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62 lines
1.7 KiB
Python
62 lines
1.7 KiB
Python
"""
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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 solution(n):
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"""Returns the number under n that generates the longest sequence using the
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formula:
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n → n/2 (n is even)
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n → 3n + 1 (n is odd)
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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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largest_number = 0
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pre_counter = 0
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for input1 in range(n):
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counter = 1
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number = input1
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while number > 1:
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if number % 2 == 0:
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number /= 2
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counter += 1
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else:
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number = (3 * number) + 1
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counter += 1
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if counter > pre_counter:
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largest_number = input1
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pre_counter = counter
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return {"counter": pre_counter, "largest_number": largest_number}
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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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("Largest Number:", result["largest_number"], "->", result["counter"], "digits")
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)
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