mirror of
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84facb78b2
* Added explanation and increased speed of the solution of problem 092
* updating DIRECTORY.md
* Added temporary fix to the failing of problem 104
* Reduced few seconds by minor improvements
* Update sol.py
Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
Co-authored-by: Christian Clauss <cclauss@me.com>
108 lines
3.0 KiB
Python
108 lines
3.0 KiB
Python
"""
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Project Euler Problem 092: https://projecteuler.net/problem=92
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Square digit chains
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A number chain is created by continuously adding the square of the digits in
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a number to form a new number until it has been seen before.
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For example,
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44 → 32 → 13 → 10 → 1 → 1
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85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89
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Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop.
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What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.
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How many starting numbers below ten million will arrive at 89?
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"""
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DIGITS_SQUARED = [sum(int(c, 10) ** 2 for c in i.__str__()) for i in range(100000)]
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def next_number(number: int) -> int:
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"""
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Returns the next number of the chain by adding the square of each digit
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to form a new number.
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For example, if number = 12, next_number() will return 1^2 + 2^2 = 5.
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Therefore, 5 is the next number of the chain.
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>>> next_number(44)
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32
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>>> next_number(10)
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1
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>>> next_number(32)
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13
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"""
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sum_of_digits_squared = 0
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while number:
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# Increased Speed Slightly by checking every 5 digits together.
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sum_of_digits_squared += DIGITS_SQUARED[number % 100000]
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number //= 100000
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return sum_of_digits_squared
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# There are 2 Chains made,
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# One ends with 89 with the chain member 58 being the one which when declared first,
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# there will be the least number of iterations for all the members to be checked.
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# The other one ends with 1 and has only one element 1.
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# So 58 and 1 are chosen to be declared at the starting.
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# Changed dictionary to an array to quicken the solution
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CHAINS: list[bool | None] = [None] * 10000000
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CHAINS[0] = True
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CHAINS[57] = False
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def chain(number: int) -> bool:
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"""
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The function generates the chain of numbers until the next number is 1 or 89.
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For example, if starting number is 44, then the function generates the
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following chain of numbers:
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44 → 32 → 13 → 10 → 1 → 1.
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Once the next number generated is 1 or 89, the function returns whether
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or not the next number generated by next_number() is 1.
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>>> chain(10)
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True
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>>> chain(58)
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False
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>>> chain(1)
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True
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"""
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if CHAINS[number - 1] is not None:
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return CHAINS[number - 1] # type: ignore
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number_chain = chain(next_number(number))
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CHAINS[number - 1] = number_chain
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while number < 10000000:
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CHAINS[number - 1] = number_chain
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number *= 10
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return number_chain
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def solution(number: int = 10000000) -> int:
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"""
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The function returns the number of integers that end up being 89 in each chain.
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The function accepts a range number and the function checks all the values
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under value number.
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>>> solution(100)
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80
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>>> solution(10000000)
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8581146
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"""
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for i in range(1, number):
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if CHAINS[i] is None:
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chain(i + 1)
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return CHAINS[:number].count(False)
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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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