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Kickstarter 2 Challenges

You'll find these challenges dispersed in the Kickstarter page for the second crowdfunding campaign of the comic book series.

All four challenges are dispersed in the "Campaign" page.

Challenge 1

For the first challenge we get the word "ROT13" and what looks like a gibberish url. ROT13 is a version of the Caesar cipher with 13 rotations.

CyberChef has a ROT13 recipe we can use to decode the URL.

Going to that URL will give us a bonus flag. One down, three to go!

Challenge 2

Next, we are given the word "HEX" and a long hexadecimal string.

CyberChef has a From Hex recipe, we use that to get to the second stage of the challenge.

The decoded string contains Z->A and another gibberish url, similar to the one in challenge 1.

Z->A must be the key to decrypting the URL. Let's try to figure out what it means.

We are only given Z->A so let's assume at every step that this is as simple as possible. We can add complications later if it does not work.

If Z becomes A, then A becomes Z.

If Z becomes A and A becomes Z, then Y becomes B and B becomes Y.

... repeat until N becomes M ...

Ok this looks like an interesting algorithm, let's try it out!

Yup, that's the solution, cool, we get a second bonus flag.

Challenge 3

The third challenge has the word BASE? and a bunch of letters and numbers next to it.

It seems quite straightforward, we need to figure out the base of the string and decode it.

Using our good friend CyberChef, we try different common From BaseXX recipes. The one for Base32 gives us a url!

The Authentication

Yay, the url has a second challenge in addition to the bonus flag.

We are given a python function, and asked to figure out the correct password to get the flag.

The python function runs the check function on the user-provided input. This check function does some transformation on the input and compares it with a secret that we have access to.

The check function will return True if the user-provided input is the flag. This means we have to reverse-engineer the following transformation:

entered = [ord(c) ^ 0x13 for c in pwd[::2]] + [ord(c) ^ 0x37 for c in pwd[1::2]]

# ok we can split this into two separate arrays:
entered_1 = [ord(c) ^ 0x13 for c in pwd[::2]]
entered_2 = [ord(c) ^ 0x37 for c in pwd[1::2]]
entered = entered_1 + entered_2 # <- concatenation

# now let's try to write it in pseudo-code:
entered_1 = (char_to_int(element) XOR 19) for each second element of the input (starting with index 0)
entered_2 = (char_to_int(element) XOR 55) for each second element of the input (starting with index 1)
entered = concatenate entered_1 and entered_2

# now let's reverse it:
key_even = element XOR 19 for each element in the first half of the secret
key_odd = element XOR 55 for each element in the second half of the secret
key_int = intertwine key_even and key_odd (key_even[0], then key_odd[0], then key_even[1] etc)
key = apply int_to_char to every element of key_int

# in real python this gives us:
key_even = [e ^ 19 for e in secret[:(len(secret)+1)//2]]
key_odd = [e ^ 55 for e in secret[(len(secret)+1)//2:]]
pwd = []
for i in range(n):
    if i % 2 == 0:
        pwd.append(chr(key_even[i // 2]))
    else:
        pwd.append(chr(key_odd[i // 2]))
print(''.join(pwd))

This gives us the flag.

Challenge 4

We are given the word XOR and a long hexadecimal string.

As we are not given the key to decipher the XOR-encrypted string, we have to be clever. The XOR cipher is reversible, meaning that ciphertext = xor(plaintext, key) and plaintext = xor(ciphertext, key) but also key = xor(ciphertext, plaintext). We know the ciphertext, so if we could figure out parts of the plaintext we could maybe figure out parts of the key.

Almost all of the other challenges for this CTF have the same base URL, tficomic.io. We can be almost sure that the plaintext begins with these letters.

Using CyberChef, we use the From Hex and XOR recipes and put tficomic.io/ in the Key box (make sure to select UTF8 in the dropdown). The output is not recognizable but it does look like the beginning is repeating every two characters! We add the To Hex recipe and see that the first two bytes are repeating: 0eca for as long as the plaintext is known.

We can deduce that the key is 0eca and repeats for the entire word.

We now replace the Key field with 0eca (make sure to select Hex in the dropdown) and disable the To Hex operation.

Here is our URL!

The Network

There is the real challenge at the given URL, as well as a bonus flag.

We are given a pcapng file, which is a network capture file we can open with Wireshark.

The challenge page tells us this is the capture of someone downloading a ZIP file, and that the password for the ZIP is on the website from which the file was downloaded.

Opening-up the pcapng with Wireshark, let's first find that ZIP file. We scroll down to the bottom, and click on the last HTTP packet, which says OK (application/zip). We right-click, Follow, HTTP Stream. In that new page we Show as Raw and then Save as.... Now we've got our zip file.

Next step: find the password. Scrolling back up to the top, we find the line that says 200 OK (text/html). We click on it, and then in the bottom-right panel click on "Reassembled TCP". We can now scroll down through the HTML content, and find the password in the <input> field.

Unzipping the file with the password, we get an image of the flag. Success!