Hi,
down votefavorite <http://crypto.stackexchange.com/questions/43612/why-do-we-try-out-all-possible-combinations-of-top-bits-in-openssl-timing-attack#> In the paper titled Remote Timing Attacks are Practical <https://crypto.stanford.edu/~dabo/papers/ssl-timing.pdf>, the authors mention the following: Initially our guess g of q lies between 25122512 (i.e. 2log2(N/2)2log2(N/2)) and 25112511 (i.e. 2log2(N/2)−12log2(N/2)−1). We then time the decryption of all possible combinations of the top few bits (typically 2-3). When plotted, the decryption times will show two peaks: one for q and one for p. We pick the values that bound the first peak, which in OpenSSL will always be q. I don't understand the following: - Does the initial guess of g start from an arbitrary range? - What's the rationale behind trying out top 2-3 bits? - What will the remaining bits be in this case? -- Thanks & Regards, Dipanjan
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