Jon Lang wrote: > Larry Wall wrote: >> On Tue, Jul 15, 2008 at 03:30:24PM +0200, Moritz Lenz wrote: >> : Today bacek++ implement complex logarithms in rakudo, and one of the >> : tests failed because it assumed the result to be on a different complex >> : plane. (log(-1i) returned 0- 1.5708i, while 0 + 3/2*1i was expected). >> : >> : Should we standardize on one complex plane (for example -pi <= $c.angle >> : < pi like Complex.angle does)? Or simply fix the test to be agnostic to >> : complex planes? >> >> Standardizing on one complex plane is the normal solution, though >> this being Perl 6, there's probably a better solution using infinite >> Junctions if we can assume them to be both sufficiently lazy and >> sufficiently intelligent... :) > > By the principle of least surprise, I'd recommend against this. Most > programmers, when they see 'sqrt(1)', will expect a return value of 1,

## Advertising

And that's what they get unless they write it as sqrt(1 + 0i). > and won't want to jump through the hurdles involved in picking '1' out > of 'any(1, -1)'. 1 and -1 aren't just separated by a complex plain, they are really distinct numbers > That said, I'm not necessarily opposed to these > functions including something like an ':any' or ':all' adverb that > causes them to return a junction of all possible answers; but this > should be something that you have to explicitly ask for. > > And even then, I'm concerned that it might very quickly get out of > hand. Consider: > > pow(1, 1/pi() ) :any - 1 > > (I think I got that right...) Not quite. Afaict the only functions that might return a junction are Complex.angle and Complex.log. But having $compl.angle > pi always yield True would be quite weird ;-) > Since pi is an irrational number, there are infinitely many distinct > results to raising 1 to the power of 1/pi. No. exp($x) is a single, well-defined value. > (All but one of them are > complex numbers, and all of them have a magnitude of 1, differing only > in their angles.) Thus, pow(1, 1/pi() ) :any would have to return a > junction of an indefinitely long lazy list. Now subtract 1 from that > junction. Do you have to flatten the list in order to do so, > subtracting one from each item in the list? Obviously we'd have to avoid that if there's any infinite list/junction involved somewhere ;-) But you do have a point that we can't really use infinite junctions unless we can ensure that we can do all sorts of arithmetics with it without loosing lazyness. And I don't think we can prove that (but I might give it it shot if I have some spare time) > Or is there a reasonable > way to modify the list generator to incorporate the subtraction? > > Or how about: > > sqrt(1):any + sqrt(1):any > > -- > > In any case, there's the matter of what to do when you only want one > answer, and not a junction of them. IMHO, we should standardize the > angles on '-pi ^.. pi'. My reasoning is as follows: if the imaginary > component is positive, the angle should be positive; if the imaginary > component is negative, the angle should be negative. If the imaginary > component is zero and the real component is not negative, the angle > should be zero. And the square root of -1 should be i, not -i; so if > the imaginary component is zero and the real component is negative, > the angle should be positive, not negative. > -- Moritz Lenz http://moritz.faui2k3.org/ | http://perl-6.de/