import numpy as np

def decrypt(ciphertext, n, m):
    ciphertext_int = [ord(i)-ord('a') for i in ciphertext]
    plaintext = ''
    for i in range(len(ciphertext_int)):
        plain_code = (ciphertext_int[i] - n - m*i) % 26
        plaintext += chr(plain_code + ord('a'))
    return plaintext

def getfreqs(text):
    freqs = np.zeros(26)
    for c in text:
        freqs[ord(c) - ord('a')] = freqs[ord(c) - ord('a')] + 1
    freqs = freqs / len(text)
    return freqs

# From wikipedia
english_freqs = [0.08167, 0.01492, 0.02782, 0.04253, 0.12702, 0.02228, 0.02015, 0.06094, 0.06996, 0.00153, 0.00772, 0.04025, 0.02406,\
 0.06749, 0.07507, 0.01929, 0.00095, 0.05987, 0.06327, 0.09056, 0.02758, 0.00978, 0.02360, 0.00150, 0.01974, 0.00074]

ciphertext = 'ryurbqatxhwhxryufxppwtjkvzwvwijkgpebudvirjqxxrhdbvoztgctnbbcdwlszfscmffkqphxstkundhfotxlqnkcrjxvoxislbzmvznwfirpfelkdre\
rjtxmrwvvdmrapjjqbhlsomjkgazfdxdcnfgmzfctqeexnowdisral'

maxdot = 0
key_n = 0
key_m = 0

for n in range(26):
    for m in range(26):
        dotprod = np.dot(english_freqs, getfreqs(decrypt(ciphertext, n, m).lower()))
        if(dotprod > maxdot):
            maxdot = dotprod
            key_n = n
            key_m = m

print("%d,%d had the highest dot product of %f" % (key_n, key_m, maxdot))
print("The plaintext is most likely %s" % decrypt(ciphertext, key_n, key_m))