solver.ml 31.4 KB
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(*-----------------------------------------------------------------------
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 ** Copyright (C) 2001, 2002 - Verimag.
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** This file may only be copied under the terms of the GNU Library General
** Public License 
**-----------------------------------------------------------------------
**
** File: solver.ml
** Main author: jahier@imag.fr
*)

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open List
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open Formula
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open Constraint
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open Util
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open Hashtbl
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open Gne
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open Value
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open Rnumsolver
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(****************************************************************************)
	  
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let (lookup: env_in -> subst list -> var_name -> Value.t option) = 
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  fun input pre vn ->  
    try Some(Hashtbl.find input vn)
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    with Not_found -> 
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      try Some(List.assoc vn pre)
      with Not_found -> None

(****************************************************************************)

type comp = SupZero | SupEqZero | EqZero | NeqZero

let rec (formula_to_bdd : env_in -> formula -> Bdd.t * bool) =
  fun input f ->
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    (** Returns the bdd of [f] where input and pre variables
      have been repaced by their values.
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      Also returns a flag that is true iff the formula depends on
      input and pre vars. If this flag is false, the formula is
      stored (cached) in a global table ([env_state.bdd_tbl_global]);
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      otherwise, it is stored in a table that is cleared at each new
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      step ([env_state.bdd_tbl]).  
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    *)
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    try (Env_state.bdd f, true)
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    with Not_found -> 
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      try (Env_state.bdd_global f, false)
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      with Not_found -> 
	let (bdd, dep) =
	  match f with 
	      Not(f1) ->
		let (bdd_not, dep) =  (formula_to_bdd input f1) in
		  (Bdd.dnot bdd_not, dep)

	    | Or(f1, f2) ->
		let (bdd1, dep1) = (formula_to_bdd input f1)
		and (bdd2, dep2) = (formula_to_bdd input f2)
		in
		  (Bdd.dor bdd1 bdd2, dep1 || dep2)

	    | And(f1, f2) -> 
		let (bdd1, dep1) = (formula_to_bdd input f1)
		and (bdd2, dep2) = (formula_to_bdd input f2)
		in
		  (Bdd.dand bdd1 bdd2, dep1 || dep2)
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	    | EqB(f1, f2) -> 
		let (bdd1, dep1) = (formula_to_bdd input f1)
		and (bdd2, dep2) = (formula_to_bdd input f2)
		in
		  (Bdd.eq bdd1 bdd2, dep1 || dep2)
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	    | IteB(f1, f2, f3) -> 
		let (bdd1, dep1) = (formula_to_bdd input f1)
		and (bdd2, dep2) = (formula_to_bdd input f2)
		and (bdd3, dep3) = (formula_to_bdd input f3) 
		in
		  ((Bdd.dor (Bdd.dand bdd1 bdd2) 
		      (Bdd.dand (Bdd.dnot bdd1) bdd3)),
		   dep1 || dep2 || dep3 )
		  
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	    | True ->  (Bdd.dtrue  (Env_state.bdd_manager ()), false)
	    | False -> (Bdd.dfalse (Env_state.bdd_manager ()), false)
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	    | Bvar(vn) ->    
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		( match (lookup input (Env_state.pre ()) vn) with 
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		      Some(B(bool)) -> 
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			if bool
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			then (Bdd.dtrue  (Env_state.bdd_manager ()), true)
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			else (Bdd.dfalse (Env_state.bdd_manager ()), true)
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		    | Some(_) -> 
			print_string (vn ^ " is not a boolean!\n");
			assert false
		    | None ->
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			if List.mem vn (Env_state.pre_var_names ())
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			then failwith 
			  ("*** " ^ vn ^ " is unknown at this stage.\n "
			   ^ "*** Make sure you have not used "
			   ^ "a pre on a output var at the 1st step, \n "
			   ^ "*** or a pre on a input var at the second step in "
			   ^ "your formula in the environment.\n ")
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			else (Bdd.ithvar (Env_state.bdd_manager ())
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				(Env_state.linear_constraint_to_index (Bv(vn)) false), 
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				false)
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		)
		
	    | Eq(e1, e2) -> 
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		let gne = expr_to_gne (Diff(e1, e2)) input in 
		  (gne_to_bdd gne EqZero)
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	    | Neq(e1, e2) -> 
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		let gne = expr_to_gne (Diff(e1, e2)) input in
		  (gne_to_bdd gne NeqZero)
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	    | SupEq(e1, e2) ->
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		let gne = expr_to_gne (Diff(e1, e2)) input in
		  (gne_to_bdd gne SupEqZero)
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	    | Sup(e1, e2)   ->
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		let gne = expr_to_gne (Diff(e1, e2)) input in
		  (gne_to_bdd gne SupZero)
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	    | InfEq(e1, e2) ->  
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		let gne = expr_to_gne (Diff(e2, e1)) input in
		  (gne_to_bdd gne SupEqZero)
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	    | Inf(e1, e2)   ->  
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		let gne =  expr_to_gne (Diff(e2, e1)) input in
		  (gne_to_bdd gne SupZero)
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	in
	  if dep
	  then 
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	    ( Env_state.set_bdd f bdd;	
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	      match f with 
		  Not(nf) -> () (* Already in the tbl thanks to the rec call *)
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		| _  -> Env_state.set_bdd (Not(f)) (Bdd.dnot bdd) 
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	    )
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	  else 
	    (* [f] does not depend on pre nor input vars *)
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	    ( Env_state.set_bdd_global f bdd ;	
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	      match f with 
		  Not(nf) -> () (* Already in the table thanks to the rec call *)
		| _  -> 
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		    Env_state.set_bdd_global (Not(f)) (Bdd.dnot bdd)
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	    );

