solved issue #43

Flexible probabilities with ProbLog syntax

prolog/pita.pl

@@ -1159,27 +1159,37 @@
 
 tab_dir(_M,'':_,[],[]):-!.
 
-tab_dir(M,H:P,[],[H:P]):-
-  M:tabled(H),!.
-
 % tab dir for decision variables
 % merge with the previous one?
 % the predicates are equal except
 % (?)::H and H:P
-tab_dir(M,HT,[],[H]):-
-  (HT = ?::H ; HT = (?)::H),
+tab_dir(M,D::H,[],[H]):-
+  (D == ? ; D == (?)),
   M:tabled(H),!.
 % tab dir for decision variables
 % merge with the previous one?
 % the predicates are equal except
 % (?)::H and '$util'(A,B)
 tab_dir(M,H,[],[H]):-
-  M:tabled(H),!,
   H=..[F|_],
-  F = utility.
+  F = utility,
+  M:tabled(H),!.
+
+tab_dir(M,H:P,[],[H:P]):-
+  M:tabled(H),!.
+
+tab_dir(M,P::H,[],[H:P]):-
+  P \== ?,
+  M:tabled(H),!.
+
 
 
-tab_dir(M,H:P,[(:- table HT)],[H1:P]):-
+% tab dir for decision variables
+% merge with the previous one?
+% the predicates are equal
+% except variable n 2 and 4.
+tab_dir(M,P::H,[(:- table HT)],[H1]):-
+  (P== ?;P == (?)),!,
   functor(H,F,A0),
   functor(PT,F,A0),
   PT=..[F|Args0],
@@ -1196,12 +1206,13 @@
   zero_clause(M,F/A0,LZ),
   assert(M:zero_clauses(LZ)).
 
-% tab dir for decision variables
+% tab dir for utility variables
 % merge with the previous one?
 % the predicates are equal
 % except variable n 2 and 4.
-tab_dir(M,HT1,[(:- table HT)],[H1]):-
-  (HT1 = ?::H ; HT1 = (?)::H),
+tab_dir(M,H,[(:- table HT)],[H1]):-
+  H=..[F|_],
+  F = utility,!,
   functor(H,F,A0),
   functor(PT,F,A0),
   PT=..[F|Args0],
@@ -1218,13 +1229,9 @@
   zero_clause(M,F/A0,LZ),
   assert(M:zero_clauses(LZ)).
 
-% tab dir for utility variables
-% merge with the previous one?
-% the predicates are equal
-% except variable n 2 and 4.
-tab_dir(M,H,[(:- table HT)],[H1]):-
-  H=..[F|_],
-  F = utility,
+
+tab_dir(M,Head,[(:- table HT)],[H1:P]):-
+  (Head=H:P;Head=P::H),!,
   functor(H,F,A0),
   functor(PT,F,A0),
   PT=..[F|Args0],
@@ -1241,6 +1248,7 @@
   zero_clause(M,F/A0,LZ),
   assert(M:zero_clauses(LZ)).
 
+
 pita_expansion(begin_of_file,_):-
   !,
   fail.