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<hazirafel> hey yo, my computer crashed can you plz re send me what you wrote before
<hazirafel> benn talking to myself, doing better
<hazirafel> oops
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<nicolagreco> are there ways to prove that a message m is of size L, given H(m) being public knowledge?
<nicolagreco> possibly something that does not involve snarks - but a simpler primitive
<nicolagreco> it might also be fine is we have a different authenticator than H(m)
<gmaxwell> what does size L mean? I mean if I take a message 'a' and then pad it with 500 \0s ... what size does it have?
* gmaxwell bbl
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<nicolagreco> gmaxwell: that is a good question, in your example it would be 1
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<Taek> nicolagreco: what is your end goal?
<nicolagreco> hey Taek (hi!) I want to promise you that I have a file and that it is of the size that I claim
<Taek> How do I know that the "file" you have isn't just randomly generated garbage?
<nicolagreco> that is totally fine
<nicolagreco> actually let me give you more context
<nicolagreco> (I am not sure this is even practical)
<nicolagreco> actually, I think this sparks more from curiosity than practical examples
<nicolagreco> I want to be able to know before a communication starts
<nicolagreco> how long will be the message that you will send me and I want you to commit to it
<nicolagreco> so that if you send me a longer message, I can drop the conversation and penalize you
<Taek> The penalty there is the tricky thing
<Taek> oh hmm
<gmaxwell> I assume the commitment is enough to do that.
<gmaxwell> okay so splitth message into blocks. length is in units of blocks. (I trust this is fine). H(m) is a tree hash. I send you "my message will be less than X blocks, it has root hash Y-- signed greg"
<gmaxwell> now I start streaming blocks to you along with the extra tree fragments so you can constantly verify the blocks that I send you match the root.
<gmaxwell> then if I send too many blocks, you can prove it by showing the X+1 member and my signed message.
<Taek> I'm struggling to see how this could be useful though? What's stopping you from sending blocks that aren't a part of the root?
<Taek> or sending data that doesn't match a signature?
<gmaxwell> yea, got me.
<gmaxwell> I'm just complying with the spec. :P
<nicolagreco> hahahah
<gmaxwell> Thats the story of cryprocurrency, 'Your scientists were so preoccupied with whether or not they could, they didn’t stop to think if they should.'
<nicolagreco> that is a perl, I will print it
<gmaxwell> (it's a quote :P )
<nicolagreco> but actually gmaxwell, if you can have a proof that the size of a file is actually L
<nicolagreco> then I can force you to prove me how long is the file of which I know the hash
<nicolagreco> before I even start receiving messages from you
<gmaxwell> I still think your 'how long' is kind of ill defined. e.g. make H() a tree hash and then you can have me show you the last block... to 'prove the length'. but I think that proof is mostly meaningless in most contexts.
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<nicolagreco> I agree on the ill defined..!
<nicolagreco> if you show me the last block of a merkle tree, how do I know it proves the length? you could always have hashed the previous blocks with different block sizes
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<Taek> that's easy enough to prevent
<Taek> just require a fixed block size, and if you get something the wrong size you have your proof
<Taek> (assuming each block is signed)
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<Taek> (but idk how to force a signature)
<nicolagreco> yup that works - although for you to convince me, we must have interaction
<nicolagreco> (which I actually proposed before)
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<nicolagreco> so I guess that should be the requirement, and the question now would be "can I prove without interaction that the file that I hold is of size L?"
<nicolagreco> any way going off to sleep now, thank you Taek and gmaxwell for the right direction
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<vega4> I havent read everything but
<vega4> it is not and it should NOT be possible to get to know the size of the message knowing its hash; if you could then the hashing function would be broken
<gmaxwell> that isn't what he asked for.
<gmaxwell> he has for you to be able to prove it, this would be done by sending additional information along side the hash.
