Python/project_euler/problem_092/sol1.py
Maxim Smolskiy 53b2926704
Enable ruff PGH003 rule ()
* Enable ruff PGH003 rule

* Fix

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Fix

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
2024-04-02 21:29:34 +02:00

106 lines
3.0 KiB
Python

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