Hence option B is correct. The given string 101100 has 6 letters and we are given 5 letter strings. 47. PDA accepts a string when, after reading the entire string, the PDA has emptied its stack. (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. Login Now 34. 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. State the pumping lemma for CFLs 45. Acceptance by empty stack only or final state only is addressed in problems 3.3.3 and 3.3.4. We will show conversion of a PDA accepting L by final state into another PDA that accepts L by empty stack, and vice-versa. 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. (1) L={ anbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. Part B – (5 × = marks) 11 (a) Design a DFA accept the following strings over the alphabets {0, 1}. Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. Not all context-free languages are deterministic. 48. 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. 33.When is a string accepted by a PDA? Explain your steps. Differentiate 2-way FA and TM? Step-1: On receiving 0 push it onto stack. Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. Answer to A PDA is given below which accepts strings by empty stack. - define], while the deterministic pda accept a proper subset, called LR-K languages. 1.1 Acceptance by Final State Let P = (Q,Σ,Γ,δ,q0,Z0,F) be a PDA. 46. If the simulation ends in an accept state, . This is not true for pda. Give an Example for a language accepted by PDA by empty stack. Nondeterminism can occur in two ways, as in the following examples. ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected. by reading an empty string . Initially, the stack holds a special symbol Z 0 that indicates the bottom of the stack. THEOREM 4.2.1 Let L be a language accepted by a … Login. However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. Define RE language. The input string is accepted by the PDA if: The final state is reached . The input string is accepted by the PDA if: The final state is reached . 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. In this NPDA we used some symbol which are given below: 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. So we require a PDA ,a machine that can count without limit. That is, the language accepted by a DFA is the set of strings accepted by the DFA. The state diagram of the PDA is q0 q1 q3 q2 M : aλ/A 44. 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 Whenever the inner automaton goes to the accepting state, it also moves to the empty-stack state with an $\epsilon$ transition. Differentiate PDA acceptance by empty stack method with acceptance by final state method. The stack is emptied by processing the b’s in q2. Define – Pumping lemma for CFL. 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. 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. Pushdown Automata A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. Simulate on 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. Which combination below expresses all the true statements about G? If some 2’s are still left and top of stack is a 0 then string is not accepted by the PDA. So we require a PDA ,a machine that can count without limit. 50. 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. The states q2 and q3 are the accepting states of M. The null string is accepted in q3. And finally when stack is empty then the string is accepted by the NPDA. Each input alphabet has more than one possibility to move next state. I only I and III only II and III only I, II and III. w describes the remaining input. In both these definitions, we employ the notions of instanta- neous descriptions (ID), and step relations $, as well as its reflexive and transitive closure, $ ∗. Classify some properties of CFL? α describes the stack contents, top at the left. 2 Example. Give an example of undecidable problem? (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. The stack is empty. 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. -NFAInput string Accept/reject 2 A stack filled with “stack symbols” ` 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. 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. We now show that this method of constructing a DFSM from an NFSM always works. 43. 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. But, it also implies that it could be the case that the string is impossible to derive. 2. The stack is empty.. Give examples of languages handled by PDA. Why a stack? 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. The language acceptable by the final state can be defined as: 2. 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. string w=aabbaaa. Also construct the derivation tree for the string w. (8) c)Define a PDA. 1 (2) Use your PDA from question 1 and the method to convert a PDA to a CFG to form an equivalent CFG. F3: It is known that the problem of determining if a PDA accepts every string is undecidable. So, x0 is done, with x = 10110. G produces all strings with equal number of a’s and b’s III. 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 Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. The empty stack is our key new requirement relative to finite state machines. 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. Go ahead and login, it'll take only a minute. For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. The language of strings accepted by a deterministic pushdown automaton is called a deterministic context-free language. 90. When is a string accepted by a PDA? If string is finished and stack is empty then string is accepted by the PDA otherwise not accepted. The examples that we generate have very few states; in general, there is so much more control from using the stack memory. 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 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’. Formal Definition. Give examples of languages handled by 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. G can be accepted by a deterministic PDA. An input string is accepted if after the entire string is read, the PDA reaches a final state. ` (4) 19.G denotes the context-free grammar defined by the following rules. (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. Differentiate recursive and non-recursively languages. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. When we say a problem is decidable? It's important to mention that the stack contents are irrelevant to the acceptance of the string. 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. Classify some techniques for Turing machine construction? The class of nondeterministic pda accept Context Free Languages [student op. When is a string accepted by a PDA? We define these notions in Sections 14.1.2 and 14.1.3. The language accepted by a PDA M, L(M), is the set of all accepted strings. 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. 89. 88. So, x'r = (01001)r = 10010. An instantaneous description is a triple (q, w, α) where: q describes the current state. is an accepting computation for the string. language of strings of odd length is regular, and hence accepted by a pda. 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. FA to Reg Lang PDA is to CFL FA to Reg Lang, PDA is to CFL PDA == [ -NFA + “a stack” ] Wh t k? Pda 1. PDA - the automata for CFLs What is? You must be logged in to read the answer. Classify some closure properties of CFL? 49. Notice that string “acb” is already accepted by PDA. Elaborate multihead TM. 87. 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