I have a prolog rule below
schedule(mary,[ma424,ma387,eng301]).
and I have a predicate
taking(X,Y):- schedule(X, [Y | L]).
and when I try to figure out what classes she's taking by typing
taking(mary,Y).
i'm getting
y=ma424
why isn't it printing out ALL of her classes
i've also tried this and other variation
taking(X,Y):- schedule(X,[X|L]),schedule(Y, [Y | L]),schedule(Y,L),X\=Y,X\=L.
but it doesnt work
how do I get it to print all the classes give the way my rule is defined
This is due to the way you defined the predicate.
taking(X,Y) :- % X takes class Y if...
schedule(X, % in the schedule for X,
[Y|L]). % Y is the first element.
Your program will not magically decide to search through the list L if you don't tell it to. To do that, use the member/2 predicate:
taking(Student, Class) :-
schedule(Student, Classes),
member(Class, Classes).
Related
I have written this code so far (which i found here in Stack Overflow and altered a bit)
unique(M,List) :-
append(X,Y,List),
member(M,X),
member(M,Y).
But it does the exact opposite of what i want.
for example, it is expected to give these results.
?- unique(1,[1,2,3]).
yes
?- unique(1,[1,2,3,1]).
no
but instead, it gives
?- unique(1,[1,2,3]).
no
?- unique(1,[1,2,3,1]).
yes
is there a way in which i can modify my code in order to give the desired results?
Your code effectively says "M is unique in List if List can be split into two parts, X and Y, and M is present in both X and Y." This is a very clever way of doing exactly the opposite of what you want.
Here's another way of asking if X is unique in List: if I take X from List, X is no