<steve+comp.lang.pyt...@pearwood.info> wrote:
> The question is, what is the largest integer number N such that every
> whole number between -N and N inclusive can be represented as a float?
> If my tests are correct, that value is 9007199254740992.0 = 2**53.
> Have I got this right? Is there a way to work out the gap between one
> float and the next?
That looks mathematically correct. The "gap" between floats is the
equivalent of a difference of 1 bit in the significand. For a
floating point number represented as (sign * c * 2 ** q), where c is
an integer, the gap between floats is equal to 2 ** q. There are 53
bits of precision in a double-precision float (technically an implicit
1 followed by 52 bits), so q becomes greater than 0 at 2 ** 53.
Steven D'Aprano writes:
> Python floats can represent exact integer values (e.g. 42.0), but above a > certain value (see below), not all integers can be represented. For > example:
> py> 1e16 == 1e16 + 1 # no such float as 10000000000000001.0
> True
> py> 1e16 + 3 == 1e16 + 4 # or 10000000000000003.0
> True
> So some integers are missing from the floats. For large enough values, > the gap between floats is rather large, and many numbers are missing:
> py> 1e200 + 1e10 == 1e200
> True
> The same applies for large enough negative values.
> The question is, what is the largest integer number N such that every > whole number between -N and N inclusive can be represented as a float?
> If my tests are correct, that value is 9007199254740992.0 = 2**53.
> Have I got this right? Is there a way to work out the gap between one > float and the next?
There is a way to find the distance between two IEEE floats in "ulps",
or "units in the last position", computable from the bit pattern using
integer arithmetic. I think it's then also possible to find the next
float by adding one.
I don't have a link at hand, I'm too tired to search at the moment,
and I'm no expert on floats, but you might find an answer by looking
for ulps.
> (I haven't tried to exhaustively check every float because, even at one > nanosecond per number, it will take over 200 days.)
Come to think of it, the difference between adjacent floats is exactly
one ulp. Just use the right unit :)
On Fri, 21 Sep 2012 17:29:13 +0000, Steven D'Aprano wrote:
> The question is, what is the largest integer number N such that every
> whole number between -N and N inclusive can be represented as a float?
> If my tests are correct, that value is 9007199254740992.0 = 2**53.
> Have I got this right? Is there a way to work out the gap between one
> float and the next?
CPython's "float" type uses C's "double". For a system where C's "double"
is IEEE-754 double precision, N=2**53 is the correct answer.
An IEEE-754 double precision value consists of a 53-bit integer whose
first bit is a "1", multiplied or divided by a power of two.
The largest 53-bit integer is 2**53-1. 2**53 can be represented as
2**52 * 2**1. 2**53+1 cannot be represented in this form. 2**53+2 can be
represented as (2**52+1) * 2**1.
For values x where 2**52 <= x < 2**53, the the interval between
representable values (aka Unit in the Last Place or ULP) is 1.0.
For 2**51 <= x < 2**52, the ULP is 0.5.
For 2**53 <= x < 2**54, the ULP is 2.0.
And so on.
> On 21 Sep 2012 17:29:13 GMT, Steven D'Aprano
> <steve+comp.lang.pyt...@pearwood.info> declaimed the following in
> gmane.comp.python.general:
>> The question is, what is the largest integer number N such that every >> whole number between -N and N inclusive can be represented as a float?
> Single precision commonly has 7 significant (decimal) digits. Double
> precision runs somewhere between 13 and 15 (decimal) significant digits
>> If my tests are correct, that value is 9007199254740992.0 = 2**53.
The expression 2 / sys.float_info.epsilon produces exactly that
number. That's probably not a coincidence.
> For an encoding of a double precision using one sign bit and an
> 8-bit exponent, you have 53 bits available for the mantissa.
If your floats have 64 bits, and you use 1 bit for the sign and 8 for
the exponent, you'll have 55 bits available for the mantissa.
> This
> ignores the possibility of an implied msb in the mantissa (encodings
> which normalize to put the leading 1-bit at the msb can on some machines
> remove that 1-bit and shift the mantissa one more place; effectively
> giving a 54-bit mantissa).
My machine has 64-bits floats, using 1 bit for the sign, 11 for the
exponent, leaving 52 for the mantissa. The mantissa has an implied
leading 1, so it's nominally 53 bits.
You can find this number in sys.float_info.mant_dig
> Something like an old XDS Sigma-6 used
> non-binary exponents (exponent was in power of 16 <> 2^4) and used
> "non-normalized" mantissa -- the mantissa could have up to three leading
> 0-bits); this affected the decimal significance...
Your Sigma-6 must have sys.float_info.radix == 16 then.
On Fri, 21 Sep 2012 17:29:13 +0000, Steven D'Aprano wrote:
> Python floats can represent exact integer values (e.g. 42.0), but above
> a certain value (see below), not all integers can be represented. For
> example:
> py> 1e16 == 1e16 + 1 # no such float as 10000000000000001.0 True py>
> 1e16 + 3 == 1e16 + 4 # or 10000000000000003.0 True
> So some integers are missing from the floats. For large enough values,
> the gap between floats is rather large, and many numbers are missing:
> py> 1e200 + 1e10 == 1e200 True
> The same applies for large enough negative values.
> The question is, what is the largest integer number N such that every
> whole number between -N and N inclusive can be represented as a float?
> If my tests are correct, that value is 9007199254740992.0 = 2**53.
> Have I got this right? Is there a way to work out the gap between one
> float and the next?
> (I haven't tried to exhaustively check every float because, even at one
> nanosecond per number, it will take over 200 days.)
technically this would be implementation dependant, although the other responses are probably accurate for most (if not all) current implementations :-)
-- Well, I'm a classic ANAL RETENTIVE!! And I'm looking for a way to
VICARIOUSLY experience some reason to LIVE!!
Steven D'Aprano <steve+comp.lang.pyt...@pearwood.info> writes:
> Have I got this right? Is there a way to work out the gap between one > float and the next?
Yes, 53-bit mantissa as people have mentioned. That tells you what ints
can be exactly represented. But, arithmetic in some situations can have
a 1-ulp error. So I wonder if it's possible that if n is large enough,
you might have something like n+1==n even if the integers n and n+1 have
distinct floating point representations.
On Fri, 21 Sep 2012 15:23:41 -0700, Paul Rubin wrote:
> Steven D'Aprano <steve+comp.lang.pyt...@pearwood.info> writes:
>> Have I got this right? Is there a way to work out the gap between one
>> float and the next?
> Yes, 53-bit mantissa as people have mentioned. That tells you what ints
> can be exactly represented. But, arithmetic in some situations can have
> a 1-ulp error. So I wonder if it's possible that if n is large enough,
> you might have something like n+1==n even if the integers n and n+1 have
> distinct floating point representations.
I don't think that is possible for IEEE 754 floats, where integer arithmetic is exact. But I'm not entirely sure, which is why I asked.
For non IEEE 754 floating point systems, there is no telling how bad the implementation could be :(
On Fri, 21 Sep 2012 15:23:41 -0700, Paul Rubin wrote:
> Steven D'Aprano <steve+comp.lang.pyt...@pearwood.info> writes:
>> Have I got this right? Is there a way to work out the gap between one
>> float and the next?
> Yes, 53-bit mantissa as people have mentioned. That tells you what ints
> can be exactly represented. But, arithmetic in some situations can have a
> 1-ulp error. So I wonder if it's possible that if n is large enough, you
> might have something like n+1==n even if the integers n and n+1 have
> distinct floating point representations.
Not for IEEE-754. Or for any sane implementation, for that matter. OTOH,
you can potentially get n != n due to the use of extended precision for
intermediate results.
For IEEE-754, addition, subtraction, multiplication, division, remainder
and square root are "exact" in the sense that the result is as if the
arithmetic had been performed with an infinite number of bits then rounded
afterwards. For round-to-nearest, the result will be the closest
representable value to the exact value.
Transcendental functions suffer from the "table-maker's dilemma", and the
result will be one of the two closest representable values to the exact
value, but not necessarily *the* closest.
>On 22 Sep 2012 01:36:59 GMT, Steven D'Aprano wrote:
>> For non IEEE 754 floating point systems, there is no telling how bad the >> implementation could be :(
> Let's see what can be found...
> IBM 360: Same as Sigma-6 (no surprise; hearsay is the Sigma was
>designed by renegade IBM folk; even down to using EBCDIC internally --
>but with a much different interrupt system [224 individual interrupt
>vectors as I recall, vs the IBM's 7 vectors and polling to find what
>device]).
The Control Data 6000/Cyber series had sign bit and 11-bit exponent, with
either a 48-bit mantissa or a 96-bit mantissa, packed into one or two
60-bit words. Values were not automatically normalized, so there was no
assumed 1 bit, as in IEEE-754.
-- Tim Roberts, t...@probo.com
Providenza & Boekelheide, Inc.
> Dennis Lee Bieber <wlfr...@ix.netcom.com> wrote:
>> On 22 Sep 2012 01:36:59 GMT, Steven D'Aprano wrote:
>>> For non IEEE 754 floating point systems, there is no telling how bad the >>> implementation could be :(
>> Let's see what can be found...
>> IBM 360: Same as Sigma-6 (no surprise; hearsay is the Sigma was
>> designed by renegade IBM folk; even down to using EBCDIC internally --
>> but with a much different interrupt system [224 individual interrupt
>> vectors as I recall, vs the IBM's 7 vectors and polling to find what
>> device]).
> The Control Data 6000/Cyber series had sign bit and 11-bit exponent, with
> either a 48-bit mantissa or a 96-bit mantissa, packed into one or two
> 60-bit words. Values were not automatically normalized, so there was no
> assumed 1 bit, as in IEEE-754.
