Prolog Family Tree Issues - prolog

I've been having a novice issue with setting up a family tree in Prolog. For some reason, I can't seem to get 'sibling' to return true.
% male(+Person)
% Specifies the Person is a male.
%
male(abraham).
male(homer).
male(bartholomew).
% female(+Person)
% Specifies the Person is a female.
%
female(mona).
female(marjorie).
female(lisa).
female(margaret).
% mother(+Parent,+Child)
% Specifies Parent is a mother of Child.
%
mother(mona,homer).
mother(marjorie,bartholomew).
mother(marjorie,lisa).
mother(marjorie,margaret).
% father(+Parent,+Child)
% Specifies Parent is a father of Child.
%
father(abraham,homer).
father(homer,bartholomew).
father(homer,lisa).
father(homer,margaret).
% sibling(+Person1,+Person2)
% Specifies Person1 is a sibling of Person2
%
sibling(X,Y) :-
father(X,Z),
mother(Y,Z).
Thanks very much in advance, this issue has been driving me nuts!

as an initial step, use an established relation (mother/2 will do, as well):
sibling(X, Y) :- father(F, X), father(F, Y), X #< Y.
yields
?- sibling(X,Y).
X = bartholomew,
Y = lisa ;
X = bartholomew,
Y = margaret ;
X = lisa,
Y = margaret ;
false.

Related

How to formulate an exception for findall/3 correctly?

I have startet to learn Prolog and I want to get a list of opponents of players with findall/3. In the Generalization I just want to add only the opponents to the list who are actually players, except the player I am asking for itself. How can I formulate this exception? I know about the negation as failure concept but I am not sure if and how I need it here.
player(irina).
player(anton).
player(michael).
opponent(irina, anton).
opponent(irina, maria).
opponent(irina, michael).
% Only opponents who are also players should be added to the list,
% except the player of interest itself
opponent(X, Y) :- X \= Y, player(Y).
% the query
?- findall(O, opponent(irina,O) , OpponentList).
% the result
% only the first three results from the facts,
% the results from the rule, a second entry of
% anton and michael are missing.
OpponentList = [anton, maria, michael].
I actually expected, that the resolution would work as follows:
opponent(irina, irina) :- irina \= irina, player(irina).
% false true
% false
% false, hence not added to the list
opponent(irina, anton) :- irina \= anton, player(anton).
% true true
% true
% true, hence added to the list
What am I missing?
Many thanks in advance!
Code:
player(irina).
player(anton).
player(michael).
opponent(irina, anton).
opponent(irina, maria).
opponent(irina, michael).
opponent(X, Y) :-
format("Called with X = ~q, Y = ~q\n",[X,Y]),
X \= Y, % this will fail if Y is still fresh because X and Y can be unified!
player(Y).
opponent_mod(X, Y) :-
format("Called with X = ~q, Y = ~q\n",[X,Y]),
player(Y), % this will set Y to various players
X \= Y. % the test for un-unifiability of two set vars may or may not succeed
% query helpfully as code:
q(Who,OpponentList) :- findall(O, opponent(Who,O) , OpponentList).
q_mod(Who,OpponentList) :- findall(O, opponent_mod(Who,O) , OpponentList).
% even better
q_mod_set(Who,OpponentList) :- setof(O, opponent_mod(Who,O) , OpponentList).
Run it:
Not expected:
?- q(irina,X).
Called with X = irina, Y = _19654
X = [anton, maria, michael].
Expected:
?- q_mod(irina,X).
Called with X = irina, Y = _20568
X = [anton, michael].
So what happens in the "not expected" case with Who = irina:
findall/3 tries to collect all values Y such that opponent(Who,O)
It finds anton, maria, michal through the facts
opponent(irina, anton).
opponent(irina, maria).
opponent(irina, michael).
It also tries the rule opponent(X, Y) :- ... with X = irina (once only). But the test X \= Y fails because X has been constrained to be irina but Y is still fresh (completely unconstrained). This means that a unification of X and Y could succeed if it were attempted (which is what is being tested, = does not as such mean "is not equal"). So X \= Y is false at this point of computation. If you move X \= Y after player(Y), then Y will have been constrained to whatever was grabbed from the player/1 facts, and this would reduce to an actual comparison of the values of X and Y.

