1. Merkle-Hellman Knapsack Cryptosystem
Merkle offered $100 award for breaking singly - iterated knapsack
Singly-iterated Merkle - Hellman KC was broken by Adi Shamir in 1982  
At the CRYPTO ’83 conference, Adleman used an Apple II computer to demonstrate Shamir’s method
Merkle offered $1000 award for breaking multiply-iterated knapsack
Multiply-iterated Merkle-Hellman knapsack was broken by Brickell in 1985

2. Classical Knapsack Problem
General 0-1 knapsack problem: given n items of different values vi and weights wi, find the most valuable subset of the items while the overall weight does not exceed a given capacity W

3. Subset-Sum Problem
Subset – Sum problem is a special case of knapsack problem when a value of each item is equal to its weight
Input: set of positive integers: A = {a1, a2, …an} and the positive integer S
Output:
TRUE, if there is a subset of A that sums to S and the subset itself
FALSE otherwise.
The subset-sum problem is NP-hard

4. Easy Knapsack Problem An easy knapsack problem is one in which set
A = {a1, a2, …an} is a super-increasing sequence
A super-increasing sequence is one in which the next term of the sequence is greater than the sum of all preceding terms:
a2 > a1, a3 > a1 + a2,…., an > a1 + a2 +…+ an-1
Example: A= {1, 2, 4, 8, …2n-1} is super-increasing sequence

5. Polynomial Time Algorithm for Easy Knapsack Problem
Input: A = {a1, …an} is super-increasing sequence, and S >0
Output: TRUE and P – binary array of n elements, P[i] =1 means: ai belongs to subset of A that sums to S, P[0] = 0 otherwise. The algorithm returns FALSE if the subset doesn’t exist
for i  n to 1   
if S  ai
then P[i]  1 and S  S - ai     else P[i]  0
if S != 0
then return (FALSE – no solution) else return (P[1], P[2], …P[n]). 

6. Example Input: A= {1, 2, 4, 8}, S = 11 Solution:
i = 4, S = 11 >= A[4] = 8, P[4]=1, S= S-A[4]=11-8=3
i=3, S=3 < A[3]=4, P[3]=0
i=2, S=3 >= A[2]=2, P[2]=1, S=S-A[2]=3-2=1
i=1, S=1 >= A[1]=1 P[1]=1, S=S-A[1]=1-1=0
Final answer: P[1]P[2]P[3]P[4]=1101

7. Merkle-Hellman Additive Knapsack Cryptosystem
Alice:
1. Constructs the Knapsack cryptosystem
2. Publishes the public key
3. Receives the ciphertext
4. Decrypts the ciphertext using private key
Bob:
Encrypts the plaintext using public key
Sends the plaintext to Alice

8. Alice Knapsack Cryptosystem Construction
Chooses A = {a1, …an} super-increasing sequence,
A is a private (easy) knapsack
a1+ …+ an = E
Chooses M - the next prime larger than E.
Chooses W that satisfies 2  W < M and
(W, M) = 1(W is relatively prime with M)
Computes Public (hard) knapsack B = {b1, ….bn}, where bi = Wai (mod M), 1  i  n
Keeps Private Key: A, W, M
Publishes Public key: B

9. Example: ALICE creates Public and Private Keys
Alice Private Key:
A= {1, 2, 4, 8} – super increasing
E = = 15 and M = 17 first prime > 15
W = 7, 2  W < 17, and (7, 17) = 1
Public Key:
(1*7) mod 17 = 7
(2*7) mod 17 = 14
(4*7) mod 17 = 28 mod 17 = 11
(8*7) mod 17= 56 mod 17 = 5
Public Key: B = {7, 14, 11, 5}

10. Bob – Encryption Process
Binary Plaintext P breaks up into sets of n elements long: P = {P1, …Pk}
For each set Pi compute
Ci is the ciphertext that corresponds to plaintext Pi
C = {C1, …Ck) is ciphertext that corresponds to the plaintext P
C is sent to Alice

