Regular expression over ∑={a,b,c} that represent all string of length
=> (a+b+c)(a+b+c)(a+b+c)
String having zero or more
=> a*
String having one or more
=> a+
All binary string.
=>(0+1)*
0 or more occurrence of either a or b or both
=> (a+b)*
1 or more occurrence of either a or b or both
=> (a+b)+
Binary number end with 0
=>(0+1)*0
Binary number end with 1
=>(0+1)*1
Binary starts and end with 1.
=> 1(0+1)*1
String starts and ends with same
=> 0(0+1)*0 or 1(0+1)*1 or a(a+b)*a or b(a+b)*b
All string of a and b starting with a
=> a(a/b)*
String of 0 and 1 end with 00
=> (0+1)*00
String end with abb
=> (a+b)*abb
String start with 1 and end with 0
=> 1(0+1)*0
[wp_ad_camp_1]
All binary string with at least 3 characters and 3rd character should be 0
=>(0+1)(0+1)0(0+1)*
Language which consist of exactly two b’s over the set ∑={a,b}
=>a*ba*ba*
∑={a,b} such that 3rd character from right end of the string is always a
=> (a+b)*a(a+b)(a+b)
Any number of a followed by any number of b followed by any number of c.
=> a*b*c*
binary string should contain at least 3 one
=> (0+1)*1(0+1)*1(0+1)*1(0+1)*
String should contain exactly Two 1’s
=> 0*10*10*
Length should be at least be 1 and at most 3
=> (0+1) + (0+1) (0+1) + (0+1) (0+1) (0+1)
binary string having number of zero should be multiple of 3
=> (1*01*01*01*)*+1*
[wp_ad_camp_1]
∑={a,b,c} where a are multiple of 3.
=> ((b+c)*a (b+c)*a (b+c)*a (b+c)*)*
binary string having Even number of 0.
=> (1*01*01*)*
binary string having Odd no. of 1.
=> 0*(10*10*)*10*
String should have odd length
=> (0+1)((0+1)(0+1))*
String should have even length
=> ((0+1)(0+1))*
String start with 0 and has odd length
=> 0((0+1)(0+1))*
String start with 1 and has even length
=> 1(0+1)((0+1)(0+1))*
Binary number having Even number of 1
=> (0*10*10*)*
String of length 6 or less
=> (0+1+^)6
String ending with 1 and not contain 00
=> (1+01)+
All string begins or ends with 00 or 11
=> (00+11)(0+1)*+(0+1)*(00+11)
All string not contains the sub-string 00
=> (1+01)* (^+0)
Language of all string containing both 11 and 00 as substring
=> ((0+1)*00(0+1)*11(0+1)*)+ ((0+1)*11(0+1)*00(0+1)*)
Language of C identifier.
=> (_+L)(_+L+D)*