I am trying to find all of the distinct entries where a person's name is john, peter or fred.
However, if there were, for example, two people called peter, I only want to display one occurrence of the name.
My code so far is as follows:
searchpeople(X) :-
people(_,[X|_]),
X=john; X=peter; X=fred.
I understand that the solution is probably something to do with cuts (having read other posts), but I cannot find an example where cuts are used when trying to retrieve X OR Y OR Z (In my case john, peter or fred).
Thanks in advance.
The problem is that you're confusing operator precedence. Just like more conventional programming languages where writing something like this
if ( A and B OR C OR D )
...
is almost certainly going to get you in trouble, your code has the exact same problem. Operator precedence and associativity causes
searchpeople(X) :-
people(_,[X|_]) ,
X=john ;
X=peter ;
X=fred .
to be parsed as if written
searchpeople(X) :-
( people(_,[X|_]) ,
X = john
) ;
( X = peter ;
X = fred
) .
Which is probably not what you intended.
While you could use parenthesis to get the effect you most likely want:
searchpeople(X) :-
people(_,[X|_]) ,
( X = john ;
X = peter ;
X = fred
) .
You would be better off splitting things up a bit:
search_people(X) :-
people(_,[X|_]) ,
desired_person(X).
desired_person(john).
desired_person(peter).
desired_person(fred).
It makes your intent clearer and easier to understand. It's also easier to debug and extend.
Related
I need help with understanding how to work with lists like that [a(b,c),a(x,d)]
change(S,K,R) changes first list [a,c,b] with values given in second list [c(a,x),c(b,y)]
?- change([a,c,b],[k(a,x),k(b,y)],R).
R = [x,c,y].
% my program but it works with second list that is of wrong list elements type k(a,x) but like [a,x] and works kinda poorly cause return True instead of R= x,c,y, if i print R value it is [y,c,x|_2826]
i call my code with ?- change([a,c,b],[a,x,b,y],R).
change([],[],[]):-!.
change([],[],R):-write(R),!.
change([H1|T1],[],[H1|R]):-change(T1,[],R),!.
change([H1|T1],[H2,H3|T2],R) :-
( ( H1==H2 , change(T1,T2,[H3|R]) )
; ( H1\==H2, change(T1,[H2,H3|T2],[H1|R]) )
).
Looks like you should use Association lists.
see SWI-Prolog manual
and the online doc
Completely new to prolog. Interesting journey so far in trying to change how I think, so appreciate any help here.
I am trying to assert facts for a pre-defined set of names. For example, assume I have a a set of people [alice, bob, ...] in one file. I would like to assert facts about these folks in other files, but want to make sure that these folks exist and that is checked when the facts are loaded/compiled(?).
For example, assume I don't have 'chuck' in the list and I make an assertion:
user: swipl app.pl
?- full_name(chuck, "Charlie Steel").
should result in an error.
What is the best way I can do this?
So, here's the code I came up with:
person(deborah).
person(tony).
read_my_file(Filename) :-
open(Filename, read, In),
read_my_file1(In),
close(In).
read_my_file1(In) :-
read(In, Term),
( Term == end_of_file
-> true
; assert_or_abort(Term),
read_my_file1(In)
).
assert_or_abort(Term) :-
( full_name(Person, Name) = Term
-> ( person(Person)
-> assertz(full_name(Person, Name))
; format(user, '~w is not a person I recognize~n', [Person])
)
; format(user, '~w is not a term I know how to parse~n', [Term])
).
The trick here is using read/2 to obtain a Prolog term from the stream, and then doing some deterministic tests of it, hence the nested conditional structure inside assert_or_abort/1. Supposing you have an input file that looks like this:
full_name(deborah, 'Deborah Ismyname').
full_name(chuck, 'Charlie Steel').
full_name(this, has, too, many, arguments).
squant.
You get this output:
?- read_my_file('foo.txt').
chuck is not a person I recognize
full_name(this,has,too,many,arguments) is not a term I know how to parse
squant is not a term I know how to parse
true.
?- full_name(X,Y).
X = deborah,
Y = 'Deborah Ismyname'.
When I run query human(Who). on the below .pl file
human(ann).
human(george).
human(mike).
I only get back Who = ann .
Instead of
Who = ann ;
Who = george ;
Who = mike.
Am using prolog 6.6.6. How do I get it to show the full list?
The answer you got was the following. Do you note the space before the dot?
Who = ann .
^ SPACE!!!
This space means: The query was aborted. Maybe you typed return. Or maybe you have a somewhat illconfigured terminal.
To better check this, try:
?- X = 1 ; X = 2 ; X = 3.
There you should get all three answers, too. If not, it is definitely your terminal
What you are seeing is the default behaviour of prolog.
The query
?- findall(Object,Goal,List).
Should work for you.
Eg.
findall(X, human(X), L).
It will populate the list with all possible answers.
Trying to create a predicate (timePeriod/2) that calculates the time period between two dates for a specific fact. I've managed to do this by myself, but face issues when 'other answers' exist in the same list (i.e. easier to explain with examples).
I have the following knowledge-base facts;
popStar('Jackson',1987,1991).
popStar('Jackson',1992,1996).
popStar('Michaels',1996,2000).
popStar('Newcastle',2000,2007).
popStar('Bowie',2008,2010).
And the following function, calculates the time between dates for a specific fact (as per below).
Predicate (timePeriod/2) -
timePeriod(PS,X) :-
bagof((Name,Start,End),popStar(Name,Start,End),PSs),X is End-Start+1)
Using Bowie as an example; it returns X=3 (which is correct).
