WEBVTT 00:00:01.034 --> 00:00:02.378 (tranquil music) 00:00:04.121 --> 00:00:06.165 - [Voiceover] Consider the following game. 00:00:06.165 --> 00:00:09.626 Eve instructs Bob to go into a room. (door creaks shut) 00:00:09.626 --> 00:00:12.838 Bob finds the room empty, except for some locks, 00:00:12.838 --> 00:00:16.508 an empty box, and a single deck of cards. 00:00:16.508 --> 00:00:18.511 Eve tells Bob to select a card 00:00:18.511 --> 00:00:22.800 from the deck and hide it as best he can. 00:00:22.800 --> 00:00:24.891 The rules are simple. 00:00:24.891 --> 00:00:27.060 Bob cannot leave the room with anything, 00:00:27.060 --> 00:00:30.021 cards and keys all stay in the room, 00:00:30.021 --> 00:00:34.735 and he can put, at most, one card in the box. 00:00:34.735 --> 00:00:38.363 Eve agrees that she has never seen the locks. 00:00:38.363 --> 00:00:42.784 He wins the game if Eve is unable to determine his card. 00:00:42.784 --> 00:00:45.120 So what is his best strategy? 00:00:45.120 --> 00:00:48.123 Well, Bob selected a card, six of diamonds, 00:00:48.123 --> 00:00:50.834 and threw it in the box. (box clicks shut) 00:00:50.834 --> 00:00:53.628 First he considered the different types of locks. 00:00:53.628 --> 00:00:58.133 Maybe he should lock the card in the box with the key. 00:00:58.133 --> 00:01:00.636 Though, she could pick locks, so he 00:01:00.636 --> 00:01:03.180 considers the combination lock. 00:01:03.180 --> 00:01:05.223 The key is on the back, so if he locks it 00:01:05.223 --> 00:01:08.977 and scratches it off, it seems like the best choice. 00:01:08.977 --> 00:01:12.063 But suddenly he realizes the problem. 00:01:12.063 --> 00:01:13.815 The remaining cards on the table 00:01:13.815 --> 00:01:15.984 leak information about his choice, 00:01:15.984 --> 00:01:18.487 since it's now missing from the deck. 00:01:18.487 --> 00:01:20.989 The locks are a decoy. (metal jangles) 00:01:20.989 --> 00:01:23.992 He shouldn't separate his card from the deck. 00:01:23.992 --> 00:01:25.494 He returns his card to the deck 00:01:25.494 --> 00:01:28.123 but can't remember the position of his card. 00:01:28.123 --> 00:01:31.998 So he shuffles the deck to randomize it. 00:01:32.244 --> 00:01:34.711 Shuffling is the best lock, because it leaves 00:01:34.711 --> 00:01:37.631 no information about his choice. 00:01:37.631 --> 00:01:42.631 His card is now equally likely to be any card in the deck. 00:01:42.678 --> 00:01:47.402 He can now leave the cards openly, in confidence. 00:01:48.183 --> 00:01:51.061 Bob wins the game, because the best Eve can do 00:01:51.061 --> 00:01:53.731 is simply guess, as he has left 00:01:53.731 --> 00:01:56.942 no information about his choice. 00:01:56.942 --> 00:01:58.998 Most importantly, even if we give Eve 00:01:58.998 --> 00:02:01.405 unlimited computational power, 00:02:01.405 --> 00:02:04.200 she can't do any better than a guess. 00:02:04.200 --> 00:02:08.502 This defines what we call "perfect secrecy." 00:02:08.662 --> 00:02:13.500 On September first, 1945, 29-year-old Claude Shannon 00:02:13.500 --> 00:02:17.504 published a classified paper on this idea. 00:02:17.504 --> 00:02:20.215 Shannon gave the first mathematical proof 00:02:20.215 --> 00:02:24.719 for how and why the one time pad is perfectly secret. 00:02:24.719 --> 00:02:27.430 Shannon thinks about encryption schemes 00:02:27.430 --> 00:02:29.850 in the following way. 00:02:29.850 --> 00:02:33.104 Imagine Alice writes a message to Bob, 20 letters long. 00:02:33.104 --> 00:02:34.021 (paper ruffling) 00:02:34.021 --> 00:02:35.522 This is equivalent to picking 00:02:35.522 --> 00:02:40.110 one specific page from the message space. 00:02:40.110 --> 00:02:42.863 The message space can be thought of as a complete 00:02:42.863 --> 00:02:47.117 collection of all possible 20 letter messages. 00:02:47.117 --> 00:02:47.826 (paper ruffling) 00:02:47.826 --> 00:02:49.119 Anything you can think of that's 00:02:49.119 --> 00:02:52.497 20 letters long is a page in this stack. 00:02:52.497 --> 00:02:55.792 Next, Alice applies a shared key, which is a list 00:02:55.792 --> 00:03:00.380 of 20 randomly generated shifts between one and 26. 00:03:00.380 --> 00:03:02.675 The key space is the complete collection 00:03:02.675 --> 00:03:06.511 of all possible outcomes, so generating a key is 00:03:06.511 --> 00:03:10.765 equivalent to selecting a page from this stack at random. 00:03:10.765 --> 00:03:13.810 When she applies the shift to encrypt the message, 00:03:13.810 --> 00:03:16.479 she ends up with a cipher text. 00:03:16.479 --> 00:03:18.607 The cipher text space represents 00:03:18.607 --> 00:03:22.697 all possible results of an encryption. 00:03:22.697 --> 00:03:25.030 When she applies the key, it maps 00:03:25.030 --> 00:03:28.617 to a unique page in this stack. 00:03:28.617 --> 00:03:30.785 Notice that the size of the message space 00:03:30.785 --> 00:03:32.537 equals the size of the key space 00:03:32.537 --> 00:03:35.790 equals the size of the cipher text space. 00:03:35.790 --> 00:03:38.501 This defines what we call "perfect secrecy," 00:03:38.501 --> 00:03:42.506 for if someone has access to a page of cipher text only, 00:03:42.506 --> 00:03:44.883 the only thing that they know is that 00:03:44.883 --> 00:03:48.387 every message is equally likely. 00:03:48.387 --> 00:03:50.555 So no amount of computational power 00:03:50.555 --> 00:03:54.017 could ever help improve a blind guess. 00:03:54.017 --> 00:03:56.636 Now the big problem, you're wondering, with the time pad, 00:03:56.645 --> 00:04:00.231 is we have to share these long keys in advance. 00:04:00.231 --> 00:04:03.360 To solve this problem, we need to relax our definition 00:04:03.360 --> 00:04:07.656 of secrecy by developing a definition of pseudo-randomness. 00:04:07.656 --> 00:04:09.123 (white noise)