updating challenges

capture_the_flag/challenges/heavy_computation/encrypt.py

@@ -0,0 +1,124 @@
+Heavy computation(400)
+Question:
+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!
+
+Given files:
+encrypt.py
+from Crypto.Util.number import bytes_to_long, long_to_bytes
+from Crypto.Cipher import AES
+from Crypto.Util.Padding import pad
+from secret import password, flag
+from hashlib import sha256
+
+NB_ITERATIONS = 10871177237854734092489348927
+e = 65538
+
+#Old N : N = 16725961734830292192130856503318846470372809633859943564170796604233648911148664645199314305393113642834320744397102098813353759076302959550707448148205851497665038807780166936471173111197092391395808381534728287101705
+
+N = 14968794114523720195251887716913440457986979987674770429103169854116498198112478103466085455257317270930523061714030307370028304505577267672733143013124254253285088080041831478700041394909740024011681885623055622400205
+
+
+def derive_key(password):
+    start = bytes_to_long(password)
+
+    #Making sure I am safe from offline bruteforce attack
+
+    for i in range(NB_ITERATIONS):
+        start = start ** e
+        start %= N
+
+    #We are never too cautious let's make it harder
+
+    key = 1
+    for i in range(NB_ITERATIONS):
+        key = key ** e
+        key %= N
+        key *= start
+        key %= N
+
+    return sha256(long_to_bytes(key)).digest()
+
+
+assert(len(password) == 2)
+assert(password.decode().isprintable())
+
+key = derive_key(password)
+IV = b"random_and_safe!"
+cipher = AES.new(key, AES.MODE_CBC,IV)
+enc = cipher.encrypt(pad(flag,16))
+
+with open("flag.enc","wb") as output_file:
+    output_file.write(enc)
+flag.enc(dump)
+    0000  f4 d8 e5 a2 ac 80 6c e9  dc c1 ef 1e d5 c4 51 7c   ......l.......Q|
+    0010  e3 d8 84 1a d7 c0 77 c9  9c b0 f6 f0 ab 13 63 b0   ......w.......c.
+    0020  f9 5e 8d cd 87 ce c7 d3  88 7a 4a 68 de a9 6f 96   .^.......zJh..o.
+    0030  77 cf 1e a7 95 a0 f8 1c  be 3a 66 f0 aa 73 2c 3e   w........:f..s,>
+Solution:
+Outline:
+(1) For preparartion, we calcurate Euler's totient of N by FactorDb( http://www.factordb.com/ ):
+
+N=5*23*61*701*3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847
+
+(2) First part of the functuon "derive_key":
+
+for i in range(NB_ITERATIONS):
+    start = start ** e
+    start %= N
+It is easily simplificated by Euler's theorem.
+
+start = pow(bytes_to_long(password),pow(e,NB_ITERATIONS,phi_N), N)
+where phi_N is the totient of N,
+
+phi_N=(5-1)* (23-1)* (61-1)* (701-1)* (3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847-1)
+
+(3) Second Part of the functuon "derive_key":
+
+    key = 1
+    for i in range(NB_ITERATIONS):
+        key = key ** e
+        key %= N
+        key *= start
+        key %= N
+We can replace it with the following (using "invert" function of gmpy2).
+
+    inv_e = gmpy2.invert(e-1,phi_N)
+    key = pow(start, (pow(e,NB_ITERATIONS,phi_N)-1) * inv_e, N)
+(4) Finally, We brute force the password and get the flag!
+
+Solver:
+solve.py
+from Crypto.Util.number import bytes_to_long, long_to_bytes
+from Crypto.Cipher import AES
+from Crypto.Util.Padding import pad
+from hashlib import sha256
+import gmpy2
+
+NB_ITERATIONS = 10871177237854734092489348927
+e = 65538
+N = 14968794114523720195251887716913440457986979987674770429103169854116498198112478103466085455257317270930523061714030307370028304505577267672733143013124254253285088080041831478700041394909740024011681885623055622400205
+
+#Using FactorDB, we have N=5*23*61*701*3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847
+phi_N = (5-1)*(23-1)*(61-1)*(701-1)*(3043975283150884175290138965903193067634156680289693153778518185326633105971710936004483047892546798724665417739250476586249010832824560305913279982496088828053414799963361876618585076997170323631281630177651847-1)
+
+def derive_key(password):
+    start = bytes_to_long(password)
+    start = pow(start,pow(e,NB_ITERATIONS,phi_N), N)
+    inv_e = gmpy2.invert(e-1,phi_N)
+    key = pow(start, (pow(e,NB_ITERATIONS,phi_N)-1) * inv_e, N)
+    return sha256(long_to_bytes(key)).digest()
+
+with open('flag.enc','rb') as f:
+    flag_enc = f.read()
+    for i in range(0x20, 0x100):
+        for j in range(0x20, 0x100):
+            password = long_to_bytes(i) + long_to_bytes(j)
+            key = derive_key(password)
+            IV = b"random_and_safe!"
+            cipher = AES.new(key, AES.MODE_CBC,IV)
+            flag = cipher.decrypt(flag_enc)
+            if(flag[0:6] == b'shkCTF'):
+                 print(flag.decode('utf-8')[0:flag.decode('utf-8').find('}')+1])
+                 break
+Flag:
+shkCTF{M4ths_0v3r_p4t13Nce_b4453d1f9f5386a1846e57a3ec95678f}

capture_the_flag/challenges/heavy_computation/flag.enc

@@ -0,0 +1,4 @@
+    0000  f4 d8 e5 a2 ac 80 6c e9  dc c1 ef 1e d5 c4 51 7c   ......l.......Q|
+    0010  e3 d8 84 1a d7 c0 77 c9  9c b0 f6 f0 ab 13 63 b0   ......w.......c.
+    0020  f9 5e 8d cd 87 ce c7 d3  88 7a 4a 68 de a9 6f 96   .^.......zJh..o.
+    0030  77 cf 1e a7 95 a0 f8 1c  be 3a 66 f0 aa 73 2c 3e   w........:f..s,>