# Lecture 3 Graph Representation for Regular Expressions

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					Lecture 6 Nondeterministic
Finite Automata (NFA)
NFA

tape

Finite Control
a   l   p   h   a   b   e   t

The tape is divided into finitely many cells.
Each cell contains a symbol in an alphabet
Σ.
a

• The head scans at a cell on the tape and
can read a symbol on the cell. In each
move, the head move to the right cell or
stop there (in a ε-move).
• The finite control has finitely many states
which form a set Q. For each move, the
state is changed according to the
evaluation of a transition function
Q
δ : Q x (Σ U {ε}) → 2 (the family of all
subsets of Q)
a                     a

q                         p
p=p1 or p2

• δ(q, a) = {p1, p2} for a in Σ means that if
control is in the state q, then the next state
should be p1 or p2, and the head moves
one cell to the right.
a                       a

q                       p
p=p1 or p2

• δ(q, ε) = {p1, p2} means that if the finite
control is in the state q, then the next state
can be be p1 or p2, and the head does not
move. This move is called a ε-move.
a
?

q

• δ(q, a) = Φ means that the NFA is stuck.
s

• There are some special states: an initial
state s and a final set F of final states.
• Initially, the NTM is in the initial state s and
the head scans the leftmost cell. The tape
holds an input string.
x

h

• When the head gets off the tape, the NFA
stops. An input string x is accepted by the
NFA if there is a computation path to make
the NFA stop at a final state.
• Otherwise, the input string is rejected.
• The NTM can be represented by
M = (Q, Σ, δ, s, F)
where Σ is the alphabet of input symbols.
• The set of all strings accepted by a NFA M
is denoted by L(M). We also say that the
language L(M) is accepted by M.
• The transition diagram of a NFA is an
alternative way to represent the DFA.
• For M = (Q, Σ, δ, s, F), the transition diagram of
M is a symbol-labeled digraph G=(V, E)
satisfying the following:

V = Q (s =      ,f=        for f \in F)

a
E={q         p | p \in δ(q, a)}.
δ   0          1       ε
s   p, s       s
p   q          s   q
q   q

0,1                          0

0              0
s                 p              q
1              ε
Theorem 1
• Every regular language can be accepted
by an NFA.
G(r)
• For each regular expression r, we can
construct a digraph G(r) with edges
labeled by symbols and ε as follows.

• If r=Φ, then

• If r≠Φ, then
Φ*

ε        ε
•   (01+111)*
•   (01)*(111)*
•   10(0+1)*00
•   10(0+1)*+(0+1)*00
•   ((01)*(10)*0*)*
10(0+1)*+(0+1)*00

10(0+1)*

(0+1)*00

1       0       (0+1)*

0
(0+1)*           0
0,1
1       0       ε         ε
ε                        0
ε       0
0,1
5
(0+1)*0(0+1)

0,1

0    0        0       0       0       0

1        1       1       1       1
R
Given an NFA M, construct an NFA accepting L(M) .

0
ε
0
1                  1
ε

0

0
ε

ε       1       0       1

0
Given an NFA M, construct an NFA accepting L(M) .

0              Is it correct?
0

0               What is the correct way?

First, turn NFA to DFA
and then do.


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