TAFL 1 (ECS-403) Unit- IV. 4.1 Push Down Automata. 4.2 The Formal Definition of Pushdown Automata. EXAMPLES for PDA. 4.3 The languages of PDA

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1 TAFL 1 (ECS-403) Unit- IV 4.1 Push Down Automata 4.2 The Formal Definition of Pushdown Automata EXAMPLES for PDA 4.3 The languages of PDA Acceptance by final state Acceptance by empty stack 4.4 CFA to PDA using CNF using GNF 4.5 PDA to CFG 4.6 Deterministic & Non Deterministic PDA 4.7 Two Stack PDA

2 TAFL 2 (ECS-403) 4.1 Pushdown Automata: A pushdown automaton is an extension of the nondeterministic finite automaton with Ԑ- transitions, which is one of the ways to define the regular languages. The pushdown automaton is essentially an Ԑ-NFA with the addition of a stack. The stack can be read, pushed, and popped only at the top, just like the "stack" data structure. We define two different versions of the pushdown automaton: one that accepts by entering an accepting state, like finite automata do, and another version that accepts by emptying its stack, regardless of the state it is in. 4.2 The Formal Definition of Pushdown Automata: Our formal notation for a pushdown automaton (PDA) involves seven components. We write the specification of a PDA P as foliows: P = (Q,, Ґ, δ, q 0, Z 0, F) Q: A finite set of states, like the states of a finite automaton. : A finite set of input symbols, also analogous to the corresponding component of a finite automaton. Ґ: A finite stack alphabet. This component, which has no finite automaton analog, is the set of symbols that we are allowed to push onto the stack. δ: The transition function. As for a finite automaton, δ governs the behavior of the automaton. Formally, δ takes as argument a triple δ(q, a, X), where:

3 TAFL 3 (ECS-403) 1. q is a state in Q. 2. a is either an input symbol in or a = Ԑ, the empty string, which is assumed not to be an input symbol. 3. X is a stack symbol, that is, a member of Ґ. The output of δ is a finite set of pairs (p, γ), where p is the new state, and γ is the string of stack symbols that replaces X at the top of the stack. For instance, if γ = Ԑ, then the stack is popped, if γ = X, then the stack is unchanged, and if γ = YZ, then X is replaced by Z and Y is pushed onto the stack. q 0 : The start state. The PDA is in this state before making any transitions. Z 0 : The start symbol. Initially, the PDA's stack consists of one instance of this symbol, and nothing else. F: The set of accepting states, or final states.

4 TAFL 4 (ECS-403) Example 1: Design PDA for a n b n where n 1. Solution: δ (q 0, a, Z 0 ) = (q 0, az 0 ) δ (q 0, a, a) = (q 0, aa) δ (q 0, b, a) = (q 1, Ԑ) δ (q 1, b, a) = (q 1, Ԑ) δ (q 1, Ԑ, Z 0 ) = (q 2, Ԑ) where P = ({q 0, q 1, q 2 }, {a, b}, {a, Z 0 }, δ, q 0, Z 0, q 2 ) We will simulate this PDA for the following string δ (q 0, aaabbb, Z 0 ) (q 0, aabbb, az 0 ) (q 0, abbb, aaz 0 ) (q 0, bbb, aaaz 0 ) (q 1, bb, aaz 0 ) (q 1, b, az 0 ) (q 1, Ԑ, Z 0 ) (q 2, Ԑ) ACCEPT STATE

5 TAFL 5 (ECS-403) Example 2: Design PDA for the language L = {a n b 2n n 1} Solution: δ (q 0, a, Z 0 ) = (q 0, aaz 0 ) δ (q 0, a, a) = (q 0, aaa) δ (q 0, b, a) = (q 1, Ԑ) δ (q 1, b, a) = (q 1, Ԑ) δ (q 1, Ԑ, Z 0 ) = (q 2, Ԑ) where P = ({q 0, q 1, q 2 }, {a, b}, {a, Z 0 }, δ, q 0, Z 0, {q 2 }) We will simulate this PDA for the following string δ (q 0, aabbbb, Z 0 ) (q 0, abbbb, aaz 0 ) (q 0, abbbb, aaaaz 0 ) (q 0, bbb, aaaz 0 ) (q 1, bb, aaz 0 ) (q 1, b, az 0 ) (q 1, Ԑ, Z 0 ) (q 2, Ԑ) ACCEPT STATE

6 TAFL 6 (ECS-403) Example 3: Design PDA to accept the language L = {w w Ԑ (a+b)* and n a (w) = n b (w)} Solution: δ (q 0, a, Z 0 ) = (q 0, az 0 ) δ (q 0, b, Z 0 ) = (q 0, bz 0 ) δ (q 0, a, a) = (q 0, aa) δ (q 0, b, b) = (q 0, bb) δ (q 0, a, b) = (q 0, Ԑ) δ (q 0, b, a) = (q 0, Ԑ) δ (q 0, Ԑ, Z 0 ) = (q 1, Z 0 ) where P = ({q 0, q 1 }, {a, b}, {a, Z 0 }, δ, q 0, Z 0, {q 1 }) We will simulate this PDA for the following string δ (q 0, aababb, Z 0 ) (q 0, ababb, az 0 ) (q 0, babb, aaz 0 ) (q 0, abb, az 0 ) (q 0, bb, aaz 0 ) (q 0, b, az 0 ) (q 0, Ԑ, Z 0 ) (q 1, Ԑ) ACCEPT STATE

