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