# Simplified DES¶

A simplified variant of the Data Encryption Standard (DES). Note that Simplified DES or S-DES is for educational purposes only. It is a small-scale version of the DES designed to help beginners understand the basic structure of DES.

AUTHORS:

• Minh Van Nguyen (2009-06): initial version

class sage.crypto.block_cipher.sdes.SimplifiedDES

This class implements the Simplified Data Encryption Standard (S-DES) described in [Sch1996]. Schaefer’s S-DES is for educational purposes only and is not secure for practical purposes. S-DES is a version of the DES with all parameters significantly reduced, but at the same time preserving the structure of DES. The goal of S-DES is to allow a beginner to understand the structure of DES, thus laying a foundation for a thorough study of DES. Its goal is as a teaching tool in the same spirit as Phan’s Mini-AES [Pha2002].

EXAMPLES:

Encrypt a random block of 8-bit plaintext using a random key, decrypt the ciphertext, and compare the result with the original plaintext:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES(); sdes
Simplified DES block cipher with 10-bit keys
sage: bin = BinaryStrings()
sage: P = [bin(str(randint(0, 1))) for i in range(8)]
sage: K = sdes.random_key()
sage: C = sdes.encrypt(P, K)
sage: plaintxt = sdes.decrypt(C, K)
sage: plaintxt == P
True


We can also encrypt binary strings that are larger than 8 bits in length. However, the number of bits in that binary string must be positive and a multiple of 8:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: bin = BinaryStrings()
sage: P = bin.encoding("Encrypt this using S-DES!")
sage: Mod(len(P), 8) == 0
True
sage: K = sdes.list_to_string(sdes.random_key())
sage: C = sdes(P, K, algorithm="encrypt")
sage: plaintxt = sdes(C, K, algorithm="decrypt")
sage: plaintxt == P
True

block_length()

Return the block length of Schaefer’s S-DES block cipher. A key in Schaefer’s S-DES is a block of 10 bits.

OUTPUT:

• The block (or key) length in number of bits.

EXAMPLES:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: sdes.block_length()
10

decrypt(C, K)

Return an 8-bit plaintext corresponding to the ciphertext C, using S-DES decryption with key K. The decryption process of S-DES is as follows. Let $$P$$ be the initial permutation function, $$P^{-1}$$ the corresponding inverse permutation, $$\Pi_F$$ the permutation/substitution function, and $$\sigma$$ the switch function. The ciphertext block C first goes through $$P$$, the output of which goes through $$\Pi_F$$ using the second subkey. Then we apply the switch function to the output of the last function, and the result is then fed into $$\Pi_F$$ using the first subkey. Finally, run the output through $$P^{-1}$$ to get the plaintext.

INPUT:

• C – an 8-bit ciphertext; a block of 8 bits

• K – a 10-bit key; a block of 10 bits

OUTPUT:

The 8-bit plaintext corresponding to C, obtained using the key K.

EXAMPLES:

Decrypt an 8-bit ciphertext block:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: C = [0, 1, 0, 1, 0, 1, 0, 1]
sage: K = [1, 0, 1, 0, 0, 0, 0, 0, 1, 0]
sage: sdes.decrypt(C, K)
[0, 0, 0, 1, 0, 1, 0, 1]


We can also work with strings of bits:

sage: C = "01010101"
sage: K = "1010000010"
sage: sdes.decrypt(sdes.string_to_list(C), sdes.string_to_list(K))
[0, 0, 0, 1, 0, 1, 0, 1]

encrypt(P, K)

Return an 8-bit ciphertext corresponding to the plaintext P, using S-DES encryption with key K. The encryption process of S-DES is as follows. Let $$P$$ be the initial permutation function, $$P^{-1}$$ the corresponding inverse permutation, $$\Pi_F$$ the permutation/substitution function, and $$\sigma$$ the switch function. The plaintext block P first goes through $$P$$, the output of which goes through $$\Pi_F$$ using the first subkey. Then we apply the switch function to the output of the last function, and the result is then fed into $$\Pi_F$$ using the second subkey. Finally, run the output through $$P^{-1}$$ to get the ciphertext.