7 TAFL 7 (ECS-403) Example 4: Design PDA to accept the language L = {w w Ԑ (a+b)* and n a (w) > n b (w)} Solution: δ (q 0, a, Z 0 ) = (q 0, az 0 ) δ (q 0, b, Z 0 ) = (q 0, bz 0 ) δ (q 0, a, a) = (q 0, aa) δ (q 0, b, b) = (q 0, bb) δ (q 0, a, b) = (q 0, Ԑ) δ (q 0, b, a) = (q 0, Ԑ) δ (q 0, Ԑ, a) = (q 1, a) where P = ({q 0, q 1 }, {a, b}, {a,b Z 0 }, δ, q 0, Z 0, {q 1 }) We will simulate this PDA for the following string δ (q 0, aababab, Z 0 ) (q 0, ababab, az 0 ) (q 0, babab, az 0 ) (q 0, abab, az 0 ) (q 0, bab, aaz 0 ) (q 0, ab, az 0 ) (q 0, b, aaz 0 ) (q 0, Ԑ, az 0 ) (q 1, a) ACCEPT STATE

8 TAFL 8 (ECS-403) Example 5: Design PDA to accept the language L = {w w Ԑ (a+b)* and n a (w) < n b (w)} Solution: δ (q 0, a, Z 0 ) = (q 0, az 0 ) δ (q 0, b, Z 0 ) = (q 0, bz 0 ) δ (q 0, a, a) = (q 0, aa) δ (q 0, b, b) = (q 0, bb) δ (q 0, a, b) = (q 0, Ԑ) δ (q 0, b, a) = (q 0, Ԑ) δ (q 0, Ԑ, b) = (q 1, b) where P = ({q 0, q 1 }, {a, b}, {a,b Z 0 }, δ, q 0, Z 0, {q 1 }) We will simulate this PDA for the following string δ (q 0, abbab, Z 0 ) (q 0, bbab, az 0 ) (q 0, bab, Z 0 ) (q 0, ab, bz 0 ) (q 0, b, Z 0 ) (q 0, Ԑ, bz 0 ) (q 1, b) ACCEPT STATE

9 TAFL 9 (ECS-403) Example 6: Design PDA to accept the language L = {a m b m c n m,n 1} Solution: δ (q 0, a, Z 0 ) = (q 0, az 0 ) δ (q 0, a, a) = (q 0, aa) δ (q 0, b, a) = (q 1, Ԑ) δ (q 1, b, a) = (q 1, Ԑ) δ (q 1, c, Z 0 ) = (q 1, Z 0 ) δ (q 1, Ԑ, Z 0 ) = (q 2, Z 0 ) where P = ({q 0, q 1, q 2 }, {a, b, c}, {a,b Z 0 }, δ, q 0, Z 0, {q 2 }) We will simulate this PDA for the following string δ (q 0, aabbccc, Z 0 ) (q 0, abbccc, az 0 ) (q 0, bbccc, aaz 0 ) (q 1, bccc, az 0 ) (q 1, ccc, Z 0 ) (q 1, cc, Z 0 ) (q 1, c, Z 0 ) (q 1, Ԑ, Z 0 ) (q 2, Z 0 ) ACCEPT STATE

10 TAFL 10 (ECS-403) Example 7: Design a PDA that accepts a string of well formed parenthesis. Consider the parenthesis is as (, ), [, ], {, }. Solution: δ (q 0, (, Z 0 ) = (q 1, ( Z 0 ) δ (q 0, [, Z 0 ) = (q 1, [ Z 0 ) δ (q 0, {, Z 0 ) = (q 1, { Z 0 ) δ (q 1, (, ( ) = (q 1, ( ( ) δ (q 1, [, [ ) = (q 1, [ [ ) δ (q 1, {, { ) = (q 1, { { ) δ (q 1, (, [ ) = (q 1, ( [ ) δ (q 1, (, { ) = (q 1, ( { ) δ (q 1, [, ( ) = (q 1, [ ( ) δ (q 1, {, ( ) = (q 1, { ( ) δ (q 1, {, [ ) = (q 1, { [ ) δ (q 1, [, { ) = (q 1, [ { ) δ (q 1, ], [ ) = (q 1, Ԑ ) δ (q 1, }, { ) = (q 1, Ԑ ) δ (q 1, Ԑ, Z 0 ) = (q 0, Z 0 ) P = ({q 0, q 1 }, {(, ), {, }, [, ]}, {(, {, [, Z 0 }, δ, q 0, Z 0, {q 0 } ) We will simulate this PDA for the following string δ (q 0, ({}[] ), Z 0 ) (q 1, {}[]), (Z 0 )

