Short and sweet challenge. The script has one fatal flaw when doing Diffie-Hellman:
xxxxxxxxxx
def generate_public_int(g, a, p):
return g ^ a % p
def generate_shared_secret(A, b, p):
return A ^ b % p
g ^ b % p
is equivalent to g ^ (b % p)
, and ^
is binary xor
in Python, not pow
. Hence, we can easily retrieve Bob's secret my doing B ^ g
, as B = g ^ (b % p)
, and b < p
, which leads to b % p = b
.
With this figured out, decrypting the flag is trivial.
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from Crypto.Cipher import AES
from Crypto.Util.Padding import pad, unpad
import hashlib
def is_pkcs7_padded(message):
padding = message[-message[-1]:]
return all(padding[i] == len(padding) for i in range(0, len(padding)))
def decrypt_flag(shared_secret: int, iv: str, ciphertext: str):
# Derive AES key from shared secret
sha1 = hashlib.sha1()
sha1.update(str(shared_secret).encode('ascii'))
key = sha1.digest()[:16]
# Decrypt flag
ciphertext = bytes.fromhex(ciphertext)
iv = bytes.fromhex(iv)
cipher = AES.new(key, AES.MODE_CBC, iv)
plaintext = cipher.decrypt(ciphertext)
if is_pkcs7_padded(plaintext):
return unpad(plaintext, 16).decode('ascii')
else:
return plaintext.decode('ascii')
p = 2410312426921032588552076022197566074856950548502459942654116941958108831682612228890093858261341614673227141477904012196503648957050582631942730706805009223062734745341073406696246014589361659774041027169249453200378729434170325843778659198143763193776859869524088940195577346119843545301547043747207749969763750084308926339295559968882457872412993810129130294592999947926365264059284647209730384947211681434464714438488520940127459844288859336526896320919633919
g = 2
A = 539556019868756019035615487062583764545019803793635712947528463889304486869497162061335997527971977050049337464152478479265992127749780103259420400564906895897077512359628760656227084039215210033374611483959802841868892445902197049235745933150328311259162433075155095844532813412268773066318780724878693701177217733659861396010057464019948199892231790191103752209797118863201066964703008895947360077614198735382678809731252084194135812256359294228383696551949882
B = 652888676809466256406904653886313023288609075262748718135045355786028783611182379919130347165201199876762400523413029908630805888567578414109983228590188758171259420566830374793540891937904402387134765200478072915215871011267065310188328883039327167068295517693269989835771255162641401501080811953709743259493453369152994501213224841052509818015422338794357540968552645357127943400146625902468838113443484208599332251406190345653880206706388377388164982846343351
iv = 'c044059ae57b61821a9090fbdefc63c5'
encrypted_flag = 'f60522a95bde87a9ff00dc2c3d99177019f625f3364188c1058183004506bf96541cf241dad1c0e92535564e537322d7'
b = g ^ B
shared_secret = A ^ b % p
print(decrypt_flag(shared_secret, iv, encrypted_flag))