INPUT:

• P – an 8-bit plaintext; a block of 8 bits

• K – a 10-bit key; a block of 10 bits

OUTPUT:

The 8-bit ciphertext corresponding to P, obtained using the key K.

EXAMPLES:

Encrypt an 8-bit plaintext block:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: P = [0, 1, 0, 1, 0, 1, 0, 1]
sage: K = [1, 0, 1, 0, 0, 0, 0, 0, 1, 0]
sage: sdes.encrypt(P, K)
[1, 1, 0, 0, 0, 0, 0, 1]


We can also work with strings of bits:

sage: P = "01010101"
sage: K = "1010000010"
sage: sdes.encrypt(sdes.string_to_list(P), sdes.string_to_list(K))
[1, 1, 0, 0, 0, 0, 0, 1]

initial_permutation(B, inverse=False)

Return the initial permutation of B. Denote the initial permutation function by $$P$$ and let $$(b_0, b_1, b_2, \dots, b_7)$$ be a vector of 8 bits, where each $$b_i \in \{ 0, 1 \}$$. Then

$P(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7) = (b_1, b_5, b_2, b_0, b_3, b_7, b_4, b_6)$

The inverse permutation is $$P^{-1}$$:

$P^{-1}(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7) = (b_3, b_0, b_2, b_4, b_6, b_1, b_7, b_5)$

INPUT:

• B – list; a block of 8 bits

• inverse – (default: False) if True then use the inverse permutation $$P^{-1}$$; if False then use the initial permutation $$P$$

OUTPUT:

The initial permutation of B if inverse=False, or the inverse permutation of B if inverse=True.

EXAMPLES:

The initial permutation of a list of 8 bits:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: B = [1, 0, 1, 1, 0, 1, 0, 0]
sage: P = sdes.initial_permutation(B); P
[0, 1, 1, 1, 1, 0, 0, 0]


Recovering the original list of 8 bits from the permutation:

sage: Pinv = sdes.initial_permutation(P, inverse=True)
sage: Pinv; B
[1, 0, 1, 1, 0, 1, 0, 0]
[1, 0, 1, 1, 0, 1, 0, 0]


We can also work with a string of bits:

sage: S = "10110100"
sage: L = sdes.string_to_list(S)
sage: P = sdes.initial_permutation(L); P
[0, 1, 1, 1, 1, 0, 0, 0]
sage: sdes.initial_permutation(sdes.string_to_list("01111000"), inverse=True)
[1, 0, 1, 1, 0, 1, 0, 0]

left_shift(B, n=1)

Return a circular left shift of B by n positions. Let $$B = (b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9)$$ be a vector of 10 bits. Then the left shift operation $$L_n$$ is performed on the first 5 bits and the last 5 bits of $$B$$ separately. That is, if the number of shift positions is n=1, then $$L_1$$ is defined as

$L_1(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9) = (b_1, b_2, b_3, b_4, b_0, b_6, b_7, b_8, b_9, b_5)$

If the number of shift positions is n=2, then $$L_2$$ is given by

$L_2(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9) = (b_2, b_3, b_4, b_0, b_1, b_7, b_8, b_9, b_5, b_6)$

INPUT:

• B – a list of 10 bits

• n – (default: 1) if n=1 then perform left shift by 1 position; if n=2 then perform left shift by 2 positions. The valid values for n are 1 and 2, since only up to 2 positions are defined for this circular left shift operation.

OUTPUT:

The circular left shift of each half of B.

EXAMPLES:

Circular left shift by 1 position of a 10-bit string:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: B = [1, 0, 0, 0, 0, 0, 1, 1, 0, 0]
sage: sdes.left_shift(B)
[0, 0, 0, 0, 1, 1, 1, 0, 0, 0]
sage: sdes.left_shift([1, 0, 1, 0, 0, 0, 0, 0, 1, 0])
[0, 1, 0, 0, 1, 0, 0, 1, 0, 0]


Circular left shift by 2 positions of a 10-bit string:

sage: B = [0, 0, 0, 0, 1, 1, 1, 0, 0, 0]
sage: sdes.left_shift(B, n=2)
[0, 0, 1, 0, 0, 0, 0, 0, 1, 1]


Here we work with a string of bits:

sage: S = "1000001100"
sage: L = sdes.string_to_list(S)
sage: sdes.left_shift(L)
[0, 0, 0, 0, 1, 1, 1, 0, 0, 0]
sage: sdes.left_shift(sdes.string_to_list("1010000010"), n=2)
[1, 0, 0, 1, 0, 0, 1, 0, 0, 0]

list_to_string(B)

Return a binary string representation of the list B.

