1
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So now we need to get the intersection of the site, little mappings into a little mapping salon and

2
00:00:06,390 --> 00:00:10,980
let the mapping do now will pass them into the Intersect mapping function.

3
00:00:11,340 --> 00:00:19,770
So for that in the interactive shell, he'll let us in sort see in dissected

4
00:00:22,200 --> 00:00:27,840
mapping is equal to the simple stuff heco dot.

5
00:00:28,350 --> 00:00:40,020
Now, let us first of all, open the simplest function in notepad and let us find out the actual material

6
00:00:40,020 --> 00:00:42,780
in the SEC to mappings that as mapping one.

7
00:00:43,140 --> 00:00:48,050
So now if you'll let us just copy paste the Intersect mappings one.

8
00:00:48,420 --> 00:00:57,980
And now here we are going to pass letter mapping one and letter mapping to in this case.

9
00:00:58,530 --> 00:01:06,690
And then if we try to say intersected mapping, we get both the things combined together.

10
00:01:07,050 --> 00:01:12,660
So now the list of potential decryption letters for any cipher letter in the intercepted mappings should

11
00:01:12,660 --> 00:01:17,820
be only the potential decryption letters that are in both letter mappings, one and two.

12
00:01:18,750 --> 00:01:24,800
So, for example, in the list of intercepted mappings, you can see for Z GUI it is just because in

13
00:01:25,050 --> 00:01:30,600
your letter mappings had envie, but that the mapping still had only so that it will build the thing

14
00:01:30,600 --> 00:01:34,020
so commonly those would be intercepted mappings.

15
00:01:34,200 --> 00:01:39,540
So now we will repeat the steps of the preceding step for them to decipher also which we have.

16
00:01:39,930 --> 00:01:48,630
So over here we would create Silvy letter mappings three and that is equal to we are using again.

17
00:01:48,750 --> 00:01:55,020
So simple sub hackle dot get the

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00:01:57,600 --> 00:02:03,180
blank side for letter mapping one.

19
00:02:07,120 --> 00:02:19,240
That is a huge difference in capital and look is so simple, we will just say copy paste it so that

20
00:02:19,680 --> 00:02:20,470
out there, OK.

21
00:02:20,770 --> 00:02:25,770
Then again, let's take Watpac two, which is equal to now.

22
00:02:25,780 --> 00:02:29,050
Let's take the makework battle.

23
00:02:30,550 --> 00:02:37,570
Or rather, let us cooperate on the whole thing and the only thing we will have to change is the vote.

24
00:02:37,660 --> 00:02:41,200
So now we will take the total vote in this case.

25
00:02:41,330 --> 00:02:49,120
OK, so let's copy paste that word here and then recreate the candidate as which is equal do.

26
00:02:49,750 --> 00:02:52,870
So let us take the vote patterns from.

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00:02:54,880 --> 00:03:03,410
And let's put here, but now we will change this to walk back to OK, now, after that let's run will

28
00:03:03,430 --> 00:03:14,350
say even in the range, and we would go on for length of your candidate SDR and here we would take Sillitoe.

29
00:03:14,350 --> 00:03:19,450
Mapping is equal to Symbol's Hucko Dot.

30
00:03:19,570 --> 00:03:26,110
Now you have the matter that is Dolittle's mapping.

31
00:03:27,130 --> 00:03:37,270
And then here we would be passing the little mapping one sorry, three in this case, the voter and

32
00:03:39,580 --> 00:03:40,360
candidate.

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00:03:40,480 --> 00:03:41,380
S.G. are.

34
00:03:42,400 --> 00:03:49,320
At the index, I so let us face the same war here.

35
00:03:51,850 --> 00:04:01,620
And let's execute this fall, so Novacek, our letter mapping three, we get the values printed.

36
00:04:01,850 --> 00:04:07,990
OK, so now again, we have to call for an intercepted mappings because the intercepted mappings is

37
00:04:07,990 --> 00:04:09,040
already there with us.

38
00:04:09,340 --> 00:04:12,910
So now what we would do is, let's see, intercepted

39
00:04:15,440 --> 00:04:15,880
mapping.

40
00:04:15,880 --> 00:04:21,250
One is equal to let's take a simple sub hackl.

41
00:04:21,790 --> 00:04:31,120
But now here we have already used the intercepted mappings and here we would pass the initial intercepted

42
00:04:31,120 --> 00:04:34,140
mappings that we have is this one.

43
00:04:35,050 --> 00:04:37,570
So let's pass that on.

44
00:04:37,570 --> 00:04:40,000
The second one, we have this little.

45
00:04:41,610 --> 00:04:45,300
Mapping three, the current one.

46
00:04:45,990 --> 00:04:54,060
OK, so now if we try to see intersected mapping one, so we get this mapping system.

