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