English version
Accueil     À propos     Téléchargement     Ressources     Contactez-nous    

Ce site est rarement mis à jour. Pour les informations les plus récentes, rendez-vous sur le nouveau site OCaml à l'adresse ocaml.org.

Browse thread
[Caml-list] equi-recursive Fold isomorphism
[ Home ] [ Index: by date | by threads ]
[ Search: ]

[ Message by date: previous | next ] [ Message in thread: previous | next ] [ Thread: previous | next ]
Date: 2002-07-28 (20:15)
From: Alain Frisch <frisch@c...>
Subject: Re: [Caml-list] equi-recursive Fold isomorphism
On Sun, 28 Jul 2002, John Max Skaller wrote:

> Hmmm: Ocaml 3.04+15 with -rectypes
> # let rec x = (1,(1,(1,x)));;
> val x : int * (int * (int * 'a)) as 'a =
>  ....
> Seem like Ocaml doesn't minimise the type, but:
> let rec y = (1,y);;
> x = y;;
> Works correctly (so it knows the two types are comparable).
> Interestingly, the answer is false

Are you sure ?  It seems that the comparison doesn't terminate,
as expected (as it does not consume memory). The manual
Equality between cyclic data structures may not terminate.

: both data structures
> consist of an infinite stream of 1's, represented by
> cycles of distinct lengths. No item by item comparison
> could reveal any distinction: the infinite tree expansions
> of the data structures are the same. Is Ocaml's answer correct?

As you know, to implement the comparison between cyclic values with the
expected behaviour, one has to use a coinductive algorithm with some kind
of memoization. This would be much slower for the non-cyclic case, and
many people expect a fast generic compare function. I would like to say
that you can always take the code in byterun/compare.c and implement your
own equirecursive generic comparison in C, but this is actually not
possible (you need the Is_young and Is_in_heap macros that are not
available). Using the mentioned Recurse module, you could do:

module I = struct
  type 'a t = Int of int | Pair of 'a * 'a

  let map f = function
    | Int i -> Int i
    | Pair (a,b) -> Pair (f a, f b)

  let equal f x y =
    match (x,y) with
      | (Int i, Int j) when i = j -> ()
      | (Pair (a1,b1), Pair (a2,b2)) -> f a1 a2; f b1 b2
      | _ -> raise Recursive.NotEqual

  let iter f x = ignore (map f x)
  let hash f x = Hashtbl.hash (map f x)
  let deep = 4

module X = Recursive.Make(I)

let cons x = let n = X.make () in X.define n x; n
let recurs f = let n = X.make () in X.define n (f n); n
let int' i = I.Int i
let int i = cons (int' i)
let pair' a b = I.Pair (a,b)
let pair a b = cons (pair' a b)

let x = recurs (fun x -> pair' (int 1) (pair (int 1) (pair (int 1) x)))
let y = recurs (fun y -> pair' (int 1) y)

let x = X.internalize x and y = X.internalize y
let () = Printf.printf "%i;%i;%b\n" (X.id x) (X.id y) (x = y)

(the equility test is specified by Recurse to run in constant time)

Side-note: be careful if you implement a language with subtyping and both
recursive types and recursive (cyclic) values. Indeed, the usual
equi-recursive definition of subtyping is not sound if you allow cyclic
values. For instance, the algorithm would say that the type ('a * 'a) as
'a is empty, even though it is inhabited by values.

-- Alain

To unsubscribe, mail caml-list-request@inria.fr Archives: http://caml.inria.fr
Bug reports: http://caml.inria.fr/bin/caml-bugs FAQ: http://caml.inria.fr/FAQ/
Beginner's list: http://groups.yahoo.com/group/ocaml_beginners