7 - wopr
Time spent: 3-4 hours
Tools used: x64dbg, ghidra, pyinstxtractor, HxD, Python with Z3
Challenge 7 was called wopr, and came with another executable that opens a terminal with a list of games to play.
The only game that is playable, however, is the one titled “GLOBAL THERMONUCLEAR WAR”. It then proceeds asking which city you want to get bombed, and then prompts you for the launch code.
Obviously I don’t know the launch code, let’s find out what it is!
Orientation
Opening the file in ghidra, and inspecting the strings reveals the following:
This seems to be an executable with some embedded Python code in here. I never worked with that before, so this is going to be interesting.
One of the strings that is interesting to me as well is "_MEIPASS2"
. A quick Google search reveals that this binary is probably an executable produced by “PyInstaller”. Another quick Google search reveals that there is a tool called pyinstxtractor that extracts the compiled Python code from these files:
python pyinstxtractor.py wopr.exe
This produces a new directory wopr.exe_extracted
containing all the files it needs to run the embedded Python program:
All the files? I did’t see a main
file or anything that looks like one as many tutorials that I could find on the internet.
At this point I started wasting a lot of time on opening all kinds of *.pyd
files, until I realized that the cleanup file was substantially bigger than any of the other bootstrapper files:
Then I revisited the output of pyinstxtractor
, and realized my mistake:
[*] Processing .\wopr.exe
[*] Pyinstaller version: 2.1+
[*] Python version: 37
[*] Length of package: 5068358 bytes
[*] Found 64 files in CArchive
[*] Beginning extraction...please standby
[+] Possible entry point: pyiboot01_bootstrap
[+] Possible entry point: pyiboot02_cleanup
[*] Found 135 files in PYZ archive
[*] Successfully extracted pyinstaller archive: .\wopr.exe
It literally stated the “cleanup” file is potentially an entrypoint. Oopsie by me!
Since these files were binary files, and not normal Python scripts, I searched Google again if there was anything like a disassembler or even a Python bytecode decompiler. Turns out, there is one called uncompyle
, but for some reason I couldn’t get it to work on this file (even after changing the header as many tutorials state). So instead, I went the hard way and wrote a little disassembler script based off another script I found on the internet:
import dis
import marshal, struct, sys, time, types
def show_file(fname):
f = open(fname, "rb")
magic = f.read(4)
moddate = f.read(4)
modtime = time.asctime(time.localtime(struct.unpack('L', moddate)[0]))
code = marshal.load(f)
show_code(code)
def show_code(code, indent=''):
indent += ' '
print(indent, "name", indent, code.co_name)
print(indent, "argcount:", code.co_argcount)
print(indent, "snlocals:", code.co_nlocals)
print(indent, "sstacksize:", code.co_stacksize)
print(indent, "sflags:", code.co_flags)
dis.disassemble(code)
print(indent,"sconsts")
for const in code.co_consts:
if type(const) == types.CodeType:
show_code(const, indent+' ')
print(indent, "names", indent, code.co_names)
print(indent, "varnames", indent, code.co_varnames)
print(indent, "freevars", indent, code.co_freevars)
print(indent, "cellvars", indent, code.co_cellvars)
print(indent, "filename", indent, code.co_filename)
def show_hex(label, h, indent):
h = h.hex()
if len(h) < 60:
print(indent, label, h)
else:
print(indent, label)
for i in range(0, len(h), 60):
print(indent, " ", h[i:i+60])
show_file(sys.argv[1])
This produced a file that starts off with a huge documentation string. Here is a small excerpt of it:
\nOnce upon a midnight dreary, while I pondered, weak and weary,\t\t \t\t\t \t\t \t \t \t\t\t \t\t\t\t \t\t\t\t\t\t \t\t \t \t\t\t \t\t \t\t\t\t \t\t\t \t\t\t\t\t \t \t \t \t\t \t\t \t \t\t\t \t\t \t \t \t \t\t \t \t \t\t \t\t \t \t \t\t\t\t \t\t\t \t \t \t\t \t\t \t \t \t\t \t\t \t\t \t \t \t\t\t \t \t \t\t \t\t\t\t \t \t \t\t\t
I then went on and started working on the painful process of decompiling the bytecode manually (yes I am stupid). My decompiled version of the script with some slight modifications were as follows. The names of each variable that the FLARE-ON team came up with were quite creative I must say:
BOUNCE = open("\\this\\key", "rb")
BINARY = # huge doc string....
