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

On Nov 6, 2:26 am, wangyong <hell...@xxxxxxx> wrote:
On 11 6 , 1 36 , William Hughes <wpihug...@xxxxxxxxxxx> wrote:



On Nov 5, 11:28 pm, wangyong <hell...@xxxxxxx> wrote:

You cannot. When c is not fixed there is no
one to one correspondence between K and M so
you cannot use the distribution on K to determine
the distribution of M.

Your problem is that you need
c fixed so there is a one to one correspondence
c not fixed so the distribution of K is uniform

to hold at the same time.
It is i ask you whether you use the proof like that. If
not ,list the probability distribution when c is fixed, your mistake
will appear.

When C is fixed we have

C=0

P(M=0|C=0)=0.9, P(M=1|C=0)=.1
P(K=0|C=0)=0.9, P(K=1|C=0)=.1

C=1

P(M=0|C=1)=0.9, P(M=1|C=1)=.1
P(K=0|C=1)=0.1, P(K=1|C=1)=.9

As noted, when C is fixed the probability
distribution on K is not uniform and
the probability distribution on M is the same
as that on K.

If C is not fixed we have

P(M=0) = P(C=0)*P(M=0|C=0) + P(C=1)*P(M=0|C=1)
= .5 * .9 + .5 * .9 = .9
P(M=1) = P(C=0)*P(M=1|C=0) + P(C=1)*P(M=1|C=1)
= .5 * .1 + .5 * .1 = .1

P(K=0) = P(C=0)*P(K=0|C=0) + P(C=1)*P(K=0|C=1)
= .5 * .9 + .5 * .1 = .5
P(K=1) = P(C=0)*P(K=1|C=0) + P(C=1)*P(K=1|C=1)
= .5 * .1 + .5 * .9 = .5

As noted when C is not fixed, the distribution on
K is uniform, but the distribution of M is not
the same as the distribution of K.

In neither case is the distribution of M uniform.

- William Hughes

When C is fixed we have

C=0

P(M=0|C=0)=0.9, P(M=1|C=0)=.1
P(K=0|C=0)=0.9, P(K=1|C=0)=.1

C=1

P(M=0|C=1)=0.9, P(M=1|C=1)=.1
P(K=0|C=1)=0.1, P(K=1|C=1)=.9

As noted, when C is fixed the probability
distribution on K is not uniform and
the probability distribution on M is the same
as that on K.
----------------------i ask how you prove,
but you just list the
result.
Is your answer foolish? how you get the result????

We start with

If M=0, and K=0 then C=0
If M=0, and K=1 then C=1
If M=1, and K=0 then C=1
If M=1, and K=1 then C=0

and M and K independent with
probability distributions

P(M=0)=.9 P(M=1)=.1
P(K=0)=.5 P(K=1)=.5

Consider the probability space
consisting of all possible triplets
(M,K,C). The possible values and
probabilities are computed using
the facts that M and K
are independent and that if M and K
are known then C is a known function
call it f(x,y)), so

P(M=X,K=Y,C=f(X,Y))
=P(M=X,K=Y)= P(M=X)*P(K=Y)

We have

P(M=0,K=0,C=0) = P(M=0)*P(K=0) = .45
P(M=0,K=1,C=1) = P(M=0)*P(K=1) = .45
P(M=1,K=0,C=1) = P(M=1)*P(K=0) = .05
P(M=1,K=1,C=0) = P(M=1)*P(K=1) = .05

Computing the conditional probabilities.

P(M=0|C=0) =
(sum of probabilities of all events where M=0 and C=0)/
(sum of probabilities of all events where C=0)

= (P(M=0,K=0,C=0))/(P(M=0,K=0,C=0) + P(M=1,K=1,C=0))
= .45 / (.45 + .05) = .9

P(K=0|C=0) =
(sum of probabilities of all events where K=0 and C=0)/
(sum of probabilities of all events where C=0)

= (P(M=0,K=0,C=0))/(P(M=0,K=0,C=0) + P(M=1,K=1,C=0))
= .45 / (.45 + .05) = .9

P(M=1|C=0) =
(sum of probabilities of all events where M=1 and C=0)/
(sum of probabilities of all events where C=0)

= (P(M=1,K=1,C=0))/(P(M=0,K=0,C=0) + P(M=1,K=1,C=0))
= .05 / (.45 + .05) = .1

P(K=1|C=0) =
(sum of probabilities of all events where K=1 and C=0)/
(sum of probabilities of all events where C=0)

= (P(M=1,K=1,C=0))/(P(M=0,K=0,C=0) + P(M=1,K=1,C=0))
= .05 / (.45 + .05) = .1


P(M=0|C=1) =
(sum of probabilities of all events where M=0 and C=1)/
(sum of probabilities of all events where C=1)

= (P(M=0,K=1,C=1))/(P(M=0,K=1,C=1) + P(M=1,K=0,C=1))
= .45 / (.45 + .05) = .9


P(K=0|C=1) =
(sum of probabilities of all events where K=0 and C=1)/
(sum of probabilities of all events where C=0)

= (P(M=1,K=0,C=1))/(P(M=0,K=1,C=1) + P(M=1,K=0,C=1))
= .05 / (.45 + .05) = .1



P(M=1|C=1) =
(sum of probabilities of all events where M=1 and C=1)/
(sum of probabilities of all events where C=1)

= (P(M=1,K=0,C=1))/(P(M=0,K=1,C=1) + P(M=1,K=0,C=1))
= .05 / (.45 + .05) = .1


P(K=1|C=1) =
(sum of probabilities of all events where K=1 and C=1)/
(sum of probabilities of all events where C=1)

= (P(M=0,K=1,C=1))/(P(M=0,K=1,C=1) + P(M=1,K=0,C=1))
= .45 / (.45 + .05) = .9


- William Hughes

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