Lec2 Regular Expression

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CSC 416

Lecture 2egu ar express on


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Regular Expressions A RE is a lan ua e for describin sim le lan ua es

and patterns.

Algorithm for using RE

1. Define a pattern: S1StudentnameS2

2. Loo

Find next patternStore StudentName into DB or encr t StudentName

Until no match

REs are used in a lications that involve file arsinand text matching.

Many Implementations have been made for RE.

Dr.Hussien Sharaf 2


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y oos rary //Pattern matching in a String < >. . // distrubuted under GPL v2 license

#include #include Please check

int main() {boost::regex pattern" "

[http://flavio.castelli.name/regexp-with-boost]

, _boost::regex_constants::perl);std::string stringa ("Searching for BsLug");

if (boost::regex_search (stringa, pattern,

boost::regex_constants::format_perl))printf ("found\n");else printf("not found\n");

Dr.Hussien Sharaf 3


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y oos rary//Substitution// Created by Flavio Castelli

#include

#include int main() Please check{boost::regex pattern

("bok",boost::regex_constants::icase|

[http://flavio.castelli.name/regexp-with-boost]

oos ::regex_cons an s::per ;std::string stringa ("Searching for bok");

"

std::string newString;

newString = boost::regex_replace (stringa,pattern, replace);

printf("The new string is:", . _

return 0; }Dr.Hussien Sharaf 4


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RE by C++ STLneeds extra feature pack#include

#include

#include

bool is_email_valid(const std::string& email) {// define a regular expression

const std::tr1::regexpattern("(\\w+)(\\.|_)?(\\w*)@(\\w+)(\\.(\\w+))+");

try to matc t e str ng w t t e regu ar express on

return std::tr1::regex_match(email, pattern); }

Please check

Dr.Hussien Sharaf 5

p: www.co eguru.com cpp cpp cpp_m c s ar c e.p p c


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RE by C++ STL#include

#include

#include

bool is_email_valid(const std::string& email) {// define a regular expression

const std::tr1::regexpattern("(\\w+)(\\.|_)?(\\w*)@(\\w+)(\\.(\\w+))+");

try to matc t e str ng w t t e regu ar express on

return std::tr1::regex_match(email, pattern);}

Please check

Dr.Hussien Sharaf 6

p: www.co eguru.com cpp cpp cpp_m c s ar c e.p p c


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RE by C++ STLstd::strin str "abc + inside brackets dfsd"

std::smatch m;

std::re ex search str m std::re ex b"_

if (m[0].matched)


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Other code samples Other sam les can be check at: Boost Library

1. http://stackoverflow.com/questions/5804453/c-regular-- - -

2. http://www.codeproject.com/KB/string/regex__.aspx

3. http://www.boost.org/doc/libs/1_43_0/more/getting_started/w

indows.html#build-from-the-visual-studio-ide STL Library

1. htt : www.code ro ect.com KB reci es rexsearch.as x

2. http://www.codeguru.com/cpp/cpp/cpp_mfc/stl/article.php/c15339 [recommended]

. . . .6922 [feature pack VS2008 Pack]

4. http://channel9.msdn.com/Shows/Going+Deep/C9-Lectures-ep an- - avave - an ar - emp a e- rary- - -o -n eo

Dr.Hussien Sharaf 8


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RE Rules -1= = *. ,

In RE, we write x

*

*. e = x, y , en =

In RE, we write (x+y)*

3. Kleenes star * means any combination of lettersof length zero or more.

Dr.Hussien Sharaf 9


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RE Rules -2,

by the following rules.

1. For every letter in , the letter written in bold is a regularexpression. is a regular expression.

2. Ifr1 and r2 are regular expressions, then so are:

1. r1

2. r1 r23. r +r

4. r1*

NOTE: r1+

is not a RE3. Nothing else is a regular expression.

