Merkle-Hellman Knapsack Cryptosystem - PowerPoint Presentation

Slide1

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 1985Slide2

Classical Knapsack Problem

General 0-1 knapsack problem: given

n

items of different values

v

i

and weights

w

i

, find the most valuable subset of the items while the overall weight does not exceed a given capacity WSlide3

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

= {

a

1

,

a

2

, …

a

n

} 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-hardSlide4

Easy Knapsack Problem

An easy knapsack problem is one in which set

A

= {

a

1

,

a

2

, …

a

n

} 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:

a

2

>

a

1

,

a

3

>

a

1

+

a

2

,….,

a

n

>

a

1

+

a

2

+…+

a

n

-1

Example: A= {1, 2, 4, 8, …2

n-

1

} is super-increasing sequenceSlide5

Polynomial Time Algorithm for Easy Knapsack Problem

Input:

A

= {

a

1

, …

a

n

} is super-increasing sequence, and

S >

0

Output: TRUE and

P

– binary array of n elements,

P

[

i

] =1 means:

a

i

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



a

i

then

P

[

i

]



1

and

S



S

-

a

i

    else

P

[

i

]



0

if

S

!= 0

then return

(FALSE – no solution)

else return

(

P

[1],

P

[2], …

P

[

n

]). Slide6

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]=1101Slide7

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 AliceSlide8

Alice Knapsack Cryptosystem Construction

Chooses

A

= {

a

1

, …

a

n

} super-increasing sequence,

A

is a private (easy) knapsack

a

1

+ …+

a

n

= 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

= {

b

1

, ….

b

n

}, where

b

i

= Wa

i

(

mod

M),

1

 i  n

Keeps Private Key:

A, W, M

Publishes Public key

: BSlide9

Example: ALICE creates Public and Private Keys

Alice Private Key:

A

= {1, 2, 4, 8} – super increasing

E =

1+2+4+8

=

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}Slide10

Bob – Encryption Process

Binary Plaintext

P

breaks up into sets of

n

elements long:

P

= {

P

1

, …

P

k

}

For each set

Pi

compute

C

i

is the ciphertext that corresponds to plaintext

P

i

C = {C1, …Ck) is ciphertext that corresponds to the plaintext PC is sent to AliceSlide11

Example Continue: Bob Encryption

Bob Encryption:

Plaintext: 1101 0101 1110

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}Slide12

Alice – Decryption Process

Computes

w

, the multiplicative inverse of

W

mod

M

:

wW

 1 (mod

M

)

The connection between easy and hard knapsacks:

Wa

i

=

b

i

(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 knapsackSlide13

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 = 1110 (7 = 1*1 + 1*2 + 1*4 + 0*8)Slide14

Final Project

For final project you will implement the additive knapsack cryptosystem encryption, decryption and cryptanalysis.

The project will be divided into 3 parts

Part I: Encryption and

Decryption– 80

% of the grade

Part II: Dynamic Programming Algorithm Implementation of Cryptanalysis (without backtracking)

–– 10

% of the grade

Part III: Dynamic Programming Algorithm Backtracking

––

10% of the

grade

Testing quiz will be done in class on Thursday, June 27

If you have difficulties to implement Part II or/and Part III you can receive up

to 3.5

points for each part for solving testing quiz by hand.Slide15

Part I – 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.Slide16

PART II: Ciphertext

Only Cryptanalytic Attack on

Merkle

-Hellman Knapsack: Dynamic Programming Algorithm

Input:

B

={b

1

, b

2

, … b

n

} – 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]

 TRUEfor 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

]

Slide17

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 – 1Output: array P, elements of P that equal to 1 construct a desired subset of B that sums to CSlide18

EXAMPLE

Input: B={1, 4, 5, 2}, C =3

Q[i-1][j-B[i]]

or

Q[i-1][j]

j = 0

j = 1

j = 2

j = 3

i = 0

TRUE

FALSE

FALSE

FALSE

i = 1

B[1] =1

TRUE

TRUE

Element is taken

FALSE

FALSE

i = 2

B[2] = 4

TRUE

TRUE

FALSE

FALSE

i = 3

B[3] = 5

TRUE

TRUE

FALSE

FALSE

i = 4

B[4] = 2

TRUE

TRUE

TRUE

TRUE

Element is takenSlide19

Part I Bonus

BONUS – 10 points: plaintext string's length is an exact multiple of the public key sequence length

BONUS – 10 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.