1. Section 2.1: Shift Ciphers and Modular Arithmetic The purpose of this section is to learn about modular arithmetic, which is one of the fundamental mathematical concepts we will need to implement the cryptographical techniques that we will study this semester. Afterwards, we will introduce basic concepts in cryptography and illustrate a basic cryptographic involving shift ciphers…

2. Section 2.1: Shift Ciphers and Modular Arithmetic Modular Arithmetic –In grade school, we first learned how to divide numbers. (Let div stand for division) –Example1: Consider 40 div 3. Determine the quotient and remainder and write the result as an equation. Answer: 40 div 3 = 13 * 3 R 1 where R stands for remainder or 13 * 3 + 1 Basic Arithmetic Properties –In the previous example we used what is called the division algorithm to obtain the answer.division algorithm –The integers are the numbers {…-4, -3, -2, -1, 0, 1, 2, 3, 4, …} –A number of primary interest in this class will be the remainder r that we obtain when we divide two numbers. Definition: We say that r is equal to b MOD m, written r = b MOD m, if r is the integer remainder of b divided by m. We define the variable m as the modulus. Example 2: Determine 25 MOD 7, 31 MOD 5, 26 MOD 2, and 5 MOD 7Example 2 –Answers: 4, 1, 0, 5…

3. Section 2.1: Shift Ciphers and Modular Arithmetic Note: In the division algorithm, the remainder r is non-negative, that is r ≥ 0. This fact means that when doing modular arithmetic we will never obtain a negative remainder. To compute b MOD m when b < 0 correctly, we must always look for the largest number that m evenly divides that is less than b. The next example illustrates this fact. Example 3: Compare computing 23 MOD 9 and -23 MOD 9.Example 3 –Answers: 5, 4…

4. Section 2.1: Shift Ciphers and Modular Arithmetic Doing Modular Arithmetic For Larger Numbers With A Calculator: –For b MOD m calculate (int)(b div m), call this q. –On the TI-83 the int function is under MATH-NUM-5: compute b – qm. This gives r. Or in one step: r = b – int(b / m) * m Examples 4-6:Examples 4-6 –Compute 1024 MOD 37 answer: 25 –Compute 500234 MOD 10301 answer: 5786 –Compute -3071 MOD 107 answer: 32…

5. Section 2.1: Shift Ciphers and Modular Arithmetic Generalization of Modular Arithmetic –In number theory, modular arithmetic has a more formal representation. This idea can be expressed with an example: –Example 7: Find solutions b to the equation b MOD 7 = 4 (solution worked below) another words, what numbers when divided by 7 have a remainder of 4. The easiest one is 4 itself. All other answers are multiples of 7 away from these. The solution set is b = {…-10, -3, 4, 11, 18, …}…

6. Section 2.1: Shift Ciphers and Modular Arithmetic Definition: Let m be a positive integer (the modulus of our arithmetic). Two integers a and b are said to be congruent modulo m if a – b is divisible by m. We write a ≡ b mod m. (note the lower case mod) Example 8: Illustrate why 25 ≡ 11 mod 7 –answer: 11 mod 7 = 4 = 25 mod 7 When the uppercase MOD is used, we are interested in only the specific integer remainder r when a number is divide by a modulus. The lowercase mod notation with the ≡ is used when we are looking for a set of numbers that have the same integer remainder when divided by a modulus. In this class, we will primarily use the MOD notation…

7. Section 2.1: Shift Ciphers and Modular Arithmetic When considering b MOD m, since 0 ≤ r < m, the only possible remainders are 0, 1, 2, 3, …, m-1. This causes the remainders to “wrap” around when performing modular arithmetic. Example 9: Make a table of y-values for the equation: y = (x + 5) MOD 9Example 9 Rules: (if a ≡ b mod m) –1. a + k ≡ (b + k) mod m –2. a – k ≡ (b – k) mod m Example 10: Make a list of five solutions to x + 7 ≡ 2 MOD 8.Example 10 –Answer: The solutions are x = (-7 + 2) mod 8 ≡ -5 mod 8. Since 3 ≡ - 5 mod 8 (3 is the remainder) then 5 solutions are 3, 11, 19, -5, -13. All solutions are of the form 3 mod 8…

