In this NPDA we used some symbol which are given below: Acceptance by empty stack only or final state only is addressed in problems 3.3.3 and 3.3.4. Nondeterminism can occur in two ways, as in the following examples. 50. When is a string accepted by a PDA? The given string 101100 has 6 letters and we are given 5 letter strings. by reading an empty string . This is not true for pda. (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. 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. 33.When is a string accepted by a PDA? This does not necessarily mean that the string is impossible to derive. 48. ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected. The stack is empty.. Give examples of languages handled by PDA. (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. Simulate on input . Differentiate recursive and non-recursively languages. 2 Example. 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. language of strings of odd length is regular, and hence accepted by a pda. 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. And finally when stack is empty then the string is accepted by the NPDA. For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. is an accepting computation for the string. string w=aabbaaa. The input string is accepted by the PDA if: The final state is reached . The language of strings accepted by a deterministic pushdown automaton is called a deterministic context-free language. Step-1: On receiving 0 push it onto stack. 90. Hence option B is correct. Each input alphabet has more than one possibility to move next state. Give an Example for a language accepted by PDA by empty stack. 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. Initially, the stack holds a special symbol Z 0 that indicates the bottom of the 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. 47. 44. The stack is empty. 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 PDA accepts a string when, after reading the entire string, the PDA has emptied its 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. (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. Differentiate PDA acceptance by empty stack method with acceptance by final state method. Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. That is, the language accepted by a DFA is the set of strings accepted by the DFA. When is a string accepted by a PDA? Give examples of languages handled by PDA. 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. -NFAInput string Accept/reject 2 A stack filled with “stack symbols” 34. 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. 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. If some 2’s are still left and top of stack is a 0 then string is not accepted by the PDA. Give an example of undecidable problem? Differentiate 2-way FA and TM? But, it also implies that it could be the case that the string is impossible to derive. Part B – (5 × = marks) 11 (a) Design a DFA accept the following strings over the alphabets {0, 1}. w describes the remaining input. Pushdown Automata A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. 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. 46. α describes the stack contents, top at the left. Whenever the inner automaton goes to the accepting state, it also moves to the empty-stack state with an $\epsilon$ transition. We will show conversion of a PDA accepting L by final state into another PDA that accepts L by empty stack, and vice-versa. 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’. Which combination below expresses all the true statements about G? Login Now So, x'r = (01001)r = 10010. Explain your steps. Elaborate multihead TM. 88. An instantaneous description is a triple (q, w, α) where: q describes the current state. Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. G produces all strings with equal number of a’s and b’s III. 2. G can be accepted by a deterministic PDA. Login. If the simulation ends in an accept state, . Classify some closure properties of CFL? Not all context-free languages are deterministic. However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. We now show that this method of constructing a DFSM from an NFSM always works. Define RE language. It's important to mention that the stack contents are irrelevant to the acceptance of the string. Go ahead and login, it'll take only a minute. F3: It is known that the problem of determining if a PDA accepts every string is undecidable. - define], while the deterministic pda accept a proper subset, called LR-K languages. You must be logged in to read the answer. 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. Why a stack? I only I and III only II and III only I, II and III. 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. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. The states q2 and q3 are the accepting states of M. The null string is accepted in q3. 87. FA to Reg Lang PDA is to CFL FA to Reg Lang, PDA is to CFL PDA == [ -NFA + “a stack” ] Wh t k? The language acceptable by the final state can be defined as: 2. Pda 1. 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