However, when there is repetition in the list, with more than one answer available, the predicate just states 'false'. Using the facts 'Jackson' as an example, I want to be able to calculate both of the time periods for both facts; at the same time.
So, if the predicate would work for both of the Jackson facts, the predicate timePeriod would state X=10.
Would really appreciate if anyone could suggest what to change in order for this to work correctly.
Thanks.
You probably don't quite understand what foreach/3 does. I don't think I fully understand foreach/3 either. I know for sure that it is not the same as say:
for each x in xs:
do foo(x)
Another thing: "tuples" in Prolog are not what you might expect, coming from a language like Python or Haskell. This: (a,b,c) is actually this: ','(a,','(b,c)). Much better is to use a flat term, the generic form would be triple(a,b,c). For a pair, the idiom is First-Second.
So, you can simplify your call to bagof/3 to this:
?- bagof(From-To, pop_star(Name, Start, End), Ts).
Name = 'Bowie',
Ts = [2008-2010] ;
Name = 'Jackson',
Ts = [1987-1991, 1992-1996] ;
Name = 'Michaels',
Ts = [1996-2000] ;
Name = 'Newcastle',
Ts = [2000-2007].
Once you have a list as above, you need to sum the differences, which would be maybe something like:
periods_total(Ps, T) :-
maplist(period_length, Ps, Ls),
sum_list(Ls, T).
period_length(From-To, Length) :-
Length is To - From + 1.
And then you can query like this:
?- bagof(From-To, pop_star('Jackson', From, To), Ps), periods_total(Ps, T).
Ps = [1987-1991, 1992-1996],
T = 10.
?- bagof(From-To, pop_star(Name, From, To), Ps), periods_total(Ps, T).
Name = 'Bowie',
Ps = [2008-2010],
T = 3 ;
Name = 'Jackson',
Ps = [1987-1991, 1992-1996],
T = 10 ;
Name = 'Michaels',
Ps = [1996-2000],
T = 5 ;
Name = 'Newcastle',
Ps = [2000-2007],
T = 8.
SWI-Prolog has a nice library to handle aggregation: it builds upon standard 'all solutions' predicates like findall/3,setof/3,bagof/3, so you should first grasp the basic of these (as Boris explained in his answer). With the library, a single query solves your problem:
timePeriod(PS,X) :-
aggregate(sum(P), B^E^(popStar(PS,B,E),P is E-B+1), X).
Trying to create a predicate (timePeriod/2) that calculates the time period between two dates for a specific fact. I've managed to do this by myself, but face issues when 'other answers' exist in the same list (i.e. easier to explain with examples).
I have the following knowledge-base facts;
popStar('Jackson',1987,1991).
popStar('Jackson',1992,1996).
popStar('Michaels',1996,2000).
popStar('Newcastle',2000,2007).
popStar('Bowie',2008,2010).
And the following function, calculates the time between dates for a specific fact (as per below).
Predicate (timePeriod/2) -
timePeriod(PS,X) :-
bagof((Name,Start,End),popStar(Name,Start,End),PSs),X is End-Start+1)
Using Bowie as an example; it returns X=3 (which is correct).
However, when there is repetition in the list, with more than one answer available, the predicate just states 'false'. Using the facts 'Jackson' as an example, I want to be able to calculate both of the time periods for both facts; at the same time.
So, if the predicate would work for both of the Jackson facts, the predicate timePeriod would state X=10.
Would really appreciate if anyone could suggest what to change in order for this to work correctly.
Thanks.
You probably don't quite understand what foreach/3 does. I don't think I fully understand foreach/3 either. I know for sure that it is not the same as say:
for each x in xs:
do foo(x)
Another thing: "tuples" in Prolog are not what you might expect, coming from a language like Python or Haskell. This: (a,b,c) is actually this: ','(a,','(b,c)). Much better is to use a flat term, the generic form would be triple(a,b,c). For a pair, the idiom is First-Second.
So, you can simplify your call to bagof/3 to this:
?- bagof(From-To, pop_star(Name, Start, End), Ts).
Name = 'Bowie',
Ts = [2008-2010] ;
Name = 'Jackson',
Ts = [1987-1991, 1992-1996] ;
Name = 'Michaels',
Ts = [1996-2000] ;
Name = 'Newcastle',
Ts = [2000-2007].
Once you have a list as above, you need to sum the differences, which would be maybe something like:
periods_total(Ps, T) :-
maplist(period_length, Ps, Ls),
sum_list(Ls, T).
period_length(From-To, Length) :-
Length is To - From + 1.
And then you can query like this:
?- bagof(From-To, pop_star('Jackson', From, To), Ps), periods_total(Ps, T).
Ps = [1987-1991, 1992-1996],
T = 10.
?- bagof(From-To, pop_star(Name, From, To), Ps), periods_total(Ps, T).
Name = 'Bowie',
Ps = [2008-2010],
T = 3 ;
Name = 'Jackson',
Ps = [1987-1991, 1992-1996],
T = 10 ;
Name = 'Michaels',
Ps = [1996-2000],
T = 5 ;
Name = 'Newcastle',
Ps = [2000-2007],
T = 8.
SWI-Prolog has a nice library to handle aggregation: it builds upon standard 'all solutions' predicates like findall/3,setof/3,bagof/3, so you should first grasp the basic of these (as Boris explained in his answer). With the library, a single query solves your problem:
timePeriod(PS,X) :-
aggregate(sum(P), B^E^(popStar(PS,B,E),P is E-B+1), X).