Re: Is there a bug in the inversion of floats

[Oliver Heimlich]

Am 27. April 2016 18:47:05 MESZ, schrieb John Swensen <address@hidden>:
>
>> On Apr 27, 2016, at 9:40 AM, JuanPi <address@hidden> wrote:
>> 
>> On Wed, Apr 27, 2016 at 6:26 PM, Oliver Heimlich <address@hidden>
>wrote:
>>> Am 27. April 2016 14:12:23 MESZ, schrieb JuanPi <address@hidden>:
>>>> I get the following results and I am not sure this is a bug or
>there
>>>> is something I do not understand in the way the floats are
>>>> represented/treated
>>>> 
>>>> format native-hex
>>>> 
>>>> v = complex (0,1)
>>>> inv(exp(v)),exp(-v), 1./exp(v)
>>>> ans = 8c06b50f284ae13f  ee0c098f54edeabfi
>>>> ans = 8c06b50f284ae13f  ee0c098f54edeabfi
>>>> ans = 8c06b50f284ae13f  ee0c098f54edeabfi
>>>> 
>>>> v = complex (1,0)
>>>> inv(exp(v)),exp(-v), 1./exp(v)
>>>> ans = 38ef2c36568bd73f
>>>> ans = 38ef2c36568bd73f
>>>> ans = 38ef2c36568bd73f
>>>> 
>>>> so far so good, but
>>>> 
>>>> inv(exp(v)),exp(-v), 1./exp(v)
>>>> ans = 21d8befb2a71c93f  545f8539d4cfd3bfi
>>>> ans = 23d8befb2a71c93f  555f8539d4cfd3bfi
>>>> ans = 23d8befb2a71c93f  555f8539d4cfd3bfi
>>>> 
>>>> (inv gives different results) and thought: ok but who uses inv to
>>>> invert numbers...
>>>> so I did
>>>> 
>>>> inv(exp(v/99)),exp(-v/99), 1./exp(v/99)
>>>> ans = f30d005d41adef3f  52ef0272867a84bfi
>>>> ans = f30d005d41adef3f  50ef0272867a84bfi
>>>> ans = f30d005d41adef3f  52ef0272867a84bfi
>>>> 
>>>> (exp(-a) gives different results) and started worrying
>>>> 
>>>> Essentially I hoped to get a definite false here, but I did not
>>>> 
>>>> t = linspace (0, 1, 100);
>>>> v = complex(0,1);
>>>> V = v * ( t - t.');
>>>> M = exp (-V) != 1 ./ exp (V);
>>>> any(M(:))
>>>> 
>>>> In this case the result is also true for any v
>>>> 
>>>> Should one desire that the result of exp(-v) is exactly the same as
>>>> 1/exp(v) for consistency?
>>> 
>>> Regarding your last question,  exp (-v) computes the value without
>intermediate rounding errors whereas 1 ./ exp (v) has to invert an
>intermediate result that has rounding errors and inversion isn't error
>free.
>>> 
>>> So I would expect exp (-v) to computer a better value in general.
>>> 
>>> Oliver
>> I guess it depends on how the interpreter parses 1 ./ exp (v), but if
>> indeed you do the the two operations they propagate more error.
>> The underlying question is what does matlab returns here...
>> 
>> t = linspace (0, 1, 100);
>> V = bsxfun (@minus,  t , t.');
>> M = exp (-V) != 1 ./ exp (V);
>> any(M(:))
>> 
>> 
>> -- 
>> JuanPi Carbajal
>> Public GnuPG key: 9C5B72BF
>> -----
>> The end of funding: "Many researchers were caught up in a web of
>> increasing exaggeration."
>> - Hans Moravec
>> 
>
>Matlab 2014b
>==========
>>> t = linspace (0, 1, 100);
>V = bsxfun (@minus,  t , t.');
>M = exp (-V) ~= 1 ./ exp (V);
>any(M(:))
>sum(M(:))
>
>
>ans =
>
>     1
>
>
>ans =
>
>        3195
>
>
>Octave 4.0.0
>=========
>>> t = linspace (0, 1, 100);
>>> V = bsxfun (@minus,  t , t.');
>>> M = exp (-V) != 1 ./ exp (V);
>>> any(M(:))
>ans = 1
>>> sum(M(:))
>ans = 2618
>
>
>
>John S.

The exp operation is not guaranteed to be correctly-rounded according to IEEE 
754 and may produce different results depending on your system.

Oliver