<vega4> sorry; thats why I said I haven't had time to read everything but the very subject sort of stroke me
<vega4> he wants the server to prove what;
<vega4> client is sending a hash, the server has a file; and what is the server to prove
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<vega4> if the server is to prove it is in posession of the data; sending part of the file to the client would give nothing as long as server sends 100% of the file. if you devide the file into blocks and send a hash of a single block. then this would work.but then you would need a lot of hashes; a sufficient number that it would be infeasible for the server to store every chunk of data repreented buy a set of these hashes. or m
<vega4> server to guess which hashes you've got. you would need a high resoluton of blocks
<vega4> then I've read about some 'phantasy' like proof-of-storage papers
<vega4> when ones qyery for the content from time to time
<vega4> but I think browdly speaking it works similarly to what I've described
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<vega4> I simply cant imagine any other way of proving pesession on bare data without random query for some chunks. and client does not want to store every chunk so he stores hashes of blocks
<vega4> and then some other parties query from time to time the server faisl to respond..ho loses stake blabla
<vega4> I do not even need to read the paper
<vega4> but that is phantasy
<vega4> PoW is the only proof so far to me
<vega4> it seems to me that pople noticed a flying bitcoin and now would also like to have a flying cat
<vega4> proof-of-a-flying-cat
<vega4> lets stick some wings to it and make it fly
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<vega4> and now... now I'm going to write a paper on 'PROOF-of-storage' no wait... I'm going to start a campaign just like Safe coin and get 20mln usd
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<FNinTak> there is forced interaction in the case of receiving what looks like a correct proof though, right?
<FNinTak> I.e. if you prove to me the size of some file, & I choose to store it, then you should send file, & possible more verbose proof
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<vega4> FNinTak: explain to me once again; maybe I'm tired
<vega4> 1) what do you mean by a forced interaction 2)what do you mean by a more verbose proof
<vega4> which one was less verbose
<vega4> ah you mean Bob is sending a File to Allice; Alice accepts the byte stream; and then Bob is able to query Allice for some parts of the file
<vega4> but isnt that what I decribed?
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<vega4> that is correct; but it has nothing to do with driving file size from hash which was the Op's topic I think
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<vega4> anyway; forget it;gmaxwell said that wasn't the case
<vega4> " are there ways to prove that a message m is of size L, given H(m) being public knowledge?" that was the case after all I think
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<vega4> its funny to see how over-compicated looking these papers might be
<vega4> Micorosft Research:)
<vega4> John Hopkins Maryland loooook at THAT
<vega4> I believe this article is the only one in the Internet where the term "Homomorphic Identification Protocol" is present
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<vega4> Endomorphicly Homomorphic Computationaly Feasible Identification Protocol Based On Factorizan in Finite Fields
<vega4> *Factorization
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<Taek> Is there a word for a hash function that doesn't offer collision resistance?
<Taek> And would such a function still be useful for things like random number generation?
<bsm1175321> Hash functions used to implement e.g. sets or hash maps generally have small output lengths only marginal collision resistance, and they're generally optimized for speed.
<bsm1175321> The tradeoff is adding another bucket vs. the time to hash. e.g. Google's FarmHash
<bsm1175321> They still need collision resistance though because if an attacker can find collisions, he turns your algorithm from O(1) to O(N) and it's a DoS vector.
<bsm1175321> I would think such hash functions would be very poor PRNGs...
<Taek> yeah but those are hash functions that don't propose to securely provide the properties of preimage resistance and partial preimage resistance
<Taek> also, preventing collisions is easy enough if you can keep a private salt
<Taek> (useful for hash tables, not necessarily for other things)
<bsm1175321> So you're proposing a hash function which has preimage resistance but poor collision resistance? Doesn't one imply the other?
