UTCTF 2020 Writeups

This is a collection of my writeups for challenges in UTCTF 2020

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IR

Challenge

We found this snippet of code on our employee's laptop.
It looks really scary. Can you figure out what it does?

Solution

We are given a file with LLVM IR code.

From these references, we can see how the language works.

• https://llvm.org/docs/LangRef.html
• https://idea.popcount.org/2013-07-24-ir-is-better-than-assembly/

And from here, we can slowly reverse each like into psuedo-C code.

ir_psuedocode.c

And from this, I can see what the code does:

• Input an char array to the function as a parameter
• In the 1st for-loop, the number 5 is added to each character
• In the 2nd for-loop, each character is XORed with the next character (ie. index i is XORed with index i+1 and stored back into index i)
• The output is then checked against the global array @check

Do it in reverse to get the flag

solve_ir.py

Flag

utflag{machine_agnostic_ir_is_wonderful}

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One_True_Problem

Challenge

One True Problem

Two of my friends were arguing about which CTF category is the best,
but they encrypted it because they didn't want anyone to see.
Lucky for us, they reused the same key; can you recover it?

Here are the ciphertexts:
    213c234c2322282057730b32492e720b35732b2124553d354c22352224237f1826283d7b0651
    3b3b463829225b3632630b542623767f39674431343b353435412223243b7f162028397a103e

Solution

Known plaintext attack on XOR cipher. By guessing "which CTF category is the best", I first did a decryption of 'cryptography'.

See comments in my Python script

Flag

utflag{tw0_tim3_p4ds}

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.PNG2

Challenge

In an effort to get rid of all of the bloat in the .png format, I'm proud to announce .PNG2!
The first pixel is #7F7F7F, can you get the rest of the image?

Solution

Read the file using a hex reader (xxd or hexdump).

PNG2 $ xxd pic.png2 | head -3
00000000: 504e 4732 7769 6474 683d 05cf 6865 6967  PNG2width=..heig
00000010: 6874 3d02 887f 7f7f 7f7f 7f7f 7f7f 7f7f  ht=.............
00000020: 7f7f 7f7f 7f7f 7f7f 7f7f 7f7f 7f7f 7f7f  ................

From here, we can see that the width=0x05cf and height=0x0288 are defined. This is followed by RGB bytes of the pixels.

Using a script, create a new image of the width and height and then fill in group of 3 bytes as the RGB pixel.

This will produce a readable PNG file.

./PNG2/solve_png2.py

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Random_ECB

Challenge

$ nc crypto.utctf.live 9003

Input a string to encrypt (input 'q' to quit):

Here is your encrypted string, have a nice day :)
48c4dddab92943c70724451780d8fddc2a7d2771570cb0d414c2b9b2285439b7

Solution

The challenge is a server which we can input a string to do AES-ECB.

The source for the encryption is as follows:

def encryption_oracle(plaintext):
    b = getrandbits(1)
    plaintext = pad((b'A' * b) + plaintext + flag, 16)
    return aes_ecb_encrypt(plaintext, KEY).hex()

Here, we see that the plaintext is appended with the flag. Hence, this is an AES-ECB Chosen Plaintext Attack.

• https://zachgrace.com/posts/attacking-ecb/
• https://crypto.stackexchange.com/questions/42891/chosen-plaintext-attack-on-aes-in-ecb-mode

However, there is a random character 'A' that is prepended to the plaintext.

We can defeat this by checking if the output is repeated.

1. Let's say we do a padding of 'A' * 15. We store the output.
2. Next, when we do a padding of 'A' * 14, if a random bit occurs, it will be equal to the previous output (repeated), so we know that the random character occurred.

Flag

utflag{3cb_w17h_r4nd0m_pr3f1x}