dJulkalendern 2022 Write-up

Posted Jan 4, 2023

By Petar Kostić 58 min read

Introduction

A Swedish friend and student at KTH university introduced me to a CTF-like puzzle in an advent calendar format called dJulkalendern. A window (i.e. puzzle) will be released every weekday at 12:15 PM (Europe/Stockholm, UTC+01:00)

The event is arranged by (I presume students and/or faculty at) KTH University

Luckily, you don’t really have to understand Swedish, as the website and puzzles are written in English. The flags are also single English words, without any pre- or postfix.

There also is a Discord server available. Other than some general text and announcement channels, the organizers provided a neat way to grant access to a separate channel, corresponding to each puzzle, if you’ve successfully solved that challenge. Access is granted through a Discord bot that asks you the “single-english-word-flag” through a DM.

The content in each Discord channel does force you to use e.g. Google translate to translate some Swedish to your native language if you’re interested in what people are saying.

As I work full-time, additionally participating in Advent of Code 2022, and these release at the start of the afternoon, I don’t do these exactly when the puzzle releases. Therefore, I won’t be ranked high on the scoreboard.

Day -1: Ready?

This window, Window -1, is a practice window of sorts. It does not count toward the final score. The solution is the last word in the notes of the 2003 case, which you can read on the Lore Page.

Navigating to https://djul.datasektionen.se/lore reveals the flag to be further.

Day 1: Mrs. Claus, more like Missing Claus

We are given a password prompt, along with a hint.

The root of all 3vil is: 1404928 1481544 1367631 1191016 1030301 970299 1560896 1157625 1367631 1331000

The hint is the number 3 in the “l33t speak” version of evil. It shows that a cube root must be performed on these numbers. The fastest way to do this was to open up a web browser’s debugger console and write some JavaScript. The resulting numbers are ASCII numbers. JavaScript’s String.fromCharCode clears that up. The .join("") was added for this write up to make it more neat.

  
"1404928 1481544 1367631 1191016 1030301 970299 1560896 1157625 1367631 1331000"
.split(' ')
.map((x) => String.fromCharCode(Math.cbrt(x))).join("")

This gives us the flag projection

Day 2: Writing Safe Code

This day’s puzzle instructs you how to use netcat to connect to a text-based adventure. This is also known as a Multi User Dungeon or “MUD”.

You traverse the rooms with e.g. go north, and find out what there is to interact with in a room by using the whatishere command. As you navigate, you collect small clues by e.g. using a stick to shove some things out of reach. The small clues are digits, arranged with underscores in order (e.g. 0 _ _ _, _ 3 _ _, _ _ 2 _)

Later, you encounter a safe asking for a pin number.

The last digit was hidden in plain sight, and I didn’t get the hint towards it. However, if you have 3 out of 4 digits, you can iterate the safe with a brute-force. The delay of ~20 seconds luckily wasn’t increasing on a failed attempt. I luckily wasn’t going for time either :)

I thought to myself “I’ll start at 1, worst case it’s a 0 and it’ll take ~200 seconds if I copy the command to my clipboard while waiting”. Of course, the correct digit was the number 9…

After opening the safe, the flag was revealed to be beaned.

A fun element was that you were allowed to set a username and see who else of the participants was in the room with you through the whoishere command. You were also able to say something in a room to communicate with them. Although I didn’t encounter someone else using that feature. Probably all too busy finding the safe pin.

Later in the Discord channel you hear the hint you missed. Something with 9 chili peppers on a tree, where another place in the game refers to the last digit to be the number of fruits on the hottest plant.

Day 5: Two Factor Authentication

This day’s puzzle gives you a text file, containing a single string of 40501 binary digits.

40501 is not divisible by 8 without leaving remainder values. It therefore cannot indicate something in the typical byte length of 8. This is where CyberChef should come in for the rescue. Can it find something that’s not gibberish for other byte lengths? Alas, no clear result. Time to take a step back.

The long character length gravitated me towards a bitmap grayscale image being the solution here. For the possible color values of 0 - 255, 0 maps to white, and 1 maps to black (or vice-versa).

To see if this was feasible, we’d need the dimensions of a potential image. We check the factors of the length 40501 through Wolfram Alpha.

Exactly two numbers that peak our interest: 101 and 401! We’re on a good track, it doesn’t seem that this is just a coincidence.

Before going through the effort of generating an image, maybe it’s readable in text form? We assume the text is written horizontally (401 x 101). This means that inserting a carriage return after every 401 characters should maybe show something. Hitting Ctrl + - a bunch in Visual Studio Code to zoom out, we get the following output.

You can, of course, also write a small program that makes something similar into an actual image, or use online tools like decode.fr/binary-image

Day 6: Obstacles

For this puzzle, we are given a video where a duck navigates some obstacles.

The clue is in this puzzle’s title “Obstacles”. If we record the height each set of 3 obstacles have, we get the following numbers.

011 120 010 210 111 012 112 202 001 200 221

We employ the browser’s built-in JavaScript console again for this puzzle. The parseInt() function parses a string argument and returns an integer of the specified radix (the base in mathematical numeral systems). In this case, base 3 makes the most sense of course.

  
[4, 15, 3, 21, 13, 5, 14, 20, 1, 18, 25]

We get a set of numbers that seem to be alphabet coded (1 = a, 2 = b, 3 = c etc.). Although String.fromCharCode() (and similar functions in most languages) don’t have ASCII tables where the A character starts at 1, you can of course add the number from where it does: 64 for uppercase or 96 for lowercase.

The final function becomes

  
"011 120 010 210 111 012 112 202 001 200 221"
.split(' ')
.map((x) => String.fromCharCode(parseInt(x, 3) + 96)).join("")

Which returns the flag documentary.

Day 7: Database

This puzzle includes a file called wishlist.sql (download) along with the hint

check digit 13

This .sql file first contains instructions to create a table called “Books”. Each book has an incrementing id, name, and isbn number.

  
CREATE TABLE Book (
  id SERIAL PRIMARY KEY,
  name TEXT NOT NULL,
  isbn BIGINT NOT NULL
);

Afterwards, there are 250 INSERT INTO commands that fills this table with books.

After researching how ISBN-13 numbers work, I quickly figured I’d probably have to check whether they’re all valid and go from there.

Not being too eager to set up a new database on my local machine, I employed sqliteonline.com to get started much more quickly. I copied and pasted all commands from the file and executed them there in the small sandbox it gives you after connecting.

I then employed the website’s export functionality to get a list of all ISBNs easily. I left the tab open to be able to go back anytime. I knew that, with some creative bulk editing functions in Visual Studio Code, I could get this in an easy format such that the numbers could be pasted into a programming language like C#.

Writing the function itself according to the specification seemed like a waste of time. I instead employed Rosetta Code’s ISBN13 check digit page to execute the following C# code locally.

  
var isbns = new List<string>()
{
    "9789432349908", "9780147209354", "9786991874419", "9788913389785", "9789527982560", "9780053807857",
    // Truncated for this write-up.
};

// Source: https://rosettacode.org/wiki/ISBN13_check_digit#C#
static bool CheckISBN13(string code)
{
    code = code.Replace("-", "").Replace(" ", "");
    if (code.Length != 13)
        return false;

    int sum = 0;
    foreach ((int index, char digit) in code.Select((digit, index) => (index, digit)))
    {
        if (char.IsDigit(digit))
            sum += (digit - '0') * (index % 2 == 0 ? 1 : 3);
        else
            return false;
    }
    return sum % 10 == 0;
}

foreach (string isbn in isbns.Where(CheckISBN13))
    Console.WriteLine($"{isbn} is Valid");

Note: Written in .NET 7, thus static void Main(string[] args) { /* ... */ } is not required.

