Pushdown Automata
Part II: PDAs and CFG
Chapter 12
1
PDAs and Context-Free Grammars
Theorem: The class of languages accepted by PDAs is
exactly the class of context-free languages.
Recall: context-free languages are languages that
can be defined with context-free grammars.
Restate theorem:
Can describe with context-free grammar
Can accept by PDA
2
Going One Way
Lemma: Each context-free language is accepted by
some PDA.
Proof (by construction):
The idea: Let the stack do the work.
Two approaches:
• Top down
• Bottom up
3
Top Down
The idea: Let the stack keep track of expectations.
Example: Arithmetic expressions
EE+T
ET
TTF
TF
F  (E)
F  id
(1)
(2)
(3)
(4)
(5)
(6)
(q, , E), (q, E+T)
(q, , E), (q, T)
(q, , T), (q, T*F)
(q, , T), (q, F)
(q, , F), (q, (E) )
(q, , F), (q, id)
(7) (q, id, id), (q, )
(8) (q, (, ( ), (q, )
(9) (q, ), ) ), (q, )
(10) (q, +, +), (q, )
(11) (q, , ), (q, )
4
A Top-Down Parser
Answering “Could G generate w?” starting from S.
The outline of M is:
M = ({p, q}, , V, , p, {q}), where  contains:
● The start-up transition ((p, , ), (q, S)).
● For each rule X  s1s2…sn. in R, the transition:
((q, , X), (q, s1s2…sn)).
● For each character c  , the transition:
((q, c, c), (q, )).
5
Example of the Construction
L = {anb*an}
(1) S  
(2) S  B
(3) S  aSa
(4) B  
(5) B  bB
input = a a b b a a
trans
0
3
6
3
6
2
5
7
5
7
4
6
6
state
p
q
q
q
q
q
q
q
q
q
q
q
q
q
0 (p, , ), (q, S)
1 (q, , S), (q, )
2 (q, , S), (q, B)
3 (q, , S), (q, aSa)
4 (q, , B), (q, )
5 (q, , B), (q, bB)
6 (q, a, a), (q, )
7 (q, b, b), (q, )
unread input
a a b b
a a b b
a a b b
a b b
a b b
b b
b b
b b
b
b
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a

stack

S
aSa
Sa
aSaa
Saa
Baa
bBaa
Baa
bBaa
Baa
aa
a

6
Bottom-Up
The idea: Let the stack keep track of what has been found.
(1) E  E + T
(2) E  T
(3) T  T  F
(4) T  F
(5) F  (E)
(6) F  id
Reduce Transitions:
(1) (p, , T + E), (p, E)
(2) (p, , T), (p, E)
(3) (p, , F  T), (p, T)
(4) (p, , F), (p, T)
(5) (p, , )E( ), (p, F)
(6) (p, , id), (p, F)
Shift Transitions
(7) (p, id, ), (p, id)
(8) (p, (, ), (p, ()
(9) (p, ), ), (p, ))
(10) (p, +, ), (p, +)
(11) (p, , ), (p, )
7
A Bottom-Up Parser
Answering “Could G generate w?” starting from w.
The outline of M is:
M = ({p, q}, , V, , p, {q}), where  contains:
● The shift transitions: ((p, c, ), (p, c)), for each c  .
● The reduce transitions: ((p, , (s1s2…sn.)R), (p, X)), for each rule
X  s1s2…sn. in G.
● The finish up transition: ((p, , S), (q, )).
8
Example of the Construction
L = {anb*an}
(1) S  
(2) S  B
(3) S  aSa
(4) B  
(5) B  bB
input = a a b b a a
trans
6
6
7
7
4
5
5
2
6
3
6
3
0
state
p
p
p
p
p
p
p
p
p
p
p
p
p
q
0 (p, , S), (q, )
1 (p, , ), (p, S)
2 (p, , B), (p, S)
3 (p, , aSa), (p, S)
4 (p, , ), (p, B)
5 (p, , Bb), (p, B)
6 (p, a, ), (p, a)
7 (p, b, ), (p, b)
unread input
a a b b
a b b
b b
b
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a



