CS 3240 – Chapter 7 Language Machine Grammar Regular Finite Automaton Regular Expression, Regular Grammar Context-Free Pushdown Automaton Context-Free Grammar Recursively Enumerable Turing Machine Unrestricted PhraseStructure Grammar CS 3240 - Introduction 2 Take an FA and add a stack A restricted form of unbounded memory Operations: Pop Push Ignore (λ) (Pop always precedes push) CS 3240 - Pushdown Automata 3 Three items per transition edge <read>,<pop>,<push> any or all could be λ A PDA accepts when: An accepting state is reached, and The stack is empty ▪ Different from book! ▪ The machine crashes when no move is possible ▪ implicit jail CS 3240 - Pushdown Automata 4 A deterministic PDA (no choices; λ is okay sometimes). The stack alphabet (Γ = {X}) can differ from the input alphabet (Σ = {a, b}) . Trace aabb… CS 3240 - Pushdown Automata 5 q0 q0 q0 q1 q1 aabb abb bb b λ λ X XX X λ Traditional notation (see book): (q0,aabb, λ) ⊢ (q0,abb,X) ⊢ (q0,bb,XX) ⊢ (q1,b,X) ⊢ (q1,λ, λ) Also: (q0,aabb,λ) ⊢* (q1,λ, λ) CS 3240 - Pushdown Automata (⇒ aabb ∈ L) 6 CS 3240 - Pushdown Automata 7 How would you do a2nbn? CS 3240 - Pushdown Automata 8 Each b must pop 2 X’s CS 3240 - Pushdown Automata 9 Is this deterministic? Trace the string abaacaaba. CS 3240 - Pushdown Automata 10 Input: (state-1, input character, pop character) Output: (state-2, push character(s)) Example: The edge (a,λ,a) on previous slide: δ(q0,a,λ) = (q0,a) Exercise: Rewrite wcwR in functional form CS 3240 - Pushdown Automata 11 Is this deterministic? Trace the string abaaaaaba. Trace the string abaabaaba. Trace the string abaaaaba. CS 3240 - Pushdown Automata 12 The grammar: S → aSb | aSbb | λ Trace aaabbbb CS 3240 - Pushdown Automata 13 Trace ababbaab CS 3240 - Pushdown Automata 14 CS 3240 - Pushdown Automata 15 Reasoning: There must be 2 b’s for every a. When we read an a, there are three possibilities: 1)Its 2 b’s have already been read 2)1 b has been read and 1 remains 3)Both b’s are yet to be read Try abababbbb. How would you do Twicea? CS 3240 - Pushdown Automata 16 CS 3240 - Pushdown Automata 17 CS 3240 - Pushdown Automata 18 CS 3240 - Pushdown Automata 19 Q: set of states ∑: input alphabet Γ: stack alphabet δ: Q x (∑∪λ) x (Γ∪λ) → finite subsets of Q x Γ* q0: start state F ⊆ Q: final states (z ⋲ Γ: optional stack start symbol) CS 3240 - Pushdown Automata 20 Not strictly necessary Can serve as a “bottom marker” to allow detection of an empty stack A convenience Can make things easier to design/understand Sometimes handy when seeking an associated CFG Can eliminate some lambda pops ▪ Handy later CS 3240 - Pushdown Automata 21 CS 3240 - Pushdown Automata 22 CS 3240 - Pushdown Automata 23 CS 3240 - Pushdown Automata 24 CS 3240 - Pushdown Automata 25 How do we know that PDAs are the correct machine model for CFLs? 1) Must show that for any CFG, there is a PDA that accepts its language 2) Must show that for any PDA, there is a CFG that generates its language The first case is easier than the second! It can be handy to go back and forth CS 3240 - Pushdown Automata 26 The leading character is read The LHS variable is popped The RHS string is pushed Initial transition pushes S See next slide CS 3240 - Pushdown Automata 27 CS 3240 - Pushdown Automata 28 CS 3240 - Pushdown Automata 29 Very trivial (Yay!) But the (2-state) PDA is non-deterministic State-2 is a final state 1) Push S on the stack; move to state-2 2) For every rule, have a transition that: pops the left-hand side and pushes the right-hand side 3) For every character, c, in Σ, have a rule that: consumes and pops c simultaneously CS 3240 - Pushdown Automata 30 S → aSb | aSbb | λ Trace aaabbbbb: (q0,aaabbbbb,λ) ⊢ (q1,aaabbbbb,S) ⊢ (q1,aaabbbbb,aSbb) ⊢ (q1,aabbbbb,Sbb) ⊢ (q1,aabbbbb,aSbbbb) ⊢ (q1,abbbbb,Sbbbb) ⊢ (q1,abbbbb,aSbbbbb) ⊢ (q1,bbbbb,Sbbbbb) ⊢ (q1,bbbbb,bbbbb) ⊢* (q1,λ,λ) CS 3240 - Pushdown Automata 31 S → aSbSbS | bSaSbS | bSbSaS | λ … And trace bbabaabbb CS 3240 - Pushdown Automata 32 S => (S) | SS | λ Derive ()(()) from the grammar. Then accept it by the machine. Then find a one-state DPDA for this language. CS 3240 - Pushdown Automata 33 We need to relate PDA movement to a CFG production rule somehow Key: consuming a character in the PDA corresponds to generating a character in the CFG The stack contents must also be part of all this CS 3240 - Pushdown Automata 34 PDAs of the form below have a natural translation to a CFG the reverse of CFG-to-PDA conversion (conveniently) call the stack start symbol S Translate each transition in state-2 into