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Articles by Bhanu Priya
Page 92 of 106
Design a push down automaton for L = {wwR | w ∈ {a, b}+}?
A pushdown automaton is used to implement a context-free grammar in the same way that we use a technique to design DFA for a regular grammar. A DFA work on a finite amount of information, where as a PDA works on an infinite amount of information.Generally, a pushdown automaton is −"Finite state machine" + "a stack"A pushdown automaton consist of three components −an input tape, a control unit, anda stack with infinite size.Now consider a problem that how to design push down automata for a given language −ProblemDesign a push down automaton which recognizes even length palindromes for L = ...
Read MoreWhat happens when a String is accepted or rejected by NPDA?
A string is accepted by an Non-deterministic Push down Automata (NPDA), if there is some path (i.e., sequence of instructions) from the start state to a final state that consumes all the letters of the string. Otherwise, the string is rejected by the NPDA.The language of an NPDA is the set of all strings that it accepts.An input string rejected by the NPDA under following conditions −If reading an input string finishes without reaching a final state.If for a current state/symbol on the stack/input symbol there is no transition.If it attempts to pop the empty stack.ExampleBuild an NPDA which recognises ...
Read MoreWhat are the restrictions of regular grammar?
A regular grammar is the one where each production takes one of the following restricted forms −B → ∧, B → w, B → A, B → wA.(Where A, B are non-terminals and w is a non-empty string of terminals.)Restrictions of regular grammarOnly one nonterminal can appear on the right-hand side of a production.Nonterminal must appear on the right end of the right-hand side.Therefore, the productions are as follows −A → aBc and S → TUThese are not part of a regular grammar, but the production A → abcA is.Things like A → aB|cC are allowed because they are actually ...
Read MoreDesign a DFA that accepts at most 3 a"s
Construct deterministic finite automata that accepts at most 3 a’s over an alphabet ∑={a,b}.At most 3 a’s means,The string contains 0 to max 3 a’s and any number of b’s.L= {Є,a,aa,aaa,ab,abb,bab,bbabaa, bbabaabbb,…..}Construct DFALet’s construct DFA step by step −Step 1Valid inputs − aaa, a, aa,ε .Step 2Valid inputs − b, ba, baa, baaa, bb, bba, bbba,…Step 3Valid input − bab, abba, abbbaa, babba,…Step 4Valid inputs − babab, aabb, aaba, bbbaaba, …Step 5Valid inputs − aaabbb, aaabab, baaaba, …Step 6InValid inputs − aaaa, aaabab, baaaba,
Read MoreExplain about CYK Algorithm for Context Free Grammar
CKY means Cocke-Kasami-Younger. It is one of the earliest recognition and parsing algorithms. The standard version of CKY can only recognize languages defined by context-free grammars in Chomsky Normal Form (CNF).It is also possible to extend the CKY algorithm to handle some grammars which are not in CNF (Hard to understand).Based on a “dynamic programming” approach −Build solutions compositionally from sub-solutionsIt uses the grammar directly.AlgorithmBegin for ( i = 1 to n do ) Vi1 { A | A → a is a production where i th symbol of x is a } for ( j = ...
Read MoreDesign a DFA accepting stringw so that the second symbol is zero and fourth is 1
ProblemConstruct DFA which accepts a string that contains second symbol is zero and fourth symbol is 1 over an alphabet ∑={0,1}.SolutionInput − 00110Output is accepted; because in the given string the second symbol is ‘0’ and the fourth symbol is ‘1’.Input − 11001Output − string is not accepted, because the second symbol is not ‘0’.Design DFA step by step as given below −Step 1 -Valid inputs − 0001Step 2 -Valid input − 1001Step 3 -Valid inputs − 0011, 1011Step 4 -Valid inputs − 00010, 10010, 00110, 00011, 10011, 00111, …Step 5 -Invalid inputs − 0101, 0100, 0010, 1100, 0000, 1000, …Step 6 -Valid inputs − 01010, 01000, 11111, 0100000, …
Read MoreExplain the intersection process of two DFA’s
According to the theorem, If L and M are two regular languages, then L ∩ M is also regular language.ExampleConstruct A∩B where A and B is given as follows −The language A ={10, 100, 00, 001, 1010, …..}The language B ={01, 1010, 10, 101, …..}AA = (QA, Σ, δA, qa, FA) AB = (QB, Σ, δB, qB, FB) A∩B=(QA x QB ,Σ, δ(qA x qB ,FA x F B )Where, δ(( p, q), a) =δL (p, a), δM (q, a))Here, QA x QB = {p, q} x {r, s} ={(p, r), (p, s), (q, r), (q, s)} Z = ...
Read MoreDesign a DFA accepting a language L having number of zeros in multiples of 3
ProblemConstruct a deterministic finite automata (DFA) that accepts a language L which has the number of zero’s is of multiple of 3 over an alphabet ∑=”{0,1}.SolutionIf input is: 000 Output is: string is acceptedBecause here the number of zero’s is multiple of 3.Designing DFAIn order to construct the DFA, follow the below mentioned steps −Step 1 -Valid inputs: 000, 000000, 09 , 012 , …Step 2 -Valid inputs: 1, 1000, 100000, …Step 3 -Valid inputs: 10100, 11000, 101100, …Step 4 -101010, 1101010, 1101110110, …Invalid inputs − 0,00,10000,01011, …
Read MoreConstruct DFA beginning with ‘a’ but does not have substring ‘aab’
ProblemGiven language to construct the deterministic finite automata (DFA) is, the strings start with ‘a’ but not contain substring ‘aab’ over alphabet ∑={a,b}.SolutionIf the input is: “baabba”The output is: string is not acceptedBecause the string does not start with ‘a’, and generating a substring ‘abb’,DFA transition diagramThe DFA transition diagram for the string beginning with ‘a’ but not having the substring as ‘aab’ is as follows −Transition tableThe transition table is as follows −STATEINPUT (a)INPUT (b)→ 01*4 (dead state)1*2*3*2*2*4 (dead state)3*1*3*4 (dead state)4 (dead state)4 (dead State)
Read MoreProve that the polynomial time reduction is from the Clique problem to the Vertex Cover problem
Vertex cover is a subset of vertices that covers all the edges in a graph. It is used to determine whether a given graph has a 3SAT to vertex cover.Clique is called a subset of vertices that are all directly connected. It determines whether a clique of size k exists in a graph.To prove − Vertex cover can be reduced to clique.ProofGiven a graph G=(V, E) and integer k.Get its complement graph G'=(V, E').Solve CLIQUE(G', |V|-k).If there is a solution, return yes. Otherwise, it returns as no.To prove this reduction, we need to show the following −If there is a ...
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