Nephew of my own father in prolog

i'm programming a family in prolog and i'm having trouble with the nephew implementation. When I ask if erick is the nephew of Alberto it returns true, when it should return false because Alberto is the father of erick, however, it does works for all the other cases that should be true. If someone could help me I would be very grateful.
My code:
man(beto).
man(fransisco).
man(alberto).
man(jaime).
man(manolo).
man(nolo).
man(lito).
man(manuel).
man(erick).
man(jesu).
man(jesus).
woman(emi).
woman(harumi).
woman(haru).
woman(yuneisi).
woman(yasmeli).
woman(mioara).
woman(elia).
woman(iza).
woman(alice).
woman(ofelia).
woman(arlet).
parent(manuel, alberto).
parent(ofelia, alberto).
parent(manuel, jaime).
parent(ofelia, jaime).
parent(manuel, manolo).
parent(ofelia, manolo).
parent(alberto, erick).
parent(alberto, beto).
parent(alberto, fransisco).
parent(emi, erick).
parent(emi, beto).
parent(manolo, nolo).
parent(manolo, arlet).
parent(nolo, lito).
parent(iza, lito).
parent(mioara, yuneisi).
parent(mioara, yasmeli).
parent(jaime, yuneisi).
parent(jaime, yasmeli).
parent(jesus_padre, jesu)
parent(jesus_padre, alice).
parent(jesus_padre, haru).
parent(harumi, haru).
parent(harumi, jesu).
parent(harumi, alice).
father(X,Y) :- parent(X,Y), man(X).
mother(X,Y) :- parent(X,Y), woman(X).
brother(X,Y) :- man(X), parent(F, X), parent(F, Y).
sister(X,Y) :- woman(X), parent(P, X), parent(P, Y).
grandpa(X,Y) :- father(F,Y), father(X,F), man(X).
grandma(X,Y) :- father(F,Y), mother(X,F), woman(X).
son(X,Y) :- father(Y,X), man(X).
nephew(X,Y) :- father(F,X), brother(F,Y).
Besides the missing dot after parent(jesus_padre, jesu) as pointed out by #LuaiGhunim there are a few other issues with your predicates. Your definition of brother/2 is too general. Nobody is his own brother but if you query your predicate you find several such instances:
?- brother(X,X).
X = beto ;
X = beto ;
X = fransisco ;
X = alberto ;
X = alberto ;
X = jaime ;
X = jaime ;
X = manolo ;
X = manolo ;
X = nolo ;
X = lito ;
X = lito ;
X = erick ;
X = erick ;
X = jesu ;
X = jesu ;
false.
You can easily remedy this by adding a goal dif/2:
brother(X,Y) :-
dif(X,Y),
man(X),
parent(F, X),
parent(F, Y).
Now the query above fails as it should:
?- brother(X,X).
false.
You'll still get a lot of pairs twice:
?- brother(X,Y).
X = beto, % <- 1st occurrence
Y = erick ; % <- 1st occurrence
X = beto,
Y = fransisco ;
X = beto, % <- 2nd occurrence
Y = erick ; % <- 2nd occurrence
.
.
.
The reason for that is that you can derive it via the mother or the father. In the above example (beto and erick) you'll get there via emi or alberto. These solutions might be redundant but they are correct. The same goes for your predicate sister/2:
?- sister(X,X).
X = haru ;
X = haru ;
X = yuneisi ;
X = yuneisi ;
X = yasmeli ;
X = yasmeli ;
X = alice ;
X = alice ;
X = arlet.
The remedy is the same as above:
sister(X,Y) :-
dif(X,Y),
woman(X),
parent(P, X),
parent(P, Y).
?- sister(X,X).
false.
?- sister(X,Y).
X = haru,
Y = jesu ;
X = haru,
Y = alice ;
X = haru,
Y = jesu ;
.
.
.
Your definition of grandma/2 and grandpa/2 on the other hand is too specific. To see this let's add the following facts to your code:
man(m1).
man(m2).
woman(w1).
woman(w2).
woman(w3).
parent(m1,w1).
parent(w1,w2).
parent(w2,w3).
Then the following queries should succeed but they fail instead:
?- grandpa(m1,w2).
false.
?- grandma(w1,w3).
false.
The reason for this is that the intermediate parent in your definition of grandpa/2 and grandma/2 is a father/2 where it should be a parent/2. Additionally the last goals (man(X) and woman(X)) are redundant since they are already covered by father/2 and mother/2 respectively. Instead, you could define the two predicates like so:
grandpa(X,Y) :-
parent(F,Y),
father(X,F).
grandma(X,Y) :-
parent(F,Y),
mother(X,F).
Now the above queries yield the desired result:
?- grandpa(m1,w2).
true.
?- grandma(w1,w3).
true.
Finally, a nephew according to the Cambridge Dictionary is a son of your sister or brother, or a son of the sister or brother of your husband or wife. Since you haven't got predicates for husbands and wives I'll stick with the a son of your sister or brother part. If you add facts for husbands and wives, you can add additional rules to cover the other part of the definition. You can write the first part of the definition in Prolog like so:
nephew(X,Y) :-
man(X),
dif(F,Y),
parent(P,F),
parent(P,Y),
parent(F,X).
If you query this predicate there's no erick/alberto solution any more:
?- nephew(erick,X).
X = jaime ;
X = manolo ;
X = jaime ;
X = manolo ;
false.
Prolog it's all about relations. Joins plays a fundamental role. So, often you can think in term of 'yields', or functions over the DB, and design/control the data access plan by means of joins (clauses, i.e. logical formulae) producing records - like SQL does. The procedural execution over the retrieved data is expressed in (almost) the same language: joins, giving us a taste of declarative programming. Anyway, here are the clauses provided by #tas in standard Prolog:
brother(X,Y) :-
man(X),
parent(F, X),
parent(F, Y),
X\=Y.
sister(X,Y) :-
woman(X),
parent(P, X),
parent(P, Y),
X\=Y.

