UIUCTF 2018 - Galactic Brain[fuck] (300pt)

April 8, 2018

Galactic Brain[fuck] (300pt)


Description: We brought uiuctfsck back, but we made it worse

nc challenges1.uiuc.tf 11338



The stack machine used a character by character comparison to validate the user provided flag and was vulnerable to a timing attack.


In the provided python file, we see a multi-machine implementation of the brainfuck programming language. At the top of the file, the flag is imported:

from flag import flag, do_check

When we give the program a flag it calls the following method:

def check_flag(user_flag):
    load_string(flag, 0)
    load_string(user_flag, 1)

    program = "[-)-(][x])[x](>)>("*len(flag) + 's'

    print("Ayyyy nice lol")

The load_string method simply puts a string into memory at a given machine index. So the first line loads the correct flag into machine 0 and the next line loads our flag into machine 1.

Then it loads a program and interprets it with the multi-machine brainfuck interpreter. Notice however, at no point is the user flag interpreted as brainfuck code, it is simply treated as a string. Last year, uiuctf had a similar problem where the user input was treated as code.

The interpret_program method looks like this:

def interpret_program(program_string):
    timeout = 8192
    state = {'machine': 0, 'ip': 0}

    while(state['ip'] < len(program_string) and timeout > 0):
            c = program_string[state['ip']]
            if c in operations.keys():
                    operations[c](state, program_string)
        except Exception as e:
            print("Well, you managed to break it...")
        state['ip'] += 1
        timeout -= 1
        if program_string[-1] == 's':
    if(timeout == 0):
        print("You used too many cycles. Sorry.")

Basically, it loops for a maximum of 8192 cycles and executes the program_string character by character. Notice also the following lines:

if program_string[-1] == 's':

The do_check method was one of the imports at the top so we don’t know what it does.

Let’s disect the validation program:

program = "[-)-(][x])[x](>)>("*len(flag) + 's'

The first section [-)-(] will loop while the current pointer in machine 0 is not zero. -)-( simpy decrements, moves to the next machine, decrements and then moves back. So if our initial flag character is f this loop will run 102 times. Then, [x])[x]( will exit if either the character in machine 0 or machine 1 is not zero. Therefore, if both of these characters are equal, they will be the same and we will move past this point. Finally, >)>( increments the stack pointer of each machine.

This program is repeated for each character of the flag and finally ends with s. Therefore, the conditional above will run and do_check() will be called at every step.

Since we are doing so much work per character, there is a noticable (and exploitable) timing difference between wrong and correct characters.

I found it very difficult to exploit the first time around due to large variations in network noise and actually gave up on it for awhile. I gave it a second shot at 4 am EST when almost everybody was asleep and it worked much better.

Flag: flag{briang}

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