Hackvent is a fun CTF, offering challenges that start off quite easy and build to much harder over the course of 24 days, with bonus points for submitting the flag within the first 24 hours for each challenge. This was the first year I made it past day 12, and I was excited to finish all the challenges with all time bonuses! I’ll break the solutions into four parts. The first is the easy challenges, days 1-7, which provided some basic image forensics, some interesting file types, an esoteric programming language, and two hidden flags.

Day 1

Challenge

HV19.01 censored
Categories: FUN
Level: easy
Author: M.

I got this little image, but it looks like the best part got censored on the way. Even the tiny preview icon looks clearer than this! Maybe they missed something that would let you restore the original content?

image-20191202162554454

Solution

I spent some time poking at this image, but wasn’t finding much. Eventually, I ran strings on it, and right at the top was a reference to how the blurring was done:

$ strings day1.jpg
 Exif
Censored by Santa!
GIMP 2.10.12
2019:12:24 00:00:00
Censored by Santa!
JFIF
E[Lj
6squ
F$;3
NH}z
...[snip]...

I figured I’d open it in Gimp and see if there were any tools to reverse the blurring. When I went to open it, the dialog showed the unblurred QRcode:

image-20191202162858506

The major hint in the original question was “Even the tiny preview icon looks clearer than this!”. I took a screenshot of it, and pasted it into KolourPaint, and zoomed in. At 1000%, the QRcode was very clear:

image-20191202162955627

I snapped a picure with my reader on my phone, and got back the flag:

Flag: HV19{just-4-PREview!}

I can also extract the preview/thumbnail using exiftool:

$ exiftool -b -ThumbnailImage day1.jpg > day1_thumb.jpg

Day 2

Challenge

HV19.02 Triangulation
Categories: FUN
Level: easy
Author: drschottky

Today we give away decorations for your Christmas tree. But be careful and do not break it.

It comes with a file, HV19.02-Triangulation.zip. Inside the zip, there’s a single file:

$ unzip -l Triangulation.zip 
Archive:  Triangulation.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
   414084  2019-09-29 21:47   Triangulation.stl
---------                     -------
   414084                     1 file

Triangulation.zip

Solution

.stl

I was not familiar with .stl files, but I learned that it is short for stereolithography, and is native to CAD software for drawing 3D systems. They can come in an ascii or a binary format. When I run file on this file, it reports data, indicating this is the binary format:

$ file Triangulation.stl 
Triangulation.stl: data

It wasn’t necessary for this challenge, but I did find convertSTL, which will translate between the two formats. When I converted to the ascii representation, I can see it is defining surfaces:

$ head Triangulation-ascii.stl 
solid 
  facet normal 3.043928E-01 -8.969041E-01 -3.207929E-01
    outer loop
      vertex 1.150800E+01 2.341580E+00 3.689600E+01
      vertex 1.144000E+01 2.300619E+00 3.694600E+01
      vertex 9.219000E+00 1.794000E+00 3.625500E+01
    endloop
  endfacet
  facet normal 9.450072E-01 6.206107E-02 3.211071E-01
    outer loop

Loading the File

The wikipedia page for stl includes a list of software that can handle STL data. I was immediately drawn to Clara.io, as it was free and online. After creating an account, I’m taken to “Your Documents”, which is empty. I’ll hit the Upload button, and give it my stl file:

image-20191202210509820

When I hit create, I see the 3D object:

image-20191202210624408

Edit

Given the hint not to break it, I suspect the flag is inside the ball. I’ll hit the “Edit Online” button to go to the editor:

image-20191202210907741

Interestingly, right away I can see something going on inside the ball in the wireframe version:

image-20191202210936065

I’ll hit “Select All” in the Tools tab, and then hit the box that is the Face tool. Then I can start selecting areas and deleting them until I get only what’s left in the middle:

What I’m left with is a thing that looks kind of like a QRcode. It is more clear in the “Realistic” view, as opposed to the Wireframe:

image-20191202211446591

Aztec Code

This looks like a QRcode, but it doesn’t have the squares in the three corners. I googled “qr code with box in middle” and looked at image results, and about 5 rows down, I saw this:

image-20191202211614707

An Aztec Code is a 2D barcode format, and zxing.org(a really good QRcode reader online) also handles it. I took a screenshot of the image like about and submitted it, but it didn’t work.