To my surprise only one value was valid: 9789176216132.

After checking the corresponding book name with the following query

  
SELECT b.name 
  FROM book b 
 WHERE b.isbn = 9789176216132

We got the flag subsequent.

Day 8: Only For Developers

The puzzle for the previous day showed the following Note

Notes: For tomorrows window, you will need to have discord available. You do not necessarily need to join our server, or any server where you can talk to other people. (But you are of course very welcome to join ours :))

Thus this day all participants were given a Discord invite link. After joining the server, a Discord bot called “Locker Lock” would prompt 3 questions in succession. The answers were subsequently checked after an answer was given to all 3 questions. For example, in order to know if you had question 2 correct, you must’ve had question 1 correct too.

Question 1: What is the street name of the agency?

Cooking Lane, found in the lore section of the website.

Question 2: What is this lock’s product number?

The lock is this very Discord Bot. By copying the Discord ID by right-clicking the bot on the right pane, we get the number by using the “Copy ID” function at the very bottom in the OS’s clipboard.

Question 3: What is tomtefar’s first name?

This one threw me off a short while. A quick google search reveals that “Tomtefar” refers to Father Christmas (Santa Claus) in modern time Swedish. However, “Santa” and other first names given to Santa Claus were not the right answer. However, I then noticed that a Dev called “Tomtefar” was also in this Discord channel. Going to his “About me” link revealed a website with a browser-based ray tracer.

Following the “Source code” link at the top of the page, we are taken to a GitHub repository

Although the handle shown for the most recent commit already contained the first name

we follow to the repository’s only contributor’s profile: Mathias Magnusson

The first name thus being Mathias.

After having all three questions correct, the bot says the following:

The locker door opens revealing a dark room full of dusty files. In the very centre of the room there’s a box with “sicily” written in large letters on the front.

and the flag is thus sicily

Day 9: In the lair

This day’s puzzle was another MUD that represented a FAT file system as a 5x5 grid of clusters. By traversing the grid using e.g. go south, one could use the command showcontent to see the contents of a file iff the current cluster is a file. There’s one more command in this MUD, getfat <index>, which gives us the index of the cell following the one provided.

The puzzle participants starts in cluster 2

01 02 03 04
06 07 08 09
11 12 13 14
16 17 18 19
21 22 23 24

While traversing the grid we encounter a README file in cluster 13 containing the following content

HELLO, I AM A VERY EVIL GOOSE.
I HAVE STORED MY CRIMES AND TARGETS IN THE CRIMES FOLDER.
If I have to sound more sophisticated, I store a list of synonyms in the synonyms folder,
which I can liberatingly use when inscribing sentences. 

Cluster 7 lists the files of the CRIMES folder.

Cluster	Crime Index	Contents
Cluster 11	CRIME A	https://www.youtube.com/watch?v=dQw4w9WgXcQ (Classic RickRoll URL)
Cluster 8	CRIME B	https://www.youtube.com/watch?v=iik25wqIuFo (Alternative RickRoll URL)
Cluster 14	CRIME C	https://www.youtube.com/watch?v=8ybW48rKBME (Alternative RickRoll URL)
Cluster 22	CRIME D	Hex code
Cluster 19	CRIME E	https://www.youtube.com/watch?v=xvFZjo5PgG0 (Alternative RickRoll URL)

showcontent in cluster 22 reveals the following content

FF D8 FF E0 00 10 4A 46 49 46 00 01 01 01 00 60 00 60 00 00 FF E1 00 68 45 78 69 66 00 00 4D 4D 00 2A 00 00 00 08 00 04 01 1A 00 05 00 00 00 01 00 00 00 3E 01 1B 00 05 00 00 00 01 00 00 00 46 01 28 00 03 00 00 00 01 00 02 00 00 01 31 00 02 00 00 00 11 00 00 00 4E 00 00 00 00 00 01 76 F6 00 00 03 E8 00 01 76 F6 00 00 03 E8 70 61 69 6E 74 2E 6E 65 74 20 34 2E 32 2E 31 34 00 00 FF DB 00 43 00 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF DB 00 43 01 FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF FF C0 00 11 08 01 00 01 00 03 01 22 00 02 11 01 03 11 01 FF C4 00 1F 00 00 01 05 01 01 01 01 01 01 00 00 00 00 00 00 00 00 01 02 03 04 05 06 07 08 09 0A 0B FF C4 00 B5 10 00 02 01 03 03 02 04 03 05 05 04 04 00 00 01 7D 01 02 03 00 04 11 05 12 21 31 41 06 13 51 61 07 22 71 14 32 81 91 A1 08 23 42 B1 C1 15 52 D1 F0 24 33 62 72 82 09 0A 16 17 18 19 1A 25 26 27 28 29 2A 34 35 36 37 38 39 3A 43 44 45 46 47 48 49 4A 53 54 55 56 57 58 59 5A 63 64 65 66 67 68 69 6A 73 74 75 76 77 78 79 7A 83 84 85 86 87 88 89 8A 92 93 94 95 96 97 98 99 9A A2 A3 A4 A5 A6 A7 A8 A9 AA B2 B3 B4 B5 B6 B7 B8 B9 BA C2 C3 C4 C5 C6 C7 C8 C9 CA D2 D3 D4 D5 D6 D7 D8 D9 DA E1 E2 E3 E4 E5 E6 E7 E8 E9 EA F1 F2 F3 F4 F5 F6 F7 F8 F9 FA FF C4 00 1F 01 00 03 01 01 01 01 01 01

When using CyberChef’s From Hex recipe on the hex code in Cluster 22, we encounter the following

ÿØÿà..JFIF.....`.`..ÿá.hExif..MM.*.................>...........F.(...........1.........N......vö...è..vö...èpaint.net 4.2.14..ÿÛ.C.ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÛ.C.ÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÿÀ.........".......ÿÄ............................	
.ÿÄ.µ................}........!1A..Qa."q.2..¡.#B±Á.RÑð$3br.	
.....%&'()*456789:CDEFGHIJSTUVWXYZcdefghijstuvwxyz.................¢£¤¥¦§¨©ª²³´µ¶·¸¹ºÂÃÄÅÆÇÈÉÊÒÓÔÕÖ×ØÙÚáâãäåæçèéêñòóôõö÷øùúÿÄ...........

Notably JFIF, but also the EXIF information paint.net, indicates that this hex code is part of an JPEG-image that needs to be reconstructed.

