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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 a^{i}b^{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 nontrivial 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.
Consider the languages L_{1},L_{2} and L_{3} as given below.
and
Which of the following statements is NOT TRUE?
Turning Machine is powerful Machine it can be used to accept all the languages (RL,CFL,CSL,RE)
Consider the following languages.
Which one of the following statements is FALSE?
L1 is regular and hence contextfree also. Regular expression for L1 is 0*1*0*. So, (D) is the false choice.
L2 is contextfree 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 contextfree as regular ∩ contextfree is contextfree. > (B) is true.
Complement of L2 is recursive as contextfree complement is always recursive (actually even contextsensitive).
Consider the following languages over the alphabet
Here, ω^{r} is the reverse of the string . Which of these languages are deterministic Contextfree languages?
C.
L3 is CFL and not DCFL as in no way we can deterministically determine the MIDDLE point of the input string.
Consider the contextfree grammars over the alphabet given below. S and T are nonterminals.
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
Consider the following languages.
Which of the following are CORRECT?
L1 is Csl,L2 is context free
L3 is not Context free and L4 is Dcfl
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?
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 .
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?
Let L be a given contextfree language over the alphabet . Construct L_{1} ,L_{2} as follows. Let
L2 = L.L
Context free languages are closed under Concatenation.
So L2 is Context Free Language.
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 }:
languages over alphabet set are uncountable so ans should be e
Consider the following grammar G with terminals start symbol S , and nonterminals
A language L is called prefixclosed if for every every prefix of X is also in L . Which of the following is FALSE?
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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