查找包含偶数个a和偶数个b的所有a和b字符串的正则表达式？

Question

Is it possible to count the number of times that a character has occured in a string using regular expression? Can any regular expression be given to find the all strings of a's and b's that contain an even number of a's and an even number of b's?

Answer 1

There is a simple enough finite state machine: it has four states: s00, s01, s10, and s11 depending on whether you have consumed an even or odd number of a s and an even or odd number of b s. The start state (also the end state) is the state reached by consuming an even number of both a s and b s. The transition function looks like this:

d(s00, a) = s10
d(s00, b) = s01
d(s10, a) = s00
d(s10, b) = s11
d(s01, a) = s11
d(s01, b) = s00
d(s11, a) = s01
d(s11, b) = s10

We can eliminate state s11 :

d(s00, a)  = s10
d(s00, b)  = s01
d(s10, a)  = s00
d(s10, ba) = s11
d(s10, bb) = s10
d(s01, b)  = s00
d(s01, aa) = s01
d(s01, ab) = s10

From this we can develop a regular expression without lookahead by tracing all possible paths through the FSM that return once to the start state, and repeating:

( a (bb|ba(aa)*ab)* (a|ba(aa)*b) | b (aa|ab(bb)*ba)* (b|ab(bb)*a) )*

(Meaningless blanks inserted to help me keep track of nesting of parentheses.) The idea is, if the first character is a you reach s10 ; then you can transition to s10 and back to s01 repeatedly via (bb|ba(aa)*ab)* , and finally return to s00 (without repeating s10 ) either via a or via ba(aa)*b . A similar pattern (just swap the occurrences of a and b ) gets you from s00 back to s00 via a string that starts with b . And you can make as many trips out and back to s00 as you like starting with either a or b .

Answer 2

Yes it is possible using this lookahead based regex:

^(?=(?:b*ab*a)*b*$)(?=(?:a*ba*b)*a*$)[ab]*$

RegEx Demo

(?=(?:b*ab*a)*b*$) is a lookahead that makes sure there are even number of a s in the input by matching 0 or more pairs of b*a sub-pattern.

Similar check is done for even no of b s in (?=(?:a*ba*b)*a*$)