project-euler/src/Python/Problem014.py
2021-07-24 10:17:26 +02:00

63 lines
1.5 KiB
Python

#!/usr/bin/env python3
"""
Created on 7 Jan 2018
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for problem 14 of Project Euler
https://projecteuler.net/problem=14
"""
from utils import timeit
def chain_length(n, terms):
length = 0
while n != 1:
if n in terms:
length += terms[n]
break
if n % 2 == 0:
n = n / 2
else:
n = 3 * n + 1
length += 1
return length
@timeit("Problem 14")
def compute():
"""
The following iterative sequence is defined for the set of positive
integers:
n → n/2 (n is even)
n → 3n + 1 (n 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?
NOTE: Once the chain starts the terms are allowed to go above one million.
"""
ans = 0
limit = 1_000_000
score = 0
terms = dict()
for i in range(1, limit):
terms[i] = chain_length(i, terms)
if terms[i] > score:
score = terms[i]
ans = i
return ans
if __name__ == "__main__":
print(f"Result for Problem 14: {compute()}")