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Brute Force 2 solution in Clear category for e+pi by StefanPochmann
from math import e, pi
from itertools import permutations, product
def checkio(n: float) -> str:
def correct(s):
try:
return abs(eval(s) - n) < 1e-10
except OverflowError:
return False
return min(
s
for i in range(3)
for j in range(not i, 3)
for a in [[None] * (2*(i+j)-1)]
for a[::2] in permutations(i*['e'] + j*['pi'])
for a[1::2] in product('+ - * / **'.split(), repeat=i+j-1)
for s in [''.join(a)]
if correct(s)
)
May 10, 2023
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