The author of this paper is [Adi Shamir](https://en.wikipedia.org/w...
In the scientists and cabinet example:
- $n = 11$ Number of scient...
It is important to note that the security of such threshold schemes...
Another good example would be:
You have a nuclear arsenal and th...
A combinatorial argument for this would be:
We need a *lock for ...
The main idea is:
- $2$ points determine a line
- $3$ points dete...
Programming R. Rivest
Techniques Editor
How to Share a Secret
Adi Shamir
Massachusetts Institute of Technology
In this paper we show how to divide data D into n
pieces in such a way that D is easily reconstructable
from any k pieces, but even complete knowledge of
k - 1 pieces reveals absolutely no information about D.
This technique enables the construction of robust key
management schemes for cryptographic systems that
can function securely and reliably even when misfor-
tunes destroy half the pieces and security breaches ex-
pose all but one of the remaining pieces.
Key Words and Phrases: cryptography, key manage-
ment, interpolation
CR Categories: 5:39, 5.6
1. Introduction
In [4], Liu considers the following problem:
Eleven scientists are working on a secret project. They wish to lock
up the documents in a cabinet so that the cabinet can be opened if
and only if six or more of the scientists are present. What is the
smallest number of locks needed? What is the smallest number of
keys to the locks each scientist must carry?
It is not hard to show that the minimal solution uses 462
locks and 252 keys per scientist. These numbers are
clearly impractical, and they become exponentially
worse when the number of scientists increases.
In this paper we generalize the problem to one in
which the secret is some data D (e.g., the safe combina-
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Author's present address: A. Shamir, Laboratory for Computer
Science, Massachusetts Institute of Technology, Cambridge, MA
02139.
This research was supported by the Office of Naval Research under
contract no. N00014-76-C-0366.
@1979 ACM 0001-0782/79/1100-0612 $00.75.
tion) and in which nonmechanical solutions (which
manipulate this data) are also allowed. Our goa ! is to
divide D into n pieces D, ..... D n in such a way that:
(1) knowledge of any k or more D i pieces makes D
easily computable;
(2) knowledge of any k- 1 or fewer Di pieces leaves D
completely undetermined (in the sense that all its
possible values are equally likely).
Such a scheme is called a
(k, n) threshold scheme.
Efficient threshold schemes can be very helpful in the
management of cryptographic keys. In order to protect
data we can encrypt it, but in order to protect the encryp-
tion key we need a different method (further encryptions
change the problem rather than solve it). The most
secure key management scheme keeps the key in a single,
well-guarded location (a computer, a human brain, or a
safe). This scheme is highly unreliable since a single
misfortune (a computer breakdown, sudden death, or
sabotage) can make the information inaccessible. An ob-
vious solution is to store multiple copies of the key at dif-
ferent locations, but this increases the danger of security
breaches (computer penetration~ betrayal, or human er-
rors). By using a (k, n) threshold scheme with n = 2k- 1
we get a very robust key management scheme: We can
recover the original key even when [n/2J = k- 1 of the n
pieces are destroyed, but our opponents cannot
reconstruct the key even when security breaches expose
[n/21 = k- 1 of the remaining k pieces.
In other applications the tradeoff is not between
secrecy and reliability, but between safety and conve-
nience of use. Consider, for example, a company that
digitally signs all its checks (see RSA [5]). If each ex-
ecutive is given a copy of the company's secret signature
key, the system is convenient but easy to misuse. If the
cooperation of all the company's executives is necessary
in order to sign each check, the system is safe but in-
convenient. The standard solution requires at least three
signatures per check, and it is easy to implement with a
(3, n) threshold scheme. Each executive is given a small
magnetic card with one Di piece, and the company's
signature generating device accepts any three of them in
order to generate (and later destroy) a temporary copy of
the actual signature key D. The device does not contain
any secret information and thus it need not be protected
against inspection. An unfaithful executive must have at
least two accomplices in order to forge the company's
signature in this scheme.
