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The Merkle–Hellman knapsack cryptosystem was one of the earliest public key cryptosystems invented by Ralph Merkle and Martin Hellman in 1978.[1] Although its ideas are elegant, and far simpler than RSA, it has been broken.[2]

## DescriptionEdit

Merkle-Hellman is an asymmetric-key cryptosystem, meaning that for communication, two keys are required: a public key and a private key. Furthermore, unlike RSA, it is one-way -- the public key is used only for encryption, and the private key is used only for decryption. Thus it is unusable for authentication by cryptographic signing.

The Merkle-Hellman system is based on the subset sum problem (a special case of the knapsack problem). The problem is as follows: given a set of numbers Template:Math and a number b, find a subset of Template:Math, which sums to b. In general, this problem is known to be NP-complete. However, if the set of numbers (called the knapsack) is superincreasing — that is, each element of the set is greater than the sum of all the numbers before it — the problem is 'easy' and solvable in polynomial time with a simple greedy algorithm.

### Key generationEdit

In Merkle-Hellman, the keys are knapsacks. The public key is a 'hard' knapsack, and the private key is an 'easy', or superincreasing, knapsack, combined with two additional numbers, a multiplier and a modulus, which were used to convert the superincreasing knapsack into the hard knapsack. These same numbers are used to transform the sum of the subset of the hard knapsack into the sum of the subset of the easy knapsack, which is solvable in polynomial time.

### EncryptionEdit

To encrypt a message, a subset of the hard knapsack is chosen by comparing it with a set of bits (the plaintext), equal in length to the key, and making each term in the public key that corresponds to a 1 in the plaintext an element of the subset, while ignoring the terms corresponding to 0 terms in the plaintext. The elements of this subset are added together, and the resulting sum is the ciphertext.

### DecryptionEdit

Decryption is possible because the multiplier and modulus used to transform the easy, superincreasing knapsack into the public key can also be used to transform the number representing the ciphertext into the sum of the corresponding elements of the superincreasing knapsack. Then, using a simple greedy algorithm, the easy knapsack can be solved using O(n) arithmetic operations, which decrypts the message.

## Mathematical methodEdit

### Key generationEdit

To encrypt n-bit messages, choose a superincreasing sequence

w = (w1, w2, ..., wn)

of n nonzero natural numbers. Pick a random integer q, such that

$q>\sum_{i = 1}^n w_i$,

and a random integer, r, such that gcd(r,q) = 1 (i.e. r and q are coprime).

q is chosen this way to ensure the uniqueness of the ciphertext. If it is any smaller, more than one plaintext may encrypt to the same ciphertext. r must be coprime to q or else it will not have an inverse mod q. The existence of the inverse of r is necessary so that decryption is possible.

Now calculate the sequence

β = (β1, β2, ..., βn)

where

βi = rwi mod q.

The public key is β, while the private key is (w, q, r).

### EncryptionEdit

To encrypt an n-bit message

α = (α1, α2, ..., αn),

where $\alpha_i$ is the i-th bit of the message and $\alpha_i \boldsymbol{\in}${0, 1}, calculate

$c = \sum_{i = 1}^n \alpha_i \beta_i.$

The cryptogram then is c.

### DecryptionEdit

In order to decrypt a ciphertext c a receiver has to find the message bits αi such that they satisfy

$c = \sum_{i = 1}^n \alpha_i \beta_i.$

This would be a hard problem if the βi were random values because the receiver would have to solve an instance of the subset sum problem, which is known to be NP-hard. However, the values βi were chosen such that decryption is easy if the private key (w, q, r) is known.

The key to decryption is to find an integer s that is the modular inverse of r modulo q. That means s satisfies the equation s r mod q = 1 or equivalently there exist an integer k such that sr = kq + 1. Since r was chosen such that gcd(r,q)=1 it is possible to find s and k by using the Extended Euclidean algorithm. Next the receiver of the ciphertext c computes

$c'\equiv cs \pmod{q}.$

Hence

$c' \equiv cs \equiv \sum_{i = 1}^n \alpha_i \beta_i s \pmod{q}.$

Because of rs mod q = 1 and βi = rwi mod q follows

$\beta_i s\equiv w_i r s\equiv w_i\pmod{q}.$

Hence

$c' \equiv \sum_{i = 1}^n \alpha_i w_i\pmod{q}.$

The sum of all values wi is smaller than q and hence $\sum_{i = 1}^n \alpha_i w_i$ is also in the interval [0,q-1]. Thus the receiver has to solve the subset sum problem

$c' = \sum_{i = 1}^n \alpha_i w_i.$

This problem is easy because w is a superincreasing sequence. Take the largest element in w, say wk. If wk > c' , then αk = 0, if wkc' , then αk = 1. Then, subtract wk×αk from c' , and repeat these steps until you have figured out α.

## ExampleEdit

First, a superincreasing sequence w is created

w = {2, 7, 11, 21, 42, 89, 180, 354}


This is the basis for a private key. From this, calculate the sum.

$\sum w = 706$

Then, choose a number q that is greater than the sum.

q = 881


Also, choose a number r that is in the range $[1, q)$ and is coprime to q.

r = 588


The private key consists of q, w and r.

To calculate a public key, generate the sequence β by multiplying each element in w by r mod q

β = {295, 592, 301, 14, 28, 353, 120, 236}


because

2 * 588 mod 881 = 295
7 * 588 mod 881 = 592
11 * 588 mod 881 = 301
21 * 588 mod 881 = 14
42 * 588 mod 881 = 28
89 * 588 mod 881 = 353
180 * 588 mod 881 = 120
354 * 588 mod 881 = 236


The sequence β makes up the public key.

Say Alice wishes to encrypt "a". First, she must translate "a" to binary (in this case, using ASCII or UTF-8)

01100001


She multiplies each respective bit by the corresponding number in β

a = 01100001
0 * 295
+ 1 * 592
+ 1 * 301
+ 0 * 14
+ 0 * 28
+ 0 * 353
+ 0 * 120
+ 1 * 236
= 1129


She sends this to the recipient.

To decrypt, Bob multiplies 1129 by r -1 mod $q$</code> (See Modular inverse)

1129 * 442 mod 881 = 372


Now Bob decomposes 372 by selecting the largest element in w which is less than 372. Then selecting the next largest element less than the difference, until the difference is $0$:

372 - 354 = 18
18 - 11 = 7
7 - 7 = 0


The elements we selected from our private key correspond to the 1 bits in the message

01100001


When translated back from binary, this "a" is the final decrypted message.

## ReferencesEdit

1. Merkle, Ralph and Martin Hellman, "Hiding information and signatures in trapdoor knapsacks," Information Theory, IEEE Transactions on, vol.24, no.5, pp. 525-530, Sep 1978 URL: http://ieeexplore.ieee.org/search/freesrchabstract.jsp?tp=&arnumber=1055927
2. Shamir, Adi, "A polynomial-time algorithm for breaking the basic Merkle - Hellman cryptosystem," Information Theory, IEEE Transactions on, vol.30, no.5, pp. 699-704, Sep 1984 URL: http://ieeexplore.ieee.org/search/freesrchabstract.jsp?tp=&arnumber=4568386
de:Merkle-Hellman-Kryptosystem