Section: Pushdown Automata

Transcription

1 Section: Pushdown Automata Ch. 7 - Pushdown Automata ADFA=(Q,Σ,δ,q 0,F) input tape a a b b a b tape head head moves current state 0 1 1

2 Modify DFA by adding a stack. New machine is called Pushdown Automata (PDA). input tape a a a a b b tape head head moves current state 0 1 a stack a Z 2

3 Definition: Nondeterministic PDA (NPDA) is defined by M=(Q,Σ, Γ, δ, q 0,z,F) where Q is finite set of states Σ is tape (input) alphabet Γ is stack alphabet q 0 is initial state z is start stack symbol (bottom of stack ) F Q is set of final states. δ:q (Σ {λ}) Γ finite subsets of Q Γ 3

4 Example of transitions δ(q 1,a,b) = {(q 3,b),(q 4,ab),(q 6,λ)} The diagram for the above transitions is: 4

5 Instantaneous Description: (q,w,u) Description of a Move: (q 1,aw,bx) (q 2,w,yx) iff Definition Let M=(Q,Σ,Γ,δ,q 0,z,F) be a NPDA. L(M)={w Σ (q 0,w,z) (p,λ,u), p F, u Γ }. The NPDA accepts all strings that start in q 0 and end in a final state. 5

6 Example: L={a n b n n 0}, Σ={a, b}, Γ={z,a} 6

7 Another Definition for Language Acceptance NPDA M accepts L(M) by empty stack: L(M)={w Σ (q 0,w,z) (p, λ, λ)} 7

8 Example: L={ww R w Σ + }, Σ={a, b}, Γ={z,a,b} 8

9 Example: L={ww w Σ }, Σ={a, b} 9

10 Examples for you to try on your own: (solutions are at the end of the handout). L={a n b m m>n,m,n>0},σ={a, b}, Γ={z,a} L={a n b n+m c m n, m > 0}, Σ={a, b, c}, L={a n b 2n n>0},σ={a, b} 10

11 Theorem Given NPDA M that accepts by final state, NPDA M that accepts by empty stack s.t. L(M)=L(M ). Proof (sketch) M=(Q,Σ,Γ,δ,q 0,z,F) Construct M =(Q,Σ,Γ,δ,q s,z,f ) 11

12 Theorem Given NPDA M that accepts by empty stack, NPDA M that accepts by final state. Proof: (sketch) M=(Q,Σ,Γ,δ,q 0,z,F) Construct M =(Q,Σ,Γ,δ,q s,z,f ) 12

13 Theorem For any CFL L not containing λ, an NPDA M s.t. L=L(M). Proof (sketch) Given (λ-free) CFL L. CFG G such that L=L(G). G in GNF, s.t. L(G)=L(G ). G =(V,T,S,P). All productions in P are of the form: 13

14 Example: Let G =(V,T,S,P), P= S asa aaa b A bbbb B b 14

15 Theorem Given a NPDA M, a NPDA M s.t. all transitions have the form δ(q i,a,a)={c 1,c 2,...c n } where c i =(q j,λ) or c i =(q j,bc) Each move either increases or decreases stack contents by a single symbol. Proof (sketch) 15

16 Theorem If L=L(M) for some NPDA M, then L is a CFL. Proof: Given NPDA M. First, construct an equivalent NPDA M that will be easier to work with. Construct M such that 1. accepts if stack is empty 2. each move increases or decreases stack content by a single symbol. (can only push 2 variables or no variables with each transition) M =(Q,Σ,Γ,δ,q 0,z,F) Construct G=(V,Σ,S,P) where V={(q i cq j ) q i,q j Q, c Γ} Goal: (q 0 zq f ) which will be the start symbol in the grammar. 16

17 Example: L(M)={aa b}, M=(Q,Σ,Γ,δ,q 0,z,F), Q={q 0,q 1,q 2,q 3 }, Σ={a, b},γ={a, z},f={}. 4 a,z;az 1 b,a; λ q λ,z; λ 0 q1 q a,a; λ λ,z;az 5 q 3 17

18 Construct the grammar G=(V,T,S,P), V={(q 0 Aq 0 ), (q 0 zq 0 ),(q 0 Aq 1 ), (q 0 zq 1 ),...} T=Σ S=(q 0 zq 2 ) P= 18

19 From transition 1(q 0 Aq 1 ) b From transition 2 (q 1 zq 2 ) λ From transition 3(q 0 Aq 3 ) a From transition 4 (q 0 zq 0 ) a(q 0 Aq 0 )(q 0 zq 0 ) a(q 0 Aq 1 )(q 1 zq 0 ) a(q 0 Aq 2 )(q 2 zq 0 ) a(q 0 Aq 3 )(q 3 zq 0 ) (q 0 zq 1 ) a(q 0 Aq 0 )(q 0 zq 1 ) a(q 0 Aq 1 )(q 1 zq 1 ) a(q 0 Aq 2 )(q 2 zq 1 ) a(q 0 Aq 3 )(q 3 zq 1 ) (q 0 zq 2 ) a(q 0 Aq 0 )(q 0 zq 2 ) a(q 0 Aq 1 )(q 1 zq 2 ) a(q 0 Aq 2 )(q 2 zq 2 ) a(q 0 Aq 3 )(q 3 zq 2 ) (q 0 zq 3 ) a(q 0 Aq 0 )(q 0 zq 3 ) a(q 0 Aq 1 )(q 1 zq 3 ) a(q 0 Aq 2 )(q 2 zq 3 ) a(q 0 Aq 3 )(q 3 zq 3 ) 19

