4 ms·

The algorithm is an implementation of the Sieve of Sundaram. 999 might have slipped in by an off by one error. Thanx for the suggestion of using is-prime, but i

by petre 9y ago

The algorithm is an implementation of the Sieve of Sundaram. 999 might have slipped in by an off by one error. Thanx for the suggestion of using is-prime, but in order to benchmark multiple languages, I need to run the same thing everywhere.

lizmat 9y ago
Turns out that even though primes are integers, in your Perl 6 version, every calculation was done by using floating point. And this was caused by the call to the subroutine. If you would do: sieve_sundaram(1000) instead of: sieve_sundaram(1e3) then it all of a sudden becomes 4x as fast. In Perl 5 you never know what you're dealing with with regards to values. In Perl 6 if you tell it to use a floating point, it will infect all calculations to be in floating point afterwards. `1e3` is a floating point value. `1000` is an integer in Perl 6. Also, you seem to have a sub-optimal algorithm: the second `foreach` doesn't need to go from `1..$n`, but can go from `$i..$n` instead. This brings down the runtime of the Perl 5 version of the code to 89 msecs for me. Since your program is not using BigInt in the Perl 5 version, it is basically using native integers. In Perl 6, all integer calculations are always BigInts, unless you mark them as native. If I adjust your Perl 6 version for this, the runtime goes down from 4671 msecs to 414 msecs for this version: sub sieve_sundaram(int $n) { my %a; my int @s = 2; my int $m = $n div 2 - 1; for 1..$n -> int $i { for $i..$n -> int $j { my int $p = $i + $j + 2 * $i * $j; if $p < $m { %a{$p} = True; } } } for 1..$m -> int $k { if ! %a{$k} { my int $q = 2 * $k + 1; @s.push($q); } } return @s; } sieve_sundaram(1000); So, about 11x faster than before. And just under 5x as slow as the Perl 5 version. I could further make this idiomatic Perl 6, but the most idiomatic version I've already mentioned: `(1..1000).grep( *.is-prime)`
petre 9y ago
Can we please continue this discussion out of this thread, on a proper Perl forum? I haven't found any contact pointers for you or Ultimatt.
- lizmat 9y ago
  https://stackoverflow.com/questions/tagged/perl6 https://stackoverflow.com/questions/tagged/perl6 perhaps? Or on IRC: #perl6 on irc.freenode.org ? Or the perl6-users mailing list?
  - petre 9y ago
    Allright. https://stackoverflow.com/questions/49607801/perl6-vs-perl5-benchmarking-using-prime-numbers https://stackoverflow.com/questions/49607801/perl6-vs-perl5-...

by petre 9y ago

lizmat 9y ago

Turns out that even though primes are integers, in your Perl 6 version, every calculation was done by using floating point. And this was caused by the call to the subroutine. If you would do: sieve_sundaram(1000) instead of: sieve_sundaram(1e3) then it all of a sudden becomes 4x as fast. In Perl 5 you never know what you're dealing with with regards to values. In Perl 6 if you tell it to use a floating point, it will infect all calculations to be in floating point afterwards. `1e3` is a floating point value. `1000` is an integer in Perl 6. Also, you seem to have a sub-optimal algorithm: the second `foreach` doesn't need to go from `1..$n`, but can go from `$i..$n` instead. This brings down the runtime of the Perl 5 version of the code to 89 msecs for me. Since your program is not using BigInt in the Perl 5 version, it is basically using native integers. In Perl 6, all integer calculations are always BigInts, unless you mark them as native. If I adjust your Perl 6 version for this, the runtime goes down from 4671 msecs to 414 msecs for this version: sub sieve_sundaram(int $n) { my %a; my int @s = 2; my int $m = $n div 2 - 1; for 1..$n -> int $i { for $i..$n -> int $j { my int $p = $i + $j + 2 * $i * $j; if $p < $m { %a{$p} = True; } } } for 1..$m -> int $k { if ! %a{$k} { my int $q = 2 * $k + 1; @s.push($q); } } return @s; } sieve_sundaram(1000); So, about 11x faster than before. And just under 5x as slow as the Perl 5 version. I could further make this idiomatic Perl 6, but the most idiomatic version I've already mentioned: `(1..1000).grep( *.is-prime)`

petre 9y ago

Can we please continue this discussion out of this thread, on a proper Perl forum? I haven't found any contact pointers for you or Ultimatt.

lizmat 9y ago
https://stackoverflow.com/questions/tagged/perl6 https://stackoverflow.com/questions/tagged/perl6 perhaps? Or on IRC: #perl6 on irc.freenode.org ? Or the perl6-users mailing list?
- petre 9y ago
  Allright. https://stackoverflow.com/questions/49607801/perl6-vs-perl5-benchmarking-using-prime-numbers https://stackoverflow.com/questions/49607801/perl6-vs-perl5-...