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* Fixes: #3869: Optimize Project Euler's Problem 74's Solution 2 * updating DIRECTORY.md Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
127 lines
3.5 KiB
Python
127 lines
3.5 KiB
Python
"""
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Project Euler Problem 074: https://projecteuler.net/problem=74
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Starting from any positive integer number
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it is possible to attain another one summing the factorial of its digits.
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Repeating this step, we can build chains of numbers.
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It is not difficult to prove that EVERY starting number
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will eventually get stuck in a loop.
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The request is to find how many numbers less than one million
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produce a chain with exactly 60 non repeating items.
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Solution approach:
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This solution simply consists in a loop that generates
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the chains of non repeating items.
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The generation of the chain stops before a repeating item
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or if the size of the chain is greater then the desired one.
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After generating each chain, the length is checked and the
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counter increases.
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"""
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factorial_cache = {}
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factorial_sum_cache = {}
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def factorial(a: int) -> int:
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"""Returns the factorial of the input a
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>>> factorial(5)
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120
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>>> factorial(6)
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720
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>>> factorial(0)
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1
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"""
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# The factorial function is not defined for negative numbers
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if a < 0:
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raise ValueError("Invalid negative input!", a)
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if a in factorial_cache:
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return factorial_cache[a]
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# The case of 0! is handled separately
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if a == 0:
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factorial_cache[a] = 1
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else:
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# use a temporary support variable to store the computation
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temporary_number = a
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temporary_computation = 1
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while temporary_number > 0:
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temporary_computation *= temporary_number
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temporary_number -= 1
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factorial_cache[a] = temporary_computation
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return factorial_cache[a]
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def factorial_sum(a: int) -> int:
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"""Function to perform the sum of the factorial
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of all the digits in a
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>>> factorial_sum(69)
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363600
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"""
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if a in factorial_sum_cache:
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return factorial_sum_cache[a]
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# Prepare a variable to hold the computation
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fact_sum = 0
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""" Convert a in string to iterate on its digits
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convert the digit back into an int
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and add its factorial to fact_sum.
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"""
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for i in str(a):
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fact_sum += factorial(int(i))
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factorial_sum_cache[a] = fact_sum
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return fact_sum
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def solution(chain_length: int = 60, number_limit: int = 1000000) -> int:
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"""Returns the number of numbers that produce
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chains with exactly 60 non repeating elements.
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>>> solution(10, 1000)
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26
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"""
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# the counter for the chains with the exact desired length
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chain_counter = 0
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for i in range(1, number_limit + 1):
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# The temporary list will contain the elements of the chain
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chain_set = {i}
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len_chain_set = 1
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last_chain_element = i
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# The new element of the chain
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new_chain_element = factorial_sum(last_chain_element)
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# Stop computing the chain when you find a repeating item
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# or the length it greater then the desired one.
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while new_chain_element not in chain_set and len_chain_set <= chain_length:
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chain_set.add(new_chain_element)
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len_chain_set += 1
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last_chain_element = new_chain_element
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new_chain_element = factorial_sum(last_chain_element)
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# If the while exited because the chain list contains the exact amount
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# of elements increase the counter
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if len_chain_set == chain_length:
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chain_counter += 1
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return chain_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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