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How can I know if a person X is descendant of a person Y given the descendancy degree?
I've tried this:
descendant(X, Y, 1) :- son(X, Y).
descendant(X, Y, Degree) :- son(X, Z) , descendant(Z, Y, Degree-1).
Where son(X, Y) returns yes if X is son of Y. If Degree == 1 it returns the correct answer but for descendant(X, Y, 2), for instance, should return yes if X is grandson of Y but returns no.
1) Naming: Does son(X,Y) mean "X is the son of Y"—or vice-versa?
son_of(X,Y) is better.
2) Exploit successor-arithmetics: We don't need to do general arithmetics here... we only need to count.
So let's start in the beginning...
child_of(abel, adam). % from source
child_of(abel, eve).
child_of(cain, adam).
child_of(cain, eve).
child_of(enoch, cain).
child_of(irad, enoch).
child_of(mehujael, irad).
child_of(methushael, mehujael).
child_of(lamech, methushael).
child_of(jabal, lamech).
child_of(jabal, adah).
child_of(jubal, lamech).
child_of(jubal, adah).
child_of(tubal_cain, lamech).
child_of(tubal_cain, zillah).
child_of(naamah, lamech).
child_of(naamah, zillah).
child_of(seth, adam).
child_of(seth, eve).
child_of(enos, seth).
child_of(kenan, enos).
child_of(mahalalel, kenan).
child_of(jared, mahalalel).
child_of(enoch, jared).
child_of(methuselah, enoch).
child_of(lamech, methuselah).
child_of(noah, lamech).
child_of(shem, noah).
child_of(ham, noah).
child_of(japheth, noah).
Based on child_of/2 we first define ancestor_of/2—this should be nothing new to you!
ancestor_of(Y, Z) :-
child_of(Z, Y). % If Z is a child of Y ...
% then Y is an ancestor of Z.
ancestor_of(X, Z) :-
child_of(Z, Y), % If Z is a child of Y ...
ancestor_of(X, Y). % and X is an ancestor of Y ...
% then X is an ancestor of Z.
Next, we add an additional parameter indicating the distance.
We use s/1 terms to represent natural numbers and add a new argument to ancestor_of/2:
ancestor_of_dist(Y, Z, s(0)) :-
child_of(Z, Y). % If Z is a child of Y ...
% then Y is an ancestor of Z with distance = 1."
ancestor_of_dist(X, Z, s(N)) :-
child_of(Z, Y), % If Z is a child of Y ...
ancestor_of_dist(X, Y, N). % and X is an ancestor of Y with distance N ...
% then X is an ancestor of Z with distance N+1.
So ... who is grandparent of whom?
?- ancestor_of_dist(X, Z, s(s(0))).
X = adam, Z = enoch
; X = eve, Z = enoch
; X = cain, Z = irad
; X = jared, Z = irad
; X = enoch, Z = mehujael
; ...
; X = lamech, Z = japheth
; false.
Prolog is not a functional language. Thus, the Degree-1 term is not interpreted and evaluated as an expression. For Prolog, Degree-1 is just a compound term with two arguments. That can be made clear using the standard write_canonical/1 predicate, which writes a term without using operator notation:
?- write_canonical(Degree-1).
-(_,1)
true.
Write instead:
descendant(X, Y, 1) :-
son(X, Y).
descendant(X, Y, Degree) :-
son(X, Z) ,
descendant(Z, Y, Degree0),
Degree is Degree0 + 1.
The is/2 standard built-in predicate unifies the left argument with the value of the arithmetic expression in the right argument.
P.S. Note that the alternative descendant/3 predicate definition I suggest will solve the problem you described but it is not an efficient definition as it is not a tail-recursive definition. I.e. the recursive call in the second clause is not the last call in the clause body. This issue can be easily solved, however, by using an accumulator.
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.
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.
Let's say I have the following:
parent(alice, charlie).
parent(bob, charlie).
parent(bob, diane).
parent(alice, diane).
parent(bob, eve).
parent(alice, eve).
% people are siblings of each other if they share a parent
% and aren't the same person.
sibling(A, B) :-
parent(X, A),
parent(X, B),
B \= A.
Now if I ask for the siblings of Diane, I get Charlie and Eve — twice, once through Bob and once through Alice. I only want each once.
I don't think I can use a cut here, as that would prevent backtracking altogether. What I would like, is a way to check if any exists.
Paraphrasing
sibling(A, B) :-
∃(parent(X, A), parent(X, B)),
B \= A.
I tried several cuts, but none worked.
I tried findall/3 on (parent(X, A), parent(X, B)) and checking if the resulting list is non-empty, but that doesn't unify A or B.
Using setof/3 as suggested below works, but I really want to find a way to incorporate it in the definition of sibling/2, instead of having to use it in the question. I'd really like to be able to do the following:
?- sibling(diane, X).
