0:00:04.710,0:00:07.650
For over 400 years,[br]the problem remained.

0:00:07.650,0:00:11.760
How could Alice design a cipher[br]that hides her fingerprint,

0:00:11.760,0:00:14.580
thus stopping the[br]leak of information?

0:00:14.580,0:00:18.150
The answer is randomness.

0:00:18.150,0:00:20.890
Imagine Alice rolled[br]a 26 sided die

0:00:20.890,0:00:23.360
to generate a long[br]list of random shifts,

0:00:23.360,0:00:26.810
and shared this with Bob[br]instead of a code word.

0:00:26.810,0:00:28.860
Now, to encrypt[br]her message, Alice

0:00:28.860,0:00:31.970
uses the list of[br]random shifts instead.

0:00:31.970,0:00:34.010
It is important that[br]this list of shifts

0:00:34.010,0:00:38.440
be as long as the message,[br]as to avoid any repetition.

0:00:38.440,0:00:41.250
Then she sends it to Bob, who[br]decrypts the message using

0:00:41.250,0:00:44.085
the same list of random[br]shifts she had given him.

0:00:46.870,0:00:49.460
Now Eve will have a problem,[br]because the resulting

0:00:49.460,0:00:53.210
encrypted message will have[br]two powerful properties.

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One, the shifts never fall[br]into a repetitive pattern.

0:00:59.350,0:01:02.860
And two, the encrypted message[br]will have a uniform frequency

0:01:02.860,0:01:04.230
distribution.

0:01:04.230,0:01:07.050
Because there is no frequency[br]differential and therefore

0:01:07.050,0:01:09.736
no leak, it is now[br]impossible for Eve

0:01:09.736,0:01:10.735
to break the encryption.

0:01:14.090,0:01:18.080
This is the strongest[br]possible method of encryption,

0:01:18.080,0:01:21.520
and it emerged towards the[br]end of the 19th century.

0:01:21.520,0:01:25.860
It is now known as[br]the one-time pad.

0:01:25.860,0:01:28.990
In order to visualize the[br]strength of the one-time pad,

0:01:28.990,0:01:32.320
we must understand the[br]combinatorial explosion

0:01:32.320,0:01:34.600
which takes place.

0:01:34.600,0:01:37.600
For example, the Caesar[br]Cipher shifted every letter

0:01:37.600,0:01:42.970
by the same shift, which was[br]some number between 1 and 26.

0:01:42.970,0:01:44.970
So if Alice was to[br]encrypt her name,

0:01:44.970,0:01:48.770
it would result in one of[br]26 possible encryptions.

0:01:48.770,0:01:52.290
A small number of possibilities,[br]easy to check them all,

0:01:52.290,0:01:55.280
known as brute force search.

0:01:55.280,0:01:58.060
Compare this to the one-time[br]pad, where each letter would

0:01:58.060,0:02:01.690
be shifted by a different[br]number between 1 and 26.

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Now think about the number[br]of possible encryptions.

0:02:04.000,0:02:08.050
It's going to be 26 multiplied[br]by itself five times, which

0:02:08.050,0:02:10.360
is almost 12 million.

0:02:10.360,0:02:13.030
Sometimes it's[br]hard to visualize,

0:02:13.030,0:02:15.850
so imagine she wrote her[br]name on a single page,

0:02:15.850,0:02:20.900
and on top of it stacked[br]every possible encryption.

0:02:20.900,0:02:24.520
How high do you[br]think this would be?

0:02:24.520,0:02:28.750
With almost 12 million[br]possible five-letter sequences,

0:02:28.750,0:02:32.110
this stack of paper[br]would be enormous,

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over one kilometer high.

0:02:35.130,0:02:38.240
When Alice encrypts her[br]name using the one-time pad,

0:02:38.240,0:02:42.240
it is the same as picking[br]one of these pages at random.

0:02:42.240,0:02:44.720
From the perspective of[br]Eve, the code breaker,

0:02:44.720,0:02:46.910
every five letter[br]encrypted word she

0:02:46.910,0:02:51.600
has is equally likely to[br]be any word in this stack.

0:02:51.600,0:02:55.240
So this is perfect[br]secrecy in action.