Would you like to inspect the original subtitles? These are the user uploaded subtitles that are being translated: 1 00:00:00.06 --> 00:00:01.08 - [Instructor] As we progress through the course, 2 00:00:01.08 --> 00:00:03.06 we'll use our newly gained knowledge 3 00:00:03.06 --> 00:00:06.00 to develop a file encryptor. 4 00:00:06.00 --> 00:00:07.08 We'll base this on the World War II 5 00:00:07.08 --> 00:00:10.00 three rotor Enigma machine. 6 00:00:10.00 --> 00:00:13.00 As we can see, the Enigma has a plug board at the front 7 00:00:13.00 --> 00:00:16.06 and a set of rotors located in slots. 8 00:00:16.06 --> 00:00:19.02 There could be up to eight rotors and for any session, 9 00:00:19.02 --> 00:00:22.06 three would be selected and placed in the machine. 10 00:00:22.06 --> 00:00:24.00 The rotor consists of wires, 11 00:00:24.00 --> 00:00:27.02 which connect two of the 26 characters of the alphabet, 12 00:00:27.02 --> 00:00:30.08 therefore providing a full alphabet substitution. 13 00:00:30.08 --> 00:00:31.08 For each rotor, 14 00:00:31.08 --> 00:00:34.08 the alphabet wheel can be set to start character. 15 00:00:34.08 --> 00:00:39.06 So a single rotor could be in one of 26 possible states. 16 00:00:39.06 --> 00:00:40.09 The way the Enigma works 17 00:00:40.09 --> 00:00:43.04 is that a character is entered on the keyboard, 18 00:00:43.04 --> 00:00:45.01 it passes through the plug board, 19 00:00:45.01 --> 00:00:47.04 and may be exchanged with another character. 20 00:00:47.04 --> 00:00:51.02 Up to 10 characters of the alphabet could be exchanged. 21 00:00:51.02 --> 00:00:53.00 The character that comes out of the plug board 22 00:00:53.00 --> 00:00:55.05 is then passed to the first rotor. 23 00:00:55.05 --> 00:00:57.09 When a character arrives at the first rotor, 24 00:00:57.09 --> 00:01:02.03 it substitutes it, and then turns the rotor one position. 25 00:01:02.03 --> 00:01:04.04 The substituted character is then passed 26 00:01:04.04 --> 00:01:05.03 to the second rotor, 27 00:01:05.03 --> 00:01:08.07 which does the same substitution using its coding. 28 00:01:08.07 --> 00:01:11.03 It would only step after a complete rotation 29 00:01:11.03 --> 00:01:14.00 of the first rotor. 30 00:01:14.00 --> 00:01:16.02 After passing through the third rotor, 31 00:01:16.02 --> 00:01:18.03 it'S reflected back through the rotors 32 00:01:18.03 --> 00:01:21.01 and reverse substituted. 33 00:01:21.01 --> 00:01:22.09 The final character is then passed 34 00:01:22.09 --> 00:01:24.02 through the plug board, 35 00:01:24.02 --> 00:01:26.05 where it may or may not be substituted 36 00:01:26.05 --> 00:01:30.00 and becomes the next character in the cipher message. 37 00:01:30.00 --> 00:01:33.00 The reflector was the cause of some weakness in Enigma, 38 00:01:33.00 --> 00:01:35.05 but it did mean that decryption was very easy. 39 00:01:35.05 --> 00:01:37.06 Just run the cipher through the machine 40 00:01:37.06 --> 00:01:40.00 and plain text appears. 41 00:01:40.00 --> 00:01:42.04 We can see then that our program will need to provide 42 00:01:42.04 --> 00:01:45.08 for a set of rotors, which have coded settings, 43 00:01:45.08 --> 00:01:48.09 a plug board, and a fixed reflector. 44 00:01:48.09 --> 00:01:51.06 We'll allow the user to select three rotors, 45 00:01:51.06 --> 00:01:54.06 set the alphabet start position for each rotor, 46 00:01:54.06 --> 00:01:57.05 and enter two plug board connections. 47 00:01:57.05 --> 00:02:00.06 We'll use an abbreviated alphabet of zero to F 48 00:02:00.06 --> 00:02:03.00 to cover each of the hexadecimal characters, 49 00:02:03.00 --> 00:02:06.01 and this will allow us to process readable messages 50 00:02:06.01 --> 00:02:08.02 as four bit nibbles. 51 00:02:08.02 --> 00:02:10.09 The program when run will read a file 52 00:02:10.09 --> 00:02:13.00 and process it nibble by nibble, 53 00:02:13.00 --> 00:02:16.03 passing each nibble through the plug board and three rotors, 54 00:02:16.03 --> 00:02:19.09 and then returning it via the plug board to be written. 55 00:02:19.09 --> 00:02:22.01 The first rotor will turn after each nibble, 56 00:02:22.01 --> 00:02:25.03 the second one, every 16 nibbles, 57 00:02:25.03 --> 00:02:29.03 and the third, once every 256 nibbles. 58 00:02:29.03 --> 00:02:31.06 The point at which the second and third rotors turn 59 00:02:31.06 --> 00:02:33.05 is determined by what is known 60 00:02:33.05 --> 00:02:35.05 as a notch in the previous wheel, 61 00:02:35.05 --> 00:02:38.09 which we'll simulate by configuring a notch position. 62 00:02:38.09 --> 00:02:41.01 One of the advantages of this design 63 00:02:41.01 --> 00:02:44.03 is that if you feed in the cipher using the same settings, 64 00:02:44.03 --> 00:02:46.00 then it reverses the ciphering 65 00:02:46.00 --> 00:02:47.04 and you get plain text out. 5289