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63 lines
2.0 KiB
Python
63 lines
2.0 KiB
Python
# -*- coding: utf-8 -*-
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"""
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Collatz conjecture: start with any positive integer n. Next term obtained from
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the previous term as follows:
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If the previous term is even, the next term is one half the previous term.
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If the previous term is odd, the next term is 3 times the previous term plus 1.
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The conjecture states the sequence will always reach 1 regardless of starting
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n.
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Problem Statement:
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The following iterative sequence is defined for the set of positive 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 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) contains
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10 terms. Although it has not been proved yet (Collatz Problem), it is thought
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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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"""
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def collatz_sequence(n):
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"""Returns the Collatz sequence for n."""
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sequence = [n]
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while n != 1:
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if n % 2 == 0:
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n //= 2
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else:
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n = 3 * n + 1
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sequence.append(n)
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return sequence
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def solution(n):
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"""Returns the number under n that generates the longest Collatz sequence.
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# The code below has been commented due to slow execution affecting Travis.
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# >>> solution(1000000)
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# {'counter': 525, 'largest_number': 837799}
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>>> solution(200)
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{'counter': 125, 'largest_number': 171}
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>>> solution(5000)
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{'counter': 238, 'largest_number': 3711}
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>>> solution(15000)
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{'counter': 276, 'largest_number': 13255}
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"""
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result = max([(len(collatz_sequence(i)), i) for i in range(1, n)])
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return {"counter": result[0], "largest_number": result[1]}
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if __name__ == "__main__":
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result = solution(int(input().strip()))
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print(
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"Longest Collatz sequence under one million is %d with length %d"
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% (result["largest_number"], result["counter"])
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)
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