bugtraq - Re: Joint encryption?

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Message-ID: <4217B6CB.7090709@comcast.net>
Date: Sat, 19 Feb 2005 16:59:39 -0500
From: John Richard Moser <nigelenki@...cast.net>
To: davids@...master.com
Cc: bugtraq@...urityfocus.com
Subject: Re: Joint encryption?

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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
> 
> 
> 

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