prof. busch - lsu1 pushdown automata pdas. prof. busch - lsu2 pushdown automaton -- pda input string...
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Prof. Busch - LSU 1
Pushdown AutomataPDAs
Prof. Busch - LSU 2
Pushdown Automaton -- PDA
Input String
Stack
States
Prof. Busch - LSU 3
Initial Stack Symbol
Stack
$
Stack
z
bottomspecial symbol
stackhead
top
Appears at time 0
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The States
q1 q2a, b c
Inputsymbol
Popsymbol
Pushsymbol
Prof. Busch - LSU 5
q1 q2a, b c
a
b top
input
stack
a
Replaceeh
$eh
$
c
Prof. Busch - LSU 6
q1 q2a, c
a a
Pushb
eh
$eh
$
bc
top
input
stack
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q1 q2a, b
a a
Popb
eh
$eh
$
top
input
stack
Prof. Busch - LSU 8
q1 q2a,
a a
No Changeb
eh
$eh
$
btop
input
stack
Prof. Busch - LSU 9
Non-Determinism
q1
q2a, b c
q3a, b c
q1 q2, b c
transition
PDAs are non-deterministic
Allowed non-deterministic transitions
Prof. Busch - LSU 10
Example PDA
,
a, a
b, a q0 q1 q2 q3
b, a
, $ $
PDA : M }0:{)( nbaML nn
Prof. Busch - LSU 11
,
a, a
b, a q0 q1 q2 q3
b, a
, $ $
}0:{)( nbaML nn
Basic Idea:
1. Push the a’s on the stack
2. Match the b’s on input with a’s on stack
3. Match found
Prof. Busch - LSU 12
a, a
b, a 0q q1 q2 q3
Execution Example:
Input
a a a b b b
currentstate
b, a
Time 0
, , $ $
Stack
$
Prof. Busch - LSU 13
a, a
b, a q0 q1 q2 q3
Input
a a a b b b
b, a
Time 1
, , $ $
Stack
$
Prof. Busch - LSU 14
a, a
b, a q0 q1 q2 q3
Input
Stack
a a a b b b
$
a
b, a
Time 2
, , $ $
Prof. Busch - LSU 15
a, a
b, a q0 q1 q2 q3
Input
Stack
a a a b b b$
aa
b, a
Time 3
, , $ $
Prof. Busch - LSU 16
a, a
b, a q0 q1 q2 q3
Input
Stack
a a a b b b
$
aaa
b, a
Time 4
, , $ $
Prof. Busch - LSU 17
a, a
b, a q0 q1 q2 q3
Input
a a a b b b
Stack
$
aaa
b, a
Time 5
, , $ $
Prof. Busch - LSU 18
a, a
b, a q0 q1 q2 q3
Input
a a a b b b$
a
Stack
b, a
Time 6
, , $ $
a
Prof. Busch - LSU 19
a, a
b, a q0 q1 q2 q3
Input
a a a b b b$
Stack
b, a
Time 7
, , $ $
a
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a, a
b, a q0 q1 q2 q3
Input
a a a b b b
b, a
Time 8
accept, , $ $
$
Stack
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A string is accepted if there is a computation such that:
All the input is consumed AND The last state is an accepting state
we do not care about the stack contentsat the end of the accepting computation
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0q
a, a
b, a q1 q2 q3
Rejection Example:
Input
a a b
currentstate
b, a
Time 0
, , $ $
Stack
$
Prof. Busch - LSU 23
0q
a, a
b, a q1 q2 q3
Rejection Example:
Input
a a b
currentstate
b, a
Time 1
, , $ $
Stack
$
Prof. Busch - LSU 24
0q
a, a
b, a q1 q2 q3
Rejection Example:
Input
a a b
currentstate
b, a
Time 2
, , $ $
Stack
$
a
Prof. Busch - LSU 25
0q
a, a
b, a q1 q2 q3
Rejection Example:
Input
a a b
currentstate
b, a
Time 3
, , $ $
Stack
$
aa
Prof. Busch - LSU 26
0q
a, a
b, a q1 q2 q3
Rejection Example:
Input
a a b
currentstate
b, a
Time 4
, , $ $
Stack
$
aa
Prof. Busch - LSU 27
0q
a, a
b, a q1 q2 q3
Rejection Example:
Input
a a b
currentstate
b, a
Time 4
, , $ $
Stack
$
aa
reject
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The string is rejected by the PDAaab
a, a
b, a q0 q1 q2 q3
b, a
, , $ $
There is no accepting computation foraab
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Another PDA example
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
PDA : M }},{:{)( bavvvML R
Prof. Busch - LSU 30
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
}},{:{)( bavvvML RBasic Idea:
1. Push v on stack
2. Guess middle of input
3. Match on input with v on stack
Rv
4. Match found
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Execution Example:
Input
Time 0
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
a ab b
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Input
a ab
Time 1
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
Prof. Busch - LSU 33
Input
Time 2
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa ab b
b
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Input
Time 3
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa ab b
b
Guess the middle of string
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Input
Time 4
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa ab b
b
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Input
Time 5
Stack
$
, $ $1q q2
bb
aa
,
,
, q0
bb
aa
,
,
a ab b a
Prof. Busch - LSU 37
Input
Time 6
Stack
$
, $ $q1
bb
aa
,
,
, q0
bb
aa
,
,
a ab b
accept
q2
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Rejection Example:
Input
Time 0
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
a b b b
Prof. Busch - LSU 39
Input
Time 1
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa b b b
Prof. Busch - LSU 40
Input
Time 2
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
a b b b
Prof. Busch - LSU 41
Input
Time 3
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
Guess the middle of string
a b b b
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Input
Time 4
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
a b b b
Prof. Busch - LSU 43
Input
Time 5
Stack
$
, $ $1q q2
bb
aa
,
,
, q0
bb
aa
,
,
aa b b b
There is no possible transition.
