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Saturday, October 26 2013 03:30

Hack.lu 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>
commands:
-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: https://ctf.fluxfingers.net/static/downloads/fluxarchiv/hacklu2013_archiv_challenge1.tar.gz

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

rev400.png

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]) {
 
							checkHashOfPassword(password);
							//encryptDecryptData(f); // Useless actually, only used to check the MAGIC_VALUE
 
							if (verifyArchiv(f)) {
								printf("Password found: %s\n", password);
								return 0;
							}
						}
		}
	}
 
	fclose(f);
 
	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.so hooker.c -ldl
awe@awe-laptop ~/hacklu/reverse/FluxArchiv1 % time LD_PRELOAD=./hooker.so ./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=./hooker.so ./archiv -l FluxArchiv.arc BADGER
LD_PRELOAD=./hooker.so ./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: https://ctf.fluxfingers.net/static/downloads/fluxarchiv/hacklu2013_archiv_challenge1.tar.gz

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:

rev500_CFG.png

# PURPLE: go to the initial offset, read next_chunk (which will be used to calculate the offset for the next chunk)
f.seek(init_offset)
next_chunk = unpack("<Q", doRC4(f.read(8)))
 
# YELLOW: read max_chunks (total number of chunks)
max_chunks = unpack("<Q", doRC4(f.read(8)))
 
# DARK GREEN: overwrite next_chunk in the file with junk
next_offset = next_chunk << 4
next_offset = next_offset + (next_offset << 6) + 0x20
fseek(next_offset)
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))
 
# LIGHT GREEN: 
for chunk_index in range(max_chunks):
    # GREY: go to the next chunk offset, read next_chunk
    f.seek(next_offset)
    next_chunk = unpack("<Q", doRC4(f.read(8)))
 
    # ORANGE: overwrite the next_chunk in the file with junk
    f.seek(next_offset) ; 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+")
 
    try:
        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
 
            f.seek(next_offset)
            f.read(8) # Skip overwritten next_chunk
 
            content = decryptRC4(f.read(0x408))
            o.write(content)
 
            next_chunk += 1
    except:
        pass
    finally:
        o.close()
 
f.close()

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.

Thursday, October 18 2012 16:01

HackYou CTF - Reverse100, Reverse200, Reverse300 Writeups

Reverse 100 - Open-Source



Download file code.c.
Simply read the source...

% ./code `python2 -c 'print 0xcafe'` 25 h4cky0u
Brr wrrr grr
Get your key: c0ffee

Flag: c0ffee

Reverse 200 - LoseYou



Download file rev200.zip.

Extract the archive to obtain task2.bin and task2.exe.
I decided to study task2.bin.
The routine to reverse is sub_80483DC.
We quicky notice this:

hackyou_reverse200.png

That's very basic, eax will contain our guessed number, ecx will contain the randomly generated number.
All we need to do is to break on the "cmp eax, ecx" at 08048503 and set eax to ecx.

(gdb) b *0x08048503
Breakpoint 1 at 0x8048503
(gdb) r
Starting program: /tmp/task2.bin 
warning: Could not load shared library symbols for linux-gate.so.1.
Do you need "set solib-search-path" or "set sysroot"?
Welcome to the LoseYou lottery!
Generating random.....
Make your guess (number 0 or 1): tg
 
Breakpoint 1, 0x08048503 in ?? ()
(gdb) set $eax=$ecx
(gdb) c
Continuing.
You... you... win??? so lucky! Grab the flag:
::: oh_you_cheat3r :::
[Inferior 1 (process 27619) exited normally]
(gdb)

Flag: oh_you_cheat3r

Reverse 300 - ashtree



Download file rev300.zip.
Once again I decided to study the ELF version: task3.bin.

Step 1 - Unpack

The binary seems to be packed by a modified UPX (at least, UPX string is replaced by LOL...).
Let's trace execution:

% strace ./task3.bin 
execve("./task3.bin", ["./task3.bin"], [/* 34 vars */]) = 0
[ Process PID=5640 runs in 32 bit mode. ]
getpid()                                = 5640
gettimeofday({1350215641, 677903}, NULL) = 0
unlink("/tmp/upxCRBOGQOAFQI")           = -1 ENOENT (No such file or directory)
open("/tmp/upxCRBOGQOAFQI", O_RDWR|O_CREAT|O_EXCL, 0700) = 3
ftruncate(3, 9036)                      = 0
old_mmap(NULL, 9036, PROT_READ|PROT_WRITE, MAP_SHARED, 3, 0) = 0xf775a000
old_mmap(0xf775d000, 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_FIXED|MAP_ANONYMOUS, -1, 0) = 0xf775d000
munmap(0xfffffffff775a000, 9036)        = 0
close(3)                                = 0
open("/tmp/upxCRBOGQOAFQI", O_RDONLY)   = 3
getpid()                                = 5640
access("/proc/5640/fd/3", R_OK|X_OK)    = 0
unlink("/tmp/upxCRBOGQOAFQI")           = 0
fcntl(3, F_SETFD, FD_CLOEXEC)           = 0
execve("/proc/5640/fd/3", ["./task3.bin"], [/* 41 vars */]) = 0
[...]

