Is this implementation tail-recursive

I read in an algorithmic book that the Ackermann function cannot be made tail-recursive (what they say is "it can't be transformed into an iteration"). I'm pretty perplex about this, so I tried and come up with this:

let Ackb m n =
  let rec rAck cont m n = 
    match (m, n) with
      | 0, n -> cont (n+1)
      | m, 0 -> rAck cont (m-1) 1
      | m, n -> rAck (fun x -> rAck cont (m-1) x) m (n-1)
  in rAck (fun x -> x) m n
;;

(it's OCaml / F# code).

My problem is, I'm not sure that this is actually tail recursive. Could you con开发者_如何学运维firm that it is? If not, why? And eventually, what does it mean when people say that the Ackermann function is not primitive recursive?

Thanks!

Yes, it is tail-recursive. Every function can be made tail-rec by an explicit transformation to Continuation Passing Style.

This does not mean that the function will execute in constant memory : you build stacks of continuations that must be allocated. It may be more efficient to defunctionalize the continuations to represent that data as a simple algebraic datatype.

Being primitive recursive is a very different notion, related to expressiveness of a certain form of recursive definition that is used in mathematical theory, but probably not very much relevant to computer science as you know it: they are of very reduced expressiveness, and systems with function composition (starting with Gödel's System T), such as all current programming languages, are much more powerful.

In term of computer languages, primtive recursive functions roughly correspond to programs without general recursion where all loop/iterations are statically bounded (the number of possible repetitions is known).

Yes.

By definition, any recursive function can be transformed into an iteration as long as it has access to an unbounded stack-like construct. The interesting question is whether it can be done without a stack or any other unbounded data storage.

A tail-recursive function can be turned into such an iteration only if the size of its arguments is bounded. In your example (and almost any recursive function that uses continuations), the cont parameter is for all means and purposes a stack that can grow to any size. Indeed, the entire point of continuation-passing style is to store data usually present on the call stack ("what to do after I return?") in a continuation parameter instead.