20 From transition 5(q 3 zq 0 ) (q 0 Aq 0 )(q 0 zq 0 ) (q 0 Aq 1 )(q 1 zq 0 ) (q 0 Aq 2 )(q 2 zq 0 ) (q 0 Aq 3 )(q 3 zq 0 ) (q 3 zq 1 ) (q 0 Aq 0 )(q 0 zq 1 ) (q 0 Aq 1 )(q 1 zq 1 ) (q 0 Aq 2 )(q 2 zq 1 ) (q 0 Aq 3 )(q 3 zq 1 ) (q 3 zq 2 ) (q 0 Aq 0 )(q 0 zq 2 ) (q 0 Aq 1 )(q 1 zq 2 ) (q 0 Aq 2 )(q 2 zq 2 ) (q 0 Aq 3 )(q 3 zq 2 ) (q 3 zq 3 ) (q 0 Aq 0 )(q 0 zq 3 ) (q 0 Aq 1 )(q 1 zq 3 ) (q 0 Aq 2 )(q 2 zq 3 ) (q 0 Aq 3 )(q 3 zq 3 ) 20

21 Recognizing aaab in M: (q 0, aaab, z) (q 0, aab, Az) (q 3,ab,z) (q 0,ab,Az) (q 3,b,z) (q 0,b,Az) (q 1,λ,z) (q 2,λ,λ) Derivation of string aaab in G: (q 0 zq 2 ) a(q 0 Aq 3 )(q 3 zq 2 ) aa(q 3 zq 2 ) aa(q 0 Aq 3 )(q 3 zq 2 ) aaa(q 3 zq 2 ) aaa(q 0 Aq 1 )(q 1 zq 2 ) aaab(q 1 zq 2 ) aaab 21

22 Definition: A PDA M=(Q,Σ,Γ,δ,q 0,z,F) is deterministic if for every q Q, a Σ {λ}, b Γ 1. δ(q, a, b) contains at most 1 element 2. if δ(q, λ, b) then δ(q, c, b)= for all c Σ Definition: L is DCFL iff DPDA M s.t. L=L(M). Examples: 1. Previous pda for {a n b n n 0} is deterministic. 2. Previous pda for {a n b m c n+m n, m > 0} is deterministic. 3. Previous pda for {ww R w Σ + },Σ ={a, b} is nondeterministic. Note: There are CFL s that are not deterministic. 22

23 L={a n b n n 1} {a n b 2n n 1} is a CFL and not a DCFL. Proof: L = {a n b n : n 1} {a n b 2n :n 1} It is easy to construct a NPDA for {a n b n : n 1} and a NPDA for {a n b 2n : n 1}. Thesetwocanbe joined together by a new start state and λ-transitions to create a NPDA forl.thus,liscfl. Now show L is not a DCFL. Assume that there is a deterministic PDA M such that L = L(M). We will construct a PDA that recognizes a language that is not a CFL and derive a contradiction. Construct a PDA M as follows: 1. Create two copies of M: M 1 and M 2. The same state in M 1 and M 2 23

24 are called cousins. 2. Remove accept status from accept states in M 1, remove initial status from initial state in M 2. In our new PDA, we will start in M 1 and accept in M Outgoing arcs from old accept states in M 1, change to end up in the cousin of its destination in M 2. This joins M 1 and M 2 into one PDA. There must be an outgoing arc since you must recognize both a n b n and a n b 2n. After reading nb s, must accept if no more b s and continue if there are more b s. 4. Modify all transitions that read a b and have their destinations in M 2 to read a c. This is the construction of our new PDA. 24

25 When we read a n b n and end up in an old accept state in M 1, then we will transfer to M 2 and read the rest of a n b 2n. Only the b s in M 2 have been replaced by c s, so the new machine accepts a n b n c n. The language accepted by our new PDA is a n b n c n. But this is not a CFL. Contradiction! Thus there is no deterministic PDA M such that L(M) =L. Q.E.D. 25

26 Example: L={a n b m m>n,m,n>0}, Σ={a, b}, Γ={z,a} a,a;aa b,a; λ b,z;z a,z;az b,a; λ b,z;z q 0 q 1 q 2 q 3 Example: L={a n b n+m c m n, m > 0}, Σ={a, b, c}, a,a;aa b,a; b,b;bb a,z;az b,a; q0 q1 q2 b,z;bz c,b; q3 q5,z;z q4 c,b; Example: L={a n b 2n n>0},σ={a, b} a,a;aaa b,a; λ a,z;aaz b,a; λ λ,z; λ q 0 q 1 q 2 q 3 26

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