EDIT: please change the article link to the most recent version (as of now still v2), it is currently pointing to the v1 version which misses the figures.
I'm still reading this, but if this checks out, this is one of the most significant discoveries in years.
Why use splines or polynomials or haphazardly chosen basis functions if you can just fit (gradient descent) your data or wave functions to the proper computational EML tree?
Got a multidimensional and multivariate function to model (with random samples or a full map)? Just do gradient descent and convert it to approximant EML trees.
Perform gradient descent on EML function tree "phi" so that the derivatives in the Schroedinger equation match.
But as I said, still reading, this sounds too good to be true, but I have witnessed such things before :)
This is amazing! I love seeing FRACTRAN-shaped things on the homepage :) This reminds me of how 1-bit stacks are encoded in binary:
A stack of zeros and ones can be encoded in a single number by keeping with bit-shifting and incrementing.
Pushing a 0 onto the stack is equivalent to doubling the number.
Pushing a 1 is equivalent to doubling and adding 1.
Popping is equivalent to dividing by 2, where the remainder is the number.
I use something not too far off for my daily a programming based on a similar idea:Rejoice is a concatenative programming language in which data is encoded as multisets that compose by multiplication. Think Fractran, without the rule-searching, or Forth without a stack.
All possible 36 distinct level-2 eml functions of one variable (the first 18 of them with entirely Real outputs, the other 18 with "intermediate" complex-valued components):
This makes a good benchmark LLMs:
``` look at this paper: https://arxiv.org/pdf/2603.21852
now please produce 2x+y as a composition on EMLs ```
Opus(paid) - claimed that "2" is circular. Once I told it that ChatGPT have already done this, finished successfully.
ChatGPT(free) - did it from the first try.
Grok - produced estimation of the depth of the formula.
Gemini - success
Deepseek - Assumed some pre-existing knowledge on what EML is. Unable to fetch the pdf from the link, unable to consume pdf from "Attach file"
Kimi - produced long output, stopped and asked to upgrade
GLM - looks ok
> A calculator with just two buttons, EML and the digit 1, can compute everything a full scientific calculator does
Reminds me of the Iota combinator, one of the smallest formal systems that can be combined to produce a universal Turing machine, meaning it can express all of computation.
derivation of -x seems wrong. we can look at the execution trace on a stack machine, but it's actually not hard to see. starting from the last node before the output, we see that the tree has the form
eml(z, eml(x, 1))
= e^z - ln(eml(x, 1))
= e^z - ln(e^x)
= e^z - x
and the claim is that, after it's expanded, z will be such that this whole thing is equal to -x. but with some algebra, this is happening only if e^z = 0,
and there is no complex number z that satisfies this equation. indeed if we laboriously expand the given formula for z (the left branch of the tree), we see that it goes through ln(0), and compound expressions.x^-1 has the same problem.
both formulae work ...sort of... if we allow ln(0) = Infinity and some other moxie, such as x / Infinity = 0 for all finite x.
I made https://github.com/nullwiz/emlvm/tree/main yesterday, for fun :^)
That's quite interesting.
Few ideas that come to my mind when reading this:
1. One should also add absolute value (as sqrt(x*x)?) as a desired function and from that min, max, signum in the available functions. Since the domain is complex some of them will be a bit weird, I am not sure.
2. I think, for any bijective function f(x) which, together with its inverse, is expressible using eml(), we can obtain another universal basis eml(f(x),f(y)) with the added constant f^-1(1). Interesting special case is when f=exp or f=ln. (This might also explain the EDL variant.)
3. The eml basis uses natural logarithm and exponent. It would be interesting to see if we could have a basis with function 2^x - log_2(y) and constants 1 and e (to create standard mathematical functions like exp,ln,sin...). This could be computationally more feasible to implement. As a number representation, it kinda reminds me of https://en.wikipedia.org/wiki/Elias_omega_coding.
4. I would like to see an algorithm how to find derivatives of the eml() trees. This could yield a rather clear proof why some functions do not have indefinite integrals in a symbolic form.
5. For some reason, extending the domain to complex numbers made me think about fuzzy logics with complex truth values. What would be the logarithm and exponential there? It could unify the Lukasiewicz and product logics.
> using EML trees as trainable circuits ..., I demonstrate the feasibility of exact recovery of closed-form elementary functions from numerical data at shallow tree depths up to 4
That's awesome. I always wondered if there is some way to do this.
For completeness, there is also Peirce’s arrow aka NOR operation which is functionally complete. Fun applications iirc VMProtect copy protection system has an internal VM based on NOR.
Quick google seach brings up https://github.com/pr701/nor_vm_core, which has a basic idea
I made a fun marimo notebook to try and derive these myself. I structured each cell in order based on the diagram at the end of the paper. It uses Sympy to determine if the function is correct or not.
https://gist.github.com/CGamesPlay/9d1fd0a9a3bd432e77c075fb8...
