Using a regular expression to search for arbitrary lengths of consecutive blocks

Question

I have a string containing ones and zeros. I want to determine if there are substrings of 1 or more characters that are repeated at least 3 times in a row. For example, the string '000' has a substring of length 1 consisting of one null character that is repeated 3 times. The string '010010010011' actually has 3 such substrings, each of which is repeated 3 times ('010', '001' and '100').

Is there a regular expression expression that can find these repeating patterns without knowing either the particular pattern or the length of the pattern? I don't care what the pattern is and what its length is, just that the string contains a 3-peat pattern.

Answer 1

(.+)\1\1 

\ may be a different character depending on your choice of language. This means that you need to match any string, and then try to match it again.

\1 means repeat the first match.

Answer 2

Something may work here, however, it will tell you if there is a pattern repeated three times, and (I don't think) cannot be expanded to let you know if there are others:

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

Violation of this:

  (.+) matches any 1 or more characters, starting anywhere in the string .*? allows any length of interposing other characters (0 or more) \1 matches whatever was captured by the (...+) parentheses .*? 0 or more of anything \1 the original pattern, again 

If you want the repetitions to appear immediately next to them, use

  /(.+)\1\1/ 

... as suggested by @Buh Buh - notation \1 vs. $1 may vary depending on your regexp system.

Answer 3

It looks weird, but it could be a solution:

 /000000000|100100100|010010010|001001001|110110110|011011011|101101101|111111111/ 

It contains all possible combinations three times. So your regex will match these numbers (i.e.):

• 10010010011
• 00010010011
• 10110110110

But not for them:

• 101010101010
• 001110111110
• 111000111000

And it doesn't matter where the sequence appears in the entire line.