mirror of
https://github.com/TheAlgorithms/Python.git
synced 2024-12-18 09:10:16 +00:00
f7ac8b5ed0
* Added doctest and more explanation about Dijkstra execution. * tests were not passing with python2 due to missing __init__.py file at number_theory folder * Removed the dot at the beginning of the imported modules names because 'python3 -m doctest -v data_structures/hashing/*.py' and 'python3 -m doctest -v data_structures/stacks/*.py' were failing not finding hash_table.py and stack.py modules. * Moved global code to main scope and added doctest for project euler problems 1 to 14. * Added test case for negative input. * Changed N variable to do not use end of line scape because in case there is a space after it the script will break making it much more error prone. * Added problems description and doctests to the ones that were missing. Limited line length to 79 and executed python black over all scripts. * Changed the way files are loaded to support pytest call. * Added __init__.py to problems to make them modules and allow pytest execution. * Added project_euler folder to test units execution * Changed 'os.path.split(os.path.realpath(__file__))' to 'os.path.dirname()' * Added Burrows-Wheeler transform algorithm. * Added changes suggested by cclauss * Fixes for issue 'Fix the LGTM issues #1024'. * Added doctest for different parameter types and negative values. * Fixed doctest issue added at last commit. * Commented doctest that were causing slowness at Travis. * Added comment with the reason for some doctest commented. * pytest --ignore
71 lines
2.1 KiB
Python
71 lines
2.1 KiB
Python
# -*- coding: utf-8 -*-
|
|
"""
|
|
Collatz conjecture: start with any positive integer n. Next term obtained from
|
|
the previous term as follows:
|
|
|
|
If the previous term is even, the next term is one half the previous term.
|
|
If the previous term is odd, the next term is 3 times the previous term plus 1.
|
|
The conjecture states the sequence will always reach 1 regardless of starting
|
|
n.
|
|
|
|
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?
|
|
"""
|
|
from __future__ import print_function
|
|
|
|
try:
|
|
raw_input # Python 2
|
|
except NameError:
|
|
raw_input = input # Python 3
|
|
|
|
|
|
def collatz_sequence(n):
|
|
"""Returns the Collatz sequence for n."""
|
|
sequence = [n]
|
|
while n != 1:
|
|
if n % 2 == 0:
|
|
n //= 2
|
|
else:
|
|
n = 3 * n + 1
|
|
sequence.append(n)
|
|
return sequence
|
|
|
|
|
|
def solution(n):
|
|
"""Returns the number under n that generates the longest Collatz sequence.
|
|
|
|
# The code below has been commented due to slow execution affecting Travis.
|
|
# >>> solution(1000000)
|
|
# {'counter': 525, 'largest_number': 837799}
|
|
>>> solution(200)
|
|
{'counter': 125, 'largest_number': 171}
|
|
>>> solution(5000)
|
|
{'counter': 238, 'largest_number': 3711}
|
|
>>> solution(15000)
|
|
{'counter': 276, 'largest_number': 13255}
|
|
"""
|
|
|
|
result = max([(len(collatz_sequence(i)), i) for i in range(1, n)])
|
|
return {"counter": result[0], "largest_number": result[1]}
|
|
|
|
|
|
if __name__ == "__main__":
|
|
result = solution(int(raw_input().strip()))
|
|
print(
|
|
"Longest Collatz sequence under one million is %d with length %d"
|
|
% (result["largest_number"], result["counter"])
|
|
)
|