Math1 Notes For Crypto

CRYPTOGRAPHY One of the most exciting applications of algebraic structures is in Cryptography. Cryptography is the science and mathematics of writing in secret codes- a practice that is familiar to us since our early years. Its practical uses range from pin codes and passwords to hiding confidential military intelligence. For our discussion, we only look at its most basic method. First, we assign values to each of the letters, as in: A = 1, B = 2, C = 3, . . ., Y = 25, Z = 0 Note that the letter Z is assigned the value of 0, because we will be working in the ring (Z26 , +26 , •26 ).

In order to encode a message, we need an “Encryption Key”. This is usually a linear function which will allow us to map the corresponding values of each letter in the message to a new value, and hence, another letter. Example: Let us encode the word WORD using the Encryption Key y = 3x + 7. Note that we have the values W = 23, O = 15, R = 18, and D = 4. For each letter, we input their value into the Encryption Key (taken in modulo 26 ) to get their corresponding letter: for W:

y = 3(23) + 7 = 76 ≡ 24 → X

for O:

y = 3(15) + 7 = 52 ≡0→Z

for R:

y = 3(18) + 7 = 61 ≡9→I

for D:

y = 3(4) + 7 = 19 ≡ 19 → S

Therefore, the encoded word is “XZIS”.

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To decode the message, we need a “Decryption Key”. This is simply the inverse function of the Encryption Key. (obtained by solving for x in terms of y 1 in the original “password”.)

So, to get the Decryption Key, we have: y = 3x + 7 → 3x = y − 7 x=

y−7 3

Note that this expression is equivalent to x = [y + (−7)] • 3−1 Meanwhile, in Z26 , −7 = 19 and 3−1 = 9, So we have the Decryption Key: x = 9(y + 19). So, to decode the word XZIS, we repeat the same process as in encryption, using the Decryption Key: Example: for X = 24

x = 9(24 + 19) = 9(17) = 153 ≡ 23 → W

Repeat this process for the other letters, to get the decoded word, “WORD”.

Exercises: 1. Encode the following words using the Encryption Key: y = 5x + 3 (a) CODE (b) ALGEBRA (c) Encode your full name using the given encryption key. 2. Encode the word MATH using the following Encryption Keys: (a) y = 3x + 11 (b) y = 7x + 6 (c) y = 11x + 4 3. Decode the following messages given that the encryption key is y = 5x + 12 (a) EP E AQD CHIN IDK ZKQXH PXIY VXKQOEDU E CZATT DIH TERK ED RQED. (b) GIM WETT NQCC EP GIM AQD XKQF HZEC.

1 Note that this is done while in the ring (Z , + , • ), so subtraction and division have different results than the usual 26 26 26 which we do in the field of real numbers.

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