Implementing-cryptography-using-python

124 Chapter 4 ■ Cryptographic Math and Frequency Analysis
will brute-force the decryption by examining the frequency analysis of each
attempt. You will know you have the correct key once you have a frequency
analysis of .065:
import urllib.request, ssl, random
# get the text of the Declaration of Independence
response = urllib.request.urlopen("https://raw.githubusercontent.com/
noidentity29/AppliedCryptoPython/master/declaration.txt", context=ssl._
create_unverified_context())
originalText = response.read()
# Strip out all non-alphabet characters
# There are a number of ways to do this but this is the simplest to
understand
modifiedText = str(originalText.strip())
modifiedText = modifiedText.replace(" ", "")
modifiedText = " ".join(modifiedText.split())
modifiedText = modifiedText.lower()
plaintext = ""
for c in modifiedText:
if (c.isalpha()):
plaintext = plaintext + c
# Encrypt the plaintext using a random Caesar shift
def shiftBy(c, n):
return chr(((ord(c) - ord('a') + n) % 26) + ord('a'))
caesar_key = random.randint(1,25)
print ("shhh the secret caesar key is: ", caesar_key)
encrypted = list(map(lambda x: shiftBy(x, caesar_key), plaintext))
# Use FA to determine Caesar key for encrypted text
normal_freqs = {'a': 0.080642499002080981, 'c': 0.026892340312538593,
'b': 0.015373768624831691,
'e': 0.12886234260657689, 'd': 0.043286671390026357, 'g': 0.019625534749730816, 'f':
0.024484713711692099, 'i': 0.06905550211598431, 'h': 0.060987267963718068, 'k':
0.0062521823678781188, 'j': 0.0011176940633901926, 'm': 0.025009719347800208, 'l':
0.041016761327711163, 'o': 0.073783151266212627, 'n': 0.069849754102356679, 'q':
0.0010648594165322703, 'p': 0.017031440203182008, 's': 0.063817324270355996, 'r':
0.06156572691936394, 'u': 0.027856851020401599, 't': 0.090246649949305979, 'w':
0.021192261444145363, 'v': 0.010257964235274787, 'y': 0.01806326249861108, 'x':
0.0016941732664605912, 'z': 0.0009695838238376564}