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. - define], while the deterministic pda accept a proper subset, called LR-K languages. language of strings of odd length is regular, and hence accepted by a pda. PDA accepts a string when, after reading the entire string, the PDA has emptied its stack. Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. 2. ` (4) 19.G denotes the context-free grammar defined by the following rules. An input string is accepted if after the entire string is read, the PDA reaches a final state. You must be logged in to read the answer. Login Now If string is finished and stack is empty then string is accepted by the PDA otherwise not accepted. Explain your steps. For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. 88. PDA - the automata for CFLs What is? It's important to mention that the stack contents are irrelevant to the acceptance of the string. The language accepted by a PDA M, L(M), is the set of all accepted strings. When is a string accepted by a PDA? Hence option B is correct. Nondeterminism can occur in two ways, as in the following examples. 87. Notice that string “acb” is already accepted by PDA. 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. Give an Example for a language accepted by PDA by empty stack. When we say a problem is decidable? 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. 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, Z) = (q0, bZ) δ(q0, b, b) = (q0, bb) δ(q0, b, a) = (q0, ε) δ(q0, a, b) = (q0, ε) δ(q0, ε, Z) = (qf, Z) Note: qf is Final State. Why a stack? Define RE language. Answer to A PDA is given below which accepts strings by empty stack. is an accepting computation for the string. ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected. G can be accepted by a deterministic PDA. The empty stack is our key new requirement relative to finite state machines. Differentiate PDA acceptance by empty stack method with acceptance by final state method. This is not true for pda. 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 Formal Definition. The examples that we generate have very few states; in general, there is so much more control from using the stack memory. So, x'r = (01001)r = 10010. ` S->ASB/ab/SS A->aA/A B->bB/A (i)Give a left most derivation of aaabb in G. Draw the associated parse tree. Part B – (5 × = marks) 11 (a) Design a DFA accept the following strings over the alphabets {0, 1}. 89. If the simulation ends in an accept state, . 46. THEOREM 4.2.1 Let L be a language accepted by a … -NFAInput string Accept/reject 2 A stack filled with “stack symbols” But, it also implies that it could be the case that the string is impossible to derive. w describes the remaining input. State the pumping lemma for CFLs 45. The stack is empty.. Give examples of languages handled by PDA. This does not necessarily mean that the string is impossible to derive. If it ends DFA A MBwB w Bw accept Theorem Proof in a Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. Login. Step-1: On receiving 0 push it onto stack. α describes the stack contents, top at the left. Initially, the stack holds a special symbol Z 0 that indicates the bottom of the stack. Which combination below expresses all the true statements about G? 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. 44. Pushdown Automata A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. The given string 101100 has 6 letters and we are given 5 letter strings. In both these deﬁnitions, we employ the notions of instanta- neous descriptions (ID), and step relations $, as well as its reﬂexive and transitive closure, $ ∗. 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. That is, the language accepted by a DFA is the set of strings accepted by the DFA. 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. So we require a PDA ,a machine that can count without limit. Acceptance by empty stack only or final state only is addressed in problems 3.3.3 and 3.3.4. 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. 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. (1) L={ anbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. Elaborate multihead TM. F3: It is known that the problem of determining if a PDA accepts every string is undecidable. If some 2’s are still left and top of stack is a 0 then string is not accepted by the PDA. The stack is emptied by processing the b’s in q2. 49. The stack is empty. The states q2 and q3 are the accepting states of M. The null string is accepted in q3. Also construct the derivation tree for the string w. (8) c)Define a PDA. The input string is accepted by the PDA if: The final state is reached . In this type of input string, at one input has more than one transition states, hence it is called non deterministic PDA and input string contain any order of ‘a’ and ‘b’. (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. Define – Pumping lemma for CFL. Classify some techniques for Turing machine construction? Can be applied to DFA, NFA, REX, PDA, CFG, TM, Informatik Theorie II (A) WS2009/10 acs-07: Decidability 4 4.1 is a decidable language ="On input , , where is a DFA and is a string: 1. The class of nondeterministic pda accept Context Free Languages [student op. So we require a PDA ,a machine that can count without limit. string w=aabbaaa. So in the end of the strings if nothing is left in the STACK then we can say that CFL is accepted in the 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 When is a string accepted by a PDA? 50. However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. Differentiate recursive and non-recursively languages. Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. 1 (2) Use your PDA from question 1 and the method to convert a PDA to a CFG to form an equivalent CFG. In this NPDA we used some symbol which are given below: Acceptance by Final State: The PDA is said to accept its input by the final state if it enters any final state in zero or more moves after reading the entire input. 2 Example. 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. An instantaneous description is a triple (q, w, α) where: q describes the current state. We deﬁne these notions in Sections 14.1.2 and 14.1.3. 47. 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. G produces all strings with equal number of a’s and b’s III. Simulate on input . The input string is accepted by the PDA if: The final state is reached . 48. 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. 33.When is a string accepted by a PDA? 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