# Piccolo con problema con il tipo float

**URL:** <https://www.ruby-forum.com/t/piccolo-con-problema-con-il-tipo-float/196481>\
**Category:** Ruby\
**Created:** [October 8, 2007, 11:06pm UTC](https://www.ruby-forum.com/t/piccolo-con-problema-con-il-tipo-float/196481 "2007-10-08T23:06:53Z")\
**Posts on this page:** 4\
**Page:** 1

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**Author:** ![matteobo77](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/m/c0e974/32.png) [@matteobo77](https://www.ruby-forum.com/u/matteobo77)\
**Post date:** [October 8, 2007, 11:06pm UTC](https://www.ruby-forum.com/t/piccolo-con-problema-con-il-tipo-float/196481/1 "2007-10-08T23:06:53Z")

</div>

Ho iniziato da poco a programmare in ruby, e stavo facendo un piccolo  
programma che, dato un sistema di disequazioni, trova i punti comuni a  
queste, impostando un problema di programmazione lineare.  
Viene chiesto in input prima il numero di disequazioni (vincoli) e  
quindi i parametri in questo modo:  
ax+by+c\>=0  
prima il coefficiente di x, quindi quello di y, il termine noto, e  
infine un quarto parametro da impostare ad 1 o a -1 per il segno della  
disequazione.  
Ho provato il seguente sistema:  
0.9x+1.5y\<=45  
x+y\<=36  
x\>=10  
y\>=0  
Quindi inserisco da input:  
4 | 0.9 | 1.5 | 45 | -1 | 1 | 1 | 36 | -1 | 1 | 0 | 10 | 1 | 0 | 1 | 0 |  
1  
I punti in comune calcolati a mano sono:  
(10,0) (36,0) (15,21) (10,24)  
E qui sorge il problema: il programma mi scarta il punto (15,21) alla  
prima disequazione (provate a mano, la verifica!) e la fatidica somma  
0.9_0.15+1.5_21-45 ridà nientepocodimenoche 7.105427357601e-015!!!  
Fatta numericamente dà 0, eppure il programma sputa fuori questo  
risultato!  
Sono ore che ci combatto, datemi una mano vi prego!  
Grazie in anticipo!

Il file principale è allegato, seguono le due piccole classi che ho  
implementato:

point.rb

class Point  
attr\_reader :x, :y  
def initialize (x,y)  
@x=x  
@y=y  
end  
end

diseq.rb

require ‘myrbs\point’

class Diseq \< Point  
attr\_reader :t,:s  
def initialize (x,y,t,s)  
super(x,y)  
@t=t  
@s=s  
end  
def solve(a,b)  
c=x_a+y_b-t  
puts c  
if(c\*s\>=0)  
o=1  
else  
o=0  
end  
o  
end  
end

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<div class="post-metadata">

**Author:** ![matteobo77](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/m/c0e974/32.png) [@matteobo77](https://www.ruby-forum.com/u/matteobo77)\
**Post date:** [October 8, 2007, 11:09pm UTC](https://www.ruby-forum.com/t/piccolo-con-problema-con-il-tipo-float/196481/2 "2007-10-08T23:09:03Z")

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I’m sorry, I’ve posted in the wrong place, could someone delete this  
post?

---

<div class="post-metadata">

**Author:** ![matteobo77](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/m/c0e974/32.png) [@matteobo77](https://www.ruby-forum.com/u/matteobo77)\
**Post date:** [October 8, 2007, 11:47pm UTC](https://www.ruby-forum.com/t/piccolo-con-problema-con-il-tipo-float/196481/3 "2007-10-08T23:47:38Z")

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On 10/8/07, Matteo M. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote:

> I’m sorry, I’ve posted in the wrong place, could someone delete this  
> post?

Nope. [ruby-talk.com](http://ruby-talk.com) is a two-way mirroring solution for a mailing list.

-austin

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<div class="post-metadata">

**Author:** ![matteobo77](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/m/c0e974/32.png) [@matteobo77](https://www.ruby-forum.com/u/matteobo77)\
**Post date:** [October 8, 2007, 11:52pm UTC](https://www.ruby-forum.com/t/piccolo-con-problema-con-il-tipo-float/196481/4 "2007-10-08T23:52:55Z")

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On 10/8/07, Matteo M. [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote:

> 0.9x+1.5y\<=45  
> 0.9_0.15+1.5_21-45 ridà nientepocodimenoche 7.105427357601e-015!!!
> 
> require ‘myrbs\point’  
> puts c  
> if(c\*s\>=0)  
> o=1  
> else  
> o=0  
> end  
> o  
> end  
> end

Hi,

your problem is that floats are not exact. They have limited  
precision, and some numbers (especially irrational and with periodic  
fractional part) cannot be stored as floats precisely.  
Imagine using decimal notation with max 5 digits: 1/3 = 0.33333 or  
0.33334. therefore 3 \* 1/3 = 3 \* 0.33333 = 0.99999 that is 0.00001  
less that expected result. Floats are the same except they use binary  
notation and more digits.

One way to avoid it is using BigDecimal instead of floats, although  
they are much slower.  
Sometimes it’s possible to live with the small differences.

Notice that in test/unit, they are checking float results to be within  
specified tolerance (assert\_in\_delta).

(My Italian is sufficient to understand your post, but not to reply 😉

Jano