[plt-scheme] [ANN] Heist: a Scheme interpreter in Ruby

2009/8/4 James Coglan <jcoglan at googlemail.com>

>
>
> 2009/8/4 James Coglan <jcoglan at googlemail.com>
>
>
>>
>> 2009/8/4 Chongkai Zhu <czhu at cs.utah.edu>
>>
>>> I just checked both r5rs and r6rs. Their definition of "/simplest/
>>> rational number" is more complex than it can be. Here's PLT's (and another
>>> system's) doc of rationalize:
>>>
>>> ----
>>> (rationalize x tolerance) ? real?
>>> x : real?
>>> tolerance : real?
>>>
>>> Among the real numbers within (abs tolerance) of x, returns the one
>>> corresponding to an exact number whose denominator is smallest. If multiple
>>> integers are within tolerance of x, the one closest to 0 is used.
>>>
>>> ----
>>>
>>> rationalize(x,dx)
>>> yields the rational number with smallest denominator that lies within dx
>>> of x.
>>>
>>> ----
>>>
>>> in which you can see only to make the denominator smallest is enough. Is
>>> this enough hint for you to come to an algorithm?
>>
>>
>> That's certainly enough of a simplification to give me an idea, though I'm
>> not sure it'll be very efficient. I'll try to implement it and post here for
>> feedback.
>>
>
> This is an attempt at finding the first rational with the smallest
> denominator that's in range. Anyone spot anything terribly wrong with it?
>

Apologies, `ceil` should have read `ceiling`:

(define (rationalize x tolerance)
  (cond [(rational? x)
          x]
        [(not (zero? (imag-part x)))
          (make-rectangular (rationalize (real-part x) tolerance)
                            (rationalize (imag-part x) tolerance))]
        [else
          (let* ([t (abs tolerance)]
                 [a (- x t)]
                 [b (+ x t)]
            (do ([i 1 (+ i 1)]
                 [z '()])
                ((number? z) z)
              (let ([p (ceiling (* a i))]
                    [q (floor (* b i))])
                (if (<= p q)
                    (set! z (/ (if (positive? p) p q)
                               i))))))]))

James
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