Announcements Final Exam Dates have been Context Free Languages announced Tuesday, November 11 12:30 – 2:30 pm PDAs and CFLs Room TBA Conflicts? Let me know. Before We Start Plan for today Exam 1 will be returned in 2nd half 1st half PDAs and CFLs 2nd half PDA homework session / Exam 1 Any questions? Languages The Language Bubble Recall. What is a language? Context Free Languages What is a class of languages? Regular Languages Finite Languages 1
Context Free Languages Context Free Grammars Context Free Languages(CFL) is the Let’s formalize this a bit: next class of languages outside of A context free grammar (CFG) is a 4-tuple: Regular Languages: (V, T, S, P) where V is a set of variables Language / grammar: Context Free T is a set of terminals Grammar V and Σ are disjoint (I.e. V ∩ Σ = ∅ ) Machine for accepting: Pushdown S ∈ V, is your start symbol Automata Context Free Grammars Pushdown Automata Let’s formalize this a bit: Production rules Input tape Of the form A → β where A ∈ V β ∈ (V ∪ ∑ ) * string with symbols from V and ∑ 1 2 We say that γ can be derived from α in one step: A → β is a rule 3 α = α 1 A α 2 5 γ = α 1 β α 2 4 α ⇒ γ State machine Stack Pushdown Automata Pushdown Automata Let’s formalize this: About this transition function δ : A pushdown automata (PDA) is a 7-tuple: During a move of a PDA: At most one character is read from the input tape M = (Q, Σ , Γ , q 0 , Z 0 , A, δ ) where λ transitions are okay Q = finite set of states The topmost character is popped from the stack Σ = tape alphabet Γ = stack alphabet (may have symbols in common The machine will move to a new state based on: w/ Σ ) The character read from the tape q 0 ∈ Q = start state The character popped off the stack Z 0 ∈ Γ = initial stack symbol The current state of the machine A ⊆ Q = set of accepting states 0 or more symbols from the stack alphabet are pushed δ = transition function onto the stack. 2
Plan for today Equivalence of CFG and PDA Show that PDAs and CFGs are Given a CFG, G, construct a PDA M, 1. equivalent. such that L(M) = L(G) Given a PDA, M, define a CGF, G such 2. that L(G) = L(M) Questions? Step 1: CFG → PDA Step 1: CFG → PDA Given: A context free grammar G Basic idea Use the stack of the PDA to simulate the Construct: A pushdown automata M derivation of a string in the grammar. Such that: Push S (start variable of G) on the stack From this point on, there are two moves the PDA can Language generated by G is the same as make: Language accepted by M. If a variable A is on the top of the stack, pop it and push 1. the right-hand side of a production A → β from G. If a terminal, a is on the top of the stack, pop it and 2. match it with whatever symbol is being read from the tape. Step 1: CFG → PDA Step 1: CFG → PDA Observations: More observations: There can be many productions that have A string will only be accepted if: a given variable on the left hand side: After a string is completely read S → λ | 0S1 | 1S0 The stack is empty In these cases, the PDA must “choose” which string to push onto the stack after pushing a variable. I.e. the PDA being constructed in non- deterministic. 3
Step 1: CFG → PDA Step 1: CFG → PDA Let’s formalize this: Define M as follows: Let G = (V, T, S, P) be a context free Q = { q 0 , q 1 , q 2 } grammar. q o will be the start state We define a pushdown automata q 1 will be where all the work gets done q 2 will be the accepting state M = (Q, Σ , Γ , δ , q 0 , z ,F) Γ = V ∪ ∑ ∪ { z } z ∉ (V ∪ ∑ ) Such that A = { q 2 } L(M) = L(G) Step 1: CFG → PDA Step 1: CFG → PDA Transition function δ is defined as follows: Transition function δ is defined as follows: δ (q 0 , λ , z ) = { (q 1 , Sz) } δ (q 1, a, a) = { (q 1 , λ )} for all terminals a To start things off, push S onto the stack and Pop and match terminal. immediately go into state 1 δ (q 1, λ , z) = { (q 2 , z) } δ (q 1, λ , A) = { (q 1 , α ) | A → α is a production of After all reading is done, accept only if stack is empty. G} for all variables A No other transitions exist for M Pop and replace variable. Step 1: CFG → PDA Step 1: CFG → PDA Example: Let’s look at an example: M = (Q, Σ , Γ , δ , q 0 , z ,F) Remember the CFG for odd length palindromes: Q = { q 0 , q 1 , q 2 } Σ = { a, b } S → a | b Γ = { a, b, S, z } S → a S a | b S b F = { q 2 } Let’s convert this to a PDA. 4
Step 1: CFG → PDA Step 1: CFG → PDA State Tape input Stack Move(s) q 1 λ S (q 1 , a) (q 1 , b) λ , Z 0 / SZ 0 λ , Z 0 / Z 0 q 1 q 2 q 0 (q 1 , aSa) (q 1 , bSb) q 1 a a (q 1 , λ ) Lots of moves q 1 b b (q 1 , λ ) (see next slide) Step 1: CFG → PDA Step 1: CFG → PDA Let’s run M on abbba Questions? (q 0 , abbba, Z) → (q 1 , abbba, SZ) → (q 1 , abbba, aSaZ) // push → (q 1 , bbba, SaZ) // match → (q 1 , bbba, bSbaZ) // push → (q 1 , bba, SbaZ) // match → (q 1 , bba, bbaZ) // push → (q 1 , ba, baZ) // match → (q 1 , a, aZ) // match → (q 1 , ε , Z) // match → (q 2 , ε , Z ) // accept Step 2: PDA → CFG Pushdown Automata Given: A pushdown automata M Strings accepted by a PDA by Final State Start at (q 0 , x, z) Define: A context free grammar G Start state q 0 X on the input tape Empty stack Such that: End with (q, λ , β ) Language accepted by M is the same as End in an accepting state (q ∈ F) Language generated by G All characters of x have been read Some string on the stack (doesn’t matter what). 5
Pushdown Automata Final State vs. Empty Stack Strings accepted by a PDA by Empty Stack The two means by which a PDA can Start at (q 0 , x, z) accept are equivalent wrt the class of Start state q 0 languages accepted X on the input tape Given a PDA M such that L = L(M), there Empty stack exists a PDA M’ such that L = N(M’) End with (q, λ , λ ) End in any state Given a PDA M such that L = N(M), there All characters of x have been read exists a PDA M’ such that L = L(M’) Stack is empty Step 2: PDA → CFG Step 2: PDA → CFG Basic idea Given: A pushdown automata M that We define variables in G to be triplets: accepts by empty stack [p, A, q] will represent a variable, that can generate all Define: A context free grammar G strings x that: Upon reading x on the PDA tape will Take you from state p to state q in the PDA and Have a “net result” of popping A off the stack Such that: Note that it may take many moves to get there. Language accepted by M is the same as Language generated by G Step 2: PDA → CFG Step 2: PDA → CFG Productions of G More Productions of G 1. For all states q in M, add the production 2. For every q, q 1 ∈ Q, a ∈Σ∪ { λ } and A ∈Γ S → [q 0 zq] If δ (q, a, A) contains (q 1 , λ ) then add Following these productions will generate all strings [qAq 1 ] → a that start at q o , and result in an empty stack. Final state is not important. Meaning you can get from q to q 1 while In other words, all strings accepted by M. popping A from the stack by reading an a. 6
a, A / B 1 …B m Step 2: PDA → CFG Step 2: PDA → CFG q q 1 One can show by induction (though we Even More Productions of G won’t) that 3. For every q, q 1 ∈ Q, a ∈Σ∪ { λ } and A ∈Γ [qAp] ⇒ * x iff (q, x, A) → * (p, λ , λ ) If δ (q, a, A) contains (q 1 , B 1 B 2 …B m ) then For every possible sequence of states q 2 , …q m+1 Add More specifically [q 0 zp] ⇒ * x and since we added [qAq m+1 ] → a[q 1 B 1 q 2 ] [q 2 B 2 q 3 ] … [q m B m q m+1 ] the productions S → [q 0 zp] for all p, then x ∈ L(G) Meaning: On the flip side S → [q 0 zp] will always be the first one way to pop A off the stack and to get from q to q m+1 is to production of any derivation of G read an a (q 0 , x, z) → * (p, λ , λ ) use some input to pop B 1 off the stack (bring you from q 1 to q 2 in the process), So x is accepted by empty stack While in q 2 , use some input to pop B 2 off the stack (bringing x ∈ L(M) you to q 3 in the process) And so on… Step 2: PDA → CFG Step 2: PDA → CFG Example Example M = (Q, Σ , Γ , δ , q 0 , z ,F) L = { 0 i 1 j | i ≥ j ≥ 1} λ ,X / λ Q = { q 0 , q 1 } 1,X / λ Σ = { 0, 1 } 1, X / λ λ ,Z / λ Γ = { X, Z } q 1 q 0 z = Z 0,Z / XZ F = ∅ 0,X / XX Step 2: PDA → CFG Step 2: PDA → CFG Corresponding CFG Corresponding CFG Type 2 productions Type 1 productions If δ (q, a, A) contains (q 1 , λ ) then add For all states q, S → [q 0 zq] [qAq 1 ] → a S → [q 0 Zq 1 ] δ (q 0 , 1, X) = (q 1 , λ ) [q 0 Xq 1 ] → 1 S → [q 0 Zq 0 ] δ (q 1 , 1, X) = (q 1 , λ ) [q 1 Xq 1 ] → 1 δ (q 1 , ε , X) = (q 1 , λ ) [q 1 Xq 1 ] → λ δ (q 1 , ε , Z) = (q 1 , λ ) [q 1 Zq 1 ] → λ 7
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