Skip to content
Switch branches/tags
Go to file
Cannot retrieve contributors at this time

White Rabbit CrackMe

It's easier to follow this write-up if you open up IDA and use the screenshots as guides to points of interest. It's a good idea to name symbols as you go through and I give several suggestions for symbol names throughout the text, however most of the screenshots are from a fresh IDA session so that they can be used independently.

Thanks to @hasherezade and Grant Willcox for the crackme.


Stage 1: White Rabbit

The crackme is ~7MB windows PE32 executable called white_rabbit.exe. A quick peek in CFF Explorer shows us that the bulk of the exe is in the resources (section .rsrc):


There are two RCDATA resources the first of which (id 101) is the biggest and looks suspiciously like a very low entropy file that's been encrypted with a Vignere cipher (the giveaway is the long-ish repeating patterns):


Running the executable produces an ASCII art rabbit, a green coloured introductory text (printed slowly, one character at a time) and a prompt for the first (of many?) passwords:


With these hints in mind, let's fire up IDA to have a peek at the code driving all this. After initial autoanalysis, IDA takes us straight to main(), which seems to have a very simple structure:


  • sub_402CC0 seems to do the initial setup
  • sub_4034D0 is probably Level1 (returning false triggers a "Nope!" message and exit)
  • sub_4036F0 is probably Level2 (similar reasoning)

Scanning through the three subs seems to verify our guesses as the first one references the rabbit ASCII art and the intro text, the second one references the string "Password#1:" and the third one has a reference to the string "Password#2:" (we could have gotten to the same conclusion by ignoring program flow and looking for references to the password prompt strings). Let's focus on the second sub (which I've renamed do_level1). After a bit of boilerplate, it calls sub_403D90 with two arguments: the address of a stack-based variable and the number 0x65 (which is 101 in decimal). IDA is helpfully pointing out that the second argument is probably being used in connection with some resource (the hResData comment comes from this argument being passed to a system call inside sub_403D90). Furthermore note that IDA has already named the stack variable nNumberOfBytesToWrite, indicating that it is used as a data size argument on some system call:


Indeed, a quick scan through sub_403D90 can confirm that it loads the resource identified by the second argument, copies its contents into some freshly allocated memory (courtesy of VirtualAlloc()) and returns a pointer to this copy in eax (return value) while putting the size of the loaded resource into the stack variable pointed to by its first argument. Thus we can conclude that do_level1() uses sub_403D90 (which we can rename to load_resource_into_buffer) to load resource 101 (the big, low entropy Vignere'd blob) into a writable memory chunk, the address of which is then kept in esi, while the size of the resource is stored into the stack variable [ebp+nNumberOfBytesToWrite].


Just looking at the call structure, you'd guess that the first call (sub_403990) prints the password prompt, the second call (sub_401000) reads the password and the third one calculates some checksum or hash that needs to be equal to 0x57585384. The dataflow in the arguments to the functions confirms these guesses as the first function is only passed pointers to read-only data, hence it probably only prints something, the second function gets given a pointer to the writable stack argument [ebp+var_438] as well as a pointer to a static (which turns out to be std::cin) and the third function gets either the data or a pointer to the same stack argument.

As an aside, the whole run-around with the compare to 0x10 and cmovnb arises from an optimization in the std::string implementation where if a string is less than 16 chars long, then it is stored directly in the std::string structure otherwise a memory block is allocated for the string and the first dword of the structure is set to point to this block. Offset 0x10 in the structure ([ebp+var_428]) holds the string's size() and offset 0x14 ([ebp+var_424]) holds the string's capacity().

Without further ado, we can rename these three functions to print_coloured_string, read_input_line and calculate_checksum respectively.

Looking into calculate_checksum doesn't get us much further: It calculates a 32-bit CRC-like checksum using the primitive polynomial 0x182F63B78. There are a couple of ways to invert this, but as the checksum only has 32-bits, it will not give us the complete password we need.

