Transcript Complexity and Computability Theory I

Complexity and Computability
Theory I
Lecture #3
Rina Zviel-Girshin
Leah Epstein
Winter 2002-2003
Overview
Finite automata
Deterministic automata
Regular languages
Regular operations
Nondeterminism
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FA vs. computer
• Both a finite automaton (or a finite-state
machine) and a real computer have a "central
processing unit" of a fixed, finite capacity.
• This is a common property of them both.
• A FA delivers no output at all.
• A FA is a language recognition device.
• A FA is a restricted model of real computers with
no memory apart from the fixed central processor.
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Deterministic finite automata
• A finite automaton is called deterministic if for
each  in  and each qj in Q there exist a unique
(qj,)=qk.
qj

qk
Example:
qj


?
qj

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qk
qi
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Extended transition function
• Notation: ’
• ’: Q*Q
• The extended transition function ’ defines
the computation of an automaton on word
w= u.
• Formal definition
’(q,u)=  (’(q,u), )=m
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Example
u

q
k
m
where ’(q,u)=  (’(q,u), )=  (k, )= m
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Language recognition
•
Let A=(Q,,,q0,F) be a finite automaton and
w=12...n a string over *. A accepts w if there
exists a sequence of states r0,r1,..rn in Q with the
following properties:
1. r0 = q0
2. (ri, i+1) = ri+1 for i=0,1,..n
3. rn  F
•
We say A recognizes language L if
L = {w| A accepts w}.
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Regular language
• A language is called “a regular language”
if some finite automaton recognizes it.
• Notation: R
• The class of languages accepted by finite
automata is identical to the class of regular
languages.
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Example
 is regular.

q0
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Example
* is regular.

q0
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Example
L={} is regular.
q0

q1

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Example
For each α in  the language Lα = {α} is
regular.

q0

q1

q2

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Regular operations
Union
Intersection
Complement
Minus
Concatenation
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Union
Theorem:
The class of regular languages is closed under the
union operation.
If L1 and L2 are regular, so is L1L2.
Basic idea:
•proof by construction: construct an automaton M
that uses the automaton A of L1 and the automaton B
of L2 and works by simulating both A and B.
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How can M simulate A and B?
• M could run w on A and if A accepts then M
accepts.
• Or it could run w on B and if B accepts then M
accepts.
But
• if w not accepted by A?
• We can't rewind the input tape.
• Once the letter is used the is no way back.
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Simulator
Another approach:
• We simulate A and B simultaneously.
• To do so we need to remember the current
states of both machines.
• M will have states in QA QB.
qa,qb
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Formal proof
Let A recognize L1 where A = (QA, , A, q0A, FA) and
B recognize L2 where B = (QB, , B, q0B, FB) .
Construct M to recognize L1 L2,
where M = ( Q, , , q0, F).
1. Q={ (rA,rB) | rA in QA and rB in QB}.
2.  is the same for both A and B. Otherwise let
construct  = AB
3.  is defined as follows: for each (rA,rB) in Q and 
in , let ((rA,rB),)=(A(rA,),B(rB,)).
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Formal proof
4. q0 is a pair (q0A,q0B)
5. F is the set of pairs in which at least one member is
an accept state of A or B.
F = { (rA,rB) | rA in FA or rB in FB}
or F = (FAQB)(QAFB).
This concludes the construction of the finite automaton
M that recognizes the union of L1 and L2.
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Formal proof
• Now we have to prove that L(M)= L1 L2.
To prove the equality we have to prove that:
• L1 L2  L(M).
and
• L(M) L1 L2.
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L1 L2  L(M).
• If x L1 L2 then x L1or x L2
• That means:
’A(q0A,x)=rA Fa or ’B(q0B,x))= rB FB
• So ’(q0,x)= ’((q0A,q0B),x)= (’A(q0A,x),
’B(q0B,x)) = (rA,rB)
where (rA,rB) (FAQB) or (rA,rB) (QAFB).
• By definition of F:
’(q0,x)=(rA,rB)  (FAQB)(QAFB)=F
• That means x L(M).
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L(M) L1 L2
• If xL(M) then ’(q0,x)= ’((q0A,q0B),x)=
(’A(q0A,x), ’B(q0B,x)) F
• By definition of F:
(’A(q0A,x), ’B(q0B,x))F means
’A(q0A,x) FA
or
’B(q0B,x)) FB
• So x belongs to L1or to L2.
• By definition of union: x L1 L2
• That means L(M) L1 L2.
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Intersection
Theorem:
The class of regular languages is closed under the
intersection operation.
If L1 and L2 are regular, so is L1L2.
Basic idea:
• proof by construction: construct an automaton M
that uses the automaton A of L1 and the
automaton B of L2 and works by simulating both
A and B.
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Formal proof
Let A recognize L1 where A = (QA, , A, q0A, FA) and
B recognize L2 where B = (QB, , B, q0B, FB) .
Construct M to recognize L1 L2,
where M = ( Q, , , q0, F).
1. Q={ (rA,rB) | rA in QA and rB in QB}.
2.  is the same for both A and B. Otherwise let
construct  = A  B
3.  is defined as follows: for each (rA,rB) in Q and 
in , let ((rA,rB),)=(A(rA,),B(rB,)).
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Formal proof
4. q0 is a pair (q0A,q0B)
5. F is the set of pairs in which each member is an
accept state of A or B (respectively).
F = { (rA,rB) | rA in FA and rB in FB}
or F = FAFB.
That concludes the construction of the finite automaton
M that recognizes intersection of L1 and L2.
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Correctness
• To prove the correctness of the construction
algorithm in a formal way we need to prove
that
• L(M)L(A)  L(B)
and
• L(A)  L(B)L(M).
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L(M)L(A)  L(B)
• Let w* and wL(M) 
’(q0,w)=’((q0A,q0B),w)=(’A(q0A,w),’B(q0B,w))

