in reply to Re^2: Finding repeat sequences. (only mostly regex)
in thread Finding repeat sequences.

Note that, based on that definition, if the first and last characters are the same, then the answer is "the string minus the last character". Which leads to: 

/^(.+?).*\1$/

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Which leads to a full solution of: 

/^((.*?).*?)\2*\1$/

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which might be horribly inefficient (at least for some cases) or might not; I haven't considered it.

- tye        

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Re^4: Finding repeat sequences. (only regex)
by BrowserUk (Patriarch) on Jun 18, 2013 at 20:21 UTC

    Nice reversal of the logic and closer:

    $s = 'aaaabaaaabaaaaabaaaab';;
$s =~ /^((.*?).*?)\2*\1$/ and print "$2/$1";;
a/aaaabaaaab

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      I think I had \1 and \2 backward. Then you just have to disallow the trivial solution: 

      $s = 'aaaabaaaabaaaaabaaaab';
$s =~ /^((.*?).*?)(?=.)\1*\2$/ and print "$2/$1";

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      (Update: Dropped the unneeded () around \1 that I had introduced while debugging. You probably also need to change .*? to .* so you get the longest solution not the shortest.)

      - tye        

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