# Shamir's Secret Sharing

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Shamir's Secret Sharing is an algorithm in cryptography created by Adi Shamir. It is a form of secret sharing, where a secret is divided into parts, giving each participant its own unique part, where some of the parts or all of them are needed in order to reconstruct the secret.

Counting on all participants to combine together the secret might be impractical, and therefore sometimes the threshold scheme is used where any $k$ of the parts are sufficient to reconstruct the original secret.

## Mathematical definition

This scheme is called $\left(k,n\right)$ threshold scheme. If $k=n$ then all participants are required to reconstruct the secret.

## Shamir's secret-sharing scheme One can draw an infinite number of polynomials of degree 2 through 2 points. 3 points are required to define a unique polynomial of degree 2. This image is for illustration purposes only — Shamir's scheme uses polynomials over a finite field, not representable on a 2-dimensional plane.

The essential idea of Adi Shamir's threshold scheme is that 2 points are sufficient to define a line, 3 points are sufficient to define a parabola, 4 points to define a cubic curve and so forth. That is, it takes $k\,\!$ points to define a polynomial of degree $k-1\,\!$ .

## Usage

### Example

The following example illustrates the basic idea. Note, however, that calculations in the example are done using integer arithmetic rather than using finite field arithmetic. Therefore the example below does not provide perfect secrecy and is not a true example of Shamir's scheme. So we'll explain this problem and show the right way to implement it (using finite field arithmetic).

#### Preparation

We wish to divide the secret into 6 parts $(n=6)\,\!$ , where any subset of 3 parts $(k=3)\,\!$ is sufficient to reconstruct the secret. At random we obtain two ($k-1$ ) numbers: 166 and 94.

Our polynomial to produce secret shares (points) is therefore:

#### Reconstruction

In order to reconstruct the secret any 3 points will be enough.

We will compute Lagrange basis polynomials:

Therefore

Recall that the secret is the free coefficient, which means that $S=1234\,\!$ , and we are done.

##### Problem

Although this method works fine, there is a security problem: Eve wins a lot of information about $S$ with every $D_{i}$ that she finds.

Suppose that she finds the 2 points $D_{0}=(1,1494)$ and $D_{1}=(2,1942)$ , she still doesn't have $k=3$ points so in theory she shouldn't have won anymore info about $S$ . But she combines the info from the 2 points with the public info: $n=6,k=3,f(x)=a_{0}+a_{1}x+\dots +a_{k-1}x^{k-1},a_{0}=S,a_{i}\in \mathbb {N}$ and she : Template:Ordered list $S\in [1046,1048,\dots ,1342,1344]$ . She now only has 150 numbers to guess from instead of a infinite number of natural numbers.

#### Solution

This problem can be fixed by using finite field arithmetic in a field of size $p\in \mathbb {P} :p>S,p>n$ .

Everyone that receives a point also has to know the value of $p$ so it's publicly known so you should choose a value for $p$ that is not too low because Eve knows $p>S\Rightarrow {}S\in {[0,1,\dots ,p-2,p-1]}$ , so the lower you choose $p$ , the lower the number of possible values Eve has to guess from to get $S$ .

You should also not choose it too high because Eve knows that the chance for $f(x){\pmod {p}}=f(x)$ increases with a higher $p$ and she can use the procedure from the original problem to guess $S$ (although now, instead of being sure of the 150 possible values, they just have a increased chance of being valid compared to the other natural numbers)

#### Javascript example

var prime = 257;

/* Split number into the shares */
function split(number, available, needed) {
var coef = [number, 166, 94], x, exp, c, accum, shares = [];
/* Normally, we use the line:
* for(c = 1, coef = number; c < needed; c++) coef[c] = Math.floor(Math.random() * (prime  - 1));
* where (prime - 1) is the maximum allowable value.
* However, to follow this example, we hardcode the values:
* coef = [number, 166, 94];
* For production, replace the hardcoded value with the random loop
* For each share that is requested to be available, run through the formula plugging the corresponding coefficient
* The result is f(x), where x is the byte we are sharing (in the example, 1234)
*/
for(x = 1; x <= available; x++) {
/* coef = [1234, 166, 94] which is 1234x^0 + 166x^1 + 94x^2 */
for(exp = 1, accum = coef; exp < needed; exp++) accum = (accum + (coef[exp] * (Math.pow(x, exp) % prime) % prime)) % prime;
/* Store values as (1, 132), (2, 66), (3, 188), (4, 241), (5, 225) (6, 140) */
shares[x - 1] = [x, accum];
}
return shares;
}

/* Gives the decomposition of the gcd of a and b.  Returns [x,y,z] such that x = gcd(a,b) and y*a + z*b = x */
function gcdD(a,b) {
if (b == 0) return [a, 1, 0];
else {
var n = Math.floor(a/b), c = a % b, r = gcdD(b,c);
return [r, r, r-r*n];
}
}

/* Gives the multiplicative inverse of k mod prime.  In other words (k * modInverse(k)) % prime = 1 for all 1 <= k < prime */
function modInverse(k) {
k = k % prime;
var r = (k < 0) ? -gcdD(prime,-k) : gcdD(prime,k);
return (prime + r) % prime;
}

/* Join the shares into a number */
function join(shares) {
var accum, count, formula, startposition, nextposition, value, numerator, denominator;
for(formula = accum = 0; formula < shares.length; formula++) {
/* Multiply the numerator across the top and denominators across the bottom to do Lagrange's interpolation
* Result is x0(2), x1(4), x2(5) -> -4*-5 and (2-4=-2)(2-5=-3), etc for l0, l1, l2...
*/
for(count = 0, numerator = denominator = 1; count < shares.length; count++) {
if(formula == count) continue; // If not the same value
startposition = shares[formula];
nextposition = shares[count];
numerator = (numerator * -nextposition) % prime;
denominator = (denominator * (startposition - nextposition)) % prime;
}
value = shares[formula];
accum = (prime + accum + (value * numerator * modInverse(denominator))) % prime;
}
return accum;
}

var sh = split(129, 6, 3) /* split the secret value 129 into 6 components - at least 3 of which will be needed to figure out the secret value */
var newshares = [sh, sh, sh]; /* pick any any selection of 3 shared keys from sh */


Some of the useful properties of Shamir's $\left(k,n\right)\,\!$ threshold scheme are:
3. Extensible: When $k\,\!$ is kept fixed, $D_{i}\,\!$ pieces can be dynamically added or deleted without affecting the other pieces.