"""
The following iterative sequence is defined for the set of positive integers:

n -> n/2 (is even)
n -> 3n + 1 (is odd)

Using the rule above and starting with 13, we generate the following sequence:
    13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1

It can be seen that this sequence (starting at 13 and finishing at 1 ) contains 10 terms.
Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?
"""

import time
from functools import cache


def timer(func):
    def wrapper(*args, **kwargs):
        start_time = time.time()
        result = func(*args, **kwargs)
        end_time = time.time()
        print(f"{func.__name__} took {end_time - start_time:.6f} seconds")
        return result

    return wrapper


def is_even(n: int) -> bool:
    return n & 1 == 0


def collatz_sequence(n: int) -> list[int]:
    sequence = [n]
    while sequence[-1] != 1:
        if is_even(n):
            n = n // 2
        else:
            n = 3 * n + 1
        sequence.append(n)
    return sequence


def longest_collatz_sequence(limit: int) -> int:
    longest_length = 0
    longest_start = 0
    for i in range(1, limit + 1):
        length = len(collatz_sequence(i))
        if length > longest_length:
            longest_length = length
            longest_start = i
    return longest_start


@cache
def chain_length(n):
    if n == 1:
        return 0
    next_term = 3 * n + 1 if n % 2 else n // 2
    return chain_length(next_term) + 1


@timer
def main_do() -> None:
    print(longest_collatz_sequence(1000000))


@timer
def main_cache() -> None:
    print(max(range(1, 1000001), key=chain_length))


if __name__ == "__main__":
    main_do()
    main_cache()