.
    >>
    >> (a+e) * (a+e) = a*a + 2*a*e + e*e. So your new error term is 2*a*e + e*e
    >> or (2*a + e) * e.

    >
    > Your explanation about floating-point precision, which I already knew
    > about but have only scanned here – so it might be flawed as well –,
    > notwithstanding, it is not clear to me at all what you are trying to prove
    > there.
    >
    > Computers (well, perhaps outside of mathematical software) do NOT compute
    > an equation as humans would do, so the binomial theorem does NOT apply.


    I think you have misunderstood. The binomial theorem always applies. Any
    time you multiply two numbers, both numbers can always be re-written as a
    sum of two numbers:

    10*5 = (6+4)*(2+3)

    So a perfect square can always be re-written in the form where the binomial
    theorem applies:

    5*5 = (2+3)*(2+3)
    25 = 2*2 + 2*3 + 3*2 + 3*3
    25 = 4 + 6 + 6 + 9
    25 = 25

    The binomial theorem is not a part of the algorithm for performing
    multiplication. It is part of the analysis of the errors that occur during
    multiplication. The actual mechanics of how bits are flipped is irrelevant.

    Any floating point number x should be considered as equal to (a+e), where a
    is the number actually wanted by the user, and e the error term forced upon
    the user by the use of binary floats. (If you're lucky, e=0.) Generally,
    both a and e are unknown, but of course their sum is known -- it's just the
    float x.

    So given a float x, when you square it you get this:

    Exact values: a*a = a**2

    Float values: x*x = (a+e)(a+e)
    = a**2 + 2*a*e + e**2

    So the error term has increased from e to (2*a*e+e**2). It is usual to
    assume that e**2 is small enough that it underflows to zero, so we have the
    error term e increasing to 2*a*e as a fairly simple estimate of the new
    error.


    > In an algorithm of the real implementation,
    >
    > (a + e) * (a + e)
    >
    > would be computed as follows:
    >
    > b := a + e
    > c := b * b
    > or
    > c := b + … + b
    >
    > [add the value of `b' to the value of `b' (b−1) times, since binary logic
    > does not support multiplication]


    What you probably mean to say is that binary hardware usually implements
    multiplication via repeated addition.

    http://en.wikipedia.org/wiki/Binary_multiplier

    If you don't mean that, I can't imagine what you are trying to say.
    Multiplication of numbers exists in any base, binary no less than decimal.


    > IOW, the error does propagate into the result indeed, but not as you
    > described. Indeed, thanks to rounding on assignment and multiplication
    > (i. e., setting register values or IEEE-754 floating-point mantissa and
    > exponent), the error will be different, probably greater than you compute
    > here.


    There may be other sources of error, particularly when multiplying two
    numbers of greatly different magnitudes, but that doesn't invalidate Thomas
    Rachel's point (which is only an estimate, of course).

    We can see how much error is actually there by using exact arithmetic:

    Error in float 1.1:

    >>> from fractions import Fraction as F
    >>> >>> a = F(11, 10)
    >>> x = F.from_float(1.1)
    >>> e = x - a
    >>> print e

    1/11258999068426240

    Error in float 1.1*1.1:

    >>> b = F(11, 10)**2
    >>> y = F.from_float(1.1**2)
    >>> f = y - b
    >>> print f

    21/112589990684262400

    which is slightly more than double e above, and slightly less than our
    estimate of 2*a*e = 11/56294995342131200

    So we can conclude that, at least for 1.1**2, Python floats are more
    accurate than we would expect from a simple application of the binomial
    theorem. (For implementations using IEEE doubles.)


    --
    Steven
     
    Steven D'Aprano, Sep 7, 2011
    #5
  6. On Wed, 07 Sep 2011 14:51:13 +1000, Steven D'Aprano
    <> declaimed the following in
    gmane.comp.python.general:


    > What you probably mean to say is that binary hardware usually implements
    > multiplication via repeated addition.
    >
    > http://en.wikipedia.org/wiki/Binary_multiplier
    >

    Technically, as shown by that article, binary multiplication is not
    what is commonly thought of "repeated addition" (the method used on old
    adding machines wherein one hit "+" /n/ times [though one normally
    entered a "0" for each place: rather than hit "+" 20 time for 123 * 20,
    on would hit 123, 0 "+", enter a 0 {giving 1230}, then hitting 2 "+"]).

    In binary most of the subparts are the results of left shifts
    (factors of 2); so "repeated" doesn't really apply... Multiple shift/add
    OTOH...
    --
    Wulfraed Dennis Lee Bieber AF6VN
    HTTP://wlfraed.home.netcom.com/
     
    Dennis Lee Bieber, Sep 7, 2011
    #6
  7. xyz

    Gelonida N Guest

    On 09/07/2011 06:51 AM, Steven D'Aprano wrote:
    11258999068426240
    >
    > Error in float 1.1*1.1:
    >
    >>>> b = F(11, 10)**2
    >>>> y = F.from_float(1.1**2)
    >>>> f = y - b
    >>>> print f

    > 21/112589990684262400
    >
    > which is slightly more than double e above, and slightly less than our
    > estimate of 2*a*e = 11/56294995342131200
    >
    > So we can conclude that, at least for 1.1**2, Python floats are more
    > accurate than we would expect from a simple application of the binomial
    > theorem. (For implementations using IEEE doubles.)



    The reason why the error is different from the 2*a*e is, that we
    encounter two problems.

    first problem is, that x = a + e
    e exists because a float does have a limited number (let's call it N) of
    digits and a has an infinite amount of non zero digits in the binary format.


    second problem is, that the result of the multiplication is not

    (a+e) * (a+e) but a 'rounded' version of it, because the floating point
    representation of the result would require about 2*N digits, whereas
    only N digits will be stored in the result.

    depending on the rounding which happened (up or down) the error will be
    bigger or smaller than the estimated one.
     
    Gelonida N, Sep 7, 2011
    #7
  8. xyz

    Terry Reedy Guest

    On 9/7/2011 12:51 AM, Steven D'Aprano wrote:

    > So given a float x, when you square it you get this:
    >
    > Exact values: a*a = a**2
    >
    > Float values: x*x = (a+e)(a+e)
    > = a**2 + 2*a*e + e**2
    >
    > So the error term has increased from e to (2*a*e+e**2). It is usual to
    > assume that e**2 is small enough that it underflows to zero, so we have the
    > error term e increasing to 2*a*e as a fairly simple estimate of the new
    > error.


    And the relative error, which is what is often important, increases from
    e/a to 2e/a.

    --
    Terry Jan Reedy
     
    Terry Reedy, Sep 7, 2011
    #8
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