11. Example Continue: Bob Encryption
Plaintext:
Bob breaks the plaintext into blocks of 4 digits (since the public key has 4 numbers)
P={(1101), (0101), (1110)}={P1, P2, P3}
Ciphertext:
For P1 you take 1101 and multiply by public key:
C1= 1*7 + 1*14 + 0*11 + 1*5 = 26
For P2 and P3 do the similar
C2 = 0*7 + 1*14 + 0*11 + 1*5 = 19
C3 = 1*7 +1*14 +1*11 + 0*5 = 32
Bob Sends Alice the following ciphertext:
C={26, 19, 32}

12. Alice – Decryption Process
Computes w, the multiplicative inverse of W mod M:
wW  1 (mod M)
The connection between easy and hard knapsacks:
Wai = bi (mod M) or wbi = ai (mod M) 1  i  n
For each Ci computes: Si = wCi (mod M)
Plaintext Pi could be found using polynomial time algorithm for easy knapsack

13. Example continue: Alice Decryption:
w = 5 – multiplicative inverse of 7 (mod 17)
5*26 (mod 17) = 11
5*19 (mod 17) = 10
5* 32 (mod 17) = 7
Plaintext:
P1 = 1101 (11 = 1*1 + 1*2 +0*4 + 1*8)
P2 = 0101 (10 = 0*1 + 1*2 + 0*4 + 1*8)
P3 = (7 = 1*1 + 1*2 + 1*4 + 0*8)

14. Programming Lab Encryption and Decryption – 100 points
Bonus: Dynamic Programming Algorithm Implementation of Cryptanalysis – 5 points
Bonus – 5 points: plaintext string's length is an exact multiple of the public key sequence length
Bonus – 5 points: the plaintext is the string of lower case letters. In this case your program first will find the binary equivalent for each letter and after that will use the regular algorithm.
Testing quiz will be done in class

15. Encryption and Decryption
Write a program that can do either encryption or decryption. The program must take two inputs.
The first input must be either 1 or 2, with 1 signaling encryption and 2 signaling decryption.
The second input depends on the first input.
In case of encryption, the second input - plaintext - is a binary string – sequence of 0’s and 1’s. You can assume that plaintext string's length is equal to public key sequence length and the maximal length of the string is 16 bits.
In case of decryption, the second input - ciphertext - is a decimal number.
Your program should generate the private key and the public key based on the knapsack cryptosystem algorithm.
Your program should output the private and public keys as well as encrypted or decrypted message accordingly. Also, print all intermediate important results to test your program for correctness.

16. PART II: Ciphertext Only Cryptanalytic Attack on Merkle-Hellman Knapsack: Dynamic Programming Algorithm
Input: B={b1, b2, … bn} – public key, C - ciphertext
Output: The binary array P – plaintext
Algorithm: Let Q[i, j] be TRUE if there is a subset of first i elements of B
that sums to j, 0 ≤ i ≤ n , 0 ≤ j ≤ C
Step 1: Computation of P
Q[0][0]  TRUE
for j = 1 to C do: Q[0][j]  FALSE
for i = 1 to n do:
for j = 0 to C do:
if (j – B[i] < 0): Q[i][j] = Q[i-1][j]
else: Q[i][j] = Q[i-1][j-B[i]] or Q[i-1][j]

17. Step 2: Backtracking
Let P be an array of n + 1 elements initialized to 0
i  n, j  C
while i > 0:
if (j – B[i]) ≥ 0):
if (Q[i-1][j-B[i]] is True):
P[i]  P[i] + 1
j  j – B[i]
i  i – 1
else: i  i – 1
Output: array P, elements of P that equal to 1 construct a
desired subset of B that sums to C

18. Q[i-1][j-B[i]] or Q[i-1][j]
EXAMPLE Input: B={1, 4, 5, 2}, C =3
j = 0
j = 1
j = 2
j = 3
i = 0
TRUE
FALSE
i = 1
B[1] =1
Element is taken
i = 2
B[2] = 4
i = 3
B[3] = 5
i = 4
B[4] = 2
Q[i-1][j-B[i]] or Q[i-1][j]