>> (f∘g)(x) ≡ f(g(x)) > > Definitions do not eliminate ambiguity.
Speaking of formal definitions, I would say they ought to. There is not much use of defining formally and ambiguously. > One issue, here, is that the definitions of f, g and x are contextual. What do you mean by 'contextual'? Is it related to the above definition of ∘ (composition)? Because f, g and x are just formal names there, locally bound within the definition – none of them is a free variable, so there is no context. (Expressed in plain language, the definition says the following: composition (∘) of f on g, where f and g are any two functions such that the domain of f includes the codomain of g, is a function which, given any argument from the domain of g, returns the result of applying f to the result of applying g to x.) > A variant on this issue is > that in the realm of general mathematics, a function that maps from A > to B can also map from C to D. And, A and C might have a non-empty > intersection and yet not have a subset relationship between them and > composition might be valid for A and not C. How can a function have two different domains without being in fact two different functions? Doesn't make sense to me. > All of these J phrases can be used to > represent composition between functions represented by f and g, if f > and g are verbs: > f at g > f@:g > f&g > f&:g > ([: f g) > f(g(x)) I assume that by 'f at g' you mean f@g ... Now, not all of the above represent a composition. f@g and f&g, in their monadic cases, are equivalent by definition, and are compositions. f@:g and f&:g (in their monadic cases) are equivalent but are not the composition of f on g; what they compose is the results of changing (the ranks of) their arguments rather than the arguments themselves. The same holds of [: f g (in its monadic case). > Note also that that last phrase > represents composition if x is a noun. It might also represent > composition if x was an appropriate adverb (for example: x=:1 :'@u') In neither case is it a composition. If x is a noun, f(g(x)) does not even represent a verb. If x is an adverb, so that g(x) is a verb, then f(g(x)) is a hook – again, not a composition. We have to stick to the mathematical meaning of composition because this is what the definition of functor depends on, and we were discussing functors ... >> The domain of a composition operator are functions themselves, >> not the data structures that are domains of those functions. >> A 'composition' that has to do with data structures is anything >> else but composition of functions. > > You seem to be saying that context is irrelevant. The context of what? In defining composition, I see no context at all. > And, some operations can be performed efficiently in compiled > haskell and are inefficient in interpreted haskell. Yes, but that is nothing to wonder at – the same can be said of any language that admits both modes of execution. And it doesn't necessarily mean that such a language, when interpreted, is slower than a language which is only interpreted. Whichever is faster would also depend, I guess, on the problem being solved, as well as on the language constructs and programming techniques used in the solutions. ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
