Automata Theory | Set 6 Last Updated : 23 Jul, 2025 Comments Improve Suggest changes Like Article Like Report Following questions have been asked in GATE CS 2010 exam. 1) Let L={w ∈ (0 + 1)*|w has even number of 1s}, i.e. L is the set of all bit strings with even number of 1s. Which one of the regular expression below represents L? (A) (0*10*1)* (B) 0*(10*10*)* (C) 0*(10*1*)*0* (D) 0*1(10*1)*10* Answer (B) Option (A) is incorrect because it cannot accept "110" Option (C) is incorrect because it accept a string with single 1. Option (D) is incorrect because it cannot accept 11101 2) Let L1 be a recursive language. Let L2 and L3 be languages that are recursively enumerable but not recursive. Which of the following statements is not necessarily true? (A) L2 - L1 is recursively enumerable. (B) L1 - L3 is recursively enumerable (C) L2 ∩ L1 is recursively enumerable (D) L2 ∪ L1 is recursively enumerable Answer (B) 3) Consider the languages L1={0i1j | i != j}, L2={0i1j | i = j}, L3 = {0i1j | i = 2j+1}, L4 = {0i1j | i != 2j}. Which one of the following statements is true? (A) Only L2 is context free (B) Only L2 and L3 are context free (C) Only L1 and L2 are context free (D) All are context free Answer (D) A Pushdown Automata can be built for all four languages. 4) Let w be any string of length n is {0,1}*. Let L be the set of all substrings of w. What is the minimum number of states in a non-deterministic finite automaton that accepts L? (A) n-1 (B) n (C) n+1 (D) 2n-1 Answer (C) We need minimum n+1 states to build NFA that accepts all substrings of a binary string. For example, following NFA accepts all substrings of "010" and it has 4 states. Please see GATE Corner for all previous year paper/solutions/explanations, syllabus, important dates, notes, etc. Please write comments if you find any of the answers/explanations incorrect, or you want to share more information about the topics discussed above. Comment More infoAdvertise with us Next Article Automata Theory | Set 2 K kartik Follow Improve Article Tags : Theory of Computation GATE-CS-2010 AT Similar Reads Automata Theory | Set 2 Questions Asked in the GATE CS 2012 Exam1) What is the complement of the language accepted by the NFA shown below? Assume â = {a} and ε is the empty string (A) Φ (B) ε (C) a (D) {a, ε} Answer (B) Explanation: The given alphabet â contains only one symbol {a} and the given NFA accepts all strings wit 3 min read Automata Theory | Set 9 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 3 min read Automata Theory | Set 4 Following questions have been asked in GATE CS 2011 exam. 1) Let P be a regular language and Q be context-free language such that Q ⊆ P. (For example, let P be the language represented by the regular expression p*q* and Q be {pnqn|n ∈ N}). Then which of the following is ALWAYS regular? (A) 2 min read Automata Theory | Set 7 These questions are for practice purpose for GATE CS Exam.Ques-1 . Consider L= {(TM) | TM is the Turing machine that halts on all input and L(TM)= L' for some undecidable language L'}. Here, (TM) is the encoding of a Turing machine as a string over alphabet {0, 1} then L is:(A) decidable and recursi 3 min read Automata Theory | Set 7 These questions are for practice purpose for GATE CS Exam.Ques-1 . Consider L= {(TM) | TM is the Turing machine that halts on all input and L(TM)= L' for some undecidable language L'}. Here, (TM) is the encoding of a Turing machine as a string over alphabet {0, 1} then L is:(A) decidable and recursi 3 min read Automata Theory | Set 5 Following questions have been asked in GATE CS 2009 exam. 1) S --> aSa| bSb| a| b ;The language generated by the above grammar over the alphabet {a,b} is the set of (A) All palindromes. (B) All odd length palindromes. (C) Strings that begin and end with the same symbol (D) All even length palindrome 3 min read Like