这个算法中的Python尾递归优化？

Question

I got the following problem: Let's consider the following encoding table:

0 -> A
1 -> B
2 -> C
...

• input: string containing a list of integers
• output: int representing the number of words that can be encoded from the input
• examples:

"0" -> 1 (word is A)
"10" -> 2 (words are BA and K)
"121" -> 3 (words are BCB, MB and BV) 

I wrote this algo

import string
sys.setrecursionlimit(100)  # limit the number of recursions to 100
# {0: 'a', 1: 'b', etc.}
dict_values = dict(zip(range(26), string.ascii_lowercase))

def count_split(list_numbers: list) -> int:
    if len(list_numbers) == 0:
        return 0

    elif len(list_numbers) == 1:
        return 1

    elif len(list_numbers) == 2:
        if int(list_numbers[0]) == 0:
            return 1

        elif 10 <= int("".join(list_numbers)) <= 25:
            return 2

        else:
            return 1

    else:  # list_numbers contains at least three elements
        if count_split(list_numbers[:2]) == 2:
            return count_split(list_numbers[1:]) + count_split(list_numbers[2:])

        else:
            return count_split(list_numbers[1:])

def get_nb_words(code: str) -> int:
    return count_split(list(code))

# print(get_nb_words("0124")) -> 3
# print(get_nb_words("124")) -> 3
# print(get_nb_words("368")) -> 1
# print(get_nb_words("322")) -> 2
# print(get_nb_words("12121212121212121212")) -> 10946

Surprisingly, this algo works on the last example "12121212121212121212" . I expected the number of recursions to be exceeded because at each step, the function count_split is called twice on a list. Thus, the number of calls is much more than 100 (even more than 1000) !

In the meanwhile I found this post on stackoverflow saying that tail recursion is not optimized in Python, so I'm a bit suprised !

Could someone explain to me why the recursion limit is not exceed in this algorithm ?

Answer 1

You care about the recursion depth, ie the maximum depth (height?) of the call stack.

Here's an empirical approach: (code that measures the depth)

import string, sys
sys.setrecursionlimit(100)  # limit the number of recursions to 100

dict_values = dict(zip(range(26), string.ascii_lowercase))
stack = []
max_depth = 0
def count_split(list_numbers: list) -> int:
    global max_depth
    stack.append(None)
    max_depth = max(max_depth, len(stack))
    if len(list_numbers) == 0:
        return 0

    elif len(list_numbers) == 1:
        return 1

    elif len(list_numbers) == 2:
        if int(list_numbers[0]) == 0:
            return 1

        elif 10 <= int("".join(list_numbers)) <= 25:
            return 2

        else:
            return 1

    else:  # list_numbers contains at least three elements
        if count_split(list_numbers[:2]) == 2:
            stack.pop()
            result = count_split(list_numbers[1:]) + count_split(list_numbers[2:])
            stack.pop(); stack.pop()
            return result

        else:
            result = count_split(list_numbers[1:])
            stack.pop()
            return result

def get_nb_words(code: str) -> int:
    return count_split(list(code))

print(get_nb_words("12121212121212121212"))
print(max_depth) # 20