Dynastic Writeup - Cyber Apocalypse 2024

→ 1 Introduction

The description of the challenge is shown below.

→ 2 Key Techniques

The key techniques employed in this writeup are:

• Manual Python source code review
• Adapting existing Python source code

→ 3 Artifacts Summary

The downloaded artifact had the following hash:

$ shasum -a256 crypto_dynastic.zip
353327ad9d0191bfc9e2622ff346236d0be7d7870b2195846fee81cfad0f3653  crypto_dynastic.zip

The zip file contained a single Python source code file, source.py, and output.txt:

$ unzip crypto_dynastic.zip
Archive:  crypto_dynastic.zip
   creating: crypto_dynastic/
  inflating: crypto_dynastic/source.py
  inflating: crypto_dynastic/output.txt

$ shasum -a256 crypto_dynastic/*
d0e599602c813536965e71233bac5b4272a2a221292c661098b7185a79e26684  crypto_dynastic/output.txt
f2079b4bbf17e3b3518707b2f658b058b38131e7bd3428b90f930d97831f848c  crypto_dynastic/source.py

→ 4 Static Analysis

→ 4.1 output.txt

output.txt contained a single line comment, followed by a line containing the flag. The flag only contained upper case letters, ’_‘,’?’ and ‘!’ characters.

Make sure you wrap the decrypted text with the HTB flag format :-]
DJF_CTA_SWYH_NPDKK_MBZ_QPHTIGPMZY_KRZSQE?!_ZL_CN_PGLIMCU_YU_KJODME_RYGZXL

→ 4.2 source.py

source.py contains Python code that ‘encrypts’ the flag as follows:

• Lines 12-13 loop through each character in the flag
• Lines 14-15 leave the character untouched if it is not an alphabetic character
• Lines 17-18 transform each character in the flag with the aid of two functions
  • to_identity_map converts the character to its integer representation, then returns the offset from 0x41 (decimal 65), which corresponds to ‘A’. For example, ‘E’ will become 4.
  • from_identity_map is then called with an offset equal to the index of the current character in the flag. For example, if the ‘E’ in the previous example is the 29th character currently being encrypted, from_identity_map will be called with 4 + 28 == 32, since string indices start from 0.
  • from_identity_map computes a % 26 (the integer remainder after dividing by 26), before adding it to 0x41 and converting the result back to a character. Continuing with the same example, 32 % 26 will be 6. Added to 0x41 and converted back to a character will become G.

from secret import FLAG
from random import randint

def to_identity_map(a):
    return ord(a) - 0x41

def from_identity_map(a):
    return chr(a % 26 + 0x41)

def encrypt(m):
    c = ''
    for i in range(len(m)):
        ch = m[i]
        if not ch.isalpha():
            ech = ch
        else:
            chi = to_identity_map(ch)
            ech = from_identity_map(chi + i)
        c += ech
    return c

with open('output.txt', 'w') as f:
    f.write('Make sure you wrap the decrypted text with the HTB flag format :-]\n')
    f.write(encrypt(FLAG))

→ 5 Implementing the decoder

decrypt.py was implemented that does the opposite of what source.py does:

• Line 7: the encrypt function has been renamed to decrypt
• Line 15 reverses the offset, passing chi - i to from_identity_map. Notably, this simple change is sufficient to reverse the ‘encryption’
• Line 19 opens output.txt in read mode
• Line 20 ignores the first line in output.txt
• Line 21 calls decrypt instead of encrypt

def to_identity_map(a):
    return ord(a) - 0x41

def from_identity_map(a):
    return chr(a % 26 + 0x41)

def decrypt(m):
    c = ''
    for i in range(len(m)):
        ch = m[i]
        if not ch.isalpha():
            ech = ch
        else:
            chi = to_identity_map(ch)
            ech = from_identity_map(chi - i)
        c += ech
    return c

with open('output.txt', 'r') as f:
    f.readline()
    print(f"HTB{{{decrypt(f.readline())}}}")

→ 6 Obtaining the flag

The script was run and the flag was obtained:

$ python3 decrypt.py
HTB{DID_YOU_KNOW_ABOUT_THE_TRITHEMIUS_CIPHER?!_IT_IS_SIMILAR_TO_CAESAR_CIPHER}

→ 7 Conclusion

The flag was submitted and the challenge was marked as pwned