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SubjectRe: [PATCH] (0/4) Entropy accounting fixes
On Mon, 19 Aug 2002, Oliver Xymoron wrote:
> On Mon, Aug 19, 2002 at 01:43:59AM -0400, Theodore Ts'o wrote:
> > On Sat, Aug 17, 2002 at 09:15:22PM -0500, Oliver Xymoron wrote:
> >
> > > Assuming the interrupt actually has a nice gamma-like distribution
> > > (which is unlikely in practice), then this is indeed true. The
> > > trouble is that Linux assumes that if a delta is 13 bits, it contains
> > > 12 bits of actual entropy. A moment of thought will reveal that
> > > binary numbers of the form 1xxxx can contain at most 4 bits of
> > > entropy - it's a tautology that all binary numbers start with 1 when
> > > you take off the leading zeros. This is actually a degenerate case of
> > > Benford's Law (http://mathworld.wolfram.com/BenfordsLaw.html), which
> > > governs the distribution of leading digits in scale invariant
> > > distributions.
> > >
> > > What we're concerned with is the entropy contained in digits
> > > following the leading 1, which we can derive with a simple extension
> > > of Benford's Law (and some Python):
> >
> > I'm not a statistician, and my last probability class was over 15
> > years ago, but I don't buy your extension of Benford's law. Sure, if
> > we take street addresses numbering from 1 to 10000, the probability
> > that the leading digit will be 1 will be roughly 30%. But the
> > distribution of the low two digits will be quite evenly distributed.
> > Put another way, by picking a house at random, and looking at the low
> > two digits, and can very fairly choose a number between 0 and 99.
>
> That's correct, and that's what's being calculated. The example Python
> code calculates the Benford distribution not in base 2, but in base
> 2^bits. Then it uses the Shannon entropy definition to calculate the
> overall entropy of that distribution. For a 12 bit number, we've got
> about 7 bits of entropy, most of it concentrated in the LSBs.

I think you have it slightly wrong there. By snipping away the first digit
from a number leaves you with, not Benford's distribution, but
uniform distribution, for which the Shannon entropy is naturally roughly
the bitcount.

Benford's distribution as presented (1 being the first digit 30% of the
time) concerns firstly base 10 and secondly randomly upper bound sets.
For base 2 the Benford's leading digit distribution naturally narrows down
to 1 being the first digit 100% of the time. Benford's law says little
about the remaining digits. Thus by snipping that first bit away leaves us
with a set that reflects the properties of the choice. Assuming it's
random, we get uniform distribution.

Wether the bit count of the smallest of the three deltas is
actually sufficient to guarantee us that amount of randomness in the
choice is another question. Like stated here already, it can be easily
fooled, and there's a strong possibility that it gets "fooled" already.

Clearly periodic but not delta-wise detectable sequences would pass the
delta checks. If interrupts get timed like, 1 3 11 50 1 3 11 50 1 3 11 50,
the deltas would indicate more entropy than there's present.

Some level of fourier analysis would be necessary to go further than what
we can with the deltas.

(For the record, I'm not one bit worried about the kernel rng, perhaps
the entropy bit count may be theoretically a bit off from time to time, but
remarks like "If someone then breaks SHA-1, he can do this and that..."
only amuze me. If someone breaks SHA-1 we're bound to change the hash
then, eh? Not mention that we'd be in deep shit not just kernelwise.)

--
Tommi Kynde Kyntola kynde@ts.ray.fi
"A man alone in the forest talking to himself and
no women around to hear him. Is he still wrong?"

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