By using the getfat 22 command, we get 0x9 (i.e. integer 9)

The contents of cluster 9, using the showcontent command, are

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

Using the getfat 9 command, we get 0x11 (i.e. integer 17)

The contents of cluster 17, using the showcontent command, are

05 14 14 90 B4 52 51 48 61 45 14 50 01 40 38 A2 92 98 99 20 6A 09 ED FA D3 3B D2 8E FF 00 D6 82 45 FE 82 8C 51 9C 0C 9E F4 B4 00 98 CD 03 BD 2D 1C 1C D0 02 74 3F CA 90 9C 0A 5F AE 73 ED 48 47 F9 34 00 DA 28 A5 A0 04 A2 8A 28 01 E0 73 CF 41 4E EB CF 6A 69 39 A5 07 02 90 03 0E 2A 3A 90 9C D3 08 A0 62 51 45 2E 29 85 84 A7 0A 4A 33 40 31 E7 A5 19 F5 FF 00 F5 53 32 68 E4 D0 20 34 94 52 81 9A 00 05 2D 3B 69 A3 6D 05 26 AC 32 8A 93 68 A5 C0 F4 A0 2E 88 E8 C1 A9 31 4B 40 73 11 6D 34 DA 9A A2 23 93 41 2D DC 4A 5A 4A 77 6F 7A 00 5E 31 FC FD 68 07 34 99 C5 28 3F AE 68 00 CE 3D E9 0F 1D B8 34 B9 E2 9A 68 00 CD 04 E6 92 96 80 12 8A 76 28 C5 00 36 8A 28 A0 07 51 49 45 00 14 EA 6F FF 00 AA 9C 29 02 76 0C 51 4B EF 47 4A 0A B8 DA 0F 4A 5C 52 50 3D D0 01 4F A6 2D 3B 34 12 04 50 9D E9 09 A7 2F 4A 10 87 51 45 14 C0 28 A2 8A 00 28 A2 93 34 00 B5 13 75 A9 33 4C 34 00 94 99 A2 8A 00 3A D1 4B 45 00 25 14 51 40 09 4A 28 A4 A0 09 29 0D 37 34 66 90 C3 14 52 D2 53 10 1A 28 CD 14 00 53 81 A6 D3 87 F2 A0 07 75 FC 29 33 40 E6 9D 8A 40 37 AF E1 4B 47 4E 94 DA 00 6F AD 19 A3 BD 25 30 0A 78 38 E0 D3 05 38 D0 03 F7 0F 5A 33 51 AF 5A 92 93 76 18 13 81 40 39 E2 82 32 31 40 18 A5 7D 04 2D 21 E9 4B 45 21 8C 53 D4 52 D0 00 14 51 D4 10 CE 86 8C D0 DD 69 2A C4 14 51 4B 40 09 4B 4E 00 1A 76 D1 40 11 8A 71 00 53 F6 8A 4C 0A 00 8C D2 53 A9 B4 00 51 47 BD 14 00 52 E2 8A 78 39 ED 40 DA B0 C0 0F 6A 78 1C 53 B1 45 02 12 8C FB 52 D1 40 0C 26 93 34 E2 29 A3 8F CE 90 09 DE 8A 28 A6 02 52 E6 92 8A 00 55 EB 52 54 43 AD 48 2A 58 FA 0E A2 93 38 A3 70 A4 02 D1 4D DC

Using the getfat 17 command, we get 0x5 (i.e. integer 5)

The contents of cluster 5, using the showcontent command, are

28 DC 3D 28 B0 59 8E A4 34 DD D9 E9 45 16 18 8D 4D A5 34 55 09 A1 29 71 49 45 31 12 2D 3A 9A BD 29 F4 00 52 52 D2 50 04 46 92 94 F5 34 01 9A 00 5C 0C 75 14 D0 2A 4D B4 D2 29 00 95 20 18 A8 C7 51 52 D3 1B 16 8A 28 A0 41 45 14 50 02 53 1B D6 9F 4D 23 34 00 CC 52 54 D8 A6 95 A0 06 62 96 97 06 92 91 6A C1 4A 29 28 A0 63 E9 36 D0 0D 3A A4 9D 86 6D A3 6D 49 49 4F 51 5C 66 28 34 A4 D3 0F 34 14 14 51 4B 4C 02 90 D2 D1 40 35 70 04 8A 76 E3 4C A5 A0 2C 3B 71 A4 C9 A4 A2 80 B2 1B DE 9E B4 CA 70 34 C9 25 A6 9A 4C D2 13 48 43 47 15 25 47 4F 07 34 C0 75 2D 36 96 80 16 92 8A 4A 00 5A 41 49 9A 75 00 2D 14 51 40 05 30 8A 7D 30 D0 34 36 8A 28 A4 58 51 45 14 00 B9 3E B4 99 3E B4 51 40 AC 25 2D 14 50 30 A2 8A 28 00 A2 8A 28 10 94 B4 94 50 02 D3 4D 2D 25 31 36 14 51 45 04 86 68 A2 8A 00 28 E9 45 14 00 F0 69 73 51 D1 93 40 12 66 93 34 DC 9A 4C 9A 06 38 D3 94 FA D2 0E 94 50 21 F9 1E B4 64 7A D3 28 A0 07 12 29 9C D3 87 26 9F 40 D3 B1 17 34 62 A5 A2 80 E6 22 C1 A3 06 A4 A5 A0 39 88 B0 68 C1 A9 68 A0 39 88 B0 7D 0D 18 3E 95 2D 14 07 31 16 0F A5 18 3E 95 25 2D 01 76 43 45 3D FB 53 28 1D C2 92 8A 3E B4 05 C2 8A 53 D7 8A 4A 09 0A 28 A2 80 12 96 8A 28 00 A2 8A 78 5C 8E B4 00 CA 29 FB 68 C5 03 B2 19 46 29 FB 68 DB 40 F4 13 B7 E1 4C A7 F6 A6 50 48 51 4A 06 4E 05 3B 61 F6 FF 00 3F 85 00 32 8A 30 68 A0 02 8A 28 A0 02 8A 28 A0 02 8A 28 A0 02 8A 28 A0 02 9C 9F 78 53 69 C9 F7 85 00 3D FA 0A 8E A4 7E 82 A3 A0 68 29 29 68 A0 41 45 14 50 01 45 2D 14 00 94 52 D1 40 05 3D 7A 0A 8E 9E 0F D2 80 1F 45 37 3F 4F D6 8C 9F F3 9A 00 5A 29 B9 A3 26 81 89 D8 FE

Using the getfat 5 command, we get 0x3 (i.e. integer 3)

