Cryptography¶
Warning
This module is intended for educational purposes only. Do not use the functions in this module for real cryptographic applications. If you wish to encrypt real data, we recommend using something like the cryptography module.
Encryption is the process of hiding a message and a cipher is a means of doing so. Included in this module are both block and stream ciphers:
Shift cipher
Affine cipher
substitution ciphers
Vigenere’s cipher
Hill’s cipher
Bifid ciphers
RSA
Kid RSA
linear-feedback shift registers (for stream ciphers)
ElGamal encryption
In a substitution cipher “units” (not necessarily single characters) of plaintext are replaced with ciphertext according to a regular system.
A transposition cipher is a method of encryption by which the positions held by “units” of plaintext are replaced by a permutation of the plaintext. That is, the order of the units is changed using a bijective function on the position of the characters to perform the encryption.
A monoalphabetic cipher uses fixed substitution over the entire message, whereas a polyalphabetic cipher uses a number of substitutions at different times in the message.
- sympy.crypto.crypto.AZ(s=None)[source]¶
Return the letters of
s
in uppercase. In case more than one string is passed, each of them will be processed and a list of upper case strings will be returned.Examples
>>> from sympy.crypto.crypto import AZ >>> AZ('Hello, world!') 'HELLOWORLD' >>> AZ('Hello, world!'.split()) ['HELLO', 'WORLD']
See also
- sympy.crypto.crypto.padded_key(key, symbols)[source]¶
Return a string of the distinct characters of
symbols
with those ofkey
appearing first. A ValueError is raised if a) there are duplicate characters insymbols
or b) there are characters inkey
that are not insymbols
.Examples
>>> from sympy.crypto.crypto import padded_key >>> padded_key('PUPPY', 'OPQRSTUVWXY') 'PUYOQRSTVWX' >>> padded_key('RSA', 'ARTIST') Traceback (most recent call last): ... ValueError: duplicate characters in symbols: T
- sympy.crypto.crypto.check_and_join(phrase, symbols=None, filter=None)[source]¶
Joins characters of
phrase
and ifsymbols
is given, raises an error if any character inphrase
is not insymbols
.- Parameters:
phrase
String or list of strings to be returned as a string.
symbols
Iterable of characters allowed in
phrase
.If
symbols
isNone
, no checking is performed.
Examples
>>> from sympy.crypto.crypto import check_and_join >>> check_and_join('a phrase') 'a phrase' >>> check_and_join('a phrase'.upper().split()) 'APHRASE' >>> check_and_join('a phrase!'.upper().split(), 'ARE', filter=True) 'ARAE' >>> check_and_join('a phrase!'.upper().split(), 'ARE') Traceback (most recent call last): ... ValueError: characters in phrase but not symbols: "!HPS"
- sympy.crypto.crypto.cycle_list(k, n)[source]¶
Returns the elements of the list
range(n)
shifted to the left byk
(so the list starts withk
(modn
)).Examples
>>> from sympy.crypto.crypto import cycle_list >>> cycle_list(3, 10) [3, 4, 5, 6, 7, 8, 9, 0, 1, 2]
- sympy.crypto.crypto.encipher_shift(msg, key, symbols=None)[source]¶
Performs shift cipher encryption on plaintext msg, and returns the ciphertext.
- Parameters:
key : int
The secret key.
msg : str
Plaintext of upper-case letters.
- Returns:
str
Ciphertext of upper-case letters.
Examples
>>> from sympy.crypto.crypto import encipher_shift, decipher_shift >>> msg = "GONAVYBEATARMY" >>> ct = encipher_shift(msg, 1); ct 'HPOBWZCFBUBSNZ'
To decipher the shifted text, change the sign of the key:
>>> encipher_shift(ct, -1) 'GONAVYBEATARMY'
There is also a convenience function that does this with the original key:
>>> decipher_shift(ct, 1) 'GONAVYBEATARMY'
Notes
ALGORITHM:
- STEPS:
Number the letters of the alphabet from 0, …, N
Compute from the string
msg
a listL1
of corresponding integers.Compute from the list
L1
a new listL2
, given by adding(k mod 26)
to each element inL1
.Compute from the list
L2
a stringct
of corresponding letters.
The shift cipher is also called the Caesar cipher, after Julius Caesar, who, according to Suetonius, used it with a shift of three to protect messages of military significance. Caesar’s nephew Augustus reportedly used a similar cipher, but with a right shift of 1.
See also
References
- sympy.crypto.crypto.decipher_shift(msg, key, symbols=None)[source]¶
Return the text by shifting the characters of
msg
to the left by the amount given bykey
.Examples
>>> from sympy.crypto.crypto import encipher_shift, decipher_shift >>> msg = "GONAVYBEATARMY" >>> ct = encipher_shift(msg, 1); ct 'HPOBWZCFBUBSNZ'
To decipher the shifted text, change the sign of the key:
>>> encipher_shift(ct, -1) 'GONAVYBEATARMY'
Or use this function with the original key:
>>> decipher_shift(ct, 1) 'GONAVYBEATARMY'
- sympy.crypto.crypto.encipher_rot13(msg, symbols=None)[source]¶
Performs the ROT13 encryption on a given plaintext
msg
.Explanation
ROT13 is a substitution cipher which substitutes each letter in the plaintext message for the letter furthest away from it in the English alphabet.
Equivalently, it is just a Caeser (shift) cipher with a shift key of 13 (midway point of the alphabet).
See also
References
- sympy.crypto.crypto.decipher_rot13(msg, symbols=None)[source]¶
Performs the ROT13 decryption on a given plaintext
msg
.Explanation
decipher_rot13
is equivalent toencipher_rot13
as bothdecipher_shift
with a key of 13 andencipher_shift
key with a key of 13 will return the same results. Nonetheless,decipher_rot13
has nonetheless been explicitly defined here for consistency.Examples
>>> from sympy.crypto.crypto import encipher_rot13, decipher_rot13 >>> msg = 'GONAVYBEATARMY' >>> ciphertext = encipher_rot13(msg);ciphertext 'TBANILORNGNEZL' >>> decipher_rot13(ciphertext) 'GONAVYBEATARMY' >>> encipher_rot13(msg) == decipher_rot13(msg) True >>> msg == decipher_rot13(ciphertext) True
- sympy.crypto.crypto.encipher_affine(msg, key, symbols=None, _inverse=False)[source]¶
Performs the affine cipher encryption on plaintext
msg
, and returns the ciphertext.- Parameters:
msg : str
Characters that appear in
symbols
.a, b : int, int
A pair integers, with
gcd(a, N) = 1
(the secret key).symbols
String of characters (default = uppercase letters).
When no symbols are given,
msg
is converted to upper case letters and all other characters are ignored.- Returns:
ct
String of characters (the ciphertext message)
Explanation
Encryption is based on the map \(x \rightarrow ax+b\) (mod \(N\)) where
N
is the number of characters in the alphabet. Decryption is based on the map \(x \rightarrow cx+d\) (mod \(N\)), where \(c = a^{-1}\) (mod \(N\)) and \(d = -a^{-1}b\) (mod \(N\)). In particular, for the map to be invertible, we need \(\mathrm{gcd}(a, N) = 1\) and an error will be raised if this is not true.Notes
ALGORITHM:
- STEPS:
Number the letters of the alphabet from 0, …, N
Compute from the string
msg
a listL1
of corresponding integers.Compute from the list
L1
a new listL2
, given by replacingx
bya*x + b (mod N)
, for each elementx
inL1
.Compute from the list
L2
a stringct
of corresponding letters.