	  (bdd, dep)
and
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  (expr_to_gne: expr -> env_in -> Gne.t) =
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  fun e input -> 
    (** Evaluates pre and input vars appearing in [e] and tranlates
      it into a so-called garded normal form. Also returns a flag
      that is true iff [e] depends on pre or input vars. *)
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    let gne =
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      match e with  
	  Sum(e1, e2) ->
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	    let gne1 = (expr_to_gne e1 input)
	    and gne2 = (expr_to_gne e2 input) 
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	    in
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	      Gne.add  gne1 gne2
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	| Diff(e1, e2) -> 
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	    let gne1 = (expr_to_gne e1 input)
	    and gne2 = (expr_to_gne e2 input) 
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	    in
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	      Gne.diff gne1 gne2
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	| Prod(e1, e2) -> 
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	    let gne1 = (expr_to_gne e1 input)
	    and gne2 = (expr_to_gne e2 input) 
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	    in
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	      Gne.mult gne1 gne2
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	| Quot(e1, e2) -> 
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	    let gne1 = (expr_to_gne e1 input)
	    and gne2 = (expr_to_gne e2 input) 
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	    in
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	      Gne.quot gne1 gne2
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	| Mod(e1, e2)  -> 
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	    let gne1 = (expr_to_gne e1 input)
	    and gne2 = (expr_to_gne e2 input) 
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	    in
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	      Gne.modulo gne1 gne2
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	| Ivar(str) ->
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	    ( match (lookup input (Env_state.pre ()) str) with 
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		  Some(N(I(i))) ->
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		    (Gne.make 
		       (Ne.make "" (I i)) 
		       (Bdd.dtrue (Env_state.bdd_manager ()))
		       true
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		    )
		| None ->
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		    (Gne.make 
		       (Ne.make str (I(1)))
		       (Bdd.dtrue (Env_state.bdd_manager ()))
		       false
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		    )
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		| Some(N(F(f))) -> 
		    print_string ((string_of_float f) 
				  ^ "is a float, but an int is expected.\n");
		    assert false
		| Some(B(f)) -> 
		    print_string ((string_of_bool f) 
				  ^ "is a bool, but an int is expected.\n");
		    assert false
	    )

	| Fvar(str) ->
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	    ( match (lookup input (Env_state.pre ()) str) with 
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		  Some(N(F(f))) ->
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		    ( Gne.make 
			(Ne.make "" (F(f))) 
			(Bdd.dtrue (Env_state.bdd_manager ()))
			true
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		    )
		| None ->
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		    ( Gne.make 
			(Ne.make str (F(1.))) 
			(Bdd.dtrue (Env_state.bdd_manager ()))
			false
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		    )
		| Some(N(I(i))) -> 
		    print_string ((string_of_int i) 
				  ^ "is an int, but a float is expected.\n");
		    assert false
		| Some(B(f)) -> 
		    print_string ((string_of_bool f) 
				  ^ "is a bool, not a float is expected.\n");
		    assert false
	    )

	| Ival(i) ->  
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	    (Gne.make 
	       (Ne.make "" (I(i))) 
	       (Bdd.dtrue (Env_state.bdd_manager ()))
	       false
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	    )

	| Fval(f) -> 
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	    ( Gne.make 
		(Ne.make "" (F(f))) 
		(Bdd.dtrue (Env_state.bdd_manager ()))
		false
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	    )