<Taek> attacker generally needs to know the input to find a collision in a reasonable amount of time
<Taek> they don't imply eachother, I know this b/c md5 offers preimage resistance but not collision resistance
<andytoshi> Taek: there are definitions in the literature for one-way functions, pseudorandom functions, pseudorandom generators, etc
<andytoshi> we usually use "hash function" as "random oracle" here because being a random oracle implies basically every property that you might need, and also it makes a lot of proofs possible
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<andytoshi> but non-random oracle proofs generally specify exactly the properties of their hash functions that they need
<andytoshi> (and they use "hash families" which are indexed sets of hash functions, and they have fixed-width inputs, and in general this area of academic crypto seems _designed_ to be totally unreadable)
<bsm1175321> Taek: Does your statement arise due to length extension attacks?
<Taek> bsm1175321: "On 30 December 2008, a group of researchers [snip] had used MD5 collisions to create an intermediate certificate authority certificate that appeared to be legitimate when checked by its MD5 hash"
<Taek> err, wrong heading
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<Taek> andytoshi: thanks :)
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<bsm1175321> Taek you led me down a rabbit hole that ended up on XOF's (eXtensible Output Functions) -- hash functions with variable sized output, like SHAKE256.
<bsm1175321> Seems to me that Merkle tree (and similar constructions) which hash together a set of data, actually NEED an XOF instead of a fixed size hash, since if you hash N inputs together, the hash's security should be 256-log(N) roughly.
<bsm1175321> That is, if you make another Merkle tree with N inputs, the probability of a collision isn't 2^-256, it's lower because of the amount of inputs.
<bsm1175321> Given the large amount of data that goes into a blockchain, perhaps an XOF is an interesting object for creating Merkle trees, MMR's/Merklix type constructions, where the Merkle root has size e.g. 256+log(N) where N is the number of data elements that went into it.
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<Taek> I don't follow. Why is the security weaker for a merkle tree?
<Taek> I don't believe that it is
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<bsm1175321> It might similarly be an interesting construction for PoW...where the PoW proof size is extended by the number of leading zeros...
<bsm1175321> Taek: maybe Merkle trees are a bad example...there is a higher probability that there was a collision *somewhere* in the tree, but I think you're right, the security of the root is unchanged.
<Alanius> Taek: in order to break a Merkle tree, you have to find a second preimage for any one node in the tree; that is easier than finding a preimage to one specific node.
<Taek> I guess the same follows for collisions as well then
<Taek> hmm maybe not. It should be equally difficult to find a collision regardless of node count
<bsm1175321> Exactly...an XOF can have that property, a fixed size output hash function cannot.
<Taek> Alanius: thanks, I had not realized that before
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<bsm1175321> And actually there are O(N*log(N)) hash computations to create a Merkle tree...so the length extension needs to be a bit larger than I said to maintain the same security.
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<Alanius> bsm1175321: how did you get to O(N*log(N))?
<bsm1175321> N/2 hashes for the first layer, log(N) depth...
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<Taek> For balanced trees it's just 2N-1
<bsm1175321> Yep Taek is right.
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<bsm1175321> N/2+N/4+N/8+...
<Taek> and for imbalanced trees you will do at most one extra hash per layer, so it's worst case 2N + log(n)
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<bsm1175321> Hey did we just find the first real-world use case for an XOF?
<Taek> No security is still fine using just normal hashes
<Taek> you need to use 256bit hashes so that you have collision resistance
<bsm1175321> What's the probability there's a hash collision, somewhere in one of Bitcoin's many merkle tree histories...it's not 128 bits.
<Taek> you can find second preimages faster, but only by a factor of log(n) I think, and your second preimage security is 256bits, so unless your hash tree is 2^2^128 nodes, you are still safe there
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<Taek> hmm. Well first off I don't think it matters for the most part unless the collision occurs in exactly the same spot on two different trees
<Taek> otherwise the resulting trees don't make any sense
<Taek> but even so, I'd imagine that there are less than 2^32 total hashes in all of bitcoin's Merkle trees
<Taek> so the probability of 2 of them being in a collision should be a lot better than 2^-128
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<Taek> I think it's 2^-192?
