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(*         *   The Coq Proof Assistant / The Coq Development Team       *)
(*  v      *   INRIA, CNRS and contributors - Copyright 1999-2018       *)
(* <O___,, *       (see CREDITS file for the list of authors)           *)
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(*    //   *    This file is distributed under the terms of the         *)
(*         *     GNU Lesser General Public License Version 2.1          *)
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(** * Zgcd_alt : an alternate version of Z.gcd, based on Euclid's algorithm *)
(** Author: Pierre Letouzey *)
(** The alternate [Zgcd_alt] given here used to be the main [Z.gcd] function (see file [Znumtheory]), but this main [Z.gcd] is now based on a modern binary-efficient algorithm. This earlier version, based on Euclid's algorithm of iterated modulo, is kept here due to both its intrinsic interest and its use as reference point when proving gcd on Int31 numbers *)
Open Scope Z_scope.
(** In Coq, we need to control the number of iteration of modulo. For that, we use an explicit measure in [nat], and we prove later that using [2*d] is enough, where [d] is the number of binary digits of the first argument. *)
Fixpoint Zgcdn (n:nat) : Z -> Z -> Z := fun a b =>
match n with
| O => 1 (* arbitrary, since n should be big enough *)
| S n => match a with
| Z0 => Z.abs b
| Zpos _ => Zgcdn n (Z.modulo b a) a
| Zneg a => Zgcdn n (Z.modulo b (Zpos a)) (Zpos a)
end
end.
Definition Zgcd_bound (a:Z) :=
match a with
| Z0 => S O
| Zpos p => let n := Pos.size_nat p in (n+n)%nat
| Zneg p => let n := Pos.size_nat p in (n+n)%nat
end.
Definition Zgcd_alt a b := Zgcdn (Zgcd_bound a) a b.
(** A first obvious fact : [Z.gcd a b] is positive. *)
(** We now prove that Z.gcd is indeed a gcd. *)
(** 1) We prove a weaker & easier bound. *)
(** 2) For Euclid's algorithm, the worst-case situation corresponds to Fibonacci numbers. Let's define them: *)
Fixpoint fibonacci (n:nat) : Z :=
match n with
| O => 1
| S O => 1
| S (S n as p) => fibonacci p + fibonacci n
end.
(** 3) We prove that fibonacci numbers are indeed worst-case: for a given number [n], if we reach a conclusion about [gcd(a,b)] in exactly [n+1] loops, then [fibonacci (n+1)<=a /\ fibonacci(n+2)<=b] *)
(** 3b) We reformulate the previous result in a more positive way. *)
(** 4) The proposed bound leads to a fibonacci number that is big enough. *)
(* 5) the end: we glue everything together and take care of situations not corresponding to [0<a<b]. *)

Lemma Zgcdn_opp n a b : Zgcdn n (-a) b = Zgcdn n a b.