INPUT:

• B – a non-empty list of bits

OUTPUT:

The binary string representation of B.

EXAMPLES:

A binary string representation of a list of bits:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: L = [0, 0, 0, 0, 1, 1, 0, 1, 0, 0]
sage: sdes.list_to_string(L)
0000110100

permutation10(B)

Return a permutation of a 10-bit string. This permutation is called $$P_{10}$$ and is specified as follows. Let $$(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9)$$ be a vector of 10 bits where each $$b_i \in \{ 0, 1 \}$$. Then $$P_{10}$$ is given by

$P_{10}(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9) = (b_2, b_4, b_1, b_6, b_3, b_9, b_0, b_8, b_7, b_5)$

INPUT:

• B – a block of 10-bit string

OUTPUT:

A permutation of B.

EXAMPLES:

Permute a 10-bit string:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: B = [1, 1, 0, 0, 1, 0, 0, 1, 0, 1]
sage: sdes.permutation10(B)
[0, 1, 1, 0, 0, 1, 1, 0, 1, 0]
sage: sdes.permutation10([0, 1, 1, 0, 1, 0, 0, 1, 0, 1])
[1, 1, 1, 0, 0, 1, 0, 0, 1, 0]
sage: sdes.permutation10([1, 0, 1, 0, 0, 0, 0, 0, 1, 0])
[1, 0, 0, 0, 0, 0, 1, 1, 0, 0]


Here we work with a string of bits:

sage: S = "1100100101"
sage: L = sdes.string_to_list(S)
sage: sdes.permutation10(L)
[0, 1, 1, 0, 0, 1, 1, 0, 1, 0]
sage: sdes.permutation10(sdes.string_to_list("0110100101"))
[1, 1, 1, 0, 0, 1, 0, 0, 1, 0]

permutation4(B)

Return a permutation of a 4-bit string. This permutation is called $$P_4$$ and is specified as follows. Let $$(b_0, b_1, b_2, b_3)$$ be a vector of 4 bits where each $$b_i \in \{ 0, 1 \}$$. Then $$P_4$$ is defined by

$P_4(b_0, b_1, b_2, b_3) = (b_1, b_3, b_2, b_0)$

INPUT:

• B – a block of 4-bit string

OUTPUT:

A permutation of B.

EXAMPLES:

Permute a 4-bit string:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: B = [1, 1, 0, 0]
sage: sdes.permutation4(B)
[1, 0, 0, 1]
sage: sdes.permutation4([0, 1, 0, 1])
[1, 1, 0, 0]


We can also work with a string of bits:

sage: S = "1100"
sage: L = sdes.string_to_list(S)
sage: sdes.permutation4(L)
[1, 0, 0, 1]
sage: sdes.permutation4(sdes.string_to_list("0101"))
[1, 1, 0, 0]

permutation8(B)

Return a permutation of an 8-bit string. This permutation is called $$P_8$$ and is specified as follows. Let $$(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9)$$ be a vector of 10 bits where each $$b_i \in \{ 0, 1 \}$$. Then $$P_8$$ picks out 8 of those 10 bits and permutes those 8 bits:

$P_8(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7, b_8, b_9) = (b_5, b_2, b_6, b_3, b_7, b_4, b_9, b_8)$

INPUT:

• B – a block of 10-bit string

OUTPUT:

Pick out 8 of the 10 bits of B and permute those 8 bits.

