float equality
guille lists
sorry if this is a very naive question (I'm new to ruby), but I
haven't found an explication yet. When comparing to floats in ruby I
came across this:
>> a = 0.1
=> 0.1
>> b = 1 - 0.9
=> 0.1
>> a == b
=> false
>> a > b
=> true
>> a < b
=> false
>> a <=> b
=> 1
I'm a bit lost here, shouldn't (0.1) and (1 - 0.9) be equals regarding
the == operator? I also found that for example 0.3 == (0.2 + 0.1)
returns false, etc.
guille
PD: I'm using ruby 1.8.6 (2008-03-03 patchlevel 114) [universal-darwin9.0]
Todd Benson
> Hi,
>
> sorry if this is a very naive question (I'm new to ruby), but I
> haven't found an explication yet. When comparing to floats in ruby I
> came across this:
>
>>> a = 0.1
> => 0.1
>>> b = 1 - 0.9
> => 0.1
>>> a == b
> => false
>>> a > b
> => true
>>> a < b
> => false
>>> a <=> b
> => 1
>
> I'm a bit lost here, shouldn't (0.1) and (1 - 0.9) be equals regarding
> the == operator? I also found that for example 0.3 == (0.2 + 0.1)
> returns false, etc.
Most people will point you to this:
http://en.wikipedia.org/wiki/Floating_point_arithmetic
There are several ways around it (using BigDecimal, Rational, Integers, etc.)
Totally off-topic, but has any one figured out exactly why 1/9
(0.111...) plus 8/9 (0.888...) is 1 instead of 0.999... :-)
Todd
The Higgs bozo
> I'm a bit lost here, shouldn't (0.1) and (1 - 0.9) be equals regarding
> the == operator?
Nope. Floating-point is always inexact. This is a computer thing, not
a Ruby thing. The issue applies to all languages everywhere which use
floating point.
Here you can see the two values are slightly different:
irb(main):001:0> 1 - 0.9 == 0.1
=> false
irb(main):002:0> [1 - 0.9].pack("D")
=> "\230\231\231\231\231\231\271?"
irb(main):003:0> [0.1].pack("D")
=> "\232\231\231\231\231\231\271?"
--
Posted via http://www.ruby-forum.com/.
Sebastian Hungerecker
> I'm a bit lost here, shouldn't (0.1) and (1 - 0.9) be equals regarding
> the == operator?
No. The result of 1 - 0.9 using floating point math is not actually 0.1. In
irb it is displayed as 0.1, but that's only because Float#inspect rounds.
Using printf you can see that the result of 1-0.9 actually is 0.09999lots:
>> printf "%.30f", 1-0.9
0.099999999999999977795539507497
0.1 itself isn't actually 0.1 either - it's
0.100000000000000005551115123126...
Because of this you should not check two floats for equality (usually you want
to check for a delta or not use floats at all). This is so because of the
inherent inaccuracy of floating point maths. See this for more information:
http://docs.sun.com/source/806-3568/ncg_goldberg.html
HTH,
Sebastian
--
Jabber: sep...@jabber.org
ICQ: 205544826
Matthew Moss
> (0.111...) plus 8/9 (0.888...) is 1 instead of 0.999... :-)
Ummm... because 1/9 + 8/9 == (1 + 8)/9 == 9/9 == 1 ?
And... because 0.999... == 1?
x = 0.999...
10x = 9.999...
(10x - x) = 9.999... - 0.999...
9x = 9
x = 1
guille lists
check deeper on google
(http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-core/9655).
So I guess that if one wants to work for example with float numbers in
the range [0,1], the best way to do it is by normalising from an
integer interval depending on the precision you want, say [0,100] for
two decimal digits precision, and so on. Is there any other better
approach? Does the use of BigDecimal impose a severe penalty on performance?
guille
The Higgs bozo
While your answer is correct, you cannot subtract infinities as shown
in your proof. Look at this:
x == 1 - 1 + 1 - 1 + 1 - 1 + 1 - ...
x == 1 - 1 + 1 - 1 + 1 - 1 + ...
----------------------------------------
2x == 1 + 0 + 0 + 0 + 0 + 0 + 0 + ...
x == 0.5
Does x == 0.5? No, because x was never a number in the first place
because the given series does not converge. Your proof appears to
work because you've already assumed 0.99999... converges, but that is
what you are trying to prove.
P.S. Euler thought the answer was x == 0.5.
Tim Pease
>
> Totally off-topic, but has any one figured out exactly why 1/9
> (0.111...) plus 8/9 (0.888...) is 1 instead of 0.999... :-)
>
I figured it out once, but I can't remember precisely how it worked.
TwP
Axel Etzold
-------- Original-Nachricht --------
> Datum: Tue, 28 Oct 2008 04:31:06 +0900
> Von: "guille lists" <guill...@gmail.com>
> An: ruby...@ruby-lang.org
> Betreff: Re: float equality
Dear guille,
you could use some approximate equality check:
class Float
def approx_equal?(other,threshold)
if (self-other).abs<threshold # "<" not exact either ;-)
return true
else
return false
end
end
end
a=0.1
b=1.0-0.9
threshold=10**(-5)
p a.approx_equal?(b,threshold)
Best regards,
Axel
--
Ist Ihr Browser Vista-kompatibel? Jetzt die neuesten
Browser-Versionen downloaden: http://www.gmx.net/de/go/browser
Bilyk, Alex
Tim Pease
On Oct 27, 2008, at 2:29 PM, Bilyk, Alex wrote:
> Mathematically 1.(0) is the same thing as 0.(9). Computers simply
> represent this to whatever precision they can.
>
Hmmm ... my humor is a little to obtuse today. I was hoping the word
"precisely" would cause a mental link to the word "precision" which is
what this problem is all about -- precision and computer
representations of floating point values.
Alas. I'll stick with my day job of writing software.
Blessings,
TwP
Matthew Moss
But 0.999... does converge, while 1 - 1 + 1 - 1 +... does not.
Matthew Moss
>>> 10x = 9.999...
>>> (10x - x) = 9.999... - 0.999...
>>> 9x = 9
>>> x = 1
>>
>> While your answer is correct, you cannot subtract infinities as shown
>> in your proof. Look at this:
>>
>> x == 1 - 1 + 1 - 1 + 1 - 1 + 1 - ...
>> x == 1 - 1 + 1 - 1 + 1 - 1 + ...
>> ----------------------------------------
>> 2x == 1 + 0 + 0 + 0 + 0 + 0 + 0 + ...
>> x == 0.5
>>
>> Does x == 0.5? No, because x was never a number in the first place
>> because the given series does not converge. Your proof appears to
>> work because you've already assumed 0.99999... converges, but that is
>> what you are trying to prove.
>
>
> But 0.999... does converge, while 1 - 1 + 1 - 1 +... does not.
On re-reading, I see that you weren't so much questioning whether
0.999... converges, but that my proof uses circular reasoning. Yeah, I
suppose that's correct.
Still, 0.999... does converge and is 1. Nyah! :p I just don't
remember the better proof I once knew.
John W Kennedy
> guille lists wrote:
>> I'm a bit lost here, shouldn't (0.1) and (1 - 0.9) be equals regarding
>> the == operator?
>
> Nope. Floating-point is always inexact.
What you mean is that binary floating point inexactly represents decimal
fractions.
> This is a computer thing, not
> a Ruby thing. The issue applies to all languages everywhere which use
> floating point.
...binary floating point.
--
John W. Kennedy
"I want everybody to be smart. As smart as they can be. A world of
ignorant people is too dangerous to live in."
-- Garson Kanin. "Born Yesterday"