Learning Outcomes
• Mahasiswa dpt menunjukkan perbedaan NFA dgn FA dan dapat
menggunakan teori tersebut..
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Outline Materi:
• Pengertian NFA
• Finite State Transduser
• Contoh aplikasi NFA
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Review: What is a computational problem?
The input is a finite string of characters from some agreed upon alphabet S.
S* = the set of finite strings composed of characters in S.
Sk = the set of strings of length k.
S0 is the empty string (length 0).
A “language” or “computational problem” L is a subset of S*.
w S* means w=w1, w2, …, wn, where wi Sfor all i.
For any machine M, we write L(M) for the language accepted by M, i.e.,
the set of strings w S* such that if M starts in state q0 and r is the state M is
in after reading wn, then r  F.
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So far: Finite Automata = Deterministic Finite Automata (DFA)
Now, Non-deterministic Finite Automata (NFA)
• Multiple arrows leaving a state for the same symbol,
• No arrows leaving a state for a given symbol,
• Arrows labeled with ∑ leaving a state.
Example:
q0
1
q1
0, 1
q2
1
aq43
0, 1
This accepts {w | w = w1, w2,…,wn-3, 1, wn-1, 1}.
Q: Why?
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a1
0
1
a1
a1
a1
a1
a1
1
a2
0
1
1
0
1
a2
a3
1
a2
1, 0
a3
------- dies an early death.
1
a4
Look at leaves. A4 (accept state)
is in there, so we accept the string.
What about  arrows? Change states before receiving new input
(causes a split on tree).
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Defn: A non-deterministic finite automaton
1. Q
is a finite set called the states,
2. ∑ is a finite set called the alphabet,
3. q0 is the start state,
4. F  Q is the set of accept states (or final states).
5.  : Q x (∑  {})  2Q
(We also write S for ∑  {} and P(Q) = 2Q. )
Note: It's possible to make a transition with out reading
any input. Also, you can go to more than one state.
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What does it formally mean for an NFA to accept a string?
Defn: An NFA accepts w over ∑ if we can rewrite
w = y1 y2 … ym where each yi  ∑ .
and we can find a sequence of states r0, ri,… rm
So that
1. r0 = q0
2. rm  F
3. ri+i   (ri, yi+1)
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Every DFA is an NFA. We’ll see the reverse is true too.
Thm: For every NFA M,  a DFA D s.t. L(M) = L (D).
Defn: “ - closure of :”
For r  Q, let
() = {p  Q :  a sequence of  transitions from state r to p}.
q0

q1
0, 1
q2
1, 
aq43
0
Ex:
(q1) = {q1}
(q2) = {q2, q3}
For R  Q, (R) =  (r), r  R.
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Thm: For every NFA M,  a DFA D s.t. L(D) = L(M)
Pf: (sketch) Let M = (Q, ∑, q0, F, ) be any NFA. We will
construct a DFA D where D = (Q´, ∑, q0´, F´, ´):
• Let Q´ = 2Q,
• For R  Q,
(i.e., R  Q´) and a  ∑{}, let
´ (R,a) = {p:  q  Q, r  R s.t. q   (r,a) and p   (q)},
• q0´ = (q0), where q0 is the start state of M,
• F´(Think
= { Sabout
 Q´ :why
S  this
F  construction
}.
works!)
x
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Thm: Regular languages are closed under union.
Pf: Let A and B be any two DFAs. We will construct
an NFA C s.t. L(C) = L(A)  L(B).
A
B
A

AB

B
(NFA)
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More formally, given MA = (Q, ∑, , q0, F), MB = (Q´, ∑, ´, q0´, F´)
(DFAs), construct NFA MC = (Q˝, ∑, ˝, q0˝, F˝):
Q˝
= Q  Q´  {q0˝},
F˝
= F  F´,
˝ (q, ) =
{  (q, ) } if q  Q
{ ´ (q, ) } if q  Q´

{q0, q0´} if q = q0˝, and = .
If w  C=L(MC), then we reached a final state p in F˝ starting
with an  transition to q0 or q0´. In the first case w  A; in the
second w  B. Therefore C  A  B.
Conversely, if w  A, then starting with an  transition to q0
leads to a final state in F  F˝, and if w  B, starting with an
 transition to q0´ leads to a state in F´  F˝. So A  B  C.
x
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Thm: Regular languages are closed under concatonation.
Pf:
A
B

A•B
A

B
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Concatonation:
We have MA = (QA, A, q0A, FA, ∑ )
MB = (QB, B, q0B, FB, ∑ )
Construct
M´ = (Q´, ´, q0´, F´, ∑ ) (This is an NFA)
Q´ = QA  QB
(Disjoint Union)
q0´ = q0A
F´ = FB
´ (q, a) = A (q,a) if q  QA \ FA
B (q,a) if q  QB
A (q,a) if q  FA and a  FA  
A (q,a) = {q, q0B} if a =  and q  FA
L(M´) = L(MA) • L(MB)
= A • B.
x
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Thm: If A is regular, A* is regular.
Pf: (sketch)



A
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