This is a straightforward generalization of the shift cipher with the added complexity of requiring 2 characters to be deciphered in order to recover the key.
See also
References
- sympy.crypto.crypto.decipher_affine(msg, key, symbols=None)[source]¶
Return the deciphered text that was made from the mapping, \(x \rightarrow ax+b\) (mod \(N\)), where
N
is the number of characters in the alphabet. Deciphering is done by reciphering with a new key: \(x \rightarrow cx+d\) (mod \(N\)), where \(c = a^{-1}\) (mod \(N\)) and \(d = -a^{-1}b\) (mod \(N\)).Examples
>>> from sympy.crypto.crypto import encipher_affine, decipher_affine >>> msg = "GO NAVY BEAT ARMY" >>> key = (3, 1) >>> encipher_affine(msg, key) 'TROBMVENBGBALV' >>> decipher_affine(_, key) 'GONAVYBEATARMY'
See also
- sympy.crypto.crypto.encipher_atbash(msg, symbols=None)[source]¶
Enciphers a given
msg
into its Atbash ciphertext and returns it.Explanation
Atbash is a substitution cipher originally used to encrypt the Hebrew alphabet. Atbash works on the principle of mapping each alphabet to its reverse / counterpart (i.e. a would map to z, b to y etc.)
Atbash is functionally equivalent to the affine cipher with
a = 25
andb = 25
See also
- sympy.crypto.crypto.decipher_atbash(msg, symbols=None)[source]¶
Deciphers a given
msg
using Atbash cipher and returns it.Explanation
decipher_atbash
is functionally equivalent toencipher_atbash
. However, it has still been added as a separate function to maintain consistency.Examples
>>> from sympy.crypto.crypto import encipher_atbash, decipher_atbash >>> msg = 'GONAVYBEATARMY' >>> encipher_atbash(msg) 'TLMZEBYVZGZINB' >>> decipher_atbash(msg) 'TLMZEBYVZGZINB' >>> encipher_atbash(msg) == decipher_atbash(msg) True >>> msg == encipher_atbash(encipher_atbash(msg)) True
See also
References
- sympy.crypto.crypto.encipher_substitution(msg, old, new=None)[source]¶
Returns the ciphertext obtained by replacing each character that appears in
old
with the corresponding character innew
. Ifold
is a mapping, then new is ignored and the replacements defined byold
are used.Explanation
This is a more general than the affine cipher in that the key can only be recovered by determining the mapping for each symbol. Though in practice, once a few symbols are recognized the mappings for other characters can be quickly guessed.
Examples
>>> from sympy.crypto.crypto import encipher_substitution, AZ >>> old = 'OEYAG' >>> new = '034^6' >>> msg = AZ("go navy! beat army!") >>> ct = encipher_substitution(msg, old, new); ct '60N^V4B3^T^RM4'
To decrypt a substitution, reverse the last two arguments:
>>> encipher_substitution(ct, new, old) 'GONAVYBEATARMY'
In the special case where
old
andnew
are a permutation of order 2 (representing a transposition of characters) their order is immaterial:>>> old = 'NAVY' >>> new = 'ANYV' >>> encipher = lambda x: encipher_substitution(x, old, new) >>> encipher('NAVY') 'ANYV' >>> encipher(_) 'NAVY'
The substitution cipher, in general, is a method whereby “units” (not necessarily single characters) of plaintext are replaced with ciphertext according to a regular system.
>>> ords = dict(zip('abc', ['\\%i' % ord(i) for i in 'abc'])) >>> print(encipher_substitution('abc', ords)) \97\98\99
References
- sympy.crypto.crypto.encipher_vigenere(msg, key, symbols=None)[source]¶
Performs the Vigenere cipher encryption on plaintext
msg
, and returns the ciphertext.Examples
>>> from sympy.crypto.crypto import encipher_vigenere, AZ >>> key = "encrypt" >>> msg = "meet me on monday" >>> encipher_vigenere(msg, key) 'QRGKKTHRZQEBPR'
Section 1 of the Kryptos sculpture at the CIA headquarters uses this cipher and also changes the order of the alphabet [R158]. Here is the first line of that section of the sculpture:
>>> from sympy.crypto.crypto import decipher_vigenere, padded_key >>> alp = padded_key('KRYPTOS', AZ()) >>> key = 'PALIMPSEST' >>> msg = 'EMUFPHZLRFAXYUSDJKZLDKRNSHGNFIVJ' >>> decipher_vigenere(msg, key, alp) 'BETWEENSUBTLESHADINGANDTHEABSENC'
Explanation
The Vigenere cipher is named after Blaise de Vigenere, a sixteenth century diplomat and cryptographer, by a historical accident. Vigenere actually invented a different and more complicated cipher. The so-called Vigenere cipher was actually invented by Giovan Batista Belaso in 1553.
This cipher was used in the 1800’s, for example, during the American Civil War. The Confederacy used a brass cipher disk to implement the Vigenere cipher (now on display in the NSA Museum in Fort Meade) [R157].
The Vigenere cipher is a generalization of the shift cipher. Whereas the shift cipher shifts each letter by the same amount (that amount being the key of the shift cipher) the Vigenere cipher shifts a letter by an amount determined by the key (which is a word or phrase known only to the sender and receiver).
For example, if the key was a single letter, such as “C”, then the so-called Vigenere cipher is actually a shift cipher with a shift of \(2\) (since “C” is the 2nd letter of the alphabet, if you start counting at \(0\)). If the key was a word with two letters, such as “CA”, then the so-called Vigenere cipher will shift letters in even positions by \(2\) and letters in odd positions are left alone (shifted by \(0\), since “A” is the 0th letter, if you start counting at \(0\)).
ALGORITHM:
INPUT:
msg
: string of characters that appear insymbols
(the plaintext)key
: a string of characters that appear insymbols
(the secret key)symbols
: a string of letters defining the alphabetOUTPUT:
ct
: string of characters (the ciphertext message)- STEPS:
Number the letters of the alphabet from 0, …, N
Compute from the string
key
a listL1
of corresponding integers. Letn1 = len(L1)
.Compute from the string
msg
a listL2
of corresponding integers. Letn2 = len(L2)
.Break
L2
up sequentially into sublists of sizen1
; the last sublist may be smaller thann1
For each of these sublists
L
ofL2
, compute a new listC
given byC[i] = L[i] + L1[i] (mod N)
to thei
-th element in the sublist, for eachi
.Assemble these lists
C
by concatenation into a new list of lengthn2
.Compute from the new list a string
ct
of corresponding letters.
Once it is known that the key is, say, \(n\) characters long, frequency analysis can be applied to every \(n\)-th letter of the ciphertext to determine the plaintext. This method is called Kasiski examination (although it was first discovered by Babbage). If they key is as long as the message and is comprised of randomly selected characters – a one-time pad – the message is theoretically unbreakable.
The cipher Vigenere actually discovered is an “auto-key” cipher described as follows.
