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

On Nov 4, 6:14 am, wangyong <hell...@xxxxxxx> wrote:
On 11 4 , 4 18 , hagman <goo...@xxxxxxxxxxxxx> wrote:



On 3 Nov., 16:06, wangyong <hell...@xxxxxxx> wrote:

On 11 2 , 6 08 , hagman <goo...@xxxxxxxxxxxxx> wrote:

[BTW, could you *plase* switch to a more standard way of quoting?
Because of your current unusual method, I always get terribly lost
about the authorship of text fragments a few levels inside]

Without looking at the encrypted text (i.e. a priori) the
keys K are uniform.
After looking at the encrypted text (i.e. a posteriori)
teh situation is different as I have quantified above.
===========Yes , it is the same to plaintext.

Exactly.
And the keys are not uniform a posteriori.
And the plaitext distribution is absolutely unchanged.
And the latter is what i scalled perfect secrecy.
Thus OPT is perfectly secret.

If you throw a fair die, the probaility of getting a 6 is 1/6.
As soon as you observe the result, the a posteriori probability that
you just got a 6 is either 1 or 0, depending on what you observe.
==yes, the condition changed.

But you seem to be unable to pin down a single step that is
invalid.
====i just point out the problem.

You don't. You merely say that you point out the problem (see, you
just did).
So I ask again: Which of my small handful of steps is not backed by
standard probability theory? Alternatively, which result from
standard probability theory that I used is wrong?

Which conditions cannot coexist?

=====the prior , the key uniform chipertext fixed, OPT. can not
coexist, \

So the three items that cannot coexist are a) "the prior",
b) "the key uniform cyphertext fixed" and
c) "OTP" ?
OTP clearly exists.
The prior (if it means an a priori distribution for plaintexts?) does
exist.
The sequence of words "the key uniform cyphertext fixed" does not
make sense to me anyway, so I don't care if that doesn't exist.

look at the papers carefully, do not just ask.

I've taen a swicft look over one of your papers (they are all the same
anyway)
and have already pointed out one error of yours (something like
mixing up P(A|B) with P(B|A)) in another post.

While it is correct that the theorem cannot be proved by example,

==yes you admit your problem.

read the sentence to its end.
I don't have to prove Shannon by example anyway.
I am satisfied with the fact that the example *disproves*
your gross deviation.

it is clear that IF you were right and observing the encrypted
message made the a-posteriori probabilities of the plaintext
message differ from the a-priori probabilities THEN you
should be able to make a "guess" of the plaintext that
would agree better than expected.
Additionally, one CAN make (as predicted by my statements)
a good "guess" (about 90% digits correct) of the key
used in my example - I just verified it from the raw data
I still have on my disk.
However, your result that the plaintext suddenly has
a different distribution can be checked by statistical
tests applied to my example and thus looks highly unlikely.
The possible conclusions are:
What all mathematicians except wangyong consider correct
arguments in probability theory is logically incorrect
but exactly describes the phenomena for the description
of which probability theory was developed in the first place.
What wangyong considers correct produces wrong descriptions
of simple testable statistical phenomena.
============your question are foolish, are unreasonable. the
uncertainty increase dont predicate incorrect. you just ask me to
guess a fixed value, or possibel value, but not a random variable.

You may view my 1000 character plaintext as a sequence of 1000
single character plaintexts.
Then 1000 is big enough a number that you are allowed to
guess a distribution instead.
But the distribution for the plaintext is still
P(M=0)=0.9 both for cyphertext 0 and for cypertext 1.
Do you want to come up with a significantly different distribution
or finally admit that no knwoledge (not even a changed distribution)
about the plaintext could be gained (an effect aka. perfect security)?

you take a mistake like shannon and OPT.

Being only as wrong as Shannon and the perfect secrecy of OTP
makes me quite satisfied.
OTOH, I see that you are totally incapable of pointing out
one single error in a short derivation and only resort to
your "Shannon made a mistake" lamenting, running logically
in circles again and again (as you do with writing essentially
the same paper again and again).
I take this as a sign that ma presence in this thread
is of no use whatsoever and shall therfore leave it.
Good luck to all others trying to succeed in this struggle.

hagman- -

- -

hagman Exactly.
And the keys are not uniform a posteriori.
And the plaitext distribution is absolutely unchanged.

=========why? the M changes as the K.

You don't. You merely say that you point out the problem (see, you
just did).

============you are denying.I have repeat a lot of times,
the answer is like the following.

