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