Pushdown Automata PDAs

1
Lecture 34

• Pushdown Automata (PDAs)
• Another example

2
Palindromes

• Let PAL be the set of palindromes over a,b
• Let PAL1 be the following related language
• wcwr w consists only of as and bs
• we add c to the input alphabet as a special
  marker character
• Strings in PAL1
• aca, bcb, abcba, aabcbaa, c
• strings not in PAL1
• aaca, aaccaa, abccba, abcb, abba
• Let PAL2 be the set of even length palindromes
• wwr w consists only of as and bs

3
PAL1

• Lets first construct a PDA for PAL1
• Basic ideas
• Have one state remember first half of string
• Have one state match second half of string to
  first half
• Transition between these two states when the
  first c is encountered

4
PDA for PAL1

• M (Q, S, G, q0, Z, A, d)
• Q q0, qm, qf
• S a, b, c
• G Z, a, b
• q0 q0
• Z Z
• A qf

5
Transition Function
Trans Current Input Top of Next Stack
State Char. Stack State
Update -------------------------------------------
------------ 1 q0 a
Z q0 aZ 2 q0
a a q0 aa 3 q0
a b q0 ab 4
q0 b Z q0
bZ 5 q0 b a
q0 ba 6 q0 b
b q0 bb 7 q0
c Z qm Z 8
q0 c a qm
a 9 q0 c b
qm b 10 qm a
a qm l 11 qm b
b qm l 12 qm
l Z qf Z
First three transitions push a on top of the
stack Second three transitions push b on the
stack Third three transitions switch state q0 to
qm No change to stack Transitions 10 and 11
match characters from first and last half of
input string
6
Notation comment
Trans Current Input Top of Next Stack
State Char. Stack State
Update -------------------------------------------
------------ 1 q0 a
Z q0 aZ 2 q0
a a q0 aa 3 q0
a b q0 ab 4
q0 b Z q0
bZ 5 q0 b a
q0 ba 6 q0 b
b q0 bb 7 q0
c Z qm Z 8
q0 c a qm
a 9 q0 c b
qm b 10 qm a
a qm l 11 qm b
b qm l 12 qm
l Z qf Z

• We might represent transition 1 in two other ways
• d(q0,a,Z) (q0, aZ)
• (q0, a, Z, q0, aZ)
• Question
• Is this PDA deterministic?

7
Computation Graph 1
Trans Current Input Top of Next Stack
State Char. Stack State
Update -------------------------------------------
------------ 1 q0 a
Z q0 aZ 2 q0
a a q0 aa 3 q0
a b q0 ab 4
q0 b Z q0
bZ 5 q0 b a
q0 ba 6 q0 b
b q0 bb 7 q0
c Z qm Z 8
q0 c a qm
a 9 q0 c b
qm b 10 qm a
a qm l 11 qm b
b qm l 12 qm
l Z qf Z
(q0, abcba, Z)
8
Computation Graph 2
Trans Current Input Top of Next Stack
State Char. Stack State
Update -------------------------------------------
------------ 1 q0 a
Z q0 aZ 2 q0
a a q0 aa 3 q0
a b q0 ab 4
q0 b Z q0
bZ 5 q0 b a
q0 ba 6 q0 b
b q0 bb 7 q0
c Z qm Z 8
q0 c a qm
a 9 q0 c b
qm b 10 qm a
a qm l 11 qm b
b qm l 12 qm
l Z qf Z
(q0, abcab, Z)
9
Computation Graph 3
Trans Current Input Top of Next Stack
State Char. Stack State
Update -------------------------------------------
------------ 1 q0 a
Z q0 aZ 2 q0
a a q0 aa 3 q0
a b q0 ab 4
q0 b Z q0
bZ 5 q0 b a
q0 ba 6 q0 b
b q0 bb 7 q0
c Z qm Z 8
q0 c a qm
a 9 q0 c b
qm b 10 qm a
a qm l 11 qm b
b qm l 12 qm
l Z qf Z
(q0, acab, Z)
10
PAL2

• Lets now construct a PDA for PAL
• What is harder this time?
• When do we switch from putting strings on the
  stack to matching?
• Example
• After seeing aab, should we switch to match mode
  or stay in stack mode?
• Solution
• Do both using nondeterminism

11
PDA for PAL2

• M (Q, S, G, q0, Z, A, d)
• Q q0, qm, qf
• S a, b
• G Z, a, b
• q0 q0
• Z Z
• A qf

12
Transition Function
Trans Current Input Top of Next Stack
State Char. Stack State
Update -------------------------------------------
------------ 1 q0 a
Z q0 aZ 2 q0
a a q0 aa 3 q0
a b q0 ab 4
q0 b Z q0
bZ 5 q0 b a
q0 ba 6 q0 b
b q0 bb 7 q0
l Z qm Z 8
q0 l a qm
a 9 q0 l b
qm b 10 qm a
a qm l 11 qm b
b qm l 12 qm
l Z qf Z
First three transitions push a on top of the
stack Second three transitions push b on the
stack Third three transitions switch state q0 to
qm Note now that one of transitions 7, 8, and 9
can be used whenever transitions 1-6 are used.
13
Computation Graph 1
Trans Current Input Top of Next Stack
State Char. Stack State
Update -------------------------------------------
------------ 1 q0 a
Z q0 aZ 2 q0
a a q0 aa 3 q0
a b q0 ab 4
q0 b Z q0
bZ 5 q0 b a
q0 ba 6 q0 b
b q0 bb 7 q0
l Z qm Z 8
q0 l a qm
a 9 q0 l b
qm b 10 qm a
a qm l 11 qm b
b qm l 12 qm
l Z qf Z
(q0, abba, Z)
14
Computation Graph 2
Trans Current Input Top of Next Stack
State Char. Stack State
Update -------------------------------------------
------------ 1 q0 a
Z q0 aZ 2 q0
a a q0 aa 3 q0
a b q0 ab 4
q0 b Z q0
bZ 5 q0 b a
q0 ba 6 q0 b
b q0 bb 7 q0
l Z qm Z 8
q0 l a qm
a 9 q0 l b
qm b 10 qm a
a qm l 11 qm b
b qm l 12 qm
l Z qf Z
(q0, aba, Z)
15
PAL

• Challenge
• Construct a PDA for PAL
• First step
• Construct a PDA for odd length palindromes
• Then
• Combine PDAs for odd length and even length
  palindromes