prolog family tree locating cousins - prolog

The Cousin Explainer
Write a program that when given a person of reference and a degree the program can tell me all of the nth degree cousins that person has. I've started by writing my facts: male/1 female/1 and child/2 these are populated with the names of my family members. I then began writing rules for cousins I figured id write a different rule for every degree so only up to the third cousins. The problems I'm having are as follows.
firstCousin_of(X,Y):-child(X,Z1),child(Y,Z2),child(Z1,Z),child(Z2,Z). I have this to find a first cousin but the 2nd cousin is more complicated which makes me think there has gotta be a way to do this recursively.
I need the rule, all_cousins(Person, Degree, ListOfCousins), which after figuring out all of the nth degree cousins edits a list of all cousins to only include those specific nth degree cousins. I'm used to imperative languages and all i can think about with this is how do I even convey to the rule which people to remove from the list from one rule to another.
lastly I have a rule, all_cousinsRemoved(Person, Degree, removed(Number, Direction), ListOfCousins) that I don't even know where to start with but ill need the all_cousins rule for this one because it does the same thing but also only includes cousins in the edited lists w=that were removed nth times either up or down.

The best solutions is make a several rules that call other rules, here an example
male(dicky).
male(randy).
male(mike).
male(don).
male(elmer).
female(anne).
female(rosie).
female(esther).
female(mildred).
female(greatgramma).
male(blair).
male(god).
female(god).
parent(don,randy).
parent(don,mike).
parent(don,anne).
parent(rosie,randy).
parent(rosie,mike).
parent(rosie,anne).
parent(elmer,don).
parent(mildred,don).
parent(esther,rosie).
parent(esther,dicky).
parent(greatgramma,esther).
parent(randy,blair).
male(mel).
male(teo).
parent(melsr,mel).
parent(melsr,teo).
american(anne).
american(X) :- ancestor(X,anne).
american(X) :- ancestor(anne,X).
relation(X,Y) :- ancestor(A,X), ancestor(A,Y).
father(X,Y) :- male(X),parent(X,Y).
father(god, _) :- male(god).
mother(X,Y) :- female(X),parent(X,Y).
son(X,Y) :- male(X),parent(Y,X).
daughter(X,Y) :- female(X),parent(Y,X).
grandfather(X,Y) :- male(X),parent(X,Somebody),parent(Somebody,Y).
aunt(X,Y) :- female(X),sister(X,Mom),mother(Mom,Y).
aunt(X,Y) :- female(X),sister(X,Dad),father(Dad,Y).
sister(X,Y) :- female(X),parent(Par,X),parent(Par,Y), X \= Y.
uncle(X,Y) :- brother(X,Par),parent(Par,Y).
cousin(X,Y) :- uncle(Unc , X),father(Unc,Y).
ancestor(X,Y) :- parent(X,Y).
ancestor(X,Y) :- parent(X,Somebody),ancestor(Somebody,Y).
brother(X,Y) :- male(X),parent(Somebody,X),parent(Somebody,Y), X \= Y.
Retrived from this repository

Related

How to print all females in prolog if my facts only contains male property

Lets suppose my db.pl file consists of only
male(Oliver)
male(james)
male(sam)
female(X) :- \+ male(X)
Now if I query,
?- male(X).
Then this will successfully return all the males.
But if I query,
?- female(X)
Then this will not. However, if I put a name in the female predicate, it correctly gives the output as yes/no.
What am I missing here? How do I get the list of all females? Or basically list of "not male".
I thought I'd give a proper answer as well. As mentioned in the comments, the fact male(Oliver). contains the variable Oliver such that e.g. also male(catherine_the_great) succeeds. But fixing this mistake will not solve the problem. Suppose we fix the facts for male and define female as above:
male(oliver)
male(james)
male(sam)
female(X) :-
\+ male(X).
Then we can not derive male(catherine_the_great) anymore but female(catherine_the_great) succeeds:
?- male(catherine_the_great).
false.
?- female(catherine_the_great).
true.
Unfortunately, by this definition, a hamburger is also female:
?- female(hamburger).
true.
What's even worse is, when we ask if there exists someone female (at all!), we get no as an answer:
?- female(X).
false.
The reason for this is that \+ is not classical negation(*) but what is called weak negation / negation as failure. \+ male(X) is true if there is no X such that male(X) is derivable. But as soon as male/1 succeeds with any solution (e.g. male(oliver)), the negation fails. In such cases the substitution male(X) is not proper substiution for female(X) but how are we supposed to get a list of all women? A possible solution is a list of persons of which we define male/female as subsets:
person(oliver).
person(james).
person(sam).
person(jackie).
person(selma).
male(oliver).
male(james).
male(sam).
female(X) :-
person(X),
\+ male(X).
When we query a list of all women, we now get:
?- female(X).
X = jackie ;
X = selma.
I'd rather use a separate predicate listing women though because it is less error prone when adding new persons.
In general, negation as failure is rarely safe. Two rules of thumb i have for myself:
whenever \+ p(X) appears as a goal, it is fully instantiated (a ground term). In our case, this is what person assures.
p(X) terminates for every ground term X. This assures termination of the negation as well.
(*) in classical logic, the rule of generalization female(jackie) → ∃X female(X) holds but here we can not derive ∃X female(X) despite being able to derive female(jackie).

