Solution to problem 14
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src/Python/Problem014.py
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src/Python/Problem014.py
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#!/usr/bin/env python3
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"""
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Created on 7 Jan 2018
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@author: David Doblas Jiménez
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@email: daviddoji@pm.me
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Solution for problem 14 of Project Euler
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https://projecteuler.net/problem=14
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"""
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from utils import timeit
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def chain_length(n, terms):
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length = 0
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while n != 1:
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if n in terms:
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length += terms[n]
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break
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if n % 2 == 0:
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n = n / 2
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else:
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n = 3 * n + 1
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length += 1
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return length
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@timeit("Problem 14")
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def compute():
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"""
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The following iterative sequence is defined for the set of positive
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integers:
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n → n/2 (n is even)
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n → 3n + 1 (n is odd)
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Using the rule above and starting with 13, we generate the following
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sequence:
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13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
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It can be seen that this sequence (starting at 13 and finishing at 1)
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contains 10 terms. Although it has not been proved yet (Collatz Problem),
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it is thought that all starting numbers finish at 1.
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Which starting number, under one million, produces the longest chain?
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NOTE: Once the chain starts the terms are allowed to go above one million.
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"""
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ans = 0
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limit = 1_000_000
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score = 0
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terms = dict()
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for i in range(1, limit):
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terms[i] = chain_length(i, terms)
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if terms[i] > score:
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score = terms[i]
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ans = i
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return ans
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if __name__ == "__main__":
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print(f"Result for Problem 14: {compute()}")
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