Re: timing attacks.

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Mike Rosing (eresrch@msn.fullfeed.com)
Wed, 24 Jun 1998 15:07:18 -0500 (CDT)


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.
> i do not see how to prove that no information is then leaked
> through timing analysis. this will make the timing data more
> subtle, perhaps to the point of uselessness, but perhaps not.

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.

> kocher provides the thought experiment of a certain squaring
> that takes a grotesquely long time. by chosing y correctly,
> you can tickle this squaring and check that the i-th bit of x is 0.
>
> however, we can take this thought experiment one step further
> and assume that for a certain x all y^x's take a longer time
> on average than other x's. then kocher's proposed solution,
> premultiply y by a random number before exponentiation, also fails.

and we can completely defeat the attack if time is accounted for
in every step. You don't need anything fancy, a few well placed nop's
will work fine.

> on the other hand, randomly splitting x will not disclose information
> _unless_ y^x' and y^{x-x'} somehow correlate, for randomly drawn x'.
> that is, they are not like y^x' and y^x'' drawn independently at random.
> assuming independence, a proof of no information leakage might be possible.
>
> do y_x' and y^{x-x'} act as independent algorithms, for random x'?
> gee ...

I'm not arguing that this might help defeat the attack, but you can
elminiate the attack completely with very little effort. The choosing of
the random x might leak info because of the time it takes the random
generator to run, unless you take the same care to count clock cycles.

Patience, persistence, truth,
Dr. mike


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The following archive was created by hippie-mail 7.98617-22 on Fri Aug 21 1998 - 17:18:59 ADT