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Articles by Bhanu Priya
Page 81 of 106
Design a TM which recognizes palindromes over = {a, b}
AlgorithmStep 1 - If there is no input, reach the final state and halt.Step 2 - If the input = “a‟, then traverse forward to process the last symbol = “a‟. Convert both a‟s to B‟.Step 3 - Move left to read the next symbol.Step 4 - If the input = “b‟, replace it by B and move right to process its equivalent “B‟ at the rightmost end.Step 5 - Convert the last ’b’ to ‘B’.Step 6 - Move left and process step 2 – 5 until there are no more inputs to process.Step 7 - If the machine reaches ...
Read MoreExplain type 2 grammar with properties
Type 2 grammars are context free grammars (CFG).All productions are of the form −A → x — where A is nonterminal, x is a string of nonterminal and terminals, A context-free grammar is equivalent to a pushdown automaton (PDA) and to context free languages.Example − Pushdown Automaton (PDA)PropertiesA grammars, G = (V, T, P, S) is said to be context free if the production rule is of the form, A → α .The transition allows A → ε [i.e., α → ε] where, A is a non terminal symbol α is any terminal or non-terminal symbol.Here, the left hand side of ...
Read MoreConstruct a TM recognizing strings of the form an bn cn| n≥1 over = {a, b, c}
AlgorithmStep 1: Process the leftmost „a‟ and replace it by „x‟.Step 2: Move right until the leftmost „b‟ is reached. Replace it by „y‟.Step 3: Move right until the leftmost „c‟ is reached. Replace it by „z‟.Step 4: Move left to reach the leftmost „a‟ and perform steps 1, 2 and 3 (n – 1) times.Step 5: Halt if there are „n‟ number of x, y, z.Turing Machine for the given language is as follows −The Turing machine, M is given by M = (Q, Σ, Γ, δ, q0, B, F)Where, Q = {q0, q1, q2, q3, q4, q5}Σ = ...
Read MoreExplain formal definition of language with examples in TOC?
The set of all strings (over terminal symbols) which can be derived from the start symbol is the language generated by the grammar G.Example 1Let grammar G be defined by the set of terminals T = {a, b}, the only non-terminal start symbol S and the set of production rules. Hence, the grammar G would be as follows −S → ∧, S → aSbOr in shorthand, it is as mentioned below −S → ∧ | aSbL(G) = {∧, ab, aabb, aaabbb, . . . }DefinitionIf G is called as a grammar with start symbol S and set of terminals T, ...
Read MoreWhat is type 3 grammar? Explain its properties
Type 3 grammars are regular grammars that describe regular / formal languages.These grammars contain production rules consisting of the following −Only one non-terminal at the left hand side, The right hand side has a single terminal and may or may not be followed by non terminals.ExampleA → ε , A → a, A → b, A → aA etc.TypesThere are two types of regular grammars namely −Right linear / Right regular grammarLeft linear / Left regular grammarLet us learn about these two types of grammar in detail.Right linear grammarThis is a regular grammar with the production rules of the formA ...
Read MoreWhat are the basic properties of products in TOC?
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 MoreExplain the technique for combining two languages in TOC?
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 MoreDesign NPDA for accepting the language L = {am b(2m) | m>=1}
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 MoreConstruct Pushdown Automata for all length palindromes
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 MoreConstruct DPDA for anbncm where, n,m≥1 in TOC
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 ...
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