Crypt function reverce challange.

[HostageOfCode]

           

			int i = 0;
            unsigned char input_buffer[BINSIZE + 1] = {0};
            unsigned char output_buffer[BINSIZE * 2] = {0};
			DWORD data_size = BINSIZE;

             for(i = 0; i < data_size; i = i + 16)
            {
                Decrypt(AESKey, input_buffer + i, output_buffer + i, 16);
            }

            for(i = 0; i < data_size - 16; i++)
            {
                output_buffer[i + 16] = output_buffer[i + 16] ^ input_buffer[i];
            }

The goal is to reverse the algorithm and obtain input_buffer if we have only the output_buffer after the xor manipulation and the AESKey outsource for the Encrypt function. Looks easy but it is not that easy after all.

Edited April 29 by HostageOfCode

[aIjundi]

Key details to solving this challenge:

1. The block size is 16 bytes.
2. The first block is only Decrypted, not xor'ed.
3. Each of the following blocks is Decrypted then xor'ed with the previous block

Solving it would boil down to Encrypting a block then xor'ing it with the next block, which would roughly be as below

unsigned char output_buffer[BINSIZE * 2] = {0};
unsigned char result_buffer[BINSIZE + 1] = {0};
DWORD data_size = BINSIZE;

for(int i = 0; i < data_size; i = i + 16)
{
    Encrypt(AESKey, result_buffer + i, output_buffer + i, 16);

    for(int j = i; j < i + 16; j++)
    {
        output_buffer[j + 16] = output_buffer[j + 16] ^ result_buffer[j];
    }
}

 

[HostageOfCode]

 

               int count = 0;
                memcpy(output_buffer, input_buffer, data_size);
                Encrypt(AESKey, output_buffer, output_buffer, 16);

                for (i = 0; i < data_size - 16; i++)
                {

                    output_buffer[i + 16] = input_buffer[i + 16] ^ output_buffer[i];

                    count++;

                    if (count == 16)
                    {

                        Encrypt(AESKey, output_buffer+ i + 1, output_buffer+ i + 1, 16);
                        count = 0;
                    }

                }

This is my solution.