stack

a
aa
baa
bbaa
Bbbaa
Bbaa
Baa
Saa
aSaa
Sa
aSa
S

9
Notice Nondeterminism
Machines constructed with the algorithm are often nondeterministic,
even when they needn't be. This happens even with trivial
languages.
Example: AnBn = {anbn: n  0}
A grammar for AnBn is:
[1] S  aSb
[2] S  
A PDA M built top-down for AnBn is:
(0)
(1)
(2)
(3)
(4)
((p, , ), (q, S))
((q, , S), (q, aSb))
((q, , S), (q, ))
((q, a, a), (q, ))
((q, b, b), (q, ))
But transitions 1 and 2 make M nondeterministic.
How about
bottom-up?
A directly constructed machine for AnBn:
10
Going The Other Way
Lemma: If a language is accepted by a pushdown automaton M, it is
context-free (i.e., it can be described by a context-free grammar).
Proof (by construction):
(We will skip this to Slide 16, as it’s rarely necessarily going this
way in practice!)
Step 1: Convert M to restricted normal form:
● M has a start state s that does nothing except push a special
symbol # onto the stack and then transfer to a state s from which
the rest of the computation begins. There must be no transitions
back to s.
● M has a single accepting state a. All transitions into a pop # and
read no input.
● Every transition in M, except the one from s, pops exactly one
symbol from the stack.
11
Converting to Restricted Normal Form
Example:
{wcwR : w  {a, b}*}
Add s and a:
Pop no more than one symbol:
12
M in Restricted Normal Form
[1]
[3]
Pop exactly one symbol:
Replace [1], [2] and [3] with:
[1] ((s, a, #), (s, a#)),
((s, a, a), (s, aa)),
((s, a, b), (s, ab)),
[2] ((s, b, #), (s, b#)),
((s, b, a), (s, ba)),
((s, b, b), (s, bb)),
[3] ((s, c, #), (f, #)),
((s, c, a), (f, a)),
((s, c, b), (f, b))
[2]
Must have one transition for
everything that could have been on
the top of the stack so it can be
popped and then pushed back on.
13
Second Step - Creating the Productions
Example:
WcWR
M=
The basic idea –
simulate a leftmost derivation of M on any input string.
14
Second Step - Creating the Productions
Example:
abcba
15
Nondeterminism and Halting
1. There are context-free languages for which no
deterministic PDA exists.
2. It is possible that a PDA may
● not halt,
● not ever finish reading its input.
3. There exists no algorithm to minimize a PDA. It is
undecidable whether a PDA is minimal.
16
Nondeterminism and Halting
It is possible that a PDA may
● not halt,
● not ever finish reading its input.
Let  = {a} and consider M =
L(M) = {a}: (1, a, ) |- (2, a, a) |- (3, , )
On any other input except a:
● M will never halt.
● M will never finish reading its input unless its input is .
Example: aa
17
Solutions to the Problem
●
For NDFSMs:
● Convert to deterministic, or
● Simulate all paths in parallel.
● For NDPDAs:
● Impossible to simulate all paths in parallel
– each path would need its won stack
● Formal solutions usually involve changing the
grammar  parse tree may not make sense
● Practical solutions that:
● Preserve the structure of the grammar, but
● Only work on a subset of the CFLs.
18
Alternative Equivalent Definitions of a
PDA (Skip to Slide 21)
Accept by accepting state at end of string (i.e., we don't
care about the stack).
From M (in our definition) we build M (in this one):
1. Initially, let M = M.
2. Create a new start state s. Add the transition:
((s, , ), (s, #)).
3. Create a new accepting state qa.
4. For each accepting state a in M do,
4.1 Add the transition ((a, , #), (qa, )).
5. Make qa the only accepting state in M.
19
Example
The balanced parentheses language
20
What About These?
● FSM plus FIFO queue (instead of stack)?
● FSM plus two stacks?
21
Comparing Regular and
Context-Free Languages
Regular Languages
● regular exprs.
● or
● regular grammars
grammars
● recognize
● = DFSMs
Context-Free Languages
● context-free
● parse
● = NDPDAs
22