a rule: ▪ c,X,Y ⟺ X → cY (i.e., <pop> → <char> <push>) ▪ (X cannot be λ) CS 3240 - Pushdown Automata 35 S → aXS | bYS | λ X → aXX | bYX | b Y → bYY | aXY | a Derive abbbaaba CS 3240 - Pushdown Automata 36 Go back and find grammars for slides 22,24-25. Also, work backward from the (()) DPDA to a grammar. CS 3240 - Pushdown Automata 37 We must also track state changes Our variables must therefore contain state and stack information Very tricky! CS 3240 - Pushdown Automata 38 Suppose X is on the stack and ‘a’ is read What can happen to X? It will be popped (and possibly put back) It may be replaced by one or more other stack symbols ▪ Which symbols will also later be popped, replaced, etc… ▪ The stack grows and shrinks and grows and shrinks … Eventually, as more input is consumed, the effect of having pushed X on the stack must be erased (or we’ll never reach an empty stack!) And the state may change many times! CS 3240 - Pushdown Automata 39 … (qi, x1x2…xk, Y1Y2…Yk) ⊢* (qj, x2…xk, Y2…Yk) ⊢* (qn, xk, Yk) ⊢* (qf, λ, λ) CS 3240 - Pushdown Automata 40 Let the symbol <qAp> represent the sequence of movements in a PDA that: starts in state q eventually ends in state p eventually removes A and its after-effects from the stack The symbol <sλf> then represents accepting a valid string (if s is the start state and f is a final state) These symbols will be our variables Because they track the machine configuration that accepts strings Just as the grammar will generate those strings CS 3240 - Pushdown Automata 41 Consider the transition δ(q,a,X) = (p,Y) This means that a is consumed, X is popped, we move directly to state p, and subsequent processing must eventually erase Y and its follow-on effects A corresponding grammar rule is: <qX?> → a<pY?> (?’s represent the same state) We don’t know where we’ll eventually end up when the stack is back to where it was before the X was first pushed All we know is we immediately go through p So we entertain all possibilities (see 3 and 4 on next slide) CS 3240 - Pushdown Automata 42 1) S → <sλf> for all final states, f 2) <qλq> → λ for all states, q These will serve as terminators 3) For transitions δ(q,a,X) = (p,Y): <qXr> → a<pYr> for all states, r 4) For transitions δ(q,a,X) = (p,Y1Y2): <qXr> → a<pY1s><sY2r> for all states, r, s And so on, for longer pushed strings CS 3240 - Pushdown Automata 43 The process requires: No multiple pops ▪ So spread them out over multiple states if needed Also, for every lambda-pop c,λ,X: Add an equivalent rule for each letter of Γ (the stack alphabet) as follows: ▪ (c,X,XX) ▪ (c,Y,XY) ▪ etc. ▪ And keep the original lambda pop!!! CS 3240 - Pushdown Automata 44 • Call the start state, s, and the final state, f. • Add the following transition on s: (a,X,XX) • 1) S → <sλf> • 2) <sλs> → λ; <fλf> → λ • Now go through each transition… CS 3240 - Pushdown Automata 45 δ(s,a,λ) = (s,X): 1. <sλs> → a<sXs> <sλf> → a<sXf> δ(s,a,X) = (s,XX): 2. <sXs> → a<sXs><sXs> <sXf> → a<sXs><sXf> <sXs> → a<sXf><fXs> <sXf> → a<sXf><fXf> CS 3240 - Pushdown Automata 46 δ(s,b,X) = (f,λ): 3. <sXs> → b<fλs> <sXf> → b<fλf> δ(f,b,X) = (f,λ): 4. <fXs> → b<fλs> <fXf> → b<fλf> CS 3240 - Pushdown Automata 47 S → <sλf> <sλs> → a<sXs> | λ <fλf> → λ <sλf> → a<sXf> <sXs> → a<sXs><sXs> <sXf> → a<sXs><sXf> | a<sXf><fXf> | b<fλf> <fXf> → b<fλf> CS 3240 - Pushdown Automata 48 S → <sλf> <sλs> → a<sXs> | λ <sλf> → a<sXf> <sXs> → a<sXs><sXs> <sXf> → a<sXs><sXf> | a<sXf><fXf> | b <fXf> → b CS 3240 - Pushdown Automata 49 S → <sλf> <sλf> → a<sXf> <sXf> → a<sXf><fXf> | b <fXf> → b CS 3240 - Pushdown Automata 50 S → a<sXf> <sXf> → a<sXf><fXf> | b <fXf> → b CS 3240 - Pushdown Automata 51 S → a<sXf> <sXf> → a<sXf>b | b CS 3240 - Pushdown Automata 52 S → aX X → aXb | b Done! (finally :-) CS 3240 - Pushdown Automata 53 Find a CFG for nb = na + 1 using the generic conversion process CS 3240 - Pushdown Automata 54 A PDA is deterministic if there are no choices: Depends on a state’s <character, pop symbol> pair: 1) δ(q,a,X) has only one choice (a could be λ), and 2) If a is not λ, then there is no δ(q,λ,X) edge, and 3) if X is not λ, then there is no δ(q,a,λ) edge. NOTE: Some PDAs are inherently nondeterministic (no deterministic equivalent, e.g., wwR) CS 3240 - Pushdown Automata 55 Revisit previous PDAs and check for determinacy CS 3240 - Pushdown Automata 56 Languages accepted by nondeterministic PDA Languages accepted by DFA Languages accepted by deterministic PDA CS 3240 - Pushdown Automata 57

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# Pushdown Automata