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Equivalence of CFGs and PDAs

Section 3.4

Mon, Oct 24, 2005

Every CFG is Equivalent to a PDA Theorem: Let G = {V, , R, S} be a context-free

grammar. There exists a PDA M such that L(M) = L(G).

Proof: Let M = {K, Σ, Γ, Δ, s, F}, where K = {s, f}. Γ = V. F = {f}. Δ = {((s, e, e), (f, S))}

{((f, e, A), (f, w)) | A w R}

{((f, a, a), (f, e)) | a }.

Every CFG is Equivalent to a PDA A typical production rule is of the form

A w. A typical tape symbol is a. It is easy to see that this PDA accepts L(G).

se, e; S

a, a; e

e, A; w

f

Example Design a PDA that accepts the language of the

grammar S AB A aA | e B aBb | e

Example

e, S; ABe, A; aAe, A; e

e, B; aBbe, B; e

se, e; S

a, a; eb, b; e

f

Example This PDA accepts the string aaaabb.

(s, aaaabb, e) (f, aaaabb, S)

(f, aaaabb, AB)

(f, aaaabb, aAB)

(f, aaabb, AB)

(f, aaabb, aAB)

(f, aabb, AB)

(f, aabb, B)

(f, aabb, aBb)

(f, abb, Bb)

Example (f, abb, aBbb)

(f, bb, Bbb)

(f, bb, bb)

(f, b, b)

(f, e, e).

Another Example Design a PDA that accepts that language of the

grammar E E + E | E * E | (E) | V V a | b | c

Another Example

e, E; E + Ee, E; E * Ee, E; (E)e, E; V

se, e; E

a, a; eb, b; ec, c; e

f

e, V; ae, V; be, V; c

(, (; e), ); e+, +; e*, *; e

Another Example Derive the string (a + b) * c.

(s, (a + b) * c, e) (f, (a + b) * c, E) (f, (a + b) * c, E * E) (f, (a + b) * c, (E) * E) (f, a + b) * c, E) * E) (f, a + b) * c, E + E) * E) (f, a + b) * c, V + E) * E) (f, a + b) * c, a + E) * E) (f, + b) * c, + E) * E) (f, b) * c, E) * E)

Another Example (f, b) * c, N) * E)

(f, b) * c, b) * E)

(f, ) * c, ) * E)

(f, * c, * E)

(f, c, E)

(f, c, V)

(f, c, c)

(f, e, e)

Nondeterminism Was nondeterminism used in the last example? Do CFGs generally lead to nondeterministic PDAs by

this method?

Another Example Another, more intuitive, PDA that accepts the

language of E E + E | E * E | (E) | V V a | b | c

qs

(a, e; e)(b, e; e)(c, e; e)

(‘(’, e; ‘(’)

(+, e; e)(*, e; e)

(‘)’, ‘)’; e)

Every PDA is Equivalent to a CFG Theorem: Let M = {K, Σ, Γ, Δ, s, F} be a push-down

automaton. There exists a CFG G such that L(G) = L(M).

Proof: Create a new start state s' and a new final state f ' and a new

symbol $. Create the transition ((s', e, e), (s, $)). Create the transitions ((f, e, $), (f ', e)) for every f F.

Every PDA is Equivalent to a CFG

Next, rewrite the PDA as an equivalent simple PDA. For every transition ((q, a, ), (p, )) where q s,

. || 2.

Modify all transitions for which || 2. Modify all transitions for which || > 2. Modify all transitions for which = e. See pp. 140 – 141 for details.

Every PDA is Equivalent to a CFG

The nonterminals in the CFG will be triples p, A, q, where p, q K and A {e, $}.

p, A, q can be interpreted as

“The problem of getting from state p to state q while popping A from the stack.”

As we write the rules of the grammar, will assume that the initial move ((s', e, e), (s, $)) has been taken.

In other words, we begin in state s with $ (the bottom marker) on the stack.

Every PDA is Equivalent to a CFG The first rule is

S s, $, f '. This represents the problem of getting from s to f ', popping

$ along the way. In other words, (s, w, $) * (f ', v, e) for some w, v Σ*.

s f 'Pop $

Every PDA is Equivalent to a CFG The second set of rules.

For each transition ((q, a, B), (r, C)) where a {e} and B, C {e}, and for each p K,

q, B, p ar, C, p. This represents replacing the problem of getting from q to

p, popping B, with the problem of getting from r to p, reading a and popping C.

q

r

pPop BPush C

Pop B

Pop C

Every PDA is Equivalent to a CFG The third set of rules.

For each transition ((q, a, B), (r, C1C2)) where a {e} and B {e}, C1, C2 , and for each p, p' K,

q, B, p ar, C1, p'p', C2, p.

q

r

pPop BPush C2

Push C1

Pop B

Pop C2

p'

Pop C1

Every PDA is Equivalent to a CFG The fourth set of rules.

For each q K,

q, e, q e.

The grammar generated by these rules has the same language as the original PDA.

