Simulate on input . You must be logged in to read the answer. string w=aabbaaa. Define – Pumping lemma for CFL. In this NPDA we used some symbol which are given below: The states q2 and q3 are the accepting states of M. The null string is accepted in q3. An input string is accepted if after the entire string is read, the PDA reaches a final state. The language of strings accepted by a deterministic pushdown automaton is called a deterministic context-free language. The language acceptable by the final state can be defined as: 2. And finally when stack is empty then the string is accepted by the NPDA. Differentiate PDA acceptance by empty stack method with acceptance by final state method. If the simulation ends in an accept state, . 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. Initially, the stack holds a special symbol Z 0 that indicates the bottom of the stack. Login Now - define], while the deterministic pda accept a proper subset, called LR-K languages. Define RE language. We now show that this method of constructing a DFSM from an NFSM always works. 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. G produces all strings with equal number of a’s and b’s III. 46. The stack is empty. Go ahead and login, it'll take only a minute. 44. 33.When is a string accepted by a PDA? 90. w describes the remaining input. Pushdown Automata A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. The input string is accepted by the PDA if: The final state is reached . Acceptance by empty stack only or final state only is addressed in problems 3.3.3 and 3.3.4. 1 (2) Use your PDA from question 1 and the method to convert a PDA to a CFG to form an equivalent CFG. 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. So, x0 is done, with x = 10110. 49. That is, the language accepted by a DFA is the set of strings accepted by the DFA. ` (4) 19.G denotes the context-free grammar defined by the following rules. 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. 2. 88. The state diagram of the PDA is q0 q1 q3 q2 M : aλ/A 50. 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 Whenever the inner automaton goes to the accepting state, it also moves to the empty-stack state with an $\epsilon$ transition. is an accepting computation for the string. 34. 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. Classify some properties of CFL? (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. Answer to A PDA is given below which accepts strings by empty stack. The examples that we generate have very few states; in general, there is so much more control from using the stack memory. The input string is accepted by the PDA if: The final state is reached . 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. 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 empty stack is our key new requirement relative to finite state machines. (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. 48. 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. Nondeterminism can occur in two ways, as in the following examples. 87. Which combination below expresses all the true statements about G? Give examples of languages handled by PDA. ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected. 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. When we say a problem is decidable? 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. 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. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. FA to Reg Lang PDA is to CFL FA to Reg Lang, PDA is to CFL PDA == [ -NFA + “a stack” ] Wh t k? Formal Definition. 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’. 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. So, x'r = (01001)r = 10010. Hence option B is correct. Not all context-free languages are deterministic. Each input alphabet has more than one possibility to move next state. Login. α describes the stack contents, top at the left. The class of nondeterministic pda accept Context Free Languages [student op. (1) L={ anbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. 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. However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. Differentiate 2-way FA and TM? 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. 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. So we require a PDA ,a machine that can count without limit. For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. 43. The given string 101100 has 6 letters and we are given 5 letter strings. This is not true for pda. 1.1 Acceptance by Final State Let P = (Q,Σ,Γ,δ,q0,Z0,F) be a PDA. This does not necessarily mean that the string is impossible to derive. Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. Classify some closure properties of CFL? Notice that string “acb” is already accepted by PDA. We define these notions in Sections 14.1.2 and 14.1.3. 47. Explain your steps. Also construct the derivation tree for the string w. (8) c)Define a PDA. language of strings of odd length is regular, and hence accepted by a pda. 2 Example. If string is finished and stack is empty then string is accepted by the PDA otherwise not accepted. So we require a PDA ,a machine that can count without limit. The stack is empty.. Give examples of languages handled by PDA. Differentiate recursive and non-recursively languages. 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 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. Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. In both these definitions, we employ the notions of instanta- neous descriptions (ID), and step relations $, as well as its reflexive and transitive closure, $ ∗. If some 2’s are still left and top of stack is a 0 then string is not accepted by the PDA. 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. Give an example of undecidable problem? Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. State the pumping lemma for CFLs 45. G can be accepted by a deterministic PDA. F3: It is known that the problem of determining if a PDA accepts every string is undecidable. We will show conversion of a PDA accepting L by final state into another PDA that accepts L by empty stack, and vice-versa. Part B – (5 × = marks) 11 (a) Design a DFA accept the following strings over the alphabets {0, 1}. ` 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. If it ends DFA A MBwB w Bw accept Theorem Proof in a When is a string accepted by a PDA? , II and III only II and III string accepted by PDA PDAs the! 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