Article View: sci.crypt
Article #2922The Sidelnikov Saga... Now he's wrong again.
From: Simon McAuliffe
Date: Tue, 26 Jan 1993 11:52
Date: Tue, 26 Jan 1993 11:52
193 lines
7961 bytes
7961 bytes
A while ago, I wrote:
> It looks like _THIS_IS_ what Dr. Sidelnikov also did, and it means one
> of three things: (ranked in order of likelihood as I see it)
>
> 1) Both Dr. Sidelnikov and I stuffed up.
>
> 2) This correlation test isn't appropriate for this situation.
>
> 3) Something funny is going on and either the RNG is badly
> implemented or IDEA has a problem.
>
> Number 1 sounds like the most likely and number 3 is just plain scary.
After looking at things more closely, I have decided that both number 1
and number 2 are true. This is not to say that a correlation test is
not suitable, but that the correlation test performed was meaningless.
The correlation formula posted recently makes good sense, but looking at
the formula for the mean, you can't help feeling something is wrong.
1
mu = -----, n >= 2
n - 1
Correlation is basically a fairly simple idea.
| . . | . | .
| . . . . | | .
| . . . . . | . . | .
. .| . . | . |.
----------+---------- ----------+---------- ----------+----------
. . .| | . . . . . | .
. . | |.. . . | .
. | | . . . | .
| | . |
Correlation ~1 Correlation ~-1 Correlation ~0
Very roughly and informally, when the data is plotted as above
correlated data tends to form fuzzy lines. A positive gradient gives a
positive correlation and a negative gradient gives a negative
correlation. In our situation the correlation is close to 0, we are
just trying to decide if it is close enough.
Okay, back to the formula for the mean. Substituting n = 2 into the
equation yields mu = 1. Really?! Lets now imagine placing two points
on our graph. The correlation measures how good a line forms between
all the points. With only two points there is a perfect line. Given
any two random (continuous) points connected by a line we must either
have a positive gradient or a negative gradient (assuming a probability
of 0 of them being horizontally or vertically adjacent). If the
gradient is positive we have a correlation of 1, and if the gradient is
negative we have a correlation of -1, therefore the expected value
(mean) of the correlation for n = 2 is
mu = P(gradient is -ve) * -1 + P(gradient is +ve) * -1
= -0.5 + 0.5
= 0
Not only does the working above show this to be true, but intuitively
you would expect random numbers to be completely uncorrelated and
therefore have a correlation value of 0. Why then does the formula give
a value of 1? This is the maximum possible value of correlation, how
can we have a maximum as the mean?!
This problem is also true for n = 3, n = 4 and upwards, it just becomes
less ridiculous for larger n.
Correlation values range from -1 to +1. A formula which is always
positive must be a little worrying in this case. Should the correlation
be biased towards positive? Why so? Why not negative? The random
variables are supposedly from a uniform distribution. The mean simply
_MUST_ be 0 for a truely random source. Not just "nearly random." For
any given random sample, we won't actually get a value of 0, but the
expected value (the mean) is _EXACTLY_ 0. If this were not true, there
would never really be any random numbers.
So I think we can assume that the given equation is not really what it
was intended to be. I don't know what Knuth's book really says because
I don't have it, but perhaps there was something lost in the posting of
it?
If we take the square of the correlation values, then the mean variation
from 0 seems to be roughly equal to this function. This hasn't been
tested of course, but there is a tendency there. Perhaps then, this is
what the equations are (the mean deviation from the mean of the squared
correlation). That's not the impression I got from the posting. Never
mind, in any case it isn't particularly useful.
I now know why my last test gave such suprising results. Time to try a
different approach.
We can still use the correlation data obtained earlier (see my last
post). Following is a list of the correlation values I used last time.
These values are still valid as they do not involve the dubious
correlation mean and standard deviation. Each correlation value is
taken from an independent sample of 1048576 bytes (524288 byte pairs).
-0.000151
0.001282
-0.001413
0.001116
0.001013
-0.002072
0.000340
-0.001034
-0.001972
-0.000230
0.000508
-0.000425
0.002383
-0.001605
0.000506
0.002695
-0.001196
0.000450
0.000016
0.002822
We now wish to construct a hypothesis test with the null hypothesis
_
H0: x = mu (ie the data is random)
and the alternative
_
H1: x =/= mu (the data is _not_ random)
The test statistic is,
_
x - mu
t = -------------
s / sqrt(n)
_
where the mean, x is
_
x = 0.0001517
and the sample standard deviation, s is
s = 0.001463
As mentioned above, the mean correlation, mu is 0 for random data.
So we now have,
t = 0.0001517
---------------------
0.001463 / sqrt(20)
= 0.4634
This test statistic should approximate a t distribution with n - 1 = 19
degrees of freedom. So, to test H0 for truth at the alpha = 5% level,
we require that
0.4634 < t (19)
0.05/2
=> 0.4634 < 2.093
Which is obviously true, so at the 5% level we do not have enough
evidence to reject H0. In fact, for this test we have an observed
significance level of 64.8% so we can clearly see that H0 cannot be
rejected and we can safely assume that the data is indeed random[ish]
(at least as far as a contiguous byte correlation test is concerned).
Of course what I have done here does not show that the random numbers
are "cryptographically strong" it only shows that Dr. Sidelnikov's claim
of "strong correlation between contiguous bytes" is false. He probably
made the same mistake I made earlier! There's nothing to say I didn't
make more mistakes of course. It wouldn't suprise me at all. That's
why you're here - to find my mistakes. :)
There may still be correlation between other bits. I believe somebody
is working on this area as we speak however. There may also be other
weaknesses in IDEA which result in weak random numbers. I imagine some
people are working in this area too.
Anyway, I hope this is of more interest than my last balls-up. And I
hope that this isn't another one... :-)
Disclaimer: As I said in my last post, I am not a statistician
by any stretch of the imagination. Don't take what
I say as gospel.
Sai (e-mail sai@tornado.welly.gen.nz)
(s-mail PO Box 3598 Wellington, New Zealand)
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Path: arc1-ca.pugleaf.net!sparky!uunet!olivea!decwrl!waikato.ac.nz!comp.vuw.ac.nz!cavebbs!tornado!sai From: sai@tornado.welly.gen.nz (Simon McAuliffe) Newsgroups: sci.crypt,alt.security.pgp Subject: The Sidelnikov Saga... Now he's wrong again. Message-ID: <D0F1XB1w165w@tornado.welly.gen.nz> Date: 26 Jan 93 11:52:00 GMT Organization: Tornado BBS, Wellington, New Zealand. Lines: 193