Then I realized that the lighter grey in my picture should be the black, and the darker blue should be the white. I opened the image in Kolour Paint, and used the paint bucket to fill in the light areas with black:

I started to do the blue areas as white, but they didn’t fill easily. I tried submitting this as is to zxing.org, and it decoded, returning the flag:

image-20191202211913740

Flag: HV19{Cr4ck_Th3_B411!}

Day 3

Challenge

HV19.03 Hodor, Hodor, Hodor
Categories: FUN
PROGRAMMING
Level: easy
Author: otauk
trolli101

image-20191202180305856

$HODOR: hhodor. Hodor. Hodor!?  = `hodor?!? HODOR!? hodor? Hodor oHodor. hodor? , HODOR!?! ohodor!?  dhodor? hodor odhodor? d HodorHodor  Hodor!? HODOR HODOR? hodor! hodor!? HODOR hodor! hodor? ! 

hodor?!? Hodor  Hodor Hodor? Hodor  HODOR  rhodor? HODOR Hodor!?  h4Hodor?!? Hodor?!? 0r hhodor?  Hodor!? oHodor?! hodor? Hodor  Hodor! HODOR Hodor hodor? 64 HODOR Hodor  HODOR!? hodor? Hodor!? Hodor!? .

HODOR?!? hodor- hodorHoOodoOor Hodor?!? OHoOodoOorHooodorrHODOR hodor. oHODOR... Dhodor- hodor?! HooodorrHODOR HoOodoOorHooodorrHODOR RoHODOR... HODOR!?! 1hodor?! HODOR... DHODOR- HODOR!?! HooodorrHODOR Hodor- HODORHoOodoOor HODOR!?! HODOR... DHODORHoOodoOor hodor. Hodor! HoOodoOorHodor HODORHoOodoOor 0Hooodorrhodor HoOodoOorHooodorrHODOR 0=`;
hodor.hod(hhodor. Hodor. Hodor!? );

Solution

A bit of Googling leads to hodor lang. Another silly language. It is basically JavaScript with certain keywords and letters replaced with variations on Hodor according to this legend.

For this challenge, it’s as simple as installing the node package:

# npm install -g hodor-lang
npm WARN npm npm does not support Node.js v10.17.0
npm WARN npm You should probably upgrade to a newer version of node as we
npm WARN npm can't make any promises that npm will work with this version.
npm WARN npm Supported releases of Node.js are the latest release of 4, 6, 7, 8, 9.
npm WARN npm You can find the latest version at https://nodejs.org/
/usr/local/bin/hodor -> /usr/local/lib/node_modules/hodor-lang/bin/hodor
/usr/local/bin/js2hd -> /usr/local/lib/node_modules/hodor-lang/bin/js2hd
+ hodor-lang@1.0.2
added 54 packages from 41 contributors in 2.636s

Now I can just run the interpreter on the code:

# hodor hodor.hd 
HODOR: \-> hodor.hd
Awesome, you decoded Hodors language! 

As sis a real h4xx0r he loves base64 as well.

SFYxOXtoMDFkLXRoMy1kMDByLTQyMDQtbGQ0WX0=

I’ll decode the base64 to get the flag:

$ echo SFYxOXtoMDFkLXRoMy1kMDByLTQyMDQtbGQ0WX0= | base64 -d
HV19{h01d-th3-d00r-4204-ld4Y}

Flag: HV19{h01d-th3-d00r-4204-ld4Y}

Day 4

Challenge

HV19.04 password policy circumvention
Categories: FUN
Level: easy
Author: DanMcFly

Santa released a new password policy (more than 40 characters, upper, lower, digit, special).

