ECE596C: Handout #9. Authentication Using Shared Secrets. Electrical and Computer Engineering, University of Arizona, Loukas Lazos
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1 ECE596C: Handout #9 Authentication Using Shared Secrets Electrical and Computer Engineering, University of Arizona, Loukas Lazos Abstract. In this lecture we introduce the concept of authentication and discuss several authentication protocols along with their vulnerabilities. Readings from Chapter 9 of D. Stinson. Authentication Authentication is the process of verifying the identity of an entity or a user. Bob has to make sure that the entity he is communicating with is Alice, i.e. that the information sent is from Alice and that it has not been modified while in transit.. Types of authentication we use in real life. Physical Attributes: appearance, voice, figure, expressions, fingerprints, eye iris.. Credentials: Trusted documents such as driver s lisences, passports, diplomatic documents, etc.. Knowledge: knowledge of PIN, passphrase, mother s maiden name, date of birth, or any secret information (usually for remote authentication. Think of possible attacks against different types of identification.. Challenge-response authentication protocols Example: Alice and Bob know each other s secret password. Alice goes to Alaska, Bob goes to Bahamas. Alice and Bob want to communicate using the Internet. Alice logs in and communicates as shown below. My password is "open" This is good enough for Bob to know that Alice is sending information OR is it? Consider the eavesdropping attacker Eve in the middle. Bob does know the password is open. But Eve also knows (from the communication the password. Bob does not know/also cannot prove that Alice was the only one who sent it. The password is only good for a one-time authentication and then it becomes public knowledge. So what can be done? ( Use key and encrypt something. ( Use key
2 ECE 596C: Cryptography for Secure Communications with Applications to Network Security and hash something. What is needed? Alice (A and Bob (B must know the shared key. Let K AB be the shared key between Alice and Bob. Here is a new protocol that can be used by A and B.. A simplistic authentication protocol using encryption Challenge R Response f ( K AB, R Generates a random number R Some notation: The entire challenge response exchange is referred to as a session. Each step within a session is referred to as a flow. The session in figure. consists of three flows. At the end of each session, the verifier in the protocol (in our case Bob decides whether the authentication was successful or not which is referred to as either accept or reject. Flows in protocol of figure.. A B: Message in clear. B A: Generated a random number R and sends in clear. A B: Computes function f(k AB,R which was agreed upon. A trivial illustration can be as follows. Let K AB = 00. Let R = 000. Let f(k AB,R = R K AB = 0. When Bob receives R K AB, he can independently compute it and verify if received R K AB? = f(kab,r. Problems with the protocol. Not symmetric. Only Bob verifies Alice, while Alice does not know if Bob is the one on the other end. Challenge R R ' Alice has no way of knowing if Bob or Eve sent the random number.
3 Handout # 9. Password guessing. If Alice and Bob are humans then passwords are a natural option for K AB (easytoremember.ifk AB isapasswordthenpasswordguessingattackispossibleforeve.i.e.to check if f( ˆK AB,R = f(k AB,R. For illustration, assume that the password, K AB = HUSKY ( Let the random R = GIBBE ( Then f(k AB,R = K AB +R mod 6 = NBTLB ( 9. So Eve can use a dictionary attack and guess HUSKY, and hence break the authentication protocol. Let s see what types of functions can we use for flow in the authentication protocl.4 Variations of the simplistic protocol Decrypts using K AB EK AB (R R Generates R and encrypts it using K AB Variation Encryption In this variation, Alice also knows that E KAB (R came from someone who knows K AB. Hopefully Bob! What if Eve send a random bit string? Alice has no way of knowing and will respond with the wrong R. Bob will reject but Alice will not know. Hence this protocol still does not authenticate both parties. Variation Hashing What if Alice and Bob used hash instead of encryption? Parallel Session Attack Within the first session, Eve asks Bob to identify himself. Eve, challenges Bob with the same challenge R that Bob sent to Alice. Bob replies with MAC K (R. Once Eve
4 4 ECE 596C: Cryptography for Secure Communications with Applications to Network Security receives the response from Bob, she is able to successfully complete the first session. To prevent the parallel session attack, Alice and Bob include their respective IDs with the challenge R, so the response to a challenge R is MAC K (ID Alice R. The security of Variation scheme relies on the following three assumptions:. The symmetric key K AB is kept secret.. Alice and have perfect random generators, i.e. there is a very small chance that the same challenge appears twice.. The MAC is secure. That is there does not exist an (ǫ,q-forger for some values of (ǫ,q (both values depend on the application. Eve can record multiple sessions and hence collect valid (x,y MAC pairs, but these pairs are not sufficient to find a collision, and/or inverse the MAC. Also the key is always kept secret so Eve cannot construct a valid MAC. Last, the probability of the same random number occurring twice is quite small and hence, Eve could not have recorded a MAC that would be used a second time..5 Formal definition of the adversary Consider the following attack depicted in figure.5. Is this a vulnerability of the authentication protocol? &!"#$%!"#$% ( ' './. - *+', *+', An adversary is considered to be active if:. The adversary creates a new message and sends it in the communication channel.. The adversary modifies a message that is already in the channel.. The adversary diverts a message already in the channel and sends it to someone else other than the intended receiver. The goal of the adversary is to force either of the two participants to decide accept in some session in which it has been active. Note that this accept has to be with respect to a message that Eve modified, injected or diverted rather than a response to a legitimate message that was passed through by an honest participant.
5 Handout # Mutual Authentication Mutual Authentication Requirements:. Both Alice and Bob decide accept in the presence of a passive adversary.. If an adversary is active on a given flow, then no participant will decide accept after that flow. The adversary can attempt the following attacks in order to break a mutual authentication scheme:. Eve impersonates Alice, trying to make Bob to accept.. Eve impersonates Bob, trying to make Alice to accept.. Eve is active in some session involving both Alice and Bob, trying to make both Alice and Bob to accept. Suppose Alice and Bob want to verify each other. Then the following is one option. R f ( K AB, R 4 R 5 f ( K AB, R Observe that the protocol requires 5 messages. Can be improved? Consider the following., R R, f ( K AB, R f ( K AB, R Any problem with the above protocol? Eve does has a few possible attacks here. One of them is the reflection attack. In the reflection attack, Eve starts a parallel session with Alice challenging
6 6 ECE 596C: Cryptography for Secure Communications with Applications to Network Security with nonce R. Alice replies and challenges Bob with nonce R. Then Eve starts a parallel session with Bob and challenges him with nonce R received from Alice. Bob replies to the challenge and challenges who he believes is Alice with nonce R. Then Eve forwards Bob s reply to Alice forcing her to decide accept. I am Bob, R accept f(r, K AB, R f(r, K AB, R f(r, K AB, R The reflection attack is possible even if the IDs of each node are included in the replies :;<= 04 78? A>B C? 7 D D FGF E :;<= 7 D? A>B C H;<= 7 D? A>B C? 7 I. Problem is due to f(k AB,R i where the K AB is common for Alice and Bob. Instead let one use K AB and other use K AB +constant (such as K AB + etc.. Any other way to rectify this problem (think of other ways of creating assymetry?
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