EXAMPLES:

Permute a 10-bit string:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: B = [1, 1, 0, 0, 1, 0, 0, 1, 0, 1]
sage: sdes.permutation8(B)
[0, 0, 0, 0, 1, 1, 1, 0]
sage: sdes.permutation8([0, 1, 1, 0, 1, 0, 0, 1, 0, 1])
[0, 1, 0, 0, 1, 1, 1, 0]
sage: sdes.permutation8([0, 0, 0, 0, 1, 1, 1, 0, 0, 0])
[1, 0, 1, 0, 0, 1, 0, 0]


We can also work with a string of bits:

sage: S = "1100100101"
sage: L = sdes.string_to_list(S)
sage: sdes.permutation8(L)
[0, 0, 0, 0, 1, 1, 1, 0]
sage: sdes.permutation8(sdes.string_to_list("0110100101"))
[0, 1, 0, 0, 1, 1, 1, 0]

permute_substitute(B, key)

Apply the function $$\Pi_F$$ on the block B using subkey key. Let $$(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7)$$ be a vector of 8 bits where each $$b_i \in \{ 0, 1 \}$$, let $$L$$ and $$R$$ be the leftmost 4 bits and rightmost 4 bits of B respectively, and let $$F$$ be a function mapping 4-bit strings to 4-bit strings. Then

$\Pi_F(L, R) = (L \oplus F(R, S), R)$

where $$S$$ is a subkey and $$\oplus$$ denotes the bit-wise exclusive-OR function.

The function $$F$$ can be described as follows. Its 4-bit input block $$(n_0, n_1, n_2, n_3)$$ is first expanded into an 8-bit block to become $$(n_3, n_0, n_1, n_2, n_1, n_2, n_3, n_0)$$. This is usually represented as follows

$\begin{split}\begin{tabular}{c|cc|c} n_3 & n_0 & n_1 & n_2 \\ n_1 & n_2 & n_3 & n_0 \end{tabular}\end{split}$

Let $$K = (k_0, k_1, k_2, k_3, k_4, k_5, k_6, k_7)$$ be an 8-bit subkey. Then $$K$$ is added to the above expanded input block using exclusive-OR to produce

$\begin{split}\begin{tabular}{c|cc|c} n_3 + k_0 & n_0 + k_1 & n_1 + k_2 & n_2 + k_3 \\ n_1 + k_4 & n_2 + k_5 & n_3 + k_6 & n_0 + k_7 \end{tabular} = \begin{tabular}{c|cc|c} p_{0,0} & p_{0,1} & p_{0,2} & p_{0,3} \\ p_{1,0} & p_{1,1} & p_{1,2} & p_{1,3} \end{tabular}\end{split}$

Now read the first row as the 4-bit string $$p_{0,0} p_{0,3} p_{0,1} p_{0,2}$$ and input this 4-bit string through S-box $$S_0$$ to get a 2-bit output.

$\begin{split}S_0 = \begin{tabular}{cc|cc} \hline Input & Output & Input & Output \\\hline 0000 & 01 & 1000 & 00 \\ 0001 & 00 & 1001 & 10 \\ 0010 & 11 & 1010 & 01 \\ 0011 & 10 & 1011 & 11 \\ 0100 & 11 & 1100 & 11 \\ 0101 & 10 & 1101 & 01 \\ 0110 & 01 & 1110 & 11 \\ 0111 & 00 & 1111 & 10 \\\hline \end{tabular}\end{split}$

Next read the second row as the 4-bit string $$p_{1,0} p_{1,3} p_{1,1} p_{1,2}$$ and input this 4-bit string through S-box $$S_1$$ to get another 2-bit output.

$\begin{split}S_1 = \begin{tabular}{cc|cc} \hline Input & Output & Input & Output \\\hline 0000 & 00 & 1000 & 11 \\ 0001 & 01 & 1001 & 00 \\ 0010 & 10 & 1010 & 01 \\ 0011 & 11 & 1011 & 00 \\ 0100 & 10 & 1100 & 10 \\ 0101 & 00 & 1101 & 01 \\ 0110 & 01 & 1110 & 00 \\ 0111 & 11 & 1111 & 11 \\\hline \end{tabular}\end{split}$

Denote the 4 bits produced by $$S_0$$ and $$S_1$$ as $$b_0 b_1 b_2 b_3$$. This 4-bit string undergoes another permutation called $$P_4$$ as follows:

$P_4(b_0, b_1, b_2, b_3) = (b_1, b_3, b_2, b_0)$

The output of $$P_4$$ is the output of the function $$F$$.

INPUT:

• B – a list of 8 bits

• key – an 8-bit subkey

OUTPUT:

The result of applying the function $$\Pi_F$$ to B.

