From: Eric Mahurin <eric_mahurin@...>
Date: 2005-05-06T21:22:44+09:00
Subject: Re: Float to Rational

Here is another simple way I came up with.  This shouldn't be
any less accurate than the to_r you have since the error in the
mantissa is at least 1 mantissa bit (I'm assuming 50 mantissa
bits).  The difference between this and what you had was you
only multiplied by 2 forcing the denominator to be a power of 2
and you waited until the numerator was exactly an integer. 
17.0/37 seems to be the float with the smallest denominator
that starts giving problems in this algorithm.  Another
possibility would be to test incrementing/decrementing the
numerator and/or denominator to see if you get a simpler
rational.  This should be valid since you know that mantissa
error translates to numerator/denominator error.

require 'mathn'

class Float
  
  Err = 2.0**-50

  def to_r
    num = self
    denominator = 1
    loop do
      # factors needed to get denominators 2-36 (sometimes 37)
      [2,3,2,5,7,2,3,11,13,2,17,19,23,5,3,29,31,2,37].each do
|f|
        if (1-num/(numerator=num.round)).abs<=Err
            return numerator/denominator
        end
        denominator *= f
        num *= f
      end
    end
  end

end


> require 'mathn'
> 
> class Numeric
>   def inverse
>     1/self
>   end
> end
> 
> class Float
>   
>   def to_r
>     n = 1
>     n *= 2 until (self*n) % 1 == 0
>     (self*n).to_i/n
>   end
>   
>   def round_to_r
>     return self.to_i if self % 1 == 0
>     n = self
>     ops = []
>     count = 0
>     until ((n%1).round - n%1).abs < 1e-8 || count > 20 ||
>           n.abs == 1.0/0.0 || n == 0.0/0.0
>       int, dec = n.divmod 1
>       ops.concat [[:+, int.to_i], [:inverse]]
>       n = 1/dec
>       count += 1
>     end
>     n = n.round
>     ops.reverse.inject(n.round){|n, op| n.send(*op)}
>   end
> end
> ----
> 
> Use Float#to_r for an exact representation of the float
> value, or
> Float#round_to_r for an extremely close representation of it.
> 
> cheers,
> Mark
> 
> 



		
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