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This isn't unique, or even the least compute way to do this. For example, let f(x,y) = 1/(x-y). This too is universal. I think there's a theorem stating for any
by SideQuark 6mo ago
This isn't unique, or even the least compute way to do this. For example, let f(x,y) = 1/(x-y). This too is universal. I think there's a theorem stating for any finite set of binary operators there is a single one replacing it.
write x#y for 1/(x-y).
x#0 = 1/(x-0) = 1/x, so you get reciprocals.
Then (x#y)#0 = 1/((1/(x-y)) - 0) = x-y, so subtraction.
it's common problem to show in any (insert various algebraic structure here ) inverse and subtraction gives all 4 elementary ops.
I haven't checked this carefully, but this note seems to give a short proof (modulo knowing some other items...) https://dmg.tuwien.ac.at/goldstern/www/papers/notes/singlebinary.pdf https://dmg.tuwien.ac.at/goldstern/www/papers/notes/singlebi...
- doctorpangloss 6mo agoyes, but are you currently experiencing both hypergraphia and chatbot AI induced psychosis while also thinking about this problem?
- SideQuark 6mo agoIt's math. You can check it yourself instead of this (and many other) thoughtless posts.
- doctorpangloss 6mo agoi'm mocking the LLM-generated scientific article you're replying to, not you. i'm agreeing with you haha
- SideQuark 6mo agoDo you claim all things you don’t understand are LLMs? This is what I mean by these and many of your other comments being extremely poor quality to the point of deliberate ignorance. The paper above was published in 2012 [1], so that’s quite a feat for an LLM. This takes about zero effort to check. Put some thought or effort into your claims; they’ll look less silly. [1] https://orcid.org/0000-0002-0438-633X https://orcid.org/0000-0002-0438-633X
- deleted 6mo ago[deleted]
- ano-ther 6mo agoGood find. It cites a paper from 1935: https://www.pnas.org/doi/10.1073/pnas.21.5.252 https://www.pnas.org/doi/10.1073/pnas.21.5.252 Here is a bit more: https://mathoverflow.net/questions/57465/can-we-unify-addition-and-multiplication-into-one-binary-operation-to-what-exte https://mathoverflow.net/questions/57465/can-we-unify-additi...
- SideQuark 6mo agoOoh, that 2nd link has a nice construction by Terry Tao giving a clear way to show infinitely many such functions exist for pretty much any set of operations.
- TimTheTinker 6mo ago> This too is universal Could that be used to derive trigonometric functions with single distinct expressions?
- SideQuark 6mo agoThe exp and ln are infinite series. Exp is roughly the infinite series for cos AND the infinite series for sin. Hiding that every op is an infinite series behind a name doesn’t make things free. It just makes even trivial ops like 1+2 vastly more work.
- freehorse 6mo agoThey are not infinite series per se. They can be represented by infinite series in several ways but there are standard ways to define them that do not involve infinite series. The logarithm in particular is not even represented by an infinite series (in form of Taylor expansion) defined in the whole complex plane. And knowledge/use of trigonometric functions greatly precedes such infinite series representations. Moreover, the point is not always numerical computation. I don’t think anybody argues that eml sounds like an efficient way to compute elementary functions numerically. It may or may not still be useful for symbolic computations. The article is about producing all elementary functions, which 1/(x-y) clearly doesn’t, as it doesn’t produce any transcendental function. Like many of such universality-style results it may not have practical applications, but may still be interesting on its own right.
- SideQuark 6mo agoAny transcendental function can be produced by arithmetic, since its complete for R. Go ahead and show how to compute exp or ln without an infinite series without circular reasoning. You can’t, since they’re transcendental. There are infinitely many ways to make these binary operators. Picking extremely high compute cost ones really doesn’t make a good basis for computation.
- freehorse 6mo ago
- marton78 6mo agoYou win the internet today!
- LolWolf 6mo agoI don't think this can do any of the "standard" constants or what we generally consider to be closed-form expressions, though ! (E.g., no e, pi, exp, log, etc.)
- SideQuark 6mo agoYes it can, by using the same infinite series that exp and ln use to compute. This one just costs less in money, hardware, energy, and is faster for basically every basic op.
- LolWolf 6mo agoI think the point is that it is _finite_. if you allow infinite expressions then the basic monomial basis or quotients thereof are “even simpler”
- SideQuark 6mo agoIt’s only finite by putting the infinite series into an operation. And the basic monomial basis is not a single binary operation capable of reproducing the set of basic arithmetic ops. If you want trivial and basic, pick Peano postulates. But that’s not what this thread was about.
