adplus-dvertising
frame-decoration

Question

A shift reduce parser suffers from

a.

shift reduce conflict only

b.

reduce reduce conflict only

c.

both shift reduce conflict and reduce reduce conflict

d.

shift handle and reduce handle conflicts

Answer: (c).both shift reduce conflict and reduce reduce conflict

Engage with the Community - Add Your Comment

Confused About the Answer? Ask for Details Here.

Know the Explanation? Add it Here.

Q. A shift reduce parser suffers from

Similar Questions

Discover Related MCQs

Q. The context free grammar for language L = {a^nb^mc^k | k = |n - m|, n≥0,m≥0,k≥0} is

Q. The number of states in a minimal deterministic finite automaton corresponding to the language L = { an | n≥4 } is

Q. Regular expression for the language L = { w ∈ {0, 1}* | w has no pair of consecutive zeros} is

Q. Consider the following two languages:

L1 = {a^n b^l a^k | n + l +k>5 }
L2 = {a^n b^l a^k |n>5, l >3, k≤ l }

Which of the following is true?

Q. LL grammar for the language L = {a^n b^m c^n+m | m≥0, n≥0} is

Q. Assume the statements S1 and S2 given as:

S1: Given a context free grammar G, there exists an algorithm for determining whether L(G) is infinite.
S2: There exists an algorithm to determine whether two context free grammars generate the same language.

Which of the following is true?

Q. Assume, L is regular language. Let statements S1 and S2 be defined as :

S1 : SQRT(L) = { x| for some y with |y| = |x|^2, xy ∈L}
S2 : LOG(L) = { x| for some y with |y| = 2^|x|, xy ∈ L}

Which of the following is true ?

Q. A regular grammar for the language L = {a^nb^m | n is even and m is even}is given by

Q. Given the following productions of a grammar :

S→ aA| aBB;
A→aaA |λ ;
B→ bB| bbC;
C→ B

Which of the following is true ?

Q. The language accepted by the nondeterministic pushdown automaton

M= ({q0, q1, q2}, {a, b}, {a, b, z}, δ, q0, z, {q2}) with transitions
δ (q0 a, z) = { (q1 a), (q2 λ)};
δ (q1, b, a) = { (q1, b)}
δ (q1, b, b) ={ (q1 b)}, δ (q1, a, b) = { (q2, λ)}

is

Q. The language L = {a^n b^n a^m b^m | n ≥ 0, m ≥ 0} is

Q. Assume statements S1 and S2 defined as :

S1 : L2-L1 is recursive enumerable where L1 and L2 are recursive and recursive enumerable respectively.
S2 : The set of all Turing machines is countable.

Which of the following is true ?

Q. Non-deterministic pushdown automaton that accepts the language generated by the grammar:

S→aSS | ab is

(A) δ(q0, λ, z) = { (q1, z)};
δ(q0, a, S) = { (q1, SS)}, (q1, B) }
δ(q0, b, B) = { (q1, λ)},
δ(q1, λ, z) = { (qf, λ)}

(B) δ(q0, λ, z) = { (q1, Sz)};
δ(q0, a, S) = { (q1, SS)}, (q1, B) }
δ(q0, b, B) = { (q1, λ)},
δ(q1, λ, z) = { (qf, λ)}

(C) δ(q0, λ, z) = { (q1, Sz)};
δ(q0, a, S) = { (q1, S)}, (q1, B) }
δ(q0, b, λ) = { (q1, B)},
δ(q1, λ, z) = { (qf, λ)}

(D) δ(q0, λ, z) = { (q1, z)};
δ(q0, a, S) = { (q1, SS)}, (q1, B) }
δ(q0, b, λ) = { (q1, B)},
δ(q1, λ, z) = { (qf, λ)}

Q. Given L1 = L(a*baa*) and L2 = L(ab*)
The regular expression corresponding to language L3 = L1/L2 (right quotient) is given by

Q. Given the production rules of a grammar G1 as

S1→AB | aaB
A→a | Aa
B→b

and the production rules of a grammar G2 as

S2→aS2bS2 | bS2aS2 | λ

Which of the following is correct statement?

Q. 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

Q. The equivalent production rules corresponding to the production rules

S→Sα1|Sα2|β1|β2 is

Q. Given a Non-deterministic Finite Automation (NFA) with states p and r as initial and final states respectively transition table as given below

The minimum number of states required in Deterministic Finite Automation (DFA) equivalent to NFA is

Q. The grammar with production rules S → aSb |SS|λ generates language L given by:

Q. A pushdown automation M = (Q, Σ, Γ, δ, q0, z, F) is set to be deterministic subject to which of the following condition(s), for every q ∈ Q, a ∈ Σ ∪ {λ} and b ∈ Γ
(s1) δ(q, a, b) contains at most one element
(s2) if δ(q, λ, b) is not empty then δ(q, c, b) must be empty for every c ∈ Σ