Dr.Hussien Sharaf 10


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RE-1

={a, b}- orma y escr e a wor s w a o owe y any

number of bs

*

*

= a = a- Give examples for words in L

{a ab abb abbb ..}



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RE-2

={a, b}- orma y escr e e anguage a con a ns

nothing and contains words where any a must be

L = (abb*)*

- ve examp es or wor s n

{ ab abb abababb ..}



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RE-3

={a, b}- orma y escr e a wor s w a o owe y one

or more bs

*

= a = a- Give examples for words in L

{ab abb abbb ..}



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RE-4

={a, b, c} - orma y escr e a wor s a s ar w an afollowed by any number of bs and then end with c.

*

= a c- Give examples for words in L

{ac abc abbc abbbc ..}



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RE-5

={a, b} - orma y escr e a wor s w ere a s any comebefore bs if any.

* *

= a- Give examples for words in L

{ a b aa ab bb aaa abb abbb bbb..}

NOTE: a*

b*

(ab)*

because first language does not contain abab but second language has.

nce s ng e s e ec e en no a s can e a e



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RE-6

={a} - orma y escr e a wor s w ere coun o a s o .

L = a(aa)* OR (aa)*a

- Give examples for words in L{a aaa aaaaa ..}



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RE-7.1

={a, b, c} - orma y escr e a wor s w ere s ng e a or ccomes in the start then odd number of bs.

*

= a+c- Give examples for words in L

{ab cb abbb cbbb ..}



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RE-7.2

={a, b, c} - orma y escr e a wor s w ere s ng e a or ccomes in the start then odd number of bs in case of a

.

L = ab(bb)* +c(bb)*


{ab c abbb cbb abbbbb ..}



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RE-8

={a, b, c} - orma y escr e a wor s w ere one or more a orone or more c comes in the start then one or more

.

L = (a+c) + b+= (aa*+cc*) bb*


{ab cb aabb cbbb ..}



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RE-9

={a, b} - orma y escr e a wor s w eng ree.

L = (a+b) 3 =(a+b) (a+b) (a+b)

- List all words in L{aaa aab aba baa abb bab bba bbb}

- What is the count of words of length 4?

16 = 24

- What is the count of words of length 44?44



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RE-10.1

={a, b}, What does L describe?

* *- = a+ a a+

ny s r ng o a s an s

Single aAny string of a's and b's

- Give examples for words in L

{a ab aab bab abb ..}



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RE-10.2

a aa

abb a ab

a a a

abba a b



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-

={a, b}

- orma y escr e a wor s w a eas wo a s.

1) L = b*ab*a(a + b)*

Start with a jungle of b's (or no b's) until we findthe first a, then more b's (or no b's), then thesecond a, then we finish up with anything.

- {abbbabb aaaaa bbbabbbbabab..}



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-

={a, b}

- orma y escr e a wor s w exac y wo a s.

1) L = b*ab*ab*

- Give examples for words in L

To make the word aab, we let the first and second b*



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RE-13.1

={a, b}

- orma y escri e a wor s wit east one a an east

one b.

1) L = (a + b)*a(a + b)* b(a + b)*= an thin a an thin b an thin

But (a+b)*a(a+b)*b(a+b)* expresses all words except

some as (at least one). bb*aa*



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RE-13.2= * * * * *

Thus: (a+b)*a(a+b)*b(a+b)* + (a+b)*b(a+b)*a(a+b)*

= * * * * *

Notice that it is necessary to write bb*aa* because* * , .

oes s mp y a

(a+b)*b(a+b)*a(a+b)*= bb*aa*??False Left side includes the word aba, which the

ex ression on the ri ht side does not.



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egu ar anguages A lan ua e that can be defined b a RE is called a

regular language.

n n can no express some anguages suc as .1. If L1 and L2 are regular languages then

L1 + L2 , L1 L2 and L1 are also regular languages.2. If L is a re ular lan ua e then L L com lement

is also a regular language.

is the RE that accepts language L.

= =



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Regular Languages 2. 1 2

L1 L

2 is also a regular language. L1 =words starting with a

r =a a+b *

L2 =words ending with a*2

L3 =L1 L2 = a(a+b)*a L3 = words that start and end with a.



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Lec2 Regular Expression

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