Computers Articles - Page 79 of 100

What are the basic properties of products in TOC?

Bhanu Priya
Updated on 15-Jun-2021 11:47:26

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It is easy to see that for any language L the following simple properties hold −L · {∧} = {∧} · L = LL · ∅ = ∅ · L = ∅Now let’s see the commutativity and associativity of the operation of concatenation.Properties of products – commutativityThe operation of concatenation is not commutative. In other words, the order matters!Given two languages L and M, it’s usually true thatL · M ≠ M · LExampleIf L = {ab, ac} and M = {a, bc, abc}, then the productL · M is the languageL · M = {aba, abbc, ababc, aca, acbc, ... Read More

Design NPDA for accepting the language L = {am b(2m) | m>=1}

Bhanu Priya
Updated on 15-Jun-2021 11:43:25

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Basically a push down automata (PDA) is as follows −“Finite state machine+ a stack”PDA has three components, which are as follows −An Input tape.A control unit.A Stack with infinite size.A PDA can be formally described as seven tuples(Q, Σ, S, δ, q0, I, F)Where, Q is finite number of statesΣ is input alphabetS is stack symbolΔ is the transition function: QX(Σ∪{e})XSXQq0 is the initial state (q0 belongs to Q)I is the initial state top symbolF is a set of accepting states (F belongs to Q)ProblemConstruct a non-deterministic PDA (NPDA) for accepting the languageL = {a^m b^{2m} | m>=1}.SolutionThe strings which ... Read More

Explain the technique for combining two languages in TOC?

Bhanu Priya
Updated on 15-Jun-2021 11:43:48

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There are three common ways of creating a new language from two languages −UnionIntersectionProductLanguages are sets of strings, so they can be combined by the usual set operations of union and intersection.IntersectionIf L1 and L2 are languages over ∑, then L1 ∩ L2 is the language of strings in both L1 and L2 .For example, If L = {aa, bb, ab} and M = {ab, aabb} then, The intersection is as follows −L ∩ M = {ab}AndUnionIf L1 and L2 are languages over the alphabet ∑, then the language L1 ∪ L2 is the language of all strings in at ... Read More

Construct Pushdown Automata for all length palindromes

Bhanu Priya
Updated on 15-Jun-2021 11:39:10

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L = {ww’ | wcw’, w={0, 1}*} where w’ is the reverse of w.This is the language of all palindromes, both odd and even over the alphabet {0, 1}.For the construction of all length palindromes, let us use the Non-deterministic push down automata (NPDA).To construct the wcw’ we need to check if the string is of odd length and if reaches the middle element ‘c’ then process it and move to the next state without making any change in stack.ExampleGiven string is 1 1 0 0 1 1 1 1 0 0 1 1Result − ACCEPTEDGiven string is: 1 0 ... Read More

Construct DPDA for anbncm where, n,m≥1 in TOC

Bhanu Priya
Updated on 15-Jun-2021 11:37:23

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DPDA is the short form for the deterministic push down automata (DPDA).ProblemConstruct DPDA for anbncm where m, n>=1SolutionSo, the strings which are generated by the given language are −L={abc, aabbc, aaabbbcc, ….}That is we have to count equal number of a’s, b’s and different number of c’sLet’s count the number of a's which is equal to the number of b's.This can be achieved by pushing a's in STACK and then we will pop a's whenever "b" comes.Then for c nothing will happen.Finally, at the end of the strings if nothing is left in the STACK then we can say that ... Read More

Construct Deterministic PDA for a^n b^n where n>=1

Bhanu Priya
Updated on 15-Jun-2021 11:34:40

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ProblemConstruct deterministic push down automata (DPDA) for anbn where n>=1.SolutionSo, the strings which are generated by the given language are as follows −L={ab, aabb, aaabbb, ….}That is we have to count equal number of a’s and b’sThis can be achieved by pushing a's in STACK and then we will pop a's whenever "b" comes.Finally at the end of the strings if nothing is left in the STACK then we can declare that language is accepted in the PDA.The transition diagram is as follows −Transition FunctionsThe transition functions are as follows −δ(q0, a, Z) = (q0, aZ)δ(q0, a, a) = (q0, ... Read More

Compare Push down automata and Linear bounded automata

Bhanu Priya
Updated on 15-Jun-2021 11:32:27

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Let us understand the push down automata (PDA) and the linear bounded automata (LBA) in the theory of computation (TOC).Push-Down AutomataA PDA can be formally described as seven tuples (Q, Σ, S, δ, q0, I, F)Where, Q is finite no of statesΣ is input alphabetS is stack symbolΔ is the transition function: QX(ΣU{e})XSXQq0 is the initial state (q0 belongs to Q)I is the initial state top symbolF is a set of accepting states (F belongs to Q)A Push-Down Automaton is a finite-state machine that is equipped with a memory device that functions as a push-down store.Push-down automata are equivalent to ... Read More

What is Context-sensitive Grammar?

Bhanu Priya
Updated on 15-Jun-2021 11:30:50

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The Context sensitive grammar (CSG) is defined as G=(V, Σ, P, S)Where, V: Non terminals or variables.Σ: Input symbols.P: Production rule.P:{αAβ → αγβ, A ϵ V, α ϵ (V∪Σ)*, β ϵ (V∪Σ)*S: Starting symbol.ExampleaS→SAa|aAaA→abcIn context sensitive grammar, there is either left context or right context (αAβ i.e. α is left context and β is right) with variables.But in context free grammar (CFG) there will be no context.For example in production ruleS →0 B S 2 ,B 0 → 0 BWe cannot replace B until we get B0.Therefore, CSG is harder to understand than the CFG.The CFG, CSG and the unrestricted ... Read More

Explain Operator grammar and precedence parser in TOC

Bhanu Priya
Updated on 15-Jun-2021 11:29:18

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If the grammar satisfies the following two conditions, then we can say that type of grammar is called as operator precedence grammar.If ε is on its RHS then there exists no production rule.If two non-terminals are adjacent to each other on its RHS then there exists no production rule.Operator Grammars have the property that no production right side is empty or has two adjacent non-terminals.ExampleE-> E A E | idA-> + | *The above grammar is not an operator grammar but we can convert that grammar into operator grammar like −E-> E + E | E * E | idThere ... Read More

How to convert CFG to Greibach Normal Form?

Bhanu Priya
Updated on 15-Jun-2021 11:26:20

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A Context Free Grammar (CFG) is said to be in Greibach Normal Form(GNF), if production rules satisfy one of the following criteria −Only a start symbol can generate ε. For example, if S is the start symbol then S → ε is in GNF.A non-terminal can generate a terminal. For example, if A is Non terminal and a is terminal then, A → a is in GNF.A non-terminal can generate a terminal followed by any number of non-terminals. For Example, S → aAS is in GNF.Case 1G1 = {S → aAB | aB, A → aA| a, B → bB ... Read More

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