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Technically, they are. There was a post on HN recently about how every model is equivalent to a decision tree.
by bheadmaster 4y ago
Technically, they are. There was a post on HN recently about how every model is equivalent to a decision tree.
- dragonwriter 4y agoI mean, if it can be specified by a computational process, that’s trivially true, right?
- bheadmaster 4y agoI'm not so sure about that. Decision trees are finite set of if-else branches, and work on finite input - much like models. I assume many of the classic arbitrary-size-input algorithms cannot be represented by a decision tree. Adding two arbitrary-sized numbers comes to mind - how would one model it as a decision tree? if (a[0] + b[0] == 0) { c[0] = 0 } else if (a[0] + b[0] == 1) { c[0] = 1 } else if ... ... } else if (a[0] + b[0] == 10) { c[0] = 0 if (a[1] + b[1] == 0) { c[1] = 1 } else ... ... } ... You can see that for arbitrary-sized input, this decision tree would have grow infinitely - which is contradictory to the finite nature of decision trees. I can't think of a proof that such a computation cannot be represented by a decision tree, and it's possible that my definitions aren't quite correct. But intuitively, I think this is the way it works. Of course, if I'm wrong, any explanation on why and how would be very welcome.