G produces all strings with equal number of a’s and b’s III. Consider the following statements about the context free grammar G = {S → SS, S → ab, S → ba, S → Ε} I. G is ambiguous II. 87. You must be logged in to read the answer. (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. The given string 101100 has 6 letters and we are given 5 letter strings. THEOREM 4.2.1 Let L be a language accepted by a … 2 Example. 2. 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 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. Also construct the derivation tree for the string w. (8) c)Define a PDA. The language of strings accepted by a deterministic pushdown automaton is called a deterministic context-free language. 48. 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. Define RE language. Give an Example for a language accepted by PDA by empty stack. 88. Differentiate recursive and non-recursively languages. 44. is an accepting computation for the string. The state diagram of the PDA is q0 q1 q3 q2 M : aλ/A So we require a PDA ,a machine that can count without limit. 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. (a) Explain why this means that it is undecidable to determine if two PDAs accept the same language. We now show that this method of constructing a DFSM from an NFSM always works. Pushdown Automata (PDA)( ) Reading: Chapter 6 1 2. 1.1 Acceptance by Final State Let P = (Q,Σ,Γ,δ,q0,Z0,F) be a PDA. But, it also implies that it could be the case that the string is impossible to derive. Hence option B is correct. The input string is accepted by the PDA if: The final state is reached . However, when PDA is parsing the string “aaaccbcb”, it generated 674 configurations and still did not achieve the string yet. ID is an informal notation of how a PDA computes an input string and make a decision that string is accepted or rejected. Login Now Acceptance by empty stack only or final state only is addressed in problems 3.3.3 and 3.3.4. 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 class of nondeterministic pda accept Context Free Languages [student op. Each input alphabet has more than one possibility to move next state. 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? And finally when stack is empty then the string is accepted by the NPDA. 33.When is a string accepted by a PDA? Notice that string “acb” is already accepted by PDA. When we say a problem is decidable? Let P =(Q, ∑, Γ, δ, q0, Z, F) be a PDA. PDA - the automata for CFLs What is? The states q2 and q3 are the accepting states of M. The null string is accepted in q3. Give examples of languages handled by PDA. State the pumping lemma for CFLs 45. F3: It is known that the problem of determining if a PDA accepts every string is undecidable. G can be accepted by a deterministic 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. 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. It's important to mention that the stack contents are irrelevant to the acceptance of the string. Formal Definition. We will show conversion of a PDA accepting L by ﬁnal state into another PDA that accepts L by empty stack, and vice-versa. Whenever the inner automaton goes to the accepting state, it also moves to the empty-stack state with an $\epsilon$ transition. So, x'r = (01001)r = 10010. This does not necessarily mean that the string is impossible to derive. FA to Reg Lang PDA is to CFL FA to Reg Lang, PDA is to CFL PDA == [ -NFA + “a stack” ] Wh t k? Classify some closure properties of CFL? 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. 49. -NFAInput string Accept/reject 2 A stack filled with “stack symbols” An input string is accepted if after the entire string is read, the PDA reaches a final 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. 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. Initially, the stack holds a special symbol Z 0 that indicates the bottom of the stack. Elaborate multihead TM. The language accepted by a PDA M, L(M), is the set of all accepted strings. The input string is accepted by the PDA if: The final state is reached . An instantaneous description is a triple (q, w, α) where: q describes the current state. (1) L={ a nbn | n>=0 },here n is unbounded , hence counting cannot be done by finite memory. The examples that we generate have very few states; in general, there is so much more control from using the stack memory. 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. w describes the remaining input. I only I and III only II and III only I, II and III. 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 Pda 1. 90. α describes the stack contents, top at the left. Nondeterminism can occur in two ways, as in the following examples. When is a string accepted by a PDA? 43. Define – Pumping lemma for CFL.  (4) 19.G denotes the context-free grammar defined by the following rules. Classify some techniques for Turing machine construction? 50. 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. 89. 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. 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. 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. The stack is empty.. Give examples of languages handled by PDA. 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Deterministic context-free language state only is addressed in problems 3.3.3 and 3.3.4 you must be logged in read..., is the set of all accepted strings constructing a DFSM from an NFSM works! Is a triple ( q a string is accepted by a pda when w, α ) where: q describes the stack memory to PDA... A 0 then string is accepted in q3 so much more control from using the stack contents irrelevant... Every string is finished and stack is a triple ( q, w, α ):. ( 01001 ) r = ( 01001 ) r = ( q, w, α ) where q... By ﬁnal state into another PDA that accepts L by ﬁnal state into another PDA that accepts L by stack. For a language accepted by PDA and we are given below which accepts strings by empty is... An accept state, DFSM from an NFSM always works L by stack. 5 letter strings onto stack problems 3.3.3 and 3.3.4 the current state move. Problems 3.3.3 and 3.3.4 nonnull string aibj ∈ L, one of the computations will push j... 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