1. 1 Codes, Ciphers, and Cryptography-Ch 3.2 Michael A. Karls Ball State University

2. 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. 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. 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. 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. 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

7. 7 Mathematics of the Enigma (cont.)

8. 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. 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. 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. 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:

12. 12 Mathematics of the Enigma (cont.)

13. 13 Mathematics of the Enigma (cont.) After each letter, check shift variables (mod 26): s 1 = (s 1 + 1) mod 26 If s 1 = 0, then s 2 = (s 2 + 1) mod 26 If s 2 = 0, then s 3 = (s 3 + 1) mod 26

14. 14 Example 1 Suppose we are given  1,  2,  3, and  as follows:  1 = (0, 15, 7, 8, 19, 23) (1, 4, 17, 2) (3, 13, 25, 5, 21, 12) (6, 9, 10, 22, 11, 14, 24, 16, 18, 20)  2 = (0, 5, 9) (1, 2, 3, 7, 6, 8, 4) (10, 11, 12, 15, 18, 13, 14, 16, 19, 17) (20, 25, 22, 21, 24, 23)  3 = (0, 10, 20, 3, 13, 23, 7, 17, 5, 15, 25, 1, 11, 21) (2, 12, 22, 4, 14, 24, 8, 18, 9, 19, 6, 16)  = (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), with s 1 = 24, s 2 = 10, and s 3 = 5. Use these initial settings to encipher the message “Get your forces ready”.

15. 15 Example 1 (cont.) Solution: Take x = 6 for G and compute with (5).

16. 16 Example 1 (cont.)

17. 17 Example 1 (cont.)

18. 18 Example 1 (cont.)

19. 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. 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