def ho(h, g):
k = bytes.fromhex(format(h, 'x')).decode()
return g.get(k, k)
def eye(face):
# Maps tabls and spaces to zeroes and ones...
def fire(wood, bounce):
# Implements RC4 encryption algorithm...
if __name__ == '__main__':
a = 1702389091
b = 482955849332
g = ho(29516388843672123817340395359, globals())
aa = exec
bb = print
a = a ^ b
b = b ^ a
a = a ^ b
for i in range(256):
try:
script = lzma.decompress(fire(eye(BINARY.encode()), bytes([i]) + BOUNCE))
exec(script)
except:
pass
We see that the main function uses the humongous documentation string as input for a decryption + decompression function. The fire
and eye
functions interpret the tabs and spaces in the docs string as zeroes and ones for a binary stream,. If it succeeds for a particular integer, it executes the new script. Therefore, to obtain I slightly modified the script to output it to a file instead of executing it directly so we can analyse it better!
The resulting script is actually responsible for running the game, including all the slick ASCII art ;). The main interesting part, however, is the following code that does the verification of the launch codes:
xor = [212, 162, 242, 218, 101, 109, 50, 31, 125, 112, 249, 83, 55, 187, 131, 206]
h = list(wrong())
h = [h[i] ^ xor[i] for i in range(16)]
# ...
target = input()
t.typewriteln("\nPREPARING NUCLEAR STRIKE FOR " + target.upper())
t.typewrite("ENTER LAUNCH CODE: ")
launch_code = input().encode()
# encoding map coordinates
x = list(launch_code.ljust(16, b'\0'))
b = 16 * [None]
# calculate missile trajectory
b[0] = x[2] ^ x[3] ^ x[4] ^ x[8] ^ x[11] ^ x[14]
b[1] = x[0] ^ x[1] ^ x[8] ^ x[11] ^ x[13] ^ x[14]
b[2] = x[0] ^ x[1] ^ x[2] ^ x[4] ^ x[5] ^ x[8] ^ x[9] ^ x[10] ^ x[13] ^ x[14] ^ x[15]
b[3] = x[5] ^ x[6] ^ x[8] ^ x[9] ^ x[10] ^ x[12] ^ x[15]
b[4] = x[1] ^ x[6] ^ x[7] ^ x[8] ^ x[12] ^ x[13] ^ x[14] ^ x[15]
b[5] = x[0] ^ x[4] ^ x[7] ^ x[8] ^ x[9] ^ x[10] ^ x[12] ^ x[13] ^ x[14] ^ x[15]
b[6] = x[1] ^ x[3] ^ x[7] ^ x[9] ^ x[10] ^ x[11] ^ x[12] ^ x[13] ^ x[15]
b[7] = x[0] ^ x[1] ^ x[2] ^ x[3] ^ x[4] ^ x[8] ^ x[10] ^ x[11] ^ x[14]
b[8] = x[1] ^ x[2] ^ x[3] ^ x[5] ^ x[9] ^ x[10] ^ x[11] ^ x[12]
b[9] = x[6] ^ x[7] ^ x[8] ^ x[10] ^ x[11] ^ x[12] ^ x[15]
b[10] = x[0] ^ x[3] ^ x[4] ^ x[7] ^ x[8] ^ x[10] ^ x[11] ^ x[12] ^ x[13] ^ x[14] ^ x[15]
b[11] = x[0] ^ x[2] ^ x[4] ^ x[6] ^ x[13]
b[12] = x[0] ^ x[3] ^ x[6] ^ x[7] ^ x[10] ^ x[12] ^ x[15]
b[13] = x[2] ^ x[3] ^ x[4] ^ x[5] ^ x[6] ^ x[7] ^ x[11] ^ x[12] ^ x[13] ^ x[14]
b[14] = x[1] ^ x[2] ^ x[3] ^ x[5] ^ x[7] ^ x[11] ^ x[13] ^ x[14] ^ x[15]
b[15] = x[1] ^ x[3] ^ x[5] ^ x[9] ^ x[10] ^ x[11] ^ x[13] ^ x[15]
if b == h:
