feat: added CryptoCTF 2023

CryptoCTF 2023/Blue Office/README.md

@@ -0,0 +1,140 @@
+# Bagels
+> The Blue Office's ingenious cipher, meticulously crafted for the prestigious CCTF, became an impenetrable enigma that left even the most seasoned cryptanalysts baffled.
+
+## About the Challenge
+We have been given a zip file (You can download the file [here]). And if you unzip the file, you will see there are 2 files called `blue_office.py` and `output.txt`. Here is the content of `blue_office.py`:
+
+```python
+#!/usr/bin/enc python3
+
+import binascii
+from secret import seed, flag
+
+def gen_seed(s):
+	i, j, k = 0, len(s), 0
+	while i < j:
+		k = k + ord(s[i])
+		i += 1
+	i = 0
+	while i < j:
+		if (i % 2) != 0:
+			k = k - (ord(s[i]) * (j - i + 1))            
+		else:
+			k = k + (ord(s[i]) * (j - i + 1))
+	
+		k = k % 2147483647
+		i += 1
+
+	k = (k * j) % 2147483647
+	return k
+
+def reseed(s):
+	return s * 214013 + 2531011
+
+def encrypt(s, msg):
+	assert s <= 2**32
+	c, d = 0, s
+	enc, l = b'', len(msg)
+	while c < l:
+		d = reseed(d)
+		enc += (msg[c] ^ ((d >> 16) & 0xff)).to_bytes(1, 'big')
+		c += 1
+	return enc
+
+enc = encrypt(seed, flag)
+print(f'enc = {binascii.hexlify(enc)}')
+```
+
+This Python script is a basic encryption program. It imports the necessary modules and contains functions to generate a seed value, reseed the seed, and perform the encryption. The gen_seed function calculates a seed value based on a given string using a combination of addition, subtraction, and modulo operations. The reseed function updates the seed value using a mathematical formula. The encrypt function takes the seed and a message, iterates through each character of the message, and performs XOR operations with the seed-derived value. The encrypted output is stored and displayed in hexadecimal format. And if you check the `output.txt` file there is an encryted flag
+
+```
+enc = b'b0cb631639f8a5ab20ff7385926383f89a71bbc4ed2d57142e05f39d434fce'
+```
+## How to Solve?
+To solve this chall, we need to brute the seed first, how? Because we know the format of the flag is `CCTF{` and we know the output of the encrypted flag. We need to try check the seed from 0 to xxxx and then check the encryted match with `b0cb631639`. After that we need to create another function to decrypt the message
+
+```python
+def decrypt(s, enc):
+    assert s <= 2**32
+    c, d = 0, s
+    msg, l = b'', len(enc)
+    while c < l:
+        d = reseed(d)
+        msg += (enc[c] ^ ((d >> 16) & 0xff)).to_bytes(1, 'big')
+        c += 1
+    return msg
+```
+
+The decrypt function takes a seed value and an encrypted message, reseeds the seed value, performs a bitwise XOR operation between each byte of the encrypted message and a derived value from the reseeded seed, and returns the decrypted message. And here the final script that i used to solve this chall
+
+```python
+#!/usr/bin/enc python3
+
+import binascii
+
+def gen_seed(s):
+        i, j, k = 0, len(s), 0
+        while i < j:
+                k = k + ord(s[i])
+                i += 1
+        i = 0
+        while i < j:
+                if (i % 2) != 0:
+                        k = k - (ord(s[i]) * (j - i + 1))
+                else:
+                        k = k + (ord(s[i]) * (j - i + 1))
+
+                k = k % 2147483647
+                i += 1
+
+        k = (k * j) % 2147483647
+        return k
+
+def reseed(s):
+        return s * 214013 + 2531011
+
+def encrypt(s, msg):
+        assert s <= 2**32
+        c, d = 0, s
+        enc, l = b'', len(msg)
+        while c < l:
+                d = reseed(d)
+                enc += (msg[c] ^ ((d >> 16) & 0xff)).to_bytes(1, 'big')
+                c += 1
+        return enc
+
+def decrypt(s, enc):
+    assert s <= 2**32
+    c, d = 0, s
+    msg, l = b'', len(enc)
+    while c < l:
+        d = reseed(d)
+        msg += (enc[c] ^ ((d >> 16) & 0xff)).to_bytes(1, 'big')
+        c += 1
+    return msg
+
+seed = 0
+
+for seeds in range(1, 2**32 + 1):
+    enc = encrypt(seeds, b"CCTF{")
+    if binascii.hexlify(enc) == b'b0cb631639':
+        seed = seeds
+        break
+
+#enc = encrypt(10364460, b"CCTF{")
+#print(f'enc = {binascii.hexlify(enc)}')
+
+enc = binascii.unhexlify(b'b0cb631639f8a5ab20ff7385926383f89a71bbc4ed2d57142e05f39d434fce')
+
+for char in range(256):
+    dec = decrypt(seed, enc)
+    if all(32 <= byte <= 126 or byte == 10 or byte == 13 for byte in dec):
+        print(f"Decrypted message: {dec.decode('utf-8')}")
+    seed += 1
+```
+
+![flag](images/flag.png)
+
+```
+CCTF{__B4ck_0r!F1c3__C1pHeR_!!}
+```

CryptoCTF 2023/README.md

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+# CryptoCTF 2023
+CTF writeup for CryptoCTF 2023. I took part in this CTF competition with the HCS team, and got 184th place out of 672 teams
+
+| Category | Challenge |
+| --- | --- |
+| Intro | [Cookies](/CryptoCTF%202023/Blue%20Office/)

README.md

@@ -62,6 +62,7 @@ List of CTF events that i have joined before
 | Africa battleCTF 2023 prequal | No | - |
 | Google CTF 2023 | Yes | [Link](/Google%20CTF%202023/) |
 | UIUCTF 2023 | Yes | [Link](/UIUCTF%202023/) |
+| CryptoCTF 2023 | Yes | [Link](/CryptoCTF%202023/) |
 
 ### Local Events
 | Event Name | Writeup Available? | Writeup Link |