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The math and definitions explained

The cryptography explained

A block cipher such as DES or AES only encrypts one block of a fixed size (56-bit for DES and 128-bit for AES), but often the data we want to encrypt (the plaintext) is larger than 56-bit or 128-bit. This problem is solved by using a mode of operation which describes how to repeatedly apply DES or AES (both encryption and decryption) on plaintexts (and ciphertexts) with size larger than one block.

Most modes of operation requires an initialization vector, denoted \( IV \) (or \( ctr \) in one of the mode of operation), which is a random bit string with the same length as a block, i.e. either 56-bit or 128-bit. The initialization vector adds randomness to the encryption, i.e. if we encrypt the same plaintext twice with the same key, but with different initialization vectors, the corresponding ciphertexts have no relation at all. It's therefore important that the initialization vector is unique for every encryption but it doesn't need to be secret, because as you may see in the following they are added to the ciphertext after encryption.

In 1980 the first four modes of operation was developed for DES (which also can be used with AES):

• The electronic codebook (ECB) mode
• The cipher block chaining (CBC) mode
• The cipher feedback (CFB) mode
• The output feedback (OFB) mode

Later in 2001 NIST (the US National Institute of Standards and Technology) added the following mode of operation for AES:

• The counter (CTR) mode

The size of the key \( K \) used in a mode of operation depends on the used block cipher, i.e. whether it is DES or AES. For DES is the key 56-bit and for AES is the key 128-, 192- or 256-bit. In the following we assume that the plaintext \( m \) can be evenly divided into \( t \) blocks \( m_{1}, m_{2}, \dots, m_{t} \).

So, what if the last block of the plaintext \( m_{t} \) is shorter than the size of one block, i.e. shorter than 56-bit or 128-bit? Because DES and AES only works on blocks of 56-bit and 128-bit respectively we have to pad \( m_{t} \) such that it has the correct length. There exists various ways of padding the blocks but as you may see in the demo we add the missing \( p \) bytes \( p \) times, all with value \( p \), at the end of \( m_{t} \). E.g. if we want to encrypt the message "Hello Alice" with AES and the length of this message is 368 bit, we can divide it into three blocks \( m_{1}, m_{2}, m_{3} \). Now \( m_{1} \) and \( m_{2} \) each are 128 bit, as they should, but \( m_{3} \) is only 112 bit, i.e. we need 16 bit. Because 16 bit is the same as 2 byte (8 bit is 1 byte) we pad \( m_{3} \) with "22", i.e. we actually encrypt the message "Hello Alice22".

In the mode of operation we present it's only the ECB and CBC modes which requires the last block padded before encryption if it's to short. The stream ciphers such as CFB, OFB and CTR mode doesn't require padding because the plaintext is XORed with a key stream.

The electronic codebook (ECB) mode corresponds to the naive use of the block cipher, i.e. given the plaintexts \( m_{1}, m_{2}, \dots, m_{t} \) each plaintext is encrypted with the same key \( K \) and produces the ciphertexts \( c_{1}, c_{2}, \dots, c_{t} \). The same method is used in decryption, i.e. \( c_{i} = E_{K}(m_{i}) \) and \( m_{i} = D_{K}(c_{i}) \) where \( E_{K} \) is the encryption algorithm and \( D_{K} \) is the decryption algorithm. Therefore, in ECB mode will the encryption of the same plaintext always result in the same ciphertext, which is a serious weakness if the plaintext space is small.

In the cipher block chaining (CBC) mode is the first ciphertext block (for encryption) equal the initialization vector \( IV \) and the remaining \( t \) ciphertext blocks are the encryption of the \( i \)-th plaintext block XORed with the previous ciphertext block, i.e. \( c_{0} = IV \) and \( c_{i} = E_{K}(c_{i-1} \oplus m_{i}) \). During decryption is the \( i \)-th plaintext block computed by first decrypting the \( i \)-th ciphertext block and then XOR it with the previous ciphertext block, i.e. \( m_{i} = D_{K}(c_{i}) \oplus c_{i-1} \). Notice that if the plaintext block \( m_{i} \) during encryption is altered then \( c_{i} \) and all subsequent ciphertext blocks will be affected. This property makes the CBC mode very useful for authentication and is used in the CBC-MAC.

The output feedback (OFB) mode uses a key stream \( z_{1}, z_{2}, \dots, z_{t} \) where the first block in the key stream \( z_{0} \) is equal the initialization vector \( IV \) and the remaining \( t \) blocks are the encryption of the previous block, i.e. \( z_{0} = IV \) and \( z_{i} = E_{K}(z_{i-1}) \). The key stream is used during encryption and decryption where the \( i \)-th ciphertext is the XOR of the \( i \)-th plaintext and the \( i \)-th key stream, i.e. \( c_{i} = m_{i} \oplus z_{i} \), and the \( i \)-th plaintext is the XOR of the \( i \)-th ciphertext and the \( i \)-th key stream, i.e. \( m_{i} = c_{i} \oplus z_{i} \). Notice that only the encryption algorithm is used which makes it suitable for small devices with limited computer power such as chip cards etc.

