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You must make a regex that matches everything except itself. For example, PPCG must not match PPCG, but must match everything else, like something, PPC, and PPCG!.

Surrounding slashes are not considered part of the regex in flavors where they are usually part of the syntax. The same is true of all delimiters.

The regex needs to match the entire string, not just a substring.

You may choose your flavor, but you can't use a flavor newer than this challenge. Please pick a flavor that is easily tested.

Shortest regex in bytes wins.

(Sandbox (deleted))

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  • 2
    \$\begingroup\$ Related \$\endgroup\$ – programmer5000 Jun 9 '17 at 1:24
  • \$\begingroup\$ Also, jimmy's solution to the previous challenge can be inverted quite trivially to make a valid answer for this one. \$\endgroup\$ – Martin Ender Jun 9 '17 at 10:54
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    \$\begingroup\$ I love how you pointed out a related question on your own question. \$\endgroup\$ – Magic Octopus Urn Jun 9 '17 at 15:04
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    \$\begingroup\$ @carusocomputing Why wouldn't they? The point of related links is to make the challenges show up in each others' sidebars. \$\endgroup\$ – Martin Ender Jun 9 '17 at 15:05
  • \$\begingroup\$ @MartinEnder would assume one would put it in the post itself, not as a comment is all I was saying ;). \$\endgroup\$ – Magic Octopus Urn Jun 9 '17 at 16:02
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PCRE flavour, 45 bytes

^(?!\^()(?R){2}\z)|\1\Q(?!\^()(?R){2}\z)|\1\Q

Try it at regex101.

This is a simple modification of jimmy23013's brilliant solution to the opposite challenge. We just need to make two changes:

  • In this challenges, the delimiters shouldn't be matched so we drop them from the regex as well. However, that could lead to infinite recursion, because then no characters would be matched before the (?R) call. To fix this, we match the ^ explicitly with \^ to have one character before we enter the recursion.
  • We need to negate the result which is as simple as wrapping everything after the ^ in a negative lookahead to ensure that the input isn't equal to the string.

For the details of this quining approach, see jimmy's amazing answer.

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  • 3
    \$\begingroup\$ Even though it's a modification of that other answer I wouldn't have had a clue about how to do it so +1 \$\endgroup\$ – TheLethalCoder Jun 9 '17 at 14:58

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