Pushdown Automata (PDA)
state
control
00011111000
input
a
a
b
a
stack
Pushdown Automata (PDA)
If the input symbol is a and
the top stack symbol is x then
q1 to q2, pop x, push y, advance read head
q1
a, x → y
q2
If a = ℇ do not advance read head
If x = ℇ do not read from stack
If y = ℇ do not write to stack
When does a PDA accept a string?
input: w1w2…wn
q0
w1
r1
w2
r2
w3
…
accept if any branch accepts
wn
qf
Pushdown Automata (PDA)
(Q, Σ, Γ, δ, q0, F)
δ: Q × Σℇ × Γℇ → 𝒫(Q × Γℇ)
q1
a, x → y
a, x
→
q2
z
q3
Theorems
Not every nondeterministic
PDA has an equivalent
deterministic PDA
A language is context-free
iff some PDA recognizes it
CFL
vs.
Regular Languages
CFL vs. Regular Languages
PDA to NFA
CFL vs. Regular Languages
NFA to PDA
q1
a
q2
CFL vs. Regular Languages
NFA to PDA
q1
a, ℇ → ℇ
q2
PDA Design
Examples
PDA Design
n
n
{0 1 | n ≥ 0}
input: 000111
n
n
{0 1 |
n ≥ 0}
stack
input: 000111
q2
n
n
{0 1 |
n ≥ 0}
stack
0, ℇ → 0
input: 000111
q2
n
n
{0 1 |
n ≥ 0}
stack
0
0, ℇ → 0
input: 000111
q2
n
n
{0 1 |
0
0
n ≥ 0}
stack
0, ℇ → 0
input: 000111
q2
n
n
{0 1 |
n ≥ 0}
0
0
0
stack
0, ℇ → 0
input: 000111
q2
n
n
{0 1 |
n ≥ 0}
1, 0 → ℇ
q3
0
0
0
stack
0, ℇ → 0
1, 0 → ℇ
input: 000111
q2
n
n
{0 1 |
n ≥ 0}
1, 0 → ℇ
q3
0
0
stack
0, ℇ → 0
1, 0 → ℇ
input: 000111
q2
n
n
{0 1 |
n ≥ 0}
1, 0 → ℇ
q3
stack
0
0, ℇ → 0
1, 0 → ℇ
input: 000111
q2
n
n
{0 1 |
n ≥ 0}
1, 0 → ℇ
q3
stack
0, ℇ → 0
1, 0 → ℇ
input: 000111_
q2
n
n
{0 1 |
n ≥ 0}
1, 0 → ℇ
q3
stack
0, ℇ → 0
1, 0 → ℇ
Does this work?
n
n
{0 1 |
n ≥ 0}
q2
0, ℇ → 0
1, 0 → ℇ
q3
1, 0 → ℇ
n
n
{0 1 |
q1
n ≥ 0}
ℇ, ℇ → $
q2
0, ℇ → 0
1, 0 → ℇ
q4
ℇ, $ → ℇ
q3
1, 0 → ℇ
PDA Design
R
*
{ww | w ∈ {0, 1} }
R
{ww |
q1
w ∈ {0,
ℇ, ℇ → $
*
1} }
q2
0, ℇ → 0
1, ℇ → 1
ℇ, ℇ → ℇ
q4
ℇ, $ → ℇ
q3
0, 0 → ℇ
1, 1 → ℇ
PDA Design
i j k
{a b c | i, j, k ≥ 0 and i = j or j = k}
i
j
k
{a b c |
q1
i, j, k ≥ 0 and i = j}
ℇ, ℇ → $
q2
a, ℇ → a
ℇ, ℇ → ℇ
q4
c, ℇ → ℇ
ℇ, $ → ℇ
q3
b, a → ℇ
{aibjck | i, j, k ≥ 0 and j = k}
r1
ℇ, ℇ → $
r2
a, ℇ → ℇ
ℇ, ℇ → ℇ
r3
b, ℇ → b
ℇ, ℇ → ℇ
r5
ℇ, $ → ℇ
r4
c, b → ℇ
{aibjck | i, j, k ≥ 0 and i=j or j = k}
ℇ, ℇ → $
a, ℇ → a
ℇ, ℇ → $
q2
r2
a, ℇ → ℇ
ℇ, ℇ → ℇ
q1
r1
r3
b, ℇ → b
ℇ, ℇ → ℇ
ℇ, ℇ → ℇ
q3
b, a → ℇ
r4
c, b → ℇ
ℇ, $ → ℇ
ℇ, $ → ℇ
q4
c, ℇ → ℇ
r5
{aibjck | i, j, k ≥ 0 and i=j or j = k}
a, ℇ → ℇ
q2
ℇ, ℇ → $
a, ℇ → a
ℇ, ℇ → $
q2
ℇ, ℇ → ℇ
q1
b, ℇ → b
q3
ℇ, ℇ → ℇ
ℇ, ℇ → ℇ
q3
b, a → ℇ
q3
c, b → ℇ
ℇ, $ → ℇ
ℇ, $ → ℇ
q4
c, ℇ → ℇ
q4
Pumping Lemma:
Regular Languages
If A is a regular language,
then there is a pumping length p st
if s ∈ A with |s| ≥ p then we can write s = xyz
so that
•
∀i ≥ 0 xyiz ∈ A
•
|y| > 0
•
|xy| ≤ p
n n
To prove {0 1 | n ≥ 0} is not
regular using the Pumping Lemma
1.
2.
3.
4.
Suppose {0n1n | n ≥ 0} is regular
Call its pumping length p
Find string s ∈ A with |s| ≥ p. Let s = 0p1p
The pumping lemma says that for some split 0p1p = xyz all the following conditions hold
•
∀i ≥ 0 xyiz ∈ A
•
|y| > 0
y is a non-empty string of 0s
•
|xy| ≤ p
To prove A is not regular
using the Pumping Lemma
1.
2.
3.
4.
Suppose A is regular
Call its pumping length p
Find string s ∈ A with |s| ≥ p
The pumping lemma says that for some split s = xyz all the following conditions hold
•
∀i ≥ 0 xyiz ∈ A
•
|y| > 0
•
|xy| ≤ p