- Define
in
. Assume that
and
have degree
and define:
 |
(18) |
Then define:
 |
(19) |
For all
, we have:
 |
(20) |
Observe that, for a given
in the range
,
the number of products
is at most
.
Indeed, the polynomials
and
have
terms.
From there, we deduce that for all
, we have:
 |
(21) |
- One can proceed as follows:
- Let
machine-word-size
odd primes such that their product
exceeds
- For all
compute
- the images
and
of
and
in
,
- the product
in
.
- Using the Chinese Remaindering Algorithm coefficient-wise,
reconstruct
from
.
- We give complexity estimates (upper bounds for the number of machine-word
operations) for each of the steps
,
and
above.
For the operations on integers, we use bounds based on classical
quadratic algorithms (not fast ones, since they were not discussed
in class).
Let
be the length of a machine-word, let
be the number of machine-words to write
and let
to write
.
-
machine-word operations
-
machine-word operations
-
machine-word operations
Marc Moreno Maza
2008-03-18