	| Ite(f, e1, e2) -> 
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	    let (add_formula_to_gne_acc : Bdd.t -> bool -> Ne.t -> 
		   Bdd.t * bool -> Gne.t -> Gne.t) = 
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	      fun bdd dep1 nexpr (c, dep2) acc -> 
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		(* Used (by a Gne.fold) to add the condition [c] to every
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		   condition of a garded expression. *)
		let _ = assert (
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		  try 		    let _ = Gne.find nexpr acc in
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		      false
		  with Not_found -> true
		) 
		in
		let new_bdd = (Bdd.dand bdd c) in
		  if Bdd.is_false new_bdd
		  then acc
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		  else Gne.add_ne nexpr new_bdd (dep1 || dep2) acc
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	    in
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	    let (bdd, depf) = formula_to_bdd input f in
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	    let bdd_not = Bdd.dnot bdd
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	    and gne_t = (expr_to_gne e1 input)
	    and gne_e = (expr_to_gne e2 input) in
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	    let gne1 = 
	      Gne.fold (add_formula_to_gne_acc bdd depf) gne_t (Gne.empty ())
	    in
	    let gne  = Gne.fold (add_formula_to_gne_acc bdd_not depf) gne_e gne1 in
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	      gne
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    in
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      gne
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and
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  (gne_to_bdd : Gne.t -> comp -> Bdd.t * bool) =
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  fun gne cmp -> 
    (** Use [cmp] to compare [gne] with 0 and returns the
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      corresponding formula.  E.g., if [gne] is bounded to
      [e1 -> c1; e2 -> c2], then [gne_to_bdd gne SupZero] returns
      (the bdd corresponding to) the formula [(c1 and (e1 > 0)) or
      (c2 and (e2 > 0))] *)
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    match cmp with
	SupZero ->
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	  ( Gne.fold 
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	      (fun nexpr (c, dep) (acc, dep_acc) -> 
		 let new_dep = dep || dep_acc 
		 and bdd = 
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		   if Ne.is_a_constant nexpr
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		   then 
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		     let cst = Ne.find "" nexpr in
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		       match cst with
			   I(i) -> 
			     if i > 0 
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			     then (Bdd.dtrue (Env_state.bdd_manager ()))
			     else (Bdd.dfalse (Env_state.bdd_manager ()))
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			 | F(f) -> 
			     if f > 0. 
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			     then (Bdd.dtrue (Env_state.bdd_manager ()))
			     else (Bdd.dfalse (Env_state.bdd_manager ()))
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		   else 
		     Bdd.ithvar 
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		       (Env_state.bdd_manager ()) 
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		       (Env_state.linear_constraint_to_index (Ineq(GZ(nexpr))) dep) 
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		 in
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		   (Bdd.dor (Bdd.dand c bdd) acc, new_dep)
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	      )
	      gne 
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	      ((Bdd.dfalse (Env_state.bdd_manager ())), false)
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	  )
      | SupEqZero ->
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	  ( Gne.fold 
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	      (fun nexpr (c, dep) (acc, dep_acc) -> 
		 let new_dep = dep || dep_acc 
		 and bdd = 
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		   if Ne.is_a_constant nexpr
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		   then 
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		     let cst = Ne.find "" nexpr in
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		       match cst with
			   I(i) -> 
			     if i >= 0 
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			     then (Bdd.dtrue (Env_state.bdd_manager ()))
			     else (Bdd.dfalse (Env_state.bdd_manager ()))
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			 | F(f) -> 
			     if f >= 0. 
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			     then (Bdd.dtrue (Env_state.bdd_manager ()))
			     else (Bdd.dfalse (Env_state.bdd_manager ()))
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		   else 
		     Bdd.ithvar 
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		       (Env_state.bdd_manager ()) 
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		       (Env_state.linear_constraint_to_index (Ineq(GeqZ(nexpr))) dep)
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		 in
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		   (Bdd.dor (Bdd.dand c bdd) acc, new_dep)
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	      )
	      gne 
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	      ((Bdd.dfalse (Env_state.bdd_manager ())), false)
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	  )
      | EqZero -> 
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	  ( Gne.fold 
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	      (fun nexpr (c, dep) (acc, dep_acc) -> 
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		 let new_dep = dep || dep_acc 
		 and bdd = 
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		   if Ne.is_a_constant nexpr
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		   then 
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		     let cst = Ne.find "" nexpr in
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		       match cst with
			   I(i) -> 
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			     if i = 0 
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			     then (Bdd.dtrue (Env_state.bdd_manager ()))
