(Polybius Cipher ) You\'ve intercepted the ciphertext 12 42 42 21 43 44 25 33 11
ID: 3875043 • Letter: #
Question
(Polybius Cipher ) You've intercepted the ciphertext 12 42 42 21 43 44 25 33 11 21 34 13 25 34 23
Based on the ciphertext you suspect a Polybius cipher, but when you decrypt you get nonsense, so the sender must have used a keyed Polybius cipher. You know that the sender is lazy and suspect that the keyword is short (length 3). By bribing the secretary of the sender you find out that the keyword contains the letter P and the other two letters of the keyword precede P in the alphabet.
Decrypt the message, and describe how you solved the problem. ________________________________________
Please explain
Explanation / Answer
Polybius Square is a substitution cipher, also known as monoalphabetical cipher. This kind of ciphers are named like that because they proceed by substitute the input letters by always the same values during all the encryption process - unlike the polyalphabetical ciphers.
First, fill a 25's boxes grid (5 by 5). There's also a 36 boxes variant which allow the user to encrypt numbers too. As you see, there's only 25 boxes in the grid, which mean that we will need to exclude one letter. This letter actually depends on which language is used to encrypt or decrypt the input. In English it's the "J" or "I" that is being excluded. Here take the "J" out. We fill the grid starting at the top left corner, ending at the bottom right. Then, we just have to replace the input's letters by the two numbers in the grid that represent its coordinates. For instance, the "A" letter will be the first in the grid, into the first line and first column. "A" will so be represented by "11" in the cipher. "B" will be replaced by "12" because it is on the first line but second column, and so on.
1
2
3
4
5
1
A
B
C
D
E
2
F
G
H
I/J
K
3
L
M
N
O
P
4
Q
R
S
T
U
5
V
W
X
Y
Z
Decryption is really easy too. We again make up the Mixed Square using the keyword, and then use the pairs of numbers to locate the plaintext letter in the grid, remembering the
We shall decrypt the message " 12 42 42 21 43 44 25 33 11 21 34 13 25 34 23" using the keyword of length 3 where first letter is P and other two alphabet precede P in the alphabet. So providing the keyword in the table
1
2
3
4
5
1
P
Q
R
A
B
2
C
D
E
F
G
3
H
I
K
L
M
4
N
O
S
T
U
5
V
W
X
Y
Z
Now we look at pairs of letters in turn.
So 12 represents “C” , 42 represents “F” , 21 represents “Q” , 43 represents “L” , 44 represents “T” , 25 represents “W”,33 represents “K”, 11 represents “P”,21 represents ”Q”,34 represents “S”,13 represents “H”,25 represents “W”,34 “S”,23 represents ”I”.
So the message will be cffqltwkpqshwhi.
1
2
3
4
5
1
A
B
C
D
E
2
F
G
H
I/J
K
3
L
M
N
O
P
4
Q
R
S
T
U
5
V
W
X
Y
Z
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