Match strings with alternating characters

This is actually impossible to do in a real regular expression with an amount of states polynomial to the alphabet size, because the expression is not a Chomsky level-0 grammar.

However, Python's regexes are not actually regular expressions, and can handle much more complex grammars than that. In particular, you could put your grammar as the following.

(..)\1*

(..) is a sequence of 2 characters. \1* matches the exact pair of characters an arbitrary (possibly null) number of times.

I interpreted your question as wanting every other character to be equal (ababab works, but abcbdb fails). If you needed only the 2nd, 4th, ... characters to be equal you can use a similar one.

.(.)(.\1)*

You could match the first [a-z] followed by capturing ([a-z]) in a group. Then repeat 0+ times matching again a-z and a backreference to group 1 to keep every second character the same.

^[a-z]([a-z])(?:[a-z]\1)*$

Explanation

• ^ Start of the string
• [a-z]([a-z]) Match a-z and capture in group 1 matching a-z
• )(?:[a-z]\1)* Repeat 0+ times matching a-z followed by a backreference to group 1
• $ End of string

Regex demo

Though not a regex answer, you could do something like this:

def all_same(string):
    return all(c == string[1] for c in string[1::2])

string = 'abababababab'
print('All the same {}'.format(all_same(string)))

string = 'ababacababab'
print('All the same {}'.format(all_same(string)))

the string[1::2] says start at the 2nd character (1) and then pull out every second character (the 2 part).

This returns:

All the same True
All the same False

This is a bit complicated expression, maybe we would start with:

^(?=^[a-z]([a-z]))([a-z]\1)+$

if I understand the problem right.

Demo