The Caesar Cipher To Decipher Codes English Language Essay
Description:
The method that I used to decipher this message is Caesar cipher. First, I tried by minus the alphabet by 1. But I could not get the answer. Then I try to minus the alphabet one by one until 7 and then I got the answer, I conclude that every letter in this message subtract by 7 is the answer.
The table used to decipher this message is show as below:
Cipher Text
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
Plain Text
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
2.
EDHBPVMVPUVOPAFVOEAMHDYHRVFHBOYHRPKANQKPOPKBPXVQKPYHMHDKBMHCHHYSBDPAIUANCHIADH
Plain Text:
creativityistodiscovernewideasnewthoughtsthatmightneverhavebeenpartofyoubefore
(creativity is to discover new ideas new thoughts that might never have been part of you before)
Description:
First, I tried to decrypt this message with Caesar Cipher. I tried to shift all the letters in this message for 25 times but I could not decipher this message.
So I tried to decipher this question with mono-alphabetic substitution method. I tried to solve it by software called “Secret Code Breaker Monoalphabetic Substitution Solver” and I got a English sentence which is readable when I add spaces to separate each word in this message.
The concept used by this software to solve the cipher text is by comparing the words in the cipher text with the English dictionary database.
The answer key given by the program to solve this question is below:
Cipher Text
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
Plain Text
o
a
b
r
c
d
l
e
f
k
h
z
v
u
s
t
g
w
p
q
y
i
j
m
n
s
The snapshot of the program:
3.
HELMADINEARTHOTSUCIUDOTYR
Plain Text :
4.
LKHYVLHYPPALGBOGLVOGLGOGLPKGPEANJFANPOGLGYDOGLPKGPVHM
HBOGLYALVUDANQAPALGBOGLGYFOHHGOGLPKGPEANJFANPOGLPKHO
GLVOGLVFJVWHPAOHHDANBOGLOGL
Plain Text :
5.
DQPTAZJWB ZKFQGM NZJYY GQDCN
Plain Text:
couldhant hefori shapp rocks
so I analysis
computing theory still rocks
6.
Plain text:
A PIGPEN GOES OINK WHEN IT WRITES
Decription:
The method that I used to decipher this message is by using pigpen cipher.
First, I do the research on what are the types of cipher text available and found out this cipher text is pigpen cipher at the following website.
http://en.wikipedia.org/wiki/Substitution_cipher
http://upload.wikimedia.org/wikipedia/commons/thumb/3/36/Pigpen_cipher_key.svg/220px-Pigpen_cipher_key.svg.png
After knowing what type of cipher text used in this question, I used the diagram above to decipher each letter in this message and got the answer.
Question 2:
a) How many possible mono-alphabetic ciphers are there? Show your reasoning.
Answer:
Mono-alphabetic cipher means that there are no two letters which are encoded to the same letter and each letter cannot be encoded as itself.
Since there are 26 alphabets, so the first alphabets will have 25 possible alphabets to be replaced. After that, the next alphabets will have 24 possible alphabets to be replaced. The range of possible alphabet is reduces accordingly. So the number of possible cipher are N(N-1)(N-2)….2*1 where N = 25 or 25! .
25!
= 25 x 24 x 23 X 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= 1.551121004 x 1025
Therefore, there are approximately 1.551121004 x 1025 possible mono-alphabetic ciphers there.
b) Assuming that it is possible to test 1 billion possible ciphers per second, calculate how long it would take test all possible mono-alphabetic ciphers.
Answer:
25! / (1 x 109)
= 1.551121004 x 1016 seconds
(1.551121004 x 1016 ) / (60x60x24x365) = 4.91857243 x 108 years
Therefore, it would take approximately 4.91857243 x 108 years to test all possible mono-alphabetic ciphers.
c) Notice that there are only 14 different letters in the encrypted message. How many possible decryptions there are of this message? (Hint: As ‘J’ does not appear in the ciphertext, it doesn’t matter what letter it is encrypted from).
Answer:
Since there are 14 letters in the message, so the number of possible letters for the first letter to be replaced is 25. Then the number of possible letters for second letters is 24 letters and so on until it reaches the 14th letter.
