Regular expressions - different string the same match

Question

I understand that the pattern r'([a-z]+)\1+' is searching for a repeated multi character pattern in the search string but I do not understand why in case k2 answer isn't 'aaaaa' (5 'a'):

import re
k1 = re.search(r'([a-z]+)\1+', 'aaaa')
k2 = re.search(r'([a-z]+)\1+', 'aaaaa')
k3 = re.search(r'([a-z]+)\1+', 'aaaaaa')
print(k1)  # <_sre.SRE_Match object; span=(0, 4), match='aaaa'>
print(k2)  # <_sre.SRE_Match object; span=(0, 4), match='aaaa'>
print(k3)  # <_sre.SRE_Match object; span=(0, 6), match='aaaaaa'>

Python 3.6.1

Answer 1

Because + is greedy.

What happens is ([a-z]+) first matches 'aaaaa', then it backtracks until \1+ matches the string, and stops. Because 'aa' is the first value of the ([a-z]+) that will let \1 successfully match, that's what it returns.