Re: [dsfjdssdfsd] Any plans for drafts or discussions on here?

[Michael Hammer <[e-mail]>]

I am not sure if we are talking past each other.

Using terms from your example, 
I have one input (trying to map to what you say) that is message A.
I have another input, call it key B.
The output of the "secure algorithm" is C.
C will be known by adversary M.

The question is whether B is sufficiently random that C cannot guess it.
Also, that M cannot easily discover A knowing C.
The strength of the algorithm is part of the assurance.
The strength of the key is the other part.
Weak key B does not adequately protect message A.

Now, being random does not guarantee that the key B is not weak, just not
easily deduced by M.
But, if B is generated from inputs B1 and B2 in such a way that it tends to
reduce the randomness 
(worse case results in very small subset of keys B), then M can brute force
B to reveal A.

One of the papers cited earlier pointed out how a complex algorithm 
actually ended up converging on a small number of values.
I would hope to avoid repeating that mistake.

Michael Hammer
Principal Engineer
[e-mail]
Mobile: [phone]
500 Yosemite Drive Suite 120
Milpitas, CA 95035 USA


-----Original Message-----
From: Paul Hoffman [mailto:[e-mail]] 
Sent: Saturday, January 25, 2014 11:35 AM
To: Michael Hammer
Cc: [e-mail]
Subject: Re: [dsfjdssdfsd] Any plans for drafts or discussions on here?

On Jan 25, 2014, at 8:16 AM, Michael Hammer <[e-mail]>
wrote:

> So, if you mix a non-random input with a random input, using a 
> deterministic algorithm, the output will be more random?
> That doesn't seem right to me.

That's because it is not right for many reasons. To start, you haven't
defined "non-random" and "more random".

A better description:

Value A has X bits that cannot be known to adversary M. Value B has Y bits
that cannot be known to M.

Securely mixing A and B into a value C whose length is greater than or equal
to (X + Y) will result in C having (X + Y) bits that cannot be known by M.
If C's length is less than (A + B), every bit in C cannot be known by M.

In your question above, the fact that B might be 0 is irrelevant to the
calculation. 

--Paul Hoffman