These questions for practice purpose for GATE CS Exam.

**Ques-1:** Consider the following two statements with respect to Countability:

**Statement-1:**If X union of ‘Y’ is uncountable, then both set ‘X’ and set ‘Y’ must be uncountable.**Statement-2:**The Cartesian product of two countable sets ‘X’ and ‘Y’ is countable.

Which of the following option is true ?

**(A)** only statement 1

**(B)** only statement 2

**(C)** both the statements are true

**(D)** none

**Explanation:**

Statement-1 is not correct, because only one set can be uncountable but need not be both.

Statement-2 is correct, because Cartesian product of two countable set is countable.

Option (B) is true.

**Ques-2:** Consider the following statements:

**X:**Given a grammar, checking if the grammar is not regular is decidable problem.**Y:**If P is regular and Q is not regular then, (P.Q) is necessarily non-regular.**Z:**Pumping lemma can be used for prove that the given language is regular.

Which of the following is true?

**(A)** only X

**(B)** only Y

**(C)** both X and Z

**(D)** both X and Y

**Explanation:**

X is decidable problem because, you can check regular grammar with the help of some productions. So, this statement is correct.

Y is not correct, for counterexample P= null and Q= {a^{n}b^{n} | n ≤ 0} then, P.Q= null which is regular.

Z is also not correct because, Pumping lemma can prove that the language is not regular but can not prove that the language is regular.

So, option (A) is true.

**Ques-3:** Consider X and Y be any two Context sensitive languages and ‘R’ be any regular language. Then which of the following is/are true?

**(A)** X union of R is regular.

**(B)** X intersection of Y is context sensitive.

**(C)** Compliment of Y is context sensitive language.

**(D)** None.

**Explanation:**

**(A)**X union of R = CSL union of R = CSL but not regular. Hence it is incorrect.**(B)**Intersection of two Context sensitive language = CSL since, Context sensitive languages are closed under intersection.**(C)**Compliment of CSL = CSL since, Context sensitive languages are closed under complimentary.

Both option (B) and (C) are correct.

**Ques-4:** Consider three decision problems X, Y, and Z. It is known that X is decidable and Y is undecidable then which is the following is true?

**(A)** Z is decidable if X is reducible to Z

**(B)** Z is undecidable if Z is reducible to Y

**(C)** Z is undecidable if Y is reducible to Z

**(D)** Z is decidable if Z i reducible to Y’s compliment.

**Explanation:**

Suppose there are two problems, A and B. If A is undecidable and reducible to B then, B is also undecidable and if B is decidable and and reducible to A then A is also decidable.

So, option (C) is correct.

**Ques-5:** Which of the following is decidable?

**(A)** A Turing machine prints specific letter.

**(B)** A Turing machine computes product of two numbers.

**(C)** An arbitrary Turing machine halts after fifty steps.

**(D)** none of the above.

**Explanation:**

Option (B) is correct since, Turing machine can compute any mathematical operation.

Option (C) is also correct since, number of steps are give so it is decidable.

Both option (B) and (C) are correct.