GATE Computer Science Engineering(CSE) 2027 Test: Identify Class Language

   has only one comparison that can be done using a DPDA. Hence, its DCFL.

Context free languages are a proper subset of Recursive Languages. ∴ it is recursive too.

We have algorithms to generate prime numbers ⇒ we can generate sequence of P for the given language, hence
strings as defined by the language definition.

So by Church Turing Thesis we can say that there exists a Turing Machine which can accept the given language.

Both languages can be accepted by a DPDA :
L1 = start pushing element X into the stack on input 'a' ... start poping X on input 'b' ... move to final state when stack empty and input = 'c'

L2 = start pushing elements XX into the stack on input 'a' ... start poping X on input 'b' ... move to final state when stack
empty and input = 'epsilon'

so (A) and (D) are False

L1 ∪ L2 is a CFL ... we can build it by having L1, L2 and an extra state ... and an 'epsilon' transition to both L1 and L2 from
that extra state.

so (C) is false

L1 ∩ L2 = Phi because we have no string aⁱb^j where i=j and i=2j for i,j>=1
and clearly L1 is not a regular language

which is context free but not regular.

All are context free.
L1 -> Push 0 on stack and when 1 comes, start popping. If stack becomes empty and 1's are remaining start pushing 1. At
end of string accept if stack is non- empty.
L2 -> Do the same as for L1, but accept if stack is empty at end of string.
L3 -> Do, the same as for L2, but for each 0, push two 0's on stack and start the stack with a 0.
L4 -> Do the same as for L1, but for each 0, push two 0's on stack
All are in fact DCFL. Pushing two 0's on stack might sound non-trivial but we can do this by pushing one symbol and going
to a new state. Then on this new state on empty symbol, push one more symbol on stack and come back.

Turning Machine is powerful Machine it can be used to accept all the languages (RL,CFL,CSL,RE)

Explanation: (B) is false.

L1 is regular, so its complement would also be regular.

L1 is a regular language of the form 0^* 1^* 0^*. L2 on the other hand is a CFL as it can be derived from the following CFG
L2 = { 0^p 1^q 0^r | p,q,r>0   And p notEqualTo r }
S -> AC|CA
C -> 0C0|B
A -> 0A|0
B -> 1B|epsilon
If coming up with a CFG for L2 is difficult, one can intuitively see that by reducing it to a simpler problem. L2 is very similar to a known CFL L3 = { a^m b^l | m notEqualTo n }

(A) L2 is context free, which is true [CORRECT]
(B) L1 intersection L2 is context free, which is again true because L1 is a regular language and L2 is a CFL. RL union CFL is always a CFL. Hence [CORRECT]
(C) Complement of L2 is recursive, which is true due to the fact that complement of a CFL is CSL for sure (Context sensitive language), which in turn (CSL) is a subset of recursive languages. Hence [CORRECT]
(D) Complement of L1 is context free but not regular, which is false due to closure laws of regular languages. Complement of a RL is always a RL. Hence [INCORRECT]

C.
L3 is CFL and not DCFL as in no way we can deterministically determine the MIDDLE point of the input string.

Since while intersection all strings produced by production aSb in G1 and bSa in G2 will be 0
so only common production will be
S-->T
T-->cT | epsilon
which is nothing but c* hence it is REGULAR and INFINITE

L1 is Csl,L2 is context free
L3 is not Context free and L4 is Dcfl

l can be recognized by a dfa. we have a dfa to accept all such strings which when interpretated as decimal number are
divisible by n. where n can be anything the dfa of such can be made by a trick.
states are equal to possible remainders

if u can see the symmetry in it. write states and make fill like q0 q1 q2 q3 ...
now it is saying that it has to always start with 1 which the above dfa will not satisfy so make it a nfa by making a
transition from q0 on zero to a new dead state. now you have a nfa reduce it which will result in a deterministic dfa .

L2 = L.L

Context free languages are closed under Concatenation.

So L2 is Context Free Language.

languages over alphabet set are uncountable so ans should be e

The given grammar generates balanced parenthesis.

Lets take a smallest string : [ [ ] ] (say x )
Prefixes of x are : [ , [ [ ,[ [ ]

BUT they don't belong to the langauge generated by the given grammar.

SO, the answer will be Option D