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