81.  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 
a.  L(abb*a) 
b.  {a} U L(abb*a) 
c.  L(ab*a) 
d.  {a} U L(ab*a) 
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Answer: (b).{a} U L(abb*a)

82.  The language L = {a^n b^n a^m b^m  n ≥ 0, m ≥ 0} is 
a.  Context free but not linear 
b.  Context free and linear 
c.  Not Context free and not linear 
d.  Not Context free but linear 
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Answer: (a).Context free but not linear

83.  Assume statements S1 and S2 defined as :
S1 : L2L1 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 ? 
a.  S1 is correct and S2 is not correct 
b.  Both S1 and S2 are correct 
c.  Both S1 and S2 are not correct 
d.  S1 is not correct and S2 is correc 
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Answer: (b).Both S1 and S2 are correct

84.  Nondeterministic 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)} 
a.  A 
b.  B 
c.  C 
d.  D 
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Answer: (b).B

85.  Given L1 = L(a*baa*) and L2 = L(ab*)
The regular expression corresponding to language L3 = L1/L2 (right quotient) is given by 
a.  a*b 
b.  a*baa* 
c.  a*ba* 
d.  None of the above 
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Answer: (c).a*ba*

86.  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? 
a.  G1 is ambiguous and G2 is not ambiguous 
b.  G1 is ambiguous and G2 is ambiguous 
c.  G1 is not ambiguous and G2 is ambiguous 
d.  G1 is not ambiguous and G2 is not ambiguous 
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Answer: (b).G1 is ambiguous and G2 is ambiguous

87.  Given a grammar : S1→Sc, S→SAA, A→aSbab, there is a rightmost derivation S1=>Sc =>SAC=>SaSbc. Thus, SaSbc is a right sentential form, and its handle is 
a.  SaS 
b.  be 
c.  Sbe 
d.  aSb 
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Answer: (d).aSb

88.  The equivalent production rules corresponding to the production rules
S→Sα1Sα2β1β2 is 
a.  S→β1  β2, A→α1A  α2A  λ 
b.  S→β1  β2  β1A  β2A, A→α1A  α2A 
c.  S→β1  β2, A→α1A  α2A 
d.  S→β1  β2  β1A  β2A, A→α1A  α2A  λ 
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Answer: (d).S→β1  β2  β1A  β2A, A→α1A  α2A  λ

89.  Given a Nondeterministic 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 
a.  5 
b.  4 
c.  3 
d.  2 
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Answer: (c).3

90.  The grammar with production rules S → aSb SSλ generates language L given by: 
a.  L = {w∈{a, b}*  na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w} 
b.  L = {w∈{a, b}*  na(w) = nb(w) and na(v) ≤ nb(v) where v is any prefix of w} 
c.  L = {w∈{a, b}*  na(w) ≠ nb(w) and na(v) ≥ nb(v) where v is any prefix of w} 
d.  L = {w∈{a, b}*  na(w) ≠ nb(w) and na(v) ≤ nb(v) where v is any prefix of w} 
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Answer: (a).L = {w∈{a, b}*  na(w) = nb(w) and na(v) ≥ nb(v) where v is any prefix of w}
