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May 16, 2022 at 11:46 comment added ratchet freak @assembly_wizard yeah I was assuming that there is no overlap where A and B could be. More formally A(?=B)C is A concatenated with intersection of B and C. (with the same reasoning)
May 14, 2022 at 23:36 comment added assembly_wizard Look-around aren't always a simple intersection: ^A*(?=AB)B matches nothing, while the intersection of ^A*AB and ^A*B matches AAB. Though I'm pretty sure they're still regular.
Mar 1, 2019 at 11:06 history edited ratchet freak CC BY-SA 4.0
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Feb 16, 2015 at 18:10 history edited ratchet freak CC BY-SA 3.0
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Feb 16, 2015 at 16:33 history edited ratchet freak CC BY-SA 3.0
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Feb 16, 2015 at 16:08 history edited ratchet freak CC BY-SA 3.0
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Feb 16, 2015 at 13:36 history migrated from programmers.stackexchange.com (revisions)
Feb 14, 2015 at 0:19 comment added Jörg W Mittag @MSalters: If you want to get really technical, (a)\1 is not a regular expression, but recognizes a regular language.
Feb 13, 2015 at 22:56 vote accept peperunas
Feb 13, 2015 at 12:40 comment added MSalters Is this necessary and sufficient? It looks to me like (a)\1, while using a backref, is equivalent to aa and therefore trivially Regular. I'm also wondering if lookahead assertions can use to recognize non-Regular languages.
Feb 13, 2015 at 11:29 history answered ratchet freak CC BY-SA 3.0