:- use_module( library( pita)).
:- pita.
:- begin_lpad.
a.
b :- forall(prob( a, _), true). % don't work
c :- findall(1, prob( a, _), _). % don't work
d :- prob( a, _). % works
e :- forall(true, true). % works
f :- findall(1, true, _). % works
:- end_lpad.
/*
(ins)?- prob(b,P).
P = 0.0. % wrong
(cmd)?- prob(c,P).
P = 0.0. % wrong
(ins)?- prob(d,P).
P = 1.0. % ok
(cmd)?- prob(e,P).
P = 1.0. % ok
(ins)?- prob(f,P).
P = 1.0. % ok
*/
Regards
Hi Frank,
findall and forall are not supported by the probabilistic semantics. They are considered as builtin predicates and evaluated as they are, so if the query involves a probabilistic atom or the meta predicate prob it doesn’t work.
Best
Fabrizio
2 Likes
So I can just have this workaround:
:- use_module( library( pita)).
:- pita.
:- dynamic data/1.
a :- prob( a, _).
b.
e :- abolish(data/1), forall( prob( a, P), assertz( data(P))).
:- begin_lpad.
a.
b :- forall( prob(a,_), true). % don't work
c :- forall( a, true). % don't work
d :- forall( b, true). % works
% e :- db(e). % no P possible
f :- forall( data(_), true).
:- end_lpad.
/*
(ins)?- e, prob(f, P).
P = 1.0.
*/
Regards.
Yes you can but what is your use case?
I have no concrete usecase, just playing around and learning.
For instance this:
:- begin_lpad.
b(x) : 0.6; b(y) : 0.4.
a : P2 :- findall( X-P, prob( b(X), P), L), postprocess(L, P2).
:- end_lpad.
I want to gather probabilities of a predicate (b) in a list and postprocess them before I set a new probability for the predicate (a).
but maybe I should write something like:
b(x,0.6).
b(y,0.4).
:- begin_lpad.
b(X) : P :- db( b(X,P)).
a : P2 :- findall( X-P, b(X, P), L), postprocess(L, P2).
:- end_lpad.
This should work.
Now I have a working example:
:- use_module( library(pita)).
:- pita.
b(x,0.6).
b(y,0.4).
:- begin_lpad.
b : P :- db( b(X,P)).
b(X) : P :- db( b(X,P)).
c :- b(X1), b(X2), X1 \== X2.
postprocess( []).
postprocess( L) : P:- true
, L = [_X-P|T]
, postprocess(T)
.
a :- findall( X-P, b(X, P), L), postprocess(L).
:- end_lpad.
/*
(ins)?- prob( a, P).
P = 0.24.
(cmd)?- prob( c, P).
P = 0.24.
(ins)?- prob( b, P).
P = 0.76.
(ins)?- prob( b(X), P).
X = x,
P = 0.6 ;
X = y,
P = 0.4.
*/
a does the same as c but a is more flexible. So I could extend b/2 on top and a would be up to date.
c in the other hand should be updated as well on every step to work correctly.
And maybe I would be able to derive the last b/2 from al other b/2’s so that the P sum is always 1.
But I encountered a little trap. I made a mistake by writing ‘b(X) : P : db( b(X,P)).’ instead of ‘b(X) : P :- db( b(X,P)).’ and I didn’t see the mistake. And I wondered why it does not work. Next time I will look if this syntactic variation can be used for something useful.
Regards.
| Frank_Schwidom
March 30 |
- | - |
friguzzi:
Yes you can but what is your use case?
Now I have a working example:
:- use_module( library(pita)). :- pita. b(x,0.6). b(y,0.4). :- begin_lpad. b : P :- db( b(X,P)). b(X) : P :- db( b(X,P)). c :- b(X1), b(X2), X1 \== X2. postprocess( []). postprocess( L) : P:- true , L = [_X-P|T] , postprocess(T) . a :- findall( X-P, b(X, P), L), postprocess(L). :- end_lpad. /* (ins)?- prob( a, P). P = 0.24. (cmd)?- prob( c, P). P = 0.24. (ins)?- prob( b, P). P = 0.76. (ins)?- prob( b(X), P). X = x, P = 0.6 ; X = y, P = 0.4. */a does the same as c but a is more flexible. So I could extend b/2 on top and a would be up to date.
c in the other hand should be updated as well on every step to work correctly.And maybe I would be able to derive the last b/2 from al other b/2’s so that the P sum is always 1.
But I encountered a little trap. I made a mistake by writing ‘b(X) : P : db( b(X,P)).’ instead of ‘b(X) : P :- db( b(X,P)).’ and I didn’t see the mistake. And I wondered why it does not work. Next time I will look if this syntactic variation can be used for something useful.
I believe b(X) : P : db( b(X,P)) will be interpreted as a fact, not as a probabilistic fact, that’s why it didn’t work
Fabrizio