Problem statement:
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. Here is my implementation written in Python, awaiting your feedback.
from time import time
from operator import itemgetter
def collatz_count(n, count={1: 1}):
"""uses cache (count) to speed up the search, returns sequence
length for the given number"""
try:
return count[n]
except KeyError:
if n % 2 == 0:
count[n] = collatz_count(n / 2) + 1
else:
count[n] = collatz_count(n * 3 + 1) + 1
return count[n]
def run_test():
"""uses the previous function, returns number that generates the
largest sequence"""
time1 = time()
items = sorted([(x, collatz_count(x)) for x in range(1, 1000000)],
key=itemgetter(1), reverse=True)
maximum = items[0]
print(f' Starting number: {maximum[0]} \n Sequence length:
{maximum[1]} \n '
f'Calculated in: {time() - time1} seconds.')
if __name__ == '__main__':
run_test()