init commit, added server and client

challs/cryptoEasy/challenge.py

@@ -0,0 +1,63 @@
+#!/usr/bin/env python3
+
+from Crypto.Cipher import AES
+from Crypto.Util.Padding import pad, unpad
+import os
+
+def factorial(x):
+    prod = 1
+    for i in range (1,x+1):
+        prod = prod * i
+    return prod
+
+a = 3
+b = 8
+p = 159043501668831001976189741401919059600158436023339250375247150721773143712698491956718970846959154624950002991005143073475212844582380943612898306056733646147380223572684106846684017427300415826606628398091756029258247836173822579694289151452726958472153473864316673552015163436466970719494284188245853583109
+g = factorial(p-1)
+
+flag = "flag{...}"
+
+def getDHkey():
+    A = pow(g,a,p)
+    B = pow(g,b,p)
+    K = pow(B,a,p)
+
+    return K
+
+def handle():
+    keyExchanged = str(getDHkey())
+    encryptedFlag = encrypt(flag.encode("utf-8"),keyExchanged)
+    print("Il messaggio crittografato è: {0}".format(encryptedFlag.hex()))
+    
+    return
+
+def fakePadding(k):
+    if (len(k) > 16):
+        raise ValueError('La tua chiave è più lunga di 16 byte')
+    else:
+        if len(k) == 16:
+            return k
+        else:
+            missingBytes = 16 - len(k)
+            for i in range(missingBytes):
+                k = ''.join([k,"0"])
+            return k
+
+def encrypt(f,k):
+    key = bytes(fakePadding(k),"utf-8")
+
+    cipher = AES.new(key, AES.MODE_ECB)
+    encryptedFlag = cipher.encrypt(pad(f, AES.block_size))
+    return encryptedFlag
+
+def decrypt(f, k):
+    
+    key = fakePadding(str(k))
+ 
+    chiave = bytes(key, "utf-8")
+    cipher = AES.new(chiave, AES.MODE_ECB)
+    decryptedFlag = cipher.decrypt(f)
+    return decryptedFlag
+
+if __name__ == "__main__":
+    handle()

challs/cryptoEasy/cryptoeasy.txt

@@ -0,0 +1,3 @@
+Diffie Hellman è così costoso computazionalmente se si usano valori particolari, come venirne fuori?
+
+Ciphertext: b5609cfbad99f1b20ec3a93b97f379d8426f934ffcb77d83ea9161fefa78d243