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* Pyupgrade to Python 3.9
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111 lines
2.7 KiB
Python
111 lines
2.7 KiB
Python
"""
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Project Euler Problem 74: https://projecteuler.net/problem=74
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The number 145 is well known for the property that the sum of the factorial of its
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digits is equal to 145:
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1! + 4! + 5! = 1 + 24 + 120 = 145
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Perhaps less well known is 169, in that it produces the longest chain of numbers that
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link back to 169; it turns out that there are only three such loops that exist:
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169 → 363601 → 1454 → 169
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871 → 45361 → 871
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872 → 45362 → 872
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It is not difficult to prove that EVERY starting number will eventually get stuck in
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a loop. For example,
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69 → 363600 → 1454 → 169 → 363601 (→ 1454)
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78 → 45360 → 871 → 45361 (→ 871)
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540 → 145 (→ 145)
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Starting with 69 produces a chain of five non-repeating terms, but the longest
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non-repeating chain with a starting number below one million is sixty terms.
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How many chains, with a starting number below one million, contain exactly sixty
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non-repeating terms?
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"""
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DIGIT_FACTORIALS = {
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"0": 1,
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"1": 1,
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"2": 2,
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"3": 6,
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"4": 24,
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"5": 120,
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"6": 720,
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"7": 5040,
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"8": 40320,
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"9": 362880,
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}
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CACHE_SUM_DIGIT_FACTORIALS = {145: 145}
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CHAIN_LENGTH_CACHE = {
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145: 0,
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169: 3,
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36301: 3,
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1454: 3,
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871: 2,
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45361: 2,
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872: 2,
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}
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def sum_digit_factorials(n: int) -> int:
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"""
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Return the sum of the factorial of the digits of n.
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>>> sum_digit_factorials(145)
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145
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>>> sum_digit_factorials(45361)
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871
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>>> sum_digit_factorials(540)
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145
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"""
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if n in CACHE_SUM_DIGIT_FACTORIALS:
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return CACHE_SUM_DIGIT_FACTORIALS[n]
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ret = sum(DIGIT_FACTORIALS[let] for let in str(n))
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CACHE_SUM_DIGIT_FACTORIALS[n] = ret
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return ret
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def chain_length(n: int, previous: set = None) -> int:
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"""
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Calculate the length of the chain of non-repeating terms starting with n.
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Previous is a set containing the previous member of the chain.
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>>> chain_length(10101)
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11
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>>> chain_length(555)
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20
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>>> chain_length(178924)
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39
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"""
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previous = previous or set()
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if n in CHAIN_LENGTH_CACHE:
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return CHAIN_LENGTH_CACHE[n]
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next_number = sum_digit_factorials(n)
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if next_number in previous:
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CHAIN_LENGTH_CACHE[n] = 0
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return 0
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else:
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previous.add(n)
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ret = 1 + chain_length(next_number, previous)
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CHAIN_LENGTH_CACHE[n] = ret
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return ret
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def solution(num_terms: int = 60, max_start: int = 1000000) -> int:
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"""
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Return the number of chains with a starting number below one million which
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contain exactly n non-repeating terms.
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>>> solution(10,1000)
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28
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"""
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return sum(1 for i in range(1, max_start) if chain_length(i) == num_terms)
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if __name__ == "__main__":
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print(f"{solution() = }")
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