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

Paul Hoffman <paul.hoffman@vpnc.org> Sat, 25 January 2014 19:35 UTC

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From: Paul Hoffman <paul.hoffman@vpnc.org>
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Date: Sat, 25 Jan 2014 11:35:11 -0800
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References: <52DD996F.3040708@cs.tcd.ie> <CAF4+nEHEWaSr3HMuGtQ=vQzuuhkTo2uNpedUTNgmT5NsWRsTfA@mail.gmail.com> <30316745-8091-46AD-95A1-407757489FF9@vpnc.org> <1737731959.20140122185149@gmail.com> <03f201cf17ee$e34ccbf0$a9e663d0$@hosed.org> <15541579.20140123214020@gmail.com> <00C069FD01E0324C9FFCADF539701DB3BBF18E51@sc9-ex2k10mb1.corp.yaanatech.com> <204592464.20140123233807@gmail.com> <00C069FD01E0324C9FFCADF539701DB3BBF18FD6@sc9-ex2k10mb1.corp.yaanatech.com> <1825449796.20140124180225@gmail.com> <00C069FD01E0324C9FFCADF539701DB3BBF1BE0E@sc9-ex2k10mb1.corp.yaanatech.com>
To: Michael Hammer <michael.hammer@yaanatech.com>
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Subject: Re: [dsfjdssdfsd] Any plans for drafts or discussions on here?
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On Jan 25, 2014, at 8:16 AM, Michael Hammer <michael.hammer@yaanatech.com> 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

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