ALGORITHM:
INPUT:
key
: a string of letters (the secret key)msg
: string of letters (the plaintext message)OUTPUT:
ct
: string of upper-case letters (the ciphertext message)- STEPS:
Number the letters of the alphabet from 0, …, N
Compute from the string
msg
a listL2
of corresponding integers. Letn2 = len(L2)
.Let
n1
be the length of the key. Append to the stringkey
the firstn2 - n1
characters of the plaintext message. Compute from this string (also of lengthn2
) a listL1
of integers corresponding to the letter numbers in the first step.Compute a new list
C
given byC[i] = L1[i] + L2[i] (mod N)
.Compute from the new list a string
ct
of letters corresponding to the new integers.
To decipher the auto-key ciphertext, the key is used to decipher the first
n1
characters and then those characters become the key to decipher the nextn1
characters, etc…:>>> m = AZ('go navy, beat army! yes you can'); m 'GONAVYBEATARMYYESYOUCAN' >>> key = AZ('gold bug'); n1 = len(key); n2 = len(m) >>> auto_key = key + m[:n2 - n1]; auto_key 'GOLDBUGGONAVYBEATARMYYE' >>> ct = encipher_vigenere(m, auto_key); ct 'MCYDWSHKOGAMKZCELYFGAYR' >>> n1 = len(key) >>> pt = [] >>> while ct: ... part, ct = ct[:n1], ct[n1:] ... pt.append(decipher_vigenere(part, key)) ... key = pt[-1] ... >>> ''.join(pt) == m True
References
- sympy.crypto.crypto.decipher_vigenere(msg, key, symbols=None)[source]¶
Decode using the Vigenere cipher.
Examples
>>> from sympy.crypto.crypto import decipher_vigenere >>> key = "encrypt" >>> ct = "QRGK kt HRZQE BPR" >>> decipher_vigenere(ct, key) 'MEETMEONMONDAY'
- sympy.crypto.crypto.encipher_hill(msg, key, symbols=None, pad='Q')[source]¶
Return the Hill cipher encryption of
msg
.- Parameters:
msg
Plaintext message of \(n\) upper-case letters.
key
A \(k \times k\) invertible matrix \(K\), all of whose entries are in \(Z_{26}\) (or whatever number of symbols are being used).
pad
Character (default “Q”) to use to make length of text be a multiple of
k
.- Returns:
ct
Ciphertext of upper-case letters.
Explanation
The Hill cipher [R159], invented by Lester S. Hill in the 1920’s [R160], was the first polygraphic cipher in which it was practical (though barely) to operate on more than three symbols at once. The following discussion assumes an elementary knowledge of matrices.
First, each letter is first encoded as a number starting with 0. Suppose your message \(msg\) consists of \(n\) capital letters, with no spaces. This may be regarded an \(n\)-tuple M of elements of \(Z_{26}\) (if the letters are those of the English alphabet). A key in the Hill cipher is a \(k x k\) matrix \(K\), all of whose entries are in \(Z_{26}\), such that the matrix \(K\) is invertible (i.e., the linear transformation \(K: Z_{N}^k \rightarrow Z_{N}^k\) is one-to-one).
Notes
ALGORITHM:
- STEPS:
Number the letters of the alphabet from 0, …, N
Compute from the string
msg
a listL
of corresponding integers. Letn = len(L)
.Break the list
L
up intot = ceiling(n/k)
sublistsL_1
, …,L_t
of sizek
(with the last list “padded” to ensure its size isk
).Compute new list
C_1
, …,C_t
given byC[i] = K*L_i
(arithmetic is done mod N), for eachi
.Concatenate these into a list
C = C_1 + ... + C_t
.Compute from
C
a stringct
of corresponding letters. This has lengthk*t
.
See also
References
- sympy.crypto.crypto.decipher_hill(msg, key, symbols=None)[source]¶
Deciphering is the same as enciphering but using the inverse of the key matrix.
Examples
>>> from sympy.crypto.crypto import encipher_hill, decipher_hill >>> from sympy import Matrix
>>> key = Matrix([[1, 2], [3, 5]]) >>> encipher_hill("meet me on monday", key) 'UEQDUEODOCTCWQ' >>> decipher_hill(_, key) 'MEETMEONMONDAY'
When the length of the plaintext (stripped of invalid characters) is not a multiple of the key dimension, extra characters will appear at the end of the enciphered and deciphered text. In order to decipher the text, those characters must be included in the text to be deciphered. In the following, the key has a dimension of 4 but the text is 2 short of being a multiple of 4 so two characters will be added.
>>> key = Matrix([[1, 1, 1, 2], [0, 1, 1, 0], ... [2, 2, 3, 4], [1, 1, 0, 1]]) >>> msg = "ST" >>> encipher_hill(msg, key) 'HJEB' >>> decipher_hill(_, key) 'STQQ' >>> encipher_hill(msg, key, pad="Z") 'ISPK' >>> decipher_hill(_, key) 'STZZ'
If the last two characters of the ciphertext were ignored in either case, the wrong plaintext would be recovered:
>>> decipher_hill("HD", key) 'ORMV' >>> decipher_hill("IS", key) 'UIKY'
See also
- sympy.crypto.crypto.encipher_bifid(msg, key, symbols=None)[source]¶
Performs the Bifid cipher encryption on plaintext
msg
, and returns the ciphertext.This is the version of the Bifid cipher that uses an \(n \times n\) Polybius square.
- Parameters:
msg
Plaintext string.
key
Short string for key.
Duplicate characters are ignored and then it is padded with the characters in
symbols
that were not in the short key.symbols
\(n \times n\) characters defining the alphabet.
(default is string.printable)
- Returns:
ciphertext
Ciphertext using Bifid5 cipher without spaces.
See also
References
- sympy.crypto.crypto.decipher_bifid(msg, key, symbols=None)[source]¶
Performs the Bifid cipher decryption on ciphertext
msg
, and returns the plaintext.This is the version of the Bifid cipher that uses the \(n \times n\) Polybius square.
- Parameters:
msg
Ciphertext string.
key
Short string for key.
Duplicate characters are ignored and then it is padded with the characters in symbols that were not in the short key.
symbols
\(n \times n\) characters defining the alphabet.
(default=string.printable, a \(10 \times 10\) matrix)
- Returns:
deciphered
Deciphered text.
Examples
>>> from sympy.crypto.crypto import ( ... encipher_bifid, decipher_bifid, AZ)
Do an encryption using the bifid5 alphabet:
>>> alp = AZ().replace('J', '') >>> ct = AZ("meet me on monday!") >>> key = AZ("gold bug") >>> encipher_bifid(ct, key, alp) 'IEILHHFSTSFQYE'
When entering the text or ciphertext, spaces are ignored so it can be formatted as desired. Re-entering the ciphertext from the preceding, putting 4 characters per line and padding with an extra J, does not cause problems for the deciphering:
>>> decipher_bifid(''' ... IEILH ... HFSTS ... FQYEJ''', key, alp) 'MEETMEONMONDAY'
When no alphabet is given, all 100 printable characters will be used:
>>> key = '' >>> encipher_bifid('hello world!', key) 'bmtwmg-bIo*w' >>> decipher_bifid(_, key) 'hello world!'
If the key is changed, a different encryption is obtained:
>>> key = 'gold bug' >>> encipher_bifid('hello world!', 'gold_bug') 'hg2sfuei7t}w'
And if the key used to decrypt the message is not exact, the original text will not be perfectly obtained:
>>> decipher_bifid(_, 'gold pug') 'heldo~wor6d!'
