-NFAInput string Accept/reject 2 A stack filled with “stack symbols” 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. Classify some closure properties of CFL? 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. The stack is emptied by processing the b’s in q2. But, it also implies that it could be the case that the string is impossible to derive. The given string 101100 has 6 letters and we are given 5 letter strings. The states q2 and q3 are the accepting states of M. The null string is accepted in q3. Pushdown Automata A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. I only I and III only II and III only I, II and III. Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. This is not true for pda. Which combination below expresses all the true statements about G? language of strings of odd length is regular, and hence accepted by a pda. This does not necessarily mean that the string is impossible to derive. Give examples of languages handled by PDA. Classify some properties of CFL? Each input alphabet has more than one possibility to move next state. Differentiate recursive and non-recursively languages. 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. Pda 1. is an accepting computation for the string. Hence option B is correct. Why a 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 FA to Reg Lang PDA is to CFL FA to Reg Lang, PDA is to CFL PDA == [ -NFA + “a stack” ] Wh t k? State the pumping lemma for CFLs 45. 89. Formal Definition. So we require a PDA ,a machine that can count without limit. α describes the stack contents, top at the left. (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. 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. An instantaneous description is a triple (q, w, α) where: q describes the current state. If string is finished and stack is empty then string is accepted by the PDA otherwise not accepted. Step-1: On receiving 0 push it onto stack. 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’. 48. PDA accepts a string when, after reading the entire string, the PDA has emptied its stack. The input string is accepted by the PDA if: The final state is reached . So we require a PDA ,a machine that can count without limit. 2 Example. 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. 2. 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. Define – Pumping lemma for CFL. That is, the language accepted by a DFA is the set of strings accepted by the DFA. When we say a problem is decidable? If some 2’s are still left and top of stack is a 0 then string is not accepted by the PDA. Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. Also construct the derivation tree for the string w. (8) c)Define a PDA. In this NPDA we used some symbol which are given below: We now show that this method of constructing a DFSM from an NFSM always works. It's important to mention that the stack contents are irrelevant to the acceptance of the string. THEOREM 4.2.1 Let L be a language accepted by a … So, x'r = (01001)r = 10010. 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. 1.1 Acceptance by Final State Let P = (Q,Σ,Γ,δ,q0,Z0,F) be a PDA. Explain your steps. Differentiate 2-way FA and TM? 44. We will show conversion of a PDA accepting L by ﬁnal state into another PDA that accepts L by empty stack, and vice-versa. If it ends DFA A MBwB w Bw accept Theorem Proof in a The input string is accepted by the PDA if: The final state is reached . Whenever the inner automaton goes to the accepting state, it also moves to the empty-stack state with an $\epsilon$ transition. However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. ` (4) 19.G denotes the context-free grammar defined by the following rules. We deﬁne these notions in Sections 14.1.2 and 14.1.3. 46. 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. Differentiate PDA acceptance by empty stack method with acceptance by final state method. Not all context-free languages are deterministic. The examples that we generate have very few states; in general, there is so much more control from using the stack memory. Give an example of undecidable problem? 88. In both these deﬁnitions, we employ the notions of instanta- neous descriptions (ID), and step relations $, as well as its reﬂexive and transitive closure, $ ∗. 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. The language of strings accepted by a deterministic pushdown automaton is called a deterministic context-free language. 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 Notice that string “acb” is already accepted 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. w describes the remaining input. When is a string accepted by a PDA? Elaborate multihead TM. Simulate on input . Nondeterminism can occur in two ways, as in the following examples. Login Now G produces all strings with equal number of a’s and b’s III. (1) L={ anbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. string w=aabbaaa. Classify some techniques for Turing machine construction? The empty stack is our key new requirement relative to finite state machines. 43. 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. 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 state diagram of the PDA is q0 q1 q3 q2 M : aλ/A The class of nondeterministic pda accept Context Free Languages [student op. The language acceptable by the final state can be defined as: 2. An input string is accepted if after the entire string is read, the PDA reaches a final state. Answer to A PDA is given below which accepts strings by empty stack. 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. 87. ` 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. 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. When is a string accepted by a PDA? F3: It is known that the problem of determining if a PDA accepts every string is undecidable. 50. Give an Example for a language accepted by PDA by empty stack. 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. Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. 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. by reading an empty string . 90. And finally when stack is empty then the string is accepted by the NPDA. 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. 49. The stack is empty.. Give examples of languages handled by PDA. You must be logged in to read the answer. 47. 1 (2) Use your PDA from question 1 and the method to convert a PDA to a CFG to form an equivalent CFG. PDA - the automata for CFLs What is? If the simulation ends in an accept state, . 34. Login. 33.When is a string accepted by a PDA? Initially, the stack holds a special symbol Z 0 that indicates the bottom of the stack. 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. Part B – (5 × = marks) 11 (a) Design a DFA accept the following strings over the alphabets {0, 1}. - define], while the deterministic pda accept a proper subset, called LR-K languages. The language accepted by a PDA M, L(M), is the set of all accepted strings. For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. Define RE language. The stack is empty. 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. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. So, x0 is done, with x = 10110. 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. (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. 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