Codegate 2018 | Miro

Do you wanna play the game? : D


Note: My original writeup was posted on our writeup repo along with the handout & solution files.

Miro was a cryptography challenge in which a Python script, client.py, and PCAP, miro.pcap are given.

client.py allowed us to connect to a maze game, where the maze was the same every time and there was a single path. At first, we could move down and right, but once we tried to move left, it gave an error. Examining client.py, we noticed the following section of code.

while 1:
    data = recv_until(tls_client, "Input : ")
        print data
	    #message
	        user_input = raw_input()

    if user_input == "u":
            print "Sorry.. Not support function.."
	            exit()
		        elif user_input == "d":
			        tls_client.send("6423e47152f145ee5bd1c014fc916e1746d66e8f5796606fd85b9b22ad333101\n")
				    elif user_input == "r":
				            tls_client.send("34660cfdd38bb91960d799d90e89abe49c1978bad73c16c6ce239bc6e3714796\n")
					        elif user_input == "l":
						        print "Sorry.. Not support function.."
							        exit()
								    else:
								            print "Invalid input!"
									            exit()

The client script didn’t just send the character representing the move we wanted to make - it sent under TLSv1.2 a 64-byte hash that represented the move.

The main issue was that the up and left moves weren’t implemented, and we didn’t know the hashes that were to be sent for them.

Looking at the PCAP, we saw that it contained data being sent under TLSv1.2. A reasonable guess given the context of the problem was that the PCAP was a capture of a game which included the left and up moves via the hashes that represent them.

We extracted the public certificate from the network capture as public.der using Wireshark’s ‘Export Packet Bytes…’ feature on the packet containing the certificate.

Information could then be extracted using openssl.

$ openssl x509 -inform-DER -in public.der -text
$ openssl x509 -inform DER -in public.der -modulus -noout
  • RSA-1025
  • Public modulus N = 0x1C20BDC017E3CAA3C579B40D439E2ECD70F12C4D7F2764784C95A3FDDBA00981BA9CE5B227ADE47B0A7A0A8ACABA4541AB95C52F6B6DE3DF9EC090C6C356445B21BE437ABE10214D0B4A398A96743BBF70C864687FB2EC929F01D6EDAB2D987FE09799AD2204A2704F33061DBF9C2E03B332F0BA1A446644C864A06CD586D480B
  • Public exponent e = 65537

Our goal was to be able to reconstruct a valid private certificate to get the data encrypted under TLSv1.2. To do this, we needed to recover p and q.

Fermat factorization was a good initial approach.

We adapted the Fermat factorization code from a Stack Overflow post and ran it with the N that we had. The following code is from fermat.py.

def isqrt(n):
    x = n
        y = (x + n // x) // 2
	    while y < x:
	            x = y
		            y = (x + n // x) // 2
			        return x

def fermat(n, verbose=False):
    a = isqrt(n)
        b2 = a*a - n
	    b = isqrt(n)
	        count = 0
		    while b*b != b2:
		            if verbose:
			                print('Trying: a=%s b2=%s b=%s' % (a, b2, b))
					        a = a + 1
						        b2 = a*a - n
							        b = isqrt(b2)
								        count += 1
									    p = a+b
									        q = a-b
										    print('a=',a)
										        print('b=',b)
											    print('p=',p)
											        print('q=',q)
												    print('pq=',p*q)
												        return p,q

fermat(0x1C20BDC017E3CAA3C579B40D439E2ECD70F12C4D7F2764784C95A3FDDBA00981BA9CE5B227ADE47B0A7A0A8ACABA4541AB95C52F6B6DE3DF9EC090C6C356445B21BE437ABE10214D0B4A398A96743BBF70C864687FB2EC929F01D6EDAB2D987FE09799AD2204A2704F33061DBF9C2E03B332F0BA1A446644C864A06CD586D480B, False)

We received the following values for p and q:

  • p = 17777324810733646969488445787976391269105128850805128551409042425916175469483806303918279424710789334026260880628723893508382860291986009694703181381742497
  • q = 17777324810733646969488445787976391269105128850805128551409042425916175469168770593916088768472336728042727873643069063316671869732507795155086000807594027

Then to craft our own private RSA certificate, we used rsatool and saved the result as private.pem.

$ python rsatool.py -p 17777324810733646969488445787976391269105128850805128551409042425916175469483806303918279424710789334026260880628723893508382860291986009694703181381742497 -q 17777324810733646969488445787976391269105128850805128551409042425916175469168770593916088768472336728042727873643069063316671869732507795155086000807594027 -o private.pem

Once we had a private key, we could then use Wireshark again to decrypt the data. We used the RSA keys list under the SSL Protocol in Preferences. However, the data didn’t seem to visibly change/be affected, so we added in a SSL debug file to look into the issue. Examining the SSL debug file, debug.txt, we could see the traffic in plaintext.

From debug.txt we recovered the two hashes for up and left. left came first in the PCAP as it was the first of the two that was used in the maze.

  • left: 27692894751dba96ab78121842b9c74b6191fd8c838669a395f65f3db45c03e2
  • up: 9de133535f4a9fe7de66372047d49865d7cdea654909f63a193842f36038d362

client.py was then modified to client_solve.py so that the maze game could be played with the left and up moves enabled.

while 1:
    data = recv_until(tls_client, "Input : ")
        print data
	    #message
	        user_input = raw_input()

    if user_input == "u":
            tls_client.send("9de133535f4a9fe7de66372047d49865d7cdea654909f63a193842f36038d362\n")
	        elif user_input == "d":
		        tls_client.send("6423e47152f145ee5bd1c014fc916e1746d66e8f5796606fd85b9b22ad333101\n")
			    elif user_input == "r":
			            tls_client.send("34660cfdd38bb91960d799d90e89abe49c1978bad73c16c6ce239bc6e3714796\n")
				        elif user_input == "l":
					        tls_client.send("27692894751dba96ab78121842b9c74b6191fd8c838669a395f65f3db45c03e2\n")
						    else:
						            print "Invalid input!"
							            exit()

Once we got through the maze, we were given the flag:

🏁 C4n_y0u_d3crypt_th3_P4ck3t?? 🏁


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