- sympy.crypto.crypto.bifid5_square(key=None)[source]¶
5x5 Polybius square.
Produce the Polybius square for the \(5 \times 5\) Bifid cipher.
Examples
>>> from sympy.crypto.crypto import bifid5_square >>> bifid5_square("gold bug") Matrix([ [G, O, L, D, B], [U, A, C, E, F], [H, I, K, M, N], [P, Q, R, S, T], [V, W, X, Y, Z]])
- sympy.crypto.crypto.encipher_bifid5(msg, key)[source]¶
Performs the Bifid cipher encryption on plaintext
msg
, and returns the ciphertext.- Parameters:
msg : str
Plaintext string.
Converted to upper case and filtered of anything but all letters except J.
key
Short string for key; non-alphabetic letters, J and duplicated characters are ignored and then, if the length is less than 25 characters, it is padded with other letters of the alphabet (in alphabetical order).
- Returns:
ct
Ciphertext (all caps, no spaces).
Explanation
This is the version of the Bifid cipher that uses the \(5 \times 5\) Polybius square. The letter “J” is ignored so it must be replaced with something else (traditionally an “I”) before encryption.
ALGORITHM: (5x5 case)
- STEPS:
Create the \(5 \times 5\) Polybius square
S
associated tokey
as follows:moving from left-to-right, top-to-bottom, place the letters of the key into a \(5 \times 5\) matrix,
if the key has less than 25 letters, add the letters of the alphabet not in the key until the \(5 \times 5\) square is filled.
Create a list
P
of pairs of numbers which are the coordinates in the Polybius square of the letters inmsg
.Let
L1
be the list of all first coordinates ofP
(length ofL1 = n
), letL2
be the list of all second coordinates ofP
(so the length ofL2
is alson
).Let
L
be the concatenation ofL1
andL2
(lengthL = 2*n
), except that consecutive numbers are paired(L[2*i], L[2*i + 1])
. You can regardL
as a list of pairs of lengthn
.Let
C
be the list of all letters which are of the formS[i, j]
, for all(i, j)
inL
. As a string, this is the ciphertext ofmsg
.
Examples
>>> from sympy.crypto.crypto import ( ... encipher_bifid5, decipher_bifid5)
“J” will be omitted unless it is replaced with something else:
>>> round_trip = lambda m, k: \ ... decipher_bifid5(encipher_bifid5(m, k), k) >>> key = 'a' >>> msg = "JOSIE" >>> round_trip(msg, key) 'OSIE' >>> round_trip(msg.replace("J", "I"), key) 'IOSIE' >>> j = "QIQ" >>> round_trip(msg.replace("J", j), key).replace(j, "J") 'JOSIE'
Notes
The Bifid cipher was invented around 1901 by Felix Delastelle. It is a fractional substitution cipher, where letters are replaced by pairs of symbols from a smaller alphabet. The cipher uses a \(5 \times 5\) square filled with some ordering of the alphabet, except that “J” is replaced with “I” (this is a so-called Polybius square; there is a \(6 \times 6\) analog if you add back in “J” and also append onto the usual 26 letter alphabet, the digits 0, 1, …, 9). According to Helen Gaines’ book Cryptanalysis, this type of cipher was used in the field by the German Army during World War I.
See also
- sympy.crypto.crypto.decipher_bifid5(msg, key)[source]¶
Return the Bifid cipher decryption of
msg
.- Parameters:
msg
Ciphertext string.
key
Short string for key; duplicated characters are ignored and if the length is less then 25 characters, it will be padded with other letters from the alphabet omitting “J”. Non-alphabetic characters are ignored.
- Returns:
plaintext
Plaintext from Bifid5 cipher (all caps, no spaces).
Explanation
This is the version of the Bifid cipher that uses the \(5 \times 5\) Polybius square; the letter “J” is ignored unless a
key
of length 25 is used.Examples
>>> from sympy.crypto.crypto import encipher_bifid5, decipher_bifid5 >>> key = "gold bug" >>> encipher_bifid5('meet me on friday', key) 'IEILEHFSTSFXEE' >>> encipher_bifid5('meet me on monday', key) 'IEILHHFSTSFQYE' >>> decipher_bifid5(_, key) 'MEETMEONMONDAY'
- sympy.crypto.crypto.encipher_bifid6(msg, key)[source]¶
Performs the Bifid cipher encryption on plaintext
msg
, and returns the ciphertext.This is the version of the Bifid cipher that uses the \(6 \times 6\) Polybius square.
- Parameters:
msg
Plaintext string (digits okay).
key
Short string for key (digits okay).
If
key
is less than 36 characters long, the square will be filled with letters A through Z and digits 0 through 9.- Returns:
ciphertext
Ciphertext from Bifid cipher (all caps, no spaces).
See also
- sympy.crypto.crypto.decipher_bifid6(msg, key)[source]¶
Performs the Bifid cipher decryption on ciphertext
msg
, and returns the plaintext.This is the version of the Bifid cipher that uses the \(6 \times 6\) Polybius square.
- Parameters:
msg
Ciphertext string (digits okay); converted to upper case
key
Short string for key (digits okay).
If
key
is less than 36 characters long, the square will be filled with letters A through Z and digits 0 through 9. All letters are converted to uppercase.- Returns:
plaintext
Plaintext from Bifid cipher (all caps, no spaces).
Examples
>>> from sympy.crypto.crypto import encipher_bifid6, decipher_bifid6 >>> key = "gold bug" >>> encipher_bifid6('meet me on monday at 8am', key) 'KFKLJJHF5MMMKTFRGPL' >>> decipher_bifid6(_, key) 'MEETMEONMONDAYAT8AM'
- sympy.crypto.crypto.bifid6_square(key=None)[source]¶
6x6 Polybius square.
Produces the Polybius square for the \(6 \times 6\) Bifid cipher. Assumes alphabet of symbols is “A”, …, “Z”, “0”, …, “9”.
Examples
>>> from sympy.crypto.crypto import bifid6_square >>> key = "gold bug" >>> bifid6_square(key) Matrix([ [G, O, L, D, B, U], [A, C, E, F, H, I], [J, K, M, N, P, Q], [R, S, T, V, W, X], [Y, Z, 0, 1, 2, 3], [4, 5, 6, 7, 8, 9]])
- sympy.crypto.crypto.rsa_public_key(*args, **kwargs)[source]¶
Return the RSA public key pair, \((n, e)\)
- Parameters:
args : naturals
If specified as \(p, q, e\) where \(p\) and \(q\) are distinct primes and \(e\) is a desired public exponent of the RSA, \(n = p q\) and \(e\) will be verified against the totient \(\phi(n)\) (Euler totient) or \(\lambda(n)\) (Carmichael totient) to be \(\gcd(e, \phi(n)) = 1\) or \(\gcd(e, \lambda(n)) = 1\).
If specified as \(p_1, p_2, \dots, p_n, e\) where \(p_1, p_2, \dots, p_n\) are specified as primes, and \(e\) is specified as a desired public exponent of the RSA, it will be able to form a multi-prime RSA, which is a more generalized form of the popular 2-prime RSA.
It can also be possible to form a single-prime RSA by specifying the argument as \(p, e\), which can be considered a trivial case of a multiprime RSA.
