NFA for basic regular expressions by S9AJ32J8

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• pg 1
```									บทท 3 Regular Languages & Regular Grammar                               3-5

Theorem     Let    r    be    regular
expression. Then there exists some
nondeterministic finite accepter that
accepts .             L(r)

Consequently,        is a regular                  L(r)

language.

MD
cc pts                 แ                                  nfa  dfa
MN
accepts L

MN     cc pts

b669607a-537a-465e-9660-c7bdcf68bc2b.doc \ 14/6/12
บทท 3 Regular Languages & Regular Grammar                       3-6

nfa accepting L(r)
                               accepting state                 1





 NFA for basic regular expressions
r={}                               r=                r=a
           a

L(MN) = {}                          L(MN) = { } L(MN)= { a}

b669607a-537a-465e-9660-c7bdcf68bc2b.doc \ 14/6/12
บทท 3 Regular Languages & Regular Grammar                                    3-7

 NFA for union
M(r1)
S



                            M(r2)
S                              

r = r1+r2 , L(MN) = L(r1+r2)

 NFA for concatenation

M(r1)
S


M(r2)           
S

r = r1 r2 , L(MN) = L(r1)L(r2)


b669607a-537a-465e-9660-c7bdcf68bc2b.doc \ 14/6/12
บทท 3 Regular Languages & Regular Grammar                      3-8

 NFA for *


M(r1)
                S                   




=
r = r1* , L(MN)  L(r1*)
.          r = 1*
1
q0               แ
แ                       t


                1               


b669607a-537a-465e-9660-c7bdcf68bc2b.doc \ 14/6/12
บทท 3 Regular Languages & Regular Grammar                        3-9

ฝ.                 accept L(c+b*a) และ
L(ab(a+cb*c))

Theorem Let L be a regular
language, then there exists a regular
expression r such that L = L(r)

cc pt
nfa                 Generalized Transition Graphs

Regular Expression

GTG                                                          edges
.
a+b

a                       c
L(r) = L(a*+a*(a+b)c*)
b669607a-537a-465e-9660-c7bdcf68bc2b.doc \ 14/6/12
บทท 3 Regular Languages & Regular Grammar                                3-10

.                                 e
d                               c

q1                   qX                    q2
a                         b

State qX (qXF, qXq0)

q1  q 2
q1  q 1
q2  q 2
q2  q1
state qX
cc pt
q1  q2                    by      taking       ae*d
q1  q1                    by      taking       ae*b
q2  q2                    by      taking       ce*d
q2  q1                    by      taking       ce*b

b669607a-537a-465e-9660-c7bdcf68bc2b.doc \ 14/6/12
บทท 3 Regular Languages & Regular Grammar                      3-11

ae*d                      ce*d        ce*b

q1                            q2

ae*b

r1                       r3            r4

q1                            q2

r2

r = r1* r2(r4 + r3 r1* r2)*

b669607a-537a-465e-9660-c7bdcf68bc2b.doc \ 14/6/12
บทท 3 Regular Languages & Regular Grammar                                   3-12

.                                 0
A
1
1
0
1           C
B
0

0                                             11*0
11*0
A                              B

0

r = ( 0 + (11*0)(11*0)*0 ) *
b669607a-537a-465e-9660-c7bdcf68bc2b.doc \ 14/6/12

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