def hackAffine(message):
print('Hacking...')
# Python programs can be stopped at any time by pressing Ctrl-C (on
# Windows) or Ctrl-D (on macOS and Linux):
print('(Press Ctrl-C or Ctrl-D to quit at any time.)')
# Brute-force by looping through every possible key:
for key in range(len(SYMBOLS)**2):
keyA = getKeyParts(key)[0]
if gcd(keyA, len(SYMBOLS)) != 1:
continue
decryptedText = decryptMessage(key, message)
if not SILENT_MODE:
print('Tried Key %s... (%s)' % (key, decryptedText[:40]))
if isEnglish(decryptedText):
# Check with the user if the decrypted key has been found:
print()
print('Possible encryption hack:')
print('Key: %s' % key)
print('Decrypted message: ' + decryptedText[:200])
print()
print('Enter D for done, or just press Enter to continue hacking:')
response = input('> ')
if response.strip().upper().startswith('D'):
return decryptedText
return None
def generateKey(keySize):
# Creates public/private keys keySize bits in size.
p = 0
q = 0
# Step 1: Create two prime numbers, p and q. Calculate n = p * q:
print('Generating p prime...')
while p == q:
p = generateLargePrime(keySize)
q = generateLargePrime(keySize)
n = p * q
# Step 2: Create a number e that is relatively prime to (p-1)*(q-1):
print('Generating e that is relatively prime to (p-1)*(q-1)...')
while True:
# Keep trying random numbers for e until one is valid:
e = random.randrange(2**(keySize - 1), 2**keySize)
if gcd(e, (p - 1) * (q - 1)) == 1:
break
# Step 3: Calculate d, the mod inverse of e:
print('Calculating d that is mod inverse of e...')
d = findModInverse(e, (p - 1) * (q - 1))
publicKey = (n, e)
privateKey = (n, d)
print('Public key:', publicKey)
print('Private key:', privateKey)
return publicKey, privateKey
def checkKeys(keyA, keyB, mode):
if keyA == 1 and mode == 'encrypt':
sys.exit('Cipher is weak if key A is 1. Choose a different key.')
if keyB == 0 and mode == 'encrypt':
sys.exit('Cipher is weak if key B is 0. Choose a different key.')
if keyA < 0 or keyB < 0 or keyB > len(SYMBOLS) - 1:
sys.exit(
'Key A must be greater than 0 and Key B must be between 0 and %s '
% (len(SYMBOLS) - 1))
if gcd(keyA, len(SYMBOLS)) != 1:
sys.exit(
'Key A (%s) and the symbol set size (%s) are not relatively prime. Choose a different key.'
% (keyA, len(SYMBOLS)))
# Chapter 13 Practice Questions
# 1. What do the following expressions evaluate to?
print(17 % 1000)
print(5 % 5)
# 2. What is the GCD of 10 and 15?
# Don't do this - imports should be at the top of the file
from books.CrackingCodesWithPython.Chapter13.cryptomath import gcd
print(gcd(10, 15))
# 3. What does spam contain after executing spam, eggs = 'hello', 'world'?
spam, eggs = 'hello', 'world'
print(spam)
# 4. The GCD of 17 and 31 is 1. Are 17 and 31 relatively prime?
if not gcd(17, 31) == 1:
print("No")
else:
print("Yes")
# 5. Why aren't 6 and 8 relatively prime?
print(gcd(6, 8))
# 6. What is the formula for the modular inverse of A mod C?
# Hint: check page 183
SYMBOLS = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz1234567890 !?.'
print(
"The modular inverse of %s mod %s is number %s such that (%s * %s) mod %s == %s"
% (SYMBOLS[0], SYMBOLS[2], SYMBOLS[34], SYMBOLS[0], SYMBOLS[34],
SYMBOLS[2], SYMBOLS[-14]))
def getRandomKey():
while True:
keyA = random.randint(2, len(SYMBOLS))
keyB = random.randint(2, len(SYMBOLS))
if gcd(keyA, len(SYMBOLS)) == 1:
return keyA * len(SYMBOLS) + keyB
# This program proves that the keyspace of the affine cipher is limited
# to less than len(SYMBOLS) ^ 2.
from books.CrackingCodesWithPython.Chapter14.affineCipher import encryptMessage, SYMBOLS
from books.CrackingCodesWithPython.Chapter13.cryptomath import gcd
message = 'Make things as simple as possible, but not simpler.'
for keyA in range(2, 80):
key = keyA * len(SYMBOLS) + 1
if gcd(keyA, len(SYMBOLS)) == 1:
print(keyA, encryptMessage(key, message))