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Now, the for loop go through all the twenty six possible cipher letters and looks at this, I followed

2
00:00:05,220 --> 00:00:10,620
a map and a list of your potential decryption letters for this or for that cipher letter.

3
00:00:10,650 --> 00:00:14,700
That is the list at the letter mappings at the index of Saiful Little One.

4
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Then it checks whether the length is equal to or the length of this list is equal to one.

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If it is, then we know that there is only one letter that this letter would decrypt to decipher letters.

6
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So.

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So then over here we are adding it to the sort of decryption letters to this all letters list and the

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00:00:33,840 --> 00:00:36,480
sort of letters always add letter mappings.

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That is a four letter and they zero because it is a list of potential decryption letters that has only

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00:00:43,020 --> 00:00:46,580
one string in appendix zero of the list.

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So after the previous for loop that has started the those that the variable should contain a list of

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all the descriptions of cipher letters.

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And at this point the program is done identifying all these old letters.

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Then it checks whether they are listed as potential decryption letters for other cipher letters and

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then remove them.

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So for doing this, then we are having another for loop, which is going through all the possible twenty

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six cipher letters and looks at this four letter mapping list of the potential decryption letters.

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So here we are seeing four five letter one.

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Illiteracy are four is one in this all four letters.

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If the length of this is not equal to one and this is in your letter mappings, what are the index of

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00:01:28,800 --> 00:01:36,240
cipher letters then we are seeing the letter mapping one at the index of five letters.

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One is not remove and then we check if it is equal to one.

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And that is the new letter is now solved then.

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So no loop again and making it loop again to prove it will again continue the loop.

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00:01:49,590 --> 00:01:56,190
So here before each side fully examines through the letters in this solve letters to check whether any

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of them exist in the list of potential decryption letters for the mapping one at the index of cipher

27
00:02:03,420 --> 00:02:03,960
letter one.

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Then we check whether the list of potential decryption letters is then solved by checking whether the

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00:02:09,840 --> 00:02:13,130
length of letter mapping is one of the index of five letters.

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One is not equal to one which is over here in this line.

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And by checking whether this letter exists in the list of potential decryption letters.

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So if both the criteria are met in this condition is returning through, then we remove that sulfur

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00:02:29,430 --> 00:02:34,590
letters from S1 from the list of potential decryption letters.

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00:02:34,830 --> 00:02:40,800
And if this removal leaves only one letter in the list of decryption letters, then we again said the

35
00:02:40,800 --> 00:02:46,800
loop again, we're able to prove in this case so that the court can remove this newly solved letter

36
00:02:46,800 --> 00:02:49,460
from the four letter mappings on the next iteration.

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00:02:49,950 --> 00:02:56,610
Now, after the Vilo has gone to a full iteration without the game being set to prove, then the program

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00:02:56,610 --> 00:03:04,170
moves beyond the loop and returns the SAIFULLAH to mapping that the letter mapping one over here and

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00:03:04,170 --> 00:03:10,320
the variable at mapping one should now contain a partial or potential full resolve cipher later mapping.

40
00:03:10,590 --> 00:03:16,680
So now let's come to testing this particular method that is remove letter from mapping function.

41
00:03:17,190 --> 00:03:19,950
So let's see this particular function in action.

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00:03:20,070 --> 00:03:25,760
Testing it in the interactive shell now will go back to Interactive Shell and where we have created

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00:03:25,760 --> 00:03:27,240
the intersected mappings.

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00:03:27,690 --> 00:03:33,690
So let us just cooperate on this function name, which is easy for us to write down over here.

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00:03:34,140 --> 00:03:37,290
Now we will see what he will see.

46
00:03:37,570 --> 00:03:47,790
Letter mapping is equal to say, simple sub hackl dot this removal of letters from mapping one function

47
00:03:48,030 --> 00:03:54,770
and in this we would be passing, say, the intersected mapping one in this case.

48
00:03:55,230 --> 00:04:02,610
And then let us try to see or let us try to call the Intersect it.

49
00:04:04,400 --> 00:04:05,630
Mapping one.

50
00:04:08,260 --> 00:04:14,560
So once we have done this, but when you remove these sort of letters from intercepted mapping, will

51
00:04:14,560 --> 00:04:16,850
this start now over here on?

