Regex Capturing: repeating two letters

Question

I'm trying to check if a word have three or more repeating 'two letters pare', But the expressing returns true only if the same letter is repeated. Why?

(([a-z])([^\1]))\1\2{2,} 
    ^      ^     ^ ^  ^
    1      2     3 4  5

1) any letter (captured set \1)
2) any character that is not set 1 (captured set \2)
3) captured \1 again
4) captured \2 again
5) at least twice

words that should return TRUE: asasasassafasf, ereeeerererere, dddddtrtrtrtruuuuuuuu

words that should return FALSE: dddddddd, rrrrrrrrrrlkajf, fffffffssssssytytfffffff

Backreferences don't work within character classes (after all they might be comprised of more than one symbol). See [General approach for (equivalent of) "backreferences within character class"?](http://stackoverflow.com/q/18242119) for the lookahead alternative. — mario, Jul 19 '14 at 12:25
You really need to give us examples. There are several interpretations to `have three or more repeating 'two letters pairs'` — zx81, Jul 19 '14 at 12:28

score 3 · Accepted Answer · answered Jul 19 '14 at 12:31

3

You can solve this using a negative lookahead assertion:

([a-z])((?!\1)[a-z])(?:\1\2){2,}

Test it live on regex101.com.

Explanation:

([a-z])  # Match and capture a single ASCII letter in group 1
(        # Match and capture...
 (?!\1)  # unless it's the same letter as the one captured before
 [a-z]   # a single ASCII letter in group 2
)        # End of group 2
(?:\1\2) # Match those two letters
{2,}     # two or more times

answered Jul 19 '14 at 12:31

Tim Pietzcker

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As you pointed out in a comment in my answer, the two letters must be different, +1 for implementing that. :) – zx81 Jul 19 '14 at 12:40

anubhava · Answer 2 · 2014-07-19T14:49:43.463

2

Based on your short description following regex should work for you:

(([a-z])((?!\2)[a-z]))\1{2}

Regex Demo

([a-z]) - matches letters a to z and capture them in a group
\2 - Back reference to group #2
(?!\2)[a-z]) - matches letters a to z that is NOT group #2
\1{2} 2 instances of back reference #1

OR using relative back reference numbering:

(([a-z])(?!\g{-1})[a-z])\1{2}

In short this regex matches 3 consecutive occurrences of a letter pair of different characters.

edited Jul 19 '14 at 14:49

answered Jul 19 '14 at 12:30

anubhava

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but also returns one letter that repeat itself like: dddddddddddddd – Sdgsdg Asgasgf Jul 19 '14 at 12:34
Thank you! but why is the negative lookahead **\2** and not \1? – Sdgsdg Asgasgf Jul 19 '14 at 12:46
It is because `\1` represents `(([a-z])((?!\2)[a-z]))` and `\2` represents `([a-z])` – anubhava Jul 19 '14 at 14:35
but if it is first, why is it \2 and not \1? – Sdgsdg Asgasgf Jul 19 '14 at 14:39
1

Whatever starts first is numbers `1`. So outer `(..)` is numbers `1` and inner `([a-z])` is numbered `2` – anubhava Jul 19 '14 at 14:43
1

Also I updated answer to show use of relative back reference numbering . – anubhava Jul 19 '14 at 14:50

Regex Capturing: repeating two letters

2 Answers2

Regex Demo