Limit number of steps when matching a Regex in python

Question

Let's say you have a program that search in a user-provided text for a user-provided RegEx.

Using the re module, how can I limit the number of steps the regex will make to prevent catastrophic backtracking.

Ideally, this SME should not hang:

regex = r"^(.*?,){11}P"

test_str = "1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21P" # Example stolen from the previous link.

matches = list(re.finditer(regex, test_str, re.MULTILINE)) # This is an iterator so we consume it by casting to a list.

Duplicate of https://stackoverflow.com/questions/11901328/how-to-timeout-function-in-python-timeout-less-than-a-second? Or https://stackoverflow.com/questions/492519/timeout-on-a-function-call ? — Wiktor Stribiżew, Jul 02 '19 at 12:40
I know how to do it -- Thanks, but I would have very liked a max number of steps than a max number of seconds. — WayToDoor, Jul 02 '19 at 12:41
But yeah, I guess I'll use that to limit the execution time. — WayToDoor, Jul 02 '19 at 12:42
It is impossible to count those steps in `re`. There is no such functionality. — Wiktor Stribiżew, Jul 02 '19 at 12:43
That's all I wanted to know, thanks. If you post that as an answer, I'll gladly accept it. — WayToDoor, Jul 02 '19 at 12:43

Limit number of steps when matching a Regex in python

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