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4 Answers

3 votes
3 votes
Yes its a regular language . if  there is a loop in the state diagram then we can say that it produce infinitely many strings (with some pattern) .

Pumping lemma can not used to prove the regularity of a language . if a language fails the pumping lemma test then we can say its not a regular language but if it pass the test then the language may be regular or may not be .
2 votes
2 votes
Concept :
If pumping lemma fails then language is not regular..
Contrapositive of this statement :
If language is regular then pumping lemma satisfied..

So every regular set satisfy Pumping Lemma property trivially.. but reverse (i.e . If Pumping Lemma satisfy then regular) may not be true..
1 votes
1 votes
Ok see ,if it accept 101 and having no loop then this language  becomes finite language that means its surely regular language , coming back to your second question it may not satisfy Pumping Lema property because Pumping Lema use for proving non regularity always use in negative sense  , means if some language satisfy its property then language is not regular but if not satisfy its property we can't say anything about language .
0 votes
0 votes
  • if language have no self loop then language must be finite. and every finite language is regular and we can make a dfa for this.

  • pumping lemma is a negativity test i.e it is used nor non regular. 

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