I need to construct a regular expression that will filter a group of words that begins and ends with the same word. For example, the life of the free will output the life of the and he was and he his the same will output he was and he. The two words have to be maximum 10 characters from each other.

2 Answers 2


With PCRE, one can do:

grep -Po '\b(\w+)\b.{1,10}\b\1\b'
  • -P enables Perl-style regular expressions using PCRE.
  • -o prints only the matched text.
  • \b marks the word boundary
  • (\w+) groups a match of word characters
  • .{1,10} matches up to 10 characters and at least 1.
  • \1 refers to the group matched earlier.

Try grep with extended Regular Expression:

$ echo "the life of the free" | grep -Eo "(\b[[:alnum:]]+\b)([[:blank:]]|[[:alnum:]]){1,10}\1"
the life of the

$ echo "he was and he his the same" | grep -Eo "(\b[[:alnum:]]+\b)([[:blank:]]|[[:alnum:]]){1,10}\1"
he was and he

Here -E means extended regexp, -o means only print the matched portion of the line, \b matches the word boundary, the character class [[:alnum:]] means all alphabetic (uppercase & lowercase) and numerical characters, [[:blank:]] means space or tab, + means one or more occurrences of the previous match, {1,10} the previous match can occur between 1 to a maximum of 10 times, \1 means match the first matched group (expressed between first pair of parentheses) i.e. \b[[:alnum:]]+\b.


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