Automata and Formal Languages Push Down Automata Sipser

Automata and Formal Languages Push Down Automata Sipser pages 109 - 114 Lecture 13 Tim Sheard 1

Push Down Automata (PDAs) are e-NFAs with stack memory. Transitions are labeled by an input symbol together with a pair of the form X/a. The transition is possible only if the top of the stack contains the symbol X After the transition, the stack is changed by replacing the top symbol X with the string of symbols a. (Pop X, then push symbols of a. )

Example PDAs can accept languages that are not regular. The following one accepts: L={0 i 1 j | 0 £ i £ j} 0, X/XX s 1, Z 0/Z 0 1, X/e 0, Z 0/XZ 0 1, X/e p e, Z 0/Z 0 q

Definition A PDA is a 6 -tuple P=(Q, S, G, d, q 0, F) where Q, S, q 0, F are as in NFAs, and • G is the stack alphabet. It is assumed that initially the stack is empty. • d: Q × Sε × Gε ¾® P(Q × G ε *) is the transition function: given a state, an input symbol (or e), and a stack symbol, Gε, it gives us a finite number of pairs (q, a), where q is the next state and a is the string of stack symbols that will replace X on top of the stack. • Recall Sε =(SÈ{e}) Gε =(GÈ{e})

In our example, the transition from s to s labeled (0, Z 0/XZ 0) corresponds to the fact (s, XZ 0) d(s, 0, Z 0). A complete description of the transition function in this example is given by d(s, 0, Z 0) = {(s, XZ 0)} d(s, 0, X) = {(s, XX)} 0, X/XX d(s, e, Z 0) = {(q, Z 0)} 0, Z /XZ 1, X/e d(s, 1, X) = {(p, e)} 1, X/e e, Z /Z s p d(p, 1, X) = {(p, e)} d(p, e, Z 0) = {(q, Z 0)} e, Z /Z d(q, 1, Z 0) = {(q, Z 0)} and d(q, a, Y)=Æ for all other possibilities. 0 1, Z 0/Z 0 0 0 q

Sipser style acceptance • Suppose a string w can be written: w 1 w 2 … wm • Wi Sε Some of the wi are allowed to be e • I. e. One may write “abc” as aebce • If there exist two sequences • s 0 r 1 … r m Q • s 0 s 1 … s m G* (The si represent the stack contents at step i) 1. r 0=q 0 and s 0 = ε (ri+1, b) d(ri, wi+1, a) 2. si = at 3. Rm F si+1 = bt

Instantaneous Descriptions and Moves of PDAs IDs (also called configurations) describe the execution of a PDA at each instant. An ID is a triple (q, w, a), with this intended meaning: • q is the current state • w is the remaining part of the input • a is the current content of the stack, with top of the stack on the left.

The relation |- describes possible moves from one ID to another during execution of a PDA. If d(q, a, X) contains (p, a), then (q, aw, Xb) |- (p, w, ab) is true for every w and b. The relation |-* is the reflexive-transitive closure of |We have (q, w, a) |-* (q', w', a') when (q, w, a) leads through a sequence (possibly empty) of moves to (q', w', a')

Automata and Formal Languages 0, X/XX s 1, Z 0/Z 0 1, X/e 0, Z 0/XZ 0 1, X/e p L, Z 0/Z 0 q Z 0/Z 0 (s, 011, z) |- (s, 11, xz) |- (P, 1, z)|-(q, 1, z) |- (q, ””, Z) (s, 011, z) |- (q, 011, Z) Lecture 13 Tim Sheard 9

Properties of |Property 1. If Then (q, x, a) |-* (p, y, b) (q, xw, ag) |-* (p, yw, bg) If you only need some prefix of the input (x) and stack (a) to make a series of transitions, you can make the same transitions for any longer input and stack. Property 2. If Then (q, xw, a) |-* (p, yw, b) (q, x, a) |-* (p, y, b) It is ok to remove unused input, since a PDA cannot add input back on once consumed.

Another notion of acceptance A PDA as above accepts the string w iff (q 0, w, ε) |-* (p, e, a) is true for some final state p and some a. (We don't care what's on the stack at the end of input. ) The language L(P) of the PDA P is the set of all strings accepted by P.

Here is the chain of IDs showing that the string 001111 is accepted by our example PDA: |||||||- (s, 001111, ε) (s, 01111, XZ 0) (s, 1111, XXZ 0) (p, 111, XZ 0) (p, 11, Z 0) (q, e, Z 0) 0, X/XX s 1, Z 0/Z 0 1, X/e 0, ε/XZ 0 1, X/e p L, Z 0/Z 0 q

The language of the following PDA is {0 i 1 j | 0 < i £ j}*. How can we prove this? 1, X, e 0, X/XX s 0, Z 0/XZ 0 p L, Z 0/Z 0 1, X/e q

Example A PDA for the language of balanced parentheses: (, Z 0/XZ 0 (, X/XX p ), X/e L, Z 0/Z 0 q

Acceptance by Empty Stack Define N(P) to be the set of all strings w such that (q 0, w, ε) |-* (q, e, e) for some state q. These are the strings P accepts by empty stack. Note that the set of final states plays no role in this definition. Theorem. A language is L(P 1) for some PDA P 1 if and only if it is N(P 2) for some PDA P 2.

Proof 1 1. From empty stack to final state. Given P 2 that accepts by empty stack, get P 1 by adding a new start state and a new final state as in the picture below. We also add a new stack symbol X 0 and make it the start symbol for P 1's stack. P 2 p 0 e, X 0/Z 0 X 0 q 0 e, X 0/e (add this transition from all states of P 2 to new state Pf) Pf

Proof 2 2. From final state to empty stack. Given P 1, we get P 2 again by adding a new start state, final state and start stack symbol. New transitions are seen in the picture. P 2 p 0 e, X 0/Z 0 X 0 q 0 e, G/ e P e, G/ e