GROUP 3 TASK 1-3Answers
TASK 1
Our group began with rewriting the entire cipher text across the whiteboard in our classroom. That way we would able to see everything much clearer as it is magnified.
From this, we identified the frequency of the symbols in the cipher text and matched them with possible letters in the alphabet which had a high frequency.
Here are our findings on the frequency of the symbols in the cipher text:
1
6
4
10
11
From this we concluded that the symbol ‘a’ in the cipher text could be the letters ‘e’ or ‘t’ as ‘e’ and ‘t’ are the two most frequent letters of the alphabet. We also observed the symbol ‘I’ in the cipher text and predicted that it could be the letter
‘a’ or ‘i’ in plaintext as it appeared on its own, meaning that it would be a one letter word in plaintext.
Next, we deduced that the word ‘the’ was a common word that could be found in the cipher text. Below are the two possibilities of symbols in the cipher text that represent the word ‘the’.
Using the aforementioned information, we have concluded that the symbol ‘a’ represents the letter ‘e’ and the epsilon symbol represents the letter ‘t’. This is because the symbols ‘a’ and the epsilon are the two most frequent symbols in the cipher. This would also mean the infinity symbol represents the letter ‘h’/
Next, we replaced all the epsilon, infinity and ‘a’ symbols with the letters ‘t’, ‘h’ and
‘e’ respectively. From this, we deduced that the word ‘elizabeth’. We also found the third symbol to represent the word Babington as the cipher text is known as the
Babington plot and we used trial and error to fit in the word into the symbols.
To deduce the rest of the words, we repeatedly used the process of trial and error as well as coming up with various possible English words that could be used to fit the sentence. After much failed attempts, we came up with the answer as follows:
“To Anthony Babington I agree that you can murder Queen Elizabeth at the earliest time. From Mary”

GROUP 3 TASK 1-3Answers
Initially, we did not get the word ‘can’ but the word ‘ban’ instead. We then deduced that the symbol used for the letter ‘c’, which is ^, was also repeated in the word ‘elizabeth’. This did not make sense to us as the letter ‘c’ in the word
‘elizabeth’ is redundant. Therefore we concluded that the encoders probably used repeated/random symbols to trick and confuse anyone who was not supposed to read it, and that this trick is only known between the encoder and recipient.
Below is a photograph of our workings and answer which we wrote on the whiteboard:

GROUP 3 TASK 1-3Answers
TASK 2
Summary of The Vigenère Cipher
The Vigenere ciphere was created by
Giovan Battista Bellaso but it was named after Blaise de Vigenère.
It is a form of polyalphabetic substitution, unlike the Caesar cipher. In fact, the Vigenere cipher makes use of more than one Caesar cipher with different shft values, thus it is impossible to use frequency analysis to decrypt a
Vigenere code
The table shown on the left is known as the a tabula recta, Vigenère square, or
Vigenère table. It is used to crack a
Vigenere code. As seen in the table, the alphabet consisting of 26 letters is written in each row.
A key is chosen and it is repeated till it reaches the length of the plain text.
E.g. when the “key is cipher”, and the plaintext is “at the earliest time”
Plaintext:
Key:
When reading of the table, the horizontal row on top should be the plain text while the vertical row on the extreme left should be the key.
So the encrypted Vigenere code will be: CB IOI VCZAPIJV BXTI
Frequency Letter Frequency Letter
A 75 N
B
C
45 O
1 P
D
E
F
G
H
I
J
K
L
3 Q
90 R
127 S
42 T
12 U
18 V
2 W
1 X
0 Y
27 Z M
0
51
32
170
15
40
63
84
1
15
24
33
101

GROUP 3 TASK 1-3Answers
Here is the encrypted cipher text of the plain text we have derived from task 1:
TO ANTHONY BABINGTON I
VW PUXYQVN IESKVVASE K
AGREE THAT YOU
IVYIV
CAN MURDER
VPPA CFW KPU QLTLTY
QUEEN ELIZABETH AT THE EARLIEST HOUR FROM MARY
YLGMC LPZBIQLXY CB
TASK 3
LOL VCZAPIJV BXTI WTWB TEIA
Analyzing letter frequency, we found that the letter ‘Q’ was repeated the most frequently and hence assuming that the text was encrypted in Caesar’s Cipher, or the Vigenère cipher with a one-letter key, we replaced the ‘Q’ with an ‘E’, the most frequently used alphabet in the English Language. Counting the number of letters in between them, we realized that it was a shift of 13, more commonly known as the
ROT13 after our presentation. Still assuming the cipher used was the Caesar’s Cipher, we replaced the rest of the alphabets with the alphabet 13 places from them.
Hence we decoded the paragraph as follows:
“Although many of the messages we send via email are trivial, there are times when we want to make sure that no one can snoop on what we are saying. One way of encrypting emails is to use a software package known as pretty good privacy which combines elements of conventional cryptography and public key encryption. It was created by Philip Zimmermann and was initially offered free on the internet discussion group usenet.
The pgp software generates a random key based on the movements of your mouse and the way you type this random key is then used to encrypt your message. The next stage is to use public key encryption but instead of using this to encrypt the message, the random key used in the previous stage is encrypted using your public key and sent along with the message which has been encrypted. Using the random key, when the recipient receives your message instead of using the private key to decrypt the message they decrypt the random key and then use that to decrypt the attached message.
The publication of pgp on usenet saw zimmermann subject to a criminal investigation by the U.S government which claimed that publishing pgp in this way violated export restrictions for cryptographic software. The restrictions were put in place because the U.S. government wants to restrict universal access to strong cryptographic techniques.
Although the cryptanalysts at the national security agency can without a doubt decrypt anything that has been encrypted with the pgp software using a short binary key, it is uncertain whether it can break messages using very long keys.”