From: Daniel DeLorme <dan-ml@...42.com>
Date: 2009-02-24T16:01:00+09:00
Subject: Re: 1.9.1 regex 6.5 times slower than 1.8.6 in at least one case

Michael Brooks wrote:
> However, in one case, when using a regular expressions (posted here by 
> someone a long time ago) which I use to determine what numbers in 
> 0..10000 are prime numbers, version 1.9.1 was at least 6.5 times slower 
> (1.8.6 = 67 secs, 1.9.1 = 457 secs).
> 
> The regular express is:
> 
>     ((("1" * self) =~ /^1$|^(11+?)\1+$/) == nil)

As the only guy who would rather use a regex rather than string slicing, 
that's disheartening news. One thing you might want to check is your 
encoding. If your default encoding is UTF8, some string operations can 
be significantly slower:

$ cat p-regex.rb
4000.times do |i|
   ((("1" * i) =~ /^1$|^(11+?)\1+$/) == nil)
end

$ time 1.8/bin/ruby -KN -v p-regex.rb
ruby 1.8.6 (2008-08-11 patchlevel 287) [i686-linux]
real	0m4.411s
user	0m4.292s
sys	0m0.032s

$ time 1.8/bin/ruby -KU -v p-regex.rb
ruby 1.8.6 (2008-08-11 patchlevel 287) [i686-linux]
real	0m4.480s
user	0m4.320s
sys	0m0.004s

$ time 1.9/bin/ruby -KN -v p-regex.rb
ruby 1.9.1p0 (2009-02-22 revision 22551) [i686-linux]
real	0m8.041s
user	0m7.980s
sys	0m0.020s

$ time 1.9/bin/ruby -KU -v p-regex.rb
ruby 1.9.1p0 (2009-02-22 revision 22551) [i686-linux]
real	0m21.709s
user	0m20.913s
sys	0m0.032s

With ascii encoding, ruby1.9 is still slower than 1.8 but at least not 
six times slower.

--
Daniel