Q. Ram and Shyam have been asked to show that a certain problem Π is NP-complete. Ram shows a polynomial time reduction from the 3-SAT problem to Π, and Shyam shows a polynomial time reduction from Π to 3-SAT. Which of the following can be inferred from these reductions ?

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Q. Nobody knows yet if P = NP. Consider the language L defined as follows. Which of the following statements is true ?

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Q. If the strings of a language L can be effectively enumerated in lexicographic (i.e., alphabetic) order, which of the following statements is true ?

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Q. Let G = ({S}, {a, b} R, S) be a context free grammar where the rule set R is S → a S b | SS | ε . Which of the following statements is true?

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Q. Define languages L0 and L1 as follows :

L0 = {< M, w, 0 > | M halts on w}
L1 = {< M, w, 1 > | M does not halts on w}

Here < M, w, i > is a triplet, whose first component. M is an encoding of a Turing Machine, second component, w, is a string, and third component, i, is a bit. Let L = L0 ∪ L1. Which of the following is true ?

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Q. The language {a^m b^n C^(m+n) | m, n ≥ 1} is

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Q. Consider the following grammar G:

S → bS | aA | b
A → bA | aB
B → bB | aS | a

Let Na(w) and Nb(w) denote the number of a's and b's in a string w respectively. The language L(G) ⊆ {a, b}+ generated by G is

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Q. L1 is a recursively enumerable language over Σ. An algorithm A effectively enumerates its words as w1, w2, w3, ... Define another language L2 over Σ Union {#} as {wi # wj : wi, wj ∈ L1, i < j}. Here # is a new symbol. Consider the following assertions.

S1 : L1 is recursive implies L2 is recursive
S2 : L2 is recursive implies L1 is recursive

Which of the following statements is true ?

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Q. Consider the languages L1 = {0^i1^j | i != j}. L2 = {0^i1^j | i = j}. L3 = {0^i1^j | i = 2j+1}. L4 = {0^i1^j | i != 2j}.

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Q. S -> aSa|bSb|a|b; The language generated by the above grammar over the alphabet {a,b} is the set of

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Q. The language L= {0^i21^i | i≥0 } over the alphabet {0,1, 2} is:

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Q. Consider the CFG with {S,A,B) as the non-terminal alphabet, {a,b) as the terminal alphabet, S as the start symbol and the following set of production rules

S --> aB S --> bA
B --> b A --> a
B --> bS A --> aS
B --> aBB A --> bAA

Which of the following strings is generated by the grammar?

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Q. Consider the following statements about the context free grammar

G = {S → SS, S → ab, S → ba, S → Ε}
I. G is ambiguous
II. G produces all strings with equal number of a’s and b’s
III. G can be accepted by a deterministic PDA.

Which combination below expresses all the true statements about G?

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Q. Consider the languages:

L1 = {a^nb^nc^m | n, m > 0}
L2 = {a^nb^mc^m | n, m > 0}

Which one of the following statements is FALSE?

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Q. Consider the languages:

L1 = {ww^R |w ∈ {0, 1}*}
L2 = {w#w^R | w ∈ {0, 1}*}, where # is a special symbol
L3 = {ww | w ∈ (0, 1}*)

Which one of the following is TRUE?

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Q. Which of the following languages are context-free?

L1 = {a^m b^n a^n b^m ⎪ m, n ≥ 1}
L2 = {a^m b^n a^m b^n ⎪ m, n ≥ 1}
L3 = {a^m b^n ⎪ m = 2n + 1}

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Q. Which one of the following statements is FALSE ?

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Q. Consider the pushdown automaton (PDA) below which runs over the input alphabet (a, b, c). It has the stack alphabet {Z0, X} where Z0 is the bottom-of-stack marker. The set of states of the PDA is (s, t, u, f} where s is the start state and f is the final state. The PDA accepts by final state. The transitions of the PDA given below are depicted in a standard manner. For example, the transition (s, b, X) → (t, XZ0) means that if the PDA is in state s and the symbol on the top of the stack is X, then it can read b from the input and move to state t after popping the top of stack and pushing the symbols Z0 and X (in that order) on the stack.

(s, a, Z0) → (s, XXZ0)
(s, ϵ, Z0) → (f, ϵ)
(s, a, X) → (s, XXX)
(s, b, X) → (t, ϵ)
(t, b, X) → (t,.ϵ)
(t, c, X) → (u, ϵ)
(u, c, X) → (u, ϵ)
(u, ϵ, Z0) → (f, ϵ)

The language accepted by the PDA is

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Q. A CFG G is given with the following productions where S is the start symbol, A is a non-terminal and a and b are terminals.

S→aS∣A
A→aAb∣bAa∣ϵ

Which of the following strings is generated by the grammar above?

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Q. Consider 2 scenarios:

C1: For DFA (ϕ, Ʃ, δ, qo, F),
if F = ϕ, then L = Ʃ*

C2: For NFA (ϕ, Ʃ, δ, qo, F),
if F = ϕ, then L = Ʃ*

Where F = Final states set
ϕ = Total states set

Choose the correct option ?

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