(* Floyd's ``the tortoise and the hare'' algorithm. *)
module TortoiseHare
use import int.Int
(* We consider a function f over an abstract type t *)
type t
logic f t : t
(* Given some x0 in t, we consider the sequence of the repeated calls
to f starting from x0. *)
logic x int : t
axiom xdef: forall n:int. n >= 0 -> x (n+1) = f (x n)
logic x0 : t = x 0
(* If t is finite, this sequence will eventually end up on a cycle.
Let us simply assume the existence of this cycle, that is
x (i + lambda) = x i, for some lambda > 0 and i large enough. *)
logic mu : int
axiom mu: mu >= 0
logic lambda : int
axiom lambda: lambda >= 1
axiom cycle: forall i:int. mu <= i -> x (i + lambda) = x i
lemma cycle_gen:
forall i:int. mu <= i -> forall k:int. 0 <= k -> x (i + lambda * k) = x i
(* The challenge is to prove that the recursive function
let rec run x1 x2 = if x1 <> x2 then run (f x1) (f (f x2))
terminates when called on x0 and (f x0).
*)
logic dist int int : int
axiom dist_def1:
forall n m: int. 0 <= n <= m ->
x (n + dist n m) = x m
axiom dist_def2:
forall n m: int. 0 <= n <= m ->
forall k: int. x (n + k) = x m -> dist n m <= k
logic r (x12 : (t, t)) (x'12 : (t, t)) =
let x1, x2 = x12 in
let x'1, x'2 = x'12 in
exists m:int.
x1 = x (m+1) and x2 = x (2*m+2) and x'1 = x m and x'2 = x (2*m) and
m < mu or (mu <= m and dist (m+1) (2*m+2) < dist m (2*m))
let rec run x1 x2 variant { (x1, x2) } with r =
{ exists m:int [x m]. x1 = x m and x2 = x (2*m) }
if x1 <> x2 then
run (f x1) (f (f x2))
{ }
end
(*
Local Variables:
compile-command: "unset LANG; make -C ../.. examples/programs/tortoise_hare.gui"
End:
*)