Segment 2.1: Shift Ciphers and Modular Arithmetic - PowerPoint PPT Presentation

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Segment 2.1: Shift Ciphers and Modular Arithmetic

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  1. Section 2.1: Shift Ciphers and Modular Arithmetic Practice HW from Barr Textbook (not to hand in) p.66 # 1, 2, 3-6, 9-12, 13, 15

  2. Modular Arithmetic • In grade school, we first learned how to divide numbers. Example 1: Consider Determine the quotient and remainder of this division and write the result as an equation.

  3. Solution:

  4. The previous example illustrates a special case of the divisionalgorithm which we state next. • Before stating this algorithm, recall that the integers are the numbers in the following set: Integers:

  5. Division algorithm • Let m be a positive integer (m > 0) and let b be any integer. Then there is exactlyone pair of integers q (called the quotient) and r (called the remainder) such that where .

  6. A number of primary interest in this class will be the remainder that we obtain the division of two numbers. We will find the remainder so often that we use a special term that is used to describe its computation. This is done in the following definition.

  7. Definition • We say that r is equal to b MOD m , written as b MOD m , if r is the integer remainder of b divided by m . We define the variable m as the modulus.

  8. Example 2: Determine 25 MOD 7, 31 MOD 5, 26 MOD 2, and 5 MOD 7. Solution:

  9. Note: • In the division algorithm, the remainder r is non-negative, that is, . This fact means that when doing modular arithmetic that we will never obtain a negativeremainder. 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.

  10. Example 3: Compare computing 23 MOD 9 with -23 MOD 9. Solution:

  11. Doing Modular Arithmetic For Larger Numbers With A Calculator • To do modular arithmetic with a calculator, we use the fact from the division algorithm that and solve for the remainder to obtain

  12. Truncated Quotient (chop digits to right of decimal) • We put this result in division tableau format as follows: Remainder

  13. Example 4: Compute 1024 MOD 37: Solution:

  14. Example 5: Compute 500234 MOD 10301 Solution:

  15. Example 6: Compute -3071 MOD 107 Solution:

  16. Generalization of Modular Arithmetic • In number theory, modular arithmetic has a more formal representation which we now give a brief description of. This idea can be expressed with the following example.

  17. Example 7: Suppose we want to find a solution to the equation b MOD 7 = 4 Solution:

  18. The numbers from Example 7 that give a remainder of 4 MOD 7 represents a congruenceclass. We define this idea more precisely in the following definition.

  19. 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 (note the lower case “mod”)

  20. Note • The previous definition can be thought of more informally as follows. We say that if a and b give the sameintegerremainder when divided by m. That is, if r = a MOD m = b MOD m.

  21. Example 8: Illustrate why . Solution:

  22. The last example illustrates that when the uppercase MOD notation is used, we are interested in only the specific integer remainder when a number is divided by a modulus. The lowercase mod notation with the notation 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.

  23. *Note: • When considering bMOD m, since , the only possible remainders are This causes the remainders to “wrap” around when performing modular arithmetic. This next example illustrates this idea.

  24. Example 9: Make a table of y values for the equation y = (x + 5)MOD 9 Solution:

  25. Modular Equation Addition Prop. • Solving equations (and congruences) if modular arithmetic is similar to solving equations in the real number system. That is, if then and for any number k.

  26. Example 10: Make a list of five solutions to mod 8 Solution:

  27. Basic Concepts of Cryptography • Cryptography is the art of transmitting information in a secret manner. We next describe some of the basic terminology and concepts we will use in this class involving cryptography.

  28. Plaintext – the actual undisguised message (usually an English 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) – process of converting ciphertext back to plaintext.

  29. Notation • represents all possible remainders in a MOD system, that is, • For representing our alphabet, we use a MOD 26 system

  30. MOD 26 Alphabet Assignment • Used to perform a one to one correspondence between the alphabet letters and the elements of this set.

  31. Monoalphabetic Ciphers • Monoalphabetic Ciphers are substitution ciphers in which the correspondents agree on a rearrangement (permutation) of the alphabet. In this class, we examine 3 basic types of monoalphabetic ciphers

  32. Types of Monoalphabetic Ciphers 1. Shift Ciphers (covered in Section 2.1) 2. Affine Ciphers (covered in Section 2.2) 3. Substitution Ciphers (covered in Section 2.3)

  33. Shift Ciphers • If x is a numerical plaintext letter, we encipher x by computing the Enciphering formula for Shift Ciphers MOD 26, where k is in Here y will be the numerical ciphertext letter. *Note:k is called the key of the cipher and represents the shift amount.

  34. Example 11: The Caesar cipher, developed by Julius Caesar is a shift cipher given by MOD 26 Note that the key k = 3. MOD 26

  35. Use the Caesar cipher to create a cipher alphabet. Then use it to encipher the message “RADFORD”. Solution: To create the cipher alphabet, we substitute the MOD 26 alphabet assignment number for each letter into the Caesar shift cipher formula and calculate the corresponding ciphertext letter number as follows:

  36. This gives the corresponding correspondence between the plain and ciphertext alphabets Using the above table, we can encipher the message “RADFORD” as follows Hence, the ciphertext is “UDGIRUG” . █

  37. In the last example we did not have to create the entire plain and ciphertext alphabets to encipher the message. We could instead just used the shift cipher formula y = (x+3)MOD 26 directly. The Caesar cipher is just a special case of a shift cipher with a key of k = 3. In a general shift cipher, the key k can be any value in a MOD 26 system, that is, any value in the set

  38. Example 12: Encipher the message “SEINFELD” using a 12 shift cipher. Solution:

  39. Deciphering Shift Ciphers • Given a key k, plaintext letter number x, and ciphertext letter number y, we decipher as follows: y = (x + k) MOD 26

  40. Deciphering formula for shift ciphers x = (y – k) MOD 26 where y is the numerical ciphertext letter, x is the numerical plaintext letter, and k is the key of the cipher (the shift amount). *Note: In the deciphering shift cipher formula, – k MOD 26 can be converted to its equivalent positive form by finding a positive remainder.