The elves can’t remember such long passwords, so they found a way to continue to use their old (bad) password:

merry christmas geeks

There’s also a zip file. It contains an .ahk file:

$ unzip -l HV19-PPC.zip 
Archive:  HV19-PPC.zip
  Length      Date    Time    Name
---------  ---------- -----   ----
      630  2019-12-03 22:34   HV19-PPC.ahk
---------                     -------
      630                     1 file

HV19-PPC.ahk

Solution

Background

The .ahk file is unicode text, with Windows line endings:

$ file HV19-PPC.ahk 
HV19-PPC.ahk: UTF-8 Unicode (with BOM) text, with CRLF line terminators

AutoHotKey is a scripting language that allows to you map various key strokes into complex operations and substitutions. Here’s this script:

::merry::
FormatTime , x,, MM MMMM yyyy
SendInput, %x%{left 4}{del 2}+{right 2}^c{end}{home}^v{home}V{right 2}{ASC 00123}
return

::christmas::
SendInput HV19-pass-w0rd
return

:*?:is::
Send - {del}{right}4h

:*?:as::
Send {left 8}rmmbr{end}{ASC 00125}{home}{right 10}
return

:*?:ee::
Send {left}{left}{del}{del}{left},{right}e{right}3{right 2}e{right}{del 5}{home}H{right 4}
return

:*?:ks::
Send {del}R3{right}e{right 2}3{right 2} {right 8} {right} the{right 3}t{right} 0f{right 3}{del}c{end}{left 5}{del 4}
return

::xmas::
SendInput, -Hack-Vent-Xmas
return

::geeks::
Send -1337-hack
return

Manual Solution

That’s eight different macros that will run when text that matches them is entered.

I’m given the text the elves want to use: merry christmas geeks.

The key here is to start typing, and watch for a match.

I’ll enter merry , and I have the first match. It saves the current date/time in the format 12 December 2019 as x. Then it sends that, followed by left 4 times, and two deletes, leaving (with the cursor before the 1 in 19):

12 December 19

Next +{right 2} will select the next two characters (19), the ^c will sent ctrl-c or copy, {end} will move to the end of the line, but then {home} will move to the front of the line, where ^v or ctrl-v will paste:

1912 December 19

Finally {home} moves to the start of the line, then there’s a V, then right two spaces, and {ASC 00123}, which is the {:

V19{12 December 19

So after entering merry (with a space on the end), I’ve got the above with the cursor after the {. I’ll continue entering christmas. When I get to chri, the input looks like:

V19{chri12 December 19

Next when I hit s, the pattern for is matches, so the is goes away, and it’s replaced by - {del}{right}4h. That becomes (cursor after the h):

V19{chr- 24h December 19

But I’ll notice the is rule doesn’t have a return…so it keeps going into the next rule for as: Send {left 8}rmmbr{end}{ASC 00125}{home}{right 10}. That gives (with the cursor between the ch):

V19{rmmbrchr- 24h December 19}

I’ll enter tm:

V19{rmmbrctmhr- 24h December 19}

Now I enter as, triggering the same rule as above from the cursor between the mh:

V19{rmmbrrmmbrctmhr- 24h December 19}}

The cursor is now ten characters in, between rr and mm. I start with space then g:

V19{rmmbrr gmmbrctmhr- 24h December 19}}

Next comes ee, which triggers a rule, Send {left}{left}{del}{del}{left},{right}e{right}3{right 2}e{right}{del 5}{home}H{right 4}, leaving the cursor just inside the {:

HV19{rmmbr,rem3mebr- 24h December 19}}

Finally, I enter ks, which triggers {del}R3{right}e{right 2}3{right 2} {right 8} {right} the{right 3}t{right} 0f{right 3}{del}c{end}{left 5}{del 4}:

HV19{R3memb3r, rem3mber - the 24th 0f December}

Flag: HV19{R3memb3r, rem3mber - the 24th 0f December}

Shortcut

Alternatively to understanding how each of the macros work, I could just install AHK on a VM and run this script. After downloading and running the installer, I’ll create a new file called HV19-4.ahk, and paste in the script. Now I’ll double click on the script to run it, and then go into Notepad and start typing:

It is important not to go too fast, or I could be typing while AHK is and mess up the cursor position.

Day 5

Challenge

HV19.05
Categories: FUN
Level: easy
Author: inik

To handle the huge load of parcels Santa introduced this year a parcel tracking system. He didn’t like the black and white barcode, so he invented a more solemn barcode. Unfortunately the common barcode readers can’t read it anymore, it only works with the pimped models santa owns. Can you read the barcode

Solution

A standard barcode is coded using if a stripe if white or black, so the size and spacing of the stripes holds the data. If I upload this code to zxing.org, it will read it as a CODE_128 format barcode, with the information “Not the solution”:

image-20191205071526142

That tells me that the information is not stored in the widths of the bars, since a different message is encoded there. The only other place it could be stored then is the colors themselves (something not typically involved in standard barcodes).

