Get multiple answers from prolog predicate

Question

Suppose I have a predicate that sometimes gives me multiple outputs. Like this -

foo(Number, Out) :- Number < 10, Out = less_than_ten.
foo(Number, Out) :- Number > 0, Out = more_than_zero.

How might I get hold of all the values for Out that foo gives in another predicate, bearing in mind it can sometimes give one and sometimes give multiple (e.g. in a list)?

Edit - not quite sure the answers I've got answer my question so I'll be more specific. Taking the above predicate, I could run the query foo(5, Out). This satisfies both rules, so if I run it in SWI-prolog I'll get this -

?- foo(5, Out).
Out = less_than_ten

Then I can enter a semi-colon to get prolog to backtrack and look for the other solution -

?- foo(5, Out).
Out = less_than_ten ;
Out = more_than_zero.

So if I was executing this predicate within another, how do I get all the valid values for Out, given Number = 5?

tas tas · Accepted Answer · 2018-11-29T07:03:39

If you are only considering integers, you can opt to use CLP(FD). Then your predicate foo might look something like this:

:- use_module(library(clpfd)).

foo(Nums) :-
   Nums ins 1..9.     % the numbers in the list Nums are 0 < X < 10

You can use this predicate to test if a list of numbers is in your desired range:

   ?- foo([1,2,3,4,5]).
yes
   ?- foo([0,1,2,3,4,5]).
no
   ?- foo([1,2,3,4,5,10]).
no

If you want to use it to generate lists of integers in that range, you have make sure, that Nums is a list in order to avoid an instantiation error. You can do that by prefixing a goal length/2 in your query:

   ?- length(Nums,_), foo(Nums).
Nums = [] ? ;          % <- no number
Nums = [_A],           % <- one number
_A in 1..9 ? ;
Nums = [_A,_B],        % <- two numbers
_A in 1..9,
_B in 1..9 ? ;
Nums = [_A,_B,_C],     % <- three numbers
_A in 1..9,
_B in 1..9,
_C in 1..9 ?
.
.
.

These answers consist of residual goals (see the CLP(FD) documentation for details). If you want to see actual numbers, you have to add a goal to label the list:

   ?- length(Nums,_), foo(Nums), label(Nums).
Nums = [] ? ;
Nums = [1] ? ;
Nums = [2] ? ;
Nums = [3] ? ;
.
.
.
Nums = [1,1] ? ;
Nums = [1,2] ? ;
Nums = [1,3] ? ;
.
.
.
Nums = [9,9] ? ;
Nums = [1,1,1] ? ;
Nums = [1,1,2] ? ;
.
.
.

Get multiple answers from prolog predicate

2 Answers