Stupid question maybe (I am no mathematician), but aren't exp and ln really primitives? Aren't they implemented in terms of +,-,/,* etc? Or do we assume that we have an infinite lookup table for all possible inputs?
This looks interesting. I haven't looked in-detail, but my first thought is - why hasn't this been found in the past? Surely, people have been interested kind of question for awhile?
> For example, exp(x)=eml(x,1), ln(x)=eml(1,eml(eml(1,x),1)), and likewise for all other operations
I read the paper. Is there a table covering all other math operations translated to eml(x,y) form?
Can someone explain how is this different from lambda calculus, it seems like you can derive the same in both. I don't understand both well enough and hence the question.
Interesting!
One thing I wonder now: NAND is symmetric while this isn't, could something similar be found where function(x, y) = function(y, x)?
I'm way too unschooled to say if it's important or not, but what really excites me is the Catalan structure ("Every EML expression is a binary tree [...] isomorphic to well-studied combinatorial objects like full binary trees and Catalan objects").
So, what happens if you take say the EML expression for addition, and invert the binary tree?
So, like brainf*ck (the esoteric programming language), but for maths?
Not sure it really compares to NAND() and the likes.
Simply because bool algebra doesn't have that many functions and all of them are very simple to implement.
A complex bool function made out of NANDs (or the likes) is little more complex than the same made out of the other operators.
Implementing even simple real functions out of eml() seems to me to add a lot of computational complexity even with both exp() and ln() implemented in hardware in O(1). I think about stuff sum(), div() and mod().
Of course, I might be badly wrong as I am not a mathematician (not even by far).
But I don't see, at the moment, the big win on this.
I like this guy. https://th.if.uj.edu.pl/~odrzywolek/homepage/index.html
This is neat, but could someone explain the significance or practical (or even theoretical) utility of it?
This could have some interesting hardware implications as well - it suggests that a large dedicated silicon instruction set could accelerate any mathematical algorithm provided it can be mapped to this primitive. It also suggests a compiler/translation layer should be possible as well as some novel visualization methods for functions and methods.
I couldn't find any information on this, but is it possible that given how nicely exponentiation and logarithms differentiate and integrate, is it possible that this operator may be useful to simplify the process of finding symbolic solutions to integrals and derivatives?
Halfway through I was imagining aliens to whom this operator comes naturally and our math is weird. By the end I found out that we might be those aliens.
What would physical EML gates be implemented in reality?
Posts like these are the reason i check HN every day
Interesting, but is the required combination of EML gates less complex than using other primitives?
“ Elementary functions, for many students epitomized by the dreaded sine and cosine, ” dreaded?
Reminds me a bit of the coolest talk I ever got to see in person: https://youtu.be/FITJMJjASUs?si=Fx4hmo77A62zHqzy
It’s a derivation of the Y combinator from ruby lambdas
eml(x, y) pronounced... "email"?
I wonder how this combines with Richardson's Theorem.
I’d be really interested in an analysis of tau in light of this discovery. Would tau fit more naturally here than pi, as it does in other examples?
How would an architecture with a highly-optimized hardware implementation of EML compare with a traditional math coprocessor?
> eml(x,y)=exp(x)-ln(y)
Exp and ln, isn't the operation its own inverse depending on the parameter? What a neat find.
Is the the same as saying everything can be made from nand gates?
Very nice, though I'm not found of the name.
What comes to my mind as an alternative which I would subjectivity finer is "axe". Think axiom or axiology.
Anyone with other suggestions? Or even remarks on this one?
Got curious to see whether SymPy could be used to evaluate the expressions, so I used Claude Code to build a quick evaluator. Numeric and symbolic results appear to agree:
nix run github:pveierland/eml-eval