			     else (Bdd.dfalse (Env_state.bdd_manager ()))
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			 | F(f) -> 
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			     if f = 0. 
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			     then (Bdd.dtrue (Env_state.bdd_manager ()))
			     else (Bdd.dfalse (Env_state.bdd_manager ()))
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		   else 
		     Bdd.ithvar 
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		       (Env_state.bdd_manager ()) 
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		       (Env_state.linear_constraint_to_index (EqZ(nexpr)) dep)
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		 in
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		   (Bdd.dor (Bdd.dand c bdd) acc, new_dep)
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	      )
	      gne 
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	      ((Bdd.dfalse (Env_state.bdd_manager ())), false)
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	  )
      | NeqZero -> 
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	  ( Gne.fold 
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	      (fun nexpr (c, dep) (acc, dep_acc) -> 
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		 let new_dep = dep || dep_acc 
		 and bdd = 
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		   if Ne.is_a_constant nexpr
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		   then 
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		     let cst = Ne.find "" nexpr in
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		       match cst with
			   I(i) -> 
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			     if i = 0 
			     then (Bdd.dfalse (Env_state.bdd_manager ()))
			     else (Bdd.dtrue (Env_state.bdd_manager ()))
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			 | F(f) -> 
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			     if f = 0. 
			     then (Bdd.dfalse (Env_state.bdd_manager ()))
			     else (Bdd.dtrue (Env_state.bdd_manager ()))
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		   else 
		     Bdd.ithvar 
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		       (Env_state.bdd_manager ()) 
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		       (Env_state.linear_constraint_to_index (EqZ(nexpr)) dep)
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		 in
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		   (Bdd.dor (Bdd.dand c (Bdd.dnot bdd)) acc, new_dep)
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	      )
	      gne 
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	      ((Bdd.dfalse (Env_state.bdd_manager ())), false)
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	  )
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(*       | EqZero ->  *)
(* 	  ( Gne.fold  *)
(* 	      (fun nexpr (c, dep) (acc, dep_acc) ->  *)
(* 		 let bdd1 =  *)
(* 		   if Ne.is_a_constant nexpr *)
(* 		   then  *)
(* 		     let cst = Ne.find "" nexpr in *)
(* 		       match cst with *)
(* 			   I(i) ->  *)
(* 			     if i >= 0  *)
(* 			     then (Bdd.dtrue (Env_state.bdd_manager ())) *)
(* 			     else (Bdd.dfalse (Env_state.bdd_manager ())) *)
(* 			 | F(f) ->  *)
(* 			     if f >= 0.  *)
(* 			     then (Bdd.dtrue (Env_state.bdd_manager ())) *)
(* 			     else (Bdd.dfalse (Env_state.bdd_manager ())) *)
(*  *)
(* 		   else  *)
(* 		     Bdd.ithvar  *)
(* 		       (Env_state.bdd_manager ())  *)
(* 		       (Env_state.linear_constraint_to_index (GeqZ(nexpr)) dep)  *)
(* 		 in *)
(* 		 let bdd2 =  *)
(* 		   if Ne.is_a_constant nexpr *)
(* 		   then  *)
(* 		     let cst = Ne.find "" nexpr in *)
(* 		       match cst with *)
(* 			   I(i) ->  *)
(* 			     if i <= 0  *)
(* 			     then (Bdd.dtrue (Env_state.bdd_manager ())) *)
(* 			     else (Bdd.dfalse (Env_state.bdd_manager ())) *)
(* 			 | F(f) ->  *)
(* 			     if f <= 0.  *)
(* 			     then (Bdd.dtrue (Env_state.bdd_manager ())) *)
(* 			     else (Bdd.dfalse (Env_state.bdd_manager ())) *)
(*  *)
(* 		   else  *)
(* 		     Bdd.ithvar  *)
(* 		       (Env_state.bdd_manager ())  *)
(* 		       (Env_state.linear_constraint_to_index  *)
(* 			  (GeqZ(Ne.neg_nexpr nexpr)) dep)  *)
(* 		 in *)
(* 		 let new_dep = dep || dep_acc  *)
(* 		 and bdd = Bdd.dand bdd1 bdd2 in  *)
(* 		   (* We transform [e1 = e2] into [e1 <= e2 ^ e1 >= e2] as the  *)
(* 		      numeric solver can not handle equalities *) *)
(* 		   (Bdd.dor (Bdd.dand c bdd) acc, new_dep) *)
(* 	      ) *)
(* 	      gne  *)
(* 	      ((Bdd.dfalse (Env_state.bdd_manager ())), false) *)
(* 	  ) *)
(*  *)
(*       | NeqZero ->  *)
(*  *)
(* 	  ( Gne.fold  *)
(* 	      (fun nexpr (c, dep) (acc, dep_acc) ->  *)
(* 		 let bdd1 =  *)
(* 		   if Ne.is_a_constant nexpr *)
(* 		   then  *)
(* 		     let cst = Ne.find "" nexpr in *)
(* 		       match cst with *)
(* 			   I(i) ->  *)
(* 			     if i > 0  *)
(* 			     then (Bdd.dtrue (Env_state.bdd_manager ())) *)
(* 			     else (Bdd.dfalse (Env_state.bdd_manager ())) *)
(* 			 | F(f) ->  *)
(* 			     if f > 0.  *)
(* 			     then (Bdd.dtrue (Env_state.bdd_manager ())) *)
(* 			     else (Bdd.dfalse (Env_state.bdd_manager ())) *)
(*  *)
(* 		   else  *)
(* 		     Bdd.ithvar  *)
(* 		       (Env_state.bdd_manager ())  *)
(* 		       (Env_state.linear_constraint_to_index (GZ(nexpr)) dep)  *)
(* 		 in *)
(* 		 let bdd2 =  *)
(* 		   if Ne.is_a_constant nexpr *)
(* 		   then  *)
(* 		     let cst = Ne.find "" nexpr in *)
(* 		       match cst with *)
(* 			   I(i) ->  *)
(* 			     if i < 0  *)
(* 			     then (Bdd.dtrue (Env_state.bdd_manager ())) *)
(* 			     else (Bdd.dfalse (Env_state.bdd_manager ())) *)
(* 			 | F(f) ->  *)
(* 			     if f < 0.  *)
(* 			     then (Bdd.dtrue (Env_state.bdd_manager ())) *)
(* 			     else (Bdd.dfalse (Env_state.bdd_manager ())) *)
(*  *)
(* 		   else  *)
(* 		     Bdd.ithvar  *)
(* 		       (Env_state.bdd_manager ())  *)
(* 		       (Env_state.linear_constraint_to_index  *)
(* 			  (GZ(Ne.neg_nexpr nexpr)) dep) *)
(* 		 in *)
(* 		 let new_dep = dep || dep_acc  *)
(* 		 and bdd = Bdd.dor bdd1 bdd2 in  *)
(* 		   (* We transform [e1 <> e2] into [e1 < e2 or e1 > e2] as the  *)
(* 		      numeric solver can not handle disequalities *) *)
(* 		   (Bdd.dor (Bdd.dand c bdd) acc, new_dep) *)
(* 	      ) *)
(* 	      gne  *)
(* 	      ((Bdd.dfalse (Env_state.bdd_manager ())), false) *)
(* 	  ) *)

