Once you've downloaded crackme.py, open it in a text editor. The contents of the file are listed below.
# Hiding this really important number in an obscure piece of code is brilliant!
# AND it's encrypted!
# We want our biggest client to know his information is safe with us.
bezos_cc_secret = "A:4@r%uL`M-^M0c0AbcM-MFE0d_a3hgc3N"
# Reference alphabet
alphabet = "!\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ"+ \
"[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~"
def decode_secret(secret):
"""ROT47 decode
NOTE: encode and decode are the same operation in the ROT cipher family.
"""
# Encryption key
rotate_const = 47
# Storage for decoded secret
decoded = ""
# decode loop
for c in secret:
index = alphabet.find(c)
original_index = (index + rotate_const) % len(alphabet)
decoded = decoded + alphabet[original_index]
print(decoded)
def choose_greatest():
"""Echo the largest of the two numbers given by the user to the program
Warning: this function was written quickly and needs proper error handling
"""
user_value_1 = input("What's your first number? ")
user_value_2 = input("What's your second number? ")
greatest_value = user_value_1 # need a value to return if 1 & 2 are equal
if user_value_1 > user_value_2:
greatest_value = user_value_1
elif user_value_1 < user_value_2:
greatest_value = user_value_2
print( "The number with largest positive magnitude is "
+ str(greatest_value) )
choose_greatest()Take note of the following:
- There is a secret in the code (bezos_cc_secret)
- The secret is encoded in ROT47
- decode_secret() can decode the secret, but is never called.
Replace choose_greatest() with decode_secret(bezos_cc_secret) to decrypt the flag.
Execute the following command to grab the flag: python3.crackme.py > flag.txt
The flag will be saved in a text file called flag.txt.