Furthermore, it can be possible to form a multi-power RSA by specifying two or more pairs of the primes to be same. However, unlike the two-distinct prime RSA or multi-prime RSA, not every numbers in the complete residue system (\(\mathbb{Z}_n\)) will be decryptable since the mapping \(\mathbb{Z}_{n} \rightarrow \mathbb{Z}_{n}\) will not be bijective. (Only except for the trivial case when \(e = 1\) or more generally,
\[e \in \left \{ 1 + k \lambda(n) \mid k \in \mathbb{Z} \land k \geq 0 \right \}\]when RSA reduces to the identity.) However, the RSA can still be decryptable for the numbers in the reduced residue system (\(\mathbb{Z}_n^{\times}\)), since the mapping \(\mathbb{Z}_{n}^{\times} \rightarrow \mathbb{Z}_{n}^{\times}\) can still be bijective.
If you pass a non-prime integer to the arguments \(p_1, p_2, \dots, p_n\), the particular number will be prime-factored and it will become either a multi-prime RSA or a multi-power RSA in its canonical form, depending on whether the product equals its radical or not. \(p_1 p_2 \dots p_n = \text{rad}(p_1 p_2 \dots p_n)\)
totient : bool, optional
If
'Euler'
, it uses Euler’s totient \(\phi(n)\) which issympy.functions.combinatorial.numbers.totient()
in SymPy.If
'Carmichael'
, it uses Carmichael’s totient \(\lambda(n)\) which issympy.functions.combinatorial.numbers.reduced_totient()
in SymPy.Unlike private key generation, this is a trivial keyword for public key generation because \(\gcd(e, \phi(n)) = 1 \iff \gcd(e, \lambda(n)) = 1\).
index : nonnegative integer, optional
Returns an arbitrary solution of a RSA public key at the index specified at \(0, 1, 2, \dots\). This parameter needs to be specified along with
totient='Carmichael'
.Similarly to the non-uniquenss of a RSA private key as described in the
index
parameter documentation inrsa_private_key()
, RSA public key is also not unique and there is an infinite number of RSA public exponents which can behave in the same manner.From any given RSA public exponent \(e\), there are can be an another RSA public exponent \(e + k \lambda(n)\) where \(k\) is an integer, \(\lambda\) is a Carmichael’s totient function.
However, considering only the positive cases, there can be a principal solution of a RSA public exponent \(e_0\) in \(0 < e_0 < \lambda(n)\), and all the other solutions can be canonicalzed in a form of \(e_0 + k \lambda(n)\).
index
specifies the \(k\) notation to yield any possible value an RSA public key can have.An example of computing any arbitrary RSA public key:
>>> from sympy.crypto.crypto import rsa_public_key >>> rsa_public_key(61, 53, 17, totient='Carmichael', index=0) (3233, 17) >>> rsa_public_key(61, 53, 17, totient='Carmichael', index=1) (3233, 797) >>> rsa_public_key(61, 53, 17, totient='Carmichael', index=2) (3233, 1577)
multipower : bool, optional
Any pair of non-distinct primes found in the RSA specification will restrict the domain of the cryptosystem, as noted in the explanation of the parameter
args
.SymPy RSA key generator may give a warning before dispatching it as a multi-power RSA, however, you can disable the warning if you pass
True
to this keyword.- Returns:
(n, e) : int, int
\(n\) is a product of any arbitrary number of primes given as the argument.
\(e\) is relatively prime (coprime) to the Euler totient \(\phi(n)\).
False
Returned if less than two arguments are given, or \(e\) is not relatively prime to the modulus.
Examples
>>> from sympy.crypto.crypto import rsa_public_key
A public key of a two-prime RSA:
>>> p, q, e = 3, 5, 7 >>> rsa_public_key(p, q, e) (15, 7) >>> rsa_public_key(p, q, 30) False
A public key of a multiprime RSA:
>>> primes = [2, 3, 5, 7, 11, 13] >>> e = 7 >>> args = primes + [e] >>> rsa_public_key(*args) (30030, 7)
Notes
Although the RSA can be generalized over any modulus \(n\), using two large primes had became the most popular specification because a product of two large primes is usually the hardest to factor relatively to the digits of \(n\) can have.
However, it may need further understanding of the time complexities of each prime-factoring algorithms to verify the claim.
See also
References
- sympy.crypto.crypto.rsa_private_key(*args, **kwargs)[source]¶
Return the RSA private key pair, \((n, d)\)
- Parameters:
args : naturals
The keyword is identical to the
args
inrsa_public_key()
.totient : bool, optional
If
'Euler'
, it uses Euler’s totient convention \(\phi(n)\) which issympy.functions.combinatorial.numbers.totient()
in SymPy.If
'Carmichael'
, it uses Carmichael’s totient convention \(\lambda(n)\) which issympy.functions.combinatorial.numbers.reduced_totient()
in SymPy.There can be some output differences for private key generation as examples below.
Example using Euler’s totient:
>>> from sympy.crypto.crypto import rsa_private_key >>> rsa_private_key(61, 53, 17, totient='Euler') (3233, 2753)
Example using Carmichael’s totient:
>>> from sympy.crypto.crypto import rsa_private_key >>> rsa_private_key(61, 53, 17, totient='Carmichael') (3233, 413)
index : nonnegative integer, optional
Returns an arbitrary solution of a RSA private key at the index specified at \(0, 1, 2, \dots\). This parameter needs to be specified along with
totient='Carmichael'
.RSA private exponent is a non-unique solution of \(e d \mod \lambda(n) = 1\) and it is possible in any form of \(d + k \lambda(n)\), where \(d\) is an another already-computed private exponent, and \(\lambda\) is a Carmichael’s totient function, and \(k\) is any integer.
However, considering only the positive cases, there can be a principal solution of a RSA private exponent \(d_0\) in \(0 < d_0 < \lambda(n)\), and all the other solutions can be canonicalzed in a form of \(d_0 + k \lambda(n)\).
index
specifies the \(k\) notation to yield any possible value an RSA private key can have.An example of computing any arbitrary RSA private key:
>>> from sympy.crypto.crypto import rsa_private_key >>> rsa_private_key(61, 53, 17, totient='Carmichael', index=0) (3233, 413) >>> rsa_private_key(61, 53, 17, totient='Carmichael', index=1) (3233, 1193) >>> rsa_private_key(61, 53, 17, totient='Carmichael', index=2) (3233, 1973)
multipower : bool, optional
The keyword is identical to the
multipower
inrsa_public_key()
.- Returns:
(n, d) : int, int
\(n\) is a product of any arbitrary number of primes given as the argument.
\(d\) is the inverse of \(e\) (mod \(\phi(n)\)) where \(e\) is the exponent given, and \(\phi\) is a Euler totient.
False
Returned if less than two arguments are given, or \(e\) is not relatively prime to the totient of the modulus.
Examples
>>> from sympy.crypto.crypto import rsa_private_key
A private key of a two-prime RSA:
>>> p, q, e = 3, 5, 7 >>> rsa_private_key(p, q, e) (15, 7) >>> rsa_private_key(p, q, 30) False
A private key of a multiprime RSA:
>>> primes = [2, 3, 5, 7, 11, 13] >>> e = 7 >>> args = primes + [e] >>> rsa_private_key(*args) (30030, 823)
See also
References
- sympy.crypto.crypto.encipher_rsa(i, key, factors=None)[source]¶
Encrypt the plaintext with RSA.
