(n**e**-1)**e==n clear solution for Checking Perfect Power by veky - python coding challenges

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(n**e**-1)**e==n solution in Clear category for Checking Perfect Power by veky

perfect_power = lambda n: any(round(n**e**-1)**e==n for e in range(n.bit_length(), 1, -1))

Aug. 22, 2023

Comments:

tamagoyaki on Oct. 26, 2023, 3:22 a.m.
An amazing idea is to use "bit_length"

veky on Oct. 26, 2023, 3:41 a.m.
It's just an integer dyadic logarithm.

StefanPochmann on March 26, 2024, 11:36 p.m.
I also like <pre class='brush: python'>n in (round(n**e**-1)**e for ...)</pre>, but that of course has the huge disadvantage of not allowing a title like yours.

veky on March 27, 2024, 1:56 a.m.
Yeah, substrings which aren't subexpressions are powerfully poetic sometimes. :-)

veky on March 27, 2024, 2:04 a.m.
<a href="https://py.checkio.org/mission/checking-perfect-power/publications/veky/python-3/cloud-number-n-in-e/?ordering=most_voted&filtering=all">This title</a> kinda works...

Mine

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