CSCI 2670
Introduction to Theory of
Computing
September 23, 2004
Agenda
• Yesterday
– Pushdown automata
• Today
– Quiz
– More on pushdown automata
– Pumping lemma for CFG’s
September 23, 2004
2
Example
•
Find  for the PDA that accepts all
strings in {0,1}* with the same
number of 0’s and 1’s
1. Need to keep track of “equilibrium
point” with a $ on the stack
2. If stack top is not $, it contains the
symbol currently dominating in the
string
September 23, 2004
4
Example
•
Find  for the PDA that accepts all
strings in {0,1}* with the same
number of 0’s and 1’s
3. Push a symbol on the stack as its read
if
1. It matches the top of the stack, or
2. The top of stack is $
4. Pop the symbol off the top of the
stack if you read a 0 and the top of
stack is 1 or vice versa
September 23, 2004
5
Example
1,$1$
1,1 11
1,00 0
1,0$ $
ε,ε$
ε,$ ε
0,$0$
0,0 00
0,11 1
0,1$ $
September 23, 2004
6
Example
1,$1$
1,1 11
1,0 ε
ε,ε$
ε,$ ε
0,$0$
0,0 00
0,1 ε
This PDA is equivalent to the one on the previous slide
September 23, 2004
7
Equivalence of PDA’s and CFG’s
Theorem: A language is context free if
and only if some pushdown automaton
recognizes it
Proved in two lemmas – one for the “if”
direction and one for the “only if”
direction
September 23, 2004
8
CFG’s are recognized by PDA’s
Lemma: If a language is context free,
then some pushdown automaton
recognizes it
Proof idea: Construct a PDA following
CFG rules
September 23, 2004
9
Constructing the PDA
• You can read any symbol in  when
that symbol is at the top of the stack
– Transitions of the form a,aε
• The rules will be pushed onto the
stack – when a variable A is on top of
the stack and there is a rule Aw,
you pop A and push w
• You can go to the accept state only if
the stack is empty
September 23, 2004
10
Idea of PDA construction for AxBz
State
control
a b
State
control
a b
x
B
z
t
A
t
September 23, 2004
11
Actual construction for AxBz
ε,Az
ε, ε B
ε, ε x
In an abuse of notation, we say (q,ε,A)=(q,xBz)
September 23, 2004
12
Constructing the PDA
• Q = {qstart, qloop, qaccept}E, where E is
the set of states used for
replacement rules onto the stack
•  (the PDA alphabet) is the set of
terminals in the CFG
•  (the stack alphabet) is the union of
the terminals and the variables and
{$} (or some suitable placeholder)
September 23, 2004
13
Constructing the PDA
•
 is comprised of several rules
1. (qstart,ε,ε)=(qloop,S$)
-
Start with placeholder on the stack and
with the start variable
2. (qloop,a,a)=(qloop,ε) for every a
-
Terminals may be read off the top of the
stack
3. (qloop,ε,A)=(qloop,w) for every rule Aw
-
Implement replacement rules
4. (qloop,ε,$)=(qaccept,ε)
-
Accept when the stack is empty
September 23, 2004
14
Example
• S  SS | (S) | ()
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
ε, $ ε
qaccept
(,(ε
),)ε
September 23, 2004
15
Example
• Read (()())
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
ε, $ ε
S
$
qaccept
(,(ε
),)ε
September 23, 2004
16
Example
• Read (()())
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
ε, $ ε
(
S
)
$
qaccept
(,(ε
),)ε
September 23, 2004
17
Example
• Read (()())
ε,SSS
S
)
$
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
(
18
Example
• Read (()())
S
S
)
$
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
(
19
Example
• Read (()())
(
)
S
)
$
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
(
20
Example
• Read (()())
)
S
)
$
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
((
21
Example
• Read (()())
ε,SSS
S
)
$
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
(()
22
Example
• Read (()())
(
)
)
$
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
(()
23
Example
• Read (()())
ε,SSS
)
)
$
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
(()(
24
Example
• Read (()())
ε,SSS
ε,S(S)
)
$
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
(()()
25
Example
• Read (()())
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
$
ε, $ ε
qaccept
(()())
26
Example
• Read (()())
ε,SSS
ε,S(S)
ε,S()
qstart
ε, ε S$
qloop
(,(ε
),)ε
September 23, 2004
ε, $ ε
qaccept
(()())
27
The pumping lemma for RE’s
• The pumping lemma for RE’s depends
on the structure of the DFA and the
fact that a state must be revisited
– Only a finite number of states
y
x
September 23, 2004
z
28
The pumping lemma for CFG’s
• What might be repeated in a CFG?
