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September 23, A closer look at the transition rule (r i+1,b) (r i,w i+1,a), where s i =at and s i+1 =bt for some a,b ε and t * –The top symbol is Pushed if a=ε and b ε Popped if a ε and b=ε Changed if a ε and b ε Unchanged if a=ε and b=ε –Symbols below the top of the stack may be considered, but not changed This is t’s role
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CSCI 2670Introduction to Theory of
Computing
September 23, 2004
September 23, 2004 2
Agenda• Yesterday
– Pushdown automata• Today
– Quiz– More on pushdown automata– Pumping lemma for CFG’s
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
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 5
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 if1. It matches the top of the stack, or2. 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 6
Example
ε,ε$
0,$0$0,0 000,11 1 0,1$ $
1,$1$1,1 111,00 0 1,0$ $
ε,$ ε
September 23, 2004 7
Example
ε,ε$
0,$0$0,0 000,1 ε
1,$1$1,1 111,0 ε
ε,$ ε
This PDA is equivalent to the one on the previous slide
September 23, 2004 8
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 9
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 10
Constructing the PDA• You can read any symbol in when
that symbol is at the top of the stack– Transitions of the form a,aε
• The rules will be pushed onto the stack – when a variable A is on top of the stack and there is a rule Aw, you pop A and push w
• You can go to the accept state only if the stack is empty
September 23, 2004 11
Idea of PDA construction for AxBz
State control
a b
At
State control
a b
xBzt
September 23, 2004 12
Actual construction for AxBz
ε,Az ε, ε B
ε, ε x
In an abuse of notation, we say (q,ε,A)=(q,xBz)
September 23, 2004 13
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 {$} (or some suitable placeholder)
September 23, 2004 14
Constructing the PDA is comprised of several rules
1. (qstart,ε,ε)=(qloop,S$)- Start with placeholder on the stack and
with the start variable2. (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 Aw- Implement replacement rules
4. (qloop,ε,$)=(qaccept,ε)- Accept when the stack is empty
September 23, 2004 15
Example• S SS | (S) | ()
qstart qloop qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
September 23, 2004 16
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S() S
$qloop
September 23, 2004 17
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
(S)$
September 23, 2004 18
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
S)$
(
September 23, 2004 19
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
SS)$
(
September 23, 2004 20
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
()S)$
(
September 23, 2004 21
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
)S)$
((
September 23, 2004 22
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
S)$
(()
September 23, 2004 23
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
())$
(()
September 23, 2004 24
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
))$
(()(
September 23, 2004 25
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
)$
(()()
September 23, 2004 26
Example• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
$
(()())
September 23, 2004 27
Example• Read (()())
qstartε, ε S$ ε, $ ε
(,(ε),)ε
ε,SSSε,S(S)ε,S()
qloop
(()())
qaccept
September 23, 2004 28
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
x
y
z
September 23, 2004 29
The pumping lemma for CFG’s
• What might be repeated in a CFG?– The variables
TR
R
u v x y z
v & y will be repeated simultaneously
September 23, 2004 30
The pumping lemma for CFG’s
TRR
u v x y z
TR
uv y
zR
x yvR
September 23, 2004 31
The pumping lemma for CFG’s
TRR
u v x y z
TR
ux
z
September 23, 2004 32
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 A2. |vy| > 03. |vxy| p
September 23, 2004 33
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 34
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 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
RR
u v x y z}At most |V|
+1 steps
Question: What do we know about uvixyiz?Answer: It is accepted by the PDA
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
RR
u v x y z}At most |V|
+1 steps
Question: What do we know about |vy|?Answer: It is non-empty (because tree has fewest possible nodes)
September 23, 2004 37
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
RR
u v x y z}At most |V|
+1 steps
Question: What do we know about |vxy|?Answer: It has at most p symbols (because at most |V|+1 steps)
September 23, 2004 38
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 np. By the pumping lemma, w=uvxyz such that |vxy|p, |vy|>0, and uvixyizA for all i=0,2,1….
September 23, 2004 39
Example (cont.)• Show A is not context free, where
A={an|n is prime}Clearly, vy=ak for some kConsider 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 40