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In my diagram everything is a computation (as I understand it): From (a, b) you compute and(a, b), from which you compute not(and(a, b)). Alternatively, from (a
by tom_mellior 6y ago
In my diagram everything is a computation (as I understand it): From (a, b) you compute and(a, b), from which you compute not(and(a, b)). Alternatively, from (a, b) you compute (not(a), not(b)), from which you compute or(not(a), not(b)). The "and" and "or" steps are applications of functions, i.e., computations. The "not" steps are also applications of a function, why aren't those computations?
I guess you might say that applying the function (not x not) is somehow a higher-order "computation on other computations", so that step might be a "path" (if that's the meaning, the term is ridiculously badly chosen). But the bottom edge is simply applying the "not" function to some binary data.
- YeGoblynQueenne 6y agoSo a path is a higher order function? But then why is not also a path?