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(****************************************************************************)
(****************************************************************************)


(* Exported *)
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let rec (is_satisfiable: env_in -> formula -> bool) = 
  fun input f -> 
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(*     let _ = if Env_state.verbose () then *)
(*       ( *)
(* 	print_string (formula_to_string f);  *)
(* 	print_string "  ...test whether this formula is satisfiable...\n"; *)
(* 	flush stdout *)
(*       ) *)
(*     in *)
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    let (bdd, _) = formula_to_bdd input f in
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      not (Bdd.is_false bdd) &&
      ( 
	try 
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	  let (n, m) = Env_state.sol_number bdd in 
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	    not ((zero_sol, zero_sol) = (n, m))
	with Not_found -> true
      )
      


(****************************************************************************)
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(****************************************************************************)


(** In the following, we call a comb the bdd of a conjunction of
 litterals (var). They provide the ordering in which litterals
 appear in the bdds we manipulate.
*)


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type var = int
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let rec (build_sol_nb_table: Bdd.t -> Bdd.t -> sol_nb * sol_nb) =
  fun bdd comb -> 
    (** Returns the relative (to which bbd points to it) number of
      solutions of [bdd] and the one of its negation. Also udpates
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      the solution number table for [bdd] and its negation, and
      recursively for all its sub-bdds.

      [comb] is a positive cube that ougth to contain the indexes of
      boolean vars that are still to be generated, and the numerical
      indexes that appears in [bdd].  
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    *)
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    let _ = assert (not (Bdd.is_cst bdd) 
		    && (Bdd.topvar comb) = (Bdd.topvar bdd)) 
    in
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    let bdd_not = (Bdd.dnot bdd) in
    let (sol_nb, sol_nb_not) =
      try
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	let (nt, ne) = Env_state.sol_number bdd 
	and (not_nt, not_ne) = Env_state.sol_number bdd_not in
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	  (* solutions numbers in the table are absolute *)
	  ((add_sol_nb nt ne), (add_sol_nb not_nt not_ne))
562
      with Not_found ->
563 564
	let (nt, not_nt) = compute_absolute_sol_nb (Bdd.dthen bdd) comb in
	let (ne, not_ne) = compute_absolute_sol_nb (Bdd.delse bdd) comb in
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	  Env_state.set_sol_number bdd (nt, ne) ;
	  Env_state.set_sol_number bdd_not (not_nt, not_ne) ;
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	  ((add_sol_nb nt ne), (add_sol_nb not_nt not_ne))
    in
      (sol_nb, sol_nb_not)
and 
  (compute_absolute_sol_nb: Bdd.t -> Bdd.t -> sol_nb * sol_nb) =
  fun sub_bdd comb -> 
573
    (* Returns the absolute number of solutions of [sub_bdd] (and its
574
       negation) w.r.t. [comb], where [comb] is the comb of the
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       father of [sub_bdd].

       The [comb] is used to know which output boolean variables are
       unconstraint along a path in the bdd. Indeed, the comb is made
       of all the boolean output var indexes plus the num contraints
       indexes that appears in the bdd; hence, if the topvar of the
       bdd is different from the topvar of the comb, it means that
       the topvar of the comb is unsconstraint and we need to
       multiply the number of solution of the branch by 2.
    *)
585
    if Bdd.is_cst sub_bdd 
586
    then
587
      let sol_nb = 
588
	if Bdd.is_true comb
589
	then one_sol
590
	else (two_power_of (List.length (Bdd.list_of_support (Bdd.dthen comb)))) 
591
      in
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	if Bdd.is_true sub_bdd
	then (sol_nb, zero_sol) 
	else (zero_sol, sol_nb)
    else 
      let topvar = Bdd.topvar sub_bdd in
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      let rec
	(count_missing_vars: Bdd.t -> var -> int -> Bdd.t * int) =
	fun comb var cpt -> 
	  (* Returns [cpt] + the number of variables occurring in [comb]
	     before reaching [var] ([var] excluded). Also returns the comb
	     whch topvar is [var]. *)
	  let _ = assert (not (Bdd.is_cst comb)) in
	  let combvar = Bdd.topvar comb in
	    if var = combvar
	    then (comb, cpt)
	    else count_missing_vars (Bdd.dthen comb) var (cpt+1)
      in
609
      let (sub_comb, missing_vars_nb) = 
610
	count_missing_vars (Bdd.dthen comb) topvar 0
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      in
      let (n0, not_n0) = build_sol_nb_table sub_bdd sub_comb in
      let factor = (two_power_of missing_vars_nb) in
	(mult_sol_nb n0 factor, mult_sol_nb not_n0 factor)
	
616

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(****************************************************************************)
(****************************************************************************)

621

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let (toss_up_one_var: var_name -> subst) =
  fun var -> 
   (* *)
    let ran = Random.float 1. in
      if (ran < 0.5) 
      then (var, B(true)) 
      else (var, B(false))


let (toss_up_one_var_index: var -> subst option) =
632
  fun var -> 
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    (* if [var] is a index that corresponds to a boolean variable,
       this fonction performs a toss and returns a substitution for
       the corresponding boolean variable. It returns [None]
       otherwise.

