[KEYGEN] Balanced Tree

keygen

(_w) #1

Hello guys, this is my first post here. I’ve been lurking around for a few months solving various challenges, and today I finally decided to create my own (instead of studying for finals) :grin:

Hope you enjoy it!

The program consists of simple username and password query. There’s a slight hint in the title but it will only help you if you come far enough :stuck_out_tongue:

Difficulty

Author Assigned Level: Wannabe

Community Assigned Level:

  • Newbie
  • Wannabe
  • Hacker
  • Wizard
  • Guru

0 voters


Goals

  1. Find a working username and password pair.
  2. Create a non-brute keygen.

Rules

  1. No patching allowed!

Binary

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Decode with: cat input | base64 -d | gunzip > a.out && chmod +x a.out

EDIT: For some reason there is some weird formatting, better version here: https://pastebin.com/BkRd6Bxn


#2

I found a solution:

$ ./a.out
keygen_me by wex-ler @ 0x00sec.org
user >> 024bdfhjlnprtvxz
pass >> kcs3gow1aeimquy
success!

I have not yet created a keygen, I might do this at some point. Thanks for the challenge; you gave a bit too much away with the title in my opinion. Now, go study for your finals :stuck_out_tongue:

EDIT:
Okay, I could not resist the temptation to write a keygen as well, here it is written in python. It is pretty simple:

#!/usr/bin/python3
import string
import random

n = 31
offset = random.randint(33, 125-n)

chars = [i for i in range(offset, offset+n)]
chars = bytes(chars).decode('ascii')

stack = []
stack.append(chars)

result = []

while len(stack) > 0:
    s = stack[0] 
    i = int(len(s) / 2)
    result.append(s[i])
    
    if len(s) > 1:
        s1 = s[:i]
        s2 = s[i+1:]
        stack.append(s1)
        stack.append(s2)
    stack = stack[1:]

print("{}".format(''.join(result[15:])))
print("{}".format(''.join(result[:15])))

The results can be validated using ./solve.py | ./a.out


#3

Was fun! :smiley:

$ ./balancedtree.elf
keygen_me by wex-ler @ 0x00sec.org
user >> Thanks_for_the_challenge!
pass >> F7N3BJR159DHLPU02468ACEGIKMOQSV
success!

Considering @Noswis already wrote a keygen for random username-password pairs, here is one where you can choose the name to generate a password for:

import string
import sys
print("Enter name to generate key for: ")
name = sys.stdin.readline() # read wanted username
cMap = (string.digits+string.ascii_uppercase+string.ascii_lowercase).translate(None,name) # remove all characters from the name from the sorted ascii characters
if len(name) <= 7 or " " in name or len(cMap) < 31: print("Name is not valid (>= 7 characters, no spaces, not too long)") # validate that this code can generate a password for the name
else: print(''.join([cMap[v] for v in [15, 7, 23, 3, 11, 19, 27, 1, 5, 9, 13, 17, 21, 25, 29, 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30]])) # use a given perfect binary search tree as offsets within the character map to generate another perfect binary tree usable as a password

(_w) #4

Did you find the solution purely based off the title or did it require more analysis? I will be more secretive with titles in the future :stuck_out_tongue:


(_w) #5

I have a couple more ideas for challenges based on algorithm’s. Do people find them fun or should I focus more on obfuscation?


#6

I personally like both approaches and considering I mostly see obfuscation (or similar) based challenges some more based on algorithms might be interesting!


#7

A bit of both. The title was definitely helpful, but I still needed to dive into the binary to find specific username/password requirements. In my opinion, it is often not worth to try to understand everything in a binary for a keygen/crackme challenge. I usually make “educated” guesses about the roles of functions and constants based on their names and try things out based on that.