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Cryptography

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    $\begingroup$ It is worth mentioning that for "for certain fractal type distributions (see this paper for details) the Shannon entropy is $H$ but the key can be brute forced in much smaller than $2^H$ guesses on average" doesn't require weird/"fractal" distributions to demonstrate. Let $K$ either sample the all 0 key with probability 1/2, or sample a uniformly random key on $k$ bits otherwise. So $K = (1/2)\cdot 0^k + (1/2) \cdot \mathsf{Unif}([2^k])$. It is straightforward to show that $H(K) \geq (1/2) H(0^k) + (1/2) H(\mathsf{Unif}([2^k])) = (1/2) H(\mathsf{Unif}([2^k]))$ is still relatively high, despite $\endgroup$ Commented Jan 14 at 23:18
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    $\begingroup$ being very easy to break for "most" keys. $\endgroup$ Commented Jan 14 at 23:18
  • $\begingroup$ @MarkSchultz-Wu, good comment, my fractal was too strong a term, I meant something like your example to be included. $\endgroup$ Commented Jan 15 at 1:19
  • $\begingroup$ @kodlu: Can you refer me to a proof of your claim that with overwhelming probability all sequences from this biased source belong to the interval defined in the terms of the Shannon entropy? $\endgroup$ Commented Jan 15 at 10:36