The next segment is a bit more interesting:


Effectively it's just: sub_403C90(resource_101_address, resource_101_size, password1). Exactly the signature we would expect from a decryptor. Looking at the sub itself, it seems to implement a standard Vignere-xor cipher: It first calculates the length of the password (arg_8) which it stores into ecx and then uses div ecx to divide the current offset in the data buffer by the length of the password and get the remainder into edx which it uses as an index into the password:


The next part of do_level1() is fairly dull: Create a random filename in the temp directory using the template wallp{hex_number}.tmp and then save the decrypted resource to it (courtesy of sub_403090 which I'm renaming to save_data_to_file). After the file has been written, do_level1() calls sub_403D20 with the newly created filename as argument:


Looking up SystemParametersInfo in MSDN reveals that 0x14 corresponds to SPI_SETDESKWALLPAPER which changes the desktop wallpaper to the supplied file.

Therefore we can conclude that resource 101 is a Vignere encrypted low-entropy image file format, and hence most likely a windows BMP file. We can use the information we've gleaned to break the Vignere cipher. The length of the password is easy to see by looking at the hexdump in CFF explorer: the repeating pattern is 17 bytes long. Looking up the specification of BMP files reveals that the header starts with 'BM' followed by the length of the bitmap in little endian format. Let's use CFF explorer to save the resource to res_101.bin and then load it up in python to play around with it (I'm using the XOR cipher implementation from the PyCrypto package because I'm lazy and I already have this installed, but it's fairly easy to find other implementations on the net or write your own version):


With the known plaintext of the BMP header we've recovered the first six characters of the password: follow. The section of the file with the repeating pattern must have contained the same (constanst) value in the plaintext. We can find this value by xor-ing a segment of that section with the part of the password we've recovered. Notice that we need to align our partial decryption with a multiple of 17 (the length of the password):


The long constant sections in the plaintext file are filled with 0x01, hence it's now trivial to recover the whole password by xoring one of the repeating segments with the known plaintext of repeated 0x01:


And now we have the first password and the decrypted wallpaper:


The do_level2() function has a very similar structure to do_level1() except that it loads resource 102, it uses a more complicated decryption function (sub_403E10), it saves the result to an executable in the temp directory with filename template good_rabbit{hex number}.exe and executes it after successful decryption.

The decryption sub seems to be using the Microsoft CryptoAPI with some pretty scary cipher choices (0x800C is CALG_SHA256 and 0x660E is CALG_AES_128):


It doesn't look like we'll be recovering the password of this one the same way we did for the Vignere cipher, therefore the bitmap that first level dropped must contain some kind of hint. There are a number of forensic / steganalysis apps and websites out there. Personally, I first hit Forensically. The Luminance Gradient analysis (with histogram equalization) quickly reveals a hidden message:


The password for level2 is IMdsSqFGLf6v_wxO. We can replicate the decryption function in the code, however it's just easier to type in the two passwords we've discovered and let white_rabbit.exe drop its good_rabbit.exe payload in our temp directory (please tell me you've been running all this in a VM).

Stage 2: Good Rabbit

The stage 2 payload good_rabbit.exe is a ~150KB PE32 executable. It doesn't have any resources and appears entirely ordinary (except for one detail, which we'll come to later). Running it doesn't seem to do much - it just hangs there and has to be killed. Time to fire up IDA.

The main function seems straightforward:


  • Check for the presence of internet connection (sub_404450 calls InternetGetConnectedState)
  • Inititalize some buffers
  • Get the system browser (sub_4043E0 reads HKCR\HTTP\shell\open\command)

The next part is a bit stranger:


sub_401A80 looks like a string constructor that's called to construct a string on freshly allocated space on the stack (the sub esp, 18h). This is the variable [ebp+lpCmdLine] which is initialized to contain the browser name and gets further filled in by sub_403360 before being handed off to WinExec.

sub_403360 looks quite complicated: Lots of stack variables, several objects with vtables being constructed, references to the iostream libraries. Two strings stand out:


Looks like it sets up the URL:


Clearly we've missed something. There are a couple of ways to make progress from this point:

