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Que. Let L be a set accepted by a non deterministic finite automaton. The number of states in non-deterministic finite automaton is |Q|. The maximum number of states in equivalent finite automaton that accepts L is
a. |Q|
b. 2|Q|
c. 2^|Q| – 1
d. 2^|Q|
Answer:2^|Q|
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The equivalent production rules corresponding to the production rules

S → Sα1 | Sα2 | β1 | β2 
Let L be any language. Define even (W)  as the  strings  obtained by extracting from W the letters in the even-numbered positions and even (L) = {even (W) | W ԑ L}. We define another language Chop (L) by removing the two leftmost symbols of every string in L given by Chop(L) = {W | v W ԑ L,with | v | = 2}.

If L is regular language then
The equivalent production rules corresponding to the production rules

S→Sα1|Sα2|β1|β2 is
A grammar G is LL(1) if and only if the following conditions hold for two distinct productions A → α | β

I. First (α) ∩ First (β) ≠ {a} where a is some terminal symbol of the grammar.
II. First (α) ∩ First (β) ≠ λ
III. First (α) ∩ Follow(A) = φ if λ є First (β)
Given a grammar : S1→Sc, S→SA|A, A→aSb|ab, there is a rightmost derivation S1=>Sc =>SAC=>SaSbc. Thus, SaSbc is a right sentential form, and its handle is
The context free grammar given by

S→XYX
X→aX|bX|λ
Y→bbb

generates the language which is defined by regular expression:
A WFF that is equivalent to the WFF x=>y is:
Which of the following is not true?
Given the following statements:

S1: Every context-sensitive language L is recursive.
S2: There exists a recursive language that is not context sensitive.

Which statement is correct?
Consider the following two languages:

L1 = {0i1j | gcd(i,j)=1}
L2 is any subset of 0*.

Which of the following is correct?
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