- LolWolf 6mo agowell, the statement is: is there a single operation, built from elementary operations, such that all _other_ elementary operations have finite representations. this preprint answers that in the affirmative otoh, (x, y) -> 1/(x-y) does not answer this question at all. you can argue that the preprint does so "via the infinite series in an operation" (which I have no idea what that means; surely if exp(x) qualifies then so must 1/(x-y) if we pick a monomial basis?) but ¯\_(ツ)_/¯ now, do I think that this is groundbreaking magical research (as I'm currently seeing on twitter) no... But it's neat!
- strbean 6mo agoI think this is the novel bit: > This includes constants such as e, pi, and i; arithmetic operations including addition, subtraction, multiplication, division, and exponentiation as well as the usual transcendental and algebraic functions.
- SideQuark 6mo agoAnd those come from the infinite series needed to compute exp and ln. They’re just as much work either way. The exp and ln way are vastly costlier for every op, including simply adding 1 and 2.
- deleted 6mo ago[deleted]
- krick 6mo agoIt's not about being costly or not, this is completely irrelevant to the point being made. eml is just some abstract function, that maps ℝ² to ℝ. Same as every other mathematical function it is only really defined by the infinite set of correspondences from one value to some other value. It is NOT exp(x) - ln(y), same as exp is not a series (as you wrongfully stated in another comment). exp can be expressed (and/or defined) as a series to a mathematician familiar with a notion of series, and eml can be expressed as exp(y) - ln(y) to a mathematician familiar with exp and ln. They can also be expressed/defined multiple other ways. I am not claiming this is better than 1/(x-y) in any way (I have no idea, maybe it isn't if you look closely enough), but you are simply arguing against the wrong thing. Author didn't claim eml to be computationally efficient (it even feels weird to say that, since computational efficiency is not a trait of a mathematical function, but of a computer architecture implementing some program) or anything else, only that (eml, 1) are enough to produce every number and function that (admittedly, somewhat vaguely defined) a scientific calculator can produce. However, I want to point out that it's weird 1/(x-y) didn't appear on that graph in Figure 1, since if it's as powerful as eml, it should have all the same connections as eml, and it's a pity Odrzywołek's paper misses it.
- 6mo ago
- lemonwaterlime 6mo agoI’m fairly certain that the difference between the approaches is that the f(x,y) function you mentioned requires limits to represent certain concepts while the eml approach is essentially a tree or a chain of computations meant to represent a model of a system.
- SideQuark 6mo agoComputing exp or ln is an infinite series, and vastly more compute. Hiding series behind a name doesn’t make them free to compute.
- lemonwaterlime 6mo agoI understand your point. The paper is more about the depth of the tree to represent and audit a model versus the raw CPU clock cycles. It takes the exponent and logarithm as given since for all practical purposes, in a scientific context, they are. To represent something like sin(x) with f(x,y) requires infinite steps. Conversely, with eml you get an exact result in around 4 using identities and such. One could argue that we do Taylor Series approximations on the hardware to represent trigonometric functions, but that highlights the key aspect of the eml approach. You can write a paper with those four steps that describes an exact model, here sin(x). And people can take that paper and optimize the result. This paper is about an auditable grammar that you can compute with.
- AlotOfReading 6mo agoThey're not series, that's just a convenient way to think about defining and calculating them. I've never found it particularly useful to deal with the series definitions either, and none of the (good) approximation methods I'm aware of actually take that approach. Moreover, EML is complete in a way that your suggested function isn't: If you take a finite combination of basis functions, can it build periodic functions? Hardy proved over a century ago that real (+,-,/,*,exp,ln) can't do this (and answering the paper's unresolved question about similar real-valued functions in the negative). EML being able to build periodic functions is a lot less surprising for obvious reasons, but still pretty neat.
- gowld 6mo agoYou forgot to create numbers. You need 0 (explicitly used in your construction), and you need another number so you generate numbers that are not 0 and infinity. The OP only needs 1 number: 1.
- applicative 6mo ago1/(x-y) has nothing remotely like the power of his operator. The guy is not stupid.
- SideQuark 6mo agoIt generates the same class of functions. Read the comments and links in this thread.
- applicative 6mo agoI did.