Two different paths from X to Y in a graph

I am stuck with the following Prolog question:
Given the edges of a graph with no cycles as facts. e.g:
edge(a, b).
edge(b, c).
edge(c, d).
edge(c, e).
...
I have to write a predicate that tests whether there are two different paths between vertices X and Y. e.g the call two_paths(a, c). should return true if there are two different paths from node a to node c. I know how to check whether there is a path between two vertices:
path(X, Y) :- edge(X, Y).
path(X, Y) :- edge(X, Z), path(Z, Y).
But how should I do this to check for two distinct paths? Thank you very much for your help.
An idea might be to create a predicate path/3 that returns the constructed path, and then query for two paths that are different. Something like:
path(X,Y,[X,Y]) :- edge(X,Y).
path(X,Y,[X|T]) :- edge(X,Z), path(Z,Y,T).
Now path(a,c,T) will show you the path:
?- path(a,c,L).
L = [a, b, c] ;
false.
Now you could construct a predicate:
two_paths(X,Y) :-
path(X,Y,La),
path(X,Y,Lb),
dif(La,Lb).
In other words, you ask Prolog to construct for you a path La, next construct for you a path Lb and then check if they are not equal (dif(La,Lb)). The first constructed Lb will be equivalent to La, but due to Prologs backtracking mechanism, it will try to construct you another path for which the condition might succeed. This is a rather pure Prolog implementation (with no cut (!), once/1, etc.). More efficient approaches exists since here you will "redo" the work in your second call.
A more efficient approach could be to construct a predicate path_avoid/3 (or path_avoid/4) where you feed the first constructed path to the predicate and thus force your program to at least at some point perform a different step from the one presented. I leave this open as a potential exercise.

How to find useless states in a finite state machine Prolog

I have been thinking of a solution for finding all the useless states defined in a particular automaton and I know that one approach is to start moving from the start state and then ask for the following state all of these stored in the Prolog predicate database. I have had no luck in writing the code very well because I am learning Prolog. I wish someone could help me out with this. Here I leave what I have.
The finite state machine structure. fa_start -> the initial state, fa_move -> the valid move from one state to another, fa_final -> the final state.
% Start State
:- assert(fa_start(fa_1, s_0)).
% Final States
:- assert(fa_final(fa_1, s_5)).
:- assert(fa_move(fa_1, s_0, s_1, a)).
:- assert(fa_move(fa_1, s_1, s_4, b)).
:- assert(fa_move(fa_1, s_0, s_2, c)).
:- assert(fa_move(fa_1, s_2, s_4, d)).
:- assert(fa_move(fa_1, s_2, s_5, e)).
:- assert(fa_move(fa_1, s_0, s_3, f)).
:- assert(fa_move(fa_1, s_3, s_5, g)).
:- assert(fa_move(fa_1, s_6, s_3, h)).
:- assert(fa_move(fa_1, s_6, s_7, i)).
:- assert(fa_move(fa_1, s_7, s_8, j)).
Now, here is the code I have been writing. The idea is to start with fa_start and then keep moving using valid fa_move until it cannot reach fa_final.
adjacent(fa, N, M):-
fa_move(fa, N, M, _).
adjacent_recursive(fa, N, L):-
fa_start(fa, N),
findall(l,adjacent(fa,N,_),L).
find_paths(fa):-
adjacent_recursive(fa,s_0,_).
Thank you in advance for all the help.
I know it's difficult to find accurate information related to my specific question. I have been working on it and finally found a solution, may it not be the best, but makes the deal. I owe the idea to this site. If you find bugs, it'd be welcome.
Now, the following code prints the useless states given a finite state automata implemented as above.
find_useless_states(FA):- retractall(utiles(_)),retractall(visited(_)),
forall(fa_final(FA,S),assert(utiles(S))),
find_useless_states2(FA).
find_useless_states2(FA):- retract(utiles(S)),
not(visited(S)), assert(visited(S)),
forall((fa_move(FA,Y,S,_), not(visited(Y))),(asserta(utiles(Y)))),
find_useless_states2(FA).
find_useless_states2(_).
difference(L1, L2, R) :- intersection(L1, L2, I),
append(L1, L2, All),
subtract(All, I, R).
find_all_states_as_list(FA,L):- findall(X,fa_move(FA,X,_,_),M),findall(Y,fa_move(FA,_,Y,_),N),merge(M,N,LL),sort(LL,L).
find_useful_states_as_list(L):- findall(X,visited(X),L).
print_useless_states(FA):- find_all_states_as_list(FA,L),find_useful_states_as_list(M), difference(L,M,R), length(R,D),write(R),nl,write(D).
Hope it helps other people. Some code ideas have been used from questions posted here in stackoverflow. I thank the people who answered those.
It is possible to use the dynamic database for all such things, but it easily gets very difficult to maintain. I'd rather compile the file with your definitions, thus avoid to perform assertz/1 manually.
uselessstate(Aut, Usl) :-
setof(S0, S^( closure0(adjacent(Aut), S0,S), fa_final(Aut,S) ), S0s),
setof(t, A^B^( adjacent(Aut, A,B), (Usl=A;Usl=B) ), _),
non_member(Usl, S0s).
unreachable(Aut, Usl) :-
setof(S, S0^( fa_start(Aut, S0), closure0(adjacent(Aut), S0,S) ), Ss),
setof(t, A^B^( adjacent(Aut, A,B), (Usl=A;Usl=B) ), _),
non_member(Usl, Ss).
The definition of closure0/3 and non_member/2 is found in another answer.