The cipher feedback (CFB) mode also uses a key stream where the \( i \)-th key stream is the encryption of the previous ciphertext block, i.e. \( z_{i} = E_{K}(c_{i-1}) \). During encryption is the first ciphertext block equal the initialization vector \( IV \) and the remaining \( t \) ciphertext blocks are the XOR of the \( i \)-th plaintext block and the \( i \)-th key stream, i.e. \( c_{0} = IV \) and \( c_{i} = m_{i} \oplus z_{i} \). Decryption is almost the same with \( m_{i} = c_{i} \oplus z_{i} \). Notice again that only the encryption algorithm is used.

The counter (CTR) mode is similar to the OFB mode, but with the key stream generated slightly different. In the CTR mode is the initialization vector called a counter vector and is denoted \( ctr \). The key stream is then the XOR of \( ctr \) and the counter \( i \), i.e. \( T_{i} = ctr \oplus i \; mod \; 2^{n} \) where \( n \) is the block size in byte. During encryption is the \( i \)-th ciphertext block equal the XOR of the \( i \)-th plaintext block and the encryption of the \( i \)-th key stream, i.e. \( c_{i} = m_{i} \oplus E_{K}(T_{i}) \). During decryption is the \( i \)-th plaintext equal the XOR of the \( i \)-th ciphertext block and the encryption of the \( i \)-th key stream, i.e. \( m_{i} = c_{i} \oplus E_{K}(T_{i}) \). Notice that the encryption algorithm is used in both encryption and decryption and the key stream is only the XOR of \( ctr \) and the counter \( i \), i.e. we can run CTR mode in parallel because the key stream doesn't depend on previously blocks.

It's recommended to use either CBC or CTR mode where CTR mode has the advantage that it's fast because it can run in parallel and it only use the encryption algorithm.

Try a demo of the encryption scheme using one of the mode of operation here.

Example

Alice needs to send an email \( m \) to Bob that contains sensitive information, and hence, she encrypts the email before she sends it to Bob. They first agree on using AES in CTR mode and the 128-bit key \( K = 01001000 \dots \).

Alice then divides the email \( m \) into 128-bit blocks. Assume that the email fits into three blocks \( m_{1}, m_{2}, m_{3}\) and the 128-bit counter vector is \( ctr = 00000110 \dots \). She then encrypts the plaintext blocks by first computing the keystream \( T_{i} = ctr \oplus i \: mode \: 2^{16} \) (where \( n = 16 \) because 128 bit is equal to 16 byte), and then encrypting the \( i \)-th plaintext block by \( c_{i} = m_{i} \oplus E_{K}(T_{i}) \):

\( \eqalign{ T_{1} &= ctr \oplus 1 \; mod \; 2^{128} = 00000110 \dots \\ c_{1} &= m_{1} \oplus E_{K}(T_{1}) = 11010101 \dots \\ T_{2} &= ctr \oplus 2 \; mod \; 2^{128} = 10000110 \dots \\ c_{2} &= m_{2} \oplus E_{K}(T_{2}) = 10100101 \dots \\ T_{3} &= ctr \oplus 3 \; mod \; 2^{128} = 10000110 \dots \\ c_{3} &= m_{3} \oplus E_{K}(T_{3}) = 10111001 \dots } \)

She then sends the ciphertext blocks \( c_{1}, c_{2}, c_{3}\) to Bob.

Bob decrypts the received ciphertext blocks by first computing the key stream just as Alice did, i.e. \( T_{i} = ctr \oplus i \: mode \: 2^{16} \), and then decrypting the \( i \)-th ciphertext block by \( m_{i} = c_{i} \oplus E_{K}(T_{i}) \):

\( \eqalign{ T_{1} &= ctr \oplus 1 \; mod \; 2^{128} = 00000110 \dots \\ m_{1} &= c_{1} \oplus E_{K}(T_{1}) = 01010100 \dots \\ T_{2} &= ctr \oplus 2 \; mod \; 2^{128} = 10000110 \dots \\ m_{2} &= c_{2} \oplus E_{K}(T_{2}) = 01100101 \dots \\ T_{3} &= ctr \oplus 3 \; mod \; 2^{128} = 10000110 \dots \\ m_{3} &= c_{3} \oplus E_{K}(T_{3}) = 01111001 \dots } \)

The plaintext is then the concatenation of the plaintext blocks, i.e. \( m = m_{1} \: \| \: m_{2} \: \| \: m_{3}\).