soucema9/tls

TLS protocol

The goal of this lab is to get familiar with underlying principals of SSL/TLS cryptographic protocol.

Task 1: Diffie–Hellman key exchange

Implement the vanilla DH algorithm. Try it with p=37 and g=5. Can you make it working with recommended values p=0x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and g=2 ?

Task 2: Diffie–Hellman key

Turn a DH secret into a key. Use sha1 to generate BLOCK_SIZE = 16 long key material.

Task 3: Bulk cipher

Ensure you have working implementation of AES in CBC mode with PKCS#7 padding. It is recommended to use BLOCK_SIZE = 16 You will need encrypt(key, iv, message) and decrypt(key, iv, encrypted_message) functions. You can check your implementation with bulk_cipher.py example.

Task 4: Implement simple SSL/TLS setup

It's time to have some fun now. Checkout tls_101.py example. Implement Agent() class such that this code executes with no errors. You might want to use DH keys to seed AES_CBC bulk cipher you have implemented before The interface for the Agent() class should support:

• sending/receiving public data (p and g)
• sending/receiving public key
• sending/receiving messages

Please, use recommended values for p and g for DH key exchange protocol.

Task 5: Man-in-the-middle

Oh, no! Looks like something is wrong here! Who the hell is Mallory? Implement MITM() class such that itls_101.py runs with no errors. The interface should support:

• sending/receiving public data (p and g)
• sending/receiving public key
• intercept_message

Task 6: RSA

RSA algorithm is the most used asymmetric encryption algorithm in the world. It is based on the principal that it is easy to multiply large numbers, but factoring large numbers is very hard. Within the TLS context it is used for both key exchange and generate signatures for security certificates (do you know why is that possible?). Let us implement this algorithm. Here are few hints:

• Please use p = 13604067676942311473880378997445560402287533018336255431768131877166265134668090936142489291434933287603794968158158703560092550835351613469384724860663783, q = 20711176938531842977036011179660439609300527493811127966259264079533873844612186164429520631818559067891139294434808806132282696875534951083307822997248459 and e=3 for the key generation procedure.
• You might want to implement your invmod function. Test it with values a=19 and m=1212393831. You should get 701912218. Your function should also correctly handles the case when a=13 and m=91
• You might want to implement functions encrypt(bytes_, ...)/decrypt(bytes_,...) and separately encrypt_int(int_, ...)/decrypt_int(int_,...)
• Please use big endian notation when transforming bytes to integer

Task 7: RSA broadcast attack

It's time to check now that despite a really complex math involved in RSA algorithm it is still might be vulnerable to a number of attacks. In this exercise we will implement the RSA broadcast attack (a.k.a simplest form of Håstad's broadcast attack) Assume yourself an attacker who was lucky enough to capture any 3 of the ciphertexts and their corresponding public keys. Check out message_captured. You also know that those ciphers a related to the same message. Can you read the message? Here are a few hints for this exercise:

• The data is encrypted using encrypt_int(public, bytes2int(message.encode())).
• Please note, that in all 3 case public keys are different

How Chinese remainder theorem is helping you here?

Task 8: Bleichenbacher's RSA attack

RSA is also used to generate digital signatures. When generating a signature the algorithm is somehow reversed: the message is first "decrypted" with a private key and then is being send over an open channel to be "encrypted" with a public key known to a client. In this exercise we are going to implement the attack that broke Firefox's TLS certificate validation about 10 years years ago. The interested reader can refer to this article.

The most widely used scheme for RSA signing at that was this: one takes the hash of the message to be signed, and then encodes it like this 00 01 FF FF ... FF FF 00 ASN.1 HASH. Where ASN.1 is a very complex binary encoding of the hash type and length. The above then is "decrypted" with RSA. FF bytes provide padding to make the message exactly as long as the modulus n.

The intuition behind the Bleichenbacher's RSA attack is that while it's impossible without private key (more specifically, without d) to find a number that elevated to e gives exactly the encoding above, one can get to an approximation, for example by taking the e-th root of the target message. If e is small enough, the approximation might be good enough to get a message like 00 01 FF 00 ASN.1 HASH GARBAGE

If the verification function fails to check that the hash is aligned at the end of the message (i.e. that there are enough FF bytes), we can fake signatures that will work with any public key using a certain small e. As you can see, n becomes completely irrelevant because exponentiation by e never wraps past the modulus.

In this exercise you will be asked to implement all the functions needed to make code rsa_bleichenbachers.py running without errors. Please, use p = 19480788016963928122154998009409704650199579180935803274714730386316184054417141690600073553930946636444075859515663914031205286780328040150640437671830139 and q = 17796969605776551869310475203125552045634696428993510870214166498382761292983903655073238902946874986503030958347986885039275191424502139015148025375449097 for the key generation procedure. e as before is 3.

Task 9: DSA

The final task of this block is pretty simple. We are going to break Digital Signature Algorithm (DSA). If the used nonce is weak than it is trivial to break the DSA.

Let us set the DSA domain parameters as follows:

Parameter	Value
p	0x800000000000000089e1855218a0e7dac38136ffafa72eda7859f2171e25e65eac698c1702578b07dc2a1076da241c76c62d374d8389ea5aeffd3226a0530cc565f3bf6b50929139ebeac04f48c3c84afb796d61e5a4f9a8fda812ab59494232c7d2b4deb50aa18ee9e132bfa85ac4374d7f9091abc3d015efc871a584471bb1
q	0xf4f47f05794b256174bba6e9b396a7707e563c5b
g	0x5958c9d3898b224b12672c0b98e06c60df923cb8bc999d119458fef538b8fa4046c8db53039db620c094c9fa077ef389b5322a559946a71903f990f1f7e0e025e2d7f7cf494aff1a0470f5b64c36b625a097f1651fe775323556fe00b3608c887892878480e99041be601a62166ca6894bdd41a7054ec89f756ba9fc9530229

You also were lucky to capture the SHA1 of a message which is H=0x2bc546792a7624fb6e972b0fb85081fd20a8a28. Knowing my public key and DSA signature

Parameter	Value
y	0x84ad4719d044495496a3201c8ff484feb45b962e7302e56a392aee4abab3e4bdebf2955b4736012f21a08084056b19bcd7fee56048e004e44984e2f411788efdc837a0d2e5abb7b555039fd243ac01f0fb2ed1dec568280ce678e931868d23eb095fde9d3779191b8c0299d6e07bbb283e6633451e535c45513b2d33c99ea17
r	548099063082341131477253921760299949438196259240
s	857042759984254168557880549501802188789837994940

can you derive my private key? Its SHA-1 fingerprint (after being converted to hex) is: 0x8f96763dea794b79094eef4717ceb5f10631d634. Implement your function recover_private_key(dsa_params, dsa_sign, H, ...) and send your code.

Hint: k is the number between 0 and 2**16

Task 10 (Bonus): DSA domain parameters

Can you implement the correct procedure of generating DSA parameters and signature for L=1024 and N=160?