How can I do an ISO 9797-1 MAC with triple DES in C#?
3des, c#, cryptography, des, hash
Solution
ISO 9797-1 MAC Algorithm 3 consists of using the first DES key to perform a CBC MAC and then only for the final block perform a full 3-DES operation.
Try this:
byte[] keybytes = ParseHex(key);
byte[] key1 = new byte[8];
Array.Copy(keybytes, 0, key1, 0, 8);
byte[] key2 = new byte[8];
Array.Copy(keybytes, 8, key2, 0, 8);
DES des1 = DES.Create();
des1.Key = key1;
des1.Mode = CipherMode.CBC;
des1.Padding = PaddingMode.None;
des1.IV = new byte[8];
DES des2 = DES.Create();
des2.Key = key2;
des2.Mode = CipherMode.CBC;
des2.Padding = PaddingMode.None;
des2.IV = new byte[8];
// MAC Algorithm 3
byte[] intermediate = des1.CreateEncryptor().TransformFinalBlock(data, 0, data.Length);
// Output Transformation 3
byte[] intermediate2 = des2.CreateDecryptor().TransformFinalBlock(intermediate, intermediate.Length - 8, 8);
byte[] result = des1.CreateEncryptor().TransformFinalBlock(intermediate2, 0, 8);
Problem
I've got a project which stipulates the following encryption rules for a 24 byte block of data. 1) Cryptography should be done using full triple DES MAC algorithm as defined in 9797-1 as MAC algorithm 3 with output transformation 3 without truncation and with DES in CBC mode as block cipher with ICV set to zeros. Last 8 bytes of encrypted data constitute the value we need. The program is saying the encryption done is wrong. Are there any other things I need to do to match the above spec? The data is a 24 byte value and output of the encryption should be 8 bytes, I guess (as per the spec). I am getting the whole 24 bytes as output :( I wrote the following code to achieve the said specification: ``` des.KeySize = 128; des.Key = ParseHex(key); des.Mode = CipherMode.CBC; des.Padding = PaddingMode.None; ICryptoTransform ic = des.CreateEncryptor(); CryptoOutput = ic.TransformFinalBlock(CryptoOutput, 0, 24); ``` I tried this also: ``` MACTripleDES des = new MACTripleDES(ParseHex(key)); byte[] CDCryptp = des.ComputeHash(CryptoOutput); ```