1
00:00:00,540 --> 00:00:07,560
If you tried to entrap this with a nine letter and or three letter now everywhere, if let's take the

2
00:00:07,560 --> 00:00:17,430
first one that is your nine letter, which is the A, B, C, D, E, F, g, h.

3
00:00:17,610 --> 00:00:22,560
I again, repeat a, b, c, d, e, f.

4
00:00:23,010 --> 00:00:35,520
And if we take another letter, another that is X, Y, Z, say again, x, y, z, x, y, z, x,

5
00:00:35,520 --> 00:00:38,250
y, z and X, Y, Z.

6
00:00:38,790 --> 00:00:42,750
OK, so now here's the first one for the on line letter.

7
00:00:42,770 --> 00:00:49,590
You would have the ciphertext, something like c d i g g.

8
00:00:49,860 --> 00:00:53,790
S l do l.

9
00:00:54,960 --> 00:01:07,200
D I, f or G, then f e white, and similarly for this three little, you would have cicu f the a m

10
00:01:08,430 --> 00:01:09,270
Airfix.

11
00:01:10,640 --> 00:01:16,620
And que F.P., then Z Y.

12
00:01:16,720 --> 00:01:24,290
And this would be your encrypted give it said now the dorkiest produced or different ciphertext are

13
00:01:24,290 --> 00:01:29,990
expected and of course the hacker will know the original message or the key, but they will see that

14
00:01:29,990 --> 00:01:40,730
over here you have this, the IG, which is repeating over here and the IG over here is repeating and

15
00:01:40,820 --> 00:01:44,190
it appears at index zero and index nine.

16
00:01:44,900 --> 00:01:49,820
So because mine minus zero is nine, the spacing between the sequence is nine, which would seem to

17
00:01:49,820 --> 00:01:52,130
indicate that the original Q was a nine letter.

18
00:01:52,610 --> 00:01:54,740
So in this case, the indication is correct.

19
00:01:55,100 --> 00:02:01,250
However, if you look at this, you have the skew if the here and the Q FBO.

20
00:02:02,200 --> 00:02:08,810
Now, if we even highlight this, it also produces the same repeat sequence that appeals at the next

21
00:02:08,830 --> 00:02:14,920
zero on an index knife, but the spacing between this sequence is also nine, indicating the Gilston

22
00:02:14,920 --> 00:02:17,740
used this civil ciphertext was also a nine letter kilo.

23
00:02:18,070 --> 00:02:21,700
But we know that the key is only three letters, long averages X, Y, Z.

24
00:02:22,150 --> 00:02:28,510
So the repeated sequence of calls with the same letters in the message that is the example are encrypted

25
00:02:28,510 --> 00:02:33,940
with the same letter of key that is A.B.C. X, Y, Z, which happens when these similar letters in the

26
00:02:33,950 --> 00:02:37,570
message and the key line up and trip to the same sequence.

27
00:02:37,930 --> 00:02:43,240
Now this alignment can happen at any multiple of the real key length, like three, six, nine, 12

28
00:02:43,240 --> 00:02:48,830
four, which is why all three letters can produce all the sequence with the spacing of nine.

29
00:02:49,330 --> 00:02:56,930
So the possible length is due to is due not just spacing, but any factor to that spacing.

30
00:02:57,220 --> 00:03:03,430
So the fact those of a nine on line three and one, therefore, if you find that it predicts the spacing

31
00:03:03,430 --> 00:03:09,640
of nine, you must consider that the key would be of either nine or three beginning Morvan one because

32
00:03:09,680 --> 00:03:11,440
we Ganassi with one little piece.

33
00:03:11,440 --> 00:03:12,510
Justices' insightful.

34
00:03:13,210 --> 00:03:19,330
Not the key is most likely to be most frequently occurring factors Nuvigil can determine by counting

35
00:03:19,660 --> 00:03:23,490
the cost eight into the most frequently occurring factors of the spacing.

36
00:03:23,890 --> 00:03:26,100
They are the most likely length of the beginning.

37
00:03:26,860 --> 00:03:31,500
Now, moving ahead, we come to getting every end letters from a particular string.

38
00:03:32,050 --> 00:03:36,970
Now that we have the possible keys of the beginner key, we can use this information to decrypt the

39
00:03:36,970 --> 00:03:38,980
message one step at a time.

40
00:03:39,010 --> 00:03:44,620
For example, let's assume the key is the silent phone and we are unable to crack the ciphertext.

41
00:03:44,620 --> 00:03:49,960
We can try again assuming the Kelland is too late because the key is cycle through.

42
00:03:49,960 --> 00:03:56,530
To encrypt the plaintext equivalent of four would mean that the starting from first letter, every fourth

43
00:03:56,530 --> 00:03:58,300
letter in the ciphertext is encrypted.

44
00:03:58,300 --> 00:03:59,650
Using the first subcu.

45
00:04:00,040 --> 00:04:05,080
Every four letters, starting from the second letter of the plain text is encrypted using the second

46
00:04:05,890 --> 00:04:06,680
answer.

47
00:04:07,040 --> 00:04:12,520
So using this information, we will form a string from the ciphertext of the letters that have been

48
00:04:12,520 --> 00:04:14,260
encrypted by the same subject.

49
00:04:14,620 --> 00:04:21,370
So first, let us identify what every fourth letter in the string would be if we started from different

50
00:04:21,370 --> 00:04:21,850
letters.

51
00:04:22,150 --> 00:04:27,430
So then basically we combine all the letters endorsing the string.

