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