Re: Confirmation of Shannon’s Mistake about Perfect Secrecy of One-time-pad

On Oct 22, 12:22 am, wangyong <hell...@xxxxxxx> wrote:


The following discussion supposes the all the
plaintexts, ciphertexts and keys are the same in length.
We give a simple example of OPT to discuss the problem, plaintext
space is M＝{0,1}, ciphertext space is C＝{0,1} and key space is K＝
{0,1}.

And note that the probability distribution on C is always the
same, no matter what the probability distribution on M.
So no observation of C can give us any information about the
probability distribution on M. In particular, observing
C=0 cannot change how likely we consider the uniform
distribution on M.


According to the information that cryptanalysts got beforehand,
they can get the prior probability of plaintext as P(M=0) = 0.9 and
P(M=1) = 0.1. Later the ciphertext C=0 is intercepted. When only
considering C=0 and the cryptosystem (regardless of the prior
probability of plaintext), we can educe that the plaintexts are
equally likely,

No. This is simply false. We can educe nothing about the
probablities of the plaintext.

for there is a one-to-one correspondence between all
the plaintexts and keys for C=0.

So what? There is no relationship between the probabilities.



- William Hughes







.

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