Password Manager - BCACTF 3.0

PwdManager.py

HASHEDPWD = '111210122915474114123027144625104141324527134638392719373948'
key = {
    'a':10,
    'b':11,
    'c':12,
    'd':13,
    'e':14,
    'f':15,
    'g':16,
    'h':17,
    'i':18,
    'j':19,
    'k':20,
    'l':21,
    'm':22,
    'n':23,
    'o':24,
    'p':25,
    'q':26,
    'r':27,
    's':28,
    't':29,
    'u':30,
    'v':31,
    'w':32,
    'x':33,
    'y':34,
    'z':35,
    '0':36,
    '1':37,
    '2':38,
    '3':39,
    '4':40,
    '5':41,
    '6':42,
    '7':43,
    '8':44,
    '0':45,
    '_':46,
    '{':47,
    '}':48
}

unhashed = input("Enter the password!")

result = ''
# The Hash
for element in unhashed:
    result += str(key[element])

if result == HASHEDPWD:
    print("That's Right! The password is the flag.")
else:
    print("That's not right!")

PwdManager.py assigns each letter number and symbol a number and then replaces the character with that number. We can solve this challenge by just doing the opposite.

solution.py

HASHEDPWD = '111210122915474114123027144625104141324527134638392719373948'
key = {
    'a':10,
    'b':11,
    'c':12,
    'd':13,
    'e':14,
    'f':15,
    'g':16,
    'h':17,
    'i':18,
    'j':19,
    'k':20,
    'l':21,
    'm':22,
    'n':23,
    'o':24,
    'p':25,
    'q':26,
    'r':27,
    's':28,
    't':29,
    'u':30,
    'v':31,
    'w':32,
    'x':33,
    'y':34,
    'z':35,
    '0':36,
    '1':37,
    '2':38,
    '3':39,
    '4':40,
    '5':41,
    '6':42,
    '7':43,
    '8':44,
    '0':45,
    '_':46,
    '{':47,
    '}':48
}

PWD = ""
HASHEDPWD = [(HASHEDPWD[i:i+2]) for i in range(0, len(HASHEDPWD), 2)]
for i in HASHEDPWD:
    PWD += list(key.keys())[list(key.values()).index(int(i))]
print(PWD)

└─$ python solve.py
bcactf{5ecure_pa55w0rd_23rj13}