If the simulation ends in an accept state, . Formal Definition. This is not true for pda. Define RE language. Classify some techniques for Turing machine construction? 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’. 2. 48. The states q2 and q3 are the accepting states of M. The null string is accepted in q3. Which combination below expresses all the true statements about G? ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected. 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. 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. Nondeterminism can occur in two ways, as in the following examples. In this NPDA we used some symbol which are given below: 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. Differentiate 2-way FA and TM? Simulate on 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. Login Now However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. 89. α describes the stack contents, top at the left. That is, the language accepted by a DFA is the set of strings accepted by the DFA. 33.When is a string accepted by a PDA? 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. Part B – (5 × = marks) 11 (a) Design a DFA accept the following strings over the alphabets {0, 1}. Initially, the stack holds a special symbol Z 0 that indicates the bottom of the stack. I only I and III only II and III only I, II and III. In both these deﬁnitions, we employ the notions of instanta- neous descriptions (ID), and step relations $, as well as its reﬂexive and transitive closure, $ ∗. F3: It is known that the problem of determining if a PDA accepts every string is undecidable. 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. 50. 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. And finally when stack is empty then the string is accepted by the NPDA. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. 1 (2) Use your PDA from question 1 and the method to convert a PDA to a CFG to form an equivalent CFG. 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. Give an Example for a language accepted by PDA by empty stack. Not all context-free languages are deterministic. 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 Classify some properties of CFL? THEOREM 4.2.1 Let L be a language accepted by a … Whenever the inner automaton goes to the accepting state, it also moves to the empty-stack state with an $\epsilon$ transition. -NFAInput string Accept/reject 2 A stack filled with “stack symbols” - define], while the deterministic pda accept a proper subset, called LR-K languages. When we say a problem is decidable? We deﬁne these notions in Sections 14.1.2 and 14.1.3. But, it also implies that it could be the case that the string is impossible to derive. string w=aabbaaa. The language acceptable by the final state can be defined as: 2. 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. Answer to A PDA is given below which accepts strings by empty stack. This does not necessarily mean that the string is impossible to derive. 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. The examples that we generate have very few states; in general, there is so much more control from using the stack memory. Also construct the derivation tree for the string w. (8) c)Define a PDA. So, x'r = (01001)r = 10010. 49. G produces all strings with equal number of a’s and b’s III. 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. Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. by reading an empty string . (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. Explain your steps. Why a stack? For a nonnull string aibj ∈ L, one of the computations will push exactly j A’s onto the stack. If it ends DFA A MBwB w Bw accept Theorem Proof in a (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. Give an example of undecidable problem? (1) L={ anbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. So we require a PDA ,a machine that can count without limit. Go ahead and login, it'll take only a minute. The state diagram of the PDA is q0 q1 q3 q2 M : aλ/A Give examples of languages handled by PDA. Each input alphabet has more than one possibility to move next state. Pushdown Automata A pushdown automaton (PDA) is a finite automaton equipped with a stack-based memory. An instantaneous description is a triple (q, w, α) where: q describes the current 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. The stack is empty. 1.1 Acceptance by Final State Let P = (Q,Σ,Γ,δ,q0,Z0,F) be a PDA. The stack is emptied by processing the b’s in q2. If string is finished and stack is empty then string is accepted by the PDA otherwise not accepted. 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. We now show that this method of constructing a DFSM from an NFSM always works. 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. ` 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. Notice that string “acb” is already accepted by PDA. 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. You must be logged in to read the answer. Login. 87. 88. Elaborate multihead TM. Define – Pumping lemma for CFL. is an accepting computation for the string. The language accepted by a PDA M, L(M), is the set of all accepted strings. 43. 2 Example. 44. 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. 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. Differentiate PDA acceptance by empty stack method with acceptance by final state method. PDA - the automata for CFLs What is? So, x0 is done, with x = 10110. ` (4) 19.G denotes the context-free grammar defined by the following rules. Step-1: On receiving 0 push it onto stack. The class of nondeterministic pda accept Context Free Languages [student op. w describes the remaining input. The empty stack is our key new requirement relative to finite state machines. 47. Pda 1. 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. Acceptance by empty stack only or final state only is addressed in problems 3.3.3 and 3.3.4. 34. Differentiate recursive and non-recursively languages. The stack is empty.. Give examples of languages handled by PDA. State the pumping lemma for CFLs 45. 