Pushdown Automata - Hampden-Sydney College

Pushdown 

Automata - 

Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Pushdown Automata - Equivalence to CFGs 

Lecture 19 

Section 2.2 

Robb T. Koether 

Hampden-Sydney College 

Wed, Oct 8, 2008

Outline 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

1 Homework Review 

2 Equivalence of PDAs and CFGs 

Proof ⇒ 

Proof ⇐ 

3 Long Example 

4 Short Example 

5 Assignment

Homework Review 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Exercise 

Let Σ = {a, (, ), [, ]}. Design a PDA whose language is 

{w | w contains balanced parentheses and brackets}. 

Short 

Example 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Solution 

The a’s can be placed anywhere. 

A ( or a [ can be read anywhere, but ( must eventually 

be balanced by ) and [ must eventually be balanced by 

]. 

Use the stack to assure that the parentheses and 

brackets are properly matched.




Equivalence 

to CFGs 

Robb T. 

Koether 

Solution 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

a, ε → ε 

(, ε → ( 

[, ε → [ 

), ( → ε 

], [ → ε 

ε, ε → $ ε, $ → ε

Equivalence of PDAs and CFGs 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Theorem (Equivalence of PDAs and CFGs) 

A language is context-free if and only if it is accepted by 

some PDA. 

Short 

Example 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Proof (⇒). 

We are given a CFG G and we will construct a PDA M. 

Let M have four states: 

Q = {q 0 , q 1 , q 2 , q 3 }. 

q 0 is the start state. 

q 3 is the accept state. 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Proof (⇒). 

The transitions are 

δ(q 0 , ε, ε) = {(q 1 , $)} 

δ(q 1 , ε, ε) = {(q 2 , S)} 

δ(q 2 , ε, A) = {(q 2 , w)}, where A → w is a rule. 

δ(q 2 , a, a) = {(q 2 , ε)}, for all a ∈ Σ. 

δ(q 2 , ε, $) = {(q 3 , ε)} 

It is clear that L(M) = L(G).




Equivalence 

to CFGs 

Robb T. 

Koether 

Proof (⇒). 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

ε, ε → $ 

A, ε → w 

a, a → ε 

ε, ε → S ε, $ → ε




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Proof (⇐). 

Given a PDA M, we must construct a grammar G that 

generates L(M). 

Modify M so that 

M has a single accept state. 

M empties its stack before accepting. 

Each transition either pushes one symbol or pops one 

symbol, but not both.

The Variables 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Proof (⇐). 

For every pair of states p and q in M, we create a 

variable A pq . 

The variable A pq is interpreted to mean 

“Get from state p with an empty stack to state q with an 

empty stack.” 

Since the stack starts and ends empty, we must push a 

symbol at the beginning and pop a symbol at the end.

The Grammar Rules 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Proof (⇐). 

There are two possibilities in going from p to q. 

The first symbol pushed does not match the last symbol 

popped. 

The first symbol pushed matches the last symbol 

popped. 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Proof (⇐). 

For the rules in Group 1, use all pairs of transitions 

where the first transition pushes a symbol and the 

second transition pops a symbol, whether or not they 

are the same symbol. 

For each such pair of transitions δ(p, a, ε) = (s, c) and 

δ(t, b, d) = (q, ε), write the rules 

A pq → A pr A rq 

for all intermediate states r ∈ Q.




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Proof (⇐). 

The Group 1 rules allow us to break the processing up 

into segments that begin and end with an empty stack, 

but during which the stack is never empty. 

Short 

Example 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Proof (⇐). 

Stack 

depth 

x, ε → a 

x, y → ε 

x, ε → z 

x, b → ε 

Short 

Example 

Assignment 

p 

r 

q 

State




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Proof (⇐), continued. 

For the rules in Group 2, use all pairs of transitions 

where the second transition pops the same symbol that 

the first transition pushes. 

For each such pair of transitions δ(p, a, ε) = (r, c) and 

δ(s, b, c) = (q, ε), write the rule 

A pq → aA rs b.




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Proof (⇐). 

Stack 

depth 

a, ε → c b, c → ε 

Short 

Example 

Assignment 

p 

r 

s 

q 

State




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Proof (⇐). 

Stack 

depth 

a, ε → c b, c → ε 

Short 

Example 

Assignment 

p 

r 

s 

q 

State




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Proof (⇐), continued. 

