Perl 6 resources

I’m writing this post in as a reference for the Perl 6 lecture I’m giving today at Penguicon, but of course it may prove generally useful as well.

Official Perl 6 website
#perl6 IRC
STD.pm6 (official Perl 6 grammar)
Perl 6 Spec
Rakudo website
Rakudo repository on github
masak’s History of Perl 6 (Only up to mid-2010, but still very interesting.)
ABC module (used as an example, interesting real-world grammar usage)
jnthn’s Perl 6 talks

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An infinite stream of “Pi”

So, after TimToady’s help with my last problem, finishing this is trivial. You just convert the Haskell code without worrying about type safety.

type LFT = (Integer, Integer, Integer, Integer) 
extr :: LFT -> Integer -> Rational 
extr (q,r,s,t) x = ((fromInteger q) * x + (fromInteger r)) / 
                           ((fromInteger s) * x + (fromInteger t)) 
unit :: LFT 
unit = (1,0,0,1) 
comp :: LFT -> LFT -> LFT 
comp (q,r,s,t) (u,v,w,x) = (q*u+r*w,q*v+r*x,s*u+t*w,s*v+t*x) 

becomes

sub extr([$q, $r, $s, $t], $x) { 
    ($q * $x + $r) / ($s * $x + $t); 
}

my $unit = [1, 0, 0, 1];

sub comp([$q,$r,$s,$t], [$u,$v,$w,$x]) {
    [$q * $u + $r * $w, 
     $q * $v + $r * $x, 
     $s * $u + $t * $w, 
     $s * $v + $t * $x];
}

And then the final piece in the puzzle,

pi = stream next safe prod cons init lfts where 
  init = unit 
  lfts = [(k, 4*k+2, 0, 2*k+1) | k<-[1..]] 
  next z = floor (extr z 3) 
  safe z n = (n == floor (extr z 4)) 
  prod z n = comp (10, -10*n, 0, 1) z 
  cons z z’ = comp z z’ 

becomes

sub pi-stream() {
    stream(-> $z { extr($z, 3).floor; },
           -> $z, $n { $n == extr($z, 4).floor; },
           -> $z, $n { comp([10, -10*$n, 0, 1], $z); },
           &comp,
           $unit, 
           (1..*).map({ [$_, 4 * $_ + 2, 0, 2 * $_ + 1] }));
}

It’s a very direct translation.

Does it work?

> my @pi := pi-stream;
> say @pi[^40].join('');
3141592653589793238468163213056056860170

Yay!

Except, according to the Joy of Pi, the first 40 digits of pi are

3.1415926535 8979323846 2643383279 502884197 # pi
3.1415926535 8979323846 8163213056 056860170 # ours

What’s going wrong? I haven’t empirically verified it yet, but I’m pretty sure the issue is Rakudo’s Ints and Rats overflowing. Which means our next post is going to have to dive back into Math::BigInt and Math::FatRat…

More Pi

So, in my previous post I started converting a spigot algorithm for calculating pi from Haskell to Perl 6. I apologize for being away for so long, but I’m back at it now.

Interestingly, while I thought the previous stream function was much clearer in p6, this time out I think I have to give the edge to Haskell.

convert :: (Integer,Integer) -> [Integer] -> [Integer] 
convert (m,n) xs = stream next safe prod cons init xs 
  where 
    init = (0%1, 1%1) 
    next (u,v) = floor (u*v*n’) 
    safe (u,v) y = (y == floor ((u+1)*v*n’)) 
    prod (u,v) y = (u - fromInteger y/(v*n’), v*n’) 
    cons (u,v) x = (fromInteger x + u*m’, v/m’) 
    (m’,n’) = (fromInteger m, fromInteger n) 

The difference comes from Haskell’s extremely elegant on-the-fly pair notation. When I translate that to p6, I get

sub convert($m, $n, @x) {
    stream(-> $u { floor($u.key * $u.value * $n); },
           -> $u, $y { $y == floor(($u.key + 1) * $u.value * $n); },
           -> $u, $y { $u.key - $y / ($u.value * $n) => $u.value * $n; },
           -> $u, $x { $x + $u.key * $m => $u.value / $m; },
           0/1 => 1/1,
           @x);
}

Even with p6’s big advantage in not having to explicitly convert integers to rationals, the pair thing makes this round a win for Haskell, IMO.

Perhaps one of the other p6 programmers out there can think of a more elegant way of handling this…

Update: They sure can! The esteemed TimToady pointed out Perl 6 can do something almost identical to the Haskell approach, skipping Pairs altogether:

sub convert($m, $n, @x) {
    stream(-> [$u, $v] { floor($u * $v * $n); },
           -> [$u, $v], $y { $y == floor(($u + 1) * $v * $n); },
           -> [$u, $v], $y { [$u - $y / ($v * $n), $v * $n]; },
           -> [$u, $v], $x { [$x + $u * $m, $v / $m]; },
           [0/1, 1/1],
           @x);
}

This version compares very well with the Haskell version, IMO!