From: "Wei Dai" <weidai@eskimo.com>
Date: Tue, 24 Jan 95 19:05:11 PST
To: cypherpunks@toad.com
Subject: analysis of Chaum's MIX continued
Message-ID: <199501250304.AA05794@mail.eskimo.com>
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Last week I wrote about a way to trace who Alice, 
using a Chaumian mix, is writing to, by calculating a score 
for each user based on the number of times the receipt of a 
message by the mix from Alice is followed by a transmit to 
the user during the output phase of that batch (I need a 
name for these occurences... any sugestions?).  Does 
this method actually work?  Well, let's see...

(Let me note that this method of traffic analysis can be 
applied to any Chaum type MIX, even if the MIX uses random 
delays instead of batches.  It can even be used on entire 
MIX-nets, by treating the MIX-net as a single large mix.  
However, I have to make several assumptions in the 
following analysis of how well this method works.  This 
doesn't mean that it won't work outside those assumptions, 
just that I don't know enough statistics to figure out how 
well it would work in a more general situation.  Maybe someone 
can give me a recommendation for a good statistics textbook?)

Let's assume:
1.  there is one mix which processes a batch every time it 
receives a certain number of messages
2.  there are N users
3.  all users send at most one message to the mix per 
batch, the probability that he will do so is S (I'm assuming 
every user sends the same number of messages per unit time 
on average, many of which can be dummies, and that the 
timing of these messages are random)
4.  all users receive at most one message from the mix per 
batch, with probability R (R<S or R>S depending on 
wheather the mix eats more dummy messages than it 
generates)

Alice and Bob (the targets of the traffic analysis) are 
simply two of those users.

5.  In a particular batch, there is a probability Q (Q<=R and Q<=S) 
that Alice will send a message to Bob.  This implies that 
for each batch in which Alice doesn't send a message to 
Bob, there is a probability of (S-Q)/(1-Q) that she will 
send some other message to the mix (which may be a dummy 
message or a message to someone else).  Similiarly, for each 
batch in which Bob doesn't receive a message from Alice, 
there is a probability of (R-Q)/(1-Q) that he will receive 
some other message from the mix.

Let T be the length of time (expressed in number of 
batches) since the start of the traffic monitoring
and let M(user) be the total number of times the receipt 
of a message by the mix from Alice is followed by a transmit 
to user during the output phase of that batch.

Note that the distribution of M is a binomial 
distribution B(T, R*S).  This means:
	mean of M = T*R*S
	standard deviation of M = sqrt(T*R*S*(1-R*S))

On the other hand,
	M(Bob) = T*Q + T(1-Q) * ((S-Q)/(1-Q)) * ((R-Q)/(1-Q))
		which simplifies to:
	M(Bob) = T * (Q + (S-Q)*(R-Q)/(1-Q))

Now, we can calculate a z-score for M(Bob) by subtracting 
from it the mean of M (this difference simplifies to 
T*Q*(1-S)*(1-R)/(1-Q) ) and dividing the difference by the 
standard deviation.  We can then find the standard normal 
probability p(z) associated with the z-score, and finally 
multiply 1-p(z) and the total number of users (N) to find how 
many users can be expected to have a larger M than Bob.  
Let's call this number A.

In conclusion:
                   T*Q*(1-S)*(1-R)/(1-Q)  
A = N * (1 - p ( ------------------------- ) )
                    sqrt(T*R*S*(1-R*S))   

It seems that as long as Q>0, S<1, and R<1, A converges to 
0 as T increases.  This means under the above assumptions, 
Bob will eventually be traced out if these 3 conditions are 
met.

Wei Dai

P.S.  If there aren't any serious mistakes in the above 
analysis, I may produce a table showing how long it would 
take for A to fall below 1 for various values of Q, R, 
S, and N.  Is there any interest in this?

For now, I've attached an Excel spreadsheet so you can try 
plugging numbers into the above formula.


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     Date:  24 Jan 1995, 18:51
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