Creating a Niece Rule in Prolog

Using a family database, I need to create a niece rule (niece(X,Y))in swi-prolog which is defined as "X is a niece of Y if X is a daughter of Y's brother or sister." This is the given database with my already designed rules:
% family DB
grandfather(don,who).
father(don,ted).
father(don,barb).
father(don,paula).
father(greg,erin).
father(greg,austin).
father(wes,alyssa).
father(ted,jessica).
father(ted,david).
%mother(ted, john).
mother(audrey,ted).
mother(audrey,barb).
mother(audrey,paula).
mother(paula,erin).
mother(paula,austin).
mother(barb,alyssa).
married(don,audrey).
married(wes,barb).
married(greg,paula).
male(don).
male(ted).
male(wes).
male(greg).
male(austin).
male(david).
female(audrey).
female(barb).
female(paula).
female(alyssa).
female(jessica).
female(erin).
parent(X,Y) :-
father(X,Y)
; mother(X,Y).
grandfather(X,Y) :-
father(X,Z),
( father(Z,Y)
; mother(Z,Y)
).
samefather(X,Y) :-
father(F,X),
father(F,Y).
samemother(X,Y) :-
mother(M,X),
mother(M,Y).
sameparent(X,Y) :-
samefather(X,Y).
sameparent(X,Y) :-
samemother(X,Y),
not(samefather(X,Y)).
couple(X,Y) :-
married(X,Y),
married(X,Y).
Here is my initial try at the niece rule:
niece(X,Y) :-
parent(F,X),
sameparent(Y,F).
My idea is to use the sameparent rule to check if Y and F are siblings and then check if F is the parent of X. This rule currently doesn't work. I'm still struggling to understand the syntax of combining multiple rules. If anyone could help me by using this same logic, it would be greatly appreciated.
Removing unnecessary rules :
parent(don,ted).
parent(don,barb).
parent(don,paula).
parent(greg,erin).
parent(greg,austin).
parent(wes,alyssa).
parent(ted,jessica).
parent(ted,david).
parent(audrey,ted).
parent(audrey,barb).
parent(audrey,paula).
parent(paula,erin).
parent(paula,austin).
parent(barb,alyssa).
male(don).
male(ted).
male(wes).
male(greg).
male(austin).
male(david).
female(audrey).
female(barb).
female(paula).
female(alyssa).
female(jessica).
female(erin).
father(X,Y) :-
male(X),
parent(X,Y).
mother(X,Y) :-
female(X),
parent(X,Y).
sameparent(X,Y) :-
parent(A,X),
parent(A,Y).
Gives you :
niece(X,Y) :-
female(X),
parent(Z,X),
sameparent(Z,Y),
Z \= Y.
Meaning line by line :
X is a niece of Y if
X is a female and
one parent of X
has a parent in common with Y
who isn't himself/herself.
This gives you :
| ?- niece(X,Y).
X = alyssa
Y = ted;
X = alyssa
Y = paula;
X = jessica
Y = barb;
X = jessica
Y = paula;
X = erin
Y = ted;
X = erin
Y = barb
Twice because in the case of your database every brothers/sisters share the same two parents.
For example, if A is the daughter of B and B is the half-brother of C, A is still the niece of C.
If you want that rule to be false and A to be the niece of C only if B and C are brothers/sisters (and not only half), you can change the niece rule to the following :
sameFather(X,Y) :-
father(A,X),
father(A,Y).
sameMother(X,Y) :-
mother(A,X),
mother(A,Y).
niece(X,Y) :-
female(X),
parent(Z,X),
sameFather(Z,Y),
sameMother(Z,Y),
Z \= Y.
Then you won't get duplicate results whith niece(X,Y).
For me it worked well.
sameparent(X,Y):- parent(A,X),parent(A,Y).
niece(X,Y) :- female(X),parent(Z,X),sameparent(Z,Y),Z=Y.

