B. Floating Point Arithmetic: Issues and Limitations
Floating-point numbers are represented in computer hardware as base 2 (binary)
fractions. For example, the decimal fraction
has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction
has value 0/2 + 0/4 + 1/8. These two fractions have identical values, the only real
difference being that the first is written in base 10 fractional notation, and the second
in base 2.
Unfortunately, most decimal fractions cannot be represented exactly as binary
fractions. A consequence is that, in general, the decimal floating-point numbers you enter
are only approximated by the binary floating-point numbers actually stored in the machine.
The problem is easier to understand at first in base 10. Consider the fraction 1/3. You
can approximate that as a base 10 fraction:
or, better,
or, better,
and so on. No matter how many digits you're willing to write down, the result will
never be exactly 1/3, but will be an increasingly better approximation to 1/3.
In the same way, no matter how many base 2 digits you're willing to use, the decimal
value 0.1 cannot be represented exactly as a base 2 fraction. In base 2, 1/10 is the
infinitely repeating fraction
0.0001100110011001100110011001100110011001100110011...
Stop at any finite number of bits, and you get an approximation. This is why you see
things like:
>>> 0.1
0.10000000000000001
On most machines today, that is what you'll see if you enter 0.1 at a Python prompt.
You may not, though, because the number of bits used by the hardware to store
floating-point values can vary across machines, and Python only prints a decimal
approximation to the true decimal value of the binary approximation stored by the machine.
On most machines, if Python were to print the true decimal value of the binary
approximation stored for 0.1, it would have to display
>>> 0.1
0.1000000000000000055511151231257827021181583404541015625
instead! The Python prompt (implicitly) uses the builtin repr()
function to obtain a string version of everything it displays. For floats, repr(float)
rounds the true decimal value to 17 significant digits, giving
repr(float) produces 17 significant digits because it turns out
that's enough (on most machines) so that eval(repr(x)) == x
exactly for all finite floats x, but rounding to 16 digits is not enough to
make that true.
Note that this is in the very nature of binary floating-point: this is not a bug in
Python, it is not a bug in your code either, and you'll see the same kind of thing in all
languages that support your hardware's floating-point arithmetic (although some languages
may not display the difference by default, or in all output modes).
Python's builtin str() function produces only 12 significant
digits, and you may wish to use that instead. It's unusual for eval(str(x))
to reproduce x, but the output may be more pleasant to look at:
It's important to realize that this is, in a real sense, an illusion: the value in the
machine is not exactly 1/10, you're simply rounding the display of the true machine
value.
Other surprises follow from this one. For example, after seeing
>>> 0.1
0.10000000000000001
you may be tempted to use the round() function to chop it
back to the single digit you expect. But that makes no difference:
>>> round(0.1, 1)
0.10000000000000001
The problem is that the binary floating-point value stored for "0.1" was
already the best possible binary approximation to 1/10, so trying to round it again can't
make it better: it was already as good as it gets.
Another consequence is that since 0.1 is not exactly 1/10, adding 0.1 to itself 10
times may not yield exactly 1.0, either:
>>> sum = 0.0
>>> for i in range(10):
... sum += 0.1
...
>>> sum
0.99999999999999989
Binary floating-point arithmetic holds many surprises like this. The problem with
"0.1" is explained in precise detail below, in the "Representation
Error" section. See The Perils of Floating Point for a more complete account of other common surprises.
As that says near the end, ``there are no easy answers.'' Still, don't be unduly wary
of floating-point! The errors in Python float operations are inherited from the
floating-point hardware, and on most machines are on the order of no more than 1 part in
2**53 per operation. That's more than adequate for most tasks, but you do need to keep in
mind that it's not decimal arithmetic, and that every float operation can suffer a new
rounding error.
While pathological cases do exist, for most casual use of floating-point arithmetic
you'll see the result you expect in the end if you simply round the display of your final
results to the number of decimal digits you expect. str()
usually suffices, and for finer control see the discussion of Pythons's %
format operator: the %g, %f and %e format codes
supply flexible and easy ways to round float results for display.
B.1 Representation Error
This section explains the ``0.1'' example in detail, and shows how you can perform an
exact analysis of cases like this yourself. Basic familiarity with binary floating-point
representation is assumed.
Representation error refers to that some (most, actually) decimal
fractions cannot be represented exactly as binary (base 2) fractions. This is the chief
reason why Python (or Perl, C, C++, Java, Fortran, and many others) often won't display
the exact decimal number you expect:
>>> 0.1
0.10000000000000001
Why is that? 1/10 is not exactly representable as a binary fraction. Almost all
machines today (November 2000) use IEEE-754 floating point arithmetic, and almost all
platforms map Python floats to IEEE-754 "double precision". 754 doubles contain
53 bits of precision, so on input the computer strives to convert 0.1 to the closest
fraction it can of the form J/2**N where J is an integer
containing exactly 53 bits. Rewriting
as
and recalling that J has exactly 53 bits (is >= 2**52 but <
2**53), the best value for N is 56:
>>> 2L**52
4503599627370496L
>>> 2L**53
9007199254740992L
>>> 2L**56/10
7205759403792793L
That is, 56 is the only value for N that leaves J with exactly 53
bits. The best possible value for J is then that quotient rounded:
>>> q, r = divmod(2L**56, 10)
>>> r
6L
Since the remainder is more than half of 10, the best approximation is obtained by
rounding up:
>>> q+1
7205759403792794L
Therefore the best possible approximation to 1/10 in 754 double precision is that over
2**56, or
7205759403792794 / 72057594037927936
Note that since we rounded up, this is actually a little bit larger than 1/10; if we
had not rounded up, the quotient would have been a little bit smaller than 1/10. But in no
case can it be exactly 1/10!
So the computer never ``sees'' 1/10: what it sees is the exact fraction given above,
the best 754 double approximation it can get:
>>> .1 * 2L**56
7205759403792794.0
If we multiply that fraction by 10**30, we can see the (truncated) value of its 30 most
significant decimal digits:
>>> 7205759403792794L * 10L**30 / 2L**56
100000000000000005551115123125L
meaning that the exact number stored in the computer is approximately equal to the
decimal value 0.100000000000000005551115123125. Rounding that to 17 significant digits
gives the 0.10000000000000001 that Python displays (well, will display on any
754-conforming platform that does best-possible input and output conversions in its C
library -- yours may not!).
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