X = charlie ;
X = eve ;
false.
or this
?sibling(X, Y).
X = charlie,
Y = diane ;
X = charlie,
Y = eve ;
X = diane,
Y = charlie ;
X = diane,
Y = eve ;
X = eve,
Y = charlie ;
X = eve,
Y = diane ;
false.
Like I said below, I have a solution for this specific case. What I would like, and what I'm setting a bounty for, is a general solution.
Instead of
sibling(A, B) :-
setof(D, X^(parent(X, A), parent(X, D)), Ds),
member(B, Ds),
B \= A.
I'd like to do
sibling(A, B) :-
exists(X^(parent(X, A), parent(X, B))),
B \= A.
which unifies A and B.
How do I define exists/1?
Using cut in Prolog is very delicate. Most cuts are essentially incorrect, but work still in certain situations. You could use a cut here, provided you want exactly one answer. But since you want the entire set, you are out of luck: You need to explore all answers to determine that set.
Fortunately, there is an elegant shortcut (pun intended) for that: setof/3. So ask
?- setof(t, sibling(diane, S), _).
For this usage of setof/3, the last argument is of no interest. It is actually [t].
For a general purpose exists/1, define
exists(XGoal) :- setof(t, XGoal, _).
This allows the use of existential quantifiers.
Here is a version using only the cut predicate !/0:
person(alice).
person(bob).
person(charlie).
person(diane).
person(eve).
parent(alice, charlie).
parent(bob, charlie).
parent(bob, diane).
parent(alice, diane).
parent(bob, eve).
parent(alice, eve).
sibling(X, Y) :-
person(X),
person(Y),
X \= Y,
sameparent(X, Y).
sameparent(X, Y) :-
parent(P, X),
parent(P, Y),
!.
We do not put the predicate !/0 in the predicate sibling/2 to allow backtracking to the goal sibling(X, Y) after the first common parent has been found. Instead we put the predicate !/0 in a new predicate sameparent/2. We also add a person predicate because the goal X \= Y requires its X and Y to be instantiated. See this document for more information.
A query:
?- sibling(diane, Y).
Y = charlie
Y = eve
Another query:
?- sibling(X, Y).
X = charlie,
Y = diane
X = charlie,
Y = eve
X = diane,
Y = charlie
X = diane,
Y = eve
X = eve,
Y = charlie
X = eve,
Y = diane
parent(alice, charlie).
parent(bob, charlie).
parent(bob, diane).
parent(alice, diane).
parent(bob, eve).
parent(alice, eve).
% people are siblings of each other if they share a parent
% and aren't the same person.
sibling(A, B) :-
setof(D, X^(parent(X, A), parent(X, D)), Ds),
member(B, Ds),
B \= A.
?- sibling(X, Y).
X = charlie,
Y = diane ;
X = charlie,
Y = eve ;
X = diane,
Y = charlie ;
X = diane,
Y = eve ;
X = eve,
Y = charlie ;
X = eve,
Y = diane ;
false.
Now I'm wondering how to extract this to a method exists/1, for general use.
So, as seen in the accepted answer, we can run
?- setof(P, (parent(P,diane), parent(P,X), X\=diane), _).
or
?- setof(P, (parent(P,diane), parent(P,X)), _), X\=diane.
to get what you want.
Hence we can define a binary predicate "exists A such that B holds":
exists(A, B):- setof( A, B, _).
and use it as
sibling_v1(A, B):- exists( P, (parent(P,A), parent(P,B)) ), B \= A.
or as
sibling_v2(A, B):- exists( P, (parent(P,A), parent(P,B), B \= A) ).
to define the desired predicates.
But it'll still run through all the possible paths to find out all the possible solutions (just reporting each once). For how could it be certain to have not missed some, otherwise?
57 ?- exists( P, (parent(P,diane), parent(P,X), X\=diane)).
X = charlie ;
X = eve.
58 ?- exists( P, (parent(P,diane), parent(P,X), writeln([P,X]), X\=diane)).
[bob,charlie]
[bob,diane]
[bob,eve]
[alice,charlie]
[alice,diane]
[alice,eve]
X = charlie ;
X = eve.
59 ?-
Suppose you have a database with the following content:
son(a, d).
son(b, d).
son(a, c).
son(b, c).
So a and b are sons of d and c. Now you want to know, given a bigger database, who is brother to who. A solution would be:
brother(X, Y) :-
son(X, P),
son(Y, P),
X \= Y.
The problem with this is that if you ask "brother(X, Y)." and start pressing ";" you'll get redundant results like:
X = a, Y = b;
X = b, Y = a;
X = a, Y = b;
X = b, Y = a;
I can understand why I get these results but I am looking for a way to fix this. What can I do?