The contents of cluster 3, using the showcontent command, are

34 CA 7F 6A 65 02 0A 94 0F 97 EB FD 7F FA D5 15 3D 9B 3C 0E 9D E8 01 FF 00 C3 81 DF FC E6 9A C0 05 F7 F5 A0 B0 38 03 A7 7F F0 A4 62 A7 B9 E3 B6 28 01 02 93 FF 00 D7 A4 2A 47 5A 79 65 38 EB C1 E9 48 4A 93 9C 9F CA 80 13 69 C0 3E B4 84 11 8C F7 A7 EF 19 1E 94 84 82 41 E7 8E D8 A0 04 DA 69 08 20 E3 BD 4B 8C 90 7D 07 4F AD 37 23 76 79 34 00 DD 87 DB F3 A3 69 F6 FC E9 DB 94 12 79 24 D4 74 00 53 93 EF 0F F3 DA 9B 4E 4F BC 28 01 EF D0 54 75 23 F6 A8 E8 00 A2 8A 28 00 A5 A4 A5 A0 02 8A 4A 5A 00 28 A2 8A 00 29 45 25 14 00 EA 33 4D A5 A0 03 34 99 A2 8A 06 1D A9 B4 F0 32 38 A4 DA 7D 28 10 DA 29 DB 5B D2 8D AD E9 40 0D A2 9D B5 BD 28 DA 7D 28 01 B4 53 B6 B7 A5 1B 5B D2 80 1B 45 3B 6B 7A 51 B5 BD 28 00 2E 71 FC CD 36 9D B4 D1 B4 D0 03 68 A7 6D 34 6D 3F E4 D0 03 69 C9 F7 87 F9 ED 46 C3 FE 4D 39 54 82 09 A0 05 7E D5 1D 3D C8 A6 50 01 4B 49 45 03 16 8A 4A 5A 00 28 A2 8A 04 14 51 45 00 14 51 4B 40 C2 8A 28 A0 03 14 51 40 A0 41 8A 4A 71 3C 53 28 00 A2 8A 28 00 A2 8A 5C 50 02 52 D1 45 00 14 51 F8 51 40 05 25 2D 14 00 51 49 45 00 2D 28 A6 D3 85 00 04 0E D4 94 A6 8A 00 4A 29 68 A0 62 51 4B 45 00 25 14 51 40 05 14 51 40 05 14 51 40 82 8A 28 A0 02 8A 4A 28 01 FD 69 B4 51 40 05 1F 8D 25 14 00 B4 71 49 45 00 2F 1E F4 71 49 45 00 2D 14 94 50 02 D2 51 45 00 14 51 45 00 2E 07 AD 02 92 96 80 0A 51 49 40 A0 05 A2 8A 28 18 52 52 D2 50 01 45 14 50 01 45 14 50 20 A2 8A 28 01 70 4D 2E D3 E9 49 C8 E8 68 CB 7A D0 02 ED 3E 94 60 D2 64 FA D1 93 EB 40 0B 83 49 B4 D2 64 FA D2 E4 D0 02 ED 3E 82 93 07 DA 8C 9F 5A 32 7D 68 00 C1 F4 14 B8 3E 82 93 27 D6

Using the getfat 3 command, we get 0x4 (i.e. integer 4)

The contents of cluster 4, using the showcontent command, are

8C 9F 5A 00 5C 1F 41 46 0F A0 A4 C9 F5 A4 C9 F5 A0 07 60 FA 0A 36 9A 6F 3E B4 73 EB 40 0B B4 D1 B4 D2 73 EB 47 3E B4 00 BB 4D 1B 4D 27 3E B4 73 EB 40 0B B4 D2 ED 34 DE 7D 69 79 F5 A0 00 82 29 05 1C FA D1 40 0B 45 14 94 0C 28 A2 8A 04 14 51 45 00 14 51 45 00 7F FF D9

Although, by length, this clearly marks the end of the file. Using the getfat 4 command returns 0xFFF (i.e. integer 4095).

Using CyberChef’s “Render Image” recipe with the Input format being Hex, we get the following output when copying and pasting the hex code in the order shown above.

Revealing the flag mango

Day 12: With Love from Julius

This day’s puzzle contains a rebus-like image. The flag is computed from solving this rebus.

We start at the top and go from left to right. The first segment being

\[((110\ 0011 + (-1)^{P(heads)} + x) \leq 50\]

The binary string $110\ 0011$ refers to the s character
$P(heads)$ refers to the chance of heads when flipping a coin, equivalent to 0.5. Evaluating $(-1)^{0.5}$ (i.e. $\sqrt{-1}$) gives us the imaginary unit $i$.

\[((s + i + x) \leq 50\]

6 is indeed smaller than 50, giving us the value true for this expression.

Beneath is a hangman guessing game that’s in progress. Using a Hangman Solver reveals the only candidate word to be zero.

Both of these are fed into what seems to be a MIL/ANSI Symbol for an OR gate. Only the top expression was true, so we pass the value true into the syntactic sugar for a Python string slice (i.e. substring).

  
>>> "true"[:2]
'tr'

We’ve completed the first top half of the rebus; Now for the bottom half.

We first encounter a heart shape added to a drawing of a city within parentheses. Outside the parentheses is a triangle. I considered the triangle to be the upper-case Greek symbol for delta ($\Delta$).

This is then divided by a drawing of a clock.

After a while, I figured that the two arrows coming from the heart shape indicate a shuffle of two sets of characters within the word love, as this would be the only thing that would make sense when concatenating it with the word city. love would thus become velo + city = velocity.

The bottom half represented the word time, giving us the mathematical formula

\[\frac{\Delta velocity}{\Delta time}\]

Which is how acceleration is calculated in physics.

Next, this word is fed into what looks to be a rotating disc with 26 segments and a coin depicting Julius Caesar. This refers to a Caesar Cipher (ROT6). The title of this day’s window also gave a hint towards this. Using the ROT13 recipe in CyberChef with an Amount of 6, we get the output giikrkxgzout.

Evaluating the Python string slice in a Python interpreter

  
>>> "giikrkxgzout"[9:]
'out'

Resulting in the flag tr + out = trout

Day 13: One Way

The puzzle of Day 13 linked to a single HTML page

  
<!DOCTYPE html>
<html>
<head>
    <title>Definitely a puzzle in a newspaper</title>
</head>
<body>
    <input placeholder="password" type="text">
    <p></p>
    <script src="index.js"></script>
    <script>
        const p = document.querySelector("p");
        const input = document.querySelector("input");
        input.addEventListener("keyup", function() {
            const valid = verifyPassword(this.value);
            p.style = `color: ${valid ? "green" : "red"}`;
            p.textContent = valid ? "Congrats!" : "Wrong password :(";
        });
    </script>
</body>
</html>

The script source index.js contained 103 lines of the following code. I’ve truncated a part of it for the purpose of this write-up. The puzzle seems to be some sort of security through obscurity function nesting, containing boolean logic, that you have to reduce.

  
// Truncated to 16 lines from 103
const zmp = s => false;
const by4 = s => s[27] == 'u' && dly(s);
const hgj = s => s[32] == 't' && efh(s) || s[10] == 'R' && sfa(s) || s[22] == 's' && t3v(s);
const kke = s => false;
const lf1 = s => false;
const az4 = s => s[21] == 'y' && js1(s);
const wjz = s => s[40] == '4' && n15(s) || s[37] == 't' && nxm(s) || s[20] == 't' && yj6(s) || s[26] == 'o' && e31(s);
const vxu = s => s[24] == 'w' && opc(s) || s[40] == 't' && rvp(s);
const yq8 = s => s[7] == 'C' && n15(s);
const tdu = s => s[7] == 'X' && e31(s) || s[9] == 'x' && n15(s) || s[23] == '9' && ybd(s) || s[10] == 'l' && yj6(s);
const sfa = s => s[24] == 'o' && e31(s) || s[40] == 'J' && glo(s);
const i2g = s => s[33] == ' ' && ovt(s) || s[14] == 'h' && hum(s) || s[31] == '3' && oxa(s) || s[8] == 'z' && tdu(s);
const hum = s => false;
const w1j = s => s[31] == 'S' && nxm(s) || s[14] == '7' && ybd(s) || s[18] == 'y' && kf9(s) || s[35] == 'Y' && lf1(s);
const bpc = s => s[8] == 'j' && n15(s) || s[29] == 'L' && yj6(s);
const verifyPassword = a70;

