$|--;
      my ($a, $b, $c, $d, $m, $n) = (0,1,1,0,0,0);
while() { if($b && $d && (my $q=int($a/$b))==int($c/$d)) {
$a = 10*($a-$b*$q);                     $c = 10*($c-$d*$q);
print $q;                                           } else {
$q = $a; $a = $c; $c = $c * (($m % 3) % 2 ? $n * 2 : 1) + $q;
$q = $b;
 $b = $d;
    $d = $d * (($m++ % 3) % 2 ? $n++ * 2 : 1) + $q; } }
         # Hit control-C when it stops printing
                # due to numeric overflow

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Replies are listed 'Best First'.
Re: e with continued fractions
by tachyon (Chancellor) on Jul 18, 2004 at 05:05 UTC

    Diabolical. You can make the ASCII quite pretty too: 

                 ;;$|++;use=>
        bigint;my($a,$b,$c,$d
     ,$m,$n)            =(0,1,1
   ,0,0,0)                 ;while                  
  (){if(                    $b&&$d                   
  &&(my$                     e=int(                    
 $a/$b))==int($c/$d)){$a=10*($a-$b*$
 e);$c=10*($c-$d*$e);print$e;}else{$
 e=$a;$
 a=$c;$
  c=$c*(
  ($m%3)
   %2?$n*                      2:1)
    +$e;$e=$               b;$b=$
       d;$d=$d*(($m++%3)%2?$n++
            *2:1)+$e;}};#e

    [download]

    cheers

    tachyon

[reply]
[d/l]

      On Perl 5.8.1 (Mac OS X 10.3.4) this doesn't seem to use bigint (as you apparently intended) unless I remove the =>.

      Update: OK, you fooled me. Good obfuscation, then!

[reply]
[d/l]

        I use 5.6.x so don't actually have bigint. This is called obfuscation after all :-) The reason it does not work is that => is just a synonym for , in Perl so it reads use, bigint Or it should. Actually that , syntax parses as an error, whereas => syntax is OK GOK? 

        C:\>type test.pl
use, strict

C:\>perl -MO=Deparse test.pl
syntax error at test.pl line 1, near "use,"
test.pl had compilation errors.

C:\>type test.pl
use => strict

C:\>perl -MO=Deparse test.pl
'???', '???';
test.pl syntax OK

C:\>

        [download]

        cheers

        tachyon

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Re: e with continued fractions
by bageler (Hermit) on Jul 19, 2004 at 20:15 UTC
    if you use bigint pragma then you won't get a numeric overflow and it will continue running for much longer :)
[reply]
Good to about 45 places; spigot algorithm
by tmoertel (Chaplain) on Jul 21, 2004 at 18:02 UTC
    The output of the obfuscated code seems to be good for about 45 places (at least on my box running Perl v5.8.3 built for i386-linux-thread-multi) before it differs from the reference values from the On-Line Encyclopedia of Integer Sequences.

    If you want more precision, you might be interested in Unbounded Spigot Algorithms for the Digits of π by Jeremy Gibbons. It's a fun read, and you can adapt the methods presented in his paper to computing e digit by digit.

[reply]

      The inaccuracy appears to be the result of roundoff error; use bigint; or use Math::BigInt ':constant'; (as suggested by tachyon and bageler) fixes that also.

      Also, my algorithm is streaming in Gibbons' sense. It represents the continued fraction as a composition of linear fractional transformations.

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