- Parameters:
i : integer
The plaintext to be encrypted for.
key : (n, e) where n, e are integers
\(n\) is the modulus of the key and \(e\) is the exponent of the key. The encryption is computed by \(i^e \bmod n\).
The key can either be a public key or a private key, however, the message encrypted by a public key can only be decrypted by a private key, and vice versa, as RSA is an asymmetric cryptography system.
factors : list of coprime integers
This is identical to the keyword
factors
indecipher_rsa()
.
Notes
Some specifications may make the RSA not cryptographically meaningful.
For example, \(0\), \(1\) will remain always same after taking any number of exponentiation, thus, should be avoided.
Furthermore, if \(i^e < n\), \(i\) may easily be figured out by taking \(e\) th root.
And also, specifying the exponent as \(1\) or in more generalized form as \(1 + k \lambda(n)\) where \(k\) is an nonnegative integer, \(\lambda\) is a carmichael totient, the RSA becomes an identity mapping.
Examples
>>> from sympy.crypto.crypto import encipher_rsa >>> from sympy.crypto.crypto import rsa_public_key, rsa_private_key
Public Key Encryption:
>>> p, q, e = 3, 5, 7 >>> puk = rsa_public_key(p, q, e) >>> msg = 12 >>> encipher_rsa(msg, puk) 3
Private Key Encryption:
>>> p, q, e = 3, 5, 7 >>> prk = rsa_private_key(p, q, e) >>> msg = 12 >>> encipher_rsa(msg, prk) 3
Encryption using chinese remainder theorem:
>>> encipher_rsa(msg, prk, factors=[p, q]) 3
- sympy.crypto.crypto.decipher_rsa(i, key, factors=None)[source]¶
Decrypt the ciphertext with RSA.
- Parameters:
i : integer
The ciphertext to be decrypted for.
key : (n, d) where n, d are integers
\(n\) is the modulus of the key and \(d\) is the exponent of the key. The decryption is computed by \(i^d \bmod n\).
The key can either be a public key or a private key, however, the message encrypted by a public key can only be decrypted by a private key, and vice versa, as RSA is an asymmetric cryptography system.
factors : list of coprime integers
As the modulus \(n\) created from RSA key generation is composed of arbitrary prime factors \(n = {p_1}^{k_1}{p_2}^{k_2}\dots{p_n}^{k_n}\) where \(p_1, p_2, \dots, p_n\) are distinct primes and \(k_1, k_2, \dots, k_n\) are positive integers, chinese remainder theorem can be used to compute \(i^d \bmod n\) from the fragmented modulo operations like
\[i^d \bmod {p_1}^{k_1}, i^d \bmod {p_2}^{k_2}, \dots, i^d \bmod {p_n}^{k_n}\]or like
\[i^d \bmod {p_1}^{k_1}{p_2}^{k_2}, i^d \bmod {p_3}^{k_3}, \dots , i^d \bmod {p_n}^{k_n}\]as long as every moduli does not share any common divisor each other.
The raw primes used in generating the RSA key pair can be a good option.
Note that the speed advantage of using this is only viable for very large cases (Like 2048-bit RSA keys) since the overhead of using pure Python implementation of
sympy.ntheory.modular.crt()
may overcompensate the theoretical speed advantage.
Notes
See the
Notes
section in the documentation ofencipher_rsa()
Examples
>>> from sympy.crypto.crypto import decipher_rsa, encipher_rsa >>> from sympy.crypto.crypto import rsa_public_key, rsa_private_key
Public Key Encryption and Decryption:
>>> p, q, e = 3, 5, 7 >>> prk = rsa_private_key(p, q, e) >>> puk = rsa_public_key(p, q, e) >>> msg = 12 >>> new_msg = encipher_rsa(msg, prk) >>> new_msg 3 >>> decipher_rsa(new_msg, puk) 12
Private Key Encryption and Decryption:
>>> p, q, e = 3, 5, 7 >>> prk = rsa_private_key(p, q, e) >>> puk = rsa_public_key(p, q, e) >>> msg = 12 >>> new_msg = encipher_rsa(msg, puk) >>> new_msg 3 >>> decipher_rsa(new_msg, prk) 12
Decryption using chinese remainder theorem:
>>> decipher_rsa(new_msg, prk, factors=[p, q]) 12
See also
- sympy.crypto.crypto.kid_rsa_public_key(a, b, A, B)[source]¶
Kid RSA is a version of RSA useful to teach grade school children since it does not involve exponentiation.
Explanation
Alice wants to talk to Bob. Bob generates keys as follows. Key generation:
Select positive integers \(a, b, A, B\) at random.
Compute \(M = a b - 1\), \(e = A M + a\), \(d = B M + b\), \(n = (e d - 1)//M\).
The public key is \((n, e)\). Bob sends these to Alice.
The private key is \((n, d)\), which Bob keeps secret.
Encryption: If \(p\) is the plaintext message then the ciphertext is \(c = p e \pmod n\).
Decryption: If \(c\) is the ciphertext message then the plaintext is \(p = c d \pmod n\).
Examples
>>> from sympy.crypto.crypto import kid_rsa_public_key >>> a, b, A, B = 3, 4, 5, 6 >>> kid_rsa_public_key(a, b, A, B) (369, 58)
- sympy.crypto.crypto.kid_rsa_private_key(a, b, A, B)[source]¶
Compute \(M = a b - 1\), \(e = A M + a\), \(d = B M + b\), \(n = (e d - 1) / M\). The private key is \(d\), which Bob keeps secret.
Examples
>>> from sympy.crypto.crypto import kid_rsa_private_key >>> a, b, A, B = 3, 4, 5, 6 >>> kid_rsa_private_key(a, b, A, B) (369, 70)
- sympy.crypto.crypto.encipher_kid_rsa(msg, key)[source]¶
Here
msg
is the plaintext andkey
is the public key.Examples
>>> from sympy.crypto.crypto import ( ... encipher_kid_rsa, kid_rsa_public_key) >>> msg = 200 >>> a, b, A, B = 3, 4, 5, 6 >>> key = kid_rsa_public_key(a, b, A, B) >>> encipher_kid_rsa(msg, key) 161
- sympy.crypto.crypto.decipher_kid_rsa(msg, key)[source]¶
Here
msg
is the plaintext andkey
is the private key.Examples
>>> from sympy.crypto.crypto import ( ... kid_rsa_public_key, kid_rsa_private_key, ... decipher_kid_rsa, encipher_kid_rsa) >>> a, b, A, B = 3, 4, 5, 6 >>> d = kid_rsa_private_key(a, b, A, B) >>> msg = 200 >>> pub = kid_rsa_public_key(a, b, A, B) >>> pri = kid_rsa_private_key(a, b, A, B) >>> ct = encipher_kid_rsa(msg, pub) >>> decipher_kid_rsa(ct, pri) 200
- sympy.crypto.crypto.encode_morse(msg, sep='|', mapping=None)[source]¶
Encodes a plaintext into popular Morse Code with letters separated by
sep
and words by a doublesep
.Examples
>>> from sympy.crypto.crypto import encode_morse >>> msg = 'ATTACK RIGHT FLANK' >>> encode_morse(msg) '.-|-|-|.-|-.-.|-.-||.-.|..|--.|....|-||..-.|.-..|.-|-.|-.-'
References
- sympy.crypto.crypto.decode_morse(msg, sep='|', mapping=None)[source]¶
Decodes a Morse Code with letters separated by
sep
(default is ‘|’) and words by \(word_sep\) (default is ‘||) into plaintext.Examples
>>> from sympy.crypto.crypto import decode_morse >>> mc = '--|---|...-|.||.|.-|...|-' >>> decode_morse(mc) 'MOVE EAST'
References
- sympy.crypto.crypto.lfsr_sequence(key, fill, n)[source]¶
This function creates an LFSR sequence.
