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5

The intend appears to be that vector is secret and the main key; and version (that, we are told, is secret) is an extension of that key (or variant selector). 1) I could say that ResetCode is a MAC of quote with key vector and version. 2) Never met this particular one. Anyone with common sense should laugh at it as snake oil if it pretends to be ...


5

Yes, there are several ways in which Mallory could pretend to be Amy. One obvious way, which doesn't even involve Amy herself in any way, would be for Mallory to perform steps 1 and 2 of the protocol normally, as if he were Amy. Then, given Betty's nonce $n_b$, Mallory can start a second, parallel instance of the protocol, again pretending to be Amy, and ...


4

A good PRNG can't be exhausted, so your argument against random bytes doesn't matter. firstly I'd like the server to be completely stateless. Sounds impossible to me. a PRNG needs state. A clock needs state. An open connection is state too You probably want to store open challenges too. This part isn't strictly necessary, but very convenient. ...


3

Using time of day is predictable. Challenges should be unpredictable to an adversary. If they were not (as you sort of say), the adversary could solve a set of challenges ahead of time and then use them in rapid succession to cause a DOS. You want to meter access to the service, which is about not only ensuring clients do some work, but that work is done at ...


3

Yes, because Mallory can use Amy and Betty to get any encrypted nonce; Amy and Betty are oracles for Mallory. She just has to send the nonce she has to encrypt to either one of them and they perform the task for her (in another "authentication attempt", using step 1 & 2). Usually you protect against this kind of situation by performing an encryption ...


3

The Wikipedia article points out a good reason for using a random challenge value: preventing replay attacks. If the hash was always the same (as the hash of the symmetric key would be), then having listened in on one challenge-response cycle, a malicious listener could pass further handshake tests.


3

Whilst your altered vector value will influence the results, this extension of your script demonstrates that typically obtaining a single quote/reset pair is enough to reveal the secret version: def find_versions(quote,reset): """ Brute force search for version keys that produce reset from quote """ versions = [] for version in range(256): ...


2

It depends on the particular scheme. In a simple challenge-response scheme, the server sends a challenge, the client "encrypts" the challenge and the password to form a response, and then the client sends the response over the wire. However, standard logins typically don't use challenge/response. The password is just sent to the server by the client. When ...


2

Reading the original paper, I figure out the question. This voting scheme employed the well-known undeniable signature scheme, proposed by Chaum and Van Antwerpen in 1989 (or Chaum 1990 or Chaum and Van Antwerpen 1991). KeyGen: The RA is a signer and has a public key $X = g^x$ and a secret key $x$ Sign: For a message $m \in \mathbb{G} = \mathbb{Z}_p$, the ...


1

Unfortunately, you are probably not going to be able to fill in the missing details, unless you have a great deal of crypto experience (which it sounds like you don't have). You could start by reading about zero-knowledge proofs. There's a lot of information on that subject available. You will need to know it before you can progress. It sounds like you ...


1

You need to first understand the security properties you are trying to achieve. Possible security goals: It shouldn't be possible to answer convincingly if you don't know the secret If you do know the secret and provide a response, an eavesdropper shouldn't be able to extract the secret from seeing the challenge and response The challenger shouldn't be ...



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