  1. Persevere with analysing this function and notice that the string that is appended to the youtube URL may also come from [ebp+var_58] which is initialised by call call to sub_403110:


In turn, this function seems to use some synchronization mechanism that relies on a global mutex handle stored in hMutex which is also referenced by a function called TlsCallback_0


  1. Alternatively, a quick look at the strings leads us to multiple error messages related to network operations that roll up to TlsCallback_0.
  2. Looking at the imports turns up references to socket, connect, etc which all roll up to TlsCallback_0
  3. Notice the presence of a TlsDirectory entry in CFF Explorer

The .tls section in PE files provides native support for Thread-Local Storage (i.e. variables that have separate instances for each running thread). The PE format implementation is described in this MSDN article. The key point here is that the TlsDirectory contains a set of entries that define callbacks which the OS triggers whenever a process is started or a thread is created or destroyed. The convention is similar to the one for DllMain. Nynaeve has a more in-depth article on TLS. Needless to say that IDA already parses that section and names the callback entries appropriately.

Which brings us to TlsCallback_0:


It seems to fill in a small buffer with random numbers and immediately raise an exception! Looking at the boilerplate at the top of the function, TlsCallback_0 sets a custom exception handler which is actually implemented inside the body of the function:

stage2-tlscallback-2 stage2-tlscallback-3 stage2-tlscallback-4

The exception filter picks up all exceptions and the handler just calls sub_403330 to fill the same buffer as earlier with random characters and then continues execution as normal. As an aside, dword_42109C is the length of the string that will be exchanged with main and that comparison is there to ensure that this part of the code does not overwrite the string twice.

Anyway, it seems that both sub_403330 and the array at byte_422AE8 are decoys, so I'll rename them to decoy_randomize_array and decoy_random_array respectively.

The rest of TlsCallback_0 is fairly simple-looking: It calls sub_404480 and uses a single character result to decode some bytes into an internal buffer. The calls read as follows:

sub_404480(&local_buf, 1337);
sub_404480(&local_buf, 1338);
sub_404480(&local_buf, 1339);

sub_404480 creates a socket and then decrypts an internet address using the code below:


A quick line of IDAPython can sort this one out:

''.join([ chr((ord(x) + 0xf3)&0xff) for x in idaapi.get_many_bytes(0x41a53c,9)])

To reveal that the address in question is The function then binds to this interface and the port number given by the second argument to the function (IDA names this hostshort because it's passed to htons before being given to bind). It then calls listen and accept to accept incoming connections and then:


It receives up to four characters into a stack-based buffer which it then passes to sub_404640, giving it the address of a global buffer (unhelpfully named buf by IDA) as a second argument. If this call returns success, it sends four characters from the global buffer to the connection as a response.

The heart of this part of the challenge is the function sub_404640, which starts off with:


The second argument (arg_4) contains a pointer to the global buffer buf which this sub keeps in ecx. Looking at the structure of the code, it seems to behave like a state machine using the first character of buf as the current state and the first character of the incoming message (pointer to by arg_0) as the input. The first segment of the function only accepts states '\0', 'E' and 'Y' as valid states, hence we can make the function easier to read by renaming the targets of the corresponding branches (IDA tip: you can rename the target of a jump instruction by putting the cursor on the location name and hitting 'N' - saves having to double-click back and forth):


And hence we have the transition diagram:

'\0' -- ['9'] -> 'Y'
'Y'  -- ['3'] -> 'E'
'E'  -- ['5'] -> 'S'

In summary, good_rabbit.exe expects to receive the three characters '935' one at a time, delivered separately to the ports 1337, 1338, 1339. It uses the received characters to decode a youtube reference which it then pops up. We can reverse engineer the part of TlsCallback_0 that decodes the youtube reference, however it's a lot easier to write a Python script that just delivers the characters to the right ports.

Just fire up good_rabbit.exe and then hit this script:

from socket import create_connection

for c, p in zip('935', range(1337, 1340)):
    conn = create_connection(('', p))
    print conn.recv(1)

to get: