Re: Analyses of Mistake on Proof about Perfect Secrecy of One-time-pad


Whatever formulae may have been presented in reference [7], either it
is wrong, or you have understood it wrong.
=====tell me why,you are just windbaggary
Your claim that there is a contradiction in the one-time pad - if that
is what you claim - is simply wrong.
=====tell me why,you are just windbaggary
Of course, the six conditions are different cases of knowledge, so if
you just mean that the case "the ciphertext is known" and the case
"the ciphertext is not known" do not coexist, that is true.

And it is precisely *because* they don't coexist that we can say:

Given a plaintext probability of p=0, 0.9 and p=1, 0.1, and a key
probability of k=0, 0.5 and k=1 0.5 because of the likelihood of the
two possible communications, and the fact that the key was made by
flipping a coin, things happen.

The person sending the message looks at the key to encipher the
message. So the probability of the key now changes for that person
from 50% each way to 100% for the correct key - and that person knows
the message to send, so the probability of the correct message is 100%
also.=====tell me why,you are just windbaggary
=============you are uneducated about cryptography, If K is known,
cryptorgarphy is useless.
This does not change the fact that we don't know the message or key,
so different cases do coexist in that different people know different
things.
==========the the posterior probaiblity is different to differnt
people???
The message turns out to be 0.

So either p=0, k=0 or p=1, k=1.

Knowing the key was made by flipping a coin, and that p=0 normally
obtains 90% of the time, what happens?
======just like you, I can say k=0 obtain 0.5 of the time , what
happens?
Shannon and I claim that what happens is:

One learns nothing about the plaintext; its probability with this new
knowledge remains 90% for 0, 10% for 1. But we've learned about the
key; having seen that the key was used on a message and it resulted in
0 for ciphertext, it must also be 90% for 0 and 10% for 1.

Since knowing the ciphertext "does not coexist" with not knowing the
ciphertext, there is no inconsistency in learning about the key, and
so we don't go back and change the _a priori_ probabilities of the
key. They are absolutely fixed and immutable - the key _was_ made by
flipping a coin, or it is not the one-time-pad. Saying otherwise
appears to have been your mistake.

The a priori key probabilities are weighting factors for the ordered
pairs, and knowing the ciphertext limits the possible ordered pairs of
plaintext and key such as to affect the chances of the plaintext not
at all - this is Shannon's proof, which I have already restated.
====you just use the probability when the ciphertext is not fixed,
If you do not admit the probabilities changes when c is fixed, the
probability of k should be 0.5, but not 0,9,
If you admit the probabilities change when c is fixed, but you are
wrong for you just get the probability distribution from that
condition.

.

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