Regular expression for strings with even number of a's and odd no of b's

Question

Im having a problem in solving the problem:- Its an assignment, i solved it, but it seems to be too long and vague, Can anyboby help me please......

Regular expression for the strings with even number of a's and odd number of b's where the character set={a,b}.

Answer 1

One way to do this is to pass it through two regular expressions making sure they both match (assuming you want to use regular expressions at all, see below for an alternative):

^b*(ab*ab*)*$
^a*ba*(ba*ba*)*$

Anything else (and, in fact, even that) is most likely just an attempt to be clever, one that's generally a massive failure.

The first regular expression ensures there are an even number of a with b anywhere in the mix (before, after and in between).

The second is similar but ensures that there's an odd number of b by virtue of the starting a*ba*.

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A far better way to do it is to ignore regular expressions altogether and simply run through the string as follows:

def isValid(s):
    set evenA to true
    set oddB to false
    for c as each character in s:
        if c is 'a':
            set evenA to not evenA
        else if c is 'b':
            set oddB to  not oddB
        else:
            return false
    return evenA and oddB

Though regular expressions are a wonderful tool, they're not suited for everything and they become far less useful as their readability and maintainability degrades.

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For what it's worth, a single-regex answer is:

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

but, if I caught anyone on my team actually using a monstrosity like that, they'd be sent back to do it again.

This comes from a paper by one Greg Bacon. See here for the actual inner workings.

Answer 2

Even-Even = (aa+bb+(ab+ba)(aa+bb)*(ab+ba))*

(Even-Even has even number of Aas and b's both)

Even a's and odd b's = Even-Even b Even-Even

This hsould work