^(?!(..+)(\1)+$)
^(?!(..+)(\1)+$)
You've solved it using the actual definition of prime numbers, no trickery needed. Well played.
FYI, you don't need brackets around the \1, so can score 286.
- the first one is represented by the group (..+) it represents the number of occurrences n between 2 and +∞
- the second one is represented by (\1)+. We will repeat the first number m times, between 1 and +∞ times.
So the result of the multiplication is n*(m+1), which cannot be a prime. We just have to take the opposite with negative lookahead. It's very beautiful indeed.
See http://regex101.com/r/qN2fQ8 or http://www.regexper.com/#^%28%3F!%28..%2B%29\1%2B%24%29 to follow the above explanations.
(I learned that for (\1)+, "Note: A repeated capturing group will only capture the last iteration. Put a capturing group around the repeated group to capture all iterations or use a non-capturing group instead if you're not interested in the data")
DB<63> print "Matches!" if (("x" x 31) !~ /^(..+)\1+$/) DB<64> print "Matches!" if (("x" x 18) !~ /^(..+)\1+$/)
(The simpler regex merely looks for non-primes by matching any number of characters which are a multiple of two numbers, n x m, i.e., those which can be factorized. n comes from (..+), m comes from \1+).