Let L = {a^f(m) | m >= 1 } where f: Z^+ -> Z^+ is monotone increasing and complies that for all element n in Z^+ there is an m开发者_运维知识库 belonging to Z^+ such that f(m+1) - f(m) >= n.
I have a compiler question. Determin开发者_开发技巧e whether {(ab)^n | n >= 0} is a regular language?
In my exam, i was supposed to write all pumping lemma conditions. that exactly what i did :开发者_开发百科
A = {0^a 1^b 2^c | a &开发者_运维技巧lt; b < c} I need to show that A is not context-free. I\'m guessing I have to use the Pumping Lemma for this, but how?The goal is to prove that for any string
Most UNIX regular expressions have, besides the usual **,+,?* oper开发者_Python百科ators a backslash operator where \\1,\\2,... match whatever\'s in the last parentheses, so for example *L=(a*)b\\1* m
This question already has answers here: Closed 10 years ago. Possible Duplicate: Why does Eclipse complain about @Override on interface methods?
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