- Parameters:
key : list
A list of finite field elements, \([c_0, c_1, \ldots, c_k].\)
fill : list
The list of the initial terms of the LFSR sequence, \([x_0, x_1, \ldots, x_k].\)
n
Number of terms of the sequence that the function returns.
- Returns:
L
The LFSR sequence defined by \(x_{n+1} = c_k x_n + \ldots + c_0 x_{n-k}\), for \(n \leq k\).
Notes
S. Golomb [G171] gives a list of three statistical properties a sequence of numbers \(a = \{a_n\}_{n=1}^\infty\), \(a_n \in \{0,1\}\), should display to be considered “random”. Define the autocorrelation of \(a\) to be
\[C(k) = C(k,a) = \lim_{N\rightarrow \infty} {1\over N}\sum_{n=1}^N (-1)^{a_n + a_{n+k}}.\]In the case where \(a\) is periodic with period \(P\) then this reduces to
\[C(k) = {1\over P}\sum_{n=1}^P (-1)^{a_n + a_{n+k}}.\]Assume \(a\) is periodic with period \(P\).
balance:
\[\left|\sum_{n=1}^P(-1)^{a_n}\right| \leq 1.\]low autocorrelation:
\[\begin{split}C(k) = \left\{ \begin{array}{cc} 1,& k = 0,\\ \epsilon, & k \ne 0. \end{array} \right.\end{split}\](For sequences satisfying these first two properties, it is known that \(\epsilon = -1/P\) must hold.)
proportional runs property: In each period, half the runs have length \(1\), one-fourth have length \(2\), etc. Moreover, there are as many runs of \(1\)’s as there are of \(0\)’s.
Examples
>>> from sympy.crypto.crypto import lfsr_sequence >>> from sympy.polys.domains import FF >>> F = FF(2) >>> fill = [F(1), F(1), F(0), F(1)] >>> key = [F(1), F(0), F(0), F(1)] >>> lfsr_sequence(key, fill, 10) [1 mod 2, 1 mod 2, 0 mod 2, 1 mod 2, 0 mod 2, 1 mod 2, 1 mod 2, 0 mod 2, 0 mod 2, 1 mod 2]
References
- sympy.crypto.crypto.lfsr_autocorrelation(L, P, k)[source]¶
This function computes the LFSR autocorrelation function.
- Parameters:
L
A periodic sequence of elements of \(GF(2)\). L must have length larger than P.
P
The period of L.
k : int
An integer \(k\) (\(0 < k < P\)).
- Returns:
autocorrelation
The k-th value of the autocorrelation of the LFSR L.
Examples
>>> from sympy.crypto.crypto import ( ... lfsr_sequence, lfsr_autocorrelation) >>> from sympy.polys.domains import FF >>> F = FF(2) >>> fill = [F(1), F(1), F(0), F(1)] >>> key = [F(1), F(0), F(0), F(1)] >>> s = lfsr_sequence(key, fill, 20) >>> lfsr_autocorrelation(s, 15, 7) -1/15 >>> lfsr_autocorrelation(s, 15, 0) 1
- sympy.crypto.crypto.lfsr_connection_polynomial(s)[source]¶
This function computes the LFSR connection polynomial.
- Parameters:
s
A sequence of elements of even length, with entries in a finite field.
- Returns:
C(x)
The connection polynomial of a minimal LFSR yielding s.
This implements the algorithm in section 3 of J. L. Massey’s article [M172].
Examples
>>> from sympy.crypto.crypto import ( ... lfsr_sequence, lfsr_connection_polynomial) >>> from sympy.polys.domains import FF >>> F = FF(2) >>> fill = [F(1), F(1), F(0), F(1)] >>> key = [F(1), F(0), F(0), F(1)] >>> s = lfsr_sequence(key, fill, 20) >>> lfsr_connection_polynomial(s) x**4 + x + 1 >>> fill = [F(1), F(0), F(0), F(1)] >>> key = [F(1), F(1), F(0), F(1)] >>> s = lfsr_sequence(key, fill, 20) >>> lfsr_connection_polynomial(s) x**3 + 1 >>> fill = [F(1), F(0), F(1)] >>> key = [F(1), F(1), F(0)] >>> s = lfsr_sequence(key, fill, 20) >>> lfsr_connection_polynomial(s) x**3 + x**2 + 1 >>> fill = [F(1), F(0), F(1)] >>> key = [F(1), F(0), F(1)] >>> s = lfsr_sequence(key, fill, 20) >>> lfsr_connection_polynomial(s) x**3 + x + 1
References
- sympy.crypto.crypto.elgamal_public_key(key)[source]¶
Return three number tuple as public key.
- Parameters:
key : (p, r, e)
Tuple generated by
elgamal_private_key
.- Returns:
tuple : (p, r, e)
\(e = r**d \bmod p\)
\(d\) is a random number in private key.
Examples
>>> from sympy.crypto.crypto import elgamal_public_key >>> elgamal_public_key((1031, 14, 636)) (1031, 14, 212)
- sympy.crypto.crypto.elgamal_private_key(digit=10, seed=None)[source]¶
Return three number tuple as private key.
- Parameters:
digit : int
Minimum number of binary digits for key.
- Returns:
tuple : (p, r, d)
p = prime number.
r = primitive root.
d = random number.
Explanation
Elgamal encryption is based on the mathematical problem called the Discrete Logarithm Problem (DLP). For example,
\(a^{b} \equiv c \pmod p\)
In general, if
a
andb
are known,ct
is easily calculated. Ifb
is unknown, it is hard to usea
andct
to getb
.Notes
For testing purposes, the
seed
parameter may be set to control the output of this routine. See sympy.core.random._randrange.Examples
>>> from sympy.crypto.crypto import elgamal_private_key >>> from sympy.ntheory import is_primitive_root, isprime >>> a, b, _ = elgamal_private_key() >>> isprime(a) True >>> is_primitive_root(b, a) True
- sympy.crypto.crypto.encipher_elgamal(i, key, seed=None)[source]¶
Encrypt message with public key.
- Parameters:
msg
int of encoded message.
key
Public key.
- Returns:
tuple : (c1, c2)
Encipher into two number.
Explanation
i
is a plaintext message expressed as an integer.key
is public key (p, r, e). In order to encrypt a message, a random numbera
inrange(2, p)
is generated and the encryped message is returned as \(c_{1}\) and \(c_{2}\) where:\(c_{1} \equiv r^{a} \pmod p\)
\(c_{2} \equiv m e^{a} \pmod p\)
Notes
For testing purposes, the
seed
parameter may be set to control the output of this routine. See sympy.core.random._randrange.Examples
>>> from sympy.crypto.crypto import encipher_elgamal, elgamal_private_key, elgamal_public_key >>> pri = elgamal_private_key(5, seed=[3]); pri (37, 2, 3) >>> pub = elgamal_public_key(pri); pub (37, 2, 8) >>> msg = 36 >>> encipher_elgamal(msg, pub, seed=[3]) (8, 6)
- sympy.crypto.crypto.decipher_elgamal(msg, key)[source]¶
Decrypt message with private key.
