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Is the random forests algorithm Turing-complete? As in, can any algorithm be represented by a given "tree" in the forest?

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Random forests are useful only for inputs of a given size, so they are not really eligible for Turing-completeness. However, any function on inputs of size $n$ can be computed by a decision tree, so in that sense the model is complete (though in practice we limit the depth of the trees).

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No, they're not Turing complete. The handle inputs of a fixed size. Also, they can't loop, so will always halt -- it follows that they don't have the ability to perform Turing-complete computation.

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  • $\begingroup$ Thanks for the helpful suggestion, @DavidRicherby -- I agree, pointing that out explicitly is a nice improvement. $\endgroup$ – D.W. Feb 22 '16 at 5:46

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