Python/project_euler/problem_188/sol1.py
Joyce bcfca67faa
[mypy] fix type annotations for all Project Euler problems (#4747)
* [mypy] fix type annotations for problem003/sol1 and problem003/sol3

* [mypy] fix type annotations for project euler problem007/sol2

* [mypy] fix type annotations for project euler problem008/sol2

* [mypy] fix type annotations for project euler problem009/sol1

* [mypy] fix type annotations for project euler problem014/sol1

* [mypy] fix type annotations for project euler problem 025/sol2

* [mypy] fix type annotations for project euler problem026/sol1.py

* [mypy] fix type annotations for project euler problem037/sol1

* [mypy] fix type annotations for project euler problem044/sol1

* [mypy] fix type annotations for project euler problem046/sol1

* [mypy] fix type annotations for project euler problem051/sol1

* [mypy] fix type annotations for project euler problem074/sol2

* [mypy] fix type annotations for project euler problem080/sol1

* [mypy] fix type annotations for project euler problem099/sol1

* [mypy] fix type annotations for project euler problem101/sol1

* [mypy] fix type annotations for project euler problem188/sol1

* [mypy] fix type annotations for project euler problem191/sol1

* [mypy] fix type annotations for project euler problem207/sol1

* [mypy] fix type annotations for project euler problem551/sol1
2021-10-12 00:33:44 +08:00

69 lines
1.8 KiB
Python

"""
Project Euler Problem 188: https://projecteuler.net/problem=188
The hyperexponentiation of a number
The hyperexponentiation or tetration of a number a by a positive integer b,
denoted by a↑↑b or b^a, is recursively defined by:
a↑↑1 = a,
a↑↑(k+1) = a(a↑↑k).
Thus we have e.g. 3↑↑2 = 3^3 = 27, hence 3↑↑3 = 3^27 = 7625597484987 and
3↑↑4 is roughly 103.6383346400240996*10^12.
Find the last 8 digits of 1777↑↑1855.
References:
- https://en.wikipedia.org/wiki/Tetration
"""
# small helper function for modular exponentiation (fast exponentiation algorithm)
def _modexpt(base: int, exponent: int, modulo_value: int) -> int:
"""
Returns the modular exponentiation, that is the value
of `base ** exponent % modulo_value`, without calculating
the actual number.
>>> _modexpt(2, 4, 10)
6
>>> _modexpt(2, 1024, 100)
16
>>> _modexpt(13, 65535, 7)
6
"""
if exponent == 1:
return base
if exponent % 2 == 0:
x = _modexpt(base, exponent // 2, modulo_value) % modulo_value
return (x * x) % modulo_value
else:
return (base * _modexpt(base, exponent - 1, modulo_value)) % modulo_value
def solution(base: int = 1777, height: int = 1855, digits: int = 8) -> int:
"""
Returns the last 8 digits of the hyperexponentiation of base by
height, i.e. the number base↑↑height:
>>> solution(base=3, height=2)
27
>>> solution(base=3, height=3)
97484987
>>> solution(base=123, height=456, digits=4)
2547
"""
# calculate base↑↑height by right-assiciative repeated modular
# exponentiation
result = base
for i in range(1, height):
result = _modexpt(base, result, 10 ** digits)
return result
if __name__ == "__main__":
print(f"{solution() = }")