\(msg = (c_{1}, c_{2})\)
\(key = (p, r, d)\)
According to extended Eucliden theorem, \(u c_{1}^{d} + p n = 1\)
\(u \equiv 1/{{c_{1}}^d} \pmod p\)
\(u c_{2} \equiv \frac{1}{c_{1}^d} c_{2} \equiv \frac{1}{r^{ad}} c_{2} \pmod p\)
\(\frac{1}{r^{ad}} m e^a \equiv \frac{1}{r^{ad}} m {r^{d a}} \equiv m \pmod p\)
Examples
>>> from sympy.crypto.crypto import decipher_elgamal >>> from sympy.crypto.crypto import encipher_elgamal >>> from sympy.crypto.crypto import elgamal_private_key >>> from sympy.crypto.crypto import elgamal_public_key
>>> pri = elgamal_private_key(5, seed=[3]) >>> pub = elgamal_public_key(pri); pub (37, 2, 8) >>> msg = 17 >>> decipher_elgamal(encipher_elgamal(msg, pub), pri) == msg True
- sympy.crypto.crypto.dh_public_key(key)[source]¶
Return three number tuple as public key.
This is the tuple that Alice sends to Bob.
- Parameters:
key : (p, g, a)
A tuple generated by
dh_private_key
.- Returns:
tuple : int, int, int
A tuple of \((p, g, g^a \mod p)\) with \(p\), \(g\) and \(a\) given as parameters.s
Examples
>>> from sympy.crypto.crypto import dh_private_key, dh_public_key >>> p, g, a = dh_private_key(); >>> _p, _g, x = dh_public_key((p, g, a)) >>> p == _p and g == _g True >>> x == pow(g, a, p) True
- sympy.crypto.crypto.dh_private_key(digit=10, seed=None)[source]¶
Return three integer tuple as private key.
- Parameters:
digit
Minimum number of binary digits required in key.
- Returns:
tuple : (p, g, a)
p = prime number.
g = primitive root of p.
a = random number from 2 through p - 1.
Explanation
Diffie-Hellman key exchange is based on the mathematical problem called the Discrete Logarithm Problem (see ElGamal).
Diffie-Hellman key exchange is divided into the following steps:
Alice and Bob agree on a base that consist of a prime
p
and a primitive root ofp
calledg
Alice choses a number
a
and Bob choses a numberb
wherea
andb
are random numbers in range \([2, p)\). These are their private keys.Alice then publicly sends Bob \(g^{a} \pmod p\) while Bob sends Alice \(g^{b} \pmod p\)
They both raise the received value to their secretly chosen number (
a
orb
) and now have both as their shared key \(g^{ab} \pmod p\)
Notes
For testing purposes, the
seed
parameter may be set to control the output of this routine. See sympy.core.random._randrange.Examples
>>> from sympy.crypto.crypto import dh_private_key >>> from sympy.ntheory import isprime, is_primitive_root >>> p, g, _ = dh_private_key() >>> isprime(p) True >>> is_primitive_root(g, p) True >>> p, g, _ = dh_private_key(5) >>> isprime(p) True >>> is_primitive_root(g, p) True
Return an integer that is the shared key.
This is what Bob and Alice can both calculate using the public keys they received from each other and their private keys.
- Parameters:
key : (p, g, x)
Tuple \((p, g, x)\) generated by
dh_public_key
.b
Random number in the range of \(2\) to \(p - 1\) (Chosen by second key exchange member (Bob)).
- Returns:
int
A shared key.
Examples
>>> from sympy.crypto.crypto import ( ... dh_private_key, dh_public_key, dh_shared_key) >>> prk = dh_private_key(); >>> p, g, x = dh_public_key(prk); >>> sk = dh_shared_key((p, g, x), 1000) >>> sk == pow(x, 1000, p) True
- sympy.crypto.crypto.gm_public_key(p, q, a=None, seed=None)[source]¶
Compute public keys for
p
andq
. Note that in Goldwasser-Micali Encryption, public keys are randomly selected.- Parameters:
p, q, a : int, int, int
Initialization variables.
- Returns:
tuple : (a, N)
a
is the inputa
if it is notNone
otherwise some random integer coprime top
andq
.N
is the product ofp
andq
.
- sympy.crypto.crypto.gm_private_key(p, q, a=None)[source]¶
Check if
p
andq
can be used as private keys for the Goldwasser-Micali encryption. The method works roughly as follows.- Parameters:
p, q, a
Initialization variables.
- Returns:
tuple : (p, q)
The input value
p
andq
.- Raises:
ValueError
If
p
andq
are not distinct odd primes.
Explanation
Pick two large primes \(p\) and \(q\).
Call their product \(N\).
Given a message as an integer \(i\), write \(i\) in its bit representation \(b_0, \dots, b_n\).
For each \(k\),
- if \(b_k = 0\):
let \(a_k\) be a random square (quadratic residue) modulo \(p q\) such that
jacobi_symbol(a, p*q) = 1
- if \(b_k = 1\):
let \(a_k\) be a random non-square (non-quadratic residue) modulo \(p q\) such that
jacobi_symbol(a, p*q) = 1
returns \(\left[a_1, a_2, \dots\right]\)
\(b_k\) can be recovered by checking whether or not \(a_k\) is a residue. And from the \(b_k\)’s, the message can be reconstructed.
The idea is that, while
jacobi_symbol(a, p*q)
can be easily computed (and when it is equal to \(-1\) will tell you that \(a\) is not a square mod \(p q\)), quadratic residuosity modulo a composite number is hard to compute without knowing its factorization.Moreover, approximately half the numbers coprime to \(p q\) have
jacobi_symbol()
equal to \(1\) . And among those, approximately half are residues and approximately half are not. This maximizes the entropy of the code.
- sympy.crypto.crypto.encipher_gm(i, key, seed=None)[source]¶
Encrypt integer ‘i’ using public_key ‘key’ Note that gm uses random encryption.
- Parameters:
i : int
The message to encrypt.
key : (a, N)
The public key.
- Returns:
list : list of int
The randomized encrypted message.
- sympy.crypto.crypto.decipher_gm(message, key)[source]¶
Decrypt message ‘message’ using public_key ‘key’.
- Parameters:
message : list of int
The randomized encrypted message.
key : (p, q)
The private key.
- Returns:
int
The encrypted message.
- sympy.crypto.crypto.encipher_railfence(message, rails)[source]¶
Performs Railfence Encryption on plaintext and returns ciphertext
- Parameters:
message : string, the message to encrypt.
rails : int, the number of rails.
- Returns:
The Encrypted string message.
Examples
>>> from sympy.crypto.crypto import encipher_railfence >>> message = "hello world" >>> encipher_railfence(message,3) 'horel ollwd'
References
- sympy.crypto.crypto.decipher_railfence(ciphertext, rails)[source]¶
Decrypt the message using the given rails
- Parameters:
message : string, the message to encrypt.
rails : int, the number of rails.
- Returns:
The Decrypted string message.
Examples
>>> from sympy.crypto.crypto import decipher_railfence >>> decipher_railfence("horel ollwd",3) 'hello world'