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shunyaCTF-2024/crypto/Echoes-Of-Encryption/README.md

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+# Echoes-Of-Encryption
+
+
+Domain : Cryptography
+
+Points : 200
+
+Solves : 35
+
+
+### Given information
+
+> In December 2022, my friend Alok's device was hacked. Upon investigation, he discovered that the breach was due to a vulnerability in the Nvidia SMC which had been recently discovered and published for research purposes on the same day he was hacked.<br><br>PS- In the end, only numbers matter to grow a plant from a seed!!
+
+
+### Solution
+
+On going through the attached `encrypt.py`, we can see that a key is being randomly generated based on a seed.
+After which it performs xor between the flag and the generated key.
+So If we can get the key and xor it with the given encrypted hex, challenge solved.
+
+On reading the description of the challenge these words stood out
+```
+the breach was due to a vulnerability in the Nvidia SMC which had been recently discovered and published for research purposes on the same day he was hacked
+```
+
+I figured this hinted to the seed to generate the key.
+Sure enough, looked up nvidia smc vuln December 2022 and reached this : https://nvd.nist.gov/vuln/detail/CVE-2022-42269 \
+`CVE-2022-42269`
+
+The seed was the cve number with the year `202242269`
+
+
+Solve script:
+
+```python
+import random
+import string
+
+def decrypt_string(input_string, seed):
+    random.seed(seed)
+    
+    encrypted_string = bytes.fromhex(input_string)
+
+
+    allowed_chars = string.ascii_letters + string.digits
+    key = ''.join(random.choices(allowed_chars, k=len(input_string)//2))
+    
+    flag = ''
+    
+    for i in range(len(encrypted_string)):
+        decrypted_char = chr(encrypted_string[i] ^ ord(key[i]))
+        flag += decrypted_char
+    return flag
+
+
+seed_value = 202242269
+input_string = "5e04610a22042638723c571e1a5436142764061f39176b4414204636251072220a35583a60234d2d28082b"
+decrypted = decrypt_string(input_string, seed_value)
+print(decrypted)
+```
+
+
+### Flag 
+
+`0CTF{alw4y5_r3ad_7he_d3scr!pti0n_c4r3fully}`
+
+

shunyaCTF-2024/crypto/Echoes-Of-Encryption/solve.py

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+import random
+import string
+
+def decrypt_string(input_string, seed):
+    random.seed(seed)
+    
+    encrypted_string = bytes.fromhex(input_string)
+
+
+    allowed_chars = string.ascii_letters + string.digits
+    key = ''.join(random.choices(allowed_chars, k=len(input_string)//2))
+    
+    flag = ''
+    
+    for i in range(len(encrypted_string)):
+        decrypted_char = chr(encrypted_string[i] ^ ord(key[i]))
+        flag += decrypted_char
+    return flag
+
+
+seed_value = 202242269
+input_string = "5e04610a22042638723c571e1a5436142764061f39176b4414204636251072220a35583a60234d2d28082b"
+decrypted = decrypt_string(input_string, seed_value)
+print(decrypted)