Shamir's Secret Sharing

Shamir's Secret Sharing (SSS) is an efficient secret sharing algorithm for distributing private information (the "secret") in such a way that no individual holds intelligible information about the secret. To achieve this, the secret is converted into parts (the "shares") from which the secret can be reassembled when a sufficient number of shares are combined but not otherwise. SSS has the unusual property of information theoretic security, meaning an adversary without enough shares cannot reconstruct the secret even with infinite time and computing capacity. A standard SSS specification for cryptocurrency wallets has been widely implemented.

High-level explanation

SSS is used to secure a secret in a distributed way, most often to secure other encryption keys. The secret is split into multiple shares, which individually do not give any information about the secret.

To unlock a secret secured by SSS a minimum number of shares are needed, called the ''threshold''. No additional information about the secret can be gained by examining any number of shares fewer than the threshold (a property called perfect secrecy)''.'' In this sense, SSS is a generalisation of the one-time pad (which can be viewed as SSS with a two-share threshold and two shares in total).

Application example

A company needs to secure their vault's code. A single person knowing the code could act dishonestly or be unavailable when the vaults needs to be opened.

SSS can be used in this situation to generate shares from the vault's code which are distributed to executives in the Company. The selected threshold and number of shares given to each executive can be selected such that the vault is accessible only by (groups of) authorized individuals. If less than the threshold of shares were compromised, these shares alone would not be enough to determine the code.

Properties and weaknesses

SSS has useful properties, but also weaknesses that mean there are some situations where it should not be used.

Useful properties include:

# Secure: The scheme has Information theoretic security.
# Minimal: The size of each piece does not exceed the size of the original data.
# Extensible: For any given threshold, shares can be dynamically added or deleted without affecting existing shares
# Dynamic: Security can be easily enhanced without changing the secret, but by changing the polynomial occasionally (keeping the same free term) and constructing new shares for the participants.
# Flexible: In organizations where hierarchy is important, each participant can be issued different numbers of shares according to their importance inside the organization. For instance, with a ''threshold'' of 3, the president could unlock the safe alone if given three shares, while three secretaries with one share each must combine their shares to unlock the safe.

Weaknesses include:

# No verifiable secret sharing: During the share reassembly process, SSS does not have a way to verify the correctness of each share being used. Verifiable secret sharing aims to verify that shareholders are honest and not submitting fake shares.
# Single point of failure: The secret must exist in one single place when it is split into shares, and again in one place when it is reassembled. These are attack points, and other schemes including multisignature eliminate at least one of these single points of failure.

History

Adi Shamir first formulated the scheme in 1979.

Mathematical principle

The essential idea of the scheme is based on the Lagrange interpolation theorem, specifically that $k\,\!$ points is enough to uniquely determine a polynomial of degree less than or equal to $k-1\,\!$. For instance, 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.

Mathematical formulation

Shamir's Secret Sharing is an ideal and perfect ''$\left(k,n\right)$-threshold scheme'' based on polynomial interpolation over finite fields. In such a scheme, the aim is to divide a secret $S$ (for example, the combination to a safe) into $n$ pieces of data $S_1,\ldots,S_n$ (known as ''shares'') in such a way that:
# Knowledge of any $k$ or more shares $S_i$ makes $S$ easily computable. That is, the complete secret $S$ can be reconstructed from any combination of $k$ shares of data.
# Knowledge of any $k-1$ or fewer shares $S_i$ leaves $S$ completely undetermined, in the sense that the possible values for $S$ seem as likely with knowledge of up to $k-1$ shares as with knowledge of $0$ shares. The secret $S$ cannot be reconstructed with fewer than $k$ shares.

If $n=k$, then every piece of the original secret $S$ is required to reconstruct the secret.

Assume that the secret $S$ can be represented as an element $a_0$ of a finite field $GF(q)$ (where $q$ is larger than the number of shares being generated). Randomly choose $k-1$ elements, $a_1,\cdots,a_\,\!$, from $GF(q)$ and construct the polynomial $f\left(x\right)=a_0+a_1x+a_2x^2+a_3x^3+\cdots+a_x^\,\!$. Compute any $n\,\!$ points out on the curve, for instance set $i=1,\ldots,n\,\!$ to find points $\left(i,f\left(i\right)\right)\,\!$. Every participant is given a point (a non-zero input to the polynomial, and the corresponding output). Given any subset of $k\,\!$ of these pairs, $a_0$ can be obtained using interpolation, with one possible formula for doing so being $a_0 = f(0) = \sum_^ y_j \prod_^ \frac$, where the list of points on the polynomial is given as k pairs of the form $(x_i, y_i)$. Note that $f(0)$ is equal to the first coefficient of polynomial $f(x)$.

