Theory of Computation (Test 3)

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Theory of Computation

Theory of Computation
| Theory of Computation |
Q.1
Which one of the following options is CORRECT given three positive integers x, y and z, and a predicate P (x) = ¬ (x = 1) ∧ ∀y (∃z (x = y * z) ⇒ (y = x) ∨ (y = 1))
A. P(x) being true means that x is a prime number
B. P(x) being true means that x is a number other than 1
C. P(x) is always true irrespective of the value of x
D. P(x) being true means that x has exactly two factors other than 1 and x
Answer : Option A
Explaination / Solution:
No Explaination.


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Q.2
Which of the following languages is generated by the given grammar? S → aS|bS|ε
A. {an bm | n,m ≥ 0}
B. {w ∈ {a,b}* | w has equal number of a 's and b's}
C. {an | n ≥ 0} ∪ {bn | n ≥ 0} ∪ {an bn | n ≥ 0}
D. {a, b}∗
Answer : Option D
Explaination / Solution:

Given grammar generates all strings of a’s and b’s including null string ∴ L = (a + b) *

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Q.3
 Which of the following decision problems are undecidable?
I. Given NFAs N1 and N2, is L(N1) ∩ L(N2) = Φ?
II. Given a CFG G = (N, Σ, P, S) and a string x ∈ Σ∗, does x ∈ L(G)?
III. Given CFGs G1 and G2, is L(G1) = L(G2)?
IV. Given a TM M, is L(M) = Φ?
A. I and IV only
B. II and III only
C. III and IV only
D. II and IV only
Answer : Option C
Explaination / Solution:

There is no known algorithm to check whether the language accepted by TM is empty. Similarly there is no algorithm to check whether language CFG’s are equivalent.

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Q.4
Which one of the following regular expressions represents the language: the set of all binary strings having two consecutive 0s and two consecutive 1s?
A. (0 + 1)∗0011(0 + 1)∗ + (0 + 1)∗1100(0 + 1)∗
B. (0 + 1)∗(00(0 + 1)∗11 + 11(0 + 1)∗00)(0 + 1)∗
C. (0 + 1)∗00(0 + 1)∗ + (0 + 1)∗11(0 + 1)∗
D. 00(0 + 1)∗11 + 11(0 + 1)∗00
Answer : Option B
Explaination / Solution:

(a) contains 00 & 11 consecutively which is not the required condition. (c) Doesn’t guaranty that both 00 & 11 will be present in the string. (d) Says string should start with 11 & ends with 00 or vice versa.

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Q.5
Consider the following context-free grammars:
G1: S → aS|B, B → b|bB 
G2: S → aA|bB, A → aA|B|ε , B → bB|ε
Which one of the following pairs of languages is generated by G1 and G2, respectively?
A.  {am bn |m > 0 or n > 0} and {am bn |m > 0 and n > 0}
B. {am bn|m > 0 and n > 0} and {am bn|m > 0 or n ≤ 0}
C. {am bn|m ≥ 0 or n > 0} and {am bn|m > 0 and n > 0}
D.  {am bn|m ≥ 0 or n > 0} and {am bn|m > 0 or n > 0}
Answer : Option D
Explaination / Solution:

Lagrange’s generated by G1 = a*b+
Lagrange’s generated by G2 = a+b*\b+

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Q.6
Consider the transition diagram of a PDA given below with input alphabet Σ = {a, b} and stack alphabet Γ = {X , Z}. Z is the initial stack symbol. Let L denote the language accepted by the PDA.
Which one of the following is TRUE? 

A. L = {an bn|n ≥ 0} and is not accepted by any finite automata
B. L = {an |n ≥ 0} ∪ {an bn|n ≥ 0} and is not accepted by any deterministic PDA 
C. L is not accepted by any Turing machine that halts on every input
D. L = {an |n ≥ 0} ∪ {an bn|n ≥ 0} and is deterministic context-free
Answer : Option D
Explaination / Solution:
No Explaination.


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Q.7
Which of the following problems are decidable? 
1) Does a given program ever produce an output? 
2) If L is context-free language, then, is  also context-free? 
3) If L is regular language, then, is  also regular? 
4) If L is recursive language, then, is  also recursive?
A. 1,2,3,4
B. 1,2
C. 2,3,4
D. 3,4
Answer : Option D
Explaination / Solution:

CFL’s are not closed under complementation. Regular and recursive languages are closed under complementation.

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Q.8
Given the language L-{ab, aa, baa}, which of the following strings are in L*? 1) abaabaaabaa 2) aaaabaaaa 3) baaaaabaaaab 4) baaaaabaa

A. 1,2 and 3
B. 2,3 and 4
C. 1,2 and 4
D. 1,3 and 4
Answer : Option C
Explaination / Solution:

L ={ab, aa, baa} Let S1 = ab , S2 = aa and S3 =baa abaabaaabaa can be written as S1S2S3S1S2 aaaabaaaa can be written as S1S1S3S1 baaaaabaa can be written as S3S2S1S2

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Q.9
Consider the following grammar.

What is FOLLOW (Q) ? 
A. {R}
B. {w}
C. {w, y}
D. {w, $}
Answer : Option C
Explaination / Solution:

FOLLOW(Q) is FIRST(R) hence
FIRST(R) = {w, ϵ} 
We add ‘w’ in FOLLOW(Q) and for ϵ we calculate FIRST(S) 
FIRST(S) ={y} 
FOLLOW(Q) is {w ,y}  

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Q.10
 Consider the following context-free grammar over the alphabet ∑ = {a,b,c} with S as the start symbol.
S → abScT|abcT
T → bT|b
Which one of the following represents the language generated by the above grammar ? 
A.
B.
C.
D.
Answer : Option B
Explaination / Solution:

The given Grammar over Σ = {a, b, c} with S as the start symbol is 
S → abScT | abcT 
T→ bT | b 
The minimum length string generated by the grammar is 1: 
S→abcT→abcb ; hence all variable greater than 1. 
Other cases
S → abScT→ ab abScT cT → ab ab abScT cT cT →........→ (ab)n (cT)n
Here T can generate any number of b’s starting with single b.  

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