       Indeed, if it happens that a numerical constraint does not
       appear along a path, we simply ignore it and hence it will not
       be added to the store.
    *)
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    let cstr = Env_state.index_to_linear_constraint var in
      match cstr with 
          Bv(vn) -> Some(toss_up_one_var vn)
645
	| _  -> None
646

647

648 649
let rec (draw_in_bdd: subst list * store -> Bdd.t -> Bdd.t -> 
	   subst list * store) = 
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  fun (sl, store) bdd comb ->
    (** Returns [sl] appended to a draw of all the boolean variables
      bigger than the topvar of [bdd] according to the ordering
      induced by the comb [comb]. Also returns the (non empty) store
      obtained by adding to [store] all the numeric constraints that
      were encountered during this draw.
656

657 658 659
      Raises the [No_numeric_solution] exception whenever no valid
      path in [bdd] leads to a satisfiable set of numeric
      constraints.  
660
    *)
661
    
662 663
    if 
      Bdd.is_true bdd
664 665 666
    then
      (* Toss the remaining bool vars. *)
      ( (List.append sl
667
	   (Util.list_map_option toss_up_one_var_index (Bdd.list_of_support comb))),
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	store )
    else
      let _ = assert (not (Bdd.is_false bdd)) in 
      let _ = assert (Env_state.sol_number_exists bdd) in
      let bddvar  = Bdd.topvar bdd in
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      let cstr = (Env_state.index_to_linear_constraint bddvar) in 
	match cstr with
675 676
	    Bv(var) -> 
	      draw_in_bdd_bool var  (sl, store) bdd comb
677 678
	  | EqZ(e) -> 
	      draw_in_bdd_eq e (sl, store) bdd comb
679 680
	  |  Ineq(ineq) ->  
	      draw_in_bdd_ineq ineq (sl, store) bdd comb
681

682 683


684 685 686
and (draw_in_bdd_bool: string -> subst list * store -> Bdd.t -> Bdd.t -> 
       subst list * store) = 
  fun var (sl, store) bdd comb ->
687
    let bddvar = Bdd.topvar bdd in
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    let topvar_comb  = Bdd.topvar comb in
      
    if
      bddvar <> topvar_comb 
    then
      (* that condition means that topvar_comb is an unconstraint
	 boolean var; hence we toss it up. *)
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      let new_sl = 
	match toss_up_one_var_index topvar_comb with
	    Some(s) -> s::sl
	  | None -> sl
      in
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	draw_in_bdd (new_sl, store) bdd (Bdd.dthen comb) 
    else 
      (* bddvar = combvar *) 
      let (n, m) = Env_state.sol_number bdd in
      let _ =
	if ((eq_sol_nb n zero_sol) && (eq_sol_nb m zero_sol))
	then raise No_numeric_solution ;
      in
      let (
	bdd1, bool1, sol_nb1,
	bdd2, bool2, sol_nb2
      ) =
	let ran = Random.float 1. 
	and sol_nb_ratio = 
	  ((float_of_sol_nb n) /. (float_of_sol_nb (add_sol_nb n m))) 
	in
	  if 
	    ran < sol_nb_ratio
	      (* 
		 Depending on the result of a toss (based on the number
		 of solution in each branch), we try the [then] or the
		 [else] branch first.  
	      *)
	  then
	    ((Bdd.dthen bdd), true, n,
	     (Bdd.delse bdd), false, m )
	  else 
	    ((Bdd.delse bdd), false, m,
	     (Bdd.dthen bdd), true, n )
      in
      let (sl1, sl2, new_comb) = (
	((var, B(bool1))::sl), 
	((var, B(bool2))::sl),
	(if Bdd.is_true comb then comb else Bdd.dthen comb) 
      )
	    
      in
	(* 
	   A solution will be found in this branch iff there exists
	   at least one path in the bdd that leads to a satisfiable
	   set of numeric constraints. If it is not the case,
	   [res_opt] is bound to [None]. 
	*)
	try 
	  draw_in_bdd (sl1, store) bdd1 new_comb
	with No_numeric_solution -> 
	  if not (eq_sol_nb sol_nb2 zero_sol)
	  then draw_in_bdd (sl2, store) bdd2 new_comb
	    (* 
	       The second branch is now tried because no path in
	       the first bdd leaded to a satisfiable set of
	       numeric constraints. 
	    *) 
	  else raise No_numeric_solution
754
      
755
and (draw_in_bdd_ineq: Constraint.ineq -> subst list * store -> Bdd.t -> Bdd.t -> 
756 757 758 759 760 761 762 763 764 765 766 767 768 769 770 771 772 773 774 775 776 777 778 779 780 781 782 783 784 785 786 787 788 789 790 791 792 793 794 795 796 797 798 799 800 801 802 803 804 805 806 807 808 809 810 811 812 813 814 815 816 817 818 819
	   subst list * store) = 
  fun cstr (sl, store) bdd comb ->
    (* 
       When we add to the store an inequality constraint GZ(ne) or
       GeqZ(ne) ([split_store]), 2 stores are returned. The first is a
       store where the inequality has been added; the second is a
       store where its negation has been added.
       
