Subway

Problem
This may seem like a regular substitution cipher, but it doesn't seem to work with a regular alphabet. (Hint: what non-alphabetic characters does the ciphertext have?)
The ciphertext is W74 5o06 8v XP32W5-{qdw_0_vepsog_vx1vwewxwedq_w7ev_wepg}
Solution
We can analyse the ciphertext and try to draw conclusions. We know that the flag begins with UMDCTF. By analysing the word lengths, I also knew that the ciphertext started with "The flag is...". For instance, the mappings that I saw are commented below.
We realise that A -> 0, B -> 1, C -> 2, and so on. However, after O, things get a bit confusing again. O -> L, P -> M, and so on. Any results greater than Z will start from 0 again.
c = "W74 5o06 8v XP32W5-{qdw_0_vepsog_vx1vwewxwedq_w7ev_wepg}"
​
# D -> 3
# C -> 2
# T -> W
# H -> 7
# E -> 4
# F -> 5
# I -> 8
# L -> O
​
m = ''
for char in c:
    if char.isnumeric():
        m += chr(ord('a') + int(char))
​
    elif char.isalpha():
​
        if char.lower() < 'o':
            offset = ord(char.lower()) + 26 - ord('o')
        else:
            offset = ord(char.lower()) - ord('o')
​
        new_char = ord('l') + offset
​
        if new_char > ord('z'):
            m += str(new_char - ord('z') - 1)
        else:
            m += chr(new_char)
​
    else:
        m += char
​
print(m)
Donnie Docker
Jump Not Easy
Last modified 8mo ago