If y = f(x, T) when y is a member of T, prove f() doesn’t copy a value from T.
This is possible. For example, many functions f(0) return 0; if we can derive a function g(x) that generates y without access to T (ie. the training data) then we can plausibly assume that f(x, T) might indeed not be copying from T.
…but can we do so?
If so, we must admit that it is possible that f(x, T) = g(x) and no copying is taking place.
So, in the past 50 years of prior work can we come up with an example of a function g(x) that generates the exact code we see coming out of these LLMs?
I’ll grant, it’s not impossible.
For example, if stable diffusion generated sin wave patterns or fractals that exhibited “deep complex structure” and it happens that some artist had written code to do that exact thing, there would (I believe) be a fair case that stable diffusion was not copying the artist, it was simply a parallel implementation of the same generative code.
However, now my scepticism kicks in.
For code written by hand, that was not generated, we are suggesting that a LLM is a parallel equivalent implementation of a human mind, and it can, with purely “learnt patterns” replicate not only the intent and structure of known code, but the exact code itself, repeatedly.
I’ll grant. It’s not impossible, and it’s very difficult to prove, but it seems fabulously, unbelievably, extraordinarily unlikely that a clean room reimplementation would be identical, repeatedly.
We’re entering into the domain of assigning probability and then doing a zero knowledge proof here, but the solution is easy.
Just prove by counter example.
Train a model that generates such code without the code as training data.
You have now won the argument.
…on the bright side, you’ve also solved the 10 million dollar question of “how do I train my LLM when I don’t have enough training data” (because that’s what you just did; train it to do something without training data), so you’re now rich and give zero ducks what I think. Congrats.