A Recursively enumerable language is the language that accepts every string otherwise not. If a language that halt on every string, then we call it as recursive language.ProblemWe need to prove that the set of all languages that are not recursively enumerable on {a} is not countable.First let see what the recursive enumerable language is −Recursive Enumerable LanguageA language L is recursively enumerable if L is the set of strings accepted by some TM.If L is a recursive enumerable language then −if w ∈ L then a TM halts in a final state, if w ∉ L then a TM ... Read More

ProblemWe have to prove that the cartesian product of a finite number of countable sets is countable.SolutionLet the X1, X2 ,…….. Xn be the countable sets.Yk= X1 * X2 * …….* Xk when k =1……. N). Thus, Yn := X1 * X2 * · · · * XnProofUsing the induction −In case k = 1 then Y1 = X1 is countable.Assuming that Yk (k ∈ n, 1 ≤ k < n) is countable;Then Yk+1 = ( X1 * X2 * …….* Xk) * Xk+1 = Yk * Xk+1 where the Yk and the Xk+1 can be called countable. Hence the ... Read More

To begin with, let us learn about the P class in the theory of computation (TOC).P-ClassThe class P consists of those problems that are solvable in polynomial time, i.e. these problems can be solved in time O(n k) in the worst-case, where k is constant.These types of problems are called tractable and others are called intractable or super polynomial.Generally, an algorithm is a polynomial time algorithm, if there exists a polynomial p(n) such that the algorithm can solve any instance of size n in a time O(p(n)).Problems requiring Ω(n 50) time to solve are essentially intractable for large n. Most ... Read More

ProblemExplain the pumping lemma for context free language by showing that the language of strings in the form xnynzn is not a context free language.SolutionPumping lemma (Context free grammar)We can prove that a particular language is not context free grammar using pumping lemma.Let’s take the concept of proof by contradictionHere we assume that language is CFGConditions of pumping lemmaFirst of all consider a string and split into 5 parts those are pqrst it must satisfy the following conditions −|qs|>=1|qrs|=n (“ n” is pumping length)pqirsit € L for different values of iLet the L be the CF language.Now we can take ... Read More

The problems for which we can’t construct an algorithm that can answer the problem correctly in the infinite time are termed as Undecidable Problems in the theory of computation (TOC).A problem is undecidable if there is no Turing machine that will always halt an infinite amount of time to answer as ‘yes’ or ‘no’.ExamplesThe examples of undecidable problems are explained below. Here, CFG refers to Context Free Grammar.Whether two CFG L and M equal − Since, we cannot determine all the strings of any CFG, we can predict that two CFG are equal or not.Given a context-free language, there is ... Read More

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