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https://github.com/drilonaliu/parallel-permutation-cipher


https://github.com/drilonaliu/parallel-permutation-cipher

cryptography cuda gpu parallel-programming permutation

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README 

        
# Parallel Permutation Cipher

 

In permutation ciphering, both in C++ and CUDA, we have a method that permutes a text based on a given permutation. This method is used for both encryption and decryption. In the sequential version, we iterate over each character and permute it, whereas in the parallel version, the i-th thread permutes the i-th character of the text. To demonstrate how the algorithm works, let's take an example.



We need to permute the following text using the given permutation {2,1,0}. When permuting the letter 'F', we take the index 5 and map it to the range [0,2], resulting in the index 2. According to the given permutation, π(2) = 0. Now, we find the first index of the block to which index 5 belongs (i.e., the block [3,5]), and we add the value obtained, π(2) = 0, to this index to get the final index 3.





	

| 0 | 1 | 2 | 3 | 4 | 5 |

|---|---|---|---|---|---|

| A | B | C | D | E | F |

| C | B | A | F | E | D |






$$5 \mod 3 = 2$$



$$j = \lfloor {\frac{5}{3}} \rfloor \times 3 + \pi(2) = 1 \times 3 + 0 = 3$$



## Kernel

```

__global__ void applyPermutation(char* text, char* permutatedText, int* permutation, int permutationLength, int textLength) {

	int i = threadIdx.x+blockIdx.x*blockDim.x;

	if (i < textLength) {

		int p = permutation[i % permutationLength];

		int j = (i / permutationLength) * permutationLength + p;

		permutatedText[j] = text[i];

	}

}

```



## Padding 



In permutation ciphering, padding is considered, which is necessary when the length of the text is not a multiple of the key length. In this case, a symbol is added to represent padding, for example, the symbol @. If we have the text "ABCDEFG," padding would occur as follows:



 

	

| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

|---|---|---|---|---|---|---|---|---|

| A | B | C | D | E | F | G | @ | @ |

| C | B | A | F | E | D | @ | @ | F |






```

string permutationEncrypt(string text, int* permutation, int permutationLength) {

 	int m = text.length() % permutationLength;

 	//Add the padd symbol if padding is needed.

 	if (m != 0) {

  		int padding = permutationLength - m;

  		for (int i = 0; i < padding; i++) {

 			  text += paddSymbol;

 	 	}

 	}

 	string encryptedText = applyPermutation(text, permutation,permutationLength);

 	return encryptedText;

}

```



When decrypting the text "ABCDEFG@@", we look at the last block. If we encounter any @ (padding symbol), we remove every character after the padding symbol (which means we remove all padding symbols) and obtain the text "ABCDEFG".



```

string permutationDecrypt(string text, int* permutation, int permutationLength) {

	int* inv_permutation = getInversePermutation(permutation, permutationLength);

	string decryptedText = applyPermutation(text, inv_permutation, permutationLength);

	/*Check if padding was added by going to the last block.

	If it has the padding symbol, remove everything after it.*/

	for (int i = text.length() - permutationLength - 1; i < text.length(); i++) {

		char letter = decryptedText[i];;

		if (letter == paddSymbol) {

			decryptedText.erase(i);

			break;

		}

	}

	

	return decryptedText;

}

```