       Whether we choose the first one or the second depends on a toss
       made according the (boolean) solution number in both branches
       of the bdd. If no solution is found into that branch, the other
       branch with the other store is tried.
    *)
    let (n, m) = Env_state.sol_number bdd in
    let _ =
      if ((eq_sol_nb n zero_sol) && (eq_sol_nb m zero_sol))
      then raise No_numeric_solution ;
    in
    let (store_plus_cstr, store_plus_not_cstr) = split_store store cstr in
    let (store1, bdd1, sol_nb1,  store2, bdd2, sol_nb2) =
      let ran = Random.float 1. in
	if 
	  ran < ((float_of_sol_nb n) /. (float_of_sol_nb (add_sol_nb n m)))
	    (* 
	       Depending on the result of a toss (based on the number
	       of solution in each branch), we try the [then] or the
	       [else] branch first.  
	    *)
	then
	  (store_plus_cstr,      (Bdd.dthen bdd), n,
	   store_plus_not_cstr,  (Bdd.delse bdd), m)
	else 
	  (store_plus_not_cstr, (Bdd.delse bdd), m,
	   store_plus_cstr, (Bdd.dthen bdd), n )
    in
    let call_choice_point _ =
      (* 
	 The second branch is tried if no path in the first bdd leaded
	 to a satisfiable set of numeric constraints.  
      *)
      if 
	not (eq_sol_nb sol_nb2 zero_sol)
      then
	if 
	  is_store_satisfiable store2
	then 
	  draw_in_bdd (sl, store2) bdd2 comb
	else
	  raise No_numeric_solution
      else
	raise No_numeric_solution
    in
      (* A solution will be found in this branch iff there exists
	 at least one path in the bdd that leads to a satisfiable
	 set of numeric constraints. If it is not the case,
	 [res_opt] is bound to [None]. *)
      if 
	is_store_satisfiable store1
      then 
	try draw_in_bdd (sl, store1) bdd1 comb 
	with No_numeric_solution -> call_choice_point ()
      else
	call_choice_point ()
    	
820
and (draw_in_bdd_eq: Ne.t -> subst list * store -> Bdd.t -> Bdd.t -> 
821
       subst list * store) = 
822
  fun ne (sl, store) bdd comb ->
823 824 825 826 827 828 829 830 831 832 833 834 835 836
    (*
      When we add to the store an equality constraint EqZ(ne)
      ([split_store_eq]), 3 stores are returned. The first is a
      store where the equilaty has been added; the second is a store
      where the inequality [ne > 0] has been added; the third is a
      store where [ne < 0] has been added.

      Whether we choose the first one on not depends on toss made
      according the (boolean) solution number in both branchs of the
      bdd. If the else branch is choose (if it is chosen in the
      first place, or if backtracking occurs because no solutions is
      found in the then branch) whether we try the second or the
      third store first is (fairly) tossed up.
    *)
837 838 839 840 841 842
    let (n, m) = Env_state.sol_number bdd in
    let _ =
      if ((eq_sol_nb n zero_sol) && (eq_sol_nb m zero_sol))
      then raise No_numeric_solution ;
    in
    let (store_plus_cstr, store_plus_not_cstr, store_plus_not_cstr2) =  
843
      split_store_eq store ne  
844 845 846 847 848 849 850 851 852 853 854 855 856 857 858 859 860 861
    in
    let (
      store1, bdd1, sol_nb1, 
      store2, bdd2, sol_nb2,
      store3, bdd3, sol_nb3
    ) =
      let ran0 = Random.float 1. 
      and ran  = Random.float 1. 
      and sol_nb_ratio = 
	((float_of_sol_nb n) /. (float_of_sol_nb (add_sol_nb n m))) 
      in
	if 
	  ran0 < 0.5
	    (* 
	       When taking the negation of an equality, we can
	       either try > or <. Here, We toss which one we
	       will try first.  
	    *)
862
	then
863 864
	  if 
	    ran < sol_nb_ratio
865
	      (*
866 867 868 869 870 871 872 873 874 875 876
		 Depending on the result of a toss (based on the number
		 of solution in each branch), we try the [then] or the
		 [else] branch first.  
	      *)
	  then
	    (store_plus_cstr,      (Bdd.dthen bdd), n,
	     store_plus_not_cstr,  (Bdd.delse bdd), m,
	     store_plus_not_cstr2, (Bdd.delse bdd), m)
	  else 
	    (store_plus_not_cstr, (Bdd.delse bdd), m,
	     store_plus_cstr, (Bdd.dthen bdd), n,
877
	     store_plus_not_cstr2, (Bdd.delse bdd), m)
878 879 880 881 882 883 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934
	else
	  if
	    ran < sol_nb_ratio
	      (* Ditto *)
	  then
	    (store_plus_cstr, (Bdd.dthen bdd), n,
	     store_plus_not_cstr2, (Bdd.delse bdd), m,
	     store_plus_not_cstr, (Bdd.delse bdd), m)
	  else 
	    (store_plus_not_cstr2, (Bdd.delse bdd), m,
	     store_plus_cstr, (Bdd.dthen bdd), n,
	     store_plus_not_cstr, (Bdd.delse bdd), m)
    in
    let call_choice_point3 _ =
      (* 
	 The third possibility is tried if no path is found in the 2
	 previous ones. Note that there only is a third one if the
	 current constraint is an equality.  
      *)
      if 
	not (eq_sol_nb sol_nb3 zero_sol)
      then
	if is_store_satisfiable store3
	then draw_in_bdd (sl, store3) bdd3 comb
	else raise No_numeric_solution
      else
	raise No_numeric_solution
    in
    let call_choice_point2 _ =
      (* 
	 The second branch is tried if no path in the first bdd
	 leaded to a satisfiable set of numeric constraints.  
      *)
      if 
	not (eq_sol_nb sol_nb2 zero_sol)
      then
	if is_store_satisfiable store2
	then draw_in_bdd (sl, store2) bdd2 comb
	else call_choice_point3 ()
      else
	call_choice_point3 ()
    in
      (* 
	 A solution will be found in this branch iff there exists
	 at least one path in the bdd that leads to a satisfiable
	 set of numeric constraints. 
      *)
      if 
	is_store_satisfiable store1
      then 
	try draw_in_bdd (sl, store1) bdd1 comb
	with No_numeric_solution -> call_choice_point2 ()
      else
	call_choice_point2 ()