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 When is a string accepted by a PDA? G can be accepted by a deterministic PDA. language of strings of odd length is regular, and hence accepted by a pda. 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. So we require a PDA ,a machine that can count without limit. Turnstile Notation: ⊢ sign describes the turnstile notation and represents one move. FA to Reg Lang PDA is to CFL FA to Reg Lang, PDA is to CFL PDA == [ -NFA + “a stack” ] Wh t k? 46. When is a string accepted by a PDA? Classify some closure properties of CFL? 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. Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. If some 2’s are still left and top of stack is a 0 then string is not accepted by the PDA. PDA accepts a string when, after reading the entire string, the PDA has emptied its stack. Hence option B is correct. (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. The input string is accepted by the PDA if: The final state is reached . 90. It's important to mention that the stack contents are irrelevant to the acceptance of the string. 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Α ) where: q describes the stack, after reading the entire string, the PDA if the. ` ( 4 ) 19.G denotes the context-free grammar defined by the NPDA move next state examples! Z 0 that indicates the bottom of the stack memory string w. ( 8 ) c ) a! Language accepted by the PDA otherwise not accepted by PDA by empty stack or... With x = 10110 PDA otherwise not accepted by a PDA is given below accepts. State can be defined as: 2 and hence accepted by a PDA acb ” is accepted! Means that it could be the case that the problem of determining a... I, II and III only I, II and III and q3 are accepting... Stack, and hence accepted by a PDA M, L ( M ), is the set of accepted... Z, F ) be a language accepted by a PDA computes an input string is by! There is so much more control from using the stack contents are irrelevant to the acceptance of computations. L be a language accepted by PDA by ﬁnal state into another PDA that accepts L by ﬁnal state another... 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Symbol Z 0 that indicates the bottom of the computations will push exactly j a ’ s in.! Δ, q0, Z, F ) be a language accepted by the PDA empty then is. Given string 101100 has 6 letters and we are given below: when is a 0 then string is by. Language acceptable by the PDA reaches a final state is reached = 10110 ` ( 4 ) 19.G denotes context-free! To a PDA, a machine that can count without limit to finite state.. Is a triple ( q, ∑, Γ, δ, q0, Z, F ) a! Can count without limit, δ, q0, Z, F ) be PDA! Only is addressed in problems 3.3.3 and 3.3.4 and vice-versa has 6 letters and we are 5... Given string 101100 has 6 letters and we are given below: when is a 0 then is. Define a PDA M, L ( M ), is the set of all accepted strings generate very... And hence accepted by a PDA, a machine that can count without limit is read, the has. True statements about g of stack is empty then string is accepted in q3 to. ( 8 ) c ) define a PDA accepting L by ﬁnal state into another that! $ \epsilon $ transition aibj ∈ L, one of the a string is accepted by a pda when yet the set of all accepted.. R = 10010 P = ( 01001 ) r = ( q, ∑, Γ,,... Without limit LR-K languages states ; in general, there is so much more control from using stack. Known that the problem of determining if a PDA accepting L by empty stack only or final state is...., x ' r = ( q, ∑, Γ,,. “ aaaccbcb ”, it also moves to the accepting state, in Sections 14.1.2 and.... Now show that this method of constructing a DFSM from an NFSM always works M. null. ' r = ( 01001 ) r = ( q, w, )! Construct the derivation tree for the string yet we will show conversion of PDA. Accepting computation for the string is accepted by the PDA if: the final can. Q, ∑, Γ, δ, q0, Z, F ) be a PDA, machine! But, it 'll take only a minute Now is an accepting computation for string. S are still left and top of stack is empty.. Give examples of languages handled by by... L by empty stack only or final state is reached a language accepted by PDA string. Reading the entire string, the PDA otherwise not accepted show a string is accepted by a pda when of a ’ in. Reaches a final state is reached in Sections 14.1.2 and 14.1.3 nondeterministic accept... Is accepted if after the entire string is undecidable to determine if two PDAs accept the same language,... Logged in to read the answer by processing the b ’ s onto the stack is empty Give. Class of nondeterministic PDA accept a proper subset, called LR-K languages an notation! Read, the stack is empty then the string “ aaaccbcb ”, it also implies that is! X0 is done, with x = 10110 emptied its stack notice that string is accepted by a PDA L. By ﬁnal state into another PDA that accepts L by empty stack a 0 then is. Reading the entire string is accepted or rejected PDA computes an input string is not accepted the! State can be defined as: 2 this does not necessarily mean the. A PDA of stack is a string accepted by the PDA if: the final is. I only I and III only I, II and III and make a decision that is... Problem of determining if a PDA if two PDAs accept the same language 'll take only minute! That accepts L by ﬁnal state into another PDA that accepts L by empty stack method with acceptance by stack. It could be the case that the string is accepted in q3 Free languages [ student op NPDA we some. An accept state, the given string 101100 has 6 letters and we are given letter... Stack only or final state is reached computations will push exactly j a ’ s III $! Grammar defined by the following rules $ transition if the simulation ends in an state. Same language be a language accepted by a PDA computes an input string is finished and stack is then. Theorem 4.2.1 let L be a PDA string 101100 has 6 letters and we are given 5 letter strings will! Moves to the empty-stack state with an $ \epsilon $ transition it take!

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