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* pre-commit: Upgrade psf/black for stable style 2023 Updating https://github.com/psf/black ... updating 22.12.0 -> 23.1.0 for their `2023 stable style`. * https://github.com/psf/black/blob/main/CHANGES.md#2310 > This is the first [psf/black] release of 2023, and following our stability policy, it comes with a number of improvements to our stable style… Also, add https://github.com/tox-dev/pyproject-fmt and https://github.com/abravalheri/validate-pyproject to pre-commit. I only modified `.pre-commit-config.yaml` and all other files were modified by pre-commit.ci and psf/black. * [pre-commit.ci] auto fixes from pre-commit.com hooks for more information, see https://pre-commit.ci --------- Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
147 lines
4.6 KiB
Python
147 lines
4.6 KiB
Python
"""
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Project Euler Problem 074: 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 a
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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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Solution approach:
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This solution simply consists in a loop that generates the chains of non repeating
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items using the cached sizes of the previous chains.
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The generation of the chain stops before a repeating item or if the size of the chain
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is greater then the desired one.
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After generating each chain, the length is checked and the counter increases.
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"""
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from math import factorial
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DIGIT_FACTORIAL: dict[str, int] = {str(digit): factorial(digit) for digit in range(10)}
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def digit_factorial_sum(number: int) -> int:
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"""
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Function to perform the sum of the factorial of all the digits in number
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>>> digit_factorial_sum(69.0)
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Traceback (most recent call last):
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...
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TypeError: Parameter number must be int
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>>> digit_factorial_sum(-1)
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Traceback (most recent call last):
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...
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ValueError: Parameter number must be greater than or equal to 0
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>>> digit_factorial_sum(0)
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1
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>>> digit_factorial_sum(69)
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363600
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"""
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if not isinstance(number, int):
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raise TypeError("Parameter number must be int")
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if number < 0:
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raise ValueError("Parameter number must be greater than or equal to 0")
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# Converts number in string to iterate on its digits and adds its factorial.
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return sum(DIGIT_FACTORIAL[digit] for digit in str(number))
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def solution(chain_length: int = 60, number_limit: int = 1000000) -> int:
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"""
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Returns the number of numbers below number_limit that produce chains with exactly
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chain_length non repeating elements.
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>>> solution(10.0, 1000)
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Traceback (most recent call last):
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...
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TypeError: Parameters chain_length and number_limit must be int
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>>> solution(10, 1000.0)
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Traceback (most recent call last):
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...
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TypeError: Parameters chain_length and number_limit must be int
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>>> solution(0, 1000)
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Traceback (most recent call last):
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...
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ValueError: Parameters chain_length and number_limit must be greater than 0
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>>> solution(10, 0)
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Traceback (most recent call last):
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...
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ValueError: Parameters chain_length and number_limit must be greater than 0
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>>> solution(10, 1000)
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26
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"""
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if not isinstance(chain_length, int) or not isinstance(number_limit, int):
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raise TypeError("Parameters chain_length and number_limit must be int")
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if chain_length <= 0 or number_limit <= 0:
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raise ValueError(
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"Parameters chain_length and number_limit must be greater than 0"
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)
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# the counter for the chains with the exact desired length
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chains_counter = 0
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# the cached sizes of the previous chains
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chain_sets_lengths: dict[int, int] = {}
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for start_chain_element in range(1, number_limit):
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# The temporary set will contain the elements of the chain
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chain_set = set()
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chain_set_length = 0
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# Stop computing the chain when you find a cached size, a repeating item or the
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# length is greater then the desired one.
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chain_element = start_chain_element
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while (
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chain_element not in chain_sets_lengths
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and chain_element not in chain_set
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and chain_set_length <= chain_length
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):
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chain_set.add(chain_element)
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chain_set_length += 1
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chain_element = digit_factorial_sum(chain_element)
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if chain_element in chain_sets_lengths:
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chain_set_length += chain_sets_lengths[chain_element]
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chain_sets_lengths[start_chain_element] = chain_set_length
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# If chain contains the exact amount of elements increase the counter
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if chain_set_length == chain_length:
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chains_counter += 1
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return chains_counter
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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print(f"{solution()}")
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