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Data Structure Algorithms Articles
Page 4 of 24
Differentiate between recognizable and decidable in the Turing machine?
When we talk about Turing machines (TM) it could accept the input, reject it or keep computing which is called loop.Now a language is recognizable if and only if a Turing machine accepts the string, when the provided input lies in the language.Also, a language can be recognizable if the TM either terminates and rejects the string or doesn't terminate at all. This means that the TM continues with the computing when the provided input doesn't lie in the language.Whereas, the language is decidable if and only if there is a machine which accepts the string when the provided input ...
Read MoreProve the equalities of regular expressions by applying properties?
ProblemProve each of the following equalities of regular expressions.a. ab*a(a + bb*a)*b = a(b + aa*b)*aa*b.b. b + ab* + aa*b + aa*ab* = a*(b + ab*).SolutionProblem 1Prove that ab*a(a + bb*a)*b = a(b + aa*b)*aa*b.Let’s take LHS , = ab*a(a + bb*a)*b Use property of (a+b)* = a*(ba*)* = ab*a (a* ((bb*a) a* )* a*b = ab* a (a*bb*a)* a*b {Associative property} = ab* (a (a*bb*a)*)a*b = ab*(aa*bb*)*aa*b = a (b*(aa*bb*)*)aa*b Use property a* (ba*)*= (a+b)* = a(b+aa*b)*aa*b = RHS Hence provedProblem 2Prove that b + ab* + aa*b + aa*ab* ...
Read MoreShow that the set of recursive languages is closed under reversal?
Consider a Language L, over an alphabet T is known as recursive enumerable if there exists a turing machine (TM) which generates a sequence of numbers T* which have precisely the members of L.Whereas L is said to be recursive if there exists a Turing Machine enlisting all members of L and stopping on each member of L as the input.Thus it is clear from the above statements that every recursive language is also recursively enumerable but the converse is not true.The precise connection between families of languages is given below.TheoremStep 1 − A language L is said to be ...
Read MoreExplain the Closure Under Kleene Star of CFL in TOC?
If L is a CFL, then L*is a CFL. Here CFL refers to Context Free Language.StepsLet CFG for L has nonterminal S, A, B, C, . . ..Change the nonterminal from S to S1.We create a new CFG for L* as follows −Include all the nonterminal S1, A, B, C, . . . from the CFG for L.Include all productions of the CFG for L.Add new nonterminal S and new productionS → S1S | ∧We can repeat last productionS → S1S → S1S1S → S1S1S1S → S1S1S1S1S → S1S1S1S1∧ → S1S1S1S1Note that any word in L* can be generated by ...
Read MoreExplain the context free language closure under concatenation?
Here CFL refers to Context Free Language. Now, let us understand closure under concatenation.Closure under ConcatenationsIf L1 and L2 are CFLs, then L1L2 is a CFL.Follow the steps given below −L1 CFL implies that L1 has CFG1 that generates it.Assume that the nonterminals in CFG1 are S, A, B, C, . . ..Change the nonterminal in CFG1 to S1, A1, B1, C1, . . ..Don’t change the terminals in the CFG1.L2 CFL implies that L2 has CFG2 that generates it.Assume that the nonterminals in CFG2 are S, A, B, C, . . ..Change the nonterminal in CFG2 to S2, A2, ...
Read MoreGenerate a Context-free grammar for the language L = {anbm| m≠n}?
A context-free grammar is a quadruple G = (N, T, P, S), Where, N is a finite set of nonterminal symbols, T is a finite set of terminal symbols, N ∩ T = ∅, P is a finite set of productions of the form A → α, Where A ∈ N, α ∈ (N ∪ T)*, S is the start symbol, S ∈ N.Construct a Context free grammar for the language, L = {anbm| m ≠n}Case 1n > m − We generate a string with an equal number of a’s and b’s and add extra a’s on the left −S ...
Read MoreGive implementation-level descriptions of a Turing machine?
A Turing machine (TM) can be formally described as seven tuples −(Q, X, ∑, δ, q0, B, F)Where, Q is a finite set of states.X is the tape alphabet.∑ is the input alphabet.δ is a transition function:δ𝛿:QxX->QxXx{left shift, right shift}.q0 is the initial state.B is the blank symbol.F is the final state.A Turing machine T recognises a string x (over ∑) if and only when T starts in the initial position and x is written on the tape, T halts in a final state.T is said to recognize a language A, if x is recognised by T and if and ...
Read MoreExplain Type-1 grammar in TOC
Chomsky Hierarchy represents the class of languages that are accepted by the different machines.Chomsky hierarchyHierarchy of grammars according to Chomsky is explained below as per the grammar types −Type 0. Unrestricted grammars Turing Machine (TM)Type 1. Context-sensitive grammars Linear Bounded Automaton (LBA)Type 2. Context-free grammars Pushdown Automaton (PDA)Type 3. Regular grammars Finite Automaton (FA)Type-1 Context Sensitive Grammar (CSG)Type 1 grammar is also known as context sensitive grammarThe context sensitive grammar is used to represent context sensitive languageThe CSG follows some rules, which are as follows −The context sensitive grammar may have more than one symbol on the left hand side ...
Read MoreExplain the balancing parenthesis of PDA
Pushdown Automata (PDA) are the finite automata (FAs), but with the ability to push and pop symbols to/from a stack.PDA accepts strings if there is a legal path from start state to acceptance state for input. Otherwise, the string is rejected.A PDA can be represented by a 7-tuple(Q, ∑, ℾ, q0, ha, ∆, δ)WhereThe PDA is to finite subsets of Q ☓ (ℾ ∪ {∆})*.Parentheses are balanced ifWhile reading string, number of opening parentheses >= number of closing parentheses.When string is read, number of opening parentheses = number of closing parentheses.Examples(())() − Balanced((()() − Not balanced)()(() − Not balancedThe context ...
Read MoreDesign a PDA which recognizes the language
ProblemGenerate the push down automata (PDA) that recognizes the language E={aibj| i is not equal to j and I is not equal to 2j}.SolutionConsider the two languages as given below −L1={aibj|i,j>=0 and i>2j}L2={aibj|i,j>=0 and iaA A->aaAb|aA|epsilonIn L2, the number of a's are less than double the number of b'sSo the CFG for L2 becomes as follows − S2->Bb|aBb B->Bb|aBb|aaBb|epsilon S->S1|S2L1: {aibj:i>2j}L2:{aibj: i
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