The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
0 votes

asked in Theory of Computation by Active (3.5k points) | 33 views
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)*
In L4  it should be a*.
u=a*,v=a* and conncatanation of both is also a*.????
@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.?
Yes i missed v is reverse of L5

1 Answer

0 votes

All are regular language.

iii) regular exp= a*b*c*

iv) regular exp=(aa)*

answered by Boss (25.3k points)
edited by


@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!

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
I am still [email protected] Poonia(dee) please explain!

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
47,894 questions
52,261 answers
67,679 users