Why do we try out all possible combinations of top bits in OpenSSL timing attack?

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Hi,

In the paper titled Remote Timing Attacks are Practical, 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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