Introduction

PatriotCTF is an online jeopardy-style CTF hosted by George Mason University's Competitive Cyber Club. All are welcome to participate, including students and security professionals. Challenges range from beginner to expert, accommodating all skill levels. In this competition, I ranked 376th out of 1,343 participants while playing solo. Although my performance wasn't as strong compared with another team, I learned valuable lessons from each challenge. In this blog, I will share my solutions and insights from the competition

Cryptography

Cipher from Hell

We were given two file, first one is encrypted flag ENCRYPTED and another file is source code that used to encrypt the flag

import math, sys

inp = input("Enter your flag... ").encode()

s = int.from_bytes(inp)

o = (
	(6, 0, 7),
	(8, 2, 1),
	(5, 4, 3)
)

c = math.floor(math.log(s, 3))

if not c % 2:
	sys.stderr.write("Error: flag length needs to be even (hint: but in what base system?)!\n")
	sys.exit(1)

ss = 0

while c > -1:
	ss *= 9
	ss += o[s//3**c][s%3]
	
	s -= s//3**c*3**c
	s //= 3
	c -= 2

open("encrypted", 'wb').write(ss.to_bytes(math.ceil(math.log(ss, 256)), byteorder='big'))

From this source code we can analyze the way the flag was encrypted. So the main point we have to reverse the process to get the flag. Below is the script I used to solve this encrypted flag

import math

hex_string = (
    "e61233fef5fc46032f8c377c3f0f731e"
    "53923da8a656be6da1661764cdb090ca"
    "bba59dede782d0792283664b"
)

encrypted_int = int(hex_string, 16)

inverse_map = {
    6: (0, 0), 0: (0, 1), 7: (0, 2),
    8: (1, 0), 2: (1, 1), 1: (1, 2),
    5: (2, 0), 4: (2, 1), 3: (2, 2)
}

curr = encrypted_int
base9_digits = []

while curr > 0:
    base9_digits.append(curr % 9)
    curr //= 9

base9_digits = base9_digits[::-1] 

left_trits = []
right_trits_reversed = []

for digit in base9_digits:
    row, col = inverse_map[digit]
    left_trits.append(row)
    right_trits_reversed.append(col)

all_trits = left_trits + right_trits_reversed[::-1]

flag_int = 0
for trit in all_trits:
    flag_int = flag_int * 3 + trit

byte_len = (flag_int.bit_length() + 7) // 8
try:
    flag = flag_int.to_bytes(byte_len, byteorder='big')
    print(f"Recovered Flag: {flag.decode('utf-8')}")
except Exception as e:
    print(f"Error decoding bytes: {e}")
    print(f"Raw bytes: {flag_int.to_bytes(byte_len, byteorder='big')}")

If we try to run the code we will get the flag as shown below

pctf{a_l3ss_cr4zy_tr1tw1s3_op3r4ti0n_f37d4b}

On Going Writeup

CTFPatriotCTFCyber Security

Writeup Patriot CTF 2025.

Introduction

Cryptography

Cipher from Hell

On Going Writeup