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

On Oct 23, 12:02 pm, wangyong <hell...@xxxxxxx> wrote:
Fair enough, I admit that I have not read all your posts and papers.
However, this particularly simple example seems like an excellent way
for a lazy person to get to the bottom of what you are talking about.
So, if you feel like explaining here in brief and simple terms what
you think is wrong with my calculation, or how the scenario that you
are envisaging differs from the one I described, then it might lead
more quickly to some better understanding. If not then no worries...

----i have given the link. what I want to say is you are lazy.

You referred me to your "other paper" and I don't know which one you
mean. Anyway, in your post at http://groups.google.com/group/sci.math/msg/ac43dd1ffe1a3ffb
you say:

<begin quote>
A necessary and sufficient condition for perfect secrecy can be found
as follows: We have by Bayes' theorem

in which:
P(M)= a priori probability of message M
PM(E)= conditional probability of cryptogram E if message M is chosen
i.e. the sum of the probabilities of all keys which produce
cryptogram
E from message M.
P(E)= probability of obtaining cryptogram E from any cause.
PE(M)= a posteriori probability of message M if cryptogram E is
intercepted.
For perfect secrecy PE(M) must equal P(M) for all E and all M.
<end quote>

I have already shown that, in the simple scenario being discussed,
PE(M) = P(M) for E = 0 ("E" was called "C" in my post). It is equally
easy to show that PE(M) = P(M) for E = 1. So, your conditions for
"perfect secrecy" are satisfied.

Where's the problem?

.

Relevant Pages

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