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16

The Futurama code: Second part

Posted on: Jul, 30th 2010
Leonardo Da Vinci
At the prior post, I did an explanation about how break the encryption of the alien alphabet 1, used subliminally in several episodes of Futurama. Today I'm going to do the same with the alien alphabet 2. Although you don't see Futurama, if you want know some mysteries of the cryptography, don't let of read this post.

The encryption method of the alien alphabet 2 is a little bit more complicated than the used at the 1, for this reason I recommend, to the ones that they haven't do yet, that they read the first post "The Futurama Code: First part" if you want understand completely this.

Jeff Westbrook it's algorithmic researcher, informatics doctor and screenwriter of the Simpson and Futurama and is the creator of the alien alphabets that appears at the series. For the alien alphabet 2, Jeff Westbrook have took a classic encryption method, that use simply a technique of substitution, and have added another substitution of alien symbols as the used at the alien alphabet 1.

The classic encryption method that I'm speaking about its known like the encryption with autokey text, invented by Gerolamo Cardano and it's a variant of the encryption of Vigenère. This method use the same text of the message like key to encrypt the same message.

To make the encryption you have to apply the next formula, letter by letter to the clear text, less to the first letter that it's considered already encrypted:

Actual letter encrypted=[(position in the alphabet of the actual letter to encrypt, beginning by zero) + (Position in the alphabet of the prior encrypted letter)) module 26

The module operation means, remainder of integer division and what it does is maintain cyclically, the results of the formula between brackets, between 0 and 25, in this way when the part of the formula between brackets it's 25, the module will result 25. With 26 will result a 0, with 27 an 1 and so on.

For those who don't like the formulas, can calculate it in a more simple way, using the "tabula recta", invented in 1508 by Johannes Trithemius. To encrypt the actual letter:

1º Search the letter in the upper row.

2º Search the prior encrypted letter in column of the left.

The place where crosses the chosen row and column is the actual encrypted letter.

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

---------------------------------------------------

A A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

B B C D E F G H I J K L M N O P Q R S T U V W X Y Z A

C C D E F G H I J K L M N O P Q R S T U V W X Y Z A B

D D E F G H I J K L M N O P Q R S T U V W X Y Z A B C

E E F G H I J K L M N O P Q R S T U V W X Y Z A B C D

F F G H I J K L M N O P Q R S T U V W X Y Z A B C D E

G G H I J K L M N O P Q R S T U V W X Y Z A B C D E F

H H I J K L M N O P Q R S T U V W X Y Z A B C D E F G

I I J K L M N O P Q R S T U V W X Y Z A B C D E F G H

J J K L M N O P Q R S T U V W X Y Z A B C D E F G H I

K K L M N O P Q R S T U V W X Y Z A B C D E F G H I J

L L M N O P Q R S T U V W X Y Z A B C D E F G H I J K

M M N O P Q R S T U V W X Y Z A B C D E F G H I J K L

N N O P Q R S T U V W X Y Z A B C D E F G H I J K L M

O O P Q R S T U V W X Y Z A B C D E F G H I J K L M N

P P Q R S T U V W X Y Z A B C D E F G H I J K L M N O

Q Q R S T U V W X Y Z A B C D E F G H I J K L M N O P

R R S T U V W X Y Z A B C D E F G H I J K L M N O P Q

S S T U V W X Y Z A B C D E F G H I J K L M N O P Q R

T T U V W X Y Z A B C D E F G H I J K L M N O P Q R S

U U V W X Y Z A B C D E F G H I J K L M N O P Q R S T

V V W X Y Z A B C D E F G H I J K L M N O P Q R S T U

W W X Y Z A B C D E F G H I J K L M N O P Q R S T U V

X X Y Z A B C D E F G H I J K L M N O P Q R S T U V W

Y Y Z A B C D E F G H I J K L M N O P Q R S T U V W X

Z Z A B C D E F G H I J K L M N O P Q R S T U V W X Y

For instance, the clear text "MATT GROENING", after applying this first step we get "MMFY EVJNAIVB" (in the "tabula recta" I have indicated how is made the encryption of the third letter).

After applying this encryption we have to replace the characters of the human language by the following alien symbols.

Symbols of the language 2

The advantage of the alien encryption 2 with regard to the 1, is that it is not vulnerable to simple cryptanalysis attacks, like the Kasiski method or the statistical attack of frequencies, because every letter of the original alphabet it's encrypted with a different symbol each time, making more difficult to decrypt.

The problem of that cryptosystem (encryption method), is that it has not key. The security of the message, is based on that, in order to decrypt the text, you have to know the method of decrypt, while with any other cryptosystem the decrypt method can be known by any person, because the text only will can be decrypted with the key. For instance, at encryption of Cesar, the decrypt method is displace the alphabet a number of places to the left. The specific number of places that we have to displace, is the key. Originally the key is three, although we can choose another value. Find the way to decrypt the cryptograms depends principally of that we know the cryptosystem. In the Cesar cryptosystem, although we know the cryptosystem, we can't decrypt it without the key.

Again, the creators of the series show this image as a Rosetta stone, with the same text encrypted and clear, that can be useful to find out the used code:

The pontiff crocodile

In the clear text some letters are repeated. Any cryptanalyst would realize that the same letter it is not encrypted always with the same symbol, although this is not sufficient to know that they are using the autokey text, simply is a clue to begin to try possible cryptosystems.

Fry, Bender and Lila browsing virtually Internet

Hera says "Codebreakers chat room".

Proffesor Farnsworth

In the screen puts AOL.

To decrypt we must use the following formula:

Actual letter decrypted=[(position in the alphabet of the actual decrypt letter)-(position in the alphabet of the prior letter encrypted)] module 26

This time the operation "module 26", makes that we have to sum 26 while we got a negative result in the part between brackets.

Obviously, here too we can use the "tabula recta", taking in the left column the prior encrypted letter and inside of its matching row, the actual encrypted letter. In the column that we find it and in the most upper row, we find the actual decrypted letter.

Perhaps you would like encrypt some messages at this site.

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