Writeup The Neverending Crypto Challenge
Introduction
Last weekend i participated together with the Ulm Security Sparrows in the TU CTF of the University of Tulsa Tandy School. A CTF, for those of you who do not know is a hacking contest where hackers break security stuff and have fun overall. I was mainly busy with the “neverending crypto challenge” and in this post I am going to document what I have done there. Many thanks to Uni Tulsa and to the Ulm Security Sparrows.
This post, together with the writeups of my comrades can also be found here.
Level1
The first hint that was given was to connect to port 24069 at the ip 146.148.102.236. Of course I did that and was greeted
Welcome to The Neverending Crypto!
Quick, find Falkor and get through this!
This is level 1, the Bookstore
Round 1. Give me some text:
So, I typed ‘a’ on my keyboard and the server responded
a encrypted is .-
What is -.. — -. - ..-. — .-. –. . - .- ..- .-. -.– -. decrypted?
Well, I am not a specialist of morse code, but i recognize morse when I see it. Nevertheless I was too slow to search on wikipedia for a table, decode that stuff and type it back. The server closed the connection. So I wrote a script to decode morse. After many trial and errors I finally succeeded and answered correctly to the server who only wanted that I decode more morse.
Correct!
Round 2. Give me some text:
a encrypted is .-
What is .- - .-. . -.– ..- …- … –. – — .-. -.- decrypted?
So I had to modify my script. I put a loop around the decoder. At first it was an endless loop, but at 50 rounds the script exploded.
- Round 50. Give me some text:
- a encrypted is .-
- What is - .-. -.– … .– .. – – .. -. –. decrypted?
- Cleartext is:try swimming
Correct!
Round complete!
TUCTF{i_wi11_n0t_5teal}
——————————————
You take the book
This is level 2 the attic
Round 1. Give me some text:
So. we finally made it to level 2. Sweet victory. This TUCTF{i_wi11_n0t_5teal}-stuff is a token which you could put in the website and then my team got 10 credits. That’s how CTFs work.
What follows is my code. The file was named scriptl1.py
ip='146.148.102.236'
port=24069
CODEL1 = { '.-' : 'a', '-...':'b', '-.-.':'c',
'-..' : 'd', '.' :'e', '..-.':'f',
'--.' : 'g', '....':'h', '..' :'i',
'.---': 'j', '-.-' :'k', '.-..':'l',
'--' : 'm', '-.' :'n', '---' :'o',
'.--.': 'p', '--.-':'q', '.-.' :'r',
'...' : 's', '-' :'t', '..-' :'u',
'...-': 'v', '.--' :'w', '-..-':'x',
'-.--': 'y', '--..':'z',
}
import socket
import re
# Level 1
sock = socket.socket()
sock.connect((ip, port))
for i in xrange(50):
print(sock.recv(256))
# Round i. Give me some text:
sock.send("a\n")
print("#######")
answer=sock.recv(256)
print("server answer:" +answer)
# a encrypted is .-
# What is ..-. .-. .- --. -- . -. - .- - .. --- -. decrypted?
# :
match=re.search('What is (.*) decrypted?', answer)
morse=match.group(1)
cleartext=''
for morsechar in morse.replace(" "," space ").split():
if morsechar == 'space':
cleartext += ' '
else:
cleartext += CODEL1[morsechar]
print("Cleartext is:" + cleartext)
sock.send(cleartext+"\n")The stuff with the space is a dirty hack. If you split a string containing multiple spaces in Python the sucker will just disregard it. So i substituted multiple spaces by “ space “, split and then appended a space when encountering the word “space”. Well, at CTFs you are sometimes under timepressure and code elegance is subordinate.
Level 2
For level 2 I had to go through level 1 every time. At least that is what the guys at the irc-channel said:
(13:35:28) uss_cryptobunny: whoisjohngalt: in the crypto challenge is there a shortcut to level2 or do i have to run through level1 every time?
(13:35:59) neptunia: ^ every time
(13:36:04) neptunia: that’s why its neverending
So i restructured my code. I made a main.py which executed the different files for the different levels. At the end it looked like this:
execfile("scriptl1.py")
execfile("scriptl2.py")
execfile("scriptl3.py")For level 2 i tried some letters. If the phrase was too long, the server said it is too long and closed the connection. It was a really slow process, since i had to go through the first level every time. I quickly discovered that the cipher was a simple monoalphabetic substitution. That is one of the easiest ciphers in existence. One simply substitutes each letter with a different letter. For example ‘a’ becomes ‘n’, ‘b’ becomes ‘o’. It was simply a matter of trying out all the letters, making a codebook and reusing the code from the first level. Oh, and did I mention putting everything in a loop with 50 rounds?
And then finally, level 3:
- Round 50. Give me some text:
- ABCDEFGH encrypted is NOPQRSTU
- What is v-|%r-‘|# decrypted?
- Cleartext is:i owe you
Correct!
Round complete!
TUCTF{c4n_s0me0ne_turn_a_1ight_0n}
——————————————
You have found The Nothing
This is level 3, The Nothing.