Example Use the algorithm to find a grammar for the language

of the following PDA.

ps

a, e; a

e, e; e

b, a; e

Modify the PDA Add the states s′ and f ′.

ps

a, e; a

e, e; e

b, a; e

s′ f ′e, e; $ e, $; e

Modify the PDA Replace the transitions with simple transitions.

ps

a, a; aaa, $; a$

e, a; ae, $; $

b, a; e

s′ fe, e; $ e, $; e

The First Kind of Production The first group:

S s, $, f

The Second Kind of Production The second group, for transitions ((q, a, B), (r, C)): ((s, e, a), (p, a))

s, a, s p, a, s s, a, p p, a, p s, a, f p, a, f

((s, e, $), (p, $)) s, $, s p, $, s s, $, p p, $, p s, $, f p, $, f

The Second Kind of Production ((p, b, a), (p, e))

p, a, s bp, e, s p, a, p bp, e, p p, a, f bp, e, f

((p, e, $), (f ′, e)) p, $, s f, e, s p, $, p f, e, p p, $, f f, e, f

The Third Kind of Production The third group, for transitions ((q, a, B), (r, C1C2)):

((s, a, a), (s, aa)) s, a, s as, a, ss, a, s s, a, s as, a, pp, a, s s, a, s as, a, ff, a, s s, a, p as, a, ss, a, p s, a, p as, a, pp, a, p s, a, p as, a, ff, a, p s, a, f ′ as, a, ss, a, f s, a, f ′ as, a, pp, a, f s, a, f ′ as, a, ff, a, f

The Third Kind of Production The third group, for transitions ((q, a, B), (r, C1C2)):

((s, a, $), (s, a$)) s, $, s as, a, ss, $, s s, $, s as, a, pp, $, s s, $, s as, a, ff, $, s s, $, p as, a, ss, $, p s, $, p as, a, pp, $, p s, $, p as, a, ff, $, p s, $, f ′ as, a, ss, $, f s, $, f ′ as, a, pp, $, f s, $, f ′ as, a, ff, $, f

The Fourth Kind of Production The fourth group:

s, e, s e p, e, p e f, e, f e

Simplify the Grammar Give each “nonterminal” a simple name.

A = s, a, s B = s, a, p C = s, a, f D = p, a, s E = p, a, p F = p, a, f G = f, a, s H = f, a, p I = f, a, f


J = s, $, s K = s, $, p L = s, $, f M = p, $, s N = p, $, p O = p, $, f P = f, $, s Q = f, $, p R = f, $, f


T = s, e, s U = s, e, p V = s, e, f W = p, e, s X = p, e, p Y = p, e, f Z = f, e, s = f, e, p = f, e, f

Organize the grammar Write the productions, using the simpler names, in

alphabetical order (except for S), grouping them according to the nonterminal on the left.

Organize the grammar S L A D | aAA | aBD | aCG B E | aAB | aBE | aCH C F | aAC | aBF | aCI D bW E bX F bY J M | aAJ | aBM | aCP K N | aAK | aBN | aCQ L O | aAL | aBO | aCR

Organize the grammar M Z N O T e X e e

Simplify the grammar This algorithm typically produces a huge list of

productions, most of which are unnecessary. We will describe a procedure that will eliminate all

unnecessary productions.

Eliminate “Dead-End” Nonterminals Identify all nonterminals that do not appear on the left

side of any production or that do not appear on the right side of any production.

Remove all productions involving those nonterminals.

Repeat this until there is no change.

Eliminate “Dead-End” Nonterminals Those nonterminals are initially G, H, I, P, Q, R, T,

U, V, W, Y, Z, . Then we also eliminate D, F, M, N. The remaining productions are

S L A aAA B E | aAB | aBE C aAC E bX J aAJ

Eliminate “Dead-End” Nonterminals

K aAK L O | aAL | aBO O X e e

Eliminate Purely Recursive Nonterminals Identify all nonterminals with the property that

whenever they appear on the left side of a production, they also appear on the right side.

Remove all productions involving those nonterminals.

Repeat this until there is no change.

Eliminate Purely Recursive Nonterminals Thus, we should eliminate productions containing A,

C, J, K. The remaining productions are

S L B E | aBE E bX L O | aBO O X e e

Eliminate Trivial Nonterminals Identify all nonterminals whose only production is to

replace them with the empty string or a single terminal.

Substitute e into all right sides in which they appear. Eliminate their productions. Repeat this until there is no change.

Eliminate Trivial Nonterminals Thus, we should initially eliminate T, X, . Then eliminate E, O. The remaining productions are

S L B b | aBb L e | aB

Eliminate Trivial Nonterminals Identify all nonterminals whose only production

replaces them with another nonterminal. Make the substitution in all right sides. Eliminate their productions. The production for the start symbol S may be handled

differently since we must retain S.

Eliminate Trivial Nonterminals The remaining productions are

S aB | e B b | aBb

This is a grammar for {anbn | n 0}.

Examples Let = {a, b}. Find a grammar for the language

{w * | in each prefix of w, #a #b}. Find a grammar for the language

{w * | #a = #b}. Find a grammar for the language

{w * | w = xxR for some x *}. Find a grammar for the language

{w * | w = xay for some x, y * with |x| = |y|}.