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$\begingroup$ I agree with everything, my confusion lay in the fact that if you view matrix multiplication relative to the size of the matrix measured in bits (meaning you write $n=2^{b}$ where $b$ is the bitlength of n) you arrive per your calculation at $O(b^j2^{kb})$ operations - more costly than the DLP, which I thought weird. $\endgroup$

polarbear
– polarbear

2014-08-11 16:32:35 +00:00
Commented Aug 11, 2014 at 16:32
$\begingroup$ @polarbear : $\:$ That's why we "view matrix multiplication relative to the size of the" $\hspace{1.2 in}$ matrices (since those are the inputs). $\;\;\;\;$ $\endgroup$

user991
– user991

2014-08-11 19:26:25 +00:00
Commented Aug 11, 2014 at 19:26

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