# ...
eye = [219, 232, 81, 150, 126, 54, 116, 129, 3, 61, 204, 119, 252, 122, 3, 209, 196, 15, 148, 173, 206, 246, 242, 200, 201, 167, 2, 102, 59, 122, 81, 6, 24, 23]
flag = fire(eye, launch_code).decode()
t.typewrite(f"CONGRATULATIONS! YOU FOUND THE FLAG:\n\n{flag}\n")
else:
t.typewrite("\nIDENTIFICATION NOT RECOGNIZED BY SYSTEM\n--CONNECTION TERMINATED--\n")
We can see that the input launch code is being fed through a huge system of linear equations with 16 variables, and then compared with a hardcoded launch code that is being decrypted upon runtime. If this equation is satisfied, the original launch code is used as a decryption key of the flag.
To solve this equation, we first need to know the final answer, that is the value of h
. To do that, we need to know what wrong()
does:
def wrong():
trust = windll.kernel32.GetModuleHandleW(None)
computer = string_at(trust, 1024)
dirty, = struct.unpack_from('=I', computer, 60)
_, _, organize, _, _, _, variety, _ = struct.unpack_from('=IHHIIIHH', computer, dirty)
assert variety >= 144
participate, = struct.unpack_from('=I', computer, dirty + 40)
for insurance in range(organize):
name, tropical, inhabitant, reader, chalk, _, _, _, _, _ = struct.unpack_from('=8sIIIIIIHHI', computer, 40 * insurance + dirty + variety + 24)
if inhabitant <= participate < inhabitant + tropical:
break
spare = bytearray(string_at(trust + inhabitant, tropical))
# ...
return hashlib.md5(spare).digest()
Yikes! It seems to be using data from the original executable, by calling the native procedure GetModuleHandleW
and then traversing the PE file structures. Time to analyse and rewrite the code so it works on files stored on disk instead:
def wrong():
original_pe = open("wopr.exe", "rb").read()
mz_header = original_pe[0:1024]
e_lfanew, = struct.unpack_from('=I', mz_header, 60)
# Find number of sections.
_, _, nr_of_sections, _, _, _, sizeof_optional_header, _ = struct.unpack_from('=IHHIIIHH', mz_header, e_lfanew)
assert sizeof_optional_header >= 144
# Find section containing entrypoint.
address_of_entrypoint, = struct.unpack_from('=I', mz_header, e_lfanew + 40)
for insurance in range(nr_of_sections):
name, virtual_size, virtual_address, raw_size, raw_address, _, _, _, _, _ = struct.unpack_from('=8sIIIIIIHHI', mz_header, 40 * insurance + e_lfanew + sizeof_optional_header + 24)
if virtual_address <= address_of_entrypoint < virtual_address + virtual_size:
break
text_section = bytearray(original_pe[raw_address:(raw_address + raw_size)])
# Obtain relocation dir.
reloc_rva, reloc_size = struct.unpack_from('=II', mz_header, e_lfanew + 0xa0)
reloc_dir = original_pe[0x00049A00:(0x00049A00+0x00001800)]
# Apply relocations.
i = 0
while i <= len(reloc_dir) - 8:
rva, size_of_block = struct.unpack_from('=II', reloc_dir, i)
if rva == 0 and size_of_block == 0:
break
slot = reloc_dir[i + 8:i + size_of_block]
for i in range(len(slot) >> 1):
item, = struct.unpack_from('=H', slot, 2 * i)
fabricate = item >> 12
if fabricate != 3: continue
item = item & 4095
ready = rva + item - virtual_address
if 0 <= ready < len(text_section):
v = struct.unpack_from('=I', text_section, ready)[0] - 0x00400000
struct.pack_into('=I', text_section, ready, v)