Round 1. Give me some text:
So, here is the code of scriptl2.py which solved the second level:
ip='146.148.102.236'
port=24069
import socket
import re
# Round complete!
# TUCTF{i_wi11_n0t_5teal}
# ------------------------------------------
# You take the book
# This is level 2 the attic
# Level 2
CODEL2 = {
'-' : ' ',
'n' : 'a',
'o' : 'b',
'p' : 'c',
'q' : 'd',
'r' : 'e',
's' : 'f',
't' : 'g',
'u' : 'h',
'v' : 'i',
'w' : 'j',
'x' : 'k',
'y' : 'l',
'z' : 'm',
'{' : 'n',
'|' : 'o',
'}' : 'p',
'~' : 'q',
' ' : 'r',
'!' : 's',
'"' : 't',
'#' : 'u',
'$' : 'v',
'%' : 'w',
'&' : 'x',
"'" : 'y',
'(' : 'z'
}
# built using many trials and assuming monoalphabetic substitution
# hopefully they did not put any numbers in
for i in xrange(50):
print(sock.recv(256))
# Round i. Give me some text:
sock.send("ABCDEFGH\n")
print("#######")
answer=sock.recv(256)
print("server answer:" +answer)
# server answer: STUVW encrypted is -`abcd
match=re.search('What is (.*) decrypted?', answer)
ciphertext=match.group(1)
cleartext=''
for cipherletter in ciphertext:
cleartext += CODEL2[cipherletter]
print("Cleartext is:" + cleartext)
sock.send(cleartext+"\n")Level 3
The third level was hard. If i remember correctly ‘b’ decrypted was ‘b’, ‘bb’ decrypted was ‘yy’, ‘bbb’ decrypted was again ‘bbb’. I had no idea what was going on. And for each trial I had to walk through the two levels before. I opened multiple shells to run through the stuff in parallel. Spaces were preserved, so there was no permutation, just substitution of characters through other characters. I also assumed that it was a deterministic encryption, since groups of characters were always substituted by the same group of characters with the same length. As i later discovered this was not the case. the nondeterminism was just very seldom.
And then I had an idea. As with all good ideas, it is not entirely clear how they form. Just that small spark of intuition. I looked at all the cleartexts I had decrypted before and discovered that there was about a hand full of them, all having to do with the neverending story. Nice. So i thought that they will perhaps be the same again and tried them all. It was made a little more difficult by the fact, that for some cleartexts there were multiple ciphertexts. For example ‘dont forget auryn’ could either encrypt to ‘erby urpi.y agpfb’ or to ‘sykg typdfg alpjk’.
After I made that codebook with the words, the rest was simple. And after 50 rounds, I saw the familiar text:
- Round 50. Give me some text:
- the nothing encrypted is ghf kyghukd
- What is ravf ghf ;pukcfrr decrypted?
- Cleartext is:save the princess
Correct!
Round complete!
TUCTF{5omething_is_b3tt3r_th4n_n0thing}
——————————————
You have run into a bad place…
This is level 4, the swamps of sadness.
Round 1. Give me some text:
Here is the code for scriptl3.py:
ip='146.148.102.236'
port=24069
import socket
import re
# Round complete!
# TUCTF{i_wi11_n0t_5teal}
# ------------------------------------------
# You take the book
# This is level 2 the attic
# Level 2
CODEL3 = {
"agpfjl vr dmype" : "atreyu vs gmork",
"ayp.fg ko imrpt" : "atreyu vs gmork",
"xaoycab" : "bastian",
"barguak" : "bastian",
"erby urpi.y agpfb" : "dont forget auryn",
"sykg typdfg alpjk" : "dont forget auryn",
"sgpfjl vr dmype" : "dtreyu vs gmork",
"tpadmfkgaguyk" : 'fragmentation',
"upaim.byaycrb" : 'fragmentation',
"duakg glpgif" : "giant turtle",
"icaby ygpyn." : "giant turtle",
"dmyper cyyi" : "gmorks cool",
"imrpto jrrn" : "gmorks cool",
"c r,. frg" : "i owe you",
"u ywf jyl" : "i owe you",
"myyk chuis" : "moon child",
"mrrb jdcne" : "moon child",
"mf ngjtepairb" : "my luckdragon",
"mj ilcespadyk" : "my luckdragon",
"pfasukd ur sakdfpylpr" : "reading is dangerours",
"p.aecbi co eabi.prgpo" : "reading is dangerours",
"ravf ghf ;pukcfrr" : "save the princess",
"oak. yd. lpcbj.oo" : "save the princess",
"ghf kyghukd" : "the nothing",
"yd. brydcbi" : "the nothing",
"ghf ypacif" : "the oracle",
"yd. rpajn." : "the oracle",
"ypf o,cmmcbi" : "try swimming",
"gpj rwummukd" : "try swimming",
"wficymf agpfjl" : "welcome atreyu",
",.njrm. ayp.fg" : "welcome atreyu"
}
# we found out, that in the previous challenges, we always had the
# same words, so we built a table
for i in xrange(50):
print(sock.recv(256))
# Round i. Give me some text:
sock.send("the nothing\n")
print("#######")
answer=sock.recv(256)
print("server answer:" +answer)
match=re.search('What is (.*) decrypted?', answer)
ciphertext=match.group(1)
cleartext = CODEL3[ciphertext]
print("Cleartext is:" + cleartext)
sock.send(cleartext+"\n")
print(sock.recv(256))
# Round complete!
# TUCTF{5omething_is_b3tt3r_th4n_n0thing}
# ------------------------------------------
# You have run into a bad place...
# This is level 4, the swamps of sadness.
# Round 1. Give me some text:
=crypto3=Level 4
Level 4 was really damn weird. The encryptions were always totally different. And I had to wait half an eternity to run through all my script to get more cleartext-ciphertext pairs. I have no idea how the code there worked. And again if I succeeded I would only get 10 points. In comparison some web sql-injection challenge yielded at least 50 points. This was basically the reason I quit and tried some other challenges. So no code beyond level 3.
I have one log which is the transcript of the communication between my program and the server, which is perhaps interesting. It can be found here.