Exact integer-valued floats

[Steven D'Aprano]

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.)

 

[Ian Kelly]

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.

Cheers,
Ian

 

[Jussi Piitulainen]

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

 

[Nobody]

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.

http://en.wikipedia.org/wiki/IEEE_754-1985

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.

 

[Dennis Lee Bieber]

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.

For an encoding of a double precision using one sign bit and an
8-bit exponent, you have 53 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). 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...

 

[Hans Mulder]

Single precision commonly has 7 significant (decimal) digits. Double
precision runs somewhere between 13 and 15 (decimal) significant digits

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.


Hope this helps,

-- HansM

 

[Paul Rubin]

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.

 

[Dennis Lee Bieber]

On 21/09/12 22:26:26, Dennis Lee Bieber wrote:

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.

Mea Culpa -- doing mental arithmetic too fast

 

[Steven D'Aprano]

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

 

[Dennis Lee Bieber]

For non IEEE 754 floating point systems, there is no telling how bad the
implementation could be

Let's see what can be found...

http://www.bitsavers.org/pdf/sds/sigma/sigma6/901713B_Sigma_6_Reference_Man_Jun71.pdf
pages 50-54
A sign bit, 7-bit offset base 16 exponent/characteristic, 24-bit
mantissa/fraction (for short; long float is 56 bit mantissa).

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]).



Motorola Fast Floating Point (software library for the Amiga, and
apparently also used on early Palm units)
Sign bit, 7-bit binary exponent, 24-bit mantissa


VAX http://nssdc.gsfc.nasa.gov/nssdc/formats/VAXFloatingPoint.htm
(really nasty looking as bits 6 is most significant, running down to bit
0, THEN bits 32-16 with 16 the least significant; extend for longer
formats)
F-float: sign bit, 8-bit exponent, 23 bits mantissa
D-float: as above but 55 bits mantissa
G-float: sign bit, 11-bit exponent, 52 bits mantissa
H-float: sign bit, 15-bit exponent, 112 bits mantissa

 

[Nobody]

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.

 

[Dave Angel]

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.

 

[Hans Mulder]

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.

Floats still have 0.0 and -0.0, even in IEEE-754.


Was Python ever ported to such unusual hardware?


-- HansM