Re: [RFC][PATCH] Make cryptoapi non-optional?

[Matt Mackall <mpm@selenic.com>]

On Fri, Aug 15, 2003 at 10:06:42AM +0200, M?ns Rullg?rd wrote:
> Andries Brouwer <aebr@win.tue.nl> writes:
> 
> >> > > entropy(x) >= entropy(x xor y)
> >> > > entropy(y) >= entropy(x xor y)
> >> > 
> >> > Is this trolling? Are you serious?
> >> 
> >> These lemma are absolutely true.
> >
> > David, did you read this line:
> >
> >> > Try to put z = x xor y and apply your insight to the strings x and z.
> >
> > Let us do it. Let z be an abbreviation for x xor y.
> >
> > The lemma that you believe in, applied to x and z, says
> >
> >  entropy(x) >= entropy(x xor z)
> >  entropy(z) >= entropy(x xor z)
> >
> > But x xor z equals y, so you believe for arbitrary strings x and y that
> >
> >  entropy(x) >= entropy(y)
> >  entropy(x xor y) >= entropy(y).
> >
> > This "lemma", formulated in this generality, is just plain nonsense.
> 
> Not quite non-sense, but it would mean that for any strings x and y, 
> 
>   entropy(x) == entropy(y),
> 
> which seems incorrect.

No, it's a premise stated at the beginning of the thread. We're
assuming perfect distribution for x and y. The problem here is that x
and y can be dependent or independent. If they're independent, then
there's no issue. If they're dependent (for instance correlated or
anticorrelated) then x^y biases toward zero or one. Which clearly has
less entropy.

-- 
Matt Mackall : http://www.selenic.com : of or relating to the moon