Secret sharing using the Chinese remainder theorem

Secret sharing schemes: several types

There are several types of secret sharing schemes. The most basic types are the so-called threshold schemes, where only the cardinality of the set of shares matters. In other words, given a secret S, and n shares, any set of t shares is a set with the smallest cardinality from which the secret can be recovered, in the sense that any set of t-1 shares is not enough to give S. This is known as a threshold access structure. We call such schemes (t,n) threshold secret sharing schemes, or t-out-of-n scheme.

Threshold secret sharing schemes differ from one another by the method of generating the shares, starting from a certain secret. The first ones are Shamir's threshold secret sharing scheme, which is based on polynomial interpolation in order to find S from a given set of shares, and George Blakley's geometric secret sharing scheme, which uses geometric methods to recover the secret S. Threshold secret sharing schemes based on the CRT are due to Mignotte and Asmuth-Bloom, they use special sequences of integers along with the CRT.

Chinese remainder theorem

Let k ⩾ 2 , m 1 , . . . , m k ⩾ 2 , and b 1 , . . . , b k ∈ Z . The system of congruences

has solutions in Z if and only if b i ≡ b j mod ( m i , m j ) for all 1 ⩽ i , j ⩽ k , where ( m i , m j ) denotes the greatest common divisor (GCD) of m_i and m_j. Furthermore, under these conditions, the system has a unique solution in Z/nZ where n = [ m 1 , . . . , m k ] , which denotes the least common multiple (LCM) of m 1 , . . . , m k .

Secret sharing using the CRT

Since the Chinese remainder theorem provides us with a method to uniquely determine a number S modulo k-many relatively prime integers m 1 , m 2 , . . . , m k , given that S < ∏ i = 1 k m i , then, the idea is to construct a scheme that will determine the secret S given any k shares (in this case, the remainder of S modulo each of the numbers m_i), but will not reveal the secret given less than k of such shares.

Ultimately, we choose n relatively prime integers m 1 < m 2 < . . . < m n such that S is smaller than the product of any choice of k of these integers, but at the same time is greater than any choice of k-1 of them. Then the shares s 1 , s 2 , . . . , s n are defined by s i = S mod   m i for i = 1 , 2 , . . . , n . In this manner, thanks to the CRT, we can uniquely determine S from any set of k or more shares, but not from less than k. This provides the so-called threshold access structure.

This condition on S can also be regarded as

Since S is smaller than the smallest product of k of the integers, it will be smaller than the product of any k of them. Also, being greater than the product of the greatest k − 1 integers, it will be greater than the product of any k − 1 of them.

There are two Secret Sharing Schemes that utilize essentially this idea, Mignotte's and Asmuth-Bloom's Schemes, which are explained below.

Mignotte's threshold secret sharing scheme

As said before, Mignotte's threshold secret sharing scheme uses, along with the CRT, special sequences of integers called the (k,n)-Mignotte sequences which consist of n integers, pairwise coprime, such that the product of the smallest k of them is greater than the product of the k − 1 biggest ones. This condition is crucial because the scheme is built on choosing the secret as an integer between the two products, and this condition ensures that at least k shares are needed to fully recover the secret, no matter how they are chosen.

Formally, let 2 ≤ k ≤ n be integers. A (k,n)-Mignotte sequence is a strictly increasing sequence of positive integers m 1 < ⋯ < m n , with ( m i , m j ) = 1 for all 1 ≤ i < j ≤ n, such that m n − k + 2 ⋯ m n < m 1 ⋯ m k . We call this range the authorized range. We build a (k,n)-threshold secret sharing scheme as follows: We choose the secret S as a random integer in the authorized range. We compute, for every 1 ≤ i ≤ n, the reduction modulo m_i of S that we call s_i, these are the shares. Now, for any k different shares s i 1 , … , s i k , we consider the system of congruences:

By the Chinese remainder theorem, since m i 1 , … , m i k are pairwise coprime, the system has a unique solution modulo m i 1 ⋯ m i k . By the construction of our shares, this solution is nothing but the secret S to recover.

Asmuth-Bloom's threshold secret sharing scheme

This scheme also uses special sequences of integers. Let 2 ≤ k ≤ n be integers. We consider a sequence of pairwise coprime positive integers m 0 < . . . < m n such that m 0 . m n − k + 2 . . . m n < m 1 . . . m k . For this given sequence, we choose the secret S as a random integer in the set Z/m₀Z.

We then pick a random integer α such that S + α . m 0 < m 1 . . . m k . We compute the reduction modulo m_i of S + α . m 0 , for all 1 ≤ i ≤ n, these are the shares I i = ( s i , m i ) . Now, for any k different shares I i 1 , . . . , I i k , we consider the system of congruences:

{ x ≡ s i 1   mod   m i 1 ⋮ x ≡ s i k   mod   m i k

By the Chinese remainder theorem, since m i 1 , . . . , m i k are pairwise coprime, the system has a unique solution S₀ modulo m i 1 . . . m i k . By the construction of our shares, the secret S is the reduction modulo m₀ of S₀.

It is important to notice that the Mignotte (k,n)-threshold secret-sharing scheme is not perfect in the sense that a set of less than k shares contains some information about the secret. The Asmuth-Bloom scheme is perfect: α is independent of the secret S and

∏ i = n − k + 2 n m i α } < ∏ i = 1 k m i m 0

Therefore α can be any integer modulo

This product of k − 1 moduli is the largest of any of the n choose k − 1 possible products, therefore any subset of k − 1 equivalences can be any integer modulo its product, and no information from S is leaked.

Example

The following is an example on the Asmuth-Bloom's Scheme. For practical purposes we choose small values for all parameters. We choose k=3 and n=4. Our pairwise coprime integers being m 0 = 3 , m 1 = 11 , m 2 = 13 , m 3 = 17 and m 4 = 19 . They satisfy the Asmuth-Bloom required sequence because 3 ⋅ 17 ⋅ 19 < 11 ⋅ 13 ⋅ 17 .

Say our secret S is 2. Pick α = 51 , satisfying the required condition for the Asmuth-Bloom scheme. Then 2 + 51 ⋅ 3 = 155 and we compute the shares for each of the integers 11, 13, 17 and 19. They are respectively 1, 12, 2 and 3. We consider one possible set of 3 shares: among the 4 possible sets of 3 shares we take the set {1,12,2} and show that it recovers the secret S=2. Consider the following system of congruences:

To solve the system, let M = 11 ⋅ 13 ⋅ 17 . From a constructive algorithm for solving such a system, we know that a solution to the system is x 0 = 1 ⋅ e 1 + 12 ⋅ e 2 + 2 ⋅ e 3 , where each e_i is found as follows:

By Bézout's identity, since ( m i , M / m i ) = 1 , there exist positive integers r_i and s_i, that can be found using the Extended Euclidean algorithm, such that r i . m i + s i . M / m i = 1 . Set e i = s i ⋅ M / m i .

From the identities 1 = 1 ⋅ 221 − 20 ⋅ 11 = ( − 5 ) ⋅ 187 + 72 ⋅ 13 = 5 ⋅ 143 + ( − 42 ) ⋅ 17 , we get that e 1 = 221 , e 2 = − 935 , e 3 = 715 , and the unique solution modulo 11 ⋅ 13 ⋅ 17 is 155. Finally, S = 155 ≡ 2 mod 3 .