Find what's not provable in Prolog

Consider the following:
link(step1, step2) .
link(step2, step3) .
link(step3, step4) .
goal(X) :- \+ link(X, _) .
I would like goal functor to represent a step that is not at the beginning of a link.
But when I try :
| ?- goal(X).
no
(instead of telling me that step4 is a solution)
The following, however, does evaluate to yes:
goal(step4).
I'm guessing that's because I'm asking prolog to find what it can't find (sigh…)
Any way I can do this?
The trouble comes with Prolog not knowing what your universe of valid values of X could be. In this particular example, you could do it by defining what a valid step is:
valid_step(X) :- link(X, _) ; link(_, X).
This would then help you tell goal what to choose from for the "universe of valid steps":
goal(X) :- valid_step(X), \+ link(X, _).
Yielding:
| ?- goal(X).
X = step4
yes
| ?-
Or, if more specifically, you really mean to find X which is present as a destination link but not a source link:
goal(X) :- link(_, X), \+ link(X, _).
It depends upon the big picture of what your facts are, what they mean, and what the semantics of goal really are.
Perhaps a more suitable way to define valid steps, if the link relation isn't what logically definesto what a valid step is, would be to make valid_step a set of facts, instead of the predicate I show above:
valid_step(step1).
valid_step(step2).
valid_step(step3).
valid_step(step4).
valid_step(step5).
So this is a simple, completely independent definition of what a valid step is, and can be used by other relations (predicates) that need this information.
It may be tempting to do:
valid_steps([step1,step2,step3,step4,step5]).
And then:
goal(X) :- valid_steps(Valid), member(X, Valid), \+ link(X, _).
But I believe the list of discrete facts is preferable.

How to express a Total Order Relation in Prolog?

What is the best way to express a total order relation in Prolog ?
For example, say I have a set of facts
person(tim)
person(ana)
person(jack)
...
and I want to express the following truth about a person's fortune: for each two persons X and Y, if not(X==Y), either X is richer than Y or Y is richer than X.
So my problem is that the richer clause should be capable of instantiating its variables and also to ensure that it is never the case that richer(X, Y) and richer(Y, X) at the same time.
Here is a better example to see what I mean:
person(tim).
person(john).
happier(tim, john).
hates(X, Y) :- person(X), person(Y), richer(Y, X).
hates(X, Y) :- person(X), person(Y), richer(X, Y), happier(Y, X).
Now the answer to the query hates(john, tim) should return true, because if richer satisfies the mentioned property, one of those two hates clauses must be true. In a resolution based inference engine I could assert the fact (richer(X, Y) V richer(Y, X)) and the predicate hates(john, tim) could be proved to be true.
I don't expect to be able to express this the same way in Prolog with the same effect. However, how can I implement this condition so the given example will work ?
Note also that I don't know who is richer: tim or john. I just now that one is richer than the other.
Thank you.
you cannot write in pure Prolog that a predicate should be a total order: that would require higher order logic since you want to declare a property about a predicate.
this is a predicate that checks if a relationship is total order on a finite set:
is_total_order(Foo,Set):-
forall(
(member(X,Set),
member(Y,Set)),
(
XY =.. [Foo,X,Y],
YX =.. [Foo,Y,X],
(call(XY);call(YX)), %checking if the relationship is total
\+ (call(XY),call(YX), X\=Y) %checking if it's an order
)
).
the operator =../2 (univ operator) creates a predicate from a list (ie: X =.. [foo,4,2]. -> X = foo(4,2)) and the predicate call/1 calls another predicate.
As you can see we use meta-predicates that operate on other predicates.
For an total order on infinite sets things get more complicated since we cannot check every single pair as we did before. I dont even think that it can be written since proving that a relationship is a total order isn't something trivial.
Still, this doesnt declare that a predicate is a total order; it just checks if it is.
So I believe your question is the best way to represent the relation between these 3 people. If you don't actually care about their wealth #'s, just their relative order, you could add predicates like this -
is_richer(tim, anna).
is_richer(anna, jack).
and then a predicate to find all people who X is richer than -
richer(X, Z) :-
is_richer(X, Z).
richer(X, Z) :-
is_richer(X, Y),
richer(Y, Z).
If your is_richer predicate contains cycles though (in this example, if you added is_richer(jack, tim)), this could explode. You need to track where you have visited in this tree.
An example run of richer would be:
?- richer(X, Y).
X=tim
Y=anna ;
X=anna
Y=jack ;
X=tim
y=jack ;
no

Resources