Finally, the Group 3 rules are of the form 

A pp → ε 

for every state p. 

Assignment

The Start Symbol 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Proof (⇐), conclusion. 

Let q start be the start state and q accept be the accept 

state. 

Then the start symbol is A qstart q accept 

. 

If we can eventually replace the start symbol with all 

terminals, then the string of terminals must be in the 

language. 

Therefore, L(G) = L(M).

Example 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Example (Convert a PDA to a CFG) 

Find a grammar for the following PDA. 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

p 

a, ε → a 

ε, ε → $ 

q 

ε, $ → ε 

b, a → ε 

r

Example 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 


The variables are A pp , A pq , A pr , A qp , A qq , A qr , A rp , A rq , 

and A rr . 

Short 

Example 

Assignment

The Grammar Rules (Group 1) 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


Then we get the Group 1 grammar rules, based on the 

transitions. 

δ(p, ε, ε) = (q, $) 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 

δ(q, ε, $) = (r, ε) 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


Find all pairs of transitions where 

The first transition pushes a symbol x. 

The second transition pops a symbol y. 

The rules: 

δ(p, ε, ε) = (q, $) 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 

δ(q, ε, $) = (r, ε)




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


The pair 

δ(p, ε, ε) = (q, $) 

δ(q, b, a) = (q, ε) 

gives the rules A pq → A px A xq for all x ∈ Q. 

That is, 

A pq → A pp A pq 

A pq → A pq A qq 

A pq → A pr A rq




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


The pair 

δ(p, ε, ε) = (q, $) 

δ(q, ε, $) = (r, ε) 

gives the rules A pr → A px A xr for all x ∈ Q. 

That is, 

A pr → A pp A pr 

A pr → A pq A qr 

A pr → A pr A rr




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


The pair 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 

gives the rules A qq → A qx A xq for all x ∈ Q. 

That is, 

A qq → A qp A pq 

A qq → A qq A qq 

A qq → A qr A rq




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


The pair 

δ(q, a, ε) = (q, a) 

δ(q, ε, $) = (r, ε) 

gives the rules A qr → A qx A xr for all x ∈ Q. 

That is, 

A qr → A qp A pr 

A qr → A qq A qr 

A qr → A qr A rr




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


We get the first group of rules. 

A pq → A pp A pq A qq → A qp A pq 

A pq → A pq A qq A qq → A qq A qq 

A pq → A pr A rq A qq → A qr A rq 

A pr → A pp A pr A qr → A qp A pr 

A pr → A pq A qr A qr → A qq A qr 

A pr → A pr A rr A qr → A qr A rr




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


We may eliminate all rules with variables that represent 

impossible situations. 

A pq → A pp A pq A qq → A qp A pq 

A pq → A pq A qq A qq → A qq A qq 

A pq → A pr A rq A qq → A qr A rq 

A pr → A pp A pr A qr → A qp A pr 

A pr → A pq A qr A qr → A qq A qr 

A pr → A pr A rr A qr → A qr A rr




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


The Group 1 rules are 





A pr → A pr A rr 



A qr → A qr A rr




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


Then we get the Group 2 grammar rules, based on the 

transitions. 

δ(p, ε, ε) = (q, $) 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 

δ(q, ε, $) = (r, ε) 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


For every stack symbol s and for every pair of 

transitions, one of which pushes s and the other of 

which pops s, we write a grammar rule 

A xy → aA zw b 

where a is the symbol read when s is pushed and b is 

the symbol read when s is popped.




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(p, ε, ε) = (q, $) 

δ(q, ε, $) = (r, ε) 

gives the rule 

A pr → εA qq ε = A qq 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(p, ε, ε) = (q, $) 

δ(q, ε, $) = (r, ε) 



Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(p, ε, ε) = (q, $) 

δ(q, ε, $) = (r, ε) 



Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(p, ε, ε) = (q, $) 

δ(q, ε, $) = (r, ε) 



Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(p, ε, ε) = (q, $) 

δ(q, ε, $) = (r, ε) 



Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 


A qq → aA qq b 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 



Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 



Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 



Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


The pair 

δ(q, a, ε) = (q, a) 

δ(q, b, a) = (q, ε) 



Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 


The Group 2 rules are 

A pr → A qq 


Short 

Example 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


Finally, the Group 3 rules are 

A pp → ε 

A qq → ε 

A rr → ε 

Assignment




Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


This gives us the grammar rules 





A pr → A pr A rr 



A qr → A qr A rr 

A pr → A qq 


A pp → ε 

A qq → ε 

A rr → ε

The Grammar 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 


The start symbol is A pr . 

