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QUESTION: 1

The language over the alphabet

Solution:

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.

QUESTION: 2

Which of the following is true for the language

Solution:

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.

QUESTION: 3

Consider the following languages.

Solution:

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 a^{i}b^{j} where i=j and i=2j for i,j>=1

and clearly L1 is not a regular language

QUESTION: 4

Let are languages as defined below:

Then L is

Solution:

which is context free but not regular.

QUESTION: 5

Consider the languages

Solution:

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.

QUESTION: 6

Consider the languages L_{1},L_{2} and L_{3} as given below.

and

Which of the following statements is NOT TRUE?

Solution:

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

QUESTION: 7

Consider the following languages.

Which one of the following statements is FALSE?

Solution:

L1 is regular and hence context-free also. Regular expression for L1 is 0*1*0*. So, (D) is the false choice.

L2 is context-free but not regular. We need a stack to count if the number of 0's before and after the 1 (1 may be absent also) are not same. So, L1 ∩ L2 is context-free as regular ∩ context-free is context-free. -> (B) is true.

Complement of L2 is recursive as context-free complement is always recursive (actually even context-sensitive).

QUESTION: 8

Consider the following languages over the alphabet

Here, ω^{r} is the reverse of the string . Which of these languages are deterministic Context-free languages?

Solution:

C.

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

QUESTION: 9

Consider the context-free grammars over the alphabet given below. S and T are non-terminals.

Solution:

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

QUESTION: 10

Consider the following languages.

Which of the following are CORRECT?

Solution:

L1 is Csl,L2 is context free

L3 is not Context free and L4 is Dcfl

QUESTION: 11

Let L consist of all binary strings beginning with a 1 such that its value when converted to decimal is divisible by 5. Which of the following is true?

Solution:

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 .

QUESTION: 12

Consider the following languages over the alphabet {0,1}.

Where x^{r} is the reverse of string x; e.g. 011^{R} =110 Which of the following is true?

Solution:

QUESTION: 13

Let L be a given context-free language over the alphabet . Construct L_{1} ,L_{2} as follows. Let

Solution:

L2 = L.L

Context free languages are closed under Concatenation.

So L2 is Context Free Language.

QUESTION: 14

Let be a one letter alphabet and be a two letter alphabet. A language over an alphabet is a set of finite length words comprising letters of the alphabet. Let L_{1} and L_{2 } be the set of languages over ∑_{1} and ∑_{2} respectively. Which of the following is true about L_{1} and L_{2 }:

Solution:

languages over alphabet set are uncountable so ans should be e

QUESTION: 15

Consider the following grammar G with terminals start symbol S , and non-terminals

A language L is called prefix-closed if for every every prefix of X is also in L . Which of the following is FALSE?

Solution:

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

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