EverCrack Open Source Cryptanalysis

Polyalphabetic substitution ciphers use two or more cipher

alphabets to encrypt a clear message. Like a monoalphabetic substitution

cipher, the identity of each clear unit is changed [while its position is

retained] but the identity of each clear unit is changed over time. This

is a new dimension in the complexity of the encryption scheme. By using multiple

cipher alphabets, the clear symbol "T" can encrypt to "X" at one point in the

cipher text, "P" at another point, and so on.

Conversely, the cipher symbol "X" can decrypt back to "T" at

one point, decrypt back to "C" at another point, and so on. This may sound

confusing, but this system is accomplished via a key.

Consider using a simple system of using two cipher alphabets to

encrypt the message: T H I S I S A S E C R E T M E S S A G E

[with only two alphabets, no real key is required, so we simply alternate be-

tween the two cipher alphabets when encrypting].

Clear Alphabet

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

Cipher Alphabet #1

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

Cipher Alphabet #2

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

Using the first cipher alphabet to encrypt the first letter

of the message, the second cipher alphabet to encrypt the second letter, and

then repeating this process

[the encrypted message is]: U J J U J U B U F E S G U O F U T C H G

[original message for comparison]: T H I S I S A S E C R E T M E S S A G E

First, lets look at the clear-to-cipher symbols: "T" gets

encrypted as "U" in both cases [it simply hit the right spot which is more

likely with a short key - in this case the key length is 2]. "I" has the same

result. "S" is encrypted as "U" four times but encrypted to "T" once. "E" is

encrypted as "F" and "G" both twice.

Now, lets look at cipher-to-clear symbols: "U" is decrypted to

"T" twice and "S" four times. "J" is decrypted to "H" at one point and "I" at

the other points.

The point here is that frequency analysis [on monographs] cannot

work here - the distribution is skewed. Polyalphabetic ciphers hide the fre-

quencies of the underlying clear message. Not only that, the patterns of the

words are no longer the same. "T H I S" is a word with no redundancy [repetitive

letters] whereas the cipher word representing it, "U J J U", appears to have two

repretitive letters.

To perform cryptanalysis on a polyalphabetic cipher you would

have to figure out the key length and then divide the message into N cipher

alphabets [where N is the key length]. In this case, the length is two, so you

would perform a frequency analysis [starting with the first letter] on every

other letter [encrypted by the first cipher alphabet] and then frequency

analysis [starting with the second letter] on every other letter [encrypted by

the second cipher alphabet]. In this case, because the message is so short, a

frequency analysis or IOC would not be very helpful.