i += size_of_block
# Hash the text section and return.
return hashlib.md5(text_section).digest()
If we run the script now, we obtain that the result is:
[0xa7, 0xbf, 0xd2, 0x9e, 0x0f, 0x01, 0x6b, 0x53, 0x68, 0x37, 0xb7, 0x60, 0x7c, 0xba, 0xb4, 0xa8]
Onto the system of equations. For this I used Z3 to solve it quickly for me:
h = [0xa7, 0xbf, 0xd2, 0x9e, 0x0f, 0x01, 0x6b, 0x53, 0x68, 0x37, 0xb7, 0x60, 0x7c, 0xba, 0xb4, 0xa8] # Originally a call to wrong()
xor = [212, 162, 242, 218, 101, 109, 50, 31, 125, 112, 249, 83, 55, 187, 131, 206]
h = [h[i] ^ xor[i] for i in range(16)]
# Define variables and constants.
b = [BitVecVal(h[i], 32) for i in range(16)]
x = [BitVec('x' + str(i), 32) for i in range(16)]
s = Solver()
# Add constriants:
s.add(b[0] == x[2] ^ x[3] ^ x[4] ^ x[8] ^ x[11] ^ x[14])
s.add(b[1] == x[0] ^ x[1] ^ x[8] ^ x[11] ^ x[13] ^ x[14])
s.add(b[2] == x[0] ^ x[1] ^ x[2] ^ x[4] ^ x[5] ^ x[8] ^ x[9] ^ x[10] ^ x[13] ^ x[14] ^ x[15])
s.add(b[3] == x[5] ^ x[6] ^ x[8] ^ x[9] ^ x[10] ^ x[12] ^ x[15])
s.add(b[4] == x[1] ^ x[6] ^ x[7] ^ x[8] ^ x[12] ^ x[13] ^ x[14] ^ x[15])
s.add(b[5] == x[0] ^ x[4] ^ x[7] ^ x[8] ^ x[9] ^ x[10] ^ x[12] ^ x[13] ^ x[14] ^ x[15])
s.add(b[6] == x[1] ^ x[3] ^ x[7] ^ x[9] ^ x[10] ^ x[11] ^ x[12] ^ x[13] ^ x[15])
s.add(b[7] == x[0] ^ x[1] ^ x[2] ^ x[3] ^ x[4] ^ x[8] ^ x[10] ^ x[11] ^ x[14])
s.add(b[8] == x[1] ^ x[2] ^ x[3] ^ x[5] ^ x[9] ^ x[10] ^ x[11] ^ x[12])
s.add(b[9] == x[6] ^ x[7] ^ x[8] ^ x[10] ^ x[11] ^ x[12] ^ x[15])
s.add(b[10] == x[0] ^ x[3] ^ x[4] ^ x[7] ^ x[8] ^ x[10] ^ x[11] ^ x[12] ^ x[13] ^ x[14] ^ x[15])
s.add(b[11] == x[0] ^ x[2] ^ x[4] ^ x[6] ^ x[13])
s.add(b[12] == x[0] ^ x[3] ^ x[6] ^ x[7] ^ x[10] ^ x[12] ^ x[15])
s.add(b[13] == x[2] ^ x[3] ^ x[4] ^ x[5] ^ x[6] ^ x[7] ^ x[11] ^ x[12] ^ x[13] ^ x[14])
s.add(b[14] == x[1] ^ x[2] ^ x[3] ^ x[5] ^ x[7] ^ x[11] ^ x[13] ^ x[14] ^ x[15])
s.add(b[15] == x[1] ^ x[3] ^ x[5] ^ x[9] ^ x[10] ^ x[11] ^ x[13] ^ x[15])
# Solve!
try:
s.check()
m = s.model()
code = [0] * 16
for i in range(16):
code[i] = int(str(m[x[i]]))
print(", ".join(map(str, code)))
print("".join(map(chr, code)))
except Z3Exception as e:
print(e)
Z3, what a beast it is, only takes one split second to spit out the correct answer:
53, 67, 48, 71, 55, 84, 89, 50, 76, 87, 73, 50, 89, 88, 77, 66
5C0G7TY2LWI2YXMB
Entering the key lets the game “simulate” the nuclear launch trajectory, and then concludes with:
*** SIMULATION COMPLETED ***
A STRANGE GAME.
THE ONLY WINNING MOVE IS
NOT TO PLAY.
CONGRATULATIONS! YOU FOUND THE FLAG:
L1n34R_4L93bR4_i5_FuN@flare-on.com