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# Simple Substitution Dictionary Hacker, http://inventwithpython.com/hacking (BSD Licensed)
import pyperclip, simpleSubKeyword, detectEnglish
SILENT_MODE = False
def main():
myMessage = r"""SJITDOPIQR: JIR RIQMUNQRO AY P WDQC QCR NRSMRQN JT A SJITDORO QJ CRMNRGT AY S. -PHAMJNR ADRMSR"""
brokenCiphertext = hackSimpleSubDictionary(myMessage)
if brokenCiphertext == None:
# hackSimpleSubDictionary() will return the None value if it was unable to hack the encryption.
print('Hacking failed. Unable to hack this ciphertext.')
else:
# The plaintext is displayed on the screen. For the convenience of the user, we copy the text of the code to the clipboard.
print('Copying broken ciphertext to clipboard:')
print(brokenCiphertext)
pyperclip.copy(brokenCiphertext)
def hackSimpleSubDictionary(message):
print('Hacking with %s possible dictionary words...' % (len(detectEnglish.ENGLISH_WORDS) * 3))
# Python programs can be stopped at any time by pressing Ctrl-C (on Windows) or Ctrl-D (on Mac and Linux)
print('(Press Ctrl-C or Ctrl-D to quit at any time.)')
tryNum = 1
# brute-force by looping through every possible key
for key in detectEnglish.ENGLISH_WORDS:
if tryNum % 100 == 0 and not SILENT_MODE:
print('%s keys tried. (%s)' % (tryNum, key))
decryptedText = simpleSubKeyword.decryptMessage(key, message)
if detectEnglish.getEnglishCount(decryptedText) > 0.20:
# Check with the user to see if the decrypted key has been found.
print()
print('Possible encryption hack:')
print('Key: ' + str(key))
print('Decrypted message: ' + decryptedText[:100])
print()
print('Enter D for done, or just press Enter to continue hacking:')
response = input('> ')
if response.upper().startswith('D'):
return decryptedText
tryNum += 1
return None
if __name__ == '__main__':
main()
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