The above snippet shows that certain 3 character const functions with the parameter s that immediately evaluate to false. Using Visual Studio Code’s Find and Replace functionality, we can reduce a lot of the code that verifies the password. Afterwards, we sort each row in order of the zero-based index number of the string parameter s.

  
const verifyPassword = a70;
const cus = s => s[0]  == 'T' && oai(s);
const hx2 = s => s[1]  == 'h' && n75(s);
const puf = s => s[2]  == 'e' && s6w(s);
const opc = s => s[3]  == ' ' && tla(s);
const a70 = s => s[4]  == 'p' && ef0(s);
const m8z = s => s[5]  == 'a' && puf(s);
const rsi = s => s[6]  == 's' && p17(s);
const etz = s => s[7]  == 's' && rsi(s);
const nag = s => s[8]  == 'w' && q24(s);
const bhw = s => s[9]  == 'o' && xu1(s);
const t6x = s => s[10] == 'r' && i2g(s);
const j1o = s => s[11] == 'd' && epw(s);
const n75 = s => s[12] == ' ' && f4j(s);
const yy9 = s => s[13] == 'f' && vxu(s);
const y65 = s => s[14] == 'o' && hgj(s);
const xu1 = s => s[15] == 'r' && v3a(s);
const s35 = s => s[16] == ' ' && nag(s);
const smg = s => s[17] == 't' && q5m(s);
const s3u = s => s[18] == 'o' && yy9(s);
const oai = s => s[19] == 'd' && t6x(s);
const atb = s => s[20] == 'a' && bhw(s);
const az4 = s => s[21] == 'y' && js1(s);
const hgj = s => s[22] == 's' && t3v(s);
const ef0 = s => s[23] == ' ' && smg(s);
const vxu = s => s[24] == 'w' && opc(s);
const tla = s => s[25] == 'i' && cqx(s);
const txk = s => s[26] == 'n' && atb(s);
const t3v = s => s[27] == 'd' && vz0(s);
const ovt = s => s[28] == 'o' && txk(s);
const js1 = s => s[29] == 'w' && etz(s);
const cqx = s => s[30] == ' ' && hx2(s);
const v3a = s => s[31] == 'i' && ea6(s);
const bz6 = s => s[32] == 's' && az4(s);
const i2g = s => s[33] == ' ' && ovt(s);
const p17 = s => s[34] == 't' && j1o(s);
const s6w = s => s[35] == 'h' && y65(s);
const q24 = s => s[36] == 'r' && s3u(s);
const vz0 = s => s[37] == 'o' && cus(s);
const epw = s => s[38] == 'u' && s35(s);
const f4j = s => s[39] == 'g' && m8z(s);
const q5m = s => s[40] == 'h' && bz6(s);
const ea6 = s => s[41] == undefined;

Reading each character from top to bottom, we get The password for todays window is through. The flag thus being through.

Day 14: The plot thickens

This day’s puzzle only contained to a small 16x11 image, called cipher.png

The title “The plot thickens” and the text in the puzzle’s story

You had a gut feeling that something was off. The whole situation smelled of feathers. You could sense that there was a plot somewhere, but who or what it was, was still shrouded in mystery.

indicate that the values need to be plotted. We are, however, working with a PNG image that also contain alpha values. What do we exactly plot?

Loading CyberChef, uploading the image as the input, and using the Extract RGBA recipe, I noticed that every 4th value (the alpha value) was 255 anyway. It seemed that the RGB values themselves would need to represent the 3D $(x,y,z)$ coordinates.

After some trial-and-error Google queries looking for a way to plot this quickly in e.g. R or Python’s matplotlib, I encountered the website franciscouzo.github.io/image_colors that extracts the RGB values as sets of coordinates, and immediately plots them in 3D for you. After rotating the camera, we encounter the following shape.

The flag thus being pikachu.

Day 15: Goose Hunting

Today’s puzzle includes a file called cipher.txt. The puzzle text contained a reference to Sparta. By querying Google for “ciphers in spartan times” it’s revealed we’re dealing with a Scytale Cipher.

.. Before that she was known as Helen of Sparta, but that’s a tale* as old as the sky.” You approach a large gate, leading into a gated community with even larger houses than those you’ve already passed. On the locked gate there is a keyboard glued onto the gate, next to a screen which asks you for the ‘longest word’. Alongside it is a long band of seemingly random letters.

I’ve also found that this cipher is functionally similar to a Caesar Box cipher

We employ CyberChef again, as it has Caesar Box Cipher functionality. A trial-and-error achieved Box Height of 28 as a parameter gives us the following text.

Dear fellow Spartan, As I sit here, armor shining and spear at the ready, I can’t help but reflect on the journey that has brought us to this point. For as long as I can remember, being a Spartan warrior has been my ultimate goal. As a child, I was taken from my family and brought to the Agoge, where I began my training. It was tough, but I was determined to become the best warrior I could be. I spent countless hours practicing with my weapons, honing my skills and building my strength. But the training wasn’t just about physical prowess. We were also taught the values that make a true Spartan: courage, discipline, and loyalty. These were the principles that would guide us in battle, and in all aspects of our lives. As I grew older and moved up the ranks, I took part in many battles, defending our city from those who would seek to do it harm. And now, here we are, preparing to face our greatest challenge yet. But I am not afraid. I know that each and every one of us is ready for this moment. We have dedicated our lives to this cause, and we will not falter. As we prepare for battle, I wanted to share with you our plan for victory. We will strike at dawn, when our enemies are at their most vulnerable. Our strategy is simple: we will divide into two groups, one attacking from the front and the other from the rear. This will catch our enemies off guard, and give us the upper hand. Once we have engaged the enemy, we will fight with all the skill and strength that our training has given us. We will not falter, we will not retreat. We will stand our ground and emerge victorious. But above all else, we must remember our discipline. We must not be swayed by anger or fear, but focus on the task at hand. If we stay true to our training and our values, victory will be ours. So let us go forth with courage and determination. We are Spartans, the finest warriors in all the land. Let us go forth, brothers in arms, and face our enemies with courage and determination. We will show them the strength of the Spartan spirit, and we will emerge victorious. Today is the day, we prove it! Yours in battle, A fellow Spartan soldier.

Using the CyberChef recipe above, it’s revealed that the longest word and thus the flag is determination.

Day 16: Too many mirrors

This day’s puzzle was another MUD. Similar to the previous MUDs, we use e.g. go east to navigate the map. Certain rooms in the map contained mirrors, aimed horizontally or vertically, that transforms the whole map according to the mirror’s orientation. We interact with the mirror to flip the map. Certain rooms also only let the player move in a certain direction (e.g. only east or west). These constraints remain put when the map is flipped by a mirror, letting the player proceed by interacting with them.

Once you’ve reached the exit in the room, you’ll be put into the next room.

If you get stuck, you can always reset the floor by using the restartfloor command.

Floor 1

The starting position contains a vertical mirror ($V_m$) to the east. By flipping the map vertically here, we can go east twice to counteract this constraint.

Similar for the southern only N room, we first flip the map horizontally by going to the northern mirror ($H_m$).

The same idea applies for the other constraint rooms and their respective mirrors. In order to solve floor 1, we are required to interact with every mirror except for the $V_m$ to the south-east.

Floor 2

This floor added an extra dynamic to the puzzle: a shiny orb that you can place in a single room such that you can ignore that room’s movement limitations. However, when flipping the map, the orb stays put.