11 TAFL 11 (ECS-403) (q 1, }[]), {(Z 0 ) (q 1, []), (Z 0 ) (q 1, ]), [(Z 0 ) (q 1, ), (Z 0 ) (q 1, Ԑ, Z 0 ) (q 0, Z 0 ) ACCEPT STATE 4.3 The languages of PDA: The Languages can be accepted by a Pushdown Automata using two approaches: a) Acceptance by final state: The PDA accepts its input by consuming it and then it enters in the final state. b) Acceptance by empty stack: On reading the input string from initial configuration for some PDA, the stack of PDA gets empty Acceptance by final state: Let P = (Q,, Ґ, δ, q 0, Z 0, F) be a PDA. Then L(P), the language accepted by P by final state, is Theorem: The language Lww r the language of strings in {0, 1}* that have the form ww R. Let us see why that statement is true. The proof is an if-and-only-if statement Proof: (IF) In this part; we have only to show that If x = ww R, then PDA leads to accept state: That is, one option the PDA has is to read w from its input and store it on its stack, in reverse. Next, it goes spontaneously to state q 1 and matches w R on the input with the same string on its stack, and finally goes spontaneously to the state q 2.

12 TAFL 12 (ECS-403) (Only If) In this part we have to show that final state is achieved only if the input is of the form x = ww R.

13 TAFL 13 (ECS-403)

14 TAFL 14 (ECS-403)

15 TAFL 15 (ECS-403)

16 TAFL 16 (ECS-403) 4.6 Deterministic PDA A DPDA is simply a pushdown automata without non-determinism. i.e. no epsilon transitions or transitions to multiple states on same input Only one state at a time DPDA not as powerful a non-deterministic PDA This machine accepts a class of languages somewhere between regular languages and context-free languages. For this reason, the DPDA is often skipped as a topic In practice the DPDA can be useful since determinism is much easier to implement.

17 TAFL 17 (ECS-403) Deterministic PDA: DPDA Allowed transitions: q a, b w 1 q 2 λ, b w q1 q2 (deterministic choices) Prof. Busch - LSU 2 Allowed transitions: a, b w 1 q 2 λ, b w 1 q 2 q 1 q 1 a, c w 2 q 3 λ, c w 2 q 3 (deterministic choices) Prof. Busch - LSU 3

18 TAFL 18 (ECS-403) Not allowed: a, b w 1 q 2 λ, b w 1 q 2 q 1 q 1 a, b w 2 q 3 a, b w 2 q 3 (non deterministic choices) Prof. Busch - LSU 4 Definition: A language L is deterministic context-free if there exists some DPDA that accepts it Example: The language n L( M ) = { a b : n 0} n is deterministic context-free Prof. Busch - LSU 6

19 TAFL 19 (ECS-403) DPDA example n L( M ) = { a b : n 0} n a, λ a b, a λ q a, λ a b, a λ 0 q 1 q λ, $ $ 2 q 3 Prof. Busch - LSU 5 Example of Non-DPDA (PDA) L( M ) = { vv R : v { a, b} * } a, λ a b, λ b a, a λ b, b λ q 0 λ, λ λ λ, $ $ q 1 q 2 Prof. Busch - LSU 7

20 TAFL 20 (ECS-403) Not allowed in DPDAs a, λ a b, λ b a, a λ b, b λ q 0 λ, λ λ λ, $ $ q 1 q 2 Prof. Busch - LSU Two Stack PDA: Let us define a two-stack pushdown automaton to be a sextuple M = (K, Σ, Γ,, s, F) K is a finite set of states, Σ is an alphabet (the input symbols), Γ is an alphabet (the stack symbols), s K is the initial state, F K is the set of final states, and

21 TAFL 21 (ECS-403), the transition relation, is a finite subset of (K x (Σ U {e}) x Γ* x Γ*) x (K x Γ* x Γ*), where the third parameter pops the first stack, the fourth parameter pops the second stack, the fifth parameter pushes a symbol onto the first stack, and the sixth parameter pushes a symbol onto the second stack. Example: L = {a n b n c n n N} This problem was not solvable with a normal PDA, but solvable with a TM and a twostack PDA Σ = {a, b, c}, Γ = {a, b}, K = {q0, q1, q2, f}, s = {q0}, F = {f} = (q0, a, Ԑ, Ԑ) = (q0, a, Ԑ) (q0, a, a, Ԑ) = (q0, aa, Ԑ) (q0, Ԑ, Ԑ, Ԑ) = (q1, Ԑ, Ԑ) (q1, b, Ԑ, Ԑ) = (q1, Ԑ, b) (q1, b, Ԑ, b) = (q1, Ԑ, bb) (q1, c, a, b) = (q2, Ԑ, Ԑ) (q2, c, a, b) = (q2, Ԑ, Ԑ) (q2, Ԑ, Ԑ, Ԑ) = (f, Ԑ, Ԑ)