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> The term is being tortured beyond endurance by being forced to describe what LLMs do. No it isn't, this is exactly the kind of process that Markov chains des
by Jensson 2y ago
> The term is being tortured beyond endurance by being forced to describe what LLMs do.
No it isn't, this is exactly the kind of process that Markov chains describe. They describe a stateless probability function that you chain over and over to produce a sequence of results, exactly like what an LLM does, they predict the next token based on precious tokens repeatedly to produce text.
There is no bending of any concepts here, anyone who studied a higher level math probability course will instantly recognize LLMs as a markov chain. It is common for markov chains to have infinite number of states, like a real position, and then moving randomly from there. You can model particles in a gas like that.
- CamperBob2 2y agoI'm not disagreeing that it's correct, I'm disagreeing that it's meaningful. Calling an LLM a "Markov model" reveals an agenda rather than an insight.
- Jensson 2y ago> Calling an LLM a "Markov model" reveals an agenda rather than an insight. What agenda? Knowing that LLM's are markov chains helps me understand and makes me better at using them, since I know every word is just calculated probabilistically based on the previous N words, the fastest way to describe that is to say it is a Markov chain. This makes it hard for them to coordinate their early words with things that will happen later, all that from just knowing it is a Markov chain. I feel the knee jerk rejection of this statement is more of an agenda, why do you feel it is so wrong to say it is a Markov chain? Are you sure there is no agenda behind those feelings?
- ben_w 2y agoThe agenda may well be "pure mathematics". I don't think "Markov chain" is a useful description despite agreeing it's an accurate one, simply because I have a preference for applied maths over pure maths and a full transition table version would be obscenely large. In this, it's also technically accurate to say that a perceptron network with one hidden layer can approximate any function: technically yes, but not in any practical sense. But if you're working in the pure maths domain, knowing that it can in principle be one, means you can apply any tools that only work on Markov chains. Given what else I've seen mathematicians do, I won't be surprised if pure maths researchers can get useful results even though the transition matrix is literally too large to fit into our universe without collapsing it into a black hole — it would hardly be the first time I've even seen mathematicians talking about quantities that large.