		
935

936 937 938 939

(* exported *)
let (draw : vn list -> vnt list -> Bdd.t -> Bdd.t -> subst list * subst list) =
  fun bool_vars_to_gen num_vnt_to_gen comb bdd ->
940 941
    (** Draw the output and local vars to be generated by the environnent. *)
    let (bool_subst_l, store) = 
942
      draw_in_bdd ([], (new_store num_vnt_to_gen)) bdd comb
943
    in
944 945 946 947 948 949
    let num_subst_l = 
      match Env_state.draw_mode () with
	  Env_state.Verteces -> draw_verteces store
	| Env_state.Edges    -> draw_edges store
	| Env_state.Inside   -> draw_inside store 
    in
950
    let subst_l = append bool_subst_l num_subst_l in
951
    let (out_vars, _) = List.split (Env_state.output_var_names ()) 
952 953 954
    in
      assert ( 
	(*  Checks that we generated all variables. *)
955 956
	let (gen_vars, _) = List.split subst_l in
	let (num_vars_to_gen, _) = List.split num_vnt_to_gen in
957
	let vars_to_gen = append bool_vars_to_gen num_vars_to_gen in
958 959 960 961 962 963 964 965
          if (sort (compare) gen_vars) = (sort (compare) vars_to_gen) 
	  then true
	  else
	    (
	      output_string stderr " \ngen vars :";
              List.iter (fun vn -> output_string stderr (vn ^ " ")) gen_vars;
	      output_string stderr " \nvar to gen:";
	      List.iter (fun vn -> output_string stderr (vn ^ " ")) vars_to_gen;
966
	      output_string stderr " \n";
967 968
	      false
	    )
969 970 971 972 973
      );
      (* Splits output and local vars. *)
      List.partition 
	(fun (vn, _) -> List.mem vn out_vars) 
	subst_l
974

975

976 977 978 979 980

(****************************************************************************)
(****************************************************************************)


981
(* Exported *)
982 983
let (solve_formula: env_in -> int -> formula -> formula -> vnt list ->
       (subst list * subst list) list option) =
984 985 986 987 988 989 990 991 992
  fun input p f bool_vars_to_gen_f num_vars_to_gen  ->
    
    let _ = if Env_state.verbose () then
      (
	print_string (formula_to_string f); 
	print_string "\n";
	flush stdout
      )
      in
993

994
    let bdd = 
995
      (* The bdd of f has necessarily been computed (by is_satisfiable) *)
996 997
      try Env_state.bdd f
      with Not_found -> Env_state.bdd_global f
998
    in
999 1000 1001 1002 1003 1004 1005 1006 1007 1008 1009 1010 1011
    let (comb0, _) = formula_to_bdd input bool_vars_to_gen_f in
    let comb = 
      (* All boolean vars should appear in the comb so that when we
	 find that such a var is missing along a bdd path, we can
	 perform a (fair) toss for it. On the contrary, if a
	 numerical contraint disappear from a bdd (eg, consider [(f
	 && false) || true]), it is not important; fairly tossing a
	 (boolean) value for a num constaint [nc] and performing a
	 fair toss in the resulting domain is equivalent to directly
	 perform the toss in the (unconstraint wrt [nc]) initial
	 domain.  
      *)
      Bdd.dand (Bdd.support bdd) comb0 
1012 1013
    in	
    let bool_vars_to_gen = Formula.support bool_vars_to_gen_f in
1014
    let _ =
1015
      if not (Env_state.sol_number_exists bdd)
1016
      then
1017 1018 1019 1020 1021
	let rec skip_unconstraint_bool_var_at_top comb v =
	  (* [build_sol_nb_table] supposes that the bdd and its comb 
	     have the same top var. 
	  *)
	  if Bdd.is_true comb then comb
1022 1023
	  else 
	    let topvar = (Bdd.topvar comb) in
1024 1025
	      if v = topvar then comb 
	      else skip_unconstraint_bool_var_at_top (Bdd.dthen comb) v
1026
	in
1027
	let comb2 = skip_unconstraint_bool_var_at_top comb (Bdd.topvar bdd) in 
1028 1029
	let _ = build_sol_nb_table bdd comb2 in
	  ()
1030
    in
1031
      try 	
1032
	Some(Util.unfold (draw bool_vars_to_gen num_vars_to_gen comb) bdd p)
1033
      with No_numeric_solution -> 
1034
	Env_state.set_sol_number bdd (zero_sol, zero_sol);
1035
	None
1036