WEBVTT 00:00:00.000 --> 00:00:04.710 00:00:04.710 --> 00:00:07.650 For over 400 years, the problem remained. 00:00:07.650 --> 00:00:11.760 How could Alice design a cipher that hides her fingerprint, 00:00:11.760 --> 00:00:14.580 thus stopping the leak of information? 00:00:14.580 --> 00:00:18.150 The answer is randomness. 00:00:18.150 --> 00:00:20.890 Imagine Alice rolled a 26 sided die 00:00:20.890 --> 00:00:23.360 to generate a long list of random shifts, 00:00:23.360 --> 00:00:26.810 and shared this with Bob instead of a code word. 00:00:26.810 --> 00:00:28.860 Now, to encrypt her message, Alice 00:00:28.860 --> 00:00:31.970 uses the list of random shifts instead. 00:00:31.970 --> 00:00:34.010 It is important that this list of shifts 00:00:34.010 --> 00:00:38.440 be as long as the message, as to avoid any repetition. 00:00:38.440 --> 00:00:41.250 Then she sends it to Bob, who decrypts the message using 00:00:41.250 --> 00:00:44.085 the same list of random shifts she had given him. 00:00:44.085 --> 00:00:46.870 00:00:46.870 --> 00:00:49.460 Now Eve will have a problem, because the resulting 00:00:49.460 --> 00:00:53.210 encrypted message will have two powerful properties. 00:00:53.210 --> 00:00:56.765 One, the shifts never fall into a repetitive pattern. 00:00:56.765 --> 00:00:59.350 00:00:59.350 --> 00:01:02.860 And two, the encrypted message will have a uniform frequency 00:01:02.860 --> 00:01:04.230 distribution. 00:01:04.230 --> 00:01:07.050 Because there is no frequency differential and therefore 00:01:07.050 --> 00:01:09.736 no leak, it is now impossible for Eve 00:01:09.736 --> 00:01:10.735 to break the encryption. 00:01:10.735 --> 00:01:14.090 00:01:14.090 --> 00:01:18.080 This is the strongest possible method of encryption, 00:01:18.080 --> 00:01:21.520 and it emerged towards the end of the 19th century. 00:01:21.520 --> 00:01:25.860 It is now known as the one-time pad. 00:01:25.860 --> 00:01:28.990 In order to visualize the strength of the one-time pad, 00:01:28.990 --> 00:01:32.320 we must understand the combinatorial explosion 00:01:32.320 --> 00:01:34.600 which takes place. 00:01:34.600 --> 00:01:37.600 For example, the Caesar Cipher shifted every letter 00:01:37.600 --> 00:01:42.970 by the same shift, which was some number between 1 and 26. 00:01:42.970 --> 00:01:44.970 So if Alice was to encrypt her name, 00:01:44.970 --> 00:01:48.770 it would result in one of 26 possible encryptions. 00:01:48.770 --> 00:01:52.290 A small number of possibilities, easy to check them all, 00:01:52.290 --> 00:01:55.280 known as brute force search. 00:01:55.280 --> 00:01:58.060 Compare this to the one-time pad, where each letter would 00:01:58.060 --> 00:02:01.690 be shifted by a different number between 1 and 26. 00:02:01.690 --> 00:02:04.000 Now think about the number of possible encryptions. 00:02:04.000 --> 00:02:08.050 It's going to be 26 multiplied by itself five times, which 00:02:08.050 --> 00:02:10.360 is almost 12 million. 00:02:10.360 --> 00:02:13.030 Sometimes it's hard to visualize, 00:02:13.030 --> 00:02:15.850 so imagine she wrote her name on a single page, 00:02:15.850 --> 00:02:20.900 and on top of it stacked every possible encryption. 00:02:20.900 --> 00:02:24.520 How high do you think this would be? 00:02:24.520 --> 00:02:28.750 With almost 12 million possible five-letter sequences, 00:02:28.750 --> 00:02:32.110 this stack of paper would be enormous, 00:02:32.110 --> 00:02:35.130 over one kilometer high. 00:02:35.130 --> 00:02:38.240 When Alice encrypts her name using the one-time pad, 00:02:38.240 --> 00:02:42.240 it is the same as picking one of these pages at random. 00:02:42.240 --> 00:02:44.720 From the perspective of Eve, the code breaker, 00:02:44.720 --> 00:02:46.910 every five letter encrypted word she 00:02:46.910 --> 00:02:51.600 has is equally likely to be any word in this stack. 00:02:51.600 --> 00:02:55.240 So this is perfect secrecy in action. 00:02:55.240 --> 00:02:55.867