Input is not consumed
Prof. Busch - LSU 44
Another computation on same string:
Input Time 0
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
a b b b
Prof. Busch - LSU 45
Input
Time 1
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
aa b b b
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Input
Time 2
Stack
$
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
ab
a b b b
Prof. Busch - LSU 47
Input
Time 3
Stack
$ab
a b b b
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
b
Prof. Busch - LSU 48
Input
Time 4
Stack
a b b b
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
$abbb
Prof. Busch - LSU 49
Input
Time 5
Stack
a b b b
$abbb
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
No accept state is reached
Prof. Busch - LSU 50
, $ $q1 q2
bb
aa
,
,
, q0
bb
aa
,
,
There is no computation that accepts string abbb
)(MLabbb
Prof. Busch - LSU 51
Pushing & Popping Strings
q1 q221, wwa
Inputsymbol
Pushstring
Popstring
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q1 q2cdfeba ,
a
btop
input
stack
a
Replacee
h he
cdf
pushstring
Example:
$ $
epopstring
top
Prof. Busch - LSU 53
q1 q2cdfeba ,
q1
q2
ea, b,
f,
,
d, c,
pop
push
Equivalent transitions
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Another PDA example
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
PDA M
)}()(:},{{)( * wnwnbawML ba
Prof. Busch - LSU 55
Time 0
Input
a ab b b
currentstate
a
$
Stack
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
Execution Example:
Prof. Busch - LSU 56
Time 1
Input
a ab b b a
$
Stack
0
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
Prof. Busch - LSU 57
Time 3
Input
a bb b a
$
Stack
a
$
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
0
Prof. Busch - LSU 58
Time 4
Input
a bb b a
$
Stack
a
1
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
Prof. Busch - LSU 59
Time 5
Input
a bb b a
$
Stack
a
11
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
Prof. Busch - LSU 60
Time 6
Input
a bb b a
$
Stack
a
1
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
1
Prof. Busch - LSU 61
Time 7
Input
a bb b a
$
Stack
a
1
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
Prof. Busch - LSU 62
Time 8
Input
a bb b a a
$
Stack
$$, q1 q2
a, $ 0$a, 0 00a,1
b, $ 1$b, 1 11b, 0
accept
Prof. Busch - LSU 63
Formalities for PDAs
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q1 q221, wwa
)},{(),,( 2211 wqwaq Transition function:
Prof. Busch - LSU 65
q1
q221, wwa
q331, wwa
)},(),,{(),,( 332211 wqwqwaq
Transition function:
Prof. Busch - LSU 66
Formal Definition
Pushdown Automaton (PDA)
),,,δ,Γ,Σ,( 0 FzqQM
States
Inputalphabet
Stackalphabet
Transitionfunction
Acceptstates
Stackstartsymbol
Initialstate
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Instantaneous Description
),,( suq
Currentstate Remaining
input
Currentstackcontents
Prof. Busch - LSU 68
a, a
b, a q0 q1 q2 q3
Input
Stack
a a a b b b$
aa
b, a
Time 4:
, , $ $
Example: Instantaneous Description
$),,( 1 aaabbbq
a
Prof. Busch - LSU 69
a, a
b, a q0 q1 q2 q3
Input
Stack
a a a b b b$
aa
b, a
Time 5:
, , $ $
Example: Instantaneous Description
$),,( 2 aabbq
a
Prof. Busch - LSU 70
We write:
$),,($),,( 21 aabbqaaabbbq
Time 4 Time 5
Prof. Busch - LSU 71
a, a
b, a q0 q1 q2 q3
b, a
, , $ $
,$),(,$),($),,($),,(
$),,($),,($),,(
,$),(,$),(
3222
111
10
qqabqaabbq
aaabbbqaaabbbqaaabbbq
aaabbbqaaabbbq
A computation:
Prof. Busch - LSU 72
,$),(,$),($),,($),,(
$),,($),,($),,(
,$),(,$),(
3222
111
10
qqabqaabbq
aaabbbqaaabbbqaaabbbq
aaabbbqaaabbbq
For convenience we write:
,$),(,$),( 30 qaaabbbq
Prof. Busch - LSU 73
Language of PDA
Language accepted by PDA :M
)},,(),,(:{)( 0 sqzwqwML f
Initial state Accept state
)(ML
Prof. Busch - LSU 74
Example:,$),(,$),( 30 qaaabbbq
a, a
b, a q0 q1 q2 q3
b, a
, , $ $
PDA :M
)(MLaaabbb
Prof. Busch - LSU 75
,$),(,$),( 30 qbaq nn
a, a
b, a q0 q1 q2 q3
b, a
, , $ $
)(MLba nn
PDA :M
Prof. Busch - LSU 76
a, a
b, a q0 q1 q2 q3
b, a
, , $ $
PDA :M
}0:{)( nbaML nnTherefore: