Shamir secret sharing and information theoretic security

R.A. Hettinga rah at shipwright.com
Tue Feb 17 18:03:07 EST 2009



Begin forwarded message:

From: Sarad AV <jtrjtrjtr2001 at yahoo.com>
Date: February 17, 2009 9:51:09 AM EST
To: cypherpunks at al-qaeda.net
Subject: Shamir secret sharing and information theoretic security

hi,


I was going through the wikipedia example of shamir secret sharing  
which says it is information theoretically secure.

http://en.wikipedia.org/wiki/Shamir%27s_Secret_Sharing

In the example in that url, they have a polynomial
f(x) = 1234 + 166.x + 94.x^2

they construct 6 points from the polynomial
(1,1494);(2,1942);(3,2578);(4,3402);(5,4414);(6,5615)

the secret here is S=1234. The threshold k=3 and the number of  
participants n=6.

If say, first two users collude then
1494 = S + c1 .1 + c2.1
1942 = S + c1 .2 + c2.2

clearly, one can start making inferences about the sizes of the  
unknown co-efficients c1 and c2 and S.

However, it is said in the URL above that Shamir secret is information  
theoretically secure

in the url below they say
http://en.wikipedia.org/wiki/Information_theoretic_security
"Secret sharing schemes such as Shamir's are information theoretically  
secure (and in fact perfectly secure) in that less than the requisite  
number of shares of the secret provide no information about the secret."

how can that be true? we already are able to make inferences.

Moreover say that, we have 3 planes intersecting at a single point in  
euclidean space, where each plane is a secret share(Blakely's scheme).  
With 2 plane equations, we cannot find the point of intersection but  
we can certainly narrow down to the line where the planes intersect.  
There is information loss about the secret.


from this it appears that Shamir's secret sharing scheme leaks  
information from its shares but why is it then considered information  
theoretically secure?

They do appear to leak information as similar to k-threshold schemes  
using chinese remainder theorem.

what am i missing?

Thanks,
Sarad.


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