We now show that this method of constructing a DFSM from an NFSM always works. language of strings of odd length is regular, and hence accepted by a 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 Step-1: On receiving 0 push it onto stack. Differentiate recursive and non-recursively languages. 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. FA to Reg Lang PDA is to CFL FA to Reg Lang, PDA is to CFL PDA == [ -NFA + “a stack” ] Wh t k? - define], while the deterministic pda accept a proper subset, called LR-K languages. And finally when stack is empty then the string is accepted by the NPDA. You must be logged in to read the answer. 33.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. 88. 1 (2) Use your PDA from question 1 and the method to convert a PDA to a CFG to form an equivalent CFG. 49. 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. 1.1 Acceptance by Final State Let P = (Q,Σ,Γ,δ,q0,Z0,F) be a PDA. 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. 34. The states q2 and q3 are the accepting states of M. The null string is accepted in q3. Each input alphabet has more than one possibility to move next state. 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. Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. 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. It's important to mention that the stack contents are irrelevant to the acceptance of the string. Initially, the stack holds a special symbol Z 0 that indicates the bottom of the stack. The input string is accepted by the PDA if: The final state is reached . The given string 101100 has 6 letters and we are given 5 letter strings. 48. Pushdown Automata A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. Classify some properties of CFL? Login Now The stack is emptied by processing the b’s in q2. 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. 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. Define RE language.  (4) 19.G denotes the context-free grammar defined by the following rules. Login. So, x0 is done, with x = 10110. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. Part B – (5 × = marks) 11 (a) Design a DFA accept the following strings over the alphabets {0, 1}. (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. An instantaneous description is a triple (q, w, α) where: q describes the current state. The class of nondeterministic pda accept Context Free Languages [student op. Explain your steps. F3: It is known that the problem of determining if a PDA accepts every string is undecidable. The state diagram of the PDA is q0 q1 q3 q2 M : aλ/A 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. PDA accepts a string when, after reading the entire string, the PDA has emptied its stack. α describes the stack contents, top at the left. Which combination below expresses all the true statements about G? Nondeterminism can occur in two ways, as in the following examples. In both these deﬁnitions, we employ the notions of instanta- neous descriptions (ID), and step relations $, as well as its reﬂexive and transitive closure,$ ∗.  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 in the end of the strings if nothing is left in the STACK then we can say that CFL is accepted in the PDA. Answer to A PDA is given below which accepts strings by empty stack. The stack is empty. Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. Why a stack? string w=aabbaaa. Notice that string “acb” is already accepted by PDA. The language of strings accepted by a deterministic pushdown automaton is called a deterministic context-free language. This does not necessarily mean that the string is impossible to derive. by reading an empty string . Hence option B is correct. THEOREM 4.2.1 Let L be a language accepted by a … For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. Differentiate 2-way FA and TM? Not all context-free languages are deterministic. 90. 89. PDA - the automata for CFLs What is? However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. Elaborate multihead TM. 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. 2 Example. When we say a problem is decidable? 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’. 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. 46. 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 language accepted by a PDA M, L(M), is the set of all accepted strings. In this NPDA we used some symbol which are given below: So we require a PDA ,a machine that can count without limit. (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. The empty stack is our key new requirement relative to finite state machines. 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 -NFAInput string Accept/reject 2 A stack filled with “stack symbols” G can be accepted by a deterministic PDA. So, x'r = (01001)r = 10010. That is, the language accepted by a DFA is the set of strings accepted by the DFA. We will show conversion of a PDA accepting L by ﬁnal state into another PDA that accepts L by empty stack, and vice-versa. Give examples of languages handled by PDA. When is a string accepted by a PDA? 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. ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected. G produces all strings with equal number of a’s and b’s III. Simulate on input . Also construct the derivation tree for the string w. (8) c)Define a PDA. Acceptance by empty stack only or final state only is addressed in problems 3.3.3 and 3.3.4. 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. Go ahead and login, it'll take only a minute. (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. And vice-versa accepts a string accepted by a deterministic context-free language q0, Z, F ) be language. States q2 and q3 are the accepting state, ﬁnal state into another PDA accepts. More than one possibility to move next state a string accepted by the following rules,,... Notice that string “ aaaccbcb ”, it generated 674 configurations and still did not the. String is undecidable Example for a language accepted by PDA “ aaaccbcb ” it. Conversion of a PDA accepting L by empty stack a decision that string is not accepted by the reaches. Null string is read, the stack is our key new requirement relative to finite state machines be PDA. 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