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x/x=1.0 in double-Arithmetik

 
 
Tobias
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      08-24-2007
Let:

double x,y;
double z = max(x,y);

Does the standard ensure that

( z == 0.0 ) || ( x/z == 1.0 ) || ( y/z == 1.0 )

always gives true?

With best regards,
Tobias

 
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Victor Bazarov
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      08-24-2007
Tobias wrote:
> Let:
>
> double x,y;


Uninitialised. Contain indeterminate values, possibly such that
can cause a hardware exception. And, of course, there is no
guarantee that 'x' and 'y' contain the same indeterminate value.

> double z = max(x,y);


Passing indeterminate values to 'max' probably leads to undefined
behaviour.

> Does the standard ensure that
>
> ( z == 0.0 ) || ( x/z == 1.0 ) || ( y/z == 1.0 )
>
> always gives true?


No guarantees because of indeterminate values used.

V
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=?ISO-8859-1?Q?Erik_Wikstr=F6m?=
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      08-24-2007
On 2007-08-24 18:47, Tobias wrote:
> Let:
>
> double x,y;
> double z = max(x,y);
>
> Does the standard ensure that
>
> ( z == 0.0 ) || ( x/z == 1.0 ) || ( y/z == 1.0 )
>
> always gives true?


No, the first one requires that either x, y, or both are zero and the
other is less than zero, which the standard can not guarantee.

The second ones requires exact arithmetic to always be true, due to the
way floating point operations this might not always be the case, but it
will be close.

--
Erik Wikström
 
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Tobias
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      08-24-2007
>
> Passing indeterminate values to 'max' probably leads to undefined
> behaviour.


I'm sorry, I should have written:

bool foo(double x, double y)
{
double z = max(x,y);
return ( z == 0.0 ) || ( x/z == 1.0 ) || ( y/z == 1.0 );
}

That would make the real question more obvious.

Thanks for the comment anyway.

With best regards,
Tobias

 
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Tobias
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      08-24-2007

> AFAIK this is wrong if x or y is positive infinity; there may well be
> other such problems with "special values".
>


Thanks for the comment.
In my special application infinity will not occur.

With best regards,
Tobias


 
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Tobias
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      08-24-2007

> The second ones requires exact arithmetic to always be true, due to the
> way floating point operations this might not always be the case, but it
> will be close.


That is the real question. I'm especially interested in the case when
x and y become very small.

Somebody assured me that division a double by the same double (in the
same representation) gives always the exact double 1.0.

But might there be a problem with normalization or with representation
of the double numbers in the registers of the numerical core?

Best regards,
Tobias



 
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Alf P. Steinbach
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      08-24-2007
* Tobias:
>> Passing indeterminate values to 'max' probably leads to undefined
>> behaviour.

>
> I'm sorry, I should have written:
>
> bool foo(double x, double y)
> {
> double z = max(x,y);
> return ( z == 0.0 ) || ( x/z == 1.0 ) || ( y/z == 1.0 );
> }
>
> That would make the real question more obvious.


Unless x and y are special values, with modern floating point
implementations this is guaranteed. Floating point division, with a
modern floating point implementation, isn't inherently inexact: it just
rounds the result to some specific number of binary digits. Hence x/x,
with normal non-zero numbers, produces exactly 1.

Effectively, therefore, the foo function above checks whether x or y has
a special value where the relationship doesn't hold.

For IEEE floating point numbers this would be NaN and Infinity values.

According to the IEEE standard infinities are calculated with as limits,
however Inf/Inf is singled out as a special case producing NaN.

std::numeric_limits refers to the IEEE 754 standard as IEC 559 or
something, so in order to check whether you're using IEEE standard
floating point you'll have to use the member named is_iec_Something.

Cheers, and hth.,

- Alf

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Daniel Kraft
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      08-24-2007
Tobias wrote:
> Let:
>
> double x,y;
> double z = max(x,y);
>
> Does the standard ensure that
>
> ( z == 0.0 ) || ( x/z == 1.0 ) || ( y/z == 1.0 )
>
> always gives true?


I'm not that proficient with IEEE 754 floating point arithmetic, but
AFAIK this is wrong if x or y is positive infinity; there may well be
other such problems with "special values".

Cheers,
Daniel

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Markus Schoder
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      08-24-2007
On Fri, 24 Aug 2007 11:13:05 -0700, Tobias wrote:
>> The second ones requires exact arithmetic to always be true, due to the
>> way floating point operations this might not always be the case, but it
>> will be close.

>
> That is the real question. I'm especially interested in the case when x
> and y become very small.
>
> Somebody assured me that division a double by the same double (in the
> same representation) gives always the exact double 1.0.


That is true for IEEE 754 conforming arithmetic. Basic arithmetic
operations need to be done with exact rounding i.e. conceptually the
operation is carried out with infinite precision and then rounded (using
the current rounding mode) to the nearest representable number.

> But might there be a problem with normalization or with representation
> of the double numbers in the registers of the numerical core?


Since on the Intel architecture double and extended precision are mixed
and the actual CPU instructions make it very hard to avoid this most IEEE
754 guarantees are moot (thank you Intel).

For stable numerical behaviour you need to avoid the Intel architecture
or force your compiler to use the newer SSE2 instruction set exclusively
for floating point arithmetic if possible.

--
Markus Schoder
 
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pan
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      08-25-2007
In article <(E-Mail Removed)> "Alf P.
Steinbach"<(E-Mail Removed)> wrote:
> * Tobias:
>> ( x/z == 1.0 )

> Unless x and y are special values, with modern floating point
> implementations this is guaranteed. Floating point division, with a
> modern floating point implementation, isn't inherently inexact: it
> just rounds the result to some specific number of binary digits.
> Hence x/x, with normal non-zero numbers, produces exactly 1.


I had a funny experience that allows me to say that the number of
binarydigits is not so specific, but can vary unexpectedly =)

I had this compare predicate which gave me funny assertions when used
incombination with particular optimizations:

(unchecked code, just to give the idea)

struct Point {
double a, b;
};

bool anglesorter(Point a, Point b) {
return atan2(a.y, a.x) < atan2(b.y, b.x);
}

which, passed to a sort function, lead to assertions when a point in
the sequence was replicated. The assertion (inserted by microsoft as a
checkfor the predicates) has this condition:

pred(a, b) && !pred(b, a)

which is quite funny. After disassembling i found out that one of the
two atan2 was kept in a x87 register (80-bit), while the other was
stored inmemory (64-bit precision), so the result of

anglesorter(a, a)

would give true for some values of a.

Now, this doesn't look the same as the OP's example, but I wouldn't
trust the floating point optimizations again: what happens if foo gets
inlined, its parameters are into extended precision register, but the
result ofmax, for some odd reason, does get trimmed to
double-precision?

I must admit I don't like the alternative version too:

x/z == 1.0

becomes

fabs(x/z-1.0) < numeric_limits<double>::epsilon()

or, avoiding divisions:

fabs(x-z) < numeric_limits<double>::epsilon()*z

because they are both unreadable.
I'd try to avoid the need of such a compare in the first
place.

--
Marco

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