1

Suppose both k and f are functions of type int -> int. If we know k(l-1)=f(l) holds for all l of type int, can we be certain that k is the function v->f(v+1) ?

I have this question while doing an exercise of functional programming: to transform a length function

let rec len xs = 
  match xs with 
    | [] -> 0 
    | x:xr -> 1 + len xr;;

to a continuation-passing version. My answer to the exercise would be

   let rec lenc xs k = 
      match xs with 
        | [] -> k 0 
        | x:xr -> lenc xr (fun v -> k(v+1))

but I am unsure if the (fun v -> k(v+1)) part can be replaced by other solutions. To know this, we would need to determine a unique "k", given that k(l-1)=f(l) holds for all function f of type int->int and l of type int?

zell
  • 9,830
  • 10
  • 62
  • 115

1 Answers1

3

Do you want mathematical proof?

(1) k(i - 1) = f(i)

(2) j = i - 1

from (1) & (2):

(3) k(j) = f(i)

from (2):

(4) i = j + 1

from (3) & (4)

(5) k(j) = f(j + 1)

q.e.d.

Of course (fun v -> k(v+1)) is your k and the inside k is f

AMieres
  • 4,944
  • 1
  • 14
  • 19