47
00:04:54,510 --> 00:05:00,050
So in this example, we are able to find a solution for the geese that have only one value in the list.

48
00:05:00,570 --> 00:05:07,800
For example, decrypts to Vibert, that aim would be Krypto in this case already because we know that

49
00:05:08,040 --> 00:05:15,840
the key decrypts to eight, we can reduce that gate or the geek must to Krypto and not a so after all,

50
00:05:15,840 --> 00:05:22,290
if it is solved that those are used by one side FALLETTA, it can be used by another because a simple

51
00:05:22,290 --> 00:05:25,160
substitution cipher and a plain text letter.

52
00:05:25,210 --> 00:05:25,740
Exactly.

53
00:05:25,740 --> 00:05:26,730
One cipher little.

54
00:05:27,150 --> 00:05:33,690
So now let's have a look at the remove solve letters from mapping function and find the solve letters

55
00:05:33,690 --> 00:05:36,860
and remove them from the list of potential decryption letters.

56
00:05:37,230 --> 00:05:42,240
We will need the intersected map that we have just created so we will not lose.

57
00:05:42,240 --> 00:05:47,850
This interactive shit is coming to the identifying solve letters and mapping.

58
00:05:48,210 --> 00:05:51,140
You have to remove some of the tools from mapping function.

59
00:05:51,570 --> 00:05:57,300
So just for any cipher letters in the letter mappings that have only one potential decryption letters.

60
00:05:57,520 --> 00:06:03,120
So these letters are considered solved, which means that they are or any other cipher letter.

61
00:06:03,120 --> 00:06:05,250
Which of a disabled veteran?

62
00:06:05,250 --> 00:06:10,120
The list of potential decryption letters can't possibly decrypt this letter.

63
00:06:10,470 --> 00:06:16,860
So this could cause a chain of chain reaction because one potential decryption letter is removed from

64
00:06:16,880 --> 00:06:20,700
other list of potential decryption letters holding just two letters.

65
00:06:20,700 --> 00:06:22,770
The result could be a new cipher.

66
00:06:23,370 --> 00:06:29,400
OK, so the program handles the situation by looping and removing the newly solved letter from the entire

67
00:06:29,400 --> 00:06:30,870
cipher letter mapping.

68
00:06:31,560 --> 00:06:32,010
Now.

69
00:06:33,300 --> 00:06:35,580
Let us do one thing, let us open our.

70
00:06:36,710 --> 00:06:38,490
Simple Subiaco function.

71
00:06:39,500 --> 00:06:42,740
So let's go to a record on this one.

72
00:06:43,130 --> 00:06:49,100
OK, so now here, if we go for the removal of letters from up, you can see this.

73
00:06:49,500 --> 00:06:51,850
OK, so over here we are having the loop.

74
00:06:52,370 --> 00:06:54,730
They invited this loop again is true.

75
00:06:54,980 --> 00:06:59,970
We are assuming that this will not go against making this false here.

76
00:07:00,260 --> 00:07:06,020
It's not because the difference for dictionaries pass with the little mappings parameter that the dictionary

77
00:07:06,020 --> 00:07:08,390
will contain the changes made in the functions.

78
00:07:08,630 --> 00:07:10,490
Even after the function returns.

79
00:07:10,640 --> 00:07:16,310
It creates a loop again variable that holds a boolean value, which determines that the code needs to

80
00:07:16,310 --> 00:07:19,390
loop again when it finds another source of letters.

81
00:07:19,910 --> 00:07:25,010
So if the loop again variable is set to truly program execution and does the VI loop at the beginning

82
00:07:25,010 --> 00:07:27,770
of the loop, it sets the loop again to force.

83
00:07:27,950 --> 00:07:32,840
And the code assumes that this is the last iteration through the value and the loop.

84
00:07:32,840 --> 00:07:38,540
Again, variable is only set to true if the program finds a new SÖLVI for later during the iteration.

85
00:07:38,660 --> 00:07:45,890
Now the next part of the code over here, OK, it creates a list of cipher letters that have exactly

86
00:07:45,890 --> 00:07:47,860
one potential decryption letters.

87
00:07:48,080 --> 00:07:51,930
And these are the sort of letters that will be removed from the mappings.

88
00:07:52,160 --> 00:07:53,660
So this is all letters.

89
00:07:53,660 --> 00:08:00,410
One, if create value for, say, for little ones in your letter are if the length of this is equal

90
00:08:00,410 --> 00:08:05,840
to one, then you're appending it to solve letters dot append method that the letter mappings with the

91
00:08:05,840 --> 00:08:06,640
index of five four.

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00:08:06,650 --> 00:08:07,970
Letter zero.