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Apr 17, 2019 at 6:33 comment added D.W. @samlaf, could be -- that exceeds my knowledge. I don't know enough about the BSS model to have an informed opinion about your question. Sorry!
Apr 17, 2019 at 3:39 comment added samlaf Would the BSS machine (from what I understand, real-valued TM) be a way to link these concepts? It seems that being BSS-complete implies both Turing-complete and Universal Approximator.
Jan 17, 2017 at 0:41 comment added D.W. @Redcoat, I don't agree, because Turing-complete talks about functions on booleans, while universal-approximator talks about functions on real numbers. At this point I suspect maybe what would be most useful would be to familiarize yourself with the precise, mathematical definition of each of those two concepts.
Jan 16, 2017 at 20:08 comment added Spacey Thanks DW! It is getting clearer - however in my mind, there does seem to be a link: For example, it seems to me then, that one cannot have a turing-machine that is NOT a universal approximator simultaneously, right? In that sense, being turing-complete, would also mean that you are a universal approximator. Would you agree with this?
Jan 16, 2017 at 20:03 comment added D.W. @Redcoat, good questions! I added another paragraph to my answer; see revised answer (last paragraph).
Jan 16, 2017 at 20:02 history edited D.W. CC BY-SA 3.0
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Jan 16, 2017 at 19:58 comment added Spacey Hmm, thanks for the answer, although I am still confused somewhat: Let me phrase a different way: How is it that a non-turing-complete machine can also be a universal approximator? Does that mean that universal-approximation is a "weaker" criteria of machines that "turing-completeness"?
Jan 16, 2017 at 19:06 history answered D.W. CC BY-SA 3.0