Bijection caml

let rec bij = function
    | (0, 0) -> 0
    | (0, y) -> bij (y-1, 0)+1
    | (x, y) -> if x <= y then bij (x-1, y)+1 else bij(x, y+1)+1;;
print_int(bij(4,1));;
(* voila tres bien *)

let rec lexico = fun (* renvoie true si l1 > l2 *)
    | [] [] -> true
    | [] l -> true
    | l [] -> false
    | (t::q) (t2::q2) -> if t > t2 then true
                        else
                            if t < t2 then false
                                else
                                    lexico q q2;;


let rec ack = fun
    | 0 p -> p+1
    | n 0 -> ack (n-1) 1
    | n p -> ack (n-1) (ack n (p-1));;
print_int(ack 4 4);;