Get only one solution for sibling pairs in family tree

I have the following Prolog code, modified from Wikipedia:
mother_child(trude, sally).
father_child(tom, sally).
father_child(tom, erica).
father_child(mike, tom).
different(X, Y) :- X \== Y.
sibling(X, Y) :- parent_child(Z, X), parent_child(Z, Y), different(X, Y).
parent_child(X, Y) :- father_child(X, Y).
parent_child(X, Y) :- mother_child(X, Y).
I get the following:
?- sibling(X, Y).
X = sally,
Y = erica ;
X = erica,
Y = sally ;
false.
Is it possible to modify my code so that ?- sibling(X, Y) to return only: X = sally, y = erica ; false.? That is, I want to eliminate instances where X = Y and Y = X (converse).
The comment by #lurker says it all. Ask yourself what it is that the query is asking:
Is there a pair X, Y such that:
X has the parent Z:
Z is a father
Z is a mother
Y has the parent Z:
Z is a father
Z is a mother
X and Y are not the same
In your current database, a child can have at most one mother and one father, so the two clauses of parent_child/2 happen to be mutually exclusive for a single child.
But you have the subgoal parent_child/2 twice in sibling/2, so for every pair of siblings, you can have either one of the two in the first, and then the other one in the second, so you have two proofs for each pair. Saying X #< Y instead of X \== Y will make sure that only one of the two possible combinations of X and Y lead to a successful proof.

Prolog: Unification or Backtracking errors in program

I have a simple knowledge base that encodes a family tree. Some important rules in this representation are as follows:
% fathers
father(michael,cathy).
father(michael,sharon).
father(charles_gordon,michael).
father(charles_gordon,julie).
father(charles,charles_gordon).
father(jim,melody).
father(jim,crystal).
father(elmo,jim).
father(greg,stephanie).
father(greg,danielle).
% mothers
mother(melody,cathy).
mother(melody,sharon).
mother(hazel,michael).
mother(hazel,julie).
mother(eleanor,melody).
mother(eleanor,crystal).
mother(crystal,stephanie).
mother(crystal,danielle).
% parents
parent(X,Y) :- father(X,Y).
parent(X,Y) :- mother(X,Y).
% men
male(michael).
male(charles_gordon).
male(charles).
male(jim).
male(elmo).
male(greg).
% women
female(cathy).
female(sharon).
female(julie).
female(hazel).
female(eleanor).
female(melody).
female(crystal).
female(stephanie).
female(danielle).
person(X) :- male(X) ; female(X).
parent(X,Y) :- father(X,Y) ; mother(X,Y). % X is parent of Y
child(X,Y) :- parent(Y,X).
elder(X,Y) :- parent(X,Y). % X is an elder of Y, meaning X is a parent or an ancestor of Y
elder(X,Y) :- parent(X,Z), elder(Z,Y).
junior(X,Y) :- child(X,Y). % X is a junior of Y, meaning X is a child or some descendant of Y
junior(X,Y) :- child(X,Z), junior(Z,Y).
I am attempting to find the nearest elder between two individuals(predicate ne(X,Y,Z)). This individual Z is the elder of both X and Y, and no junior of Z is also an elder of BOTH X and Y.
My attempt looks like this:
ne(X,Y,Z) :- person(X),
person(Y),
X \= Y,
elder(Z,X),
elder(Z,Y),
junior(A,Z),
not(elder(A,X)),
not(elder(A,Y)).
but this is somehow incorrect, because whenever I run ?- ne(stephanie,cathy,Z). i get
Z = jim ;
Z = jim ;
Z = jim ;
Z = jim ;
Z = elmo ;
Z = elmo ;
Z = elmo ;
Z = elmo ;
Z = eleanor ;
Z = eleanor ;
Z = eleanor ;
Z = eleanor ;
but i'm only supposed to get one answer, and I can't figure out what's wrong. Thanks!
from this graph
seems this answer is correct
?- ne(stephanie,cathy,A).
A = eleanor ;
A = jim.
here is my attempt to ne/3
ne(X,Y,Z) :-
setof(A, (
elder(A, X),
elder(A, Y),
X \= Y,
\+ (elder(A, T), elder(T, X) , elder(T, Y) ))
, As), member(Z, As).
not sure it's the better way...
Setof/3 (joined with member/2) is used to eliminate duplicate answers, since we get
?- aggregate(count,A^ne(stephanie,cathy,A),N).
N = 32.
with this core logic
ne(X,Y,A) :-
elder(A, X),
elder(A, Y),
X \= Y,
\+ (elder(A, T), elder(T, X) , elder(T, Y)).
note variable A replaces locally the original Z
edit
I didn't took in account the savvy comment by #Boris, but after removing the duplicate parent/2 definition, the setof/3+member/2 trick become useless.

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