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

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