should be finished now

RSAEncrypter/rsa_encrypter.py

@@ -3,104 +3,22 @@
 from Crypto.Util.number import getPrime, bytes_to_long, long_to_bytes
 import random
 import math
-import decimal
 
-decimal.setcontext(decimal.Context(prec=2000))
-message = open("./flag.txt").read().encode('utf-8')
+message = open("./flag.txt").read().encode('utf-8')  
 
-def find_invpow(x,n):
-    high = 1
-    while high ** n < x:
-        high *= 2
-    low = high//2
-    while low < high:
-        mid = (low + high) // 2
-        if low < mid and mid**n < x:
-            low = mid
-        elif high > mid and mid**n > x:
-            high = mid
-        else:
-            return mid
-    return mid + 1
-
-def inv(a, m) :  
-
-    m0 = m  
-    x0 = 0
-    x1 = 1
-
-    if (m == 1) :  
-        return 0
-    while (a > 1) :
-        q = a // m  
-
-        t = m  
-        m = a % m  
-        a = t  
-
-        t = x0  
-
-        x0 = x1 - q * x0  
-
-        x1 = t  
-    if (x1 < 0) :  
-        x1 = x1 + m0  
-
-    return x1  
-
-
-def findMinX(num, rem, k) :  
-
-    # Compute product of all numbers  
-    prod = 1
-    for i in range(0, k) :  
-        prod = prod * num[i]  
-
-    # Initialize result  
-    result = 0
-
-    # Apply above formula  
-    for i in range(0,k):  
-        pp = prod // num[i]  
-        result = result + rem[i] * inv(pp, num[i]) * pp  
-
-
-    return result % prod  
 
 def encode():
     n = getPrime(256)*getPrime(256)
-    # randNum = random.randrange(0,3)
-    # if randNum==0:
-    #     n = 99,991 * 11,113
-    # elif randNum==1:
-    #     n = 13,537 * 10,159
-    # else:
-    #     n = 12037 * 10061
-
     rsa_key = RSA.construct((n,3))
     cipher = PKCS1_OAEP.new(rsa_key)
     ciphertext1 = cipher.encrypt(message)
     ciphertext = pow(bytes_to_long(message), 3, n)
     return (ciphertext, n)
 
-# Driver method  
-num = [0, 0, 0]  
-rem = [0, 0, 0]  
-k = 3
-count = 0
-
 print("Return format: (ciphertext, modulus)")
-t = encode()
-num[count] = t[1]
-rem[count] = t[0]
-count+=1
-#sent = input("Did you recieve the message? (y/n) ")
-while count<3:#sent=='n':
-    t = encode()
-    num[count] = t[1]
-    rem[count] = t[0]
-    count+=1
-    #sent = input("How about now? (y/n) ")
-
-print(findMinX(num,rem,k)==bytes_to_long(message))
-print(long_to_bytes(find_invpow(decimal.Decimal(findMinX(num, rem, k)),3)).decode('utf-8'))
+print(encode())
+sent = input("Did you recieve the message? (y/n) ")
+while sent=='n':
+    print(encode())
+    sent = input("How about now? (y/n) ")
+print("Message acknowledged.")

RSAEncrypter/solve.txt

@@ -1,3 +1,82 @@
-Look at this link: https://cims.nyu.edu/~regev/teaching/lattices_fall_2004/ln/rsa
-Or just plug in to this website: https://chineseremaindertheorem.com/calculator.php 
-and take the cube root (forget the modulus) then decode utf8 to get the flag
+from Crypto.PublicKey import RSA
+from Crypto.Cipher import PKCS1_OAEP
+from Crypto.Util.number import getPrime, bytes_to_long, long_to_bytes
+import random
+import math
+import decimal
+
+decimal.setcontext(decimal.Context(prec=2000))
+
+def find_invpow(x,n):
+    high = 1
+    while high ** n < x:
+        high *= 2
+    low = high//2
+    while low < high:
+        mid = (low + high) // 2
+        if low < mid and mid**n < x:
+            low = mid
+        elif high > mid and mid**n > x:
+            high = mid
+        else:
+            return mid
+    return mid + 1
+
+def inv(a, m) :  
+
+    m0 = m  
+    x0 = 0
+    x1 = 1
+
+    if (m == 1) :  
+        return 0
+    while (a > 1) :
+        q = a // m  
+
+        t = m  
+        m = a % m  
+        a = t  
+
+        t = x0  
+
+        x0 = x1 - q * x0  
+
+        x1 = t  
+    if (x1 < 0) :  
+        x1 = x1 + m0  
+
+    return x1  
+
+
+def findMinX(num, rem, k) :  
+
+    # Compute product of all numbers  
+    prod = 1
+    for i in range(0, k) :  
+        prod = prod * num[i]  
+
+    # Initialize result  
+    result = 0
+
+    # Apply above formula  
+    for i in range(0,k):  
+        pp = prod // num[i]  
+        result = result + rem[i] * inv(pp, num[i]) * pp  
+
+
+    return result % prod
+
+# Driver method  
+num = [0, 0, 0]  
+rem = [0, 0, 0]  
+k = 3
+
+num[0] = put the first modulus here
+rem[0] = put the first ciphertext here
+num[1] = put the second modulus here
+rem[1] = put the second ciphertext here
+num[2] = put the third modulus here
+rem[2] = put the third ciphertext here
+
+print(findMinX(num,rem,k)==bytes_to_long(message))
+print(long_to_bytes(find_invpow(decimal.Decimal(findMinX(num, rem, k)),3)).decode('utf-8'))