This post is a compilation of various information security questions and topics that are likely to be tested in the InfSi2 exam at HSR. I'm expanding this post step by step while I'm reviewing the contents of this semester. I've done something similar with the subject "Automata and languages" last semester and it helped me a lot.
1. Cryptographical strength
Hover over the algorithm to see its solution, or press the button below to show all the solutions. I made a special Jekyll plugin for this, however since this blog is no longer Jekyll based it's not working anymore - tw 25.01.2018
|AES-GCM-128||"128"||AES Galois Counter Mode (GCM)|
|DH ECP-256||"128"||Elliptic Curve Diffie Hellman (ECP)|
|DH ECP-384||"192"||Elliptic Curve Diffie Hellman (ECP)|
|DH ECP-512||"256"||Elliptic Curve Diffie Hellman (ECP)|
|DH MODP 4096||"256"||DH More Modular Exponential|
|ECDSA-256||"128"||Elliptic Curve Digital Signature (DSA)|
|ECDSA-384||"192"||Elliptic Curve Digital Signature (DSA)|
|ECDSA-512||"256"||Elliptic Curve Digital Signature (DSA)|
2. True Random numbers
Here are two sources that provide random numbers:
Q: What can be said about source a?
Source a is reliable, but it returns 1 more often then 0. So it has a bias towards 1. Sources of randomness that have a bias aren't ideal.
Q: What can be said about source b?
Source b had a good distribution between 1 and 0. However there are many gaps in the data stream which is also not ideal.
Q: How can source a & b be used to build a good random number generator?
The random values can be incorporated in a entropy pool via a hash function (e.g. SHA1, SHA256). The pool can then be used as a random seed for a good pseudo random number generator.
3. Quantum Key Cryptography
Quantum Key Cryptography uses entangled Photons to distribute a secure key between two parties. It has ben successfully demonstrated by three independent research groups. What makes Quantum Key Cryptography special is that due to the laws of quantum physics it is possible to detect when a eavesdropper intercepts part of a quantum key. So the compromised parts of the data can be discarded.
Fundamental Laws of Quantum Physics
- One cannot take a measurement without perturbing the system.
- One cannot determine simultaneously the position and the momentum of a particle with arbitrarily high accuracy.
- One cannot simultaneously measure the polarization of a photon in the vertical-horizontal basis and simultaneously in the diagonal basis.
- One cannot draw pictures of individual quantum processes.
- One cannot duplicate an unkown quantum state.
BB84 Quantum Key Distribution Protocol 
Alice creates a random bit 0 or 1 and then randomly selects one of two bases (rectilinear or diagonal) to transmit it in. Alice then sends a single photon in the state specified to Bob, using the quantum channel. This process is then repeated. Alice records the state, basis and time of each photon sent.
Since Bob doesn't know the basis the photons were encoded in, he selects a basis at random to measure in, either rectilinear or diagonal. He does this for each photon, recording the time, measurement basis and result.
After Bob has measured all the photons, he communicates with Alice over the public channel. Alice tells Bob the basis each photon was sent in, and Bob the basis each was measured in. Both discard the photon measurements where Bob used a different basis, which is about half on average. The remaining bits are used as a shared key.
To check for a eavesdropper Alice and Bob now compare a certain subset of their remaining bit strings. If someone has gained any information about the photons polarization, this introduces errors in Bobs measurements. If more then a certain amount (p) bits differ Bob and Alice abort the key and try again, possibly with a different quantum channel.
Photon Yield vs Transmission Distance
Attenuation in a monomode fiber with λ = 1550nm: 0.2db/km
|Distance||dB||Survival rate of photons|
|50km||10dB||1 out of 10 photons survive|
|100km||20dB||1 out of 100 photons survive|
|150km||30dB||1 out of 1000 photons survive|
Q: Which of the following key bits are valid?
Hover over the Validity cells to see the solution, or press the button below to show all the solutions.
|Key Bit Alice||0||1||1||1||0||1||0||1|
|Key Bit Bob||0||1||1||1||1||0||0||1|
Q: With the BB84 protocol Alice sends 10'000 key bits to Bob to produce a 256bit AES key. What happens if Bob moves and the distance between Alice and Bob increases from 50km to 100km?
- Because of the absorption a lot less key bits reach their target. At 50km distance 1 of 10 reach their target. At 100km only 1 of 100 reach their target.
- With 100bits (10'000/100) it's no longer possible to produce a 256bit AES key.
Q: With what method can Alice and Bob check whether a eavesdropper steals or inserts photons?
They can insert random decoy states. These decoy states are sent with a much lower output rate so that there is a different statistical distribution of the received photons. An eavesdropper can't differentiate between the two different types of photons.
4. Generating keys with Pseudo Random Functions
A HMAC-based pseudo random function expands a secret and a public initial seed into a key stream of arbitrary length. This can be achieved iterating a SHA-1 or SHA-256 based HMAC function controlled by a secret key using the following algorithm:
v(0) = seed v(i) = HMAC(Key, V(i-1)) key stream = V(1)|V(2)|V(3)|...
The key stream is formed by concatenating the MAC values V(i) up to the desired output length.
A HMAC-SHA256 based pseudo random function is loaded with a 256bit key that is generated by a quantum source. The PRF is initialized with a 512bit seed and generates now 1536bit key material for two 256bit AES session keys and two 512bit
HMAC-SHA512 session keys.
Q: How can the four keys be cracked through a brute-force attack most easily?
The brute force attack should be aimed at the pseudo random function with 256bit key because all the other keys are derived from that.
So the 1536bit key material doesn't have more then 256bit entropy. It's best to use the first 256bit AES session key for brute force attempts.
And then to check whether there's a plain text result through decryption and histogram generation.
Q: How many tries are needed for a brute force attack in the worst case?
In the worst case 2256 tries are needed.
MACsec stands for Media Access Layer Security, it's a technology used on the Data Link Layer.
|PAE||Port Access Entity|
|CAK||Connectivity Association Key||Pre-shared secret, strong long-term key.|
|SC||Secure Channel||Each PAE sets up a secure channel.|
|SA||Secure Association||On every SC there's an SA.|
|SAK||Secure Association Key||Short-term key.|
|MKA||MACsec Key Agreement Protocol|
|MKPDU||MACsec Key Agreement Protocol Data Unit||carried via EAPOL|
|EAP||Extended Authentication Protocol|
|EAPOL||EAP over LAN|
Q: What's the requirement that a PAE (e.g. ethernet port) can join a CA?
The PAE needs the CAK (shared secret) of the CA group.
Q: How many session keys (SAK) are needed by a CA with N PAEs so that everybody can communicate with everybody?
Each PAE has its own SAK that can be used to communicate with its N-1 peers. So N SAKs are required.
Q: How are the required keys (SAK) generated in a CA with N PAEs using the MACsec Key Agreement Protocol (MKA)?
The N PAEs chose a key server which generates random SAK keys. Every SAK is encrypted with the CAK and distributed to the N PAEs.
Q: How are the MKPDUs transported over ethernet or wifi?
They are carried via the EAPOL protocol (EAP over LAN).
Q: What's the default crypto suite of MACsec and what's its cryptographic strength?
The MKA Key Derivation Function (KDF) is a pseudo random function (PRF) based on AES-CMAC with a 128 or 256bit key.