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asked in Theory of Computation by Active (2.3k points) | 29 views
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all are regular..!

regular exprsn for i.)  (aa)*(bb)*[even+even=even]+(aa)*a(bb)*b[odd+odd=even]

                            ii.)    (aa)*a(bb)*[odd+even=odd]+(aa)* b(bb)*[even+odd=odd]

                           iii.)    a*b*c*

                          iv.)    (aa)*
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In L4  it should be a*.
u=a*,v=a* and conncatanation of both is also a*.????
+1
@abhishek how can you accept 1,3,5 a's if you take u as aa than v is reverse of it so it will be aa too is that not so.?
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Yes i missed v is reverse of L5

1 Answer

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All are regular language.

iii) regular exp= a*b*c*

iv) regular exp=(aa)*

answered by Boss (24k points)
edited by
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@abhishekmehta4u

@Shubham Shukla 6... my doubt is in L1 and L2 we have to perform unbounded counting against number of a's and b's then how it became regular though I understood the regular expressions given by you!

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here why to count regular exprsn will take care of it only languages falling in this exprsn would be accepted..regualr exrsn can generate infinte languages so why you thinking about unbounded here we dont need to do any comaprsion or checking...even and odd checking possible by regukar exprsn so thats why abve 2 are regular
0
I am still [email protected] Poonia(dee) please explain!


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