Homework SourseA ciphertext C TEBKFKQEBZLROPBLCERJXKBSBKQP is known to be
A ciphertext
C = “TEBKFKQEBZLROPBLCERJXKBSBKQP” is known to be enciphered using the Caesar cipher. Decipher the ciphertext C by using brute force attack. Please show your work.
Solution
We can hack the Caesar cipher by using a cryptanalytic technique called “brute-force”. Because our code breaking program is so effective against the Caesar cipher, you shouldn’t use it to encrypt your secret information.
Ideally, the ciphertext would never fall into anyone’s hands. But Kerckhoffs’s Principle (named after the19th-century cryptographer Auguste Kerckhoffs) says that a cipher should still be secure even if everyone else knows how the cipher works and has the ciphertext (that is, everything except the key). This was restated by the 20th century mathematician Claude Shannon as Shannon’s Maxim: “The enemy knows the system.”
The Brute-Force Attack
Nothing stops a cryptanalyst from guessing one key, decrypting the ciphertext with that key, looking at the output, and if it was not the correct key then moving on to the next key. The technique of trying every possible decryption key is called a brute-force attack. It isn’t a very sophisticated hack, but through sheer effort (which the computer will do for us) the Caesar cipher can be broken.
The Ceasar cipher is the widely known cipher in which letters are shifted using the formula:
ci = E(pi) = pi + k
Given the ciphertext TEBKFKQEBZLROPBLCERJXKBSBKQP. A Statistical Analysis of Frequency shows us that B is the most common letter in the ciphertext. Using the table in page 29 of [2] (basically this table shows that e is the most common letter in plain English) we can find the correspondence B = e in plaintext. From this we realize that B has been shifted 3 position so k = 3 with high probability.
Given that k = 3 we construct the following table.
Plaintext
a
b
c
d
e
f
g
h
i
j
k
l
m
Ciphertext
D
E
F
G
H
I
J
K
L
M
N
O
P
Plaintext
n
o
p
q
r
s
t
u
v
w
x
y
z
Ciphertext
Q
R
S
T
U
V
W
X
Y
Z
A
B
C
Now, using this table to find the correspondences for each letter of the ciphertext with the plaintext we can decipher the message. For example T in the ciphertext is q in plaintext. By doing the same with each letter we have:
TEBKFKQEBZLROPBLCERJXKBSBKQP
Inserting spaces in all the appropriate places we have the final plaintext
There are 28 chars in this text.
The no of columns are probably a factor of this number 28
Transposition (or anagram) ciphers are where the letters are jumbled up together. Instead of replacing characters with other characters, this cipher just changes the order of the characters. This means that the giveaway for a transposition cipher is that frequency analysis shows that the constituent letters are what would be expected in a standard text (eg. e is the most common English letter). What typically happens is that the text to be encrypted is arranged in a number of columns. These columns are then reordered resulting in encrypted text eg. (1, 2, 3, 4, 5) -> (4, 5, 3, 2, 1). To decrypt you need to workout the number of columns - this is usually based on a common factor of the total number of characters in the text - and then rearrange the columns.
tebkf kqebz lropb lcerj xkbsb kqp
| Plaintext | a | b | c | d | e | f | g | h | i | j | k | l | m | |
| Ciphertext | D | E | F | G | H | I | J | K | L | M | N | O | P | |
| Plaintext | n | o | p | q | r | s | t | u | v | w | x | y | z | |
| Ciphertext | Q | R | S | T | U | V | W | X | Y | Z | A | B | C |
