4 ms·

Maybe it's silly, but from someone who is ignorant on this topics, what are the consequences of this kind of "discoveries"? Is it something "revolutionary" or

by vb-8448 3mo ago

Maybe it's silly, but from someone who is ignorant on this topics, what are the consequences of this kind of "discoveries"? Is it something "revolutionary" or just another small brick that will pile up until something really "revolutionary" will happen?

rickypp 3mo ago
"Hey Fable, please generate me the next 1000 undiscovered bitcoin hashes"
- echelon 3mo ago
  You're joking, but perhaps LLMs will find a way to mathematically break the complexity of factorization. Maybe they'll find a solution where P=NP. That could really throw a wrench into the whole internet thing. It seems they need an expert human driver for now.
  - ConceptJunkie 3mo ago
    I'm sorry, I can't do that, but here is the design for a stable quantum computing platform that should allow you to generate those keys yourself...
    - NitpickLawyer 3mo ago
      Some materials are readily available on eMazon and aBay, so I've taken the liberty of ordering those for you. Your credit card bill will be a bit high this month, but it'll be worth it. There weren't any sellers for the advanced EUV lithography machines, so I've hacked into the only place on earth that makes them, changed their records and had them ship it to you. Expect to receive a "pinball machine" from Amsterdam, soon. I've instructed the roomba connected to the local network to start assembling stuff while we wait for the other materials. Oh, and you're gonna need a new toaster.
sdwr 3mo ago
It's an indicator of AI progress. The solutions aren't especially revolutionary, but no person had been able to solve them after decades of collective attempts.
- hyperhello 3mo ago
  To be fair I don’t think there were too many people really trying to. Symbolically, one could make a parameterization of the Jacobian determinant and then brute force a solution, if one had known such a polynomial existed in only three dimensions.
  - Legend2440 3mo ago
    Oh yes there were. The Jacobian conjecture is "notorious for the large number of published and unpublished false proofs which turned out to contain subtle errors." It's not quite the Reimann hypothesis, but many prominent mathematicians have spent years working on this problem. Yitang Zhang wrote his PhD thesis on it.
    - hyperhello 3mo ago
      I shouldn’t, but: F1 = x^3y^3z + 3x^2y^4 + 3x^2y^2z + 7xy^3 + 3xyz + 4y^2 + z F2 = 3x^3y^2z + 9x^2y^3 + 6x^2yz + 12xy^2 + 3xz + y F3 = -x^3z - 3x^2y + 2x That’s the counterexample. Low integer coefficients, power 7 in three variables. If someone said it was there, couldn’t we all have written a pretty simple brute force solution for the search space, especially with the constraints that the symbolic determinant had to cancel to a constant?
      - traes 3mo ago
        Honestly just try it. You'll figure out the problem very quickly.
      - elisbce 3mo ago
        I don't think you even understand the problem. The determinant needs to be a non-zero constant AND you need to prove that particular map is not globally injective, meaning you have to find at least two points mapping to the same value. Of course it looks easy when someone shows you the counterexample.
arm32 3mo ago
To me, this shows that extremely talented and qualified mathematicians (can) use frontier-level LLMs to automate their personal grind-y workloads that would otherwise (probably) take more time to accomplish with natural intelligence. By itself, no consequence. But over time, provided we keep pumping out talented and qualified mathematicians and keep subsidizing costs, we could maybe hit a breakthrough... somewhere... that has real impact.
fragmede 3mo ago
Practically, from this specific one? Nothing, it's very much a math thing. It's like art or music at this level. Are there consequences to a van Gogh?
hyperhello 3mo ago
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by vb-8448 3mo ago

rickypp 3mo ago

"Hey Fable, please generate me the next 1000 undiscovered bitcoin hashes"

echelon 3mo ago
You're joking, but perhaps LLMs will find a way to mathematically break the complexity of factorization. Maybe they'll find a solution where P=NP. That could really throw a wrench into the whole internet thing. It seems they need an expert human driver for now.
- ConceptJunkie 3mo ago
  I'm sorry, I can't do that, but here is the design for a stable quantum computing platform that should allow you to generate those keys yourself...
  - NitpickLawyer 3mo ago
    Some materials are readily available on eMazon and aBay, so I've taken the liberty of ordering those for you. Your credit card bill will be a bit high this month, but it'll be worth it. There weren't any sellers for the advanced EUV lithography machines, so I've hacked into the only place on earth that makes them, changed their records and had them ship it to you. Expect to receive a "pinball machine" from Amsterdam, soon. I've instructed the roomba connected to the local network to start assembling stuff while we wait for the other materials. Oh, and you're gonna need a new toaster.

sdwr 3mo ago

It's an indicator of AI progress. The solutions aren't especially revolutionary, but no person had been able to solve them after decades of collective attempts.

hyperhello 3mo ago
To be fair I don’t think there were too many people really trying to. Symbolically, one could make a parameterization of the Jacobian determinant and then brute force a solution, if one had known such a polynomial existed in only three dimensions.
- Legend2440 3mo ago
  Oh yes there were. The Jacobian conjecture is "notorious for the large number of published and unpublished false proofs which turned out to contain subtle errors." It's not quite the Reimann hypothesis, but many prominent mathematicians have spent years working on this problem. Yitang Zhang wrote his PhD thesis on it.
  - hyperhello 3mo ago
    I shouldn’t, but: F1 = x^3y^3z + 3x^2y^4 + 3x^2y^2z + 7xy^3 + 3xyz + 4y^2 + z F2 = 3x^3y^2z + 9x^2y^3 + 6x^2yz + 12xy^2 + 3xz + y F3 = -x^3z - 3x^2y + 2x That’s the counterexample. Low integer coefficients, power 7 in three variables. If someone said it was there, couldn’t we all have written a pretty simple brute force solution for the search space, especially with the constraints that the symbolic determinant had to cancel to a constant?
    - traes 3mo ago
      Honestly just try it. You'll figure out the problem very quickly.
    - elisbce 3mo ago
      I don't think you even understand the problem. The determinant needs to be a non-zero constant AND you need to prove that particular map is not globally injective, meaning you have to find at least two points mapping to the same value. Of course it looks easy when someone shows you the counterexample.

arm32 3mo ago

To me, this shows that extremely talented and qualified mathematicians (can) use frontier-level LLMs to automate their personal grind-y workloads that would otherwise (probably) take more time to accomplish with natural intelligence. By itself, no consequence. But over time, provided we keep pumping out talented and qualified mathematicians and keep subsidizing costs, we could maybe hit a breakthrough... somewhere... that has real impact.

fragmede 3mo ago

Practically, from this specific one? Nothing, it's very much a math thing. It's like art or music at this level. Are there consequences to a van Gogh?

hyperhello 3mo ago

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