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The following archive was created by hippie-mail 7.98617-22 on Fri Aug 21 1998 - 17:19:02 ADT
burt rosenberg (burt@passaic.cs.miami.edu)
Wed, 24 Jun 1998 20:45:35 -0400 (EDT)
>
>
> On Wed, 24 Jun 1998, burt rosenberg wrote:
>
> >
> > this solution, to fill time, is most likely faster, and possibly
> > simpler, but it doesn't fully cover the calculation's tracks.
>
> How do you mean it doesn't cover the calculations tracks? Every
> y^x will take the exact same time no matter what x is. This is slower,
> not faster, but it's a defense.
>
say we do:
t1 = multiply( s, s ) ; // the square
t2 = multiply( s, y ) ; // the multiply, just in case
if ( bit(i,x) ) s = t2 ;
else s = t1 ;
to keep the times for one stage the same, but s is now
either y*s*s or s*s, and it is possible that in the next
stage (y*s*s)^2 gives a different timing than (s*s)^2.
actually, this is covered in a parenthetical remark in the
orig. paper. even w/ this care, it cannot be said that
timing{y^x} is not a function of x, hence can possibly
be leaking info. about x.
-burt
The following archive was created by hippie-mail 7.98617-22 on Fri Aug 21 1998 - 17:19:02 ADT