I wrote a small Python program to take a look at the pixels:

#!/usr/bin/env python3

import sys
from PIL import Image


im = Image.open('157de28f-2190-4c6d-a1dc-02ce9e385b5c.png')

width, _ = im.size

for i in range(width):
    pix = im.getpixel((i, 5))
    print([chr(x) if x < 128 else x for x in pix])

For each pixel, I’ll print the three RGB values, showing it as a character if it’s less than 128, or the int value otherwise. When I run this, the output looks like:

[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
['s', 'P', 'X']
['s', 'P', 'X']
['s', 'P', 'X']
['s', 'P', 'X']
['s', 'P', 'X']
['s', 'P', 'X']
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
['t', 'Y', '8']
['t', 'Y', '8']
['t', 'Y', '8']
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
[255, 255, 255]
['l', 'P', 'Y']
['l', 'P', 'Y']
['l', 'P', 'Y']
['l', 'P', 'Y']
['l', 'P', 'Y']
['l', 'P', 'Y']
['l', 'P', 'Y']
['l', 'P', 'Y']
['l', 'P', 'Y']
[255, 255, 255]

Since I already decided that the width of a given stripe doesn’t matter, I’m going to update to ignore repeated lines by saving the previous line and not printing if it matches the current:

#!/usr/bin/env python3

import sys
from PIL import Image


im = Image.open('157de28f-2190-4c6d-a1dc-02ce9e385b5c.png')

width, _ = im.size

prev = ''
for i in range(width):
    pix = im.getpixel((i, 5))
    if not pix == prev:
        print([chr(x) if x < 128 else x for x in pix])
    prev = pix

Now I get:

[255, 255, 255]
['s', 'P', 'X']
[255, 255, 255]
['t', 'Y', '8']
[255, 255, 255]
['l', 'P', 'Y']
[255, 255, 255]
['m', 'E', 'I']
[255, 255, 255]
['r', '1', 'O']
[255, 255, 255]
['y', '3', 'F']
[255, 255, 255]
['s', 'P', '0']
[255, 255, 255]
['e', 'Q', 'Z']
[255, 255, 255]
['g', '8', 'P']
[255, 255, 255]
['z', '9', '4']
[255, 255, 255]
['u', 'L', 'S']
[255, 255, 255]
['h', 'T', '8']
[255, 255, 255]
['e', 'G', 'H']
[255, 255, 255]
['z', '0', 'V']
[255, 255, 255]
['a', 'X', '1']
[255, 255, 255]
...[snip]...

It seems like the rows of 255s (white) are likely just spacing. I’ll remove those. Everything else is coming out ASCII, which is nice. I’ll try just grouping all those characters into a stream and seeing if it means anything. I’ll change the output for characters greater than 128 to an empty string, and joing the results into a string, appending it to output, and print it at the end:

#!/usr/bin/env python3

import sys
from PIL import Image


im = Image.open('157de28f-2190-4c6d-a1dc-02ce9e385b5c.png')

width, _ = im.size

prev = ''
output = ''
for i in range(width):
    pix = im.getpixel((i, 5))
    if not pix == prev:
        output += ''.join(([chr(x) if x < 128 else '' for x in pix]))
    prev = pix

print(output)

When I run this, I get a nice ASCII string:

$ python3 solve_5.py 
sPXtY8lPYmEIr1Oy3FsP0eQZg8Pz94uLShT8eGHz0VaX1g09lO{tODlJ1gJfxIfzIigUcjSuiHliQtv6_v0tsMosI_aQgiI3e4twS_b9atE_uGSh4Pa8TlN_qVRcV3lPaf6dq5erGrcO}wXSfL1q00vV9eW0nJOgWMx2Ze3Ek20o3EaS3lRNtUFy8PvB6eBE

This isn’t the flag, but I see both { and } in there. I decided to look at each of the RBG seperately. I’ll set output to a list of three strings, and for each pixel, append each character to a different string:

#!/usr/bin/env python3

import sys
from PIL import Image


im = Image.open("157de28f-2190-4c6d-a1dc-02ce9e385b5c.png")

width, _ = im.size

prev = ""
output = ["", "", ""]
for i in range(width):
    pix = im.getpixel((i, 5))
    if not pix == prev:
        output = [o + chr(p) if p < 128 else o for p, o in zip(pix, output)]
    prev = pix

print("\n".join(output))

When I run this, I see the flag in the output:

$ python3 solve_5.py 
stlmrysegzuhezagltlgxzgjiivvssaiewbtuhalqclfqrcwfqvengxekoaltyve
PYPE13PQ89LTG0X0OOJJIIUSHQ60MIQI4S9EG48NVVP65GOXL0VWJW2323SRU8BB
X8YIOF0ZP4S8HV19{D1fficult_to_g3t_a_SPT_R3ader}S1090OMZE0E3NFP6E

It’s not clear to me what the strings before or after it are, but I can definitely extract a flag there.

Flag: HV19{D1fficult_to_g3t_a_SPT_R3ader}

Day 6

Challenge

HV19.06
Categories: CRYPTO
FUN
Level: easy
Author: T.B.

I’m given a text, with weird individual characters in italics:

image-20191205202239031

Solution

There’s a couple hints in there. The first is that the piece is talking about Francie Bacon, and that it’s a crypto challenge. If I Google Francis Bacon cryptography, the first hit is interesting:

image-20191206172111540

The Bacon Cipher is typically a series of a and b in groups of five that make letters. But there’s not reason it can’t be anything else with two options. This leads to the first thing I noticed in the challenge. The weird italics in certain spots.

Because I have to log in to hackvent, and I didn’t want to bother with that in my script, I saved the page as page.html. Then I used re to find the block of text I was interested in. The italics stuff stops towards the end, but I grabbed that entire section. At first I tried to apply the Bacon cipher to all characters, and it didn’t work. Then I tried just doing it to A-Z, and it worked!

#!/usr/bin/env python3

import re
import string


with open("page.html", "r") as f:
    page = f.read().replace("\n", " ")

in_text = re.findall(r'<p>(<em>F.*.)</p><pre class="jss1269">', page)[0]

bacon_in = ""
ital = False

while in_text:
    if in_text.startswith("<em>"):
        in_text = in_text[4:]
        ital = True
    elif in_text.startswith("</em>"):
        in_text = in_text[5:]
        ital = False
    elif in_text[0] not in string.ascii_letters:
        in_text = in_text[1:]
    elif not ital:
        bacon_in += "a"
        in_text = in_text[1:]
    else:
        bacon_in += "b"
        in_text = in_text[1:]

lookup = {
    "aaaaa": "A",
    "aaaab": "B",
    "aaaba": "C",
    "aaabb": "D",
    "aabaa": "E",
    "aabab": "F",
    "aabba": "G",
    "aabbb": "H",
    "abaaa": "I",
    "abaab": "J",
    "ababa": "K",
    "ababb": "L",
    "abbaa": "M",
    "abbab": "N",
    "abbba": "O",
    "abbbb": "P",
    "baaaa": "Q",
    "baaab": "R",
    "baaba": "S",
    "baabb": "T",
    "babaa": "U",
    "babab": "V",
    "babba": "W",
    "babbb": "X",
    "bbaaa": "Y",
    "bbaab": "Z",
}

bacon_in = bacon_in[: (len(bacon_in) // 5) * 5]

result = ""
for i in range(0, len(bacon_in), 5):
    result += lookup[bacon_in[i : i + 5]]

print(result)

When I run this, I get a message:

$ ./solve_day6.py 
SANTALIKESHISBACONBUTALSOTHISBACONTHEPASSWORDISHVXBACONCIPHERISSIMPLEBUTCOOLXREPLACEXWITHBRACKETSANDUSEUPPERCASEFORALLCHARACTERAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

I’ll add spaces:

SANTA LIKES HIS BACON BUT ALSO THIS BACON THE PASSWORD IS HVXBACONCIPHERISSIMPLEBUTCOOLX REPLACE X WITH BRACKETS AND USE UPPERCASE FOR ALL CHARACTER

This tells me to take the flag and change the Xs to {}, giving:

Flag: HV19{BACONCIPHERISSIMPLEBUTCOOL}

Hidden 1

Challenge

HV19.H1 Hidden One
Categories: FUN
Level: novice
Author: hidden

Sometimes, there are hidden flags. Got your first?

Solution

This challenge showed up on the list at the same time as Day 6, so it seems related. The data at the bottom of the page is particularly interesting:

image-20191208134304911

If I click on the little clipboard at the top right:

image-20191208134330818

When I paste that, I can see there’s extra whitespace on the end of each line:

$ xxd textarea.txt 
00000000: 426f 726e 3a20 4a61 6e75 6172 7920 3232  Born: January 22
00000010: 0920 2020 2020 0920 0920 2020 0920 2020  .     . .   .   
00000020: 0920 0920 2020 2020 2020 0920 2020 2020  . .       .     
00000030: 0920 2009 2020 0a44 6965 643a 2041 7072  .  .  .Died: Apr
00000040: 696c 2039 2020 2009 2020 0920 0920 2020  il 9   .  . .   
00000050: 2009 2020 0920 2020 2020 2009 2020 2009   .  .      .   .
00000060: 0920 2009 2020 0a4d 6f74 6865 723a 204c  .  .  .Mother: L
00000070: 6164 7920 416e 6e65 2020 2009 0920 0920  ady Anne   .. . 
00000080: 2020 0920 2020 0920 2020 2020 2009 2020    .   .      .  
00000090: 0920 2020 2020 2009 2020 0a46 6174 6865  .      .  .Fathe
000000a0: 723a 2053 6972 204e 6963 686f 6c61 7309  r: Sir Nicholas.
000000b0: 2009 2020 2020 2020 0909 2020 2020 0920   .      ..    . 
000000c0: 2020 2009 2020 0920 2009 2020 2020 2020     .  .  .      
000000d0: 0920 2020 2020 200a 5365 6372 6574 733a  .      .Secrets:
000000e0: 2075 6e6b 6e6f 776e 2020 2020 2020 0920   unknown      . 
000000f0: 0920 2009 2009 2020 2020 0920 2020 2009  .  . .    .    .
00000100: 2020 2009 2020 2020 2020 2009 2020 0a       .       .  .

At first my mind went to the wopr challenge from Flare-On 2019. There, tabs and spaces were used as 1s and 0s to make binary. But that didn’t work here. I also looked at the Whitespace esoteric language, pasting the text into interpreters like this and this. But it didn’t work. Eventually, I found stegsnow. I installed it (apt install stegsnow) and gave it a run:

$ stegsnow textarea.txt 
Warning: residual of 5 bits not output
iYB     a6i%&P

That didn’t work. But then I tried with the -C options for compression. If something is hidden with -C, it must be decoded with -C as well. This returned the flag:

$ stegsnow -C textarea.txt 
HV19{1stHiddenFound}

Flag: HV19{1stHiddenFound}

Day 7

Challenge

HV19.07 Santa Rider
Categories: FUN
Level: easy
Author: inik

Santa is prototyping a new gadget for his sledge. Unfortunately it still has some glitches, but look for yourself.

I’m given a zip file, which includes the video, 3DULK2N7DcpXFg8qGo9Z9qEQqvaEDpUCBB1v.mp4.

3DULK2N7DcpXFg8qGo9Z9qEQqvaEDpUCBB1v.mp4

Solution

I downloaded the embedded .mp4 and got the one in the archive - they are the same:

$ md5sum *.mp4
db154b453cc4ee3f8cb375a3d0b8017c  13e4f1a0-bb71-44ec-be54-3f5f23991033.mp4
db154b453cc4ee3f8cb375a3d0b8017c  3DULK2N7DcpXFg8qGo9Z9qEQqvaEDpUCBB1v.mp4

I started to look at the lights. They start going back and forth 6 times, and then there’s a random seeming pattern in the middle. I used ffmpeg to pull out the frames:

$ ffmpeg -i 3DULK2N7DcpXFg8qGo9Z9qEQqvaEDpUCBB1v.mp4 3DULK2N7DcpXFg8qGo9Z9qEQqvaEDpUCBB1v%04d.jpg -hide_banner
Input #0, mov,mp4,m4a,3gp,3g2,mj2, from '3DULK2N7DcpXFg8qGo9Z9qEQqvaEDpUCBB1v.mp4':
  Metadata:
    major_brand     : isom
    minor_version   : 512
    compatible_brands: isomiso2avc1mp41
    encoder         : Lavf58.20.100
  Duration: 00:00:22.59, start: 0.000000, bitrate: 925 kb/s
    Stream #0:0(und): Video: h264 (High) (avc1 / 0x31637661), yuv420p(tv, bt709/unknown/bt709), 1280x720 [SAR 1:1 DAR 16:9], 914 kb/s, 30 fps, 30 tbr, 15360 tbn, 60 tbc (default)
    Metadata:
      handler_name    : VideoHandler
    Stream #0:1(und): Audio: aac (LC) (mp4a / 0x6134706D), 48000 Hz, stereo, fltp, 2 kb/s (default)
    Metadata:
      handler_name    : SoundHandler
Stream mapping:
  Stream #0:0 -> #0:0 (h264 (native) -> mjpeg (native))
Press [q] to stop, [?] for help
[swscaler @ 0x55e7c0050200] deprecated pixel format used, make sure you did set range correctly
Output #0, image2, to '3DULK2N7DcpXFg8qGo9Z9qEQqvaEDpUCBB1v%04d.jpg':
  Metadata:
    major_brand     : isom
    minor_version   : 512
    compatible_brands: isomiso2avc1mp41
    encoder         : Lavf58.29.100
    Stream #0:0(und): Video: mjpeg, yuvj420p(pc), 1280x720 [SAR 1:1 DAR 16:9], q=2-31, 200 kb/s, 30 fps, 30 tbn, 30 tbc (default)
    Metadata:
      handler_name    : VideoHandler
      encoder         : Lavc58.54.100 mjpeg
    Side data:
      cpb: bitrate max/min/avg: 0/0/200000 buffer size: 0 vbv_delay: -1
frame=  677 fps=148 q=24.8 Lsize=N/A time=00:00:22.56 bitrate=N/A speed=4.93x    
video:13280kB audio:0kB subtitle:0kB other streams:0kB global headers:0kB muxing overhead: unknown

I’m left with 677 images. I’ll scan through them to find where the random pattern starts.

At 273, I come across this:

Since there are eight lights, my first thought it bits. I could look at different bit orders, but if I assume the lowest bits to the right, this is 0x48. Looking at the ASCII table, that’s H. Moving ahead until the lights change, there’s a few that show the next pattern:

That’s 0x56, or V.

Here’s the rest of the frames:

FrameImageHexASCII
2730x48H
2760x56V
2790x311
2820x399
2850x7B{
2880x311
2920x6Dm
2950x5F_
2980x61a
3020x6Cl
3050x73s
3080x300
3110x5F_
3140x77w
3170x300
3200x72r
3230x6Bk
3260x311
3290x6En
3320x67g
3350x5F_
3380x300
3410x6En
3440x5F_
3470x61a
3500x5F_
3530x72r
3560x333
3590x6Dm
3620x300
3650x74t
3680x333
3710x5F_
3740x63c
3770x300
3800x6En
3830x74t
3860x72r
3890x300
3920x6Cl
3950x7D}

Putting that all together, I get the flag:

Flag: HV19{1m_als0_w0rk1ng_0n_a_r3m0t3_c0ntr0l}

Hidden 2

Challenge

HV19.H2 Hidden Two
Categories: FUN
Level: novice
Author: inik

Again a hidden flag.

Solution

This popped up along side day 7. I noticed when working on day 7 that I could both download the mp4 from the page and as a resource. When I unzipped the downloaded file, I got a mp4 that was identical to the one I got from the page.

I’ll also notice that every time I download an object from this framework, it comes named as a GUID (ie, 13e4f1a0-bb71-44ec-be54-3f5f23991033.mp4). In previous challenges, I either got the file directly named as a GUID and I had to rename it to what it should have been, or it was in a zip with an expressive name.

So what’s the meaning of 3DULK2N7DcpXFg8qGo9Z9qEQqvaEDpUCBB1v.mp4. Turns out that name is base58 encoded. If I drop it in here, I get a flag:

Flag: HV19{Dont_confuse_0_and_O}