Example calculation

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 to make the idea easier to understand. Therefore the example below does not provide perfect secrecy and is not a proper example of Shamir's scheme. The next example will explain the problem.

Preparation

Suppose that the secret to be shared is 1234 $(S=1234)\,\!$.

In this example, the secret will be split into 6 shares $(n=6)\,\!$, where any subset of 3 shares $(k=3)\,\!$ is sufficient to reconstruct the secret. $k-1=2$ numbers are taken at random. Let them be 166 and 94.

: This yields coefficients $(a_0=1234;a_1=166;a_2=94),\,\!$ where $a_0$ is the secret

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

: $f(x)=1234+166x+94x^2\,\!$

Six points $D_=(x, f(x))$ from the polynomial are constructed as:

: $D_0=(1, 1494);D_1=(2, 1942);D_2=(3, 2578);D_3=(4, 3402); D_4= (5, 4414);D_5=(6, 5614)\,\!$

Each participant in the scheme receives a different point (a pair of $x\,\!$ and $f(x)\,\!$). Because $D_$ is used instead of $D_x$ the points start from $(1, f(1))$ and not $(0, f(0))$. This is necessary because $f(0)$ is the secret.

Reconstruction

In order to reconstruct the secret, any 3 points are sufficient

Consider using the 3 points$\left(x_0,y_0\right)=\left(2,1942\right);\left(x_1,y_1\right)=\left(4,3402\right);\left(x_2,y_2\right)=\left(5,4414\right)\,\!$.

Computing the Lagrange basis polynomials:

: $\ell_0(x)=\frac\cdot\frac=\frac\cdot\frac=\fracx^2-\fracx+\frac\,\!$

: $\ell_1(x)=\frac\cdot\frac=\frac\cdot\frac=-\fracx^2+\fracx-5\,\!$

: $\ell_2(x)=\frac\cdot\frac=\frac\cdot\frac=\fracx^2-2x+\frac\,\!$

Using the formula for polynomial interpolation, $f(x)$ is:

:
\begin
f(x) & =\sum_^2 y_j\cdot\ell_j(x) \\ pt& =y_0\ell_0(x)+y_1\ell_1(x)+y_2\ell_2(x) \\ pt& =1942\left(\fracx^2-\fracx+\frac\right) + 3402\left(-\fracx^2+\fracx-5\right) + 4414\left(\fracx^2-2x+\frac\right) \\ pt& =1234+166x+94x^2
\end

Recalling that the secret is the free coefficient, which means that $S=1234\,\!$, and the secret has been recovered.

Computationally efficient approach

Using polynomial interpolation to find a coefficient in a source polynomial $S=f(0)$ using Lagrange polynomials is not efficient, since unused constants are calculated.

Considering this, an optimized formula to use Lagrange polynomials to find $f(0)$ is defined as follows:

: $f(0) = \sum_^ y_j \prod_^ \frac$

Problem of using integer arithmetic

Although the simplified version of the method demonstrated above, which uses integer arithmetic rather than finite field arithmetic, works, there is a security problem: Eve gains 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 does not have $k=3$ points, so in theory she should not have gained any more information about $S$. But she could combine the information from the 2 points with the public information: $n=6, k=3, f(x)=a_0+a_1x+\cdots+a_x^, a_0=S, a_i\in\mathbb$. Doing so, Eve could perform the following algebra:

# Fill the formula for $f(x)$ with $S$ and the value of $k: f(x)=S+a_1x+\cdots+a_x^\Rightarrowf(x)=S+a_1x+a_2x^2$
# Fill (1) with the values of $D_0$'s $x$ and $f(x): 1494=S+a_1+a_1^2\Rightarrow1494=S+a_1+a_2$
# Fill (1) with the values of $D_1$'s $x$ and $f(x): 1942=S+a_2+a_2^2\Rightarrow1942=S+2a_1+4a_2$
# Subtract (3)-(2): $(1942-1494)=(S-S)+(2a_1-a_1)+(4a_2-a_2)\Rightarrow448=a_1+3a_2$ and rewrite this as $a_1=448-3a_2$. Eve knows that $a_2\in\mathbb$ so she starts replacing $a_2$ in (4) with 0, 1, 2, 3, ... to find all possible values for $a_1$:
## $a_2=0\rightarrowa_1=448-3\times0=448$
##$a_2=1\rightarrowa_1=448-3\times1=445$
##$a_2=2\rightarrowa_1=448-3\times2=442$
##$\,\,\,\,\,\,\,\,\,\vdots$
##$a_2=148\rightarrowa_1=448-3\times148=4$
##$a_2=149\rightarrowa_1=448-3\times149=1$
#After checking $a_2=149$, she stops because would get negative values for $a_1$ with larger values of $a_2$ (which is impossible because $a_1\in\mathbb$). Eve can now conclude a_2\in ,1,\dots,148,149/math>
#Now, Eve can replace $a_1$ by (4) in (2): $1494=S+(448-3a_2)+a_2\RightarrowS=1046+2a_2$. Now, replacing $a_2$ in (6) by the values found in (5), she gets S\in 046+2\times0,1046+2\times1,\dots,1046+2\times148,1046+2\times149/math> which leads her to the information: S\in 046,1048,\dots,1342,1344
Eve now only has 150 numbers to guess from instead of an infinite quantity of natural numbers.

