family relation in prolog :descendantTree - prolog

I want recursively store a family relation in a list structure. I have the following code:
descendantTree(X,[X]) :-
\+ parent(X,_).
descendantTree(X,L) :-
forall(parent(X,Y), descendantTree(Y,YL)),
append([X],[YL],L).
parent(john, paul).
parent(john, mary).
parent(paul, henry).
parent(paul, june).
parent(mary, adam).
parent(henry, helen).
What I expected is like:
L =[john,[paul,[henry,[helen]],[june]],[mary,[adam]]]
But actually it just return:
L = [john, _9250].
Could you tell me why?

The forall/2 predicate implements a generate and test loop using negation. It can be defined as:
% for all solutions of Generate, Test is true
forall(Generate, Test) :-
\+ (Generate, \+ Test)).
The use of negation means that no bindings will be returned when a call succeeds. In your case, that means that you're calling:
append([X],[_],L)
hence the result you get:
L = [john, _9250].
Look instead into the all-solutions family of predicates: bagof/3, setof/3, and findall/3 and then consider editing your question with your findings.

Related

unexpected backtracking in simple prolog example

The following is a simple prolog program.
% parent facts
parent(john, jane).
parent(john, james).
parent(sally, jane).
parent(martha, sally).
parent(deirdre, martha).
% ancestor recursive definition
ancestor(X,Y) :- parent(X,Y).
ancestor(X,Y) :- parent(X,A), ancestor(A,Y).
Using SWI-Prolog, specifically swish.swi-prolog.org, the following simple query produces the correct answer true, but also appears to backtrack and then say false.
?- ancestor(john, jane).
true
false
This puzzles me, and is unexpected.
I tried tracing:
trace, ancestor(john, jane).
Call:ancestor(john,jane)
Call:parent(john,jane)
Exit:parent(john,jane)
Exit:ancestor(john,jane)
1true
Redo:parent(john,jane)
Fail:parent(john,jane)
Redo:ancestor(john,jane)
Call:parent(john,_696)
Exit:parent(john,jane)
Call:ancestor(jane,jane)
Call:parent(jane,jane)
Fail:parent(jane,jane)
Redo:ancestor(jane,jane)
Call:parent(jane,_698)
Fail:parent(jane,_698)
Fail:ancestor(jane,jane)
Redo:parent(john,_696)
Exit:parent(john,james)
Call:ancestor(james,jane)
Call:parent(james,jane)
Fail:parent(james,jane)
Redo:ancestor(james,jane)
Call:parent(james,_698)
Fail:parent(james,_698)
Fail:ancestor(james,jane)
Fail:ancestor(john,jane)
What puzzles me is that prolog seems to redo: parent(john, jane). Question: Why is this?
My own (admittedly student) understanding is that prolog will only backtrack to try additional values for variables unbound at query time. Here there are no variables in the first rule of the recursive definition that are unbound at query time: ancestor(X,Y) :- parent(X,Y).
I could understand prolog trying new values for the variables in the second rule of the recursive definition: ancestor(X,Y) :- parent(X,A), ancestor(A,Y). Here prolog could try new values for A.
UPDATE - I tried removing the recursive part of the definition:
ancestor(X,Y) :- parent(X,Y).
% ancestor(X,Y) :- parent(X,A), ancestor(A,Y).
The query still results in backtracking.
?- ancestor(john, jane).
true
false
.. although the trace is shorter.
trace, ancestor(john, jane).
Call:ancestor(john,jane)
Call:parent(john,jane)
Exit:parent(john,jane)
Exit:ancestor(john,jane)
1true
Redo:parent(john,jane)
Fail:parent(john,jane)
Fail:ancestor(john,jane)
false
This focusses on the unexpected behaviour more sharply. Why does prolog backtrack to a point which has no unbound variables?
This can be boiled down to:
parent(john, jane).
parent(john, james).
parent(sally, jane).
?- parent(john, jane).
true ;
false.
The choicepoint happens because there is no index on the combination of both arguments.
The easiest way to solve this is to use tabling, by adding to the code:
:- table parent/2.
:- table ancestor/2.
... which improves determinism.

Expanding Prolog clauses without parameters

I am writing a program that transforms other programs by expanding predicates. I usually do this using clause/2, but it doesn't always expand a predicate if it has no parameters:
:- set_prolog_flag('double_quotes','chars').
:- initialization(main).
main :- clause(thing,C),writeln(C).
% this prints "true" instead of "A = 1"
thing :- A = 1.
Is it possible to expand predicates that have no parameters?
Some general remark: Note that this code is highly specific to SWI. In other systems which are ISO conforming you can only access definitions via clause/2, if that predicate happens to be dynamic.
For SWI, say listing. to see what is happening.
?- assert(( thing :- A = 1 )).
true.
?- listing(thing).
:- dynamic thing/0.
thing.
true.
?- assert(( thing :- p(A) = p(1) )).
true.
?- assert(( thing(X) :- Y = 2 )).
true.
?- listing(thing).
:- dynamic thing/0.
thing.
thing :-
p(_)=p(1).
:- dynamic thing/1.
thing(_).
true.
It all looks like some tiny source level optimization.