25 x 24 x 23 X 22 x 21 x 20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12
= 3.885885152 x 1017
So there are approximately 3.885885152 x 1017 different ciphers.
d) Assuming, as before, than you can test 1 billion possible codes per second, calculate how long it would take to test all possible decryptions of this message.
Answer:
3.885885152 x 1017 / (1x 109)
= 3.885885152 x 108 seconds
(3.885885152 x 108) / (60 x 60 x 24 x 365)
= 12.32206 years
Therefore, it would takes approximately 12.32206 years to test all possible decryptions of this message.
e) How many possible decryptions of this message are there if you are told the
following partially complete table?
Answer:
Since 5 letters is already known, so the remaining number of possible decryption for the 6th letter is 20, 7th letters is 19 and so on until 14th which is 7.
20 x 19 x 18 x 17 x 16 x 15 x 14 x 13 x 12 x 11 x 10 x 9 x 8 x 7
= 3.379030567 x 1015
f) If you are told that the message probably contains the name of an Australian city and a university, how much more of the table can you complete? Can you completely decipher the message? Why or why not? Explain your answer.
Answer:
There are five letters which have already given as below:
Encrypted letter
X
F
Y
Z
W
Plaintext letter
M
E
L
B
I
From the table above, possibly the name of an Australian city is “MELBOURNE” and it fit exactly as “XFYZABCDF”.
So, from the information above. I can conclude the table below:
Encrypted letter
X
F
Y
Z
W
A
B
C
D
F
Plaintext letter
M
E
L
B
I
O
U
R
N
E
Now I substitute the plaintext letter into the encrypted message.
P
X
W
Q
W
U
Z
T
U
F
V
W
D
X
F
Y
Z
A
B
C
D
F
M
I
I
B
E
I
N
M
E
L
B
O
U
R
N
E
“PXWQ” seem to be the name of the university in Melbourne which is “RMIT”.
So from the result above, I try to decipher the message with the following attempt to make the message make sense:
Attempt 1: Replace “P” as “R”
Attempt 2: Replace “U” as “S”.
Attempt 3: Replace “Q” as “T”.
Attempt 4: Replace “T” as “A”
Attempt 5: Replace “V” as “D”
And now I will get the result and below:
RMIT IS BASED IN MELBOURNE
HOWEVER, from the result above, R is encoded into 2 letters above which is “P” and “C”.
Since there are no letter can encoded to 2 letters, So I concluded that this message can’t be completely decipher.
Question 3:
There are a number of open source encryption softwares/systems available over the Internet. Which system(s)/software(s) would you recommend and why?
Answer:
There is a number of open source encryption software or systems available over the Internet that provides data protection such as TrueCrypt by TrueCrypt Foundation, DiskCryptor by ntldr, Crosscrypt by Stefan Scherrer and FreeOTFE by Sarah Dean. Based on the research I did from the internet, I found that the TrueCrypt is the most popular open source encryption software which runs on all of the major platforms such as Windows XP/Vista/7, Mac OS X and Linux. However, I would like to recommend FreeOTFE because of its distinct advantages over other open sources encryption software.
The first reason why I recommend FreeOTFE is because it is highly portable. Unlike other encryption system, FreeOTFE offers FreeOTFE Explorer which allows FreeOTFE volumes to be accessed without installing it or any device drivers. Besides that, administrator rights are not required for the user to use this portable FreeOFTE. This feature makes it ideal and convenient for use with the removable media or USB drives so that user can use it to encrypt their data with any supported computer. For example, user can use FreeOTFE in internet cafes where computer are available but not granted with administrator rights. Unlike other software which administrator rights and installation is required to run the software, so I think FreeOTFE is a better choice compare to them.
Besides that, FreeOTFE supports a lot of ciphers and hash algorithms. The ciphers that it implements include AES, Blowfish, CAST5/CAST6, DES/Triple DES, MARS, RC6, Serpent and Twofish.
It supports hash algorithms such as MD5, RIPEMD320, SHA512, Tiger, Whirlpool and many more.
Its cipher modes supported include XTS, LRW and cipher-block chaining (CBC) which including XTS-AES-128 and XTS-AES-256. Moreover, it also offers security tokens or smartcards supported for extra security. Compares to TrueCrypt, TrueCrypt only supports SHA-1, RIPEMD-160, Whirlpool hash algorithms and fewer ciphers than FreeOTFE.
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