A New Formula for the Determinant

https://arxiv.org/abs/2301.06586

by scentoni • 4 years ago

95 33 4 years ago

arxiv.org

scentoni• 4 years ago

A new explicit formula for the determinant that contains superexponentially fewer terms than the usual Leibniz formula

esperent• 4 years ago

What does "superexponentially fewer" mean?

quchen3• 4 years ago

»More than exponential«.

The standard formula for determinants of an n×n matrix (equation 1 in the paper) is the sum of n! terms, one term for each permutation of {1..n}.

The paper details a method (equation 11) in which the sum is instead over all what they call »partial partitions of {1..n}«, denoted PP(n). I don’t know the size of PP(n), but if their claim is correct then PP(n) has exponentially less elements than S(n) (which has n! elements). I don’t see the value of |PP(n)| in the paper, but if it’s O(n^42), that would be an example of something that’s superexponentially smaller/faster. (n! is approximately n^n/e^n, the Stirling Formula.)

ur-whale7• 4 years ago

Very interesting, but ... another paper without code.

beyondCritics• 4 years ago

It's about higher math dude.

ur-whale7• 4 years ago

It's about higher math dude.

It's about a formula (another word for algorithm) for computing the determinant, it's not like they shat a new theorem in category theory.

And however higher-mathy that may be, a link to a chunk of python code implementing the formula certainly would not have been detrimental to the paper.

Unless of course all of the other higher math folks would look down upon a despicable math researcher that would dare spoil a perfectly honorable higher math research paper by adding disgustingly usable code to it.

Think of the insult!

People could even get down to running the code just to check that the formula actually works !

Ivory tower much ?

kzrdude• 4 years ago

I'm a layman, but this seems to be a very approachable presentation of a paper.

Section titles are approachable and it has figures and graphics with geometric ideas.

psychphysic• 4 years ago

And equation 12 shows an expansion of the formula given in eq11.

vore• 4 years ago

The formula is literally in equation 11.

zython• 4 years ago

The code is trivial and left as an exercise to the reader. /s

quchen3• 4 years ago

This, but unironically. Their formula (11) involves basics (sums and products), the only nontrivial part would be the partial partitions, and for that the first paragraph of chapter 2 explains how to get it from ordinary partitions, which is a standard function.

ur-whale7• 4 years ago

The code is trivial and left as an exercise to the reader. /s

Yeah, the margin was probably not wide enough.

zuzatm2• 4 years ago

I'd argue that the results of this paper are not meant to be implemented, hence no code required.

abetusk• 4 years ago

I'm sorry, I'm having a hard time parsing this paper. It looks like even calculating "partial partitions" is exponential. Have they found a nice polynomial time algorithm to find determinants or is this just an exponential algorithm but with fewer components than what is "traditional"?

FYI, The Bareiss algorithm can calculate the determinant in polynomial time, including bounding the bit complexity to polynomial space [0].

[0] https://en.wikipedia.org/wiki/Bareiss_algorithm

steppi• 4 years ago

From my read of the paper. This new formula also gives an exponential time algorithm. It’s not of interest as a practical method for computing determinants. It does however have interesting geometric and combinatorial properties and they used it to give new bounds for the tensor rank and Waring rank of a determinant respectively. You can think of these ranks loosely as measures of the “complexity” of a determinant (but not in the sense of computational complexity). These ranks are of interest in pure mathematics but I can’t really say more about them because it’s been too many years since I’ve studied related math.

robinhouston3• 4 years ago

Oh wow! I wasn’t expecting to see this on HN. I’m one of the authors of the paper, and I’m happy to answer any questions. (I’m sorry I didn’t see it earlier, it was posted in the middle of my night.)

First of all, to clear up a possible misunderstanding: this new formula will not help you to compute determinants more quickly. The number of terms still grows a lot faster than polynomial, and if you just want to compute the determinant, there are polynomial-time algorithms for that.

The best theoretical results on the complexity of computing determinants, that I know of, are in [0]. In practice you'd probably use Gaussian elimination, or maybe Bareiss for integers.

Someone complained that there’s no code in the paper. Since it isn’t really useful for practical computation, I didn’t think code would be useful, but in fact it did start as a Mathematica program when I first discovered it. Adam Goucher (one of my co-authors) wrote a blog post on it [1] within a few days of its discovery, which links to my original Mathematica notebook.

(Edit: ur-whale points out that the link to the Mathematica notebook no longer works. I have now republished it at [2].)

[0] https://users.cs.duke.edu/~elk27/bibliography/04/KaVi04_2697...

[1] https://cp4space.hatsya.com/2022/07/02/a-combinatorial-proof...

[2] https://www.wolframcloud.com/obj/robin.houston/Published/det...

ur-whale7• 4 years ago

Someone complained that there’s no code in the paper.

That was me, apparently much to the dislike of a number of HNers :)

Thank you very much for the pointer to the notebook although it is not easy to access, you seem to need a Wolfram account to get to it.

[EDIT]: "Sorry, you do not have permission to access this item." after I logged into Wolfram ...

Since it isn’t really useful for practical computation, I didn’t think code would be useful

This is a very common misconception.

There is a large category of people out there (software engineers among other) that find it much easier to read code than reading abstract math formulas full of weird greek letters and which typically pre-suppose what the reader does and doesn't know (things that are considered "trivial" by the author and not worth mentioning in the paper is often a huge obstacle for the reader to proceed with understanding what goes on).

Code is explicit to the degree that a machine can execute it, and therefore often way easier as a path to understanding an idea.

Publishing code also has the minor benefits of actually exposing things that are claimed to work but often don't - again because a machine can execute it whereas a math formula in a PDF does not have that nice property.

robinhouston3• 4 years ago

Oh, what a pain. Sorry. It looks as though Wolfram Cloud removes published notebooks after 60 days unless you pay extra.

I’ve just republished it at https://www.wolframcloud.com/obj/robin.houston/Published/det...

I’d be interested to know whether the code _does_ make it easier for you to understand.

otikik• 4 years ago

Also keep in mind that there’s people out there that will take your code and make it 40.363.757.265% more efficient (perhaps not in this case)

https://youtu.be/c33AZBnRHks

VyseofArcadia2• 4 years ago

My experience in math unfortunately is that code is just about unpublishable. I'd love to be able to just have relevant code in a paper, but editors will reject it.

They'll also reject you for having too much discussion and explanation. Now that almost all "papers" are PDFs and we don't have to worry about printing costs, I worry that the expected brevity in math is just elitism. Sure, if you're one of the six experts in the sub sub subfield the paper concerns itself with, the brevity is welcome, but for everyone else...

VyseofArcadia2• 4 years ago

I'm a former academic whose research area was matrix theory. I left for the dark side and do software engineering these days.

I'm just jazzed to see a math paper in my wheelhouse on the front page of HN.

zuzatm2• 4 years ago

Minor comment: the improvement over the state-of-the-art is in the second order term of the exponent.

Their formula express det. with B_n = exp( n ln n - n lnln n - O(n)) terms.

The old formula had exp(n ln n - O(n)) terms.

It's a nice bound, but won't change much for non-theoretical results.

A New Formula for the Determinant

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