For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. 34. Why a stack? Which combination below expresses all the true statements about G? Differentiate recursive and non-recursively languages. -NFAInput string Accept/reject 2 A stack filled with “stack symbols” equiv is any set containing a final state of ND because a string takes M equiv to such a set if and only if it can take ND to one of its final states. Also construct the derivation tree for the string w. (8) c)Define a PDA. Nondeterminism can occur in two ways, as in the following examples. Pda 1. The stack is empty. But, it also implies that it could be the case that the string is impossible to derive. G produces all strings with equal number of a’s and b’s III. 87. 47. FA to Reg Lang PDA is to CFL FA to Reg Lang, PDA is to CFL PDA == [ -NFA + “a stack” ] Wh t k? To convert this to an empty stack acceptance PDA, I add the two states, one before the previous start state, and another state after the last to empty the stack. 46. This is not true for pda. Thereafter if 2’s are finished and top of stack is a 0 then for every 3 as input equal number of 0’s are popped out of stack. Accepted Language & Decided Language - A TM accepts a language if it enters into a final state for any input string w. A language is recursively enumerable (generated by Type-0 grammar) if it is acce Notice that string “acb” is already accepted by PDA. Since pda languages are closed under union it su ces to construct a pda for the language f x˙1y˙2z j x;y;z 2 fa;bg ;jxj = jzj;˙1;˙2 2 fa;bg;˙1 6= ˙2 g. 5 As a consequence, the DPDA is a strictly weaker variant of the PDA and there exists no algorithm for converting a PDA to an equivalent DPDA, if such a DPDA exists. 49. In this NPDA we used some symbol which are given below: If some 2’s are still left and top of stack is a 0 then string is not accepted by the PDA. (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. We will show conversion of a PDA accepting L by final state into another PDA that accepts L by empty stack, and vice-versa. Each transition is based on the current input symbol and the top of the stack, optionally pops the top of the stack, and optionally pushes new symbols onto the stack. ` (4) 19.G denotes the context-free grammar defined by the following rules. In both these definitions, we employ the notions of instanta- neous descriptions (ID), and step relations $, as well as its reflexive and transitive closure, $ ∗. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. Give an example of undecidable problem? So in the end of the strings if nothing is left in the STACK then we can say that language is accepted in the PDA. w describes the remaining input. 1.1 Acceptance by Final State Let P = (Q,Σ,Γ,δ,q0,Z0,F) be a PDA. (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. An instantaneous description is a triple (q, w, α) where: q describes the current state. 50. We now show that this method of constructing a DFSM from an NFSM always works. string w=aabbaaa. 1 (2) Use your PDA from question 1 and the method to convert a PDA to a CFG to form an equivalent CFG. Define RE language. Differentiate 2-way FA and TM? Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. Then L(P), the language accepted by P by final state, is L(P) = {w|(q0,w,Z0) ∗ ` (q, ,α)} for some state q ∈ F and any stack string α. (1) L={ anbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. The language accepted by a PDA M, L(M), is the set of all accepted strings. Each input alphabet has more than one possibility to move next state. Example 1 : This DFA accepts {} because it can go from the initial state to the accepting state (also the initial state) without reading any symbol of the alphabet i.e. We have designed the PDA for the problem: STACK Transiton Function δ(q0, a, Z) = (q0, aZ) δ(q0, a, a) = (q0, aa) δ(q0, b, a) = (q1, ε) δ(q1, b, a) = (q1, ε) δ(q1, ε, Z) = (qf, Z) Note: qf is Final State. If string is finished and stack is empty then string is accepted by the PDA otherwise not accepted. If the simulation ends in an accept state, . Formal Definition. 48. α describes the stack contents, top at the left. You must be logged in to read the answer. Login. Classify some properties of CFL? Step-1: On receiving 0 push it onto stack. Classify some closure properties of CFL? PDA - the automata for CFLs What is? Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. An input string is accepted if after the entire string is read, the PDA reaches a final state. Go ahead and login, it'll take only a minute. Simulate on input . Problem – Design a non deterministic PDA for accepting the language L = {: m>=1}, i.e., L = {abb, aabbbb, aaabbbbbb, aaaabbbbbbbb, .....} In each of the string, the number of a’s are followed by double number of b’s. is an accepting computation for the string. The states q2 and q3 are the accepting states of M. The null string is accepted in q3. Give an Example for a language accepted by PDA by empty stack. It's important to mention that the stack contents are irrelevant to the acceptance of the string. 89. And finally when stack is empty then the string is accepted by the NPDA. Our First PDA Consider the language L = { w ∈ Σ* | w is a string of balanced digits } over Σ = { 0, 1} We can exploit the stack to our advantage: Whenever we see a 0, push it onto the stack. i j b, C pop k b, C push(D) i j Λ, C pop k b, C push(D) Acceptance: A string w is accepted by a PDA if there is a path from the start state to a final state such that the input symbols on the path edges concatenate to w. Otherwise, w is rejected. 43. I only I and III only II and III only I, II and III. Whenever we see a 1, pop the corresponding 0 from the stack (or fail if not matched) When input is consumed, if the stack is empty, accept. The stack is emptied by processing the b’s in q2. The state diagram of the PDA is q0 q1 q3 q2 M : aλ/A However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. 88. So we require a PDA ,a machine that can count without limit. 2. 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