NFA’s versus FA’s
„
Show that: For any FA
M = (Q, Σ, δ, q0, F), there exists an NFA
M1 = (Q1, Σ, δ1, q1, F1) such that
L(M) = L(M1).
NFA’s versus FA’s
„
NFA’s may look more powerful than FA’s,
but they are not!
„
Theorem: For any NFA
M = (Q, Σ, δ, q0, F), there exists a FA
M1 = (Q1, Σ, δ1, q1, F1) such that
L(M) = L(M1).
1
NFA’s versus FA’s
„
What is the real difference between a FA
and a NFA?
NFA’s versus FA’s
„
The idea is that the states of the FA
correspond to a set of states of the NFA.
The
DFA keeps track of the set of states that
the NFA can be in.
Construct an FA transition diagram equivalent
to the following NFA transition diagram.
a
q0
a
b
a
q2
q1
b a
2
Proof of Theorem NFA = DFA
„
Let M = (Q, Σ, q0, F, δ) be a NFA. Define an
equivalent FA M1 as follows:
M1 = (Q1, Σ, q1, F1, δ1) such that
Q1
= P(Q)
q1 = { q0 }
A1 = { q∈ Q1 | q I F ≠ ∅ }
for all q∈Q1 and a∈Σ, δ1(q, a) = U p∈q δ(p, a)
We still have to prove that L(M) = L(M1).
Proof of Theorem NFA = DFA
„
The proof is by induction. The main problem
is to formulate what we are going to prove in
the proper way.
example, try to prove that for all x ∈ Σ*:
x∈L(M) iff x∈ L(M1) by induction on the length of
x.
For
„
This statement is clearly true, and it implies that L(M)
= L(M1). But if you try to prove it, you’ll get stuck!
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Proof of Theorem NFA = DFA
example, try to prove that for all x ∈Σ*.
δ*1 (q1, x) = δ*(q0, x) by structural induction on
x.
For
„
This implies that L(M) = L(M1).
NFA to DFA
„
Given an NFA M it can be converted to a DFA
M1.
1.
2.
3.
Determine the states in M1. If M has n states then M1
has 2n states.
Label each of M1’s states with the corresponding
subset of M’s states.
Determine M1’s start and accepting states.
4.
Start state is the set of states reachable from M’s start state
via the ε arrows plus the M’s start state.
Determine M1’s transition function.
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NFA to DFA example
0,1
q0
1
q1
0,1
q2
0,1
q3
How many states could the DFA have?
What are the start states?
What are the accept states?
Convert this NFA to a DFA.
NFA-ε’s vs. NFA’s
„
Theorem: For any NFA-ε
M = (Q, Σ, q0, F, δ), there exists a NFA
M1 = (Q1, Σ, q1, F1, δ1) such that L(M) = L(M1).
5
NFA-ε’s vs NFA’s
„
Convert the following NFA-ε to an NFA.
C
0
A
ε
0
ε
0
B
ε
D
NFA-ε’s vs. NFA’s vs. DFA’s
„
NFA-ε’s, NFA’s and FA’s are all
equivalent.
6
Kleene’s Theorem
„
Kleene’s Theorem: A language L over the
alphabet Σ is regular if and only if there is a
FA that accepts L.
Theorem: Any regular language can be
accepted by an FA.
„ Theorem: The language accepted by any FA
is regular.
„
Corollary
„
Corollary: A language is regular if and only
if some nondeterministic finite automaton
recognizes it.
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Closure: Union
„
Theorem: The class of regular languages
is closed under the union operation.
Can
now prove using NFAs rather than
having to prove using a Cartesian product
construction.
Closure: Union
„
Proof:
N1 = (Q1, Σ, δ1, q1, F1) recognize A1 and
N2 = (Q2, Σ, δ2, q2, F2) recognize A2
Construct N = (Q, Σ, δ, q, F)
Let
Q = {q0} U Q1 U Q2
„ The state q0 is the start state of N
„ The accept states F = F1 U F2
„ Define δ so that for any q∈ Q and an a ∈ Σε
δ(q,a) = δ1(q,a) q∈Q1
δ2(q,a) q ∈Q2
„
{q1,q2} q = q0 and a = ε
q = q0 and a ≠ ε
∅
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Closure: Concatenation
„
Theorem: The class of regular languages
is closed under the concatenation
operation.
Closure: Concatenation
„
Proof:
N1 = (Q1, Σ, δ1, q1, F1) recognize A1 and
N2 = (Q2, Σ, δ2, q2, F2) recognize A2
Construct N = (Q, Σ, δ, q, F)
Let
Q = Q1 U Q2
„ The state q1 is the start state of N
„ The accept states F = F2
„ Define δ so that for any q∈ Q and an a ∈ Σε
δ(q,a) = δ1(q,a)
q∈Q1 and q∉F1
δ1(q,a)
q ∈F1 and a ≠ ε
„
δ1(q,a) U {q2} q = F1 and a = ε
δ2(q,a)
q = Q2
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Closure: Kleene Star
„
Theorem: The class of regular languages
is closed under the Kleene star operation.
Closure: Kleene Star
„
Proof:
N1 = (Q1, Σ, δ1, q1, F1) recognize A1
Construct N = (Q, Σ, δ, q, F) to recognize A*1
Let
Q = {q0} U Q1
„ The state q0 is the start state of N
„ The accept states F = {q0} U F1
„ Define δ so that for any q ∈ Q and an a ∈ Σε
„
δ(q,a) = δ1(q,a)
δ1(q,a)
δ1(q,a) U {q1}
{q1}
∅
q∈Q1 and q∉F1
q ∈F1 and a ≠ ε
q = F1 and a = ε
q = q0 and a = ε
q = q0 and a ≠ ε
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