Tuesday 19 March 2013

JUNE 2012 - PAPER III

11. X.25 is ________ Network.
(A) Connection Oriented Network
(B) Connection Less Network
(C) Either Connection Oriented or Connection Less
(D) Neither Connection Oriented nor Connection Less

Ans:-A

Explanation:-
X.25 is a connection oriented protocol.


12. Which of the following can be used for clustering of data ?
(A) Single layer perception
(B) Multilayer perception
(C) Self organizing map
(D) Radial basis function

Ans:-C


13. Which of the following is scheme to deal with deadlock ?

(A) Time out
(B) Time in
(C) Both (A) & (B)
(D) None of the above

Ans:-A


14. If the pixels of an image are shuffled then the parameter that may change is
(A) Histogram
(B) Mean
(C) Entropy
(D) Covariance

Ans:-D


15. The common property of functional language and logical programming language :
(A) Both are declarative
(B) Both are based on λ-calculus
(C) Both are procedural
(D) Both are functional

Ans:-A


16. Given the following statements :
(i) The power of deterministic finite state machine and non- deterministic finite state machine are same.
(ii) The power of deterministic pushdown automaton and non- deterministic pushdown automaton are same.
Which of the above is the correct statement(s) ?
(A) Both (i) and (ii)
(B) Only (i)
(C) Only (ii)
(D) Neither (i) nor (ii)

Ans:-B

Explanation:-
The answer is B. But why is it so?. A very good explanation is given in the book "Theory of computation" by A.A.Puntambekar.
We all know that finite machine is of two types. One is deterministic finite state machine and the other one non deterministic finite state machine. Both these machine accept regular language only. So the power of DFA = NFA. So the first statement is true.
Next comes the question of pushdown automaton and the power of deterministic and non-deterministic being the same. PDA has more power than FA because PDA has a memory and so can accept large class of languages than FA. PDA accepts the language of context free grammar. The power of DPDA is less than NPDA because NPDA accepts a larger class of context free language.
Turing machines can accept a more large class of language. Therefore it is the most powerful computational model. The power of deterministic and non deterministic turing machine is the same.


17. LetQ(x,y)denote “x+y=0” and let there be two quantifications given as
(i) ∃y∀x Q(x, y)
(ii) ∀x∃y Q(x, y)
where x & y are real numbers. Then which of the following is valid ?
(A) (i) is true & (ii) is false.
(B) (i) is false & (ii) is true.
(C) (i) is false & (ii) is also false.
(D) both (i) & (ii) are true.

Ans:-B

Explanation:-
The symbol ∀ is called the universal quantifier. The universal qualification of P(x) is the statement "P(x) for all values x in the universe" which is written as ∀xP(x).
The symbol ∃ is called the existential quantifier and represents the phrase "there exists" or "for some". The existential quantification of P(x) is the statement "P(x) for some values x in the universe which is written as ∃xP(x).
A nested quantifier is one where two quantifiers are nested if one is within the scope of the other.
The order of nested universal quantifiers in a statement without other quantifiers can be changed without changing the meaning of the quantified statement.
The order of nested existential quantifiers in a statement without other quantifiers can be changed without changing the meaning of the quantified statement.
But the order of nested universal and existential quantifier is important. The order cannot be changed without affecting the meaning.
Let us consider the quantification ∀x∃y Q(x, y). The domain is real numbers. Q(x,y) is x+y=0. The quantification can be explained this way. Since the universal quantification comes first, it has to be understood as, for all real numbers x there is a real number y such that x+y=0, which is actually true. We are saying that y is the additive inverse of x.
Now let us consider the quantification ∃y∀x Q(x, y). Since the existential quantification comes first, it has to be understood as there is a real number y such that for all real numbers x, x+y=0 , which is false.
So, the quantification (i) is false and (ii) is true.


No comments:

Post a Comment