Pushdown Automata

Part II: PDAs and CFG

Chapter 121

PDAs and Context-Free Grammars

Theorem: The class of languages accepted by PDAs is exactly the class of context-free languages.

Recall: context-free languages are languages that can be defined with context-free grammars.

Restate theorem:

Can describe with context-free grammar

Can accept by PDA

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Going One Way

Lemma: Each context-free language is accepted by some PDA.

Proof (by construction):

The idea: Let the stack do the work.

Two approaches:

• Top down

• Bottom up

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Top Down The idea: Let the stack keep track of expectations.

Example: Arithmetic expressions

E E + TE TT T FT FF (E)F id

(1) (q, , E), (q, E+T) (7) (q, id, id), (q, )(2) (q, , E), (q, T) (8) (q, (, ( ), (q, )(3) (q, , T), (q, T*F) (9) (q, ), ) ), (q, )(4) (q, , T), (q, F) (10) (q, +, +), (q, )(5) (q, , F), (q, (E) ) (11) (q, , ), (q, )(6) (q, , F), (q, id)

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A Top-Down Parser

Answering “Could G generate w?” starting from S.

The outline of M is:

M = ({p, q}, , V, , p, {q}), where contains: ● The start-up transition ((p, , ), (q, S)).

● For each rule X s1s2…sn. in R, the transition: ((q, , X), (q, s1s2…sn)).

● For each character c , the transition: ((q, c, c), (q, )). 5

Example of the Construction L = {anb*an}

0 (p, , ), (q, S) (1) S 1 (q, , S), (q, )(2) S B 2 (q, , S), (q, B)(3) S aSa 3 (q, , S), (q, aSa)(4) B 4 (q, , B), (q, )(5) B bB 5 (q, , B), (q, bB)

6 (q, a, a), (q, )input = a a b b a a 7 (q, b, b), (q, )trans state unread input stack

p a a b b a a 0 q a a b b a a S3 q a a b b a a aSa6 q a b b a a Sa3 q a b b a a aSaa6 q b b a a Saa2 q b b a a Baa5 q b b a a bBaa7 q b a a Baa5 q b a a bBaa7 q a a Baa4 q a a aa6 q a a6 q 6

Bottom-Up

(1) E E + T(2) E T(3) T T F (4) T F(5) F (E)(6) F id

Reduce Transitions:(1) (p, , T + E), (p, E)(2) (p, , T), (p, E)(3) (p, , F T), (p, T)(4) (p, , F), (p, T)(5) (p, , )E( ), (p, F)(6) (p, , id), (p, F)

Shift Transitions(7) (p, id, ), (p, id) (8) (p, (, ), (p, () (9) (p, ), ), (p, )) (10) (p, +, ), (p, +) (11) (p, , ), (p, )

The idea: Let the stack keep track of what has been found.

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A Bottom-Up Parser Answering “Could G generate w?” starting from w.

The outline of M is:

M = ({p, q}, , V, , p, {q}), where contains:

● The shift transitions: ((p, c, ), (p, c)), for each c .

● The reduce transitions: ((p, , (s1s2…sn.)R), (p, X)), for each rule X s1s2…sn. in G.

● The finish up transition: ((p, , S), (q, )).

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Example of the Construction L = {anb*an}

0 (p, , S), (q, ) (1) S 1 (p, , ), (p, S)(2) S B 2 (p, , B), (p, S)(3) S aSa 3 (p, , aSa), (p, S)(4) B 4 (p, , ), (p, B)(5) B bB 5 (p, , Bb), (p, B)

6 (p, a, ), (p, a)input = a a b b a a 7 (p, b, ), (p, b)trans state unread input stack

p a a b b a a 6 p a b b a a a6 p b b a a aa7 p b a a baa7 p a a bbaa4 p a a Bbbaa5 p a a Bbaa5 p a a Baa2 p a a Saa6 p a aSaa3 p a Sa6 p aSa3 p S0 q 9

Notice Nondeterminism Machines constructed with the algorithm are often nondeterministic,

even when they needn't be. This happens even with trivial languages.

Example: AnBn = {anbn: n 0}

A grammar for AnBn is: A PDA M built top-down for AnBn is:

(0) ((p, , ), (q, S))[1] S aSb (1) ((q, , S), (q, aSb))[2] S (2) ((q, , S), (q, ))

(3) ((q, a, a), (q, )) (4) ((q, b, b), (q, ))

But transitions 1 and 2 make M nondeterministic.

A directly constructed machine for AnBn:

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How about bottom-up?

Going The Other WayLemma: If a language is accepted by a pushdown automaton M, it is

context-free (i.e., it can be described by a context-free grammar).

Proof (by construction): (We will skip this to Slide 16, as it’s rarely necessarily going this

way in practice!)

Step 1: Convert M to restricted normal form:

● M has a start state s that does nothing except push a special symbol # onto the stack and then transfer to a state s from which the rest of the computation begins. There must be no transitions back to s.

● M has a single accepting state a. All transitions into a pop # and

read no input.

● Every transition in M, except the one from s, pops exactly one symbol from the stack.

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Converting to Restricted Normal Form

Example:

{wcwR : w {a, b}*}

Add s and a:

Pop no more than one symbol:

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M in Restricted Normal Form

[1] ((s, a, #), (s, a#)), ((s, a, a), (s, aa)), ((s, a, b), (s, ab)), [2] ((s, b, #), (s, b#)), ((s, b, a), (s, ba)), ((s, b, b), (s, bb)), [3] ((s, c, #), (f, #)), ((s, c, a), (f, a)), ((s, c, b), (f, b))

Must have one transition for everything that could have been on the top of the stack so it can be popped and then pushed back on.

Pop exactly one symbol: Replace [1], [2] and [3] with:

[1]

[2]

[3]

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Second Step - Creating the Productions

Example: WcWR

M =

The basic idea –

simulate a leftmost derivation of M on any input string.14

Second Step - Creating the Productions

Example: abcba

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Nondeterminism and Halting

1. There are context-free languages for which no deterministic PDA exists.

2. It is possible that a PDA may ● not halt, ● not ever finish reading its input.

3. There exists no algorithm to minimize a PDA. It is undecidable whether a PDA is minimal.

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Nondeterminism and Halting It is possible that a PDA may ● not halt, ● not ever finish reading its input.

Let = {a} and consider M =

L(M) = {a}: (1, a, ) |- (2, a, a) |- (3, , )

On any other input except a: ● M will never halt. ● M will never finish reading its input unless its input is .

Example: aa 17

Solutions to the Problem

● For NDFSMs: ● Convert to deterministic, or ● Simulate all paths in parallel.

● For NDPDAs: ● Impossible to simulate all paths in parallel – each path would need its won stack ● Formal solutions usually involve changing the grammar parse tree may not make sense ● Practical solutions that: ● Preserve the structure of the grammar, but ● Only work on a subset of the CFLs.

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Alternative Equivalent Definitions of a PDA (Skip to Slide 21)

Accept by accepting state at end of string (i.e., we don't care about the stack).

From M (in our definition) we build M (in this one):

1. Initially, let M = M.2. Create a new start state s. Add the transition:

((s, , ), (s, #)).3. Create a new accepting state qa. 4. For each accepting state a in M do,

4.1 Add the transition ((a, , #), (qa, )).5. Make qa the only accepting state in M.

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Example

The balanced parentheses language

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● FSM plus FIFO queue (instead of stack)?

● FSM plus two stacks?

What About These?

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Comparing Regular and Context-Free Languages

Regular Languages Context-Free Languages● regular exprs. ● or● regular grammars ● context-free grammars● recognize ● parse● = DFSMs ● = NDPDAs

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