52
00:04:16,870 --> 00:04:24,280
Initially we had an M and B, but now here we just have been this, which is what we predicted from

53
00:04:24,280 --> 00:04:25,030
the case.

54
00:04:25,030 --> 00:04:25,600
It would be.

55
00:04:25,840 --> 00:04:33,850
So now each side has just one potential decryption letter so we can use this at the mappings to start

56
00:04:33,850 --> 00:04:34,510
decrypting.

57
00:04:34,930 --> 00:04:40,370
Now, again, we will return to the interactive shell example one more time so we will not lose this.

58
00:04:40,370 --> 00:04:44,990
We will keep this open now coming to a more simple sub function.

59
00:04:45,340 --> 00:04:52,290
So if we go back to our interactive shell and let us open our file again.

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00:04:52,780 --> 00:04:55,720
So let's go back to you.

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00:04:55,730 --> 00:04:58,710
Drive your simple substitution.

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00:04:59,080 --> 00:05:02,840
Now, if we go back to over hack simple sub one function.

63
00:05:03,100 --> 00:05:09,790
So now that you have seen how the function that is inside letter mapping one, your add letters to mapping

64
00:05:09,790 --> 00:05:15,940
one, your intersected mappings one and remove some of letters from mapping one have manipulated the

65
00:05:15,940 --> 00:05:18,160
cipher letter mapping that you have passed them.

66
00:05:18,580 --> 00:05:22,360
So now let us use them in our particular program to decrypt the function.

67
00:05:22,420 --> 00:05:29,080
OK, so now here we are going to define the victory or symbols of one function, which will take a parameter

68
00:05:29,080 --> 00:05:33,910
of message start in that we will see intercepted maps.

69
00:05:33,910 --> 00:05:40,470
One is equal to calling the blank Saifullah mappings cipher wordlist Asgeir in which we are stored in

70
00:05:40,480 --> 00:05:42,250
the known Letrozole space.

71
00:05:42,250 --> 00:05:48,940
But now here we are creating a new cipher letter mapping that, restoring the intersected map.

72
00:05:48,940 --> 00:05:55,280
One variable and this variable will eventually hold intersected mappings for each of the cipher words.

73
00:05:55,720 --> 00:06:03,280
Now after this, we basically remove any known little characters from the messages and reuse the objects.

74
00:06:03,280 --> 00:06:09,160
Object in the known Letrozole space pattern matches in a string that is in the letter or provide space

75
00:06:09,160 --> 00:06:09,710
character.

76
00:06:09,910 --> 00:06:17,020
Then we have this sub function you can see over here if you just move a bit now in this sub matter,

77
00:06:17,020 --> 00:06:18,070
on this sub function.

78
00:06:18,400 --> 00:06:23,200
Now this is basically called on a regular expression and it has taken two arguments.

79
00:06:23,500 --> 00:06:29,320
The function as the string in the second argument for matches, and it replaces those matches with the

80
00:06:29,320 --> 00:06:30,810
string and the astonishment.

81
00:06:31,120 --> 00:06:33,910
Then it returns a string with all these replacements.

82
00:06:34,090 --> 00:06:41,290
And in the example, this method tells the program to go through the uppercase string and replace all

83
00:06:41,290 --> 00:06:43,870
the known little characters with the blank string.

84
00:06:44,080 --> 00:06:49,030
Now, this makes this subroutines a string with all punctuation numbered characters removed.

85
00:06:49,270 --> 00:06:54,510
And this string is story for word, list of variable of which we have created.

86
00:06:54,670 --> 00:06:56,980
Then we execute this forward list.

87
00:06:56,980 --> 00:07:05,470
Variable should contain a list of uppercase letters and the for loop assigns each letter in the message

88
00:07:05,470 --> 00:07:07,420
loosely cipher word variable.

89
00:07:07,750 --> 00:07:12,830
And inside this little code creates a blank map gets deciphered.

90
00:07:12,850 --> 00:07:18,190
All candidates are the candidate letters to decipher letter mappings and then intersects this mapping

91
00:07:18,190 --> 00:07:20,330
with the intersected map.

92
00:07:20,510 --> 00:07:24,270
Now we will see more in detail about this in the next session.

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00:07:24,310 --> 00:07:25,550
That's it from this session.

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00:07:25,570 --> 00:07:26,500
Thank you very much.