MadeEasy Test Series: Theory Of Computation - Identify Class Language

Question

If L1 = { a^n | n ≥ 0 } and L2 = { b^n | n ≥ 0 }, Consider

then L1 . L2 wil be

a) (ab)^n

b) a^n b^n

c) b^n a^n

d)b^m a^n

e)  { a^m b^n | m ≥ 0, n ≥ 0 }

why answer is d why not b ????

Comments

L1.L2=a*b*     so d is not correct not b also

---

D) is not correct.

L1.L2 = a*b* = {a^mbⁿ | m,n>=0}

Answer 1

'n' of  L1 is different from the 'n' of L2 (i.e these two 'n' are not equal), So due to this difference L1.L2 = {a^mb^n :m>=0,n>=0} will be the answer. 

Comments

got it thanks bro ..!

Answer 2

L1 and L2 are both regular languages, therefore concatenation of both i.e L1.L2 is also regular language. And this language is "Number of a's followed by number of b's".

Answer 3

option e is right

Answer 4

no option is correct:- D must be like this

D. a^mbⁿ and m,n >=0   which also represent a*b*...

L1={null,a,aa,aaa,......}=a*

L2={null,b,bb,bbb,......}=b*

Comments

now one option is correct , tell me one thing please if it should be a^n b^n then whats wrong here ....

---

aⁿbⁿ is CFL, as count of 'a' = count of 'b' ,but a^mbⁿ is regular, because count of 'a' and 'b' does not depend on each other.

---

got it according to you we  , lets say one example we use stack for a^n b^n (FOR CFL)??

correct??