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)
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David Schwartz wrote:
>>The authentication works as below:
>>

[...]

>
> There's a ludicrously simple and incredibly brilliant way to do this. For a polynomial of order N, you need N points on the polynomial to find the equation that describes the polynomial. So if you want to share a secret amount 28 people such that any 15 are needed to know it, just make the secret the coefficients of a 15th order polynomial and compute 28 points that satisfy the polynomial.
>
> So, for the 28/15 example, pick 15 random coefficients (C1, C2, C3, ...), and then your 28 pieces of the key (K1 ... K25) are the solutions to:
>
> Kx = C1 + C2 * x + C3 * x^2 + C3 * x^3 ... C15 * x^14
>

Math check here, wikipedia seems to think:

If the players store their shares on insecure computer servers, an
attacker could hack in and steal the shares. If it is not practical to
change the secret, the uncompromised (Shamir-style) shares can be
renewed. The dealer generates a new random polynomial with constant term
zero and calculates for each remaining player a new ordered pair, where
the x-coordinates of the old and new pairs are the same. Each player
then adds the old and new y-coordinates to each other and keeps the
result as the new y-coordinate of the secret.

-- http://en.wikipedia.org/wiki/Secret_sharing

I'm confused: "polynomial with constant term zero" "The dealer encodes
the secret as the curve's y-intercept"

now after playing with my calculator I suddenly remember something about
any polynomial's Y intercept being the constant term. See, if all the
coefficients are 999999999999999999999999999999999999999999999999,
multiplied by X=0, the result is still 0.

Is wikipedia wrong here? Or do I have a misunderstanding of "constant
term"?

> For x=1 to 28.
>
> With any 15 solutions to the equation above, you can compute C1 through C15. With any 14, you can't even get started.
>
> DS
>
>
>
> DS
>
>
>

- --
All content of all messages exchanged herein are left in the
Public Domain, unless otherwise explicitly stated.

Creative brains are a valuable, limited resource. They shouldn't be
wasted on re-inventing the wheel when there are so many fascinating
new problems waiting out there.
-- Eric Steven Raymond
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[ reply ]
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)
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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