>Maybe a year - or two model releases - from now, the AI assistant will be undeniably stronger than Tao, and not an equal anymore.

we're kind of well past that (in my opinion), if you consider that this is the same ai assistant that can help you with a recipe, diagnose a weird sound in your car, help with biology homework, translate languages, and so on.

even in math alone, i think its indisputably already stronger than Tao, considering it has approximately this much depth in ~all of the math subfields.

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eh_why_not 7 hours ago | parent | next [-]

Agreed, it's stronger "horizontally". But I also think that we're not far away from it being stronger vertically; i.e. superior to Tao, in that such turn-by-turn guidance by him in solving sophisticated and difficult problems will not be necessary for long.

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noosphr 3 hours ago | parent | prev | next [-]

It only has that depth because you prompt it towards the experts that understand that depth.

An example that happened 10 minutes ago: contracts in racket, it kept arguing that you can't use -> in a contract of a function with a rest argument. I had to mention ... explicitly that it wrote the code correctly.

	▲	larsfaye 4 minutes ago \| parent [-]
		Yeah...I know I should be massively impressed, and I am to a degree, but isn't this what we should expect? LLMs can pick up on patterns that no human can see. In that sense, they really are a type of "search engine", but I use that term loosely. Tao is using them as a way to sift and sort theories and information that the model has a plethora of training data on. It's like being able to converse with the codex of humanity, and extract data via an algorithm that is highly adept at sorting through it.

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lpd59 5 hours ago | parent | prev [-]

To be fair, Tao's specialty is not algebraic geometry.

	▲	buzzin_ 5 hours ago \| parent [-]
		Interestingly enough, ChatGPT started his answer like this only once: "This is exactly the question I would ask next. My impression is: Most standard invariants are...." And this was a response to this prompt: "Is there a chance of an indirect argument of X ~ A^3 coming from computing some invariant of X that forces it to be A^3? (I am not all that expert in algebraic geometry but I'm thinking like degree or Betti numbers or something.)"