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Computational complexity of modular exponentiation
ID: computational-complexity-of-modular-exponentiation
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Computational complexity of modular exponentiation
by
Ciro Santilli
37
Updated
2025-07-16
math.stackexchange.com/questions/2382011/computational-complexity-of-modular-exponentiation-from-rosens-discrete-mathem
mentions:
b
n
m
o
d
m
(1)
can be calculated in:
l
o
g
(
m
)
2
l
o
g
(
n
)
(2)
Remember that
l
o
g
(
m
)
and
l
o
g
(
n
)
are the
lengths
in
bits
of
m
and
n
, so in terms of the
length
in
bits
M
and
N
we
'
d
get:
M
2
N
(3)
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