Write-up for FlareOn7 challenge #1 - fidler

3 min

Description

We are introduced to the challenge with the following message:

Welcome to the Seventh Flare-On Challenge!

This is a simple game. Win it by any means necessary and the victory screen will reveal the flag. Enter the flag here on this site to score and move on to the next level.

This challenge is written in Python and is distributed as a runnable EXE and matching source code for your convenience. You can run the source code directly on any Python platform with PyGame if you would prefer.

I did not even bother to install PyGame and went straight into the only interesting file fidler.py, which contains all the logic of the program.

Analysis

It is a small and very easy to follow python program. Within a quick glimpse, we inmediately bump into the following two functions: decode_flag and victory_screen.

def decode_flag(frob):
    last_value = frob
    encoded_flag = [1135, 1038, 1126, 1028, 1117, 1071, 1094, 1077, 1121, 1087, 1110, 1092, 1072, 1095, 1090, 1027,
                    1127, 1040, 1137, 1030, 1127, 1099, 1062, 1101, 1123, 1027, 1136, 1054]
    decoded_flag = []

    for i in range(len(encoded_flag)):
        c = encoded_flag[i]
        val = (c - ((i%2)*1 + (i%3)*2)) ^ last_value
        decoded_flag.append(val)
        last_value = c

    return ''.join([chr(x) for x in decoded_flag])


def victory_screen(token):
    screen = pg.display.set_mode((640, 160))
    clock = pg.time.Clock()
    heading = Label(20, 20, 'If the following key ends with @flare-on.com you probably won!',
                    color=pg.Color('gold'), font=pg.font.Font('fonts/arial.ttf', 22))
    flag_label = Label(20, 105, 'Flag:', color=pg.Color('gold'), font=pg.font.Font('fonts/arial.ttf', 22))
    flag_content_label = Label(120, 100, 'the_flag_goes_here',
                               color=pg.Color('red'), font=pg.font.Font('fonts/arial.ttf', 32))

    controls = [heading, flag_label, flag_content_label]
    done = False

    flag_content_label.change_text(decode_flag(token))
    
    # --snip--

The function decode_flag runs a fairly easy decryption routine to retrive the flag, depending on the parameter frob received. This function will be called from victory_screen function passing as argument the same token parameter it received, without applying any modification to it.

If we look for calls to victory_screen function, we find the following snippet of code under game_screen function:

def game_screen():
    # --snip--
    target_amount = (2**36) + (2**35)
    if current_coins > (target_amount - 2**20):
        while current_coins >= (target_amount + 2**20):
            current_coins -= 2**20
        victory_screen(int(current_coins / 10**8))
        return
    # --snip--

We see that the token argument passed to victory_screen will be equal to int(current_coins / 10**8). Thus, we need to obtain which value(s) of current_coints will satisfy the required conditions to arrive to this call to victory_screen.

Obtain token value

Observe that to enter the body of the first if, we need that:

current_coins > 2^36 + 2^35 - 2^20 = 103078166528

Then, it will enter a while loop that decreases the value of current_coins by 2^20 until the following condition is met:

current_coins < 2^36 + 2^35 + 2^20 = 103080263680

Thus, we get that current_coins must satisfy:

103078166528 < current_coins < 103080263680

Notice that current_coins is then divided by 10^8 and casted to int type. This will essentially remove the 8 least significant digits of current_coins, meaning that any current_coins value within the valid interval will lead to the token argument having a value of 1030. Thus, our token must be 1030.

Obtain flag

To get the flag, we can simply extract the decode_flag function into a new python file and print the value it returns when passing 1030 as argument.

def decode_flag(frob):
    last_value = frob
    encoded_flag = [1135, 1038, 1126, 1028, 1117, 1071, 1094, 1077, 1121, 1087, 1110, 1092, 1072, 1095, 1090, 1027,
                    1127, 1040, 1137, 1030, 1127, 1099, 1062, 1101, 1123, 1027, 1136, 1054]
    decoded_flag = []

    for i in range(len(encoded_flag)):
        c = encoded_flag[i]
        val = (c - ((i%2)*1 + (i%3)*2)) ^ last_value
        decoded_flag.append(val)
        last_value = c

    return ''.join([chr(x) for x in decoded_flag])

print(decode_flag(1030))
$ python decoder.py
[email protected]

So, we obtained the flag [email protected] and can move onto the next challenge!

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