7CCSMCIS Cryptography and Information Security
Caesar Cipher: Exercise
Use the following relative frequencies in an English text of 1000 letters:
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 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
73 | 9 | 30 | 44 | 130 | 28 | 16 | 35 | 74 | 2 | 3 | 35 | 25 | 78 | 74 | 27 | 3 | 77 | 63 | 93 | 27 | 13 | 16 | 5 | 19 | 1 |
to decide the most likely shift used to obtain:
K DKVO DYVN LI KX SNSYD, PEVV YP CYEXN KXN PEBI, CSQXSPISXQ XYDRSXQ.
Don’t just brute force but proceed strategically. Tally the frequencies of letters in the ciphertext.
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 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1 | 2 | 4 | 3 | 3 | 4 | 1 | 4 | 1 | 4 | 3 | 1 | 6 | 4 | 7 | 5 |
As X
appears in the ciphertext 7 times and E
is the most frequent letter in the given table, it is reasonable to assume that X = E
. If X = E
, we have a shift of
$ -19 \bmod 26 $
. This would result in our cipher -> plaintext looking like this:
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 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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 |
Which gives this plaintext for the ciphertext:
R KRCV KFCU SP RE ZUZFK, WLCC FW JFLEU REU WLIP, JZXEZWPZEX EFKYZEX.
which, doesn’t make any sense. Systematically you can then apply X = ?
where ?
is the next most frequent letter from the given table. This gives X = T
and finally X = N
$ -10 \bmod 26 $
where you’ll find:
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 |
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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 |
Which gives this plaintext for the ciphertext:
A TALE TOLD BY AN IDIOT, FULL OF SOUND AND FURY, SIGNIFYING NOTHING.
The Playfair Cipher: Exercise
Use the keyword “CHARLES” to encrypt the plaintext
MEET ME AT HAMMERSMITH BRIDGE TONIGHT
First we construct our
$ 5 \times 5 $
matrix with the keyword CHARLES
:
We can then split the plaintext into pairs with X
s to fill repeated characters when necessary:
ME ET ME AT HA MX ME RS MI TH BR ID GE TO NI GH TX
We can then cipher this to get:
GD DO GD RQ AR KY GD HD NK PR DA MS OG UP GK IC QY
And decrypt to get:
ME ET ME AT HA MX ME RS M(I/J) TH BR (I/J)D GE TO N(I/J) GH TX
MEET ME AT HAMXMERSM(I/J)TH BR(I/J)DGE TON(I/J)GHTX
MEET ME AT HAMMERSMITH BRIDGE TONIGHT
Viginère Cipher: Exercise
Use the tableu and keyword RELATIONS
to encrypt TO BE OR NOT TO BE THAT IS THE QUESTION
.
Steps:
- Find
$ x $
value for each character in key by using it’s index in the alphabet. i.e.
R = 17
,E = 4
- Perform Caesar cipher for each character in plaintext with each $ x $ value, repeating key when necessary.
Key: | R | E | L | A | T | I | O | N | S |
---|---|---|---|---|---|---|---|---|---|
$ x $ | 17 | 4 | 11 | 0 | 19 | 8 | 14 | 13 | 18 |
Encrypting TO BE OR NOT BE THAT IS THE QUESTION
gives:
Key: RE LA TI ONS RE LA TION SR ELA TIONSREL
Plain-text: TO BE OR NOT TO BE THAT IS THE QUESTION
Cipher-text: KS ME HZ BBL KS ME MPOG AJ XSE JCSFLZSY
To decrypt:
Cipher-text: KS ME HZ BBL KS ME MPOG AJ XSE JCSFLZSY
Key: RE LA TI ONS RE LA TION SR ELA TIONSREL
Plain-text: TO BE OR NOT TO BE THAT IS THE QUESTION
The Viginère cipher presents an improvement over the Caesar cipher as it places a positional dependency on the cipher-text. It is isn’t immune to frequency attacks as you can see it is possible to guess the length of the key where there are reptitions in the cipher-text (there is ME
encrypted for LA
twice).
The Churchyard Cipher (Simplified): Exercise
What kind of cipher is it?
This is a mono-alphabetic cipher as each letter is mapped to one symbol/character in the ciphertext alphabet.
Why is it so difficult to break? (Especially without the hint!)
It is really difficult to break as it uses a ciphertext alphabet that we are not as familiar with.
What is the plaintext message?
So, after reading about the pigpen cipher and the tic-tac-toe hint; it’s clear that we can place our alphabet inside 3 tic-tac-toe grids and use the surrounding borders as an identifier for each letter; like so:
We can guess that the dots refer to which grid to use, but which one? We can easily find out by writing down each combination (there are only 3!):
I E D E D B E T D E A B H
R N M N M K N R M N J K Q
_ W V W V T W _ V W S T _
From here, there is only one sensical phrase: REMEMBER DEATH
.
What is the key?
From the above, this makes our key:
One-time pad: Exercise
Given two distinct cipher-texts that have used the same one-time pad, what technique could an attacker use to break them?
The attacker could perform the same pad operation on the cipher-texts to obtain some information about the key. Ideally he’d need to intercept more cipher-texts with the same one-time pad. See coursework from QMUL (using XORs).