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https://github.com/The-Art-of-Hacking/h4cker.git
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125 lines
4.9 KiB
Python
125 lines
4.9 KiB
Python
Heavy computation(400)
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Question:
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A friend of mine handed me this script and challenged me to recover the flag. However, I started running it on my school cluster and everything is burning now... Help me please!
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Given files:
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encrypt.py
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from Crypto.Util.number import bytes_to_long, long_to_bytes
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from Crypto.Cipher import AES
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from Crypto.Util.Padding import pad
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from secret import password, flag
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from hashlib import sha256
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NB_ITERATIONS = 10871177237854734092489348927
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e = 65538
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#Old N : N = 16725961734830292192130856503318846470372809633859943564170796604233648911148664645199314305393113642834320744397102098813353759076302959550707448148205851497665038807780166936471173111197092391395808381534728287101705
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N = 14968794114523720195251887716913440457986979987674770429103169854116498198112478103466085455257317270930523061714030307370028304505577267672733143013124254253285088080041831478700041394909740024011681885623055622400205
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def derive_key(password):
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start = bytes_to_long(password)
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#Making sure I am safe from offline bruteforce attack
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for i in range(NB_ITERATIONS):
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start = start ** e
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start %= N
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#We are never too cautious let's make it harder
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key = 1
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for i in range(NB_ITERATIONS):
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key = key ** e
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key %= N
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key *= start
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key %= N
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return sha256(long_to_bytes(key)).digest()
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assert(len(password) == 2)
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assert(password.decode().isprintable())
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key = derive_key(password)
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IV = b"random_and_safe!"
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cipher = AES.new(key, AES.MODE_CBC,IV)
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enc = cipher.encrypt(pad(flag,16))
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with open("flag.enc","wb") as output_file:
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output_file.write(enc)
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flag.enc(dump)
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0000 f4 d8 e5 a2 ac 80 6c e9 dc c1 ef 1e d5 c4 51 7c ......l.......Q|
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0010 e3 d8 84 1a d7 c0 77 c9 9c b0 f6 f0 ab 13 63 b0 ......w.......c.
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0020 f9 5e 8d cd 87 ce c7 d3 88 7a 4a 68 de a9 6f 96 .^.......zJh..o.
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0030 77 cf 1e a7 95 a0 f8 1c be 3a 66 f0 aa 73 2c 3e w........:f..s,>
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Solution:
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Outline:
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(1) For preparartion, we calcurate Euler's totient of N by FactorDb( http://www.factordb.com/ ):
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N=5*23*61*701*3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847
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(2) First part of the functuon "derive_key":
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for i in range(NB_ITERATIONS):
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start = start ** e
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start %= N
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It is easily simplificated by Euler's theorem.
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start = pow(bytes_to_long(password),pow(e,NB_ITERATIONS,phi_N), N)
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where phi_N is the totient of N,
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phi_N=(5-1)* (23-1)* (61-1)* (701-1)* (3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847-1)
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(3) Second Part of the functuon "derive_key":
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key = 1
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for i in range(NB_ITERATIONS):
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key = key ** e
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key %= N
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key *= start
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key %= N
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We can replace it with the following (using "invert" function of gmpy2).
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inv_e = gmpy2.invert(e-1,phi_N)
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key = pow(start, (pow(e,NB_ITERATIONS,phi_N)-1) * inv_e, N)
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(4) Finally, We brute force the password and get the flag!
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Solver:
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solve.py
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from Crypto.Util.number import bytes_to_long, long_to_bytes
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from Crypto.Cipher import AES
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from Crypto.Util.Padding import pad
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from hashlib import sha256
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import gmpy2
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NB_ITERATIONS = 10871177237854734092489348927
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e = 65538
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N = 14968794114523720195251887716913440457986979987674770429103169854116498198112478103466085455257317270930523061714030307370028304505577267672733143013124254253285088080041831478700041394909740024011681885623055622400205
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#Using FactorDB, we have N=5*23*61*701*3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847
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phi_N = (5-1)*(23-1)*(61-1)*(701-1)*(3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847-1)
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def derive_key(password):
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start = bytes_to_long(password)
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start = pow(start,pow(e,NB_ITERATIONS,phi_N), N)
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inv_e = gmpy2.invert(e-1,phi_N)
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key = pow(start, (pow(e,NB_ITERATIONS,phi_N)-1) * inv_e, N)
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return sha256(long_to_bytes(key)).digest()
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with open('flag.enc','rb') as f:
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flag_enc = f.read()
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for i in range(0x20, 0x100):
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for j in range(0x20, 0x100):
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password = long_to_bytes(i) + long_to_bytes(j)
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key = derive_key(password)
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IV = b"random_and_safe!"
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cipher = AES.new(key, AES.MODE_CBC,IV)
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flag = cipher.decrypt(flag_enc)
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if(flag[0:6] == b'shkCTF'):
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print(flag.decode('utf-8')[0:flag.decode('utf-8').find('}')+1])
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break
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Flag:
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shkCTF{M4ths_0v3r_p4t13Nce_b4453d1f9f5386a1846e57a3ec95678f}
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