8. Section 2.1: Shift Ciphers and Modular Arithmetic Basic Concepts of Crytography –Cryptography - is the art of transmitting information in a secret manner. –Plaintext – the undisguised message that we want to send. –Ciphertext – the secret disguised message that is transmitted. –Encryption (encipherment) – the process of converting plaintext to ciphertext. –Decryption (decipherment) – the process of converting ciphertext back to plaintext. Notation: Z m = {0, 1, 2, … m-1}. (the remainders) –Z 26 = {0, 1, 2, 3, …, 25} –We use Z 26 to represent our alphabet. In setting up a one to one correspondence we have the Alphabet Assignment…Alphabet Assignment…

9. Section 2.1: Shift Ciphers and Modular Arithmetic Monoalphabetic Ciphers – Substitution ciphers. The correspondents (those sending messages) agree upon a rearrangement (permutation) of the alphabet. –We will consider 3 basic ciphers of this type: Shift cipher – section 2.1 Affine cipher – section 2.2 Substitution cipher – section 2.3…

10. Section 2.1: Shift Ciphers and Modular Arithmetic Shift Ciphers: All the plaintext letters are assigned a corresponding letter that has been shifted k units. –For example if k = 3 (the Caesar Cipher) then the plaintext letter A becomes the Ciphertext letter D, B becomes E, and so on. –The Shift Cipher Let x be a numerical plaintext letter (ex: x = 0 when the plaintext letter is A) Let k be an element of Z 26. k is called the key of the cipher and represents the shift amount. Then y = (x + k) mod 26, where y is the numerical ciphertext letter. –The Caesar Cipher tableCaesar Cipher table Encipher the message “Radford”.Radford –Note: In ASCII the letter A has an ASCII number of 65. We could write a program that does the shift cipher for us. The basic formula would be y = (x – 65 + 3) MOD 26 (where x is the numerical ASCII value of the letter). Everything works great except for the letters X, Y, and Z. We would need a control structure to ensure that these last three letters become A, B, and C, respectively. –The Caesar Cipher is just a special case of the shift cipher with the key k = 3. In general k can be any number in the set {0, 1, 2, …, 25}. (Of course for k = 0, the cipher text and plaintext are the same.) –Example 12: Encipher the message SEINFELD using a 12 shift cipher. (link to solution below) Answer…Answer

11. Section 2.1: Shift Ciphers and Modular Arithmetic Deciphering Shift Ciphers –Given a key k, plaintext letter number x, and ciphertext letter number y: y = (x + k) MOD 26(note: y = y MOD 26) Subtract k from both sides (y – k) MOD 26 = x MOD 26 = x. Therefore, x = (y – k) MOD 26 Note: - k can be replaced with + m where m + k = 26. –Example 13: Suppose k = 3 and the ciphertext = “YLUJLQLD”. Decipher the message.Decipher the message –Example 14: Decipher the message “EVZCJZXDFE” that was enciphered using a 17 shift cipher…Decipher the message

12. Section 2.1: Shift Ciphers and Modular Arithmetic As the last two examples illustrate, one must know the key k used in a shift cipher when deciphering a message. This leads to an important question. How can we decipher a message in a shift cipher if we do not know the key k? Cryptanalysis is the process of trying to break a cipher by finding its key. Cryptanalysis in general is not an easy problem. The more secure a cipher is, the harder it is to cryptanalyze. We will soon see that a shift cipher is not very secure and is relatively easy to break. Methods for Breaking a Shift Cipher –Since x = (y – k) Mod 26, we can test all 26 possibilities. One of them will work. –Frequency analysis: Uses the fact that the most frequently occurring letters in ciphertext produced by shift cipher has a good chance of corresponding to the most frequently occurring letters in the standard English alphabet. The most frequently occurring letters in English are E, T, A, O, I, N, S. Frequency Table Two Letter Combinations Three Letter Combinations…!Three Letter Combinations