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by vrighter 2 days ago
"far less deterministic" does not make sense. it's either deterministic, or it isn't. llms aren't. And to anyone who starts spouting the mantra that you can just set the temperature to zero and i just don't know what "deterministic means", I invite you to predict an llms output before you submit the prompt. I know the compiler will respect my code's semantics. You can't do that with any model.
1 comments

Have you seen the shenanigans modern C compilers are up to?

If you think your compiler will respect your code's semantics, you might be in for a rude awakening.

And there are certainly ways of influencing an LLM to be "more deterministic".

So the statement not only makes sense, it is fully inclusive of your claimed boundary condition of "completely vs. not at all". Last I checked, "not at all deterministic" is, in fact, "far less deterministic" than "completely deterministic".

YMMV.

¯\_(ツ)_/¯

The compiler does respect my code's semantics, provided I obey the contract and avoid UB. If I do, it does that perfectly. I compile it a million times I get the exact same semantics every time (yes, stuff like register allocation might be nondeterministic, but NEVER the semantics).

You can save your prompt and every time you use it, you get a different result with different semantics, different structure and different bugs. The compiler then respects and preserves that broken code's semantics perfectly, like it does my own.

"deterministic" isn't a spectrum. Yes, 0 on a scale of 0 to 1 is technically "far less" but now you're just being pedantic.