52
00:04:27,920 --> 00:04:30,650
It will reveal every fourth.

53
00:04:31,630 --> 00:04:34,220
So as we have taken this particular key.

54
00:04:34,840 --> 00:04:43,180
So if we have to identify every fourth letter selected, this is the first one, then your fourth letter

55
00:04:43,180 --> 00:04:47,280
is this similarly, then we come to the next fourth letter.

56
00:04:48,070 --> 00:04:50,450
Let's come to the next fourth letter here.

57
00:04:51,010 --> 00:04:52,630
Then again, you have the E.

58
00:04:53,020 --> 00:04:54,190
You have C v.

59
00:04:57,810 --> 00:05:02,890
Similar to this, something like, again, then you have an.

60
00:05:04,350 --> 00:05:05,700
Then you have said.

61
00:05:07,520 --> 00:05:13,030
See, I after all you have are then you have each.

62
00:05:14,730 --> 00:05:23,910
After which you are having sex again, then sick, then you have the X, this is the second one and

63
00:05:23,910 --> 00:05:25,950
then you have the X.

64
00:05:27,020 --> 00:05:39,350
Then you have a note also again in here, also similarly here also you have seen the then you have your

65
00:05:39,350 --> 00:05:39,750
key.

66
00:05:40,440 --> 00:05:45,490
See, I am the last one that see you.

67
00:05:46,190 --> 00:05:52,610
Now, let us highlight this and let's make a difference.

68
00:05:55,980 --> 00:05:57,740
Let's make it bold enough.

69
00:05:58,010 --> 00:05:59,640
So this is all we have highlighted.

70
00:06:00,290 --> 00:06:05,390
So next, we can find out every little, starting with the second little icon, like we have done the

71
00:06:05,630 --> 00:06:06,170
same one.

72
00:06:06,470 --> 00:06:12,590
So again, then you will have this one one letter just besides the one which we have selected now.

73
00:06:13,520 --> 00:06:16,600
So in this way, we can find out the letters there.

74
00:06:16,610 --> 00:06:18,790
That is the first letter we have the second letter.

75
00:06:18,800 --> 00:06:19,880
We have the third letter.

76
00:06:19,890 --> 00:06:22,250
We have a default letter we have now.

77
00:06:22,250 --> 00:06:29,180
After this, we will go ahead and understand the next concept and this that is using the frequency analysis

78
00:06:29,180 --> 00:06:30,080
to break it up.

79
00:06:30,650 --> 00:06:36,470
Now, if it gets the correct length length, each of the four strings we created in the previous position

80
00:06:36,470 --> 00:06:39,400
over here would have been encrypted with one subcu.

81
00:06:39,590 --> 00:06:46,370
So this means that when a string is decrypted with the correct safety and undergoes a frequency analysis,

82
00:06:46,370 --> 00:06:50,110
decrypted letters are likely to have a high English frequency med school.

83
00:06:50,600 --> 00:06:58,040
So using that, we can again create another example like us, which produces decryption with the closest

84
00:06:58,040 --> 00:07:01,760
frequency matched to English, are most likely to be the real suffix.

85
00:07:02,360 --> 00:07:08,240
Like, for example, I n.w. an X result in the highest frequency school for the fourth string.

86
00:07:08,510 --> 00:07:14,480
So note that these scores are low in general because this isn't enough ciphertext to give us a large

87
00:07:14,720 --> 00:07:19,420
sample of text, but they will or they work well enough for this particular example here.

88
00:07:20,000 --> 00:07:25,420
So the next step is to repeat this process for other three strings to find out their most likely suffixes.

89
00:07:26,210 --> 00:07:31,820
Now, because there are five possible subpoenas for the fall, subject to 40 seconds and one for the

90
00:07:31,820 --> 00:07:36,340
third one and five for the fourth one, total number of combinations becomes 50.

91
00:07:36,650 --> 00:07:39,810
So in other words, we need to brute force 50 possibilities.

92
00:07:39,860 --> 00:07:43,580
But this is much better than brute force and true somewhere around four.

93
00:07:43,580 --> 00:07:46,410
Fifty six or nine sixty seven possible.

94
00:07:46,650 --> 00:07:52,450
So now our task has now not narrowed down a list of possible subjects.

95
00:07:52,790 --> 00:07:56,690
So this difference becomes even greater if the beginning is longer.

96
00:07:57,290 --> 00:08:01,810
So now we will go ahead for writing down a source code for beginner hacking program.

97
00:08:02,120 --> 00:08:05,620
So we'll create it in a new file editor and create a new file.

98
00:08:05,630 --> 00:08:12,320
And don't forget to keep your check English, your frequency analysis demo, your beginning for demo.

99
00:08:12,320 --> 00:08:18,770
And if I put files in the same direction, I want to see the current file and then you will execute

100
00:08:18,770 --> 00:08:21,940
and understand what I was trying to created from this.

101
00:08:22,490 --> 00:08:27,920
So in the next session we would start creating with the Code of Virginia Hackel.

102
00:08:28,610 --> 00:08:34,970
That is second example, how to hack Virginia's for using this particular concept that we have done

103
00:08:34,970 --> 00:08:35,420
manually.

104
00:08:35,420 --> 00:08:38,610
Now we would implement the same thing into other python hacking.

105
00:08:39,020 --> 00:08:40,170
That's it from the session.

106
00:08:40,190 --> 00:08:41,080
Thank you very much.