– The variables
T
R
R
u
v
x
y z
v & y will be repeated simultaneously
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The pumping lemma for CFG’s
T
R
T
R
R
u v
x
y z
u
v v
September 23, 2004
R
R
x
z
y y
30
The pumping lemma for CFG’s
T
R
T
R
R
u v
x
u
y z
September 23, 2004
x
z
31
The pumping lemma for CFL’s
Theorem: If A is a context-free
language, then there is a number p
(the pumping length) where, if s is
any string in A of length at least p,
then s may be divided into five
pieces s=uvxyz satisfying the
conditions:
1. For each i  0, uvixyiz  A
2. |vy| > 0
3. |vxy|  p
September 23, 2004
32
Finding the pumping length of a CFL
• Let b equal the longest right-hand
side of any rule (assume b > 1)
– Each node in the parse tree has at most
b children
– At most bh nodes are h steps from the
start node
• Let p equal b|V|+2, where |V| is the
number of variables (could be huge!)
– Tree height is at least |V|+2
September 23, 2004
33
Proving the pumping lemma for CFG’s
• Let s be any string in A with length at
least p = b|V|+2 and let  be a parse
tree for s with the fewest possible
nodes
– The height of  is at least |V|+2
– The longest path in  has a length of at
least |V|+2
• Leaf is a terminal and at least |V|+1 variables
– Some variable appears twice in the path
September 23, 2004
34
Proving the pumping lemma for CFG’s
• Let R be the last variable to appear
twice on this path
– The last two occurrences of R are both
at most |V|+1 steps from the leaf
T
R
R
}
At most |V|+1
steps
u v
x
y z
Question: What do we know about uvixyiz?
Answer: It is accepted by the PDA
September 23, 2004
35
Proving the pumping lemma for CFG’s
• Let R be the last variable to appear
twice on this path
– The last two occurrences of R are both
at most |V|+1 steps from the leaf
T
R
R
}
At most |V|+1
steps
u v
x
y z
Question: What do we know about |vy|?
Answer: It is non-empty (because tree has fewest
possible nodes)
September 23, 2004
36
Proving the pumping lemma for CFG’s
• Let R be the last variable to appear
twice on this path
– The last two occurrences of R are both
at most |V|+1 steps from the leaf
T
R
R
}
At most |V|+1
steps
u v
x
y z
Question: What do we know about |vxy|?
Answer: It has at most p symbols (because at most
|V|+1 steps)
September 23, 2004
37
Example
• Show A is not context free, where
A={an|n is prime}
Proof: Assume A is context-free and
let p be the pumping length of A. Let
w=an for any np. By the pumping
lemma, w=uvxyz such that |vxy|p,
|vy|>0, and uvixyizA for all i=0,2,1….
September 23, 2004
38
Example (cont.)
• Show A is not context free, where
A={an|n is prime}
Clearly, vy=ak for some k
Consider the string uvn+1xyn+1z
This string add n copies of ak to w –
i.e., this is an+nk
Since the exponent is n(1+k) this in not
in A, which contradicts the pumping
lemma. Therefore, A is not context
free.
September 23, 2004
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40