<Taek> (2^256 / 2^32 * 2^32)
<Taek> dammit. order of operations is hard ==> 2^256 / (2^32*2^32)
<bsm1175321> Taek: Why that? This is a https://en.wikipedia.org/wiki/Birthday_attack
<Taek> The birthday attack means that given n hashes, there's an n^2 chance of a collision
<Taek> n^2 / (n^hash_size)
<bsm1175321> Yes, so with a 256 bit hash function you'll have 50% probability of finding a collision after computing 2^128 hashes.
<bsm1175321> Now I wish to hold 2^128 constant as I add hashes to a Merkle tree, by extending the output size of the hashes.
<Taek> right. But in Bitcoin there are only (generously) 2^32 hashes total
<bsm1175321> So if I generate one new hash, the probability that it has a collision with an existing hash is 2^(128-32) = 2^(96). Therefore I should be using a hash function today that is larger by 64 bits = 320 bits, to hold the 128 bit hash-collision security factor constant.
<Taek> no, the probabilitiy that it has a collision with an existing hash is 2^256 / (num existing hashes)
<Taek> If you have 2^128 hashes and zero collisions, the probability of getting a collision on your next hash is 2^-128
<bsm1175321> which is a lot larger than the birthday problem number of 2^128...hence I'm confused.
<bsm1175321> Oh I see.
<bsm1175321> Taek: no, if you have 2^128 256-bit hashes and zero collisions, the probability of getting a collision on your next hash is 50%. That's the birthday problem.
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<Taek> nope
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<Taek> if you have 2^128 hashes and you haven't checked if you've had a collision yet, the probability that you have a collision is 50%
<bsm1175321> Ah I see
<Taek> because each of the 2^128 hashes has a 2^-128 chance of colliding with another hash
<bsm1175321> The constraint of having already checked that there's no collisions is important.
<Taek> right
<bsm1175321> That's not actually the case here...no one is checking that intermediate hashes in Bitcoin's Merkle trees don't already have collisions.
<Taek> right but we also don't actually have 2^128 of them yet either
<bsm1175321> So the probability that there's *already* a collision in the 2^32 hashes is 2^-96. (and that number is the same for any new one I add)
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<Taek> we've only got 2^32, each of which has at most (actually less, b/c of merkle tree structure) 2^-(256-32) chance of colliding with another
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<Taek> so each of the 2^32 hashes has a 2^-224 chance of collision. Which means 2^-192 chance total of collision in the existing tree
<Taek> assuming 2^32 hashes ofc
<bsm1175321> Your argument is not compatible with the birthday calculation. The sqrt enters because I can have a collision with *any* other hash.
<bsm1175321> Following your logic, if I had 128 hashes, I'd have a 2^-128 probability of finding a collision, and that's not the case. It's 50% by the birthday attack problem.
<bsm1175321> *if I had 2^128 hashes...
<Taek> you'd have 2^-128 probability of collision per hash
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<bsm1175321> Per hash, times 2^128.
<Taek> right
<Taek> so when it's 2^32, you've got 2^32 chances which are each 2^-224 likely to be a collision
<bsm1175321> And that's your 192...
<Taek> right
<bsm1175321> Ok I think I agree.
<bsm1175321> Then we'd want to length-extend the hash output back to 2^-256...
<Taek> no, it's fine. 2^-128 is considered secure
<bsm1175321> So the required hash size to retain a fixed collision parameter is S+2*log(N) where S is the size of the starting hash function with no elements in Merkle trees, and N is the number of hashed elements so far.
<bsm1175321> Taek: I know. I'm just interested abstractly in what's required to fix this security parameter.
<bsm1175321> One might make the argument the other way...you could get away with a 16-byte hash function for speed, if you length-extended it this way.
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<MaxSan> has anyone ran any blockchain analysis on the spam that happens on the network?
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