And it's been a long time (about 39 years), but as I recall the CDC 6400
(at least) had no integer multiply or divide. You had to convert to
float first. The other oddity about the CDC series is it's the last
machine I've encountered that used ones-complement for ints, with two
values for zero.
True. Seymour wanted all of the integer instructions to be combinatorial logic, rather than iterative. Fortunately, since the floating point binary point was to the right, it was trivial to pack integers to float, do a floating computation, then unpack back to integer.
Apologize in advance for top-posting. My Xoom makes bottom-posting awkward.
--
Tim Roberts, t...@probo.com
Providenza & Boekelheide, Inc.
> Dennis Lee Bieber <wlfr...@ix.netcom.com> wrote:
>> On 22 Sep 2012 01:36:59 GMT, Steven D'Aprano wrote:
>>> For non IEEE 754 floating point systems, there is no telling how bad the
>>> implementation could be :(
>> Let's see what can be found...
>> IBM 360: Same as Sigma-6 (no surprise; hearsay is the Sigma was
>> designed by renegade IBM folk; even down to using EBCDIC internally --
>> but with a much different interrupt system [224 individual interrupt
>> vectors as I recall, vs the IBM's 7 vectors and polling to find what
>> device]).
> The Control Data 6000/Cyber series had sign bit and 11-bit exponent, with
> either a 48-bit mantissa or a 96-bit mantissa, packed into one or two
> 60-bit words. Values were not automatically normalized, so there was no
> assumed 1 bit, as in IEEE-754.
And it's been a long time (about 39 years), but as I recall the CDC 6400
(at least) had no integer multiply or divide. You had to convert to
float first. The other oddity about the CDC series is it's the last
machine I've encountered that used ones-complement for ints, with two
values for zero.
> On 09/22/2012 05:05 PM, Tim Roberts wrote:
>> Dennis Lee Bieber <wlfr...@ix.netcom.com> wrote:
>>> On 22 Sep 2012 01:36:59 GMT, Steven D'Aprano wrote:
>>>> For non IEEE 754 floating point systems, there is no telling how bad the >>>> implementation could be :(
>>> Let's see what can be found...
>>> IBM 360: Same as Sigma-6 (no surprise; hearsay is the Sigma was
>>> designed by renegade IBM folk; even down to using EBCDIC internally --
>>> but with a much different interrupt system [224 individual interrupt
>>> vectors as I recall, vs the IBM's 7 vectors and polling to find what
>>> device]).
>> The Control Data 6000/Cyber series had sign bit and 11-bit exponent, with
>> either a 48-bit mantissa or a 96-bit mantissa, packed into one or two
>> 60-bit words. Values were not automatically normalized, so there was no
>> assumed 1 bit, as in IEEE-754.
> And it's been a long time (about 39 years), but as I recall the CDC 6400
> (at least) had no integer multiply or divide. You had to convert to
> float first.
You didn't have to convert if your ints would fit in 48 bits.
If that was the case, then using the float multiply and divide
instructions would do the trick.
> The other oddity about the CDC series is it's the last machine I've
> encountered that used ones-complement for ints, with two values for zero.
> On 09/22/2012 05:05 PM, Tim Roberts wrote:
>> Dennis Lee Bieber <wlfr...@ix.netcom.com> wrote:
>>> On 22 Sep 2012 01:36:59 GMT, Steven D'Aprano wrote:
>>>> For non IEEE 754 floating point systems, there is no telling how bad the >>>> implementation could be :(
>>> Let's see what can be found...
>>> IBM 360: Same as Sigma-6 (no surprise; hearsay is the Sigma was
>>> designed by renegade IBM folk; even down to using EBCDIC internally --
>>> but with a much different interrupt system [224 individual interrupt
>>> vectors as I recall, vs the IBM's 7 vectors and polling to find what
>>> device]).
>> The Control Data 6000/Cyber series had sign bit and 11-bit exponent, with
>> either a 48-bit mantissa or a 96-bit mantissa, packed into one or two
>> 60-bit words. Values were not automatically normalized, so there was no
>> assumed 1 bit, as in IEEE-754.
> And it's been a long time (about 39 years), but as I recall the CDC 6400
> (at least) had no integer multiply or divide. You had to convert to
> float first. The other oddity about the CDC series is it's the last
> machine I've encountered that used ones-complement for ints, with two
> values for zero.
Well, for what it is worth… the DEC Laboratory INstrument Computer (LINC-8, sort of a forced acronym because I believe they were built to specs issued by Lincoln Labs.) and the later DEC PDP-12 (which incorporated the LINC-8 instruction set, along with the PDP-8 basic set) also did ones-complement integer arithmetic. I never used a LINC-8, but I worked for several years around PDP-12s. (They also had a build-in CRT and a MUX'd analog-to-digital converter with CPU instructions for driving both directly.) As I remember, they maxed out at 32k (12-bit) words of RAM. I don't know when they were discontinued, but there were still some PDP-12s in use as late as 1988 when I lost track.
