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¶
Bases:
sage.structure.sage_object.SageObject
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 keyK
. The decryption process of S-DES is as follows. Let P be the initial permutation function, P−1 the corresponding inverse permutation, ΠF the permutation/substitution function, and σ the switch function. The ciphertext blockC
first goes through P, the output of which goes through ΠF using the second subkey. Then we apply the switch function to the output of the last function, and the result is then fed into ΠF 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 bitsK
– a 10-bit key; a block of 10 bits
OUTPUT:
The 8-bit plaintext corresponding to
C
, obtained using the keyK
.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 keyK
. The encryption process of S-DES is as follows. Let P be the initial permutation function, P−1 the corresponding inverse permutation, ΠF the permutation/substitution function, and σ the switch function. The plaintext blockP
first goes through P, the output of which goes through ΠF using the first subkey. Then we apply the switch function to the output of the last function, and the result is then fed into ΠF 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 bitsK
– a 10-bit key; a block of 10 bits
OUTPUT:
The 8-bit ciphertext corresponding to
P
, obtained using the keyK
.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 (b0,b1,b2,…,b7) be a vector of 8 bits, where each bi∈{0,1}. ThenP(b0,b1,b2,b3,b4,b5,b6,b7)=(b1,b5,b2,b0,b3,b7,b4,b6)The inverse permutation is P−1:
P−1(b0,b1,b2,b3,b4,b5,b6,b7)=(b3,b0,b2,b4,b6,b1,b7,b5)INPUT:
B
– list; a block of 8 bitsinverse
– (default:False
) ifTrue
then use the inverse permutation P−1; ifFalse
then use the initial permutation P
OUTPUT:
The initial permutation of
B
ifinverse=False
, or the inverse permutation ofB
ifinverse=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
byn
positions. Let B=(b0,b1,b2,b3,b4,b5,b6,b7,b8,b9) be a vector of 10 bits. Then the left shift operation Ln is performed on the first 5 bits and the last 5 bits of B separately. That is, if the number of shift positions isn=1
, then L1 is defined asL1(b0,b1,b2,b3,b4,b5,b6,b7,b8,b9)=(b1,b2,b3,b4,b0,b6,b7,b8,b9,b5)If the number of shift positions is
n=2
, then L2 is given byL2(b0,b1,b2,b3,b4,b5,b6,b7,b8,b9)=(b2,b3,b4,b0,b1,b7,b8,b9,b5,b6)INPUT:
B
– a list of 10 bitsn
– (default: 1) ifn=1
then perform left shift by 1 position; ifn=2
then perform left shift by 2 positions. The valid values forn
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 P10 and is specified as follows. Let (b0,b1,b2,b3,b4,b5,b6,b7,b8,b9) be a vector of 10 bits where each bi∈{0,1}. Then P10 is given by
P10(b0,b1,b2,b3,b4,b5,b6,b7,b8,b9)=(b2,b4,b1,b6,b3,b9,b0,b8,b7,b5)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 P4 and is specified as follows. Let (b0,b1,b2,b3) be a vector of 4 bits where each bi∈{0,1}. Then P4 is defined by
P4(b0,b1,b2,b3)=(b1,b3,b2,b0)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 P8 and is specified as follows. Let (b0,b1,b2,b3,b4,b5,b6,b7,b8,b9) be a vector of 10 bits where each bi∈{0,1}. Then P8 picks out 8 of those 10 bits and permutes those 8 bits:
P8(b0,b1,b2,b3,b4,b5,b6,b7,b8,b9)=(b5,b2,b6,b3,b7,b4,b9,b8)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 ΠF on the block
B
using subkeykey
. Let (b0,b1,b2,b3,b4,b5,b6,b7) be a vector of 8 bits where each bi∈{0,1}, let L and R be the leftmost 4 bits and rightmost 4 bits ofB
respectively, and let F be a function mapping 4-bit strings to 4-bit strings. ThenΠF(L,R)=(L⊕F(R,S),R)where S is a subkey and ⊕ denotes the bit-wise exclusive-OR function.
The function F can be described as follows. Its 4-bit input block (n0,n1,n2,n3) is first expanded into an 8-bit block to become (n3,n0,n1,n2,n1,n2,n3,n0). 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=(k0,k1,k2,k3,k4,k5,k6,k7) 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 p0,0p0,3p0,1p0,2 and input this 4-bit string through S-box S0 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 p1,0p1,3p1,1p1,2 and input this 4-bit string through S-box S1 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 S0 and S1 as b0b1b2b3. This 4-bit string undergoes another permutation called P4 as follows:
P4(b0,b1,b2,b3)=(b1,b3,b2,b0)The output of P4 is the output of the function F.
INPUT:
B
– a list of 8 bitskey
– an 8-bit subkey
OUTPUT:
The result of applying the function ΠF to
B
.EXAMPLES:
Applying the function ΠF 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 keyK
.INPUT:
K
– a 10-bit secret key of this Simplified DESn
– (default: 1) ifn=1
then return the first subkey based onK
; ifn=2
then return the second subkey. The valid values forn
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 keyK
.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 (b0,b1,b2,b3,b4,b5,b6,b7) be a vector of 8 bits, where each bi∈{0,1}. Then the switch function σ is given byσ(b0,b1,b2,b3,b4,b5,b6,b7)=(b4,b5,b6,b7,b0,b1,b2,b3)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 inB
.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]