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

On 11 5 , 10 41 , William Hughes <wpihug...@xxxxxxxxxxx> wrote:
On Nov 4, 9:31 pm, wangyong <hell...@xxxxxxx> wrote:





On 11 5 , 10 25 , William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Nov 4, 9:17 pm, wangyong <hell...@xxxxxxx> wrote:

On 11 5 , 9 57 , William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Nov 4, 8:52 pm, wangyong <hell...@xxxxxxx> wrote:

On 11 4 , 8 47 , William Hughes <wpihug...@xxxxxxxxxxx> wrote:

On Nov 4, 6:44 am, wangyong <hell...@xxxxxxx> wrote:> On 11 3 , 10 18 , William Hughes <wpihug...@xxxxxxxxxxx> wrote:

If you start with a situation where the plaintext might

not be uniformly distributed, then you the key cannot
be uniformly distributed for a fixed cyphertext.

Note *for a fixed cyphertext*. The reply

=======no, if c is not fixed,

is irrlevant. The condition (repeated many
times by you), is that the cyphertext is fixed).

- William Hughes

note c==ciphertext

Yes, and the condition (repeated many times
by you) is that c is fixed. The condition
"c is not fixed" is irrelevant.

- William Hughes- -

- -

if c is not fixed,we have both
the key uniformly distributed and the plaintext not uniformly
distributed at the same time.
I never think
if c is fixed,we have both
the key uniformly distributed and the plaintext not uniformly
distributed at the same time.

You start with the plaintext not uniformly distributed.
You claim that the key is uniformly distributed and
c is fixed.

- William Hughes- -

- -

look carefully.

On looking carefully I note:
the plaintext is not uniform, the key is not uniform, c is fixed.
No contradiction.

- William Hughes- -

====================================are your proof is like the
following,If so, my reply is correct, you just use the probabiltiy
when the c is not fixed.

In your favourite example of plaintext probabilities
P(M=0) = 0.9 P(M=1) = 0.1
P(K=0) = 0.5 P(K=1) = 0.5
the following events (plaintext,key,cyphertext) occur with
the following probabilities:
(0,0,0) 0.45
(0,1,1) 0.45
(1,0,1) 0.05
(1,1,0) 0.05
Now you insist on using a "fixed" cyphertext.
There are two possibilities for a fixed cyphertext; let's treat
one after the other:
1) C=0:
Note: This cyphertext occurs with probaility 0.45 + 0.05 = 0.5
The conditional probability of plaintext M=0 is 0.45/(0.45+0.05) =
0.9
The conditional probability of key K=0 is 0.45/(0.45+0.05) = 0.9
Just divide "hit probabilities by "allowed" probabilities.
2) C=1:
Note: This cyphertext occurs with probaility 0.45 + 0.05 = 0.5
The conditional probability of plaintext M=0 is 0.45/(0.45+0.05) =
0.9
The conditional probability of key K=0 is 0.05/(0.45+0.05) = 0.1
Just divide "hit probabilities by "allowed" probabilities.


Conclusion

.

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