Solution using finite field arithmetic

Geometrically this attack exploits the fact that the order of the polynomial is known and thus gives information into the paths the polynomial take between known points. This reduces possible values of unknown points since the points must lie on a smooth curve, and the polynomial must have coefficients that are natural numbers.

This problem can be fixed by using finite field arithmetic. A field of size $p\in\mathbb:p>S,p>n$ is used. The figure shows a polynomial curve over a finite field. In contrast to a smooth curve it appears disorganised and disjointed.

In practice this is only a small change. A prime $p$ must be chosen that is bigger than the number of participants and every $a_i$ (including $a_0=S$). The points on the polynomial must also be calculated as $(x, f(x)\bmod p)$ instead of $(x, f(x))$.

Everybody who receives a point must also know the value of $p$, so it is considered to be publicly known. Therefore, one should select a value for $p$ that is not too low to prevent attacks where somebody guesses every possible value for $S$.

For this example, choose $p=1613$, so the polynomial becomes $f(x)=1234+166x+94x^2\bmod$ which gives the points: $(1,1494);(2,329);(3,965);(4,176);(5,1188);(6,775)$

This time Eve doesn't gain any information when she finds a $D_x$ (until she has $k$ points).

Suppose again that Eve finds $D_0=\left(1,1494\right)$ and $D_1=\left(2,329\right)$, and the public information is: $n=6, k=3, p=1613, f(x)=a_0+a_1x+\dots+a_x^\mod, a_0=S, a_i\in\mathbb$. Attempting the previous attack, Eve can:

# Fill the $f(x)$-formula with $S$ and the value of $k$ and $p$: $f(x)=S+a_1x+\dots+a_x^\mod1613\Rightarrowf(x)=S+a_1x+a_2x^2-1613m_x: m_x\in\mathbb$
# Fill (1) with the values of $D_0$'s $x$ and $f(x): 1494=S+a_1 1+a_2 1^2-1613m_1\Rightarrow 1494=S+a_1+a_2-1613m_1$
# Fill (1) with the values of $D_1$'s $x$ and $f(x): 1942=S+a_2+a_2^2-1613m_2\Rightarrow1942=S+2a_1+4a_2-1613m_2$
# Subtracts (3)-(2): $(1942-1494)=(S-S)+(2a_1-a_1)+(4a_2-a_2)+(1613m_1-1613m_2)\Rightarrow448=a_1+3a_2+1613(m_1-m_2)$ and rewrites this as $a_1=448-3a_2-1613(m_1-m_2)$
# Using $a_2\in\mathbb$ so she starts replacing $a_2$ in (4) with 0, 1, 2, 3, ... to find all possible values for $a_1$:
## $a_2=0\rightarrowa_1=448-3\times0-1613(m_1-m_2)=448-1613(m_1-m_2)$
## $a_2=1\rightarrowa_1=448-3\times1-1613(m_1-m_2)=445-1613(m_1-m_2)$
##$a_2=2\rightarrowa_1=448-3\times2-1613(m_1-m_2)=442-1613(m_1-m_2)$
##$\,\,\,\,\,\,\,\,\,\vdots$
This time she is not able to stop because $(m_1-m_2)$ could be any integer modulo $p$ (even negative if $m_2>m_1$) so there are $p$ possible values for $a_1$. She knows that 48,445,442,\ldots/math> always decreases by 3, so if $1613$ was divisible by $3$ she could conclude a_1\in , 4, 7, \ldots/math>. However, $p$ is prime she can not conclude this. Thus, using a finite field avoids this possible attack.

