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

 From: James Coglan (jcoglan at googlemail.com) Date: Tue Aug 4 10:00:56 EDT 2009 Previous message: [plt-scheme] [ANN] Heist: a Scheme interpreter in Ruby Next message: [plt-scheme] [ANN] Heist: a Scheme interpreter in Ruby Messages sorted by: [date] [thread] [subject] [author]

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?

(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 (ceil  (* a i))]
[q (floor (* b i))])
(if (<= p q)
(set! z (/ (if (positive? p) p q)
i))))))]))

James
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