Threshold schemes are ideally suited to applications in
which a group of mutually suspicious individuals with
conflicting interests must cooperate. Ideally we would
like the cooperation to be based on mutual consent, but
the veto power this mechanism gives to each member can
paralyze the activities of the group. By properly choos-
ing the k and n parameters we can give any sufficiently
large majority the authority to take some action while
giving any sufficiently large minority the power to block
it.
612
Communications November 1979
of Volume 22
the ACM Number 11
2. A Simple (k, n)
Threshold Scheme
Our scheme is based on polynomial' interpolation:
given k points in the 2-dimensional plane (x,, y,) .....
(xk, Yk). with distinct xi's , there is one and only one
polynomial
q(x)
of degree k - 1 such that q(x)
=yi
for all
i. Without loss of generality, we can assume that the data
D is (or can be made) a number. To divide it into pieces
D~, we pick a random k-1 degree polynomial
q(x)=ao+alx+ ... ak_ixk-~
in which
ao=D ,
and
evaluate:
D~ = q(1) ..... D i = q(i) ..... D n = q(n).
Given any subset of k of these D~ values (together with
their identifying indices), we can find the coefficients of
q(x) by interpolation, and then evaluate
D=q(O).
Knowledge of just k- 1 of these values, on the other
hand, does not suffice in order to calculate D.
To make this claim more precise, we use modular
arithmetic instead of real arithmetic. The set of integers
modulo a prime number p forms a field in which inter-
polation is possible. Given an integer valued data D, we
pick a prime p which is bigger than both D and n. The
coefficients a~ ..... ak_~ in
q(x)
are randomly chosen
from a uniform distribution over the integers in [0, p),
and the values D~ ..... Dn are computed modulo p.
Let us now assume that k-1 of these n pieces are
revealed to an opponent. For each candidate value D' in
[0, p) he can construct one and only one polynomial
q '(x) of degree k- 1 such that q '(0) =D' and q '(0 =D~
for the k- 1 given arguments. By construction, these p
possible polynomials are equally likely, and thus there is
abolutely nothing the opponent can deduce about the
real value of D.
Efficient
O(n
log 2 n) algorithms for polynomial evalu-
ation and interpolation are discussed in [1] and [3], but
even the straightforward quadratic algorithms are fast
enough for practical key management schemes. If the
number D is long, it is advisable to break it into shorter
blocks of bits (which are handled separately) in order to
avoid multiprecision arithmetic operations. The blocks
cannot be arbitrarily short, since the smallest usable
value of p is n + 1 (there must be at least n + 1 distinct
arguments in [0, p) to evaluate
q(x)
at). However, this is
not a severe limitation since sixteen bit modulus (which
can be handled by a cheap sixteen bit arithmetic unit)
suffices for applications with up to 64,000 D~ pieces.
Some of the useful properties of this (k, n) threshold
scheme (when compared to the mechanical locks and
keys solutions) are:
(1) The size of each piece does not exceed the size of the
original data.
(2) When k is kept fixed, D~ pieces can be dynamically
added or deleted (e.g., when executives join or leave
the company) without affecting the other D i pieces.
(A piece is deleted only when a leaving executive
makes it completely inaccessible, even to himself.)
(3) It is easy to change the D i pieces without changing
the original data D--all we need is a new polynomial
q(x)
with the same free term. A frequent change of
this type can greatly enhance security since the pieces
exposed by security breaches cannot be accumulated
unless all of them are values of the same edition of
the
q(x)
polynomial.
(4) By using tuples of polynomial values as Di pieces, we
can get a hierarchical scheme in which the number of
pieces needed to determine D depends on their im-
portance. For example, if we give the company's
president three values of
q(x),
each vice-president
two values of
q(x),
and each executive one value of
q(x),
then a (3, n) threshold scheme enables checks to
be signed either by any three executives, or by any
two executives one of whom is a vice-president, or by
the president alone.
A different (and somewhat less efficient) threshold
scheme was recently developed by G.R. Blakley [2].
