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CS 301 - Lecture 11 Nondeterministic Pushdown Automata
Fall 2008
Review
– Alphabets, strings, languages
– Deterministic Finite and Nondeterministic Automata – Equivalence of NFA and DFA – Regular Expressions – Regular Grammars – Properties of Regular Languages – Languages that are not regular and the pumping lemma
– Context Free Grammars – Derivations: leftmost, rightmost and derivation trees – Parsing and ambiguity – Simplifying Grammars and Normal Forms
– Nondeterministic Pushdown Automata
Pushdown Automata PDAs
Pushdown Automaton -- PDA
Input String Stack States
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2
The States
q1 q2 a, b → c
Input symbol Pop symbol Push symbol
q1 q2 a, b → c a b
top input stack
a
Replace
e h $ e h $ c q1 q2 a, λ → c a a
Push
b e h $ e h $ b c
top input stack
q1 q2 a, b → λ a a
Pop
b e h $ e h $
top input stack
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3
q1 q2 a, λ → λ a a
No Change
b e h $ e h $ b
top input stack
q1 q2 λ → $ , a a a
Pop
$
top input stack A Possible Transition empty
q1 q2 c b a → , a
input A Bad Transition The automaton Halts in state and Rejects the input string Empty stack
q1
HALT
q1 q2 c a → λ , a
input A Bad Transition The automaton Halts in state and Rejects the input string Empty stack
q1
HALT
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4
q1 q2 z y x → ,
No transition is allowed to be followed When the stack is empty Empty stack
q1 q2 b a → $ , a a
Pop
$
top input stack A Good Transition
b
Non-Determinism
q1 q2 a, b → c q3 a, b → c q1 q2 λ, b → c transition − λ
These are allowed transitions in a Non-deterministic PDA (NPDA)
NPDA: Non-Deterministic PDA
Example:
λ, λ → λ a, λ → a b, a → λ q0 q1 q2 q3 b, a → λ λ, $ → $
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5
a, λ → a b, a → λ
q
q1 q2 q3
Execution Example: Input
a a a b b b
current state
b, a → λ
Time 0
λ, λ → λ λ, $ → $
Stack
$ a, λ → a b, a → λ q0 q1 q2 q3
Input
a a a b b b b, a → λ
Time 1
λ, λ → λ λ, $ → $
Stack
$ a, λ → a b, a → λ q0 q1 q2 q3
Input Stack
a a a b b b $ a b, a → λ
Time 2
λ, λ → λ λ, $ → $ a, λ → a b, a → λ q0 q1 q2 q3
Input Stack
a a a b b b $ a a b, a → λ
Time 3
λ, λ → λ λ, $ → $
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6
a, λ → a b, a → λ q0 q1 q2 q3
Input Stack
a a a b b b $ a a a b, a → λ
Time 4
λ, λ → λ λ, $ → $ a, λ → a b, a → λ q0 q1 q2 q3
Input
a a a b b b
Stack
$ a a a b, a → λ
Time 5
λ, λ → λ λ, $ → $ a, λ → a b, a → λ q0 q1 q2 q3
Input
a a a b b b $ a
Stack
b, a → λ
Time 6
λ, λ → λ λ, $ → $ a a, λ → a b, a → λ q0 q1 q2 q3
Input
a a a b b b $
Stack
b, a → λ
Time 7
λ, λ → λ λ, $ → $ a
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7
a, λ → a b, a → λ q0 q1 q2 q3
Input
a a a b b b b, a → λ
Time 8 accept
λ, λ → λ λ, $ → $ $
Stack A string is accepted if there is a computation such that: All the input is consumed AND The last state is a final state At the end of the computation, we do not care about the stack contents The input string is accepted by the NPDA:
aaabbb
a, λ → a b, a → λ q0 q1 q2 q3 b, a → λ λ, λ → λ λ, $ → $
} : { ≥ = n b a L
n n is the language accepted by the NPDA:
a, λ → a b, a → λ q0 q1 q2 q3 b, a → λ
In general,
λ, λ → λ λ, $ → $
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8
Another NPDA example
λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , ,
NPDA M
} { ) (
R
ww M L =
Execution Example: Input Time 0 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a b b
Input
a a b
Time 1 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a b
Input Time 2 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a a b b b
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Input Time 3 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a a b b b
Guess the middle
Input Time 4 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a a b b b
Input Time 5 Stack
$ λ, $ → $
1
q
q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a b b a
Input Time 6 Stack
$ λ, $ → $ q1 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a b b
accept
q2
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Rejection Example: Input Time 0 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a b b b
Input Time 1 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a b b b
Input Time 2 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a b a b b b
Input Time 3 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a b
Guess the middle
a b b b
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Input Time 4 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a b a b b b
Input Time 5 Stack
$ λ, $ → $
1
q
q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a b b b
There is no possible transition. Input is not consumed Another computation on same string: Input Time 0 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a b b b
Input Time 1 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a a b b b
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Input Time 2 Stack
$ λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , a b a b b b
Input Time 3 Stack
$ a b a b b b λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , b
Input Time 4 Stack
a b b b λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , , $ a b b b
Input Time 5 Stack
a b b b $ a b b b λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , ,
No final state is reached
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λ, $ → $ q1 q2 b b a a → → λ λ , , λ, λ → λ q0 λ λ → → b b a a , ,
There is no computation that accepts string abbb
) (M L abbb∉
A string is rejected if there is no computation such that: All the input is consumed AND The last state is a final state At the end of the computation, we do not care about the stack contents In other words, a string is rejected if in every computation with this string: The input cannot be consumed OR The input is consumed and the last state is not a final state OR The stack head moves below the bottom of the stack
Another NPDA example
q0
NPDA M
} 1 : { ) ( − ≥ = m n b a M L
m n
λ λ λ → → → $ , , , b a b a a
This NPDA generates: and for each substring u
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14
Execution Example: Input Time 0 Stack
a a
b
λ λ λ → → → $ , , , b a b a a
q0 $
Input Time 1 Stack
a a
b
q0 a $
λ λ λ → → → $ , , , b a b a a Input Time 2 Stack
a a
b
q0 a a $
λ λ λ → → → $ , , , b a b a a Input Time 3 Stack
a a
b
q0 a
accept
$
λ λ λ → → → $ , , , b a b a a
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Rejection example: Input Time 0 Stack
a
b
b
q0
b
$
Input Time 1 Stack
q0 a a
b
b b
$
λ λ λ → → → $ , , , b a b a a Input Time 2 Stack
q0 a
b
b b
$
λ λ λ → → → $ , , , b a b a a Input Time 3 Stack
q0 a
b
b b
λ λ λ → → → $ , , , b a b a a
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Input Time 4 Stack
q0
Halt and Reject
a
b
b b
λ λ λ → → → $ , , , b a b a a
Pushing Strings
q1 q2 a, b → w
Input symbol Pop symbol Push string
q1 q2 a, b → cdf a b
top input stack
a
Push
e h h e c d f
pushed string Example:
$ $
What’s Next
– Linz Chapter 1,2.1, 2.2, 2.3, (skip 2.4), 3, 4, 5, 6.1, 6.2, (skip 6.3), 7.1, and 7.2 – JFLAP Chapter 1, 2.1, (skip 2.2), 3, 4, 5, 6, 7
- Next Lecture Topics from Chapter 7.2
– Pushdown Automata and Context Free Grammars
- Quiz 2 in Recitation on Wednesday 10/1
– Covers Linz 3, 4 and JFLAP 3, 4 – Closed book, but you may bring one sheet of 8.5 x 11 inch paper with any notes you like. – Quiz will take the full hour
– Homework Due Thursday