Prolog will always try to find every possible solution available for your statements considering your set of truths. The expansion works as depth-first search:
son(a, d).
son(b, d).
son(a, c).
son(b, c).
brother(X, Y) :-
son(X, P),
son(Y, P),
X \= Y.
brother(X, Y)
_______________________|____________________________ [son(X, P)]
| | | |
X = a, P = d X = b, P = d X = a, P = c X = a, P = b
| | | |
| ... ... ...
|
| (X and P are already defined for this branch;
| the algorithm now looks for Y's)
|__________________________________________ [son(Y, d)]
| |
son(a, d) -> Y = a son(b, d) -> Y = b
| |
| | [X \= Y]
X = a, Y = a -> false X = a, Y = b -> true
|
|
solution(X = a, Y = b, P = d)
But, as you can see, the expansion will be performed in all the branches, so you'll end up with more of the same solution as the final answer. As pointed by #Daniel Lyons, you may use the setof built-in.
You may also use the ! -- cut operator -- that stops the "horizontal" expansion, once a branch has been found to be valid, or add some statement that avoids the multiple solutions.
For further information, take a look at the Unification algorithm.
First, I would advise against updating the Prolog database dynamically. For some reasons, consider the article
"How to deal with the Prolog dynamic database?".
You could use a combination of the builtin setof/3 and member/2, as #DanielLyons has suggested in his answer.
As yet another alternative, consider the following query which uses setof/3 in a rather unusual way, like this:
?- setof(t,brother(X,Y),_).
X = a, Y = b ;
X = b, Y = a.
You can eliminate one set with a comparison:
brother(X, Y) :-
son(X, P),
son(Y, P),
X \= Y, X #< Y.
?- brother(X, Y).
X = a,
Y = b ;
X = a,
Y = b ;
false.
Since X and Y will be instantiated both ways, requiring X be less than Y is a good way to cut the solutions in half.
Your second problem is that X and Y are brothers by more than one parent. The easiest solution here would be to make your rules more explicit:
mother(a, d).
mother(b, d).
father(a, c).
father(b, c).
brother(X, Y) :-
mother(X, M), mother(Y, M),
father(X, F), father(Y, F),
X \= Y, X #< Y.
?- brother(X, Y).
X = a,
Y = b ;
false.
This method is very specific to this particular problem, but the underlying reasoning is not: you had two copies because a and b are "brothers" by c and also by d—Prolog was right to produce that solution twice because there was a hidden variable being instantiated to two different values.
A more elegant solution would probably be to use setof/3 to get the solutions. This can work even with your original code:
?- setof(X-Y, (brother(X, Y), X #< Y), Brothers).
Brothers = [a-b].
The downside to this approach is that you wind up with a list rather than Prolog generating different solutions, though you can recover that behavior with member/2.
This should work. But I think it can be improved (I am not a Prolog specialist):
brother(X, Y) :-
son(X, P1),
son(Y, P1),
X #< Y,
(son(X, P2), son(Y, P2), P1 #< P2 -> false; true).
If you're using Strawberry Prolog compiler,you won't get all the answers by typing this:
?- brother(X, Y),
write(X), nl,
write(Y), nl.
In order to get all the answers write this:
?- brother(X, Y),
write(X), nl,
write(Y), nl,
fail.
I hope it helps you.:)
I got to an answer.
% Include the dictionary
:- [p1]. % The dictionary with sons
:- dynamic(found/2).
brother(X, Y) :-
% Get two persons from the database to test
son(X, P),
son(Y, P),
% Test if the two persons are different and were not already used
testBrother(X, Y).
% If it got here it's because there is no one else to test above, so just fail and retract all
brother(_, _) :-
retract(found(_, _)),
fail.
testBrother(X, Y) :-
X \= Y,
\+found(X, Y),
\+found(Y, X),
% If they were not used succed and assert what was found
assert(found(X, Y)).
It always returns fails in the end but it succeeds with the following.
brother(X, Y). % Every brother without repetition
brother('Urraca', X). % Every brother of Urraca without repetition
brother('Urraca', 'Sancho I'). % True, because Urraca and Sancho I have the same father and mother. In fact, even if they only had the same mother or the same father it would return true. A little off context but still valid, if they have three or more common parents it would still work
It fails with the following:
brother(X, X). % False because it's the same person
brother('Nope', X). % False because not is not even in the database
brother('Nope', 'Sancho I'). % False, same reason
So like this I can, for example, ask: brother(X, Y), and start pressing ";" to see every brother and sister without any repetition.
I can also do brother(a, b) and brother(b, a), assuming a and b are persons in the database. This is important because some solutions would use #< to test things and like so brother(b, a) would fail.
So there it is.