Long Example 

Short 

Example 

Assignment

The Grammar 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


The complete grammar is 

A pr → A pp A pr | A pq A qr | A pr A rr | A qq 

A pp → ε 

A pq → A pp A pq | A pq A qq 

A qq → aA qq b | A qq A qq | ε 

A qr → A qq A qr | A qr A rr 

A rr → ε.

Example 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 


Find a grammar for the language of the following PDA. 

(, ε → ( ), ( → ε 

ε, ε → $ 

a, ε → ε 

ε, $ → ε 

+, ε → ε 

Assignment

Example 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


First, we need to modify the PDA: 

(, ε → ( 

), ( → ε 

a, ε → ( r ε, ( → ε 

ε, ε → $ 

ε, $ → ε 

p 

q 

s 

ε, ( → ε t +, ε → ( 

u

Example 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


The transitions are 

Pushes 

Pops 

δ(p, ε, ε) = (q, $) δ(r, ε, () = (s, ε) 

δ(q, (, ε) = (q, () δ(s, ), () = (s, ε) 

δ(q, a, ε) = (r, () δ(t, ε, () = (q, ε) 

δ(s, +, ε) = (t, () δ(s, ε, $) = (u, ε)

Group 1 Rules 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


For the Group 1 rules, there are 16 combinations of a 

transition that pushes with a transition that pops. 

However, there are only three beginning states {p, q, s} 

and three ending states {s, q, u}. 

So there are only 9 sets of rules: 

A ps → A px A xs for all x ∈ Q 

A pq → A px A xq for all x ∈ Q 

A pu → A px A xu for all x ∈ Q 

A qs → A qx A xs for all x ∈ Q 

A qq → A qx A xq for all x ∈ Q 

A qu → A qx A xu for all x ∈ Q 

A ss → A sx A xs for all x ∈ Q 

A sq → A sx A xq for all x ∈ Q 

A su → A sx A xu for all x ∈ Q

Group 1 Rules 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


That gives 54 rules. 

We can eliminate the ones containing variables of the 

form A xp for x ≠ p and A ux for x ≠ u. 

That eliminates 12 rules, leaving 42. 

A ps → A px A xs for all x ∈ Q, x ≠ u 

A pq → A px A xq for all x ∈ Q, x ≠ u 

A pu → A px A xu for all x ∈ Q 

A qs → A qx A xs for all x ∈ Q, x ≠ p, u 

A qq → A qx A xq for all x ∈ Q, x ≠ p, u 

A qu → A qx A xu for all x ∈ Q, x ≠ p 

A ss → A sx A xs for all x ∈ Q, x ≠ p, u 

A sq → A sx A xq for all x ∈ Q, x ≠ p, u 

A su → A sx A xu for all x ∈ Q, x ≠ p

Group 2 Rules 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


There is 1 combination of transitions that pushes and 

pops $. 

There are 9 combinations of transitions that push and 

pop (. 

So there are 10 rules in Group 2. 

A pu → A qs 

A qs → (A qr 

A qs → (A qs ) 

A qq → (A qt 

A qs → aA rr 

A qs → aA rs ) 

A qq → aA rt 

A ss → +A tr 

A ss → +A ts ) 

A sq → +A tt

Group 3 Rules 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 


There are 6 rules in Group 3. 

A pp → ε 

A qq → ε 

A rr → ε 

A ss → ε 

A tt → ε 

A uu → ε

Assignment 



Equivalence 

to CFGs 

Robb T. 

Koether 

Homework 

Review 

Equivalence 

of PDAs and 

CFGs 

Proof ⇒ 

Proof ⇐ 

Long Example 

Short 

Example 

Assignment 

Homework 

Read Section 2.2, pages 115 - 123. 

Exercises 11, 12, page 129. 

Find a grammar for the language 

L = {w | w contains an equal number of a’s and b’s} 

with PDA 

a, ε → a 

a, b → ε 

b, a → ε b, ε → b 

a > b 

a, ε → $ b, ε → $ 

b. $ → ε 

a. $ → ε 

a < b

Pushdown Automata - Hampden-Sydney College

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