## Solution script
```python=
SQUARE_SIZE = 6
def generate_square(alphabet):
assert len(alphabet) == pow(SQUARE_SIZE, 2) #can be ignored here
matrix = []
for i, letter in enumerate(alphabet):
if i % SQUARE_SIZE == 0: #i%6==0
row = []
row.append(letter)
if i % SQUARE_SIZE == (SQUARE_SIZE - 1):
matrix.append(row)
return matrix
def get_index(letter, matrix):
for row in range(SQUARE_SIZE):
for col in range(SQUARE_SIZE):
if matrix[row][col] == letter:
return (row, col)
print("letter not found in matrix.")
exit()
def decrypt_pair(pair, matrix):
#print("pair=",pair)
p1 = get_index(pair[0], matrix) #(row,col)
p2 = get_index(pair[1], matrix) #(row,col)
if p1[0] == p2[0]: #same row
return matrix[p1[0]][(p1[1] - 1) % SQUARE_SIZE] + matrix[p2[0]][(p2[1] - 1) % SQUARE_SIZE]
elif p1[1] == p2[1]: #same col
return matrix[(p1[0] - 1) % SQUARE_SIZE][p1[1]] + matrix[(p2[0] - 1) % SQUARE_SIZE][p2[1]]
else: #neither
return matrix[p1[0]][p2[1]] + matrix[p2[0]][p1[1]]
def decrypt_string(s, matrix):
result = ""
for i in range(0, len(s), 2):
print("now i=",i)
result += decrypt_pair(s[i:i + 2], matrix)
return result
alphabet = "meiktp6yh4wxruavj9no13fb8d027c5glzsq"
m = generate_square(alphabet) #key table
enc_msg="y7bcvefqecwfste224508y1ufb21ld"
msg=decrypt_string(enc_msg,m)
print(msg)
```
## Output
```shell
┌──(kali㉿kali)-[~/code]
└─$ nc mercury.picoctf.net 6057
Here is the alphabet: meiktp6yh4wxruavj9no13fb8d027c5glzsq
Here is the encrypted message: y7bcvefqecwfste224508y1ufb21ld
What is the plaintext message? wd9bukbspdtj7skd3kl8d6oa3f03g0
Congratulations! Here's the flag: 2e71b99fd3d07af3808f8dff2652ae0e
```
Note that the output doesn't have to be wrapped with `picoCTF{}`
## REFs
- https://www.geeksforgeeks.org/playfair-cipher-with-examples/
- https://www.runoob.com/python3/python3-assert.html