Things like (?!..) are called assertions. I like to think of assertions as patterns that match the space between two characters. So (?!a) means that it matches the space, where the next character is not 'a'.
So in this case, your regex matches an empty string (since it doesn't match any character at all between ^ and $, only spaces.)
> ^((?!(.)(.)\3\2).)∗$
> 1. Why are my backrefs still \3 and \2?
Well, you're referencing the second and third opening parenthesis.
> but it counted in my second example above
Maybe that's because your second example is wrong? I mean, I don't really know what you are trying to achieve in your second example.
By the way, my solution for abba is
^(?!.*(.)(.)\2\1)
I guess you'll will be able to figure out what this regex means by now :)
So in this case, your regex matches an empty string (since it doesn't match any character at all between ^ and $, only spaces.)
Well, you're referencing the second and third opening parenthesis. Maybe that's because your second example is wrong? I mean, I don't really know what you are trying to achieve in your second example.By the way, my solution for abba is
I guess you'll will be able to figure out what this regex means by now :)