regular expressions 2
TRANSCRIPT
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ILMU KOMPUTASI
STMIK AMIKOM Purwokerto
Teknik Informatika
Fall 2006 Costas Busch - RPI 1
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Fall 2006 Costas Busch - RPI 2
Regular Expressions
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Fall 2006 Costas Busch - RPI 3
Regular Expressions
Regular expressions (RE)
Menjelaskan tentang bahasa regular (regular languages)
Contoh:
Menjelaskan language
*)( cba
,...,,,,,*, bcaabcaabcabca
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Fall 2006 Costas Busch - RPI 4
Recursive Definition
,,
1
1
21
21
*
r
r
rr
rr
Regular expressions
Primitif Regular expressions:
2r1rUntuk regular expressions and
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Fall 2006 Costas Busch - RPI 5
Contoh :
)(* ccbaRegular expression:
baBukan Regular expression:
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Fall 2006 Costas Busch - RPI 6
Bahasa dari Regular Expressions
: bahasa dari regular expression
Contoh:
rL r
,...,,,,,*)( bcaabcaabcacbaL
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Fall 2006 Costas Busch - RPI 7
Defenisi :
Untuk primitif regular expressions:
aaL
L
L
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Defenisi (cont’d)
Untuk Regular expressions dan
1r 2r
2121 rLrLrrL
2121 rLrLrrL
** 11 rLrL
11 rLrL
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Fall 2006 Costas Busch - RPI 9
Contoh :
Regular expression: *aba
*abaL *aLbaL
*aLbaL
*aLbLaL
*aba
,...,,,, aaaaaaba
,...,,,...,,, baababaaaaaa
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Fall 2006 Costas Busch - RPI 10
Contoh :
Regular expression
bbabar *
,...,,,,, bbbbaabbaabbarL
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Fall 2006 Costas Busch - RPI 11
Contoh :
Regular expression bbbaar **
}0,:{ 22 mnbbarL mn
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Fall 2006 Costas Busch - RPI 12
Contoh :
Regular expression *)10(00*)10( r
)(rL = { semua strings berisi substring 00 }
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Fall 2006 Costas Busch - RPI 13
Contoh :
Regular expression )0(*)011( r
)(rL = { semua strings tanpa substring 00 }
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Ekuivalensi Regular Expressions
Defenisi:
Regular expressions dan
akan ekuivalen jika
1r 2r
)()( 21 rLrL
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Fall 2006 Costas Busch - RPI 15
Contoh :
L = { semua strings tanpa substring 00 }
)0(*)011(1 r
)0(*1)0(**)011*1(2 r
LrLrL )()( 211r 2rdan
adalah regular
expressions ekuivalen
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Fall 2006 Costas Busch - RPI 16
Regular Expressions dan
Regular Languages
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Fall 2006 Costas Busch - RPI 17
Teorema
Languages
Dibangkitkan dgn
Regular Expressions
Regular
Languages
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Fall 2006 Costas Busch - RPI 18
Languages
Dibangkitkan dgn
Regular Expressions
Regular
Languages
Languages
Dibangkitkan dgn
Regular Expressions
Regular
Languages
Bukti:
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Fall 2006 Costas Busch - RPI 19
Bukti - Part 1
r
)(rL
Untuk regular expression
Bahasa adalah regular
Languages
Dibangkitkan dgn
Regular Expressions
Regular
Languages
Pembuktian dgn induksi pada ukuran r
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Fall 2006 Costas Busch - RPI 20
Basis Induksi
Primitif Regular Expressions: ,,
NFAs yg sesuai :
)()( 1 LML
)(}{)( 2 LML
)(}{)( 3 aLaML
Regular
languages
a
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Hypothesis Induktif
Misalkan :
Bahwa untuk regular expressions and ,
dan adalah regular languages
1r 2r
)( 1rL )( 2rL
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Langkah Induktif
Kita akan buktikan :
1
1
21
21
*
rL
rL
rrL
rrL
Adalah regular
Languages
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Fall 2006 Costas Busch - RPI 23
Berdasarkan defenisi regular expressions:
11
11
2121
2121
**
rLrL
rLrL
rLrLrrL
rLrLrrL
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)( 1rL )( 2rL
Dengan hipotesis induktif diketahui:
dan adalah regular languages
Regular languages tertutup untuk operasi:
*1
21
21
rL
rLrL
rLrL Union
Concatenation
Star
Diketahui juga :
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Maka :
** 11
2121
2121
rLrL
rLrLrrL
rLrLrrL
Adalah regular
languages
)())(( 11 rLrL Hal yg biasa dr regular language
(dgn induksi hypothesis)
End of Proof-Part 1
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Gunakan regular closure operasi ini,
Dpt di buat rekursif NFA yg diterima M)()( rLML
Contoh: 21 rrr
)()( 11 rLML
)()( 22 rLML
)()( rLML
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Untuk regular language dimana
regular expression dengan
Proof - Part 2
Languages
Dibangkitkan dgn
Regular Expressions
Regular
Languages
Lr LrL )(
Dapat dibuat NFA ekuivalensi dari bahasa
untuk regular expression nya
L
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Karena adalah regular, ada NFA dari
yg menerima nya :
L M
LML )(
Dengan satu state akhir (1 accepting state)
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Diagram transisi yg ekuivalen
di mana fungsi transisi adalah ekspresi reguler
M
Contoh:
a
ba,
c
Ma
ba
c
Diagram Transisi yg sesuai
Secara umum
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Contoh lain:
ba a
b
b
0q 1q 2q
ba,a
b
b
0q 1q 2q
b
bFungsi transisi
adalah regular
expressions
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Fall 2006 Costas Busch - RPI 31
State hsl reduksi:
ba a
b
b
0q 1q 2q
b
0q 2q
babb*
)(* babb
Fungsi transisi
adlh regular
expressions
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Hasil Regular Expression:
0q 2q
babb*
)(* babb
*)(**)*( bbabbabbr
LMLrL )()(
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Secara Umum
Mereduksi state:
iq q jq
a b
cd
e
iq jq
dae* bce*
dce*
bae*
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0q fq
1r
2r
3r4r
*)*(* 213421 rrrrrrr
LMLrL )()(
Hasil regular expression nya :
Dgn mengulangi proses sampai dua state
Yg tersisa, diagram transisi hsl adlh:
Diagram awal Diagram hasil
End of Proof-Part 2
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Representasi Standar untuk Regular Languages
Regular Languages
DFAs
NFAs Regular
Expressions