# 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.
from common import max_by
from functools import lru_cache
@lru_cache(maxsize=None)
def chain_size(n):
return 1 if n == 1 else 1 + chain_size(n // 2 if n % 2 == 0 else 3 * n + 1)
print(max_by(chain_size, range(1, 1000000)))
from common import max_by
from decimal import *
getcontext().prec = 2000
def is_cycle(target, string, start):
length = len(target)
limit = len(string)
for i in range(start, limit, length):
if i + length > limit: return True
t = string[i:i + length]
if t != target: return False
return True
def find_cycle(string):
for start in range(len(string)):
for i in range(start + 1, (len(string) - start) // 2 + 1):
target = string[start:i]
if is_cycle(target, string, start): return target
return None
def cycle_len(n):
cycle = find_cycle(str(Decimal(1) / Decimal(n))[2:])
r = len(cycle) if cycle else 0
return r
print(max_by(cycle_len, range(2, 1000)))