Rooms 3, 7, and 13 force you to use the $V_m$ to the west first in order to be able to reach the horizontal mirror. Afterwards the orb must me placed in room 7 as it’s relative to room 5. Then, after interacting with the horizontal mirror in room 14 to be able to go past room 9, we go to towards the exit. To “make the orb go” from room 7 to room 5 we flip the vertical mirror in room 11 along the way towards the exit.

Floor 3

You open a door and find yourself in a relatively well-lit but small space with a ceiling made out of wood and the ground being mostly covered by sand. On the wall in front of you there are two signs. One pointing left saying A and HOTEL AURORE and one pointing right saying B and PALAIS DE LA KASBAH. To your left there is what seems to be a driveway acending upwards.

To the west: Mid CT
To the east: A Ramp

This immediately reminded me of the map de_dust2 from the online first-person shooter Counter-Strike. This map contains a corner called Goose close to A Ramp and A site. If we go east towards A Ramp, we follow-up with go north to go towards Goose.

YOU HAVE FOUND THE EVIL GOOSE. here is the password: cultural

The flag thus being cultural.

Day 19: In the interrogation room

This day’s puzzle contained a link with a file called decompiler.dll, with the puzzle’s text being

“As I suspected. Quite a few files that would have run, were it not for my expert knowledge of bash… Let’s see what we got here. dAnkan says while opening a compiled C#-file, getting ready to decrypt it.

The first thing I tried was using JetBrains dotPeek. However, the following was returned.

  
// Decompiled with JetBrains decompiler
// Type: Program
// Assembly: decompiler, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null
// MVID: 4E4DE787-C7B3-4A7D-E49B331E62C0
// Assembly location: C:\Users\Petar Kostic\Downloads\decompiler.dll

using System;

Console.WriteLine("the decompiler is too smart! :((");

Fortunately, right clicking the output in dotPeek allows us to see Microsoft’s Common Intermediate Language (CIL)

  
// Type: Program 
// Assembly: decomplier, Version=1.0.0.0, Culture=neutral, PublicKeyToken=null
// MVID: 4E4DE787-C7B3-4A7D-A7DC-E49B331E62C0
// Location: C:\Users\Petar Kostic\Downloads\decomplier.dll
// Sequence point data from decompiler

.class private auto ansi beforefieldinit
  Program
    extends [System.Runtime]System.Object
{
  .custom instance void [System.Runtime]System.Runtime.CompilerServices.CompilerGeneratedAttribute::.ctor()
    = (01 00 00 00 )

  .method private hidebysig static void
    '<Main>$'(
      string[] args
    ) cil managed
  {
    .entrypoint
    .maxstack 8

    IL_0000: ldstr        ".entrypoint\r\n    // Code size       98 (0x62)\r\n    .maxstack  3\r\n    .locals init (int32[] V_0,\r\n             int32 V_1,\r\n             int32[] V_2,\r\n             int32 V_3,\r\n             int32 V_4)\r\n    IL_0000:  ldc.i4.s   10\r\n    IL_0002:  newarr     [System.Runtime]System.Int32\r\n    IL_0007:  stloc.0\r\n    IL_0008:  ldloc.0\r\n    IL_0009:  ldc.i4.0\r\n    IL_000a:  ldc.i4.0\r\n    IL_000b:  stelem.i4\r\n    IL_000c:  ldloc.0\r\n    IL_000d:  ldc.i4.1\r\n    IL_000e:  ldc.i4.2\r\n    IL_000f:  stelem.i4\r\n    IL_0010:  ldloc.0\r\n    IL_0011:  ldc.i4.2\r\n    IL_0012:  ldc.i4.s   -3\r\n    IL_0014:  stelem.i4\r\n    IL_0015:  ldloc.0\r\n    IL_0016:  ldc.i4.3\r\n    IL_0017:  ldc.i4.s   -11\r\n    IL_0019:  stelem.i4\r\n    IL_001a:  ldloc.0\r\n    IL_001b:  ldc.i4.4\r\n    IL_001c:  ldc.i4.s   17\r\n    IL_001e:  stelem.i4\r\n    IL_001f:  ldloc.0\r\n    IL_0020:  ldc.i4.5\r\n    IL_0021:  ldc.i4.s   -18\r\n    IL_0023:  stelem.i4\r\n    IL_0024:  ldloc.0\r\n    IL_0025:  ldc.i4.6\r\n    IL_0026:  ldc.i4.s   17\r\n    IL_0028:  stelem.i4\r\n    IL_0029:  ldloc.0\r\n    IL_002a:  ldc.i4.7\r\n    IL_002b:  ldc.i4.s   -11\r\n    IL_002d:  stelem.i4\r\n    IL_002e:  ldloc.0\r\n    IL_002f:  ldc.i4.8\r\n    IL_0030:  ldc.i4.s   13\r\n    IL_0032:  stelem.i4\r\n    IL_0033:  ldloc.0\r\n    IL_0034:  ldc.i4.s   9\r\n    IL_0036:  ldc.i4.s   -17\r\n    IL_0038:  stelem.i4\r\n    IL_0039:  ldc.i4.s   112\r\n    IL_003b:  stloc.1\r\n    IL_003c:  nop\r\n    IL_003d:  ldloc.0\r\n    IL_003e:  stloc.2\r\n    IL_003f:  ldc.i4.0\r\n    IL_0040:  stloc.3\r\n    IL_0041:  br.s       IL_005b\r\n\r\n    IL_0043:  ldloc.2\r\n    IL_0044:  ldloc.3\r\n    IL_0045:  ldelem.i4\r\n    IL_0046:  stloc.s    V_4\r\n    IL_0048:  nop\r\n    IL_0049:  ldloc.1\r\n    IL_004a:  ldloc.s    V_4\r\n    IL_004c:  add\r\n    IL_004d:  stloc.1\r\n    IL_004e:  ldloc.1\r\n    IL_004f:  conv.u2\r\n    IL_0050:  call       void [System.Console]System.Console::Write(char)\r\n    IL_0055:  nop\r\n    IL_0056:  nop\r\n    IL_0057:  ldloc.3\r\n    IL_0058:  ldc.i4.1\r\n    IL_0059:  add\r\n    IL_005a:  stloc.3\r\n    IL_005b:  ldloc.3\r\n    IL_005c:  ldloc.2\r\n    IL_005d:  ldlen\r\n    IL_005e:  conv.i4\r\n    IL_005f:  blt.s      IL_0043\r\n\r\n    IL_0061:  ret"
    IL_0005: pop

    // [9 1 - 9 54]
    IL_0006: ldstr        "the decompiler is too smart! :(("
    IL_000b: call         void [System.Console]System.Console::WriteLine(string)
    IL_0010: ret

  } // end of method Program::'<Main>$'

  .method public hidebysig specialname rtspecialname instance void
    .ctor() cil managed
  {
    .maxstack 8

    IL_0000: ldarg.0      // this
    IL_0001: call         instance void [System.Runtime]System.Object::.ctor()
    IL_0006: ret

  } // end of method Program::.ctor
} // end of class Program

The long string on line 22 that starts with IL_0000: ldstr. Here a string is loaded with a set of instructions. We substitute the lines within the range [19 - 28] (i.e. .entrypoint through IL_0010: ret) with the instructions from the string.

