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Data Structure Algorithms Articles
Page 10 of 24
Explain Union process in DFA
The union process in the deterministic finite automata (DFA) is explained below −If L1 and If L2 are two regular languages, their union L1 U L2 will also be regular.For example, L1 = {an | n > O} and L2 = {bn | n > O}L3 = L1 U L2 = {an U bn | n > O} is also regular.ProblemDesign a DFA over an alphabet {a, b} where the start and end are of different symbols.SolutionThere are two different types of languages are formed for a given condition −L1={ab, aab, abab, abb, …….}L1={ab, aab, abab, abb, …….}Here, L1= starts ...
Read MoreConstruct DFA of a string in which the second symbol from RHS is ‘a’
A Deterministic Finite automaton (DFA) is a 5-tuplesM=(Q, Σ, δ, q0, F)Where, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemDesign a finite automaton where the second symbol from the right hand side is ‘a’.SolutionThe language for a given string over an alphabet {a, b} is −L={aa, abaa, abbab, aaabab, ………}ExampleInput − aabaOutput − Not acceptedBecause the second letter from right hand side is not aInput − aabbabOutput − AcceptedIt is complicated to directly construct ...
Read MoreConstruct an NFA accepting strings with an even number of 0s or an odd number of 1s
Non-deterministic finite automata (NFA) also have five states which are same as DFA, but with different transition function, as shown follows −δ: Q X Σ → 2QWhere, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemConstruct NFA over an alphabet Σ={0, 1}.SolutionDesign two separate machines for the two conditions, as given below −NFA accepting only odd number of 1’sNFA accepting only even number of 0’sNFA accepting only odd number of 1’s over an alphabet Σ= ...
Read MoreDesign a DFA machine accepting odd numbers of 0’s or even numbers of 1’s
A Deterministic Finite automaton (DFA) is a five tuplesM=(Q, Σ, δ, q0, F)Where, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemConstruct a DFA machine accepting odd numbers of 0’s or even numbers of 1’s.SolutionDesign two separate machines for the two conditions over an alphabet Σ={0, 1} −DFA accepting only odd number of 0’sDFA accepting only even number of 1’sDFA accepting only odd number of 1’s over an alphabet Σ={0, 1} The language L= ...
Read MoreDesign a DFA of a string with at least two 0’s and at least two 1’s
A Deterministic Finite automaton (DFA) is a 5-tuplesM=(Q, Σ, δ, q0, F)Where, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.ProblemConstruct DFA of a string with at least two 0’s and at least two 1’s.SolutionThe language generated based on the given condition over the alphabet Σ ={0, 1) is −L={0011, 001011, 0001010, 0011001, 010101, ……}The given language accepts at least two zero’s means it can accept two or more than two zero’s and at least ...
Read MoreConstruct DFA for strings not ending with "THE"
A Deterministic Finite automaton (DFA) is a five tuplesM=(Q, Σ, δ, q0, F)Where, Q : Finite set called states.Σ : Finite set called alphabets.δ : Q × Σ → Q is the transition function.q0 ϵ Q is the start or initial state.F : Final or accept state.Accept Strings which are not ending with "THE"Observe whether the given string is ending with “the” or not.The different notations of “the” which are avoided in the end of the string are as follows − "tHE", "thE", "THE", "ThE", "THe", "The", "tHe" and "the"These all strings are not accepted from alphabet (A-Z)Let the ...
Read MoreExplain the operation of DFA with block diagram in TOC
The finite automata (FA) is represented as follows in the theory of computation (TOC) −Components of FAThe different components of finite automata are as follows −Input tapeThe input tape has a left end and extends to the right end.It is divided into blocks and each block containing a single symbol from the input alphabet Σ.The end block of the tape contains the end markers ₵ at the left end and the end marker $ at the right end.The absence of end markers indicates that the tape is of infinite length.The left-to-right sequence of symbols between the two end markers in ...
Read MoreDesign a TM that perform right shift over ∑ = {0, 1}
ProblemShift the input string right by one place and design a Turing Machine (TM) that can perform right shift over ∑ = {0, 1}.SolutionRefer an algorithm given below to design a TM −Step 1 − Move right side to the last character from the initial character.Step 2 − If the character is „0‟, replace it with „B‟ and move one step right to replace the immediate „B‟ to „0‟.Step 3 − If the character is „1‟, replace it with „B‟ and move one step right to replace the immediate „B‟ to „1‟.Step 4 − After processing as above, move left one ...
Read MoreDesign a TM that increments a binary number by 1
Turing’s machine is more powerful than both finite automata and pushdown automata. These machines are as powerful as any computer we have ever built.Formal Definition of Turing MachineA Turing machine can be formally described as seven tuples(Q, X, Σ, δ, q0, B, F)Where, Q is a finite set of states.X is the tape alphabetΣ is the input alphabetδ is a transition function: δ:QxX→QxXx{left shift, right shift}q0 is the initial stateB is the blank symbolF is the final state.A Turing Machine (TM) is a mathematical model which consists of an infinite length tape divided into cells on which input is given. ...
Read MoreDesign Turing Machine to reverse string consisting of a’s and b’s
Our aim is to design a Turing machine (TM) to reverse a string consisting of a’s and b’s over an alphabet {a, b}.ExampleInput − aabbabOutput − babbaaAlgorithmStep 1: Move to the last symbol, replace x for a or x for b and move right to convert the corresponding B to „a‟ or „b‟ accordingly. Step 2: Move left until the symbol left to x is reached. Step 3: Perform step 1 and step 2 until „B‟ is reached while traversing left. Step 4: Replace every x to B to make the cells empty since the reverse ...
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