csci 2670 introduction to theory of computing september 22, 2005
TRANSCRIPT
CSCI 2670Introduction to Theory of
Computing
September 22, 2005
September 22, 2005
Agenda
• Yesterday– Pushdown automata
• Today– Equivalence of pushdown automata
and CFG’s– Pumping lemma for CFG’s
September 22, 2005
Announcements
• Matrix Reloaded tonight!– 6:30 in Boyd 328
• Free popcorn• 25 cent Coke• Pizza will be ordered at cost if people
want
September 22, 2005
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
We will only do the “only if” step – i.e., show that every context-free language has an associated PDA
September 22, 2005
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 22, 2005
CFG’s are recognized by PDA’s
Format of the new PDA
qstart qloop qacceptε, ε S$ ε, $ ε
a,aε
ε,Aw
Start by pushing the start variable and stack bottom marker
Have a transition for each rule replacing the variable with its right hand side
Have a transition that allows us to read each alphabet symbol if it is at the top of the stack
Finish only if the stack is empty
September 22, 2005
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 22, 2005
Idea of PDA construction for AxBz
State control
a b
At
State control
a b
xBzt
September 22, 2005
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 22, 2005
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 {$} (or some suitable placeholder)
September 22, 2005
Constructing the PDA
is comprised of several rules1. (qstart,ε,ε)=(qloop,S$)
- 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 Aw- Implement replacement rules
4. (qloop,ε,$)=(qaccept,ε)- Accept when the stack is empty
September 22, 2005
Example
• S SS | (S) | ()
qstart qloop qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()S$
qloop
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
(S)$
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
S)$
(
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
SS)$
(
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
()S)$
(
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
)S)$
((
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
S)$
(()
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
())$
(()
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
))$
(()(
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
)$
(()()
September 22, 2005
Example
• Read (()())
qstart qacceptε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
$
(()())
September 22, 2005
Example
• Read (()())
qstartε, ε S$ ε, $ ε
(,(ε
),)ε
ε,SSS
ε,S(S)
ε,S()
qloop
(()())
qaccept
September 22, 2005
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 22, 2005
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
September 22, 2005
The pumping lemma for CFG’s
TR
R
u v x y z
TR
uv y
z
R
x yv
R
September 22, 2005
The pumping lemma for CFG’s
TR
R
u v x y z
TR
ux
z
September 22, 2005
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 22, 2005
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 22, 2005
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 22, 2005
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 22, 2005
September 22, 2005