Following questions have been asked in GATE CS 2012 exam.

**1) What is the complement of the language accepted by the NFA shown below? Assume ∑ = {a} and ε is the empty string **

(A) Φ

(B) ε

(C) a

(D) {a, ε}

Answer (B)

The given alphabet ∑ contains only one symbol {a} and the given NFA accepts all strings with any number of occurrences of ‘a’. In other words, the NFA accepts a+. Therefore complement of the language accepted by automata is empty string.

**2) Given the language L = {ab, aa, baa}, whih 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 (C)

Any combination of strings in set {ab, aa, baa} will be in L*.

….**1)** “abaabaaabaa” can be partitioned as a combination of strings in set {ab, aa, baa}. The partitions are “ab aa baa ab aa”

….**2)** “aaaabaaaa” can be partitioned as a combination of strings in set {ab, aa, baa}. The partitions are “aa ab aa aa”

….**3)** “baaaaabaaaab” cannot be partitioned as a combination of strings in set {ab, aa, baa}

….**4)** “baaaaabaa” can be partitioned as a combination of strings in set {ab, aa, baa}. The partitions are “baa aa ab aa”

**3) Which of the following problems are decidable?
….1) Does a given program ever produce an output?
….2) If L is a context-free language, then is L’ (complement of L) also context-free?
….3) If L is a regular language, then is L’ also regular?
….4) If L is a recursive language, then, is L’ also recursive?**

(A) 1, 2, 3, 4

(B) 1, 2,

(C) 2, 3, 4

(D) 3, 4

Answer (D)

….**1)** Is a variation of Turing Machine Halting problem and it is undecidable.

….**2)** Context Free Languages are not closed under intersection and complement. See this for details.

….**3)** Complement of Regular languages is also regular. Then a DFA that accepts the complement of L, i.e. ∑* – L, can be obtained by swapping its accepting states with its non-accepting states.

….**4)** Recursive Languages are closed under complement. See this for details.

**4) Consider the set of strings on {0,1} in which, every substring of 3 symbols has at most two zeros. For examples, 001110 and 011001 are in the language, but 100010 is not. All strings of length less than 3 are also in the language. A partially completed DFA that accepts this language is shown below.**

**The missing arcs in the DFA are**

Answer (D)

State ‘q’ is trap state. All other states are accept states. In state 00, DFA must move to ‘q’ for input symbol 0. All (non-trap) states indicate names indicate the characters seen before reaching that particular state. Option (D) is the only option that follow these rules.

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