On Wed, 14 Jul 1999, Lucas Wiman wrote:
> All,
> I recieved a message pointing out a possible error in my FAQ:
>
> > Speaking of Q2.6, I've heard that with Crandall's DWT, the subtraction
> > 2 step costs nothing at all. It's done automatically within the
> > transformation. Try checking this with George Woltman.
>
> Is this true?
>
> -Lucas
>From reading the code, as far as I can see, the subtraction cost it the
instructions:
fld MINUS2 ; Start normalization process
fadd BIGVAL ; with a BIGVAL-2.0 carry!
once every iteration, which is very little, though not for free.
The really big save in the algorithm is that the mod operation is not only
for free, but halves the size of the convolution as well, making
everything come out nice and fast.
--
Henrik Olsen, Dawn Solutions I/S URL=http://www.iaeste.dk/~henrik/
Thomas Daggert to Lucifer:
I have my soul, and I have my faith. What do you have... angel?
The Prophecy
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