#toc #regular languages

Question

Why this is not a regular language pls explain?

Comments

I am not able to get this question....Anybody please explain more!

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@abhishekmehta4u....please explain this

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well well i dnn thinks its regular

for exmple: if take k=3 and n=2 and m any value lets take m=1

now string which should be accepted should be a(bccc)(bccc) as now you need to do check for occurence of same string bccc two times that would be done only by the use of stack...so its non regular....correct me i am wrong..!

Answer 1

This is not regular becoz we need to save the count of c becoz if we put k=1 then every n times it should be 1 similarly if k=2 then every n times it should be 2.

while generating a string we will to keep the value of m,n and k

Now let us suppose m=1, n=2 and k=1

Then the string will be abcbc. That means every time c will be 1

That's why it is not regular

if k=2 and m,n remains the same then string will be abccbcc. That means everytime c will be 2

That means we will have to remmeber the count of c.

Comments

@sourabh yes we cannot . . . I proved it with pumping lemma too

@srestha (c⁺​​​​​​) ² is this regular ? 

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@sreshta

a⁺(bc⁺)⁺ $\neq$ a^m(bc^k)ⁿ 

if we take abcbcc, it is accepted by a⁺(bc⁺)⁺ but not by a^m(bc^k)ⁿ. So k's count should be maintained for every n. That is why a^m(bc^k)ⁿ is not regular. 

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ok,yes thnks @veerendra