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For regular language a*b*, is there a superset of it which is non-regular?


For regular language a*b*, is there a superset of it which is non-regular?

By : hunter_pdx
Date : November 21 2020, 11:01 PM
should help you out Yes, there is. Take B=bnabn (where n>0), that is the language of palindromes that start with b and contain a single a in the middle. This language is non-regular because the pumping lemma doesn't apply to it. It is also completely disjoint with A because every word in B contains the ba substring, while none of the words in A do.
A union of A and B is therefore a non-regular superset of A.
code :


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Non regular context-free language and infinite regular sublanguages

Non regular context-free language and infinite regular sublanguages


By : sanket
Date : March 29 2020, 07:55 AM
around this issue L = {0^n 1^n : n natural} is non-regular context free.
M = 2*3* is infinite regular.
Does an algorithm exist which can determine whether one regular language matches any input another regular language matc

Does an algorithm exist which can determine whether one regular language matches any input another regular language matc


By : user2268767
Date : March 29 2020, 07:55 AM
it should still fix some issue Any regular expression can be linked to a DFA - you can minimize the DFA and since the minimal form is unique, you can decide whether two expressions are equivalent. Dani Cricco pointed out the Hopcroft O(n log n) algorithm. There is another improved algorithm by Hopcroft and Craft which tests the equivalence of two DFAs in O(n).
For a good survey on the matter and an interesting approach to this, I reccomend the paper Testing the Equivalence of Regular Languages, from arXiv.
Proving that a language is regular by giving a regular expression

Proving that a language is regular by giving a regular expression


By : user2350335
Date : March 29 2020, 07:55 AM
should help you out I am stumped by this practice problem (not for marks): , This is kind of ugly, but it should work:
code :
ε U ( (aa) U (bb) U ((ab) U (ba) (ab) U (ba)) )*
11*011*0
Does every regular language have a proper regular superset? or proper subset?

Does every regular language have a proper regular superset? or proper subset?


By : Nagaraj
Date : March 29 2020, 07:55 AM
this will help
Does every regular language have a proper regular superset? or proper subset?
Algorithmic complexity of Regular Languages in Extended Regular Language frameworks

Algorithmic complexity of Regular Languages in Extended Regular Language frameworks


By : Paul Lyons
Date : March 29 2020, 07:55 AM
wish of those help Java's implementation of regular expression uses the NFA approach. Here is a link that explains it clearly.
Basically, a badly written but still correct regular expression can cause the engine to perform badly. For example, given the expression (a+a+)+b and the string aaaaaaaaaaaaaaa. It can take a while (depending on your machine, from several seconds to minutes) to figure out that there is no match.
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