The binary is unpacking itself into a file "/tmp/upxCRBOGQOAFQI", which is a randomly generated name. Then it is unlinked and executed by execve().

The unlink() function is guaranteed to unlink the file from the file system hierarchy but keep the file on disk until all open instances of the file are closed.

Once the execution is over, the file is deleted, so we need to find the file name, and copy the file elsewhere before it's unlinked. Let's toggle a breakpoint before execve() then.

(gdb) b *0x004024dd
Breakpoint 1 at 0x4024dd
(gdb) r
Starting program: /tmp/task3.bin 
 
Breakpoint 1, 0x004024dd in ?? ()
(gdb) x/s $ebx
0xffffd494:	"/proc/12767/fd/7"
(gdb)
% file /proc/12767/fd/7
/proc/12767/fd/7: broken symbolic link to `/tmp/upxC5RF3NOAMO5 (deleted)'

Ok! Now we can work on the unpacked binary.

Step 2 - Keygen

Our routine is sub_8048617.
Several conditions must be met:

  • - argv[1] = username
  • - argv[2] = password
  • - username != 'hackyou'
  • - len(password) == 14
  • - password[4] == '-' and password[8] == '-'
  • - sub_804838C(username, password) == True
  • - sub_804844B(username, password[5:]) == True
  • - sub_804850A(username, password[10:]) == True

Although our username must not be 'hackyou', if you take a look at sub_80485F0, the goodboy message, you'll notice it prints "Great! Now submit the license key for 'hackyou'".
sub_804838C, sub_804844B and sub_804850A all proceed in the same way, which finally compares individually 4 bytes of the given password parameter:

hackyou_rev300_ida.png

Our password character is in edx, the expected one is in eax. Lazy as i am, i didn't go much deeper :)

All we need now is to break on:

.text:0804842B                 cmp     eax, edx

... dump eax, and fix edx so that it will be equal to eax. Same for the two last routines. Let's create a basic pythonGDB script to do the work for us.

import gdb
 
passwd = ""
passwd_len = 0
 
def callback_username_condition():
    gdb.execute("set $eax=1")
 
def callback_compare_password():
    global passwd, passwd_len
 
    gdb.execute("set $edx=$eax")
 
    if passwd_len in (4, 9):
        passwd += '-'
        passwd_len += 1
 
    passwd += chr(gdb.parse_and_eval("$eax"))
    passwd_len += 1
 
    print "[+]", passwd
 
class HitBreakpoint(gdb.Breakpoint):
    def __init__(self, loc, callback):
        super(HitBreakpoint, self).__init__(
            loc, gdb.BP_BREAKPOINT, internal=False
        )
        self.callback = callback
 
    def stop(self):
        self.callback()
 
        return False
 
HitBreakpoint("*0x08048665", callback_username_condition)
HitBreakpoint("*0x0804842B", callback_compare_password)
HitBreakpoint("*0x080484EA", callback_compare_password)
HitBreakpoint("*0x080485A9", callback_compare_password)

And run it...

(gdb) source script.py 
Breakpoint 1 at 0x8048665
Breakpoint 2 at 0x804842b
Breakpoint 3 at 0x80484ea
Breakpoint 4 at 0x80485a9
(gdb) r hackyou 0123-4567-8910
Starting program: /tmp/unpacked_rev300 hackyou 0123-4567-8910
warning: Could not load shared library symbols for linux-gate.so.1.
Do you need "set solib-search-path" or "set sysroot"?
[+] k
[+] ke
[+] kec
[+] kecc
[+] kecc-h
[+] kecc-ha
[+] kecc-hac
[+] kecc-hack
[+] kecc-hack-y
[+] kecc-hack-yo
[+] kecc-hack-yo0
[+] kecc-hack-yo0u
Great! Now submit the license key for 'hackyou'
[Inferior 1 (process 14524) exited with code 01]
(gdb)

Flag: kecc-hack-yo0u

Sunday, March 25 2012 23:33

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

From: Jessica <jessica@megacortek.com>
To: w3pwnz <w3pwnz@megacortek.com>
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
http://sci.nuitduhack.com/EgZ8sv12.
---

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.

Regards,
Jessica.


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:
AAAAAAA
[...]
[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)) ])
5c1t33k

Bazinga \o/

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

From: Jessica <jessica@megacortek.com>
To: w3pwnz <w3pwnz@megacortek.com>
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.

Regards,
Jessica.


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:
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 patch.py 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")
'zApli8oW'

Let’s use it:

~/ndh2012$ nc sciteek.nuitduhack.com 4001
Sciteek protected storage #1
Enter your password: zApli8oW
<PSP version="1.99">
<MOTD>
<![CDATA[
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.
]]>
</MOTD>
<FLAG>
ea1670464251ea3b65afd624d9b17cd7
</FLAG>
<ERROR>
An unexpected error occured: PSP-UNK-ERR-001> application closed.
</ERROR>
</PSP>