# Affine Cipher by Johnathan Nop
# Affine cipher is a substitution cipher using a specific of function at mod 26. Each letter in the alphabet correspond to the numbers from 0-25, with A=0, B=1, C=2, etc. The function given for encoding is c = a*p+b (mod 26), where both a and "b" must be between 1 and 26. Along with that, a cannot be a number that have the same factors as 26. Use the encoding function to encode the message. To decode the message, the function given is p = a^-1(c-b), where a^-1 is the multiplicative inverse of a at modulus 26, and c is the number that was encoded from the function above. 

# For the upcoming examples, let "a" = 7 and "b" = 11.(a)How can we take the secret message "MATH" to a numerical form?(b)How can the (numerical) plaintext message be encrypted?(c)How do we find a^-1?(d)Using what was found in part (c), how can the message encrypted in part (b) be decrypted?(e)How do we make a table of letters, numerical forms, and encrypted numerical forms, for encryption and decryption purposes?

%r
encrypt = function(){
count = 0
plaintext = c("A", "B", "C", "D", "E", "F","G","H","I","J","K","L","M,","N","O","P","Q","R","S","T","U","V","W","X","Y","Z")
for (i in 0:(length(plaintext) - 1)){
    print(c(plaintext[i+1], count))
    count = count + 1
}
}
encrypt()

# M = 12, A = 0, T = 19, H = 7

# Note that a = 7, and b = 11. The equation for encoding is c = (a*p + b) (mod 26). So to encrypt "MATH," take each individual numerical value as p and solve for c.
Mod(7*12 +11, 26)
Mod(7*0 +11, 26)
Mod(7*19 +11, 26)
Mod(7*7 +11, 26)

# Looking at the table above, 17 = R, 11 = L, 14 = O, and 8 = I. The encrypted message is "RLOI." This is the message I would send to whoever I was communicating with. 

# Now that we know how to encrypt a message, how does the receiver decrypt the message? That is when the finding the multiplicative inverse of a is needed. (Normally brute force is the easiest, but Sage can help us.)
Mod(1/7,26)

Mod(7*15,26)

#Thus the multiplicative inverse of 7 is 15. 

# Recall that the function to decrypt is p = a^-1(c-b) (mod 26). Now that we have a^-1 = 15, we can now use the decryption function to decrypt the message "RLOI".
Mod(15*(17-11),26)
Mod(15*(11-11),26)
Mod(15*(14-11),26)
Mod(15*(8-11),26)

# Note that the number should convert back to the original numerical value of the original plaintext. 12 = M, 0 = A, 19 = T, 7 = H, thus the message is "MATH," which was the original message at the start.

%r
encrypt = function(){
a = 7
b = 11
count = 0
plaintext = c("A", "B", "C", "D", "E", "F","G","H","I","J","K","L","M,","N","O","P","Q","R","S","T","U","V","W","X","Y","Z")
for (i in 0:(length(plaintext) - 1)){
    print(c(plaintext[i+1], count, (a*count + b) %% 26))
    count = count + 1
}
}
encrypt()