Mutually referential Prolog rules in family tree

I've seen various implementations of family trees in Prolog, but I haven't found one that does what I would like to do, which is to define child and parent by referring to each other.
I want to do this because sometimes I have a fact that someone is a child of someone, at other times I have the fact that someone is the parent of someone. From either of these kinds of facts, I'd like to be able to ask who are parents and who are children.
My attempt at coding this is :-
parent(mary, fred).
child(john, peter).
child(paul, peter).
parent(P, C).
parent(P, C) :- child(C, P).
child (C, P).
child(C, P) :- parent(P, C).
This seems to work ok except that it will continue to give me the same results duplicated over and over. Eg :-
[3] ?- parent(peter, X).
true ;
X = john ;
X = paul ;
true ;
X = john ;
X = paul ;
true
Is there a way I can get it to stop after it has given me the full set of results once ?
More generally, is this kind of definition a weird thing to want to do (because of the mutual recursion) ? I want to be able to have facts of either parent or child, as reported, but also to infer the "opposite" relationship from these.
Thanks !
The issue with your program is that you are merging predicates.
parent/2 and child/2 are facts, you should not name a rule like a fact that is already defined in your program.
Rename the rules and all will work. Also, the base clause in your rules should add a condition, that match the fact, something like this:
parent(mary, fred).
child(john, peter).
child(paul, peter).
isparent(P, C):- parent(P, C).
isparent(P, C):- child(C, P).
ischild(C, P):- child(C, P).
ischild(C, P) :- parent(P, C).
Now the query:
?- isparent(peter, X).
X = john
X = paul
Also, don't use the complement rule in your condition, that's not necessary and will avoid you the recursion

How to define `not_a_parent()` predicate?

How to create such logic in GNU Prolog? How to define not_a_parent() predicate?
parent(john,chris).
parent(mary,chris).
not_a_parent(X) :- \+ parent(X,Y).
The interesting answer to the similar question is What is the logical 'not' in Prolog?. But I do not see how to implement it here.
This worked for me:
parent(john,chris).
parent(mary,chris).
parent(mary,suzanne).
parent(suzanne,jane).
parent(suzanne,peter).
parent(peter,rose).
parent(jerry,rose).
parent(jane,carl).
not_a_parent(NonParents) :- setof(Z,Y^parent(Y,Z),SetOfChildren),
findNonParents(SetOfChildren, NonParents, []),!.
findNonParents([],A,A).
findNonParents([H|SetOfChildren], NonParents, A):-
not(call(parent(H,_))),
findNonParents(SetOfChildren,NonParents,[H|A]).
findNonParents([_|SetOfChildren], NonParents, A):-
findNonParents(SetOfChildren,NonParents,A).
Result for querying not_a_parent(NonParents) is:
?- not_a_parent(NonParents).
NonParents = [rose, chris, carl].
You need to enumerate all the persons somehow. For example
not_a_parent(X) :- ( X = john ; X = mary ; X = chris ), \+ parent(X,_).
In any real program you would probably have some simple way to get all persons, and then you can do
not_a_parent(X) :- person(X), \+ parent(X,_).
You need to get all the instances of parent/2 such that X never unifies parent(X,_). findall(Template, Goal, Instances) executes the Goal until it fails and populate the list Instances with the terms that unify the Template. That way, if the list Instances is empty then parent(X,_) does not exist.
So your predicate would be like:
not_a_parent(X):- findall(_, parent(X,_) , []).

Prolog - 2 ways for progenitor predicate implementation

Given a set of facts that represent parent-child relationships via the predicate parent/2, what are the differences when defining the relation "progenitor"(ancestor) with the predicates pred1/2 and pred2/2 as shown below?
pred1(X,Z) :- parent(X,Z).
pred1(X,Z) :- parent(X,Y), pred1(Y,Z).
pred2(X,Z) :- parent(X,Z).
pred2(X,Z) :- parent(Y,Z), pred2(X,Y).
The major difference are the termination properties of both, provided there are some cycles in the relation. To understand this, I will use a failure-slice which cuts down the program to its essence w.r.t. termination.
pred1(X,Z) :- false, parent(X,Z).
pred1(X,Z) :- parent(X,Y), pred1(Y,Z). % Z has no influence
pred2(X,Z) :- false, parent(X,Z).
pred2(X,Z) :- parent(Y,Z), pred2(X,Y). % X has no influence
For pred1/2, the second argument has no influence on termination at all, whereas in pred2/2, the first argument has no influence. To see this, try the original definitions with the fact:
parent(a,a).
?- pred1(b, _), false.
false. % terminates
?- pred2(b, _), false.
loops.
?- pred1(_, b), false.
loops.
?- pred2(_, b), false.
false. % terminates
See closure/3 for a general approach to avoid such loops.
For completeness, here is another variation of transitive closure which has some conceptual advantages:
pred3(X,Z) :- parent(X,Z).
pred3(X,Z) :- pred3(X,Y), pred3(Y,Z).
Alas, its termination properties are worst. In fact, it never terminates, as is testified by the following fragment:
pred3(X,Z) :- false, parent(X,Z).
pred3(X,Z) :- pred3(X,Y), false, pred3(Y,Z).
In this fragment, the first argument is only passed on. So, no matter, what the arguments are, the program will always loop!
?- pred3(b,b), false.
loops.

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