In class, we learned the Kocher Timing Attacke on modular exponentiation. For determining whether the (e.g., LSB) is 0 or 1, we specified two values:

$T_1 = T_M - (TMult(1, y) + (\omega -1)ExpTMult)$

$T_0 = T_M - \omega ExpTMult$

where $T_M$ is the timing for $y^x\ mod\ m$, $TMult$ the timing for the multiplication, $\omega$ the Hamming weight of the exponent and $ExpTMult$ the expected timing value for a multiplication. We calculated these values for every sample. Then, if $\Sigma_{sample} T_1\ <\ \Sigma_{sample} T_0$ the LSB is likely 1. Similar for the following bits.

Question: What happens after a incorrect guess?

The linked paper says (Sec. 3):

after an incorrect exponent bit guess, no more meaningful correlations are observed.

So, it is $\Sigma T_1 \approx \Sigma T_0 $ for the following bits?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.