solver.ml 31.2 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 (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 (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 (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))
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      with Not_found ->
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	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 -> 
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    (* Returns the absolute number of solutions of [sub_bdd] (and its
566
       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.
    *)
577
    if Bdd.is_cst sub_bdd 
578
    then
579
      let sol_nb = 
580
	if Bdd.is_true comb
581
	then one_sol
582
	else (two_power_of (List.length (Bdd.list_of_support (Bdd.dthen comb)))) 
583
      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
601
      let (sub_comb, missing_vars_nb) = 
602
	count_missing_vars (Bdd.dthen comb) topvar 0
603 604 605 606 607
      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)
	
608

609

610 611 612
(****************************************************************************)
(****************************************************************************)

613

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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) =
624
  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)
637
	| _  -> None
638

639

640

641
let (is_a_numeric_constraint : Constraint.t -> bool) =
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  fun cstr -> 
    match cstr with
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	Bv(_) -> false
      | GZ(_)   -> true 
      | GeqZ(_) -> true
647
      | EqZ(_)  -> true
648 649


650

651 652
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.
659

660 661 662
      Raises the [No_numeric_solution] exception whenever no valid
      path in [bdd] leads to a satisfiable set of numeric
      constraints.  
663
    *)
664
    
665 666
    if 
      Bdd.is_true bdd
667 668 669
    then
      (* Toss the remaining bool vars. *)
      ( (List.append sl
670
	   (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
676 677
      let cstr = (Env_state.index_to_linear_constraint bddvar) in 
	match cstr with
678 679
	    Bv(var) -> 
	      draw_in_bdd_bool var  (sl, store) bdd comb
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	  | EqZ(e) -> 
	      draw_in_bdd_eq e (sl, store) bdd comb
	  |  _ ->  
683
	      draw_in_bdd_ineq cstr (sl, store) bdd comb
684

685 686


687 688 689
and (draw_in_bdd_bool: string -> subst list * store -> Bdd.t -> Bdd.t -> 
       subst list * store) = 
  fun var (sl, store) bdd comb ->
690
    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
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and (draw_in_bdd_ineq: Constraint.t -> subst list * store -> Bdd.t -> Bdd.t -> 
	   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 ()
    	
823
and (draw_in_bdd_eq: Ne.t -> subst list * store -> Bdd.t -> Bdd.t -> 
824
       subst list * store) = 
825
  fun ne (sl, store) bdd comb ->
826 827 828 829 830 831 832 833 834 835 836 837 838 839
    (*
      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.
    *)
840 841 842 843 844 845
    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) =  
846
      split_store_eq store ne  
847 848 849 850 851 852 853 854 855 856 857 858 859 860 861 862 863 864
    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.  
	    *)
865
	then
866 867
	  if 
	    ran < sol_nb_ratio
868
	      (*
869 870 871 872 873 874 875 876 877 878 879
		 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,
880
	     store_plus_not_cstr2, (Bdd.delse bdd), m)
881 882 883 884 885 886 887 888 889 890 891 892 893 894 895 896 897 898 899 900 901 902 903 904 905 906 907 908 909 910 911 912 913 914 915 916 917 918 919 920 921 922 923 924 925 926 927 928 929 930 931 932 933 934 935 936 937
	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 ()


		
938

939 940 941 942

(* 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 ->
943 944
    (** Draw the output and local vars to be generated by the environnent. *)
    let (bool_subst_l, store) = 
945
      draw_in_bdd ([], (new_store num_vnt_to_gen)) bdd comb
946
    in
947 948 949 950 951 952
    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
953
    let subst_l = append bool_subst_l num_subst_l in
954
    let (out_vars, _) = List.split (Env_state.output_var_names ()) 
955 956 957
    in
      assert ( 
	(*  Checks that we generated all variables. *)
958 959
	let (gen_vars, _) = List.split subst_l in
	let (num_vars_to_gen, _) = List.split num_vnt_to_gen in
960
	let vars_to_gen = append bool_vars_to_gen num_vars_to_gen in
961 962 963 964 965 966 967 968
          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;
969
	      output_string stderr " \n";
970 971
	      false
	    )
972 973 974 975 976
      );
      (* Splits output and local vars. *)
      List.partition 
	(fun (vn, _) -> List.mem vn out_vars) 
	subst_l
977

978

979 980 981 982 983

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


984
(* Exported *)
985 986 987
let (solve_formula: env_in -> int -> formula -> formula -> vnt list ->
       (subst list * subst list) list option) =
  fun input p f bool_vars_to_gen_f num_vars_to_gen ->
988
    let bdd = 
989
      (* The bdd of f has necessarily been computed (by is_satisfiable) *)
990 991
      try Env_state.bdd f
      with Not_found -> Env_state.bdd_global f
992
    in
993 994 995 996 997 998 999 1000 1001 1002 1003 1004 1005
    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 
1006 1007
    in	
    let bool_vars_to_gen = Formula.support bool_vars_to_gen_f in
1008
    let _ =
1009
      if not (Env_state.sol_number_exists bdd)
1010
      then
1011 1012 1013 1014 1015
	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
1016 1017
	  else 
	    let topvar = (Bdd.topvar comb) in
1018 1019
	      if v = topvar then comb 
	      else skip_unconstraint_bool_var_at_top (Bdd.dthen comb) v
1020
	in
1021
	let comb2 = skip_unconstraint_bool_var_at_top comb (Bdd.topvar bdd) in 
1022 1023
	let _ = build_sol_nb_table bdd comb2 in
	  ()
1024
    in
1025
      try 	
1026
	Some(Util.unfold (draw bool_vars_to_gen num_vars_to_gen comb) bdd p)
1027
      with No_numeric_solution -> 
1028
	Env_state.set_sol_number bdd (zero_sol, zero_sol);
1029
	None
1030