[HN Gopher] Group actions and hashing unordered multisets (2021)
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Group actions and hashing unordered multisets (2021)
Author : KqAmJQ7
Score : 59 points
Date : 2024-06-23 13:07 UTC (3 days ago)
(HTM) web link (www.jeremykun.com)
(TXT) w3m dump (www.jeremykun.com)
| contravariant wrote:
| Interesting, not often you see a non-associative variant of
| commutativity. It confused me for a bit that 'h' itself is not
| commutative, but summing arbitrary sequences _is_ order
| independent provided you start with the same seed and sum from
| left to right.
|
| Edit: Not sure the definition of phi is right though, once you
| have h(a, h*(T + S)) you're pretty much stuck since the
| commutativity doesn't allow you to rearrange things from that
| point. I think I understand the gist, you want to start
| accumulating from a different seed, except that h(a, h*(T)) is
| not the hash of T if you replace the seed with a. You'd need
| something like: h_s*({}) = s h_s*(T +
| {x}) = h_s*(h_s(T), h(x)) phi(T) = (a => h_a*(T))
|
| then commutativity could be written
| h_(h_s*({a}))({b}) = h_(h_s*({b}))({a})
|
| which is slightly more symmetric, but maybe not better.
| rjeli wrote:
| There's a nice writeup on group hashes here:
| https://cronokirby.com/posts/2021/07/on_multi_set_hashing/
|
| In particular, if you choose a group where discrete log is hard
| (such as prime order elliptic curves), multiset hashing falls out
| for free
| ihm wrote:
| I think there are a few errors here. First there is afaict no
| reason the image of phi has to break up into power-of-two cyclic
| groups.
|
| Second and more importantly, it seems very difficult to start
| with the decomposition into cyclic groups and then choose a map
| from the multiset group into the permutation group that
| corresponds to the given decomposition in a good way.
|
| Relatedly, the isomorphism between the image of phi (i.e., the
| action of accumulating hashes) and the decomposition into cyclic
| groups may be difficult to compute, which can make finding
| collisions infeasible for an attacker when they could do it
| easily if given the explicit representation.
|
| So overall the conclusion that "you might as well make this
| forced structure explicit, and just pick the block structure you
| want to use in advance" seems incorrect.
|
| The blog post someone linked on multiset hashing with elliptic
| curves proves the foregoing points. The cyclic groups do not have
| power-of-two orders and the group action is very complicated even
| though the description in terms of elliptic curve addition is
| quite simple.
| contravariant wrote:
| I think they skip a few steps, but in this derivation im phi is
| exactly Z/2^nZ so every subgroup should have cardinality 2^k
| for some k.
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