GATE 1991 Subject – Theory Of Computation Topic -Finite Automata.<\/p>\n
[fvplayer id=”152″]<\/p>\n
Finite Automata (FA) is a key topic in Theory of Computation (TOC)<\/strong> and is used to recognize Regular Languages<\/strong>. It is one of the fundamental models of computation.<\/p>\n A Finite Automaton (FA)<\/strong> is formally defined as a 5-tuple (Q, \u03a3, \u03b4, q\u2080, F)<\/strong>, where:<\/p>\n Q<\/strong> \u2192 A finite set of states.<\/p>\n<\/li>\n \u03a3<\/strong> \u2192 A finite set of input symbols (Alphabet).<\/p>\n<\/li>\n \u03b4 (Transition Function)<\/strong> \u2192 Defines transitions between states.<\/p>\n<\/li>\n q\u2080 (Start State)<\/strong> \u2192 The initial state.<\/p>\n<\/li>\n F (Final States)<\/strong> \u2192 A set of accepting states.<\/p>\n<\/li>\n<\/ul>\n Finite Automata can be classified into two main types:<\/p>\n Both DFA and NFA recognize the same class of languages (Regular Languages).<\/strong><\/p>\n States: Explanation:<\/strong><\/p>\n Start at If Stay in Any string ending in States: Explanation:<\/strong><\/p>\n Start at If If Once in Q:<\/strong> Which of the following statements is true? Correct Answer: (A) Every NFA can be converted to an equivalent DFA.<\/strong> Finite Automata<\/strong> recognizes Regular Languages<\/strong>. \u00a0Do you need more solved GATE questions or examples?<\/strong><\/p>\n In the GATE 1991 Computer Science Engineering (CSE)<\/strong> exam, Question 17(b)<\/strong> under the Theory of Computation<\/strong> section focused on Finite Automata<\/strong>.<\/span> The question required constructing a Non-Deterministic Finite Automaton (NFA)<\/strong> for a specified language and determining the number of states in the minimized equivalent Deterministic Finite Automaton (DFA)<\/strong>.<\/span><\/p>\n The task involves two main steps:<\/span><\/p>\n Designing an NFA<\/strong>: Create an NFA that accepts a particular language.<\/span><\/p>\n<\/li>\n Converting to a DFA<\/strong>: Transform the NFA into an equivalent DFA and minimize it to find the least number of states required.<\/span><\/p>\n<\/li>\n<\/ol>\n This process typically employs the powerset construction<\/strong> method to convert the NFA to a DFA, followed by DFA minimization techniques to reduce the number of states.<\/span><\/p>\n\u00a0Definition of Finite Automata<\/strong><\/h3>\n
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\u00a0Types of Finite Automata<\/strong><\/h3>\n
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\n Type<\/strong><\/th>\n Description<\/strong><\/th>\n Recognizes<\/strong><\/th>\n<\/tr>\n<\/thead>\n\n \n DFA (Deterministic Finite Automata)<\/strong><\/td>\n Exactly one transition per input symbol from each state<\/td>\n Regular Languages<\/td>\n<\/tr>\n \n NFA (Nondeterministic Finite Automata)<\/strong><\/td>\n Multiple transitions for the same input symbol or \u03b5-moves<\/td>\n Regular Languages<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n \u00a0Example of DFA<\/strong><\/h3>\n
Language:<\/strong> Strings over
{0,1}<\/code> that end with 1<\/code>.<\/h3>\nDFA Representation:<\/strong><\/h3>\n
{q0, q1}<\/code>
Alphabet: {0,1}<\/code>
Start State: q0<\/code>
Final State: q1<\/code>
Transitions:<\/p>\n\n\n
\n State<\/th>\n Input 0<\/code><\/th>\nInput 1<\/code><\/th>\n<\/tr>\n<\/thead>\n\n\n q0<\/strong><\/td>\n q0<\/td>\n q1<\/td>\n<\/tr>\n \n q1<\/strong><\/td>\n q1<\/td>\n q1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n \n
q0<\/code>.<\/p>\n<\/li>\n1<\/code> is encountered, move to q1<\/code>.<\/p>\n<\/li>\nq1<\/code> if more 1<\/code>s are encountered.<\/p>\n<\/li>\n1<\/code> is accepted.<\/p>\n<\/li>\n<\/ul>\n\u00a0Example of NFA<\/strong><\/h3>\n
Language:<\/strong> Strings containing
10<\/code> as a substring.<\/h3>\nNFA Representation:<\/strong><\/h3>\n
{q0, q1, q2}<\/code>
Alphabet: {0,1}<\/code>
Start State: q0<\/code>
Final State: q2<\/code>
Transitions:<\/p>\n\n\n
\n State<\/th>\n Input 0<\/code><\/th>\nInput 1<\/code><\/th>\n<\/tr>\n<\/thead>\n\n\n q0<\/strong><\/td>\n q0<\/td>\n q1<\/td>\n<\/tr>\n \n q1<\/strong><\/td>\n q2<\/td>\n –<\/td>\n<\/tr>\n \n q2<\/strong><\/td>\n q2<\/td>\n q2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n \n
q0<\/code>.<\/p>\n<\/li>\n1<\/code> appears, move to q1<\/code>.<\/p>\n<\/li>\n0<\/code> follows, move to q2<\/code> (accepting state).<\/p>\n<\/li>\nq2<\/code>, stay there for any input.<\/p>\n<\/li>\n<\/ul>\n\u00a0GATE 1991 Question on Finite Automata<\/strong><\/h3>\n
(A)<\/strong> Every NFA can be converted to an equivalent DFA.
(B)<\/strong> NFA can recognize more languages than DFA.
(C)<\/strong> There exists an NFA for which no equivalent DFA exists.
(D)<\/strong> DFA and NFA recognize different sets of languages.<\/p>\n
Explanation:<\/strong> DFA and NFA are equivalent in power<\/strong>, meaning any language recognized by an NFA can also be recognized by a DFA. The subset construction algorithm<\/strong> is used to convert an NFA to an equivalent DFA.<\/p>\n\u00a0Summary<\/strong><\/h3>\n
DFA<\/strong> has unique transitions, while NFA<\/strong> allows multiple transitions.
Both DFA and NFA are equivalent in power<\/strong> but differ in complexity.
Used in compilers, text searching, and pattern recognition.<\/strong><\/p>\n\ud83e\udde0 Understanding the Problem<\/h3>\n
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\ud83d\udcda Detailed Solution<\/h3>\n