Matching first and last three characters of regex (including overlap) - regex

Question

I am trying to put together a regex expression that matches a word (only one per line) that starts and ends with the same three characters.
I was able to write a solution for words that are at least 6 characters long (meaning there is no overlap), but I am unsure how to do it for overlapping starts and ends such as "heheh".
This is what I have, nice and simple:
^(...).*\1$
I am inclined to believe that this might have something with lookahead and lookbehind but I am not sure.
Any help would be appreciated, thank you!

Answer 1

You will need lookarounds since they are non-consuming patterns, i.e. the regex index is not advanced when the lookaround pattern is matched.
For example, you may do this with GNU grep:
grep -P '^(?=(...)).+\1$' file
grep -P '^(?=(\S{3})).+\1$' file # To avoid counting in spaces
grep -P '^(?=(\w{3})).+\1$' file # Or only allowing letters/digits/underscores
grep -P '^(?=(\p{L}{3})).+\1$' file # Or only allowing letters
See the regex demo
Details
^ - start of string
(?=(...)) - a positive lookahead with a capturing group inside that matches any 3 chars
.+ - any 1+ chars other than line break chars as many as possible
\1 - Group 1 value
$ - the end of string.
To extract words, you may use \w shorthand (that matches letters, digits and underscores) and word boundaries \b:
grep -oP '\b(?=(\w{3}))\w+\1\b' file
See another demo.
Details
\b - a word boundary (start of word here, because it is followed with word chars)
(?=(\w{3})) - a positive lookahead making sure there are 3 word chars while capturing them into Group 1
\w+ - 1+ word chars (not 0 or more because otherwise a 3-char word would be matched)
\1 - Group 1 value
\b - end of word here (as it is preceded with word chars).