BugTraq
Joint encryption? Feb 18 2005 07:42AM
John Richard Moser (nigelenki comcast net) (7 replies)
Re: Joint encryption? Feb 20 2005 12:09PM
Ruud H.G. van Tol (rvtol isolution nl)
Re: Joint encryption? Feb 20 2005 06:21AM
Valdis Kletnieks vt edu (1 replies)
Re: Joint encryption? Feb 20 2005 06:00PM
John Richard Moser (nigelenki comcast net)
RE: Joint encryption? Feb 19 2005 08:13PM
David Schwartz (davids webmaster com) (1 replies)
Re: Joint encryption? Feb 19 2005 09:59PM
John Richard Moser (nigelenki comcast net)
Re: Joint encryption? Feb 19 2005 07:21PM
Gandalf The White (gandalf digital net)
Re: Joint encryption? Feb 19 2005 04:32PM
Damian Menscher (menscher uiuc edu) (1 replies)
Re: Joint encryption? Feb 19 2005 05:04PM
John Richard Moser (nigelenki comcast net)
Re: Joint encryption? Feb 19 2005 10:44AM
devnull Rodents Montreal QC CA (1 replies)
[As usual when I write to bugtraq, the from address in the headers
simply discards mail, so I don't have to deal with all the broken
autoresponder mail that would otherwise land on me. To reach me, use
the address in the signature.]

> The problem is that I need a guaranteed way to create data for any
> valid N and M where N >= 3 > M >= 2 in which access to M fragments of
> the key (each fragment is encrypted) can be used to gain access to
> the rest of the fragments, which in turn allows any selection of M
> users to authenticate and gain physical access to the key.

You don't need the "...used to gain access to the rest of the
fragments..." part.

This is called "secret splitting", and there are a number of schemes by
which you can split a secret into N shares, any M of which can
reconstruct the secret, but any M-1 of which can discover nothing about
the secret. One of the simplest, at least to my mind, is based on
polynomials over a finite field. A handful of secret-splitting
schemes, including this one, are described in Schneier's _Applied
Cryptography_ (and doubtless many other places); the rest of this
message is a brief description of this technique.

Input: a secret S, and N and M as above.
Choose a prime P, larger than S.
Let c[0] be S.
Choose random values less than P for c[1] through c[M-1].
For i from 1 through N, compute (all arithmetic mod P)
h[i] = sum(j=0..M-1) (c[j] i^j) [^ is exponentiation]
Share #i is then the triple <P,i,h[i]>.

How the shares are stored is up to those charged with protecting them;
they can store them encrypted if they want. Only the h[i] value needs
to be protected.

Now, given any M shares, you can set up M equations

h[i] = sum(j=0..M-1) (c[j] i^j) (mod P, of course)

for the i and h[i] values in the shares. (Of course, if the P values
in the shares aren't all equal, at least one of the shares has been
corrupted.) This is a system of M linear equations in M unknowns (the
c[] values). Given how the coefficients of this system were chosen
(the i^j values), they will be linearly independent and the system thus
has a unique solution (since P is prime, division mod P works and
Gaussian elimination can be performed). Solve it, and c[0] will be the
secret. (You can throw away c[1] through c[M-1]; they were randomly
chosen at split time and carry no information.)

But if you have fewer than M shares, you can set up at most M-1
equations. Such a system is not solvable, and since we're working in
the finite field Z_P, you actually cannot discover anything about S; by
introducing a fictitious additional share with a suitable h[] value,
you can arrange to make c[0] come out to any value you please.

If you have more than M shares, the system is overdetermined. You can
pick any M of the shares, reconstruct the c[] values, and check that
what you get agrees with the redundant shares. (You actually don't
*have* to check, but it allows you to catch some cases of corrupted
shares.)

I've written software that implements this. See
ftp.rodents.montreal.qc.ca:/mouse/local/src/secretsplit/.

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[ reply ]
Re: Joint encryption? Feb 19 2005 12:24PM
John Richard Moser (nigelenki comcast net) (1 replies)
Re: Joint encryption? Feb 21 2005 08:02PM
peter zulu (peterzulu gmail com)
Re: Joint encryption? Feb 19 2005 10:24AM
Casper Dik Sun COM (1 replies)
Re: Joint encryption? Feb 19 2005 12:17PM
John Richard Moser (nigelenki comcast net) (1 replies)
Re: Joint encryption? Feb 21 2005 11:42AM
Robert C. Helling (R Helling damtp cam ac uk)


 

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