EXAMPLES:

Applying the function $$\Pi_F$$ to an 8-bit block and an 8-bit subkey:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: B = [1, 0, 1, 1, 1, 1, 0, 1]
sage: K = [1, 1, 0, 1, 0, 1, 0, 1]
sage: sdes.permute_substitute(B, K)
[1, 0, 1, 0, 1, 1, 0, 1]


We can also work with strings of bits:

sage: B = "10111101"
sage: K = "11010101"
sage: B = sdes.string_to_list(B); K = sdes.string_to_list(K)
sage: sdes.permute_substitute(B, K)
[1, 0, 1, 0, 1, 1, 0, 1]

random_key()

Return a random 10-bit key.

EXAMPLES:

The size of each key is the same as the block size:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: key = sdes.random_key()
sage: len(key) == sdes.block_length()
True

sbox()

Return the S-boxes of simplified DES.

EXAMPLES:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: sbox = sdes.sbox()
sage: sbox[0]; sbox[1]
(1, 0, 3, 2, 3, 2, 1, 0, 0, 2, 1, 3, 3, 1, 3, 2)
(0, 1, 2, 3, 2, 0, 1, 3, 3, 0, 1, 0, 2, 1, 0, 3)

string_to_list(S)

Return a list representation of the binary string S.

INPUT:

• S – a string of bits

OUTPUT:

A list representation of the string S.

EXAMPLES:

A list representation of a string of bits:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: S = "0101010110"
sage: sdes.string_to_list(S)
[0, 1, 0, 1, 0, 1, 0, 1, 1, 0]

subkey(K, n=1)

Return the n-th subkey based on the key K.

INPUT:

• K – a 10-bit secret key of this Simplified DES

• n – (default: 1) if n=1 then return the first subkey based on K; if n=2 then return the second subkey. The valid values for n are 1 and 2, since only two subkeys are defined for each secret key in Schaefer’s S-DES.

OUTPUT:

The n-th subkey based on the secret key K.

EXAMPLES:

Obtain the first subkey from a secret key:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: key = [1, 0, 1, 0, 0, 0, 0, 0, 1, 0]
sage: sdes.subkey(key, n=1)
[1, 0, 1, 0, 0, 1, 0, 0]


Obtain the second subkey from a secret key:

sage: key = [1, 0, 1, 0, 0, 0, 0, 0, 1, 0]
sage: sdes.subkey(key, n=2)
[0, 1, 0, 0, 0, 0, 1, 1]


We can also work with strings of bits:

sage: K = "1010010010"
sage: L = sdes.string_to_list(K)
sage: sdes.subkey(L, n=1)
[1, 0, 1, 0, 0, 1, 0, 1]
sage: sdes.subkey(sdes.string_to_list("0010010011"), n=2)
[0, 1, 1, 0, 1, 0, 1, 0]

switch(B)

Interchange the first 4 bits with the last 4 bits in the list B of 8 bits. Let $$(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7)$$ be a vector of 8 bits, where each $$b_i \in \{ 0, 1 \}$$. Then the switch function $$\sigma$$ is given by

$\sigma(b_0, b_1, b_2, b_3, b_4, b_5, b_6, b_7) = (b_4, b_5, b_6, b_7, b_0, b_1, b_2, b_3)$

INPUT:

• B – list; a block of 8 bits

OUTPUT:

A block of the same dimension, but in which the first 4 bits from B has been switched for the last 4 bits in B.

EXAMPLES:

Interchange the first 4 bits with the last 4 bits:

sage: from sage.crypto.block_cipher.sdes import SimplifiedDES
sage: sdes = SimplifiedDES()
sage: B = [1, 1, 1, 0, 1, 0, 0, 0]
sage: sdes.switch(B)
[1, 0, 0, 0, 1, 1, 1, 0]
sage: sdes.switch([1, 1, 1, 1, 0, 0, 0, 0])
[0, 0, 0, 0, 1, 1, 1, 1]


We can also work with a string of bits:

sage: S = "11101000"
sage: L = sdes.string_to_list(S)
sage: sdes.switch(L)
[1, 0, 0, 0, 1, 1, 1, 0]
sage: sdes.switch(sdes.string_to_list("11110000"))
[0, 0, 0, 0, 1, 1, 1, 1]