APSET JULY 2012 QUESTION PAPER WITH ANSWER KEY COMPUTER SCIENCE

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Automata Theory


21.Palindromes can’t be recognized by any FSA because
  1. FSA cannot remember arbitrarily large amount of information
  2. FSA cannot deterministically fix the midpoint
  3. Even if the mid point is known an FSA cannot find whether the second half of the string matches the first half
  4. all of the above
22.Let Σ = {abcde}. The number of strings in Σ* of length 4 such that no symbol is used more than once in a string is
  1. 360
  2. 120
  3. 35
  4. 36
23.Which of the following denotes Chomskian hiearchy?
  1. REG ⊂ CFL ⊂ CSL ⊂ type0
  2. CFL ⊂ REG ⊂ type0 ⊂ CSL
  3. CSL ⊂ type0 ⊂ REG ⊂ CFL
  4. CSL ⊂ CFL ⊂ REG ⊂ type0
24.A language L is accepted by a FSA iff it is
  1. CFL
  2. CSL
  3. recursive
  4. regular
25.Which of the following regular expressions denotes a language comprising of all possible strings over Σ = {ab} of length n where n is a multiple of 3.
  1. (a + b + aa + bb + aba + bba)*
  2. (aaa + bbb)*
  3. ((a + b)(a + b)(a + b))*
  4. (aaa + ab + a) + (bbb + bb + a)
26.A language is represented by a regular expression (a)*(a + ba). Which of the following string does not belong to the regular set represented by the above expression.
  1. aaa
  2. aba
  3. ababa
  4. aa
27.Which of the following is not primitive recursive but partially recursive?
  1. McCarthy’s function
  2. Riemann function
  3. Ackermann’s function
  4. Bounded function
28.Consider the following right-linear grammar G = (NTPSN = {S}
P : S → aS|aAT = {ab}
A → bA|b

Which of the following regular expression denotes L(G)?
  1. (a + b)*
  2. a(ab)*b
  3. aa*bb*
  4. a*b*
29.Which of the following strings is not generated by the following grammar? S → SaSbS|ε
  1. aabb
  2. abab
  3. aababb
  4. aaabb
30.Consider the following NFSA
The automaton accepts
  1. all words of the form {(ab)na|n ≥ 1}
  2. all words that end with a and ε
  3. all words that end with a and not ε
  4. all words containing substring ba




    Answers:

    21.d
    22.b
    23.a
    24.d
    25.c
    26.c
    27.c
    28.c
    29.d
    30.c






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3 comments:

  1. For the question no.22, you have provided answer as 120, but calculation seems to provide answer 360.

    We need to find possible combinations with length 4, and without repetition.
    so according to Multiplication Principle,
    first place can have 6 choices {a,b,c,d,e,f)
    second place can have 5 choices (b,c,d,e,f)
    third place can have 4 choices (c,d,e,f)
    fourth place can have 3 choices (d,e,f).

    applying the multiplication principle : 6*5*4*3 = 360.

    if i am wrong please do correct me. Thank you.

    ReplyDelete
  2. I guess you read the question wrong. The choices are only {a,b,c,d,e}

    ReplyDelete