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