Related Articles

# Knapsack Encryption Algorithm in Cryptography

• Difficulty Level : Basic
• Last Updated : 04 Jun, 2020

Knapsack Encryption Algorithm is the first general public key cryptography algorithm. It is developed by Ralph Merkle and Mertin Hellman in 1978. As it is a Public key cryptography, it needs two different keys. One is Public key which is used for Encryption process and the other one is Private key which is used for Decryption process. In this algorithm we will two different knapsack problems in which one is easy and other one is hard. The easy knapsack is used as the private key and the hard knapsack is used as the public key. The easy knapsack is used to derived the hard knapsack.

For the easy knapsack, we will choose a Super Increasing knapsack problem. Super increasing knapsack is a sequence in which every next term is greater than the sum of all preceding terms.

Example –

```{1, 2, 4, 10, 20, 40} is a super increasing as
1<2, 1+2<4, 1+2+4<10, 1+2+4+10<20 and 1+2+4+10+20<40.```

Derive the Public key

• Step-1:
Choose a super increasing knapsack {1, 2, 4, 10, 20, 40} as the private key.
• Step-2:
Choose two numbers n and m. Multiply all the values of private key by the number n and then find modulo m. The value of m must be greater then the sum of all values in private key, foe example 110. And the number n should have no common factor with m, foe example 31.
• Step-3:
Calculate the values of Public key using m and n.

```1x31 mod(110) = 31
2x31 mod(110) = 62
4x31 mod(110) = 14
10x31 mod(110) = 90
20x31 mod(110) = 70
40x31 mod(110) = 30```

Thus, our public key is {31, 62, 14, 90, 70, 30}
And Private key is {1, 2, 4, 10, 20, 40}.

Now take an example for understanding the process of encryption and decryption.

Example –
Lets our plain text is 100100111100101110.

1. Encryption :
As our knapsacks contain six values, so we will split our plain text in a groups of six:

`100100  111100  101110 `

Multiply each values of public key with the corresponding values of each group and take their sum.

```100100  {31, 62, 14, 90, 70, 30}
1x31+0x62+0x14+1x90+0x70+0x30 = 121

111100  {31, 62, 14, 90, 70, 30}
1x31+1x62+1x14+1x90+0x70+0x30 = 197

101110  {31, 62, 14, 90, 70, 30}
1x31+0x62+1x14+1x90+1x70+0x30 = 205 ```

So, our cipher text is 121 197 205.

2. Decryption :
The receiver receive the cipher text which has to be decrypt. The receiver also knows the values of m and n.
So, first we need to find the , which is multiplicative inverse of n mod m i.e.,

```n x mod(m) = 1
31 x mod(110) = 1 = 71```

Now, we have to multiply 71 with each block of cipher text take modulo m.

`121 x 71 mod(110) = 11 `

Then, we will have to make the sum of 11 from the values of private key {1, 2, 4, 10, 20, 40} i.e.,
1+10=11 so make that corresponding bits 1 and others 0 which is 100100.
Similarly,

```197 x 71 mod(110) = 17
1+2+4+10=17 = 111100

And, 205 x 71 mod(110) = 35
1+4+10+20=35 = 101110 ```

After combining them we get the decoded text.
100100111100101110 which is our plain text.

Attention reader! Don’t stop learning now. Get hold of all the important CS Theory concepts for SDE interviews with the CS Theory Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up