Presentation on theme: "1 Codes, Ciphers, and Cryptography-Ch 3.2 Michael A. Karls Ball State University."— Presentation transcript:
1 Codes, Ciphers, and Cryptography-Ch 3.2 Michael A. Karls Ball State University
2 Main Ingredients of the Enigma Keyboard (with A-Z) Light bulbs (with A-Z) Plugboard (with A-Z) Rotors (with 0-25) Reflector (with 0-25)
3 User Interface The keyboard and light bulbs were the user interface for entering and reading messages. The plugboard was used to increase the number of keys (settings).
4 Rotors Each rotor was connected with a series of wires that created a permutation cipher. An electric signal would enter the rotor at one position and the permutation would cause the signal to leave at a different position on the far side of the rotor.
5 Reflector The reflector rotor at the end was a fixed product of disjoint two-cycles (i.e. a permutation), such as = (0, 12)(1, 9)(2, 3) (4, 24)(5, 18)(6, 23)(7, 8) (10, 25)(11, 13)(14, 21) (15, 17)(16, 19)(20, 22) Note that -1 = , so x = x for all x.
6 Mathematics of the Enigma If we ignore the plugboard settings, the Enigma is a product of permutations! x plain yz w a b c d cipher reflectorrotor 3rotor 2rotor 1
8 Equation (1) is a model for the Enigma if we encrypt one letter! Recall that the rotors move like an odometer, with rotor 2 changing when rotor 1 changes from 25 to 0 and rotor 3 changing in a similar fashion.
9 Mathematics of the Enigma (cont.) To account for the rotation of the rotors, we introduce a “shift” variable for each rotor, which keeps track of how far a rotor has shifted from its initial position! Shift variables: s 1 rotor 1 s 2 rotor 2 s 3 rotor 3 Note: For s 2 to increase, we need s 1 to change from 25 to 0 and for s 3 to increase, we need s 2 to change from 25 to 0.
10 Mathematics of the Enigma (cont.) To account for the shift s 1, instead of computing we compute Instead of computing the inverse of to get x, i.e. we need to solve (2) for x in terms of y.
11 Mathematics of the Enigma (cont.) Applying 1 -1 to both sides of (2) yields: Adding - s 1 mod 26 to both sides of (3), We will use functions of form (2) and (4) to account for the moving rotors! To encrypt, we actually compute the following:
19 Example 1 (cont.) Thus, d = 15, which implies G is encrypted as P. Finally, update shift variables. s 1 = (s 1 + 1) mod 26 = (24+1) mod 26 = 25 s 2 = 10 s 3 = 5 Repeat with x = 4 for E, … Homework: Finish enciphering message!
20 References for Enigma Notes and Photos D. W. Hardy and C. L. Walker, Applied Algebra, Codes, Ciphers, and Discrete Algorithms, Prentice Hall, New Jersey, 2003 http://jproc.ca/crypto/enigma.html http://math.arizona.edu/~dsl/enigma.htm
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