if wL(M) then ’(q0,w)=’((q0A,q0B),w)F 
(’A(q0A,w),’B(q0B,w))F=(FAFB) 
’A(q0A,w)FA and ’B(q0B,w)FB 
means wL(A) and wL(B)  wL(A)L(B)
L(M)L(A)  L(B).
Q.E.D.
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L(A)  L(B)L(M)
• Let w* and w L(A)  L(B) 
wL(A) and wL(B)
’A(q0A,w)FA and ’B(q0B,w)FB 
’(q0,w)=’((q0A,q0B),w)=(’A(q0A,w),’B(q0B,
w))(FAFB)=F 
means wL(M) L(A)  L(B)L(M).
Q.E.D.
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Complement
Theorem:
The class of regular languages is closed under the
complement operation.
If L is regular, so is “not L” (the complement
language of L).
Basic idea:
• proof by construction: construct an automaton M
that uses the automaton A of L and rejects all
words A accepts and vice versa.
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Formal proof
Let A recognize L where A = (QA, , A, q0A, FA) .
Construct M to recognize L,
where M = ( Q, , , q0, F).
1. Q= QA
2.  = A
3.  is defined as follows: for each rA in Q and  in
, let (rA, )=A(rA,)
4. q0 =q0A
5. F = { rA | rA not in FA } or F = QA - FA.
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The set minus operation
Theorem:
The class of regular languages is closed under
the minus operation.
If L1 and L2 are regular, so is L1-L2.
Formal proof:
L1-L2 = L1(notL2)
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Concatenation
Theorem:
The class of regular languages is closed under the
concatenation operation.
If L1 and L2 are regular, so is L1L2.
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Concatenation
Basic idea:
• To construct M we need to break a word w into
two pieces u and v and check if u is in L1 and v is
in L2.
• But how to break an input into the pieces?
• To solve this problem we need to use a new
approach called nondeterminism.
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Nondeterminism
• In a nondeterministic machine several "next
states" are possible.
• The automaton may choose any of these legal next
states.

qj

qk
qi
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Nondeterminism
• It also allows to change a current state
without reading an input.
qj

qk
This situation called an  - transition
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DFA vs. NFA
• Every deterministic finite automaton DFA
is immediately a nondeterministic finite
automaton NFA.
• The opposite is not true.
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Example
• Construct a nondeterministic finite
automaton for the following language:
L = { w1 | w over *={0,1}*}
q0

q1
0,1
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Any Questions?
• More about non-determinism on next
lecture.
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