From P(M = x)·P (K = (x y)) = P(M = x) ·2-n, we can

see the condition that the ciphertext y is a fixed value is never
considered when computing P(M = x C = y). We can get that result by
reductio ad absurdum. Suppose for fixed y, if P (K = (x y))=2-n (that
is used in the proof, but indeed it is wrong. It is used just to get
wrong conclusion), we can get P(M = x C = y)= 2-n because there is a
one-to-one correspondence between all the plaintexts and keys for the
fixed ciphertext in OTP. But it is obviously wrong, for the prior
probabilities of all plaintexts are seldom equally likely. So P(M =
x)·P (K = (x y)) stand for the joint probability of x and y when y is
not fixed. But Shannon thought of the posterior probability as the
probability of plaintext when ciphertext had been intercepted, we can
see that there is a presupposition in P(M = x C = y) that y is fixed,
but in P(M = x), P (K = (x y)) and P(C=y), y is not fixed, otherwise
we can get obviously wrong results. In such way, the Bayes's formula
was misused for the probability was not on the same presupposition
and
the equation does not come into existence.
In OTP there are complex and crytic conditions that influence the
probability of plaintext, key and ciphertext, so it is essential to
cognize all the conditions and carefully use probability theory. The
proof did not realize the crytic condition that ciphertext was a
fixed
value (even though unknown) rather than a random variable.

So I ask again: Which of my small handful of steps is not backed by
standard probability theory? Alternatively, which result from
standard probability theory that I used is wrong?
========you just deny. The conditions cannot coexist, how can you use
conditional probability. just misuse it.

Which conditions cannot coexist?
=====the prior , the key uniform chipertext fixed, OPT. can not
coexist, \

So the three items that cannot coexist are a) "the prior",
b) "the key uniform cyphertext fixed" and
c) "OTP" ?
OTP clearly exists.
The prior (if it means an a priori distribution for plaintexts?) does
exist.
The sequence of words "the key uniform cyphertext fixed" does not
make sense to me anyway, so I don't care if that doesn't exist.
==================the key uniform, cyphertext fixed

look at the papers carefully, do not just ask.

I've taen a swicft look over one of your papers (they are all the
same
anyway)
and have already pointed out one error of yours (something like
mixing up P(A|B) with P(B|A)) in another post.
=======where???? , you have already pointed out????? when???

read the sentence to its end.
I don't have to prove Shannon by example anyway.
I am satisfied with the fact that the example *disproves*
your gross deviation.
==============you are just windbaggary

You may view my 1000 character plaintext as a sequence of 1000
single character plaintexts.
Then 1000 is big enough a number that you are allowed to
guess a distribution instead.
But the distribution for the plaintext is still
P(M=0)=0.9 both for cyphertext 0 and for cypertext 1.
Do you want to come up with a significantly different distribution
or finally admit that no knwoledge (not even a changed distribution)
about the plaintext could be gained (an effect aka. perfect
security)?

=======your example are useless to prove or disprove the problems
discuss.displacing the problem to cover up your mistake seems to be
your strongpoint

you take a mistake like shannon and OPT.

Being only as wrong as Shannon and the perfect secrecy of OTP
makes me quite satisfied.
OTOH, I see that you are totally incapable of pointing out
one single error in a short derivation and only resort to
your "Shannon made a mistake" lamenting, running logically
in circles again and again (as you do with writing essentially
the same paper again and again).

=============I have pointed out ,but you just deny.
I take this as a sign that ma presence in this thread
is of no use whatsoever and shall therfore leave it.
Good luck to all others trying to succeed in this struggle.

---------------I see. your All tricks have been exhausted.
dicussion just exposed your mistakes.

Please send me the Chinese version if you have one.
I cannot be convinced here, even after I read your paper in detail and
all your replies.

Jiang Bian
jxbian@xxxxxxxx
Univ. Of AR at LIT

.

Relevant Pages

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  • Re: Confirmation of Shannons Mistake about Perfect Secrecy of One-time-pad
    ... And the plaitext distribution is absolutely unchanged. ... As soon as you observe the result, the a posteriori probability that ... The sequence of words "the key uniform cyphertext fixed" does not ... message made the a-posteriori probabilities of the plaintext ...
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    ... And the plaitext distribution is absolutely unchanged. ... As soon as you observe the result, the a posteriori probability that ... The sequence of words "the key uniform cyphertext fixed" does not ... message made the a-posteriori probabilities of the plaintext ...
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