We paste this into SharpLab.io, converting from IL to C#, to get the following output.

  
using System;
using System.Runtime.CompilerServices;   

[CompilerGenerated]
internal class Program
{
    public static void Main(string[] args)
    {
        int[] array = new int[10];
        array[0] = 0;
        array[1] = 2;
        array[2] = -3;
        array[3] = -11;
        array[4] = 17;
        array[5] = -18;
        array[6] = 17;
        array[7] = -11;
        array[8] = 13;
        array[9] = -17;
        int num = 112;
        int[] array2 = array;
        int num2 = 0;
        while (num2 < array2.Length)
        {
            int num3 = array2[num2];
            num += num3;
            Console.Write((char)num);
            num2++;
        }
    }
}

Running this .cs file through .NET Fiddle or using dotnet run, we get the flag productive.

Day 20: Just another passenger

This day’s puzzle shows some instructions, along with an image

WoRks evEn foR youR tYpEWriTeR

Bury The CAPITAL. Take It DOWN, Take It DOWN, Is What’s RIGHT.
The lower rise up, rise left.

The image shows 5 instances of a 7-segment display where each segment displays a character or a number.

For the first instruction
1. We first get a the set of uppercase characters (i.e. capitals) from the instructions: WRERRYEWTR
2. We perform a Keyboard Shift Cipher of Down -> Down -> Right on the characters: CBVBBMVCNB
For the second instruction
1. We get the set of lowercase characters oksevnfoyoutprie
2. We perform a Keyboard Shift Cipher of Up -> Left on the characters: 8uq2dge858649372

Together, we get the sequence CBVBBMVCNB8uq2dge858649372

If you’d like to solve the 7-segmented displays more easily, convert all characters in the sequence (and image) to uppercase or lowercase, and use geocachingtoolbox.com to get each character of the flag.

Now we color the segments of character that contain a character in the sequence in the color red.

Revealing the flag jolly.

Day 21: Head(set) Hunt!

This day’s puzzle contains a link to the website https://sentor.djul.datasektionen.se, containing a loginform and a hint towards the website’s availability through a Tor hidden service.

Snooping around in the Chrome Debugger, I noticed that assets were being served from the files folder.

Navigating to https://sentor.djul.datasektionen.se/files/ shows the index of /files, revealing more files than loaded by the index page.

todo.txt contains the following with markdown checkbox syntax. The star marks which tasks have been completed by the developer.

[*] change admin password to "bomberman"
[*] learn how to exit vim:wqq:qq:w
[ ] fix directory listing, but how????
[ ] I lost my tor server files, and could only recover the ed25519 public key. The tor server is still up and running, but I lost the hostname too. Call IT support, perhaps they can recover it...

Although it would seem to be too easy, using the credentials admin and bomberman on the website, we get the form validation message

ERROR: user login not permitted from the clearnet. Go away!

We therefore have to complete this developer’s work by deriving the tor hidden service hostname from an ed25519 public key. Conveniently, a .zip file is hosted in the image above containing the following files.

Filename	Size	Contents
`hs_ed25519_public_key`	64 Bytes	Binary
`hs_ed25519_secret_key`	1 Byte	Empty
`hostname`	0 Bytes	Empty

The contents of the public key contain strange characters as it must be opened in binary format for reading.

In true CTF-style, we are not going to reverse engineer the ed25519 specification. Instead, we first take way too long querying Google if somebody has done this work for us. Fortunately, I encountered a the repository github.com/cmehay/pytor that contains a file called onion.py (link) that seems to have functions that we need to derive the hostname. Based on the format shown in the inheritence structure, our public key’s header specifies that it is an Onion V3 address.

  
from base64 import b32encode
from hashlib import sha3_256


def get_onion_str(_pub) -> str:
    version_byte = b"\x03"

    def checksum(pubkey):
        checksum_str = ".onion checksum".encode("ascii")
        return sha3_256(checksum_str + _pub + version_byte).digest()[:2]

    return (b32encode(_pub + checksum(_pub) + version_byte).decode().lower())


with open('hs_ed25519_public_key', mode='rb') as f:
    key_data = f.read()    
    
    # Only get the relevant part by removing the header.
    key_data = key_data.replace(b"== ed25519v1-public: type0 ==\x00\x00\x00", b"")

    print(get_onion_str(key_data))

Running the above Python code gives us the Onion V3 address. Finally, we append .onion

hw44qvorlbwlf7binb4ko4edms6wtj26dkdmju75exwmh56dnjlzylqd.onion

Navigating to this hidden server using the Tor browser, we enter the username admin and password bomberman to reveal the prompt.

Welcome admin - the flag is tabletennis

The flag thus being tabletennis.

Day 22: Virtual Pilot

This day’s puzzle text instructs us to do the following

What you need to do is find an evil duck somewhere - I think it’s hidden somewhere not visible on the map.

and

“It’s like I always say, Never Open Styled Quacking Lemons - make lemonade with them instead.” dAnkan says, sounding very satisfied with the idiom.

With the capitalization referring to the use of NoSQL for this puzzle.

The text then links us to cparta.djul.datasektionen.se, which contains an OpenStreetMap with a set of lakes containing ducks around the world. Where are these ducks coming from though?

Opening the network tab in the Chrome Debugger shows a GraphQL endpoint cparta.djul.datasektionen.se/graphql. Navigating to this URL shows us the Apollo Studio portal. This enables us to skip introspection steps and manually firing request via e.g. Postman.

Clicking on Query your server brings us to studio.apollographql.com/sandbox/explorer with the sandbox URL specified to https://cparta.djul.datasektionen.se/graphql.

Going from the root, we encounter 4 types of queries that can be performed

countries: [Country] with no arguments
country: Country with the argument query: JSON!
duck: Duck with the argument id: ID!
pond: Pond with the argument name: String!

We can also learn from the queries that are already being performed by the front-end itself. Inspecting the Network tab in the Chrome Debugger, identifiying the request, and selecting the payload tab when clicking on a random pond, reveals the following GraphQL query.

  
query Pond($name: String!) {
    pond(name:$name) {
        ducks {
            ... on FancyDuck {
                id
                name
                family {
                    name
                }
                color
            }
            ... on Mallard {
                id
                name
                family {
                    name
                }
                ability
            }
        }
    }
}

Where $name is e.g. Vänern

We’re looking for an evil duck. What can we get to know about ducks?

The introspection feature in Apollo Studio shows that there’s a type EvilDuck with the fields

family: [Duck]!
flag: String!
id: ID!
name: String!

The flag, exactly what we’re looking for.

The puzzle text specified that it would involve NoSQL. The only place where that might occur is the JSON! argument for the Country query. This is (probably) backed by MongoDB on the server.

Partly inferred from the example query above, we construct a new query. Apollo Server also provides code completion for the available fields by pressing Ctrl + Space.

  
query Country($query: JSON!) {
    country(query: $query) {
        ponds {
            ducks {
                ... on EvilDuck {
                    id
                    name
                    family {
                        id
                        name
                    }
                    flag
                    __typename
                }
            }
        }
        name
    }
}

Now we need to know what to query for. After much trial-and-error and a hint from another participant that would like to stay anonymous, I came to the conclusion that I’d have to query ducks that are not in the set of countries known to us. This is mainly because the ‘country query’ cannot return ducks within multiple countries. It will always default to the first index in the result set, in order, which is Sweden.