Usage examples

Shamir's Secret Sharing is used in:

* Vault password manager by Hashicorp
* Passwordless authentication by Secret Double Octopus
* Key Recovery in PreVeil's email encryption
* Wallet management of cryptocurrencies

Python code

"""
The following Python implementation of Shamir's Secret Sharing is
released into the Public Domain under the terms of CC0 and OWFa:
https://creativecommons.org/publicdomain/zero/1.0/
http://www.openwebfoundation.org/legal/the-owf-1-0-agreements/owfa-1-0

See the bottom few lines for usage. Tested on Python 2 and 3.
"""

from __future__ import division
from __future__ import print_function

import random
import functools

# 12th Mersenne Prime
# (for this application we want a known prime number as close as
# possible to our security level; e.g. desired security level of 128
# bits -- too large and all the ciphertext is large; too small and
# security is compromised)
_PRIME = 2 ** 127 - 1
# The 13th Mersenne Prime is 2**521 - 1

_RINT = functools.partial(random.SystemRandom().randint, 0)

def _eval_at(poly, x, prime):
"""Evaluates polynomial (coefficient tuple) at x, used to generate a
shamir pool in make_random_shares below.
"""
accum = 0
for coeff in reversed(poly):
accum *= x
accum += coeff
accum %= prime
return accum

def make_random_shares(secret, minimum, shares, prime=_PRIME):
"""
Generates a random shamir pool for a given secret, returns share points.
"""
if minimum > shares:
raise ValueError("Pool secret would be irrecoverable.")
poly = ecret+ RINT(prime - 1) for i in range(minimum - 1) points = i, _eval_at(poly, i, prime))
for i in range(1, shares + 1) return points

def _extended_gcd(a, b):
"""
Division in integers modulus p means finding the inverse of the
denominator modulo p and then multiplying the numerator by this
inverse (Note: inverse of A is B such that A*B % p

1). This can
be computed via the extended Euclidean algorithm
http://en.wikipedia.org/wiki/Modular_multiplicative_inverse#Computation
"""
x = 0
last_x = 1
y = 1
last_y = 0
while b != 0:
quot = a // b
a, b = b, a % b
x, last_x = last_x - quot * x, x
y, last_y = last_y - quot * y, y
return last_x, last_y

def _divmod(num, den, p):
"""Compute num / den modulo prime p

To explain this, the result will be such that:
den * _divmod(num, den, p) % p

num
"""
inv, _ = _extended_gcd(den, p)
return num * inv

def _lagrange_interpolate(x, x_s, y_s, p):
"""
Find the y-value for the given x, given n (x, y) points;
k points will define a polynomial of up to kth order.
"""
k = len(x_s)
assert k

len(set(x_s)), "points must be distinct"
def PI(vals): # upper-case PI -- product of inputs
accum = 1
for v in vals:
accum *= v
return accum
nums = [] # avoid inexact division
dens = []
for i in range(k):
others = list(x_s)
cur = others.pop(i)
nums.append(PI(x - o for o in others))
dens.append(PI(cur - o for o in others))
den = PI(dens)
num = sum( divmod(nums[i* den * y_s[i">.html" ;"title="divmod(nums[i">divmod(nums[i* den * y_s[i% p, dens[i], p)
for i in range(k)])
return (_divmod(num, den, p) + p) % p

def recover_secret(shares, prime=_PRIME):
"""
Recover the secret from share points
(points (x,y) on the polynomial).
"""
if len(shares) < 3:
raise ValueError("need at least three shares")
x_s, y_s = zip(*shares)
return _lagrange_interpolate(0, x_s, y_s, prime)

def main():
"""Main function"""
secret = 1234
shares = make_random_shares(secret, minimum=3, shares=6)

print('Secret: ',
secret)
print('Shares:')
if shares:
for share in shares:
print(' ', share)

print('Secret recovered from minimum subset of shares: ',
recover_secret(shares 3)
print('Secret recovered from a different minimum subset of shares: ',
recover_secret(shares 3:)

if __name__

'__main__':
main()

See also

* Secret sharing
* Secure multi-party computation
* Lagrange polynomial
* Homomorphic secret sharing – A simplistic decentralized voting protocol.
* Two-man rule
* Partial Password

References

Further reading

*.

*{{citation, last=Benzekki, first=K., title= A Verifiable Secret Sharing Approach for Secure MultiCloud Storage, publisher=Springer, journal=In Ubiquitous Networking, series=Lecture Notes in Computer Science, location=Casablanca, volume=10542, year=2017, pages=225–234, doi=10.1007/978-3-319-68179-5_20, isbn=978-3-319-68178-8.

External links

Shamir's Secret Sharing
in the Crypto++ library

Shamir's Secret Sharing Scheme (ssss)
– a GNU GPL implementation

sharedsecret
– implementation in Go

s4
- online shamir's secret sharing tool utilizing HashiCorp's shamir secret sharing algorithm

Secret sharing
Information-theoretically secure algorithms
Articles with example JavaScript code
Articles with example Python (programming language) code