# Digital Dragons CTF 2023 (Qualifying Round) ## **Digital\_Dragons\_CTF\_2023** ## **CRYPTOGRAPHY WRITEUP** ### **Author:** * Pham Quoc Trung ### **Used Language:** * Python3 ### **Problem Solving:** #### MD5 Hash Collision Generate the collision and uncover the two inputs, submit them to reveal the flag. The flag is a unique string that signifies your victory in this conquest. Note : The flag format is typically **CTF{...}** where the content inside the curly braces is the solution to the challenge. Giao diện của challenge là một trang web cho phép người dùng up lên 2 file: ![Screenshot 2023-08-09 001106](https://github.com/AcceleratorHTH/CTF-Writeup/assets/86862725/30d91581-d8c7-4c49-a2c8-649a3d0bafa0) Theo như mình thấy thì trang web sẽ so sánh 2 mã hash của 2 file mình up lên, nếu chúng giống nhau thì sẽ trả về flag. Vì vậy mình thử up cùng 1 file lên và kết quả trả về như sau: ![Screenshot 2023-08-09 001427](https://github.com/AcceleratorHTH/CTF-Writeup/assets/86862725/fb28fa9e-cb2a-4cc4-881a-93f35c4f608a) Có vẻ không được (đương nhiên là thế rồi :v). Mình thử để xem bài có handle được trường hợp đó không chứ theo như giao diện trang web thì ở đây mã hash được sử dụng là MD5, thứ mà đã xuất hiện collision (xung đột). Nôm na là với hai chuỗi dữ liệu khác nhau, ta lại tạo ra được cùng một giá trị băm MD5. Mình sẽ sử dụng 2 file *erase.exe* và *hello.exe* trong link [này](https://www.mscs.dal.ca/~selinger/md5collision/) để khai thác được lỗ hổng MD5 collision và lấy flag ![Screenshot 2023-08-09 002103](https://github.com/AcceleratorHTH/CTF-Writeup/assets/86862725/290bf080-f866-4fdb-a447-ea6d7f932310) Flag: CTF{Md5\_h@$h\_c0ll15i0n\_1z\_3zy!!!} #### MooMoo Can you crack this unpredictable secret language used between two drug cartels? Note : The flag format is typically **flag{...}** where the content inside the curly braces is the solution to the challenge. Attachment: *MooMoommmoooo\_1.txt* ``` OOOMoOMoOMoOMoOMoOMoOMoOMoOMMMmoOMMMMMMmoOMMMMOOMOomOoMoOmoOmoomOo MMMmoOMMMMMMmoOMMMMOOMOomOoMoOmoOmoomOoMMMmoOMMMMMMmoOMMMMOOMOomOo MoOmoOmooOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoO MMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMM moOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoO MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOO MOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoomOoOOO moOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoO MoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOO MOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomOomOomOoMMMmoO moOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoo mOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOO MOomoOMoOmOomoomoOMoOMoOMoOMoomOoOOOmoOOOOmOomOomOoMMMmoOmoOMMMMOO MOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoO MoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMM MOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoo mOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOO MOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoO MMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomOomOomOo MMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoomOoOOOmoOOOOmOomOoMMM moOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoO MoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMM MOOMOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoO MoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMM moOmoOMMMMOOMOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOo moomoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOo moomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoO MoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOo MMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMM MOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomOomOomOoMMM moOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOo mOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOOMOomoO MoOmOomoomoOMoOMoomOoOOOmoOOOOmOomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOo moomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoO OOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoO mOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOO mOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOo moomOomOomOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoomOoOOO moOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOO MOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoo mOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOO MOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoO MoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOomOoMMMmoOmoOMMMMOOMOomoO MoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoomOo OOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOo moOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoO MoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOomOoMMM moOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoO MoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOo MMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoO MoOMoomOoOOOmoOOOOmOomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomOomOomOo MMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoomOoOOOmoOOOOmOomOo MMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoO MoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoO MMMMOOMOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoO MoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoO moOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoO MoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoO MMMMOOMOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoO MoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOo mOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoO MoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoo mOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoomOoOOOmoOOOOmOo mOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoO MoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOo moomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOO MOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoO MMMMOOMOomoOMoOmOomoomOomOomOoMMMmoOmoOmoOMMMMOOMOomoOMoOmOomoomoO MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOo MMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoO MoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMM MOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoomOo OOOmoOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOo moOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOO moOOOOmOomOoMMMmoOMMMMOOMOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoO MoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoomOoOOOmoOOOOmOomOoMMMmoOMMMMOO MOomoOMoOmOomoomOomOoMMMmoOmoOMMMMOOMOomoOMoOmOomoomOomOomOoMMMmoO moOmoOMMMMOOMOomoOMoOmOomoomoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoOMoO MoOMoomOo ``` Bài này sử dụng một ngôn ngữ khá hay ho mà mình đã từng sử dụng để tạo ra một challenge trong giải HolaCTF 2023. Các bạn có thể xem write-up của mình tại [đây](https://github.com/AcceleratorHTH/CTF-Writeup/tree/main/Hola%20CTF%202023). Nôm na thì nó là COW programming language, ngôn ngữ con bò :v. Mình sẽ sử dụng trang web [này](https://www.cachesleuth.com/cow.html) để decode nó ![Screenshot 2023-08-09 002931](https://github.com/AcceleratorHTH/CTF-Writeup/assets/86862725/4ace3dd7-596a-4dac-a3bb-694e55784851) Flag: flag{c0ngratulat10ns\_y0u\_l3arnt\_c0w\_lang} #### Xory Xory Xory yes pappa, eating FLAG? No papa… nc 114.119.185.67 1338 Attachment: *chal.py* ```python3 import random def xory(msg, key): ct = [] for i in range(len(msg)): ct.append(msg[i] ^ key[i%len(key)]) return bytes(ct) KEY = random.randbytes(5) FLAG = open('/app/flag.txt', 'rb').read() cipher = xory(FLAG, KEY) print(cipher.hex()) ``` Bài này chỉ đơn giản là sử dụng XOR để mã hóa dữ liệu với KEY được tạo ra thông qua thuật toán random. Nhận thấy form của flag là "flag{", đủ 5 ký tự, mình sẽ XOR cipher với nó và lấy 5-bytes đầu thu được làm KEY xem sao ```python3 cipher = "885efce6b5b602cfdeffdb6deab2fa856dace7919702e8dea58002eade85be73e08b" cipher = bytes.fromhex(cipher) def xory(msg, key): ct = [] for i in range(len(msg)): ct.append(msg[i] ^ key[i%len(key)]) return bytes(ct) flag = b'flag{' KEY = xory(cipher, flag)[:5] print(xory(cipher, KEY)) ``` Và mình ra được flag lun :v Có vẻ đây vẫn là một chal khá basic. Flag: flag{X0R\_15\_w34k\_1f\_y0u\_kn0w\_KPA} #### Trial Challenge - Before Qualifying Round This code challenges participants to decrypt a given block cipher ciphertext using a specific decryption algorithm. Participants are required to understand the decryption process, implement the necessary functions, and apply their knowledge of bitwise operations and block ciphers to obtain the original plaintext. Note : The flag format is typically **flag{...}** where the content inside the curly braces is the solution to the challenge. Attachment: *BlockCIpher.py* ```python3 import struct import sys import base64 def process_data(data): padding = 4 - len(data) % 4 if padding != 0: data += "\x00" * padding result = [] blocks = struct.unpack("I" * (len(data) // 4), data.encode()) for block in blocks: result.append(block ^ block >> 16) output = b'' for block in result: output += struct.pack("I", block) return output def main(): if len(sys.argv) != 2: print("Usage: {} data".format(sys.argv[0])) sys.exit(1) data = sys.argv[1] encoded_output = process_data(data) b64_encoded_output = base64.b64encode(encoded_output).decode() print(b64_encoded_output) if __name__ == "__main__": main() #BwthZ0gbM2wOVW1lbyswX0Btc1ktOnJ5DVR9AA== ``` Ở đây, flag bị chia ra thành từng khối 4-bytes, mỗi khối được mã hóa bằng cách XOR khối đó với khối đó dịch phải 16-bits. Vậy phải giải mã như nào đây? Ở đây, do khối được dịch phải 16-bits mà 1 khối là 32-bits, ta có thể thấy 16-bits đầu của khối ta thu được sẽ là 16-bits cuối của khối ban đầu sau khi dịch phải. Để dễ hình dung thì: Giả sử: block: **0110011101100001**0000101100000111\ block>>16: 0000000000000000**0110011101100001**\ \=> XOR: **0110011101100001**0110110001100110 Vậy ta hoàn toàn có thể tạo ra block>>16 từ kết quả và xor nó với kết quả để lấy được block ban đầu nhờ tính đối xứng của XOR. Ghép các khối đó lại, ta sẽ có được flag. Dưới đây là code python cho bài này ```python3 import base64 from Crypto.Util.number import * def rev_xor_shift_right_func(data): data = format(data, "032b") data_shift = '0'*16 + data[:16] res = [] for i in range(32): res.append(int(data_shift[i]) ^ int(data[i])) return int("".join(str(i) for i in res), 2) def split_block(data,block_size): return list(int.from_bytes(data[i:i+block_size],'little') for i in range(0,len(data),block_size)) def join_blocks(blocks): return b''.join(block.to_bytes((block.bit_length() + 7) // 8, 'little') for block in blocks) cipher = "BwthZ0gbM2wOVW1lbyswX0Btc1ktOnJ5DVR9AA==" cipher = base64.b64decode(cipher) flag = join_blocks([rev_xor_shift_right_func(block) for block in split_block(cipher, 4)]) print(flag) ``` Flag: flag{w3lc0me\_t0\_34sY\_CrypT} **© 2023,Pham Quoc Trung. 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