[racket] (floor (/ ....))

From: Tim Brown (tim.brown at timb.net)
Date: Tue Jun 11 08:44:33 EDT 2013

Why are round/floor/ceiling/truncate limited to real numbers?

What stops them from being extended to all numbers?

Mathematica defines complex floor with:

> Floor <http://reference.wolfram.com/mathematica/ref/Floor.html> applies
> separately to real and imaginary parts of complex numbers

Of course, that might just be Mathematica's own idea of what a complex
floor is.

Or am I too simply lazy to define the following?

(define (complex-floor x)
  (make-rectangular (floor (real-part x)) (floor (imag-part x))))


On 10 June 2013 17:18, Bradley Lucier <lucier at math.purdue.edu> wrote:

> Re:
>> FWIW, I would have at least written:
>>         ((qty (in-range 0 (add1 (min (floor (/ weight-left weight))
>>                                      (floor (/ volume-left volume)))))))
> I have now seen the
> (floor (/ a b))
> idiom a number of times, and wonder why people prefer it to
> (quotient a b)
> Normally, to calculate (/ a b) where a and b are exact integers requires
> one to calculate (gcd a b) to put the fraction into lowest terms
> p/q=(quotient a (gcd a b))/(quotient b (gcd a b)); then, to calculate
> (floor p/q), one must calculate (quotient p q).
> For large integers of size  $N$ bits, (gcd a b) takes $O(N\log^2(N))$
> fixnum operations, where quotient takes $O(N\log N)$ operations.  This
> assumes Fourier-based methods for bignum multiplication; for more direct
> methods, the difference in operation count is larger.
> In any case, the (floor (/ ...)) idiom takes noticeably more time than
> (quotient ...).  If one knows that a and b are positive integers, they give
> the same results.
> This is an argument not to use (floor (/ ...)).  Are there arguments in
> favor of this idiom?
> Brad
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| Tim Brown <tim.brown at timb.net> | M:+44(0)7771714159 | H:+44(0)1372747875
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