### Re: Shamir secret sharing and information theoretic security

On Feb 17, 2009, at 6:03 PM, R.A. Hettinga wrote: Begin forwarded message: From: Sarad AV jtrjtrjtr2...@yahoo.com Date: February 17, 2009 9:51:09 AM EST To: cypherpu...@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 It is. Knowing some of the coefficients, or some constraints on some of the coefficients, is just dual to knowing some of the points. Neither affects the security of the system, because the coefficients *aren't secrets* any more than the values of f() at particular points are. They are *shares* of secrets, and the security claim is that without enough shares, you have no information about the remaining shares. The argument for information-theoretic security is straightforward: An n'th degree polynomial is uniquely specified if you know its value at n+1 points - or, dually, if you know n+1 coefficients. On the other hand, *any* set of n+1 points (equivalently, any set of n+1 coefficients) corresponds to a polynomial. Taking a simple approach where the secret is the value of the polynomial at 0, given v_1, v_2, ..., v_n and *any* value v, there is a (unique) polynomial of degree at most n with p(0) = v and p(i) = v_i for i from 1 to n. Dually, the value p(0) is exactly the constant term in the polynomial. Given any fixed set of values c_1, c_2, ..., c_n, and any other value c there is obviously a polynomial p(x) = Sum_{0 to n}(c_i x^i), where c_0 = c, and indeed p(0) = c. Or ... in terms of your problem: Even if I give you, not just a pair of linear equations in c1, c2, and S, but the actual values c1 and c2 - the constant term (the secret) can still be anything at all. The description above is nominally for polynomials over the reals. It works equally for polynomials over any field - like the integers mod some prime, for example. For a finite field, there is an obvious interpretation of probability (the uniform probability distribution), and given that, no information can be interpreted in terms of the difference between your a priori and a posterio estimates of the probability that p(0) takes on any particular value, the values of p(1), ..., p(n) (and that differences is exactly 0). Because there can be no uniform probability distribution over all the reals, you can't state things in quite the same way, and information theoretic security is a bit of a vague notion. Then again, no one does computations over the reals. FP values - say, IEEE single precision - aren't a field and there are undoubtedly big biases if you try to use Shamir's technique there. (It should work over infinite-precision rationals.) -- Jerry 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. - The Cryptography Mailing List Unsubscribe by sending unsubscribe cryptography to majord...@metzdowd.com - The Cryptography Mailing List Unsubscribe by sending unsubscribe cryptography to majord...@metzdowd.com

### Re: Shamir secret sharing and information theoretic security

Is it possible that the amount of information that the knowledge of a sub-threshold number of Shamir fragments leaks in finite precision setting depends on the finite precision implementation? For example, if you know 2 of a 3 of 5 splitting and you also know that the finite precision setting in which the fragments will be used is IEEE 32-bit floating point or GNU bignum can you narrow down the search for the key relative to knowing no fragments and nothing about the finite precision implementation? - The Cryptography Mailing List Unsubscribe by sending unsubscribe cryptography to majord...@metzdowd.com

### Re: Shamir secret sharing and information theoretic security

Is it possible that the amount of information that the knowledge of a sub-threshold number of Shamir fragments leaks in finite precision setting depends on the finite precision implementation? For example, if you know 2 of a 3 of 5 splitting and you also know that the finite precision setting in which the fragments will be used is IEEE 32-bit floating point or GNU bignum can you narrow down the search for the key relative to knowing no fragments and nothing about the finite precision implementation? No, not really. Think of this simple example. We will have two shares, x and y. Let's work mod 10 to make it simple. The secret value v will be x + y mod 10. The shares are created by choosing a random value for x, and then setting y to be v - x mod 10. So for example if you want to share v = 5, and x is 9, then y will be 6: 9 + 6 = 5 mod 10. Suppose that you happen to know from other information that the secret value v is either 1 or 2. Now you learn a share value x = 5. How much have you learned about v? Nothing: you can deduce that y is either 6 or 7, but you have no way of knowing which. Whatever x had turned out to be, there would be a y value corresponding to each possible v value. Learning a share tells you nothing about v, and in general Shamir sharing, learning all but one of the needed shares similarly tells you nothing about the secret. Hal Finney - The Cryptography Mailing List Unsubscribe by sending unsubscribe cryptography to majord...@metzdowd.com

### Re: Shamir secret sharing and information theoretic security

On Tue, 17 Feb 2009, R.A. Hettinga wrote: 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 ... The scheme is defined over a finite field *not* over the integers. When Shamir's scheme is run over a finite field, it is information theoretically secure. - The Cryptography Mailing List Unsubscribe by sending unsubscribe cryptography to majord...@metzdowd.com