Text from Tim & Guido discussing floating point arithmetic and what users

freddrake · freddrake · commit 417d667dd925 · 2001-06-08T16:24:58.000Z
need to understand about the binary &amp; decimal fp, so that representation
weirdness is documented somewhere.  This makes it easier to repond to "bug"
reports caused by user confusion &amp; ignorance of the issues.

This closes SF patch #426208.
diff --git a/Doc/tut/tut.tex b/Doc/tut/tut.tex
@@ -4085,4 +4085,269 @@ \section{Commentary \label{commentary}}
 be useful.
 
 
+\chapter{Floating Point Arithmetic:  Issues and Limitations
+         \label{fp-issues}}
+
+Floating-point numbers are represented in computer hardware as
+base 2 (binary) fractions.  For example, the decimal fraction
+
+\begin{verbatim}
+0.125
+\end{verbatim}
+
+has value 1/10 + 2/100 + 5/1000, and in the same way the binary fraction
+
+\begin{verbatim}
+0.001
+\end{verbatim}
+
+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:
+
+\begin{verbatim}
+0.3
+\end{verbatim}
+
+or, better,
+
+\begin{verbatim}
+0.33
+\end{verbatim}
+
+or, better,
+
+\begin{verbatim}
+0.333
+\end{verbatim}
+
+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
+
+\begin{verbatim}
+0.0001100110011001100110011001100110011001100110011...
+\end{verbatim}
+
+Stop at any finite number of bits, and you get an approximation.  This
+is why you see things like:
+
+\begin{verbatim}
+>>> 0.1
+0.10000000000000001
+\end{verbatim}
+
+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
+
+\begin{verbatim}
+>>> 0.1
+0.1000000000000000055511151231257827021181583404541015625
+\end{verbatim}
+
+instead!  The Python prompt (implicitly) uses the builtin
+\function{repr()} function to obtain a string version of everything it
+displays.  For floats, \code{repr(\var{float})} rounds the true
+decimal value to 17 significant digits, giving
+
+\begin{verbatim}
+0.10000000000000001
+\end{verbatim}
+
+\code{repr(\var{float})} produces 17 significant digits because it
+turns out that's enough (on most machines) so that
+\code{eval(repr(\var{x})) == \var{x}} exactly for all finite floats
+\var{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.
+
+Python's builtin \function{str()} function produces only 12
+significant digits, and you may wish to use that instead.  It's
+unusual for \code{eval(str(\var{x}))} to reproduce \var{x}, but the
+output may be more pleasant to look at:
+
+\begin{verbatim}
+>>> print str(0.1)
+0.1
+\end{verbatim}
+
+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 \emph{display} of the true machine value.
+
+Other surprises follow from this one.  For example, after seeing
+
+\begin{verbatim}
+>>> 0.1
+0.10000000000000001
+\end{verbatim}
+
+you may be tempted to use the \function{round()} function to chop it
+back to the single digit you expect.  But that makes no difference:
+
+\begin{verbatim}
+>>> round(0.1, 1)
+0.10000000000000001
+\end{verbatim}
+
+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:
+
+\begin{verbatim}
+>>> sum = 0.0
+>>> for i in range(10):
+...     sum += 0.1
+...
+>>> sum
+0.99999999999999989
+\end{verbatim}
+
+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
+\citetitle[http://www.lahey.com/float.htm]{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.  \function{str()} usually suffices, and for
+finer control see the discussion of Pythons's \code{\%} format
+operator: the \code{\%g}, \code{\%f} and \code{\%e} format codes
+supply flexible and easy ways to round float results for display.
+
+
+\section{Representation Error
+         \label{fp-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.
+
+\dfn{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, \Cpp,
+Java, Fortran, and many others) often won't display the exact decimal
+number you expect:
+
+\begin{verbatim}
+>>> 0.1
+0.10000000000000001
+\end{verbatim}
+
+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 \var{J}/2**\var{N} where \var{J} is an integer containing
+exactly 53 bits.  Rewriting
+
+\begin{verbatim}
+ 1 / 10 ~= J / (2**N)
+\end{verbatim}
+
+as
+
+\begin{verbatim}
+J ~= 2**N / 10
+\end{verbatim}
+
+and recalling that \var{J} has exactly 53 bits (is \code{>= 2**52} but
+\code{< 2**53}), the best value for \var{N} is 56:
+
+\begin{verbatim}
+>>> 2L**52
+4503599627370496L
+>>> 2L**53
+9007199254740992L
+>>> 2L**56/10
+7205759403792793L
+\end{verbatim}
+
+That is, 56 is the only value for \var{N} that leaves \var{J} with
+exactly 53 bits.  The best possible value for \var{J} is then that
+quotient rounded:
+
+\begin{verbatim}
+>>> q, r = divmod(2L**56, 10)
+>>> r
+6L
+\end{verbatim}
+
+Since the remainder is more than half of 10, the best approximation is
+obtained by rounding up:
+
+\begin{verbatim}
+>>> q+1
+7205759403792794L
+\end{verbatim}
+
+Therefore the best possible approximation to 1/10 in 754 double
+precision is that over 2**56, or
+
+\begin{verbatim}
+7205759403792794 / 72057594037927936
+\end{verbatim}
+
+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:
+
+\begin{verbatim}
+>>> .1 * 2L**56
+7205759403792794.0
+\end{verbatim}
+
+If we multiply that fraction by 10**30, we can see the (truncated)
+value of its 30 most significant decimal digits:
+
+\begin{verbatim}
+>>> 7205759403792794L * 10L**30 / 2L**56
+100000000000000005551115123125L
+\end{verbatim}
+
+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!).
+
 \end{document}
