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. Classify some techniques for Turing machine construction? 87. PDA accepts a string when, after reading the entire string, the PDA has emptied its stack. Explanation – Here, we need to maintain the order of a’s and b’s.That is, all the a’s are are coming first and then all the b’s are coming. In both these definitions, we employ the notions of instanta- neous descriptions (ID), and step relations $, as well as its reflexive and transitive closure, $ ∗. 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. When is a string accepted by a 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 Define – Pumping lemma for CFL. 48. Classify some properties of CFL? We define these notions in Sections 14.1.2 and 14.1.3. Notice that string “acb” is already accepted by PDA. So, x'r = (01001)r = 10010. (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. This does not necessarily mean that the string is impossible to derive. 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. 43. G produces all strings with equal number of a’s and b’s III. Why a stack? Elaborate multihead TM. 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. The class of nondeterministic pda accept Context Free Languages [student op. However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. The state diagram of the PDA is q0 q1 q3 q2 M : aλ/A 33.When is a string accepted by a PDA? ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected. When we say a problem is decidable? Nondeterminism can occur in two ways, as in the following examples. Differentiate PDA acceptance by empty stack method with acceptance by final state method. Part B – (5 × = marks) 11 (a) Design a DFA accept the following strings over the alphabets {0, 1}. We will show conversion of a PDA accepting L by final state into another PDA that accepts L by empty stack, and vice-versa. (d) the set of strings over the alphabet {a, b} containing at least three occurrences of three consecutive b's, overlapping permitted (e.g., the string bbbbb should be accepted); (e) the set of strings in {O, 1, 2} * that are ternary (base 3) representa­ tions, leading zeros permitted, of numbers that are not multiples of four. Acceptance by empty stack only or final state only is addressed in problems 3.3.3 and 3.3.4. Differentiate recursive and non-recursively languages. ` (4) 19.G denotes the context-free grammar defined by the following rules. The stack is emptied by processing the b’s in q2. 34. Simulate on input . So, the given PDA is accepting all strings of of the form x0x'r or x1x'r or xx'r, where x'r is the reverse of the 1's complement of x. The language accepted by a PDA M, L(M), is the set of all accepted strings. 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. The input string is accepted by the PDA if: The final state is reached . The empty stack is our key new requirement relative to finite state machines. ` 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. Initially, the stack holds a special symbol Z 0 that indicates the bottom of the stack. When is a string accepted by a PDA? 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. Pushdown Automata A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. And finally when stack is empty then the string is accepted by the NPDA. string w=aabbaaa. THEOREM 4.2.1 Let L be a language accepted by a … Also construct the derivation tree for the string w. (8) c)Define a PDA. An instantaneous description is a triple (q, w, α) where: q describes the current state. G can be accepted by a deterministic 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 So we require a PDA ,a machine that can count without limit. is an accepting computation for the string. The language acceptable by the final state can be defined as: 2. Formal Definition. Define RE language. 88. The examples that we generate have very few states; in general, there is so much more control from using the stack memory. 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. - define], while the deterministic pda accept a proper subset, called LR-K languages. F3: It is known that the problem of determining if a PDA accepts every string is undecidable. 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. Not all context-free languages are deterministic. 47. 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. The stack is empty.. Give examples of languages handled by 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 (2) Use your PDA from question 1 and the method to convert a PDA to a CFG to form an equivalent CFG. Each input alphabet has more than one possibility to move next state. Give an example of undecidable problem? I only I and III only II and III only I, II and III. It's important to mention that the stack contents are irrelevant to the acceptance of the string. That is, the language accepted by a DFA is the set of strings accepted by the DFA. 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. State the pumping lemma for CFLs 45. 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. For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. 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. An input string is accepted if after the entire string is read, the PDA reaches a final state. Pda 1. Hence option B is correct. 46. 1.1 Acceptance by Final State Let P = (Q,Σ,Γ,δ,q0,Z0,F) be a PDA. You must be logged in to read the answer. Whenever the inner automaton goes to the accepting state, it also moves to the empty-stack state with an $\epsilon$ transition. w describes the remaining input. Which combination below expresses all the true statements about G? 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’. language of strings of odd length is regular, and hence accepted by a pda. 49. Give an Example for a language accepted by PDA by empty stack. 2. If string is finished and stack is empty then string is accepted by the PDA otherwise not accepted. Classify some closure properties of CFL? 2 Example. So, x0 is done, with x = 10110. 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. 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 the simulation ends in an accept state, . The states q2 and q3 are the accepting states of M. The null string is accepted in q3. Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. Go ahead and login, it'll take only a minute. Login Now 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. by reading an empty string . Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. The language of strings accepted by a deterministic pushdown automaton is called a deterministic context-free language. Explain your steps. (1) L={ anbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. PDA - the automata for CFLs What is? So we require a PDA ,a machine that can count without limit. 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