picoCTF General Skills Guide

fixme2.py

Name: fixme2.py
Description: Fix the syntax error in this Python script to print the flag. Download Python script
Author: LT 'syreal' Jones
Tags: Easy, General Skills, Beginner picoMini 2022, Python
Challenge from: Beginner picoMini 2022
Files: fixme2.py
Hints:
1. Are equality and assignment the same symbol?
2. To view the file in the webshell, do: $ nano fixme2.py
3. To exit nano, press Ctrl and x and follow the on-screen prompts.
4. The str_xor function does not need to be reverse engineered for this challenge.

Theory

According to the description, to get the flag we have to fix a python code, should be easy since python tells you exactly what is wrong in the code with clarity, and even in what line it's happening, so it should be easy. Can't do anything else in theory, so let's pass on to there.

Solution

First let's download the code and run it to see what's wrong:

import random

def str_xor(secret, key):
    #extend key to secret length
    new_key = key
    i = 0
    while len(new_key) < len(secret):
        new_key = new_key + key[i]
        i = (i + 1) % len(key)        
    return "".join([chr(ord(secret_c) ^ ord(new_key_c)) for (secret_c,new_key_c) in zip(secret,new_key)])

flag_enc = chr(0x15) + chr(0x07) + chr(0x08) + chr(0x06) + chr(0x27) + chr(0x21) + chr(0x23) + chr(0x15) + chr(0x58) + chr(0x18) + chr(0x11) + chr(0x41) + chr(0x09) + chr(0x5f) + chr(0x1f) + chr(0x10) + chr(0x3b) + chr(0x1b) + chr(0x55) + chr(0x1a) + chr(0x34) + chr(0x5d) + chr(0x51) + chr(0x40) + chr(0x54) + chr(0x09) + chr(0x05) + chr(0x04) + chr(0x57) + chr(0x1b) + chr(0x11) + chr(0x31) + chr(0x0d) + chr(0x5f) + chr(0x05) + chr(0x40) + chr(0x04) + chr(0x0b) + chr(0x0d) + chr(0x0a) + chr(0x19)

flag = str_xor(flag_enc, 'enkidu')

# Check that flag is not empty
if flag = "":
  print('String XOR encountered a problem, quitting.')
else:
  print('That is correct! Here\'s your flag: ' + flag)



  File "main.py", line 19
    if flag = "":
            ^
SyntaxError: invalid syntax

So it seems that there is only one equal sign when comparing on an if statement instead of two. We can just put that other equal sign and run it normally:

import random

def str_xor(secret, key):
    #extend key to secret length
    new_key = key
    i = 0
    while len(new_key) < len(secret):
        new_key = new_key + key[i]
        i = (i + 1) % len(key)        
    return "".join([chr(ord(secret_c) ^ ord(new_key_c)) for (secret_c,new_key_c) in zip(secret,new_key)])

flag_enc = chr(0x15) + chr(0x07) + chr(0x08) + chr(0x06) + chr(0x27) + chr(0x21) + chr(0x23) + chr(0x15) + chr(0x58) + chr(0x18) + chr(0x11) + chr(0x41) + chr(0x09) + chr(0x5f) + chr(0x1f) + chr(0x10) + chr(0x3b) + chr(0x1b) + chr(0x55) + chr(0x1a) + chr(0x34) + chr(0x5d) + chr(0x51) + chr(0x40) + chr(0x54) + chr(0x09) + chr(0x05) + chr(0x04) + chr(0x57) + chr(0x1b) + chr(0x11) + chr(0x31) + chr(0x0d) + chr(0x5f) + chr(0x05) + chr(0x40) + chr(0x04) + chr(0x0b) + chr(0x0d) + chr(0x0a) + chr(0x19)

flag = str_xor(flag_enc, 'enkidu')

# Check that flag is not empty
if flag == "":
  print('String XOR encountered a problem, quitting.')
else:
  print('That is correct! Here\'s your flag: ' + flag)



That is correct! Here's your flag: picoCTF{3qu4l1ty_n0t_4551gnm3nt_f6a5aefc}

There we go! That's the flag.

I rated this level as "good"! :3