EML Evaluator — eml(x, y) = exp(x) - ln(y)
Based on arXiv:2603.21852v2 by A. Odrzywołek
Constants
------------------------------------------------------------------------------
1 K=1 d=0 got 1 expected 1 sym=ok num=ok [simplify]
e K=3 d=1 got 2.718281828 expected 2.718281828 sym=ok num=ok [simplify]
0 K=7 d=3 got 0 expected 0 sym=ok num=ok [simplify]
-1 K=17 d=7 got -1 expected -1 sym=ok num=ok [simplify]
2 K=27 d=9 got 2 expected 2 sym=ok num=ok [simplify]
-2 K=43 d=11 got -2 expected -2 sym=ok num=ok [simplify]
1/2 K=51 d=15 got 0.5 expected 0.5 sym=ok num=ok [simplify]
-1/2 K=67 d=17 got -0.5 expected -0.5 sym=ok num=ok [simplify]
2/3 K=103 d=19 got 0.6666666667 expected 0.6666666667 sym=ok num=ok [simplify]
-2/3 K=119 d=21 got -0.6666666667 expected -0.6666666667 sym=ok num=ok [simplify]
sqrt2 K=85 d=21 got 1.414213562 expected 1.414213562 sym=ok num=ok [simplify]
i K=75 d=19 got i expected i sym=ok num=ok [i²=-1, simplify]
pi K=153 d=29 got 3.141592654 expected 3.141592654 sym=ok num=ok [simplify]
Unary functions (x = 7/3)
------------------------------------------------------------------------------
exp(x) K=3 d=1 got 10.3122585 expected 10.3122585 sym=ok num=ok [simplify]
ln(x) K=7 d=3 got 0.8472978604 expected 0.8472978604 sym=ok num=ok [simplify]
-x K=17 d=7 got -2.333333333 expected -2.333333333 sym=ok num=ok [simplify]
1/x K=25 d=8 got 0.4285714286 expected 0.4285714286 sym=ok num=ok [simplify]
x - 1 K=11 d=4 got 1.333333333 expected 1.333333333 sym=ok num=ok [simplify]
x + 1 K=27 d=9 got 3.333333333 expected 3.333333333 sym=ok num=ok [simplify]
2x K=67 d=17 got 4.666666667 expected 4.666666667 sym=ok num=ok [simplify]
x/2 K=51 d=15 got 1.166666667 expected 1.166666667 sym=ok num=ok [simplify]
x^2 K=41 d=10 got 5.444444444 expected 5.444444444 sym=ok num=ok [simplify]
sqrt(x) K=59 d=16 got 1.527525232 expected 1.527525232 sym=ok num=ok [simplify]
Binary operations (x = 7/3, y = 5/2)
------------------------------------------------------------------------------
x + y K=27 d=9 got 4.833333333 expected 4.833333333 sym=ok num=ok [simplify]
x - y K=11 d=4 got -0.1666666667 expected -0.1666666667 sym=ok num=ok [simplify]
x * y K=41 d=10 got 5.833333333 expected 5.833333333 sym=ok num=ok [simplify]
x / y K=25 d=8 got 0.9333333333 expected 0.9333333333 sym=ok num=ok [simplify]
x ^ y K=49 d=12 got 8.316526261 expected 8.316526261 sym=ok num=ok [simplify]I hope this was presented at SIGBOVIK.
I guess you folks don't know about iota & jot: https://en.wikipedia.org/wiki/Iota_and_Jot
Zero will also be handy in definitions: `0=eml(1,eml(eml(1,1),1))`.
And i is obviously `sqrt(-1)`
Could this be used to prove e+pi is transcendental?
Looks like he bruteforced all combinations of two mathematical operations no ?
there ought to be a special section on HN entitled "things that will make you feel thoroughly inadequate".
Next step is to build an analog scientific calculator with only EML gates
The paper somehow seems to be missing the most interesting part, i.e. the optimal constructions of functions from eml in a readable format.
Here is my attempt. I think they should be optimal up to around 15 eml.nodrs, the latter might not be:
# 0
1=1
# 1
exp(x)=eml(x,1)
e-ln(x)=eml(1,x)
e=exp(1)
# 2
e-x=e-ln(exp(x))
# 3
0=e-e
ln(x)=e-(e-ln(x))
exp(x)-exp(y)=eml(x,exp(exp(y)))
# 4
id(x)=e-(e-x)
inf=e-ln(0)
x-ln(y)=eml(ln(x),y)
# 5
x-y=x-ln(exp(y))
-inf=e-ln(inf)
# 6
-ln(x)=eml(-inf,x)
ln(ln(x))=ln(ln(x))
# 7
-x=-ln(exp(x))
-1=-1
x^-1=exp(-ln(x))
ln(x)+ln(y)=e-((e-ln(x))-ln(y))
ln(x)-ln(y)=ln(x)-ln(y) # using x - ln(y)
# 8
xy=exp(ln(x)+ln(y))
x/y=exp(ln(x)-ln(y))
# 9
x + y = ln(exp(x))+ln(exp(y))
2 = 1+1
# 10
ipi = ln(-1)
# 13
-ipi=-ln(-1)
x^y = exp(ln(x)y)
# 16
1/2 = 2^-1
# 17
x/2 = x/2
x2 = x2
# 20
ln(sqrt(x)) = ln(x)/2
# 21
sqrt(x) = exp(ln(sqrt(x)))
# 25
sqrt(xy) = exp((ln(x)+ln(y))/2)
# 27
ln(i)=ln(sqrt(-1))
# 28
i = sqrt(-1)
-pi^2 = (ipi)(ipi)
# 31
pi^2 = (ipi)(-ipi)
# 37
exp(xi)=exp(xi)
# 44
exp(-xi)=exp(-(xi))
# 46
pi = (ipi)/i
# 90+x?
2cos(x)=exp(xi)+exp(-xi))
# 107+x?
cos(x) = (2cos(x))/2
# 118+x?
2sin(x)=(exp(x*i)-exp(-xi))/i # using exp(x)-exp(y)
# 145+x?
sin(x) = (2sin(x))/2
# 217+3x?
tan(x) = 2sin(x)/(2cos(x))
Whoa, this is huge!
My dearest congrats to the author in case s/he shows around this site ^^.
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/singlebi...