From: Michael Ulm <michael.ulm@...>
Date: 2007-08-31T15:25:17+09:00
Subject: Re: 0.06 == 0.06 returns false in Ruby?

Pe単a wrote:
--snip--
> irb(main):010:0> (0.05+0.01) - 0.06
> => 6.93889390390723e-18
> 
> as mentioned by Dan, careful on comparing floats. And as to any
> precision subject, there is what we call significant digits..
> 
> this floating problem is a faq and is very surprising on such a
> very high language such as ruby.  can we address this? maybe create
> flag like $EPSILON=0 or something, or may flag to revert to rational
> or bigdeci like $FLOAT_PROCESSOR=RATIONAL...
> 

Unfortunately, there is no easy solution to this problem. Here is a
catalog of often proposed solutions and why they do not work:

1 (proposed by doug meyer in this thread) Always use
  (x-y).abs < Float::EPSILON
  as a test for equality.

  This won't work because the rounding error easily can get bigger than
  Float::EPSILON, especially when dealing with numbers that are bigger
  than unity. e.g.
  y = 100.1 + 0.3
  y - 100.4 # => -1.421e-14, while Float::EPSILON = 2.22e-16

2 Always use 
  (x-y).abs < (x.abs + y.abs) * Float::EPSILON) 
  as a test for equality.

  Better than the first proposal, but won't work if the rounding error
  gets too large after a complex computation.
  In addition, (1) and (2) suffer from the problem that x==y and y==z do
  not imply x==z.

3 Use Bigdezimal

  This only shifts the problem a few decimal places down, and tests for
  equality will fail as with the normal floats.

4 Use Rationals

  Works if you only have to deal with rational operations. But doesn't
  solve the following
  x = sqrt(2)
  y = x + 1
  x + 0.2 == y - 0.8  # => false
  In addition, rational arithmetic can produce huge numbers pretty fast,
  and this will slow down computations enormously. 

5 Use a symbolic math package

  This could in theory solve the issue with equality, but in practice there
  is no way to decide that two symbolic representations of a number are the
  same, like
  1 / (sqrt(2) - 1) == sqrt(2) + 1
  Also, very, very slow.

6 Use interval arithmetic

  Gives you strict bounds on your solution, but can't answer x==y.


Summing up, when using floating point arithmetic there is no one true way.
There is no substitute for understanding numbers and analyzing your problem.

HTH,

Michael