Received April 1979; revised September 1979
References
1. Aho, A., Hopcroft, J., and Ullman, J. The Design and Analysis
of Computer AIgorithms. Addison-Wesley, Reading, Mass., 1974.
2. Blakley, G.R. Safeguarding cryptographic keys. Proc. AFIPS
1979 NCC, Vol. 48, Arlington, Va., June 1979, pp. 313-317.
3. Knuth, D. The Art of Computer Programming, Vol. 2:
SeminumericalAlgorithms. Addison-Wesley, Reading, Mass., 1969.
4. Liu, C.L. Introduction to Combinatorial Mathematics. McGraw-
Hill, New York, 1968.
5. Rivest, R., Shamir, A., and Adleman, L. A method for obtaining
digital signatures and public-key cryptosystems. Comm. A CM 21, 2
(Feb. 1978), 120-126.
LThe polynomials can be replaced by any other collection of func-
tions which are easy to evaluate and to interpolate.
613
Communications November 1979
of Volume 22
the ACM Number 11
In the scientists and cabinet example:
- $n = 11$ Number of scientists
- $k = 6$ Threshold of scientists needed to be present in order to open the safe
A combinatorial argument for this would be:
We need a *lock for each possible combination of 6 scientists out of the 11 total*:
$$ \dbinom{11}{6} = 462\ locks $$
Each scientist should have the keys for all the locks that correspond to a combination of scientists in which he is included. That is:
$$ \dbinom{10}{5} = 252\ keys $$
*Each scientist therefore gets exactly 252 keys*. Therefore, if you get 6 scientists together, they will be able to open all the locks as each one of the locks will correspond to a combination of scientists in which at least one of the present 6 is included. If you get just 5 scientists (or less) they will not be able to open some of the locks - one lock will correspond to the combination of the 6 remaining scientists that are not present.
It is important to note that the security of such threshold schemes is dependent on each of the pieces $D_{i}$ being held by different entities that are at least to an extent somewhat independent from each other. Say if you are using a threshold scheme to activate a nuclear arsenal, and all of the generals involved in the threshold scheme live in the same block, it would be easier for an attacker to compromise such scheme.
The author of this paper is [Adi Shamir](https://en.wikipedia.org/wiki/Adi_Shamir) an Israeli cryptographer. Shamir is one of the most important and prolific cryptographers of our age. Shamir received a BSc in Mathematics from the Tel Aviv University in 1973 and a PhD in Computer Science from the Weismann Institute in Rehovot, Israel in 1977. He did research at MIT from 1977 to 1980 and there he co-invented the **RSA** public key encryption algorithm (the S in RSA stands for Shamir). In 2002 he was awarded the Turing Award.
Another good example would be:
You have a nuclear arsenal and there is a key that activates the arsenal. That key is going to be shared amongst 8 high ranking generals. You want to make sure at least 4 of your 8 high ranking generals are present in order to activate the nuclear arsenal. This way you can afford to loose up to 4 generals and still be able to use the nuclear arsenal and, at the same time, even if one general goes crazy and decides to activate the nuclear arsenal himself he can’t do it (he would need the support of 3 other generals).
The question is:
How do you go about sharing / splitting the key that activates the nuclear arsenal?
The main idea is:
- $2$ points determine a line
- $3$ points determine a quadratic
- $\ldots$
- $k$ points determine a degree $(k-1)$ curve
The secret is the $y$ intercept. Let’s look at an example for a line.
Alice wants to split a secret between Bob, Carlos and David.
The secret is 8.
Alice chooses a secret line $f$ such that $f(0) = 8$ and gives Bob, Carlos and David one point along the curve (other than $(0,8)$).
Given just one point, Bob, Carlos and David cannot know what the secret function is and therefore have no way of finding out what the secret is. However if Bob and Carlos get together they can interpolate $f$ and easily compute the secret value $f(0) = 8$.
This is a $(2, 3)$ threshold system. You have 3 participants ($n$) and you need at least 2 ($k$) in order to get the secret.