hyperop
pip install hyperop
reduce(lambda x, y: self.lower(y, x), [a]*b)
from hyperop import hyperop
H1 = hyperop(1)
print H1(2,3), H1(3,2), H1(5,4)
# >> 5, 5, 8
H3 = hyperop(3)
print H3(2,3), H3(3,2), H3(5,4)
# >> 8, 9, 243
from math import log
H = hyperop(4)
print H(2,5)
>>> 200352993040684646497....45587895905719156736
print log(log(log(log(H(2,5),2.0),2.0),2.0),2.0) == 2
>>> True
H4 = hyperop(4)
print H4(2**0.5, 200)
# >> 2.0
hyperop
is almost surely the first python library to offer support for it.it's kinda big
Larger than all the atoms of the Universe.
Larger than all arrangements of the atoms of the Universe.
This isn't Graham's number.
def GrahamsNumber():
# This may take awhile...
g = 6
for n in range(1,64+1):
g = hyperop(g)(3,3)
return g
Comp. sci. technical question:
given infinite memory and time, will python fail?