today at 3:57 AM
> It would be good to have a formalised proof of the o(n log n) result, for example in Lean. As far as I can tell, one does not yet exist.
The result is fully formalized in Lean by @platypii and myself (on Mathlib's TM2 machine), with ongoing improvements to the exponent (currently at 1-2^-14.38): https://github.com/platypii/integer-mult-bounds-lean
Contributions are welcome!
today at 3:04 AM
> The new result does not directly imply new complexity bounds for polynomial multiplication over a fixed finite field, say multiplication in F2[x]. The state of the art for this problem is an algorithm that multiplies polynomials of degree n in time O(n log n 4^log*(n)) [HvdH19].
@platypii and I obtained a Lean proof of the same improvement for DFT over prime-order finite fields (yielding polynomial multiplication via the convolution theorem): https://github.com/platypii/integer-mult-bounds-lean/blob/ma...
today at 2:50 AM
Note in this case the title says "O(n log n)", but the important part is this is "o(n log n)", not just "O(n log n)'. The former is a strict subset of the latter. We already knew how to multiply two n-bit integers in O(n log n) time, via the FFT.
today at 3:12 AM
Roughly speaking, O(n log n) means "grows no faster than k(n log n) for some k", whereas o(n log n) means grows strictly more slowly* than k(n log n) for all* choices of k.
today at 2:55 AM
For those of us uninformed, what’s the difference between O and o?
today at 3:17 AM
Knuth originally defining them https://dl.acm.org/doi/10.1145/1008328.1008329
today at 3:07 AM
f(n) = O(g(n)) means as n increases, f(n)/g(n) must not grow beyond a constant
f(n) = o(g(n)) means as n increases, f(n)/g(n) must shrink
today at 3:29 AM
Nitpick:
s/must shrink/must eventually shrink, after finitely many n/
Equivalently, must go to zero in the limit.
today at 4:02 AM
depends on what you think shrink means. If you mean decreasing, that's false even eventually.
today at 2:59 AM
Big O basically drops constants
O(2n+1) = O(2n) = O(n)
o(2n+1) > o(2n) > o(n)
Both define upper boundary, but big O can read like "the function is generally proportional to ___" and little o can read like "the function is never more than _____"
today at 3:13 AM
that's not the convention I know... (from CLRS algorithms book).
Big O is sort of like <=. little o is sort of like <.
https://dl.acm.org/doi/10.1145/1008328.1008329
today at 3:21 AM
yes, GP is slightly off, but the idea of <= and < is accurate.
O(n) is: "there exists a K such that, for sufficiently large n, K*n is always larger".
o(n) is: "for every K, for sufficiently large n, K*n is always larger". (So o(2n+1) and o(2n) and o(n) are the same thing.)
today at 2:59 AM
If f is O(g) and o(h) then
Lim f/g -> C a constant
Lim f/h -> 0today at 3:29 AM
> Lim f/g -> C a constant.
This isn't quite correct. O(g) doesn't require the limit to exist.
today at 3:03 AM
Lmao! If someone is asking the difference between big and little o notation this is absolutely not the right way to clarify it.
If you were joking then maybe you got me
today at 3:22 AM
I do hope you can find and reattach your ass; that does sound painful.
today at 3:02 AM
Both are upper bounds. But small o is strictly smaller than upper bound. Big O is not greater than upper bound.
https://stackoverflow.com/questions/1364444/difference-betwe...
today at 2:58 AM
ty, fixed. (ahh, copy paste.)
today at 4:07 AM
> Executive summary (longer version) > The bottom line is that I think the claimed result is plausible. I have not actually read the paper.
today at 3:44 AM
> I have not actually read the paper. In fact, there are three closely related papers [TC26a], [TC26b], [TC26c], and I have not read any of them. I tried, but soon gave up. The math(s) dump does not appear to be strong on exposition. Instead, I asked Fable to explain [TC26a] to me. Fable seems to understand the argument and thinks that it is correct.
this does not feel like a healthy place to be
today at 4:07 AM
Fable was able to help a human understand the paper though, so the situation is quite a bit better than if we just had a lean proof that was apparently correct but for which all attempts at explanation were "hallucinations". There doesn't seem to be an apparently obstacle to a human fully understanding the proof in several months of focused study.
today at 3:51 AM
Why not? If mathematicians in the field cannot understand the papers, but tools at our disposal can, why not use them? What is "unhealthy" about that? Just seems like it's the reality of the world now.
today at 3:56 AM
Fable is the "gadget" of language ELI5.
Honestly it is not that much different from the divine inspiration if you have ever experienced it.
I do not know how you view the world or walk through it but language in general is simply subject to interpretation to me, the arrangement of words for the sentence is inspiration in and of itself - though the majority of time it amounts to minimum jumps.
today at 3:30 AM
Awesome write up and intro by a fellow Sydney-sider, kept me hooked! As someone who's not a maths wizard, the technical details flew by my head but the prose allowing me to empathise deeply with those in the field was very enlightening. I never realised how big of a deal the maths dump was and how much it has shaken up the field. I guess we are all just trying to figure out this whole "intelligence on tap" thing together.
today at 3:41 AM
You don’t need to be a math wizard. Just upload the paper to an AI and ask, “Explain this like I’m five.” Then, with each follow-up prompt, increase the age and level of expertise you want it to assume.
today at 4:02 AM
That algorithm won't converge anytime soon. A pre-conditioner (i.e. at least some apprentice wizardy) helps.
today at 3:17 AM
Really enjoy the writing style here.