A context-free grammar (CFG) consisting of a finite set of grammar rules is a quadruple (V, T, P, S)Where, V is a variable (non terminals).T is a set of terminals.P is a set of rules, P: V→ (V ∪ T)*, i.e., the left-hand sides of the production rule. P does have any right context or left context.S is the start symbol.By using the rules of any language, we can derive any strings in that language.For language a* CFG is as follows −S -> aSS -> ɛHere, S are the variables.a and ɛ terminals.S is the start symbol.By using these rules, ... Read More

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 More

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 More

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 More

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 More

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 More

If L1 and L2 are CFLs, then their union L1 + L2 is a CFL.Here CFL refers to Context Free Language.L1 CFL implies that L1 has a CFG, let it is 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 a CFG, Let it is CFG2, that generates it.Assume that the nonterminals in CFG2 are S, A, B, C, . . ..Change the nonterminal in CFG2 to S2, A2, ... Read More

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 More

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 More

A Deterministic Finite automata (DFA) is a five tuplesM=(Q, ∑, δ, q0, F)Where, Q − Finite set called states.∑ − Finite set called alphabets.δ − Q × ∑ → Q is the transition function.q0 ∈ Q is the start or initial state.F − Final or accept state.Let’s see the worst case number of states in DFA for the language A∩B and A*Let A and B be the two states, |A| = number of states = nA|B| = number of states = nBDFA = |A∩B|   =nA.nB|A ∪ B| =nA.nB|A*|=3/4 2nA|AB| = nA (2nB-2nB-1)NFAThe non-deterministic finite automata (NFA) also have five states ... Read More