Python/project_euler/problem_074/sol1.py

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"""
Project Euler Problem 74: https://projecteuler.net/problem=74
The number 145 is well known for the property that the sum of the factorial of its
digits is equal to 145:
1! + 4! + 5! = 1 + 24 + 120 = 145
Perhaps less well known is 169, in that it produces the longest chain of numbers that
link back to 169; it turns out that there are only three such loops that exist:
169 363601 1454 169
871 45361 871
872 45362 872
It is not difficult to prove that EVERY starting number will eventually get stuck in
a loop. For example,
69 363600 1454 169 363601 ( 1454)
78 45360 871 45361 ( 871)
540 145 ( 145)
Starting with 69 produces a chain of five non-repeating terms, but the longest
non-repeating chain with a starting number below one million is sixty terms.
How many chains, with a starting number below one million, contain exactly sixty
non-repeating terms?
"""
DIGIT_FACTORIALS = {
"0": 1,
"1": 1,
"2": 2,
"3": 6,
"4": 24,
"5": 120,
"6": 720,
"7": 5040,
"8": 40320,
"9": 362880,
}
CACHE_SUM_DIGIT_FACTORIALS = {145: 145}
CHAIN_LENGTH_CACHE = {
145: 0,
169: 3,
36301: 3,
1454: 3,
871: 2,
45361: 2,
872: 2,
}
def sum_digit_factorials(n: int) -> int:
"""
Return the sum of the factorial of the digits of n.
>>> sum_digit_factorials(145)
145
>>> sum_digit_factorials(45361)
871
>>> sum_digit_factorials(540)
145
"""
if n in CACHE_SUM_DIGIT_FACTORIALS:
return CACHE_SUM_DIGIT_FACTORIALS[n]
ret = sum(DIGIT_FACTORIALS[let] for let in str(n))
CACHE_SUM_DIGIT_FACTORIALS[n] = ret
return ret
2022-11-15 13:55:14 +00:00
def chain_length(n: int, previous: set | None = None) -> int:
"""
Calculate the length of the chain of non-repeating terms starting with n.
Previous is a set containing the previous member of the chain.
>>> chain_length(10101)
11
>>> chain_length(555)
20
>>> chain_length(178924)
39
"""
previous = previous or set()
if n in CHAIN_LENGTH_CACHE:
return CHAIN_LENGTH_CACHE[n]
next_number = sum_digit_factorials(n)
if next_number in previous:
CHAIN_LENGTH_CACHE[n] = 0
return 0
else:
previous.add(n)
ret = 1 + chain_length(next_number, previous)
CHAIN_LENGTH_CACHE[n] = ret
return ret
def solution(num_terms: int = 60, max_start: int = 1000000) -> int:
"""
Return the number of chains with a starting number below one million which
contain exactly n non-repeating terms.
>>> solution(10,1000)
28
"""
return sum(1 for i in range(1, max_start) if chain_length(i) == num_terms)
if __name__ == "__main__":
print(f"{solution() = }")