# Antony Vennard

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bio website vennard.org.uk location United Kingdom age 25 member for 2 years, 8 months seen Jan 25 at 22:33 profile views 127

# 89 Comments

 Jan12 comment Two mutually untrusted parties want to exchange data: how to ensure each one gets the data it needs? Hi @David, I merged the duplicate cross-posted question into this one as they're identical and there were good answers there - however, you've answered both! Two answers isn't a problem, just letting you know so if you want to make any edits/amendments in light of the merge, you can. Jan12 comment Two mutually untrusted parties want to exchange data: how to ensure each one gets the data it needs? @IlmariKaronen now done. Jan10 comment RSA cracking: The same message is sent to two different people problem I know it's just an example, but to extend Thomas' point, also note the modulus itself is prime, allowing you to decrypt a single message by computing the inverse of $e$ - in other words, you don't actually need the same message to be sent to two different people in this scenario - the inverse of $e=13$ is $d=137$ and you can compute $127^{137} \equiv 10 \mod(179)$. Jan7 comment How can I break REDSHIRT / REDSHIRT2 encryption? For what it's worth, whilst not originally asked here I'm reasonably happy this is an OK question. It's about a used in-the-wild algorithm, albeit not a very good algorithm by the answers given. The key part that line in the FAQ was to address questions that say: "here is some data 0x123456 0x123456 0x123456 can anyone decode it?". What this question is really asking is "can data encrypted by REDSHIRT be decrypted trivially?". That said, if anyone disagrees feel free to start a question on meta. Jan5 comment Cycle attack on RSA @EmilioFerrucci well the order of $e$ will be $k$ since $e^k = 1$ is the solution you're trying to find. $e$ can't be a subgroup on its own as it doesn't have an identity element wrt multiplication. I think you might be thinking of the notes here - finding $k$ is at least as hard as factoring unless $ord(\mathbb{Z}_{\phi(n)}^{\times})$ is known. Of course, finding that order is going to be hard in and of itself - you don't know how the group is generated in the first place (remember $\phi(n)$ is (part of the) the private key). Dec21 comment Does this block cipher mode allow for decryption? @poncho I see what you mean - I was probably a bit hasty on the close, sorry. Have edited your interpretation in, feel free to answer. Dec13 comment How do I solve this RSA instance for m? Welcome to crypto! Your question was migrated here as it is more on topic here than on SO. Plus we have TeX! If you want to register your question will be associated with your account and you can receive notifications about it and reputation from it as normal, and be able to comment on it. Dec13 comment What is the importance of Modular arithmetic in cryptography? I'd add that the Rijndael S-Box is designed to be the set of multiplicative inverses under a finite field $GF(2^8)$ - so whilst modular arithmetic is not strictly necessary for a symmetric cipher it certainly does find use in them. Nov30 comment How can I make my cipher show the avalanche effect? I've removed comments along the lines of "don't design your own cipher" - here on crypto it's perfectly acceptable to try, although you should understand it is all at your own risk of course :) I've also edited the question a little to focus more on the avalanche effect in the absence of the relevant cipher constructions. If anyone feels that is unnecessary, feel free to roll back and or improve on what I've done. Nov30 comment Is it a good idea to use bitwise XOR on a set of MD5 sums? If I understand this right - you're looking to use this method to ensure that two different aggregate values imply that the total source data remains unique? I.e. you're wondering if xoring md5 hashes will result in collisions over the data you're aggregating? If so, it might be worth adding that to the question perhaps? Just a thought - the more detail on these sort of things the better the answers, generally. Nov23 comment Security of simple xor and s-box cipher? @Polynomial for generally talking about function notation, have a read of the wikipedia article. It provides an awful lot of detail. In this case, $E$ is a function that takes $K$ and $M$ and produces a result $M$, where $K$ and $M$ are the domains available to the function as opposed to specific function operations, and M is the range. Have a read of codomains for more examples of this type of notation in use. Nov23 comment Security of simple xor and s-box cipher? @Polynomial Thanks. I think it now works - general question with specific example. I'm even more happy with that way than the way we had it before. Thanks for bearing with us and very sorry for the noise - but when we see a question with a lot of potential we'd rather make the effort to help it out than not. Nov23 comment Security of simple xor and s-box cipher? @Polynomial I think it works with the two questions separately. I think it's perfectly fine to link them together in the question (like "I've asked about my sbox design here {link}"). Nov23 comment Security of simple xor and s-box cipher? As is, I'm itching to revert that edit (preferred option) or close whilst we work out how best to accommodate this. Don't get me wrong, I want to accommodate this question, I just don't think one huge open question is the way to go. Specifically, from the FAQ: "Your questions should be reasonably scoped. If you can imagine an entire book that answers your question, youâ€™re asking too much." Nov23 comment Security of simple xor and s-box cipher? Guys - I realise there's a need to discuss ciphers sometimes, but asking to review an entire cipher is very broad and likely to end up in discussion; that's why Paulo asked for it to be split up in the first place. SE is not about having discussions as a format. I think we'd broken the essential components down into two separate questions nicely. The danger is that allowing "please review my cipher" generally means we'll have 20 such questions in a week. If you want to debate this, meta is the place for it. Nov19 comment How does a chosen ciphertext attack work, with a simple example? Hello, welcome to crypto.se. I've sent you a message in chat too, but just in case, if you register your account with us, you'll be able to pick up your response notifications and reputation here too. Oct28 comment Why would you expect to find a collision in a hash function after approximately $\sqrt{n}$ hashes? Hello, welcome to Crypto - you'll need to register your account here to pick up your question and rep! :) Oct26 comment How can rainbow tables be used for a dictionary attack? @ThomasPornin Ok, I'll make it more impressive...! That's not a bad way of looking at it, but I like to remember that it is possible to build a rainbow table incorporating all possible salts too (it's just more $\mathbb{P}$) it just becomes so costly we're not yet in a position to make it happen. I sort of alluded to the building cost of hashtables when mentioning slow hashing, but I didn't have any specific. Mind if I incorporate yours? Oct26 comment How can rainbow tables be used for a dictionary attack? @ThomasPornin ah, thanks, I've used the wrong formula, looked that bit up quite hastily. Will fix. Also, yes, you could compress it down, but even so, a 400GB table for just 6-digit passwords is huge. And we haven't added in 5 and 4 digit passwords either; it only counts passwords of exactly 6 digits. Oct11 comment Encrypt-then-MAC Confidentiality, Integrity and Authenticity @JohnBlack yes. In general $m$ should be the encrypted output (which may or may not include IVs depending on the protocol etc) and $k$ the MAC key. So then you calculate your MAC from the encrypted data, which prepends the MAC according to HMAC. Then to verify, the recipient can use the same encrypted data and the key - calculate the MAC and compare. If it isn't valid, they discard the message.