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> therefore the layered topology must have some mathematical superiority Isn't it just that backpropagation on the layered topology is relatively straightforwa
by dnautics 7y ago
> therefore the layered topology must have some mathematical superiority
Isn't it just that backpropagation on the layered topology is relatively straightforward?
That's not to say you can't write a backpropagation on an arbitrary digraph, but as you get to more and more complex digraphs, things will get harder.
I could be wrong on this.
- dodobirdlord 7y ago> Isn't it just that backpropagation on the layered topology is relatively straightforward? That's not to say you can't write a backpropagation on an arbitrary digraph... Moreover, any arbitrary digraph can be expressed as a layered topology (possibly with a lot of 0-weights). Since there's no fundamental difference you might as well work with whatever's easiest to compute with.