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Regular expressions with backreferences over unary alphabet

Submitted by: @import:stackexchange-cs··
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alphabetbackreferenceswithregularexpressionsunaryover

Problem

Setting:

  • regular expressions with backreferences



  • unary language (1-symbol alphabet)



Is the following problem decidable in this setting:

  • Given a regular expression with backreferences, does it define a regular language?



For example, (aa+)\1 defines a regular language, while (aa+)\1+ doesn't. Can we decide which one is the case?

For concreteness, "regular expressions with backreferences" here refer to e.g. the following subset of the usual Perl-compatible regular expressions:

  • a matches character a (the only character in the alphabet)



  • X* matches 0 or more occurrences of X



  • X|Y matches X or Y



  • parentheses can be used for grouping and capturing



  • \1. \2, etc. match the same string as the 1st, 2nd, etc. pair of parentheses



We can also use the normal shorthands e.g. X+ = XX*.

Solution

Evidence against the effective decidability of the problem is provided by the construction in the proof of Theorem 9 in my paper On Practical Regular Expressions: You could determine if there are finitely many Fermat primes.

Context

StackExchange Computer Science Q#64372, answer score: 4

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