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Lecture 6: The fundamental theorem
Let
L
⊆
Σ
∗
. Then the following three statements are equiv
alent.
1.
L
is accepted by some DFA.
2.
L
is accepted by some NFA.
3.
L
can be represented by a regular expression
Proof
: We will prove the following: i) 1
⇔
2, ii) 2
⇔
3.
i) 1
⇔
2
1
⇒
2: A DFA is an NFA, by de±nition.
1
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View Full Document The fundamental theorem
To prove that 2
⇒
1
: This is a constructive proof.
First, let’s work on a simple NFA that does not contain
e
edges.
q
01
q
q
2
b
a
b
a
It is possible to go from one state to several states after con
suming one input symbol. Hence, the states now are subsets of
K
:
δ
ab
{
q
0
}
{
q
1
}∅
{
q
1
}
∅{
q
0
,q
2
}
{
q
0
2
}
{
q
0
1
{
q
0
1
}
{
q
1
}{
q
0
2
}
∅
∅∅
How do we decide the ±nal states?
2
What if the given NFA has some
e
edges?
Idea: On reading
a
∈
Σ, the DFA
M
±
imitates a move of the
NFA
M
on
a
, possibly follows a ±nite number of
e
moves.
s > q1 > q2 > q3 > q4 > q5 > q6 > q7 > q8
e
eaee
b
e
e
Initial state is
{
s, q
1
,q
2
}
.
δ
(
{
s, q
1
2
}
,a
)=
{
q
3
4
5
}
δ
(
{
q
3
4
5
}
,b
{
q
6
7
8
}
3
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View Full Document The fundamental theorem
Defne
E
(
q
) = the set oF states
reachable
From
q
, without
reading any input symbols
=
{
p
∈
K

(
q, e
)
±
∗
M
(
p, e
)
}
Example:
2
3
4
0
1
5
6
7
e
e
e
e
e
e
a
b
e
e
9
10
a
8
b
b
E
(0) =
{
0
,
1
,
2
,
4
,
7
}
E
(1) =
{
1
,
2
,
4
}
E
(2) =
{
2
}
E
(3) =
{
3
,
6
,
1
,
2
,
4
,
7
}
E
(4) =
{
4
}
E
(5) =
{
5
,
6
,
7
,
1
,
2
,
4
}
E
(6) =
{
6
,
7
,
1
,
2
,
4
}
E
(7) =
{
7
}
E
(8) =
{
8
}
E
(9) =
{
9
}
E
(10) =
{
10
}
4
The fundamental theorem
Given an NFA
M
=
{
K,
Σ
,
Δ
,s,F
}
.
We want to construct an equivalent DFA that accepts the same
language:
M
±
=
{
K
±
,
Σ
,δ
±
,s
±
,F
±
}
•
K
±
=2
K
•
s
±
=
E
(
s
)
•
F
±
=
{
Q
∈
K
±

Q
∩
F
²
=
∅}
•
for all
Q
∈
K
±
,
σ
∈
Σ,
δ
±
(
Q, σ
)=
S
q
∈
Q
{
E
(
p
):(
q, σ, p
)
∈
Δ
}
.
Note:
δ
±
(
Q, σ
)mayequa
ls
∅
for some
Q
∈
K
±
,
σ
∈
Σ.
Some
Q
’s are unreachable thus are irrelevant, so we start
the construction from
E
(
s
) and only introduce a new state
when it is reachable from the start state (equivalently, con
struct the transition table row by row).
5
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View Full Document Refer to the given NFA:
a
b
E
(0) =
{
0
,
1
,
2
,
4
,
7
}
E
(3)
∪
E
(9)
=
{
3
,
6
,
1
,
2
,
4
,
7
,
9
}
E
(5)
=
{
5
,
6
,
7
,
1
,
2
,
4
}
{
3
,
6
,
1
,
2
,
4
,
7
,
9
}
E
(3)
∪
E
(9)
=
{
3
,
6
,
1
,
2
,
4
,
7
,
9
}
E
(5)
∪
E
(10)
=
{
5
,
6
,
7
,
1
,
2
,
4
,
10
}
{
5
,
6
,
7
,
1
,
2
,
4
}
E
(9)
∪
E
(3)
=
{
3
,
6
,
1
,
2
,
4
,
7
,
9
}
E
(5)
=
{
5
,
6
,
7
,
1
,
2
,
4
}
{
5
,
6
,
7
,
1
,
2
,
4
,
10
}
E
(9)
∪
E
(3)
=
{
3
,
6
,
1
,
2
,
4
,
7
,
9
}
E
(5)
∪
E
(10)
=
{
5
,
6
,
7
,
1
,
2
,
4
,
8
}
{
5
,
6
,
7
,
1
,
2
,
4
,
8
}
E
(9)
∪
E
(3)
=
{
3
,
6
,
1
,
2
,
4
,
7
,
9
}
E
(5)
=
{
5
,
6
,
7
,
1
,
2
,
4
}
5, 6, 7, 1, 2, 4
8
5, 6, 7, 1, 2, 4
10
0, 1, 2, 4, 7
a
a
b
a
b
b
a
a
b
3, 6, 1, 2, 4
5, 6, 7, 1, 2, 4
9
7,
6
To be rigorous, we should also show that (i)
M
±
is deterministic
and (ii)
L
(
M
)=
L
(
M
±
). We only sketch the proof here (refer
to textbook for details).
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This note was uploaded on 09/22/2011 for the course COMP 272 taught by Professor Prof.tai during the Spring '10 term at HKUST.
 Spring '10
 Prof.Tai

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