Question

The problem

I’m trying to create a regex in which we can check if all letters present in some reference set are present in some other string, but only in odd numbers (1, 3, 5, …).

Here is a (very) crude image of the DFA representing the problem:

My (broken) solution

I started using a finite set, {a, b}, so I would essentially check “are there both an odd number of as and an odd number of bs in the string?”

Unfortunately I did not get far on my own. I first read this thread, which is remarkably similar to this concept, but was not able to glean an answer from (aa|bb|(ab|ba)(aa|bb)*(ba|ab))*(b|(ab|ba)(bb|aa)*a). (I understand how it works, but not how to convert it to check odd numbers for both items present.)

Here is what I’ve come up with so far: ^((ab|ba)(bb|aa)?|(bb|aa)?(ab|ba))+$. This basically checks if there is ab or ba followed by bb or aa or nothing, which would result in ab, ba, abaa, abbb, baaa, or babb. (It also does the reverse of this, checking the double-letter first.) This can then repeat, indefinitely. The problem I have is I cannot seem to adjust it to match the string bbaaba without also matching bbaa.

Additionally, the method above can not be dynamically adjusted to account for {a, b, c}, for example, though I’m willing to forgo this to solve the initial problem.

Testing

Here are my test strings and the desired output, with the reasons in parentheses:

"ba"      # True (1a, 1b)
"abbb"    # True (1a, 3b)
"bbba"    # True (1a, 3b)
"bbab"    # True (1a, 3b)
"ababab"  # True (3a, 3b)
"bbaaba"  # True (3a, 3b)
"abb"     # False (2b)
"aabb"    # False (2a, 2b)
"aabba"   # False (2b)
""        # False (0a, 0b is "even")
"a"       # False (0b is "even")
"b"       # False (0a is "even")

Question

So, is this possible through regex? Or are regular expressions more limited than a DFA? I am aware that it can be done through a basic loop, but this isn’t what I’m going for.

Answer 1

Here’s one way to do it, using lookaheads to assert each condition in turn.

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

Here’s a demo with your examples. (The \ns in the demo are for presentation purposes. Also, you can drop the (.*)$ if you only need to test the match, not capture.)

I will be adding an explanation shortly.

---

Explanation

We only need to look at one half:

(?=  [^a]*a  (?:[^a]*a[^a]*a)  *  [^a]*$  )
|    |       |                 |  |
|    |       |                 |  Only accept non-'a's to the end.
|    |       |                 |
|    |       |                 Zero or more of these pairs of 'a's.
|    |       |
|    |       Strictly a pair of 'a's.
|    |
|    Find the first 'a'.
|
Use a lookahead to assert multiple conditions.