Writing the $query variable thus becomes

  
{
    "query": "{\"name\": {\"$nin\": [\"Norway\", \"Sweden\", \"Denmark\", \"United States\", \"Canada\", \"Germany\", \"Ireland\", \"Scotland\", \"England\", \"India\", \"China\", \"Ukraine\", \"Madagascar\", \"Colombia\", \"Brazil\", \"Peru\"]}}"
}

Giving us the result

  
{
  "data": {
    "country": {
      "ponds": [
        {
          "ducks": [
            {
              "id": "5592ad76-d26c-48ec-9d75-54ddcfd4045b",
              "name": "Dark Lord of Aitken basin",
              "family": [
                {
                  "id": "0f055c94-53cd-4990-bc87-daadc896de48",
                  "name": "Satan"
                }
              ],
              "flag": "monstrosity",
              "__typename": "EvilDuck"
            }
          ]
        }
      ],
      "name": "Moon"
    }
  }
}

The flag thus being monstrosity.

Additional insights

This puzzle took me way longer than the previous ones. A major waste of time was caused by being unsuccessful to write queries like

  
{
    "query": "{\"ponds.ducks\":{\"$elemMatch\":{\"__typename\":\"EvilDuck\"}}}"
}

and the subsequent backtracking after (because I thought I missed something). After solving the puzzle, I asked this in the Discord channel. The following answer was provided.

“A query like that isn’t possible because MongoDB doesn’t actually store the ponds in the countries and the ducks in those ponds. Everything is referred to by their respective IDs and resolved properly on the backend. In this case ponds.ducks wouldn’t make much sense, since ponds is a list of IDs (which on the backend is turned in to a list of type Pond)” – nnewram, cparta (sponsor) on Discord

Having barely touched MongoDB and GraphQL in the past, we learned something!.

Day 23: Pulling the plug

The last puzzle that counts towards points for the leaderboard is Day 23. This puzzle is another MUD game. This time, however, you’re able to specify your seperate room name that shouldn’t be easy to guess by another participant, indicating that you’re able to connect to the room with multiple netcat connections by having multiple terminal windows open.

The map consits of a number of rooms that can contain one of the following entities.

Trap door, which opens and remains open in the room (i.e. makes the unaccessible again) when a player leaves the room after triggering the trapdoor.
Door, which can be in the state ‘Open’ or ‘Closed’.
Gate, which only opens when all trap doors are opened
Pressure plate, which flips the state of ALL doors either ‘Open’ or ‘Closed’. (similar to a boolean negation)

I then tested the multi-user functionality, and found the following constraints

At most 3 terminals can be connected at once to a single dungeon instance.
Multiple players can only be in the start and ending rooms!
There are players inside door rooms. Going on the plate would crush someone, and we cant have that.
Disconnecting and reconnecting leaves the room state in tact, but returns that player to the start ($S$) position. (Hinted to me by anonymous participant)

Below is a map of this dungeon.

The red colored squares, containing the character T indicates a room with a Trapdoor
The brows colored squares, containing the character D, and their respective initial states O for Open or C for Closed as subscript, indicate a room with a Door.
The golden colored squares, containing the character G indicates a room with a Gate.
The blue colored squares, containing the character P indicates a room with a Pressure plate.

The goal is to have all 3 connected clients operate a jigsaw in the Exit room to reveal the flag. In order to get three players to the exit, we have to trigger all trapdoors and juggle the Pressure plates to open and close doors.

Using the map above, a solution can be achieved by following these instructions.

Player 1 traverses the sequence $11{\text -}5 \rightarrow 12{\text -}5 \rightarrow 12{\text -}6 \rightarrow 12{\text -}7$, triggering two trapdoors, and disconnects and reconnects.
Player 1 triggers a pressure plate by following the sequence $10{\text -}4 \rightarrow 10{\text -}3 \rightarrow 10{\text -}2 \rightarrow 9{\text -}2$, triggering 1 trap door.
Player 2 traverses the sequence $10{\text -}6 \rightarrow 10{\text -}7 \rightarrow 9{\text -}7 \rightarrow 8{\text -}7 \rightarrow 7{\text -}7$ $\rightarrow 6{\text -}7 \rightarrow 5{\text -}7 \rightarrow 4{\text -}7 \rightarrow 3{\text -}7 \rightarrow 2{\text -}7 \rightarrow 1{\text -}7 \rightarrow 1{\text -}6 \rightarrow 2{\text -}6 \rightarrow 3{\text -}6 \rightarrow 3{\text -}5 \rightarrow 2{\text -}5$, triggering 10 trap doors by disconnecting and reconnecting afterwards.
Player 2 goes west twice by traversing $9{\text -}5 \rightarrow 8{\text -}5$.
Player 1 steps off $9{\text -}2$ pressure plate by going to $9{\text -}3$.
Player 2 again goes west twice by traversing $7{\text -}5 \rightarrow 6{\text -}5$.
Player 1 step back on $9{\text -}2$ pressure plate, triggering the trapdoor on $9{\text -}3$.
Player 2 traverses the sequence $5{\text -}5 \rightarrow 4{\text -}5 \rightarrow 5{\text -}5 \rightarrow 5{\text -}4 \rightarrow 5{\text -}3 \rightarrow 4{\text -}3 \rightarrow 5{\text -}3$ $\rightarrow 6{\text -}3 \rightarrow 6{\text -}2 \rightarrow 6{\text -}1 \rightarrow 5{\text -}1 \rightarrow 4{\text -}1 \rightarrow 4{\text -}2 \rightarrow 4{\text -}1$, triggering 5 trapdoors and thus clearing them all.
Player 3, from the starting postiion $S$, traverses the path towards $13{\text -}2$, through the gate.
Player 1 disconnects and reconnects, now being at position $S$.
Player 3 continues towards $14{\text -}5$, triggering the pressure plate.
Player 1 continues towards $13{\text -}2$.
Player 3 goes towards the exit, leaving the pressure plate.
Player 1 continues towards $14{\text -}5$, triggering the pressure palte.
Player 2 disconnects and reconnects, now being at position $S$, and follows the paths towards the exit with the help of Player 1. The process is similar in how Player 3 helped Player 1.

After all three players interact with the jigsaw in the Exit room, the flag is revealed to be prescription

Day 24: Not so fast!

The last puzzle (that doesn’t count towards any leaderboard points) was a Google Form for providing feedback. At the end of the form, the flag was listed: improvability.

Completion Times

Day	Completion Time	Rank
-1	day 01 at 15:37:47	221
01	day 01 at 15:42:27	145
02	day 02 at 13:04:39	89
05	day 05 at 20:32:47	101
06	day 08 at 22:53:46	102
07	day 09 at 00:01:38	87
08	day 09 at 00:30:36	77
09	day 11 at 22:21:44	58
12	day 12 at 16:08:33	59
13	day 13 at 13:49:16	39
14	day 14 at 19:50:38	44
15	day 15 at 12:48:55	7
16	day 16 at 14:47:06	43
19	day 19 at 12:59:47	6
20	day 20 at 21:16:47	31
21	day 21 at 21:01:17	27
22	day 23 at 01:16:04	19
23	day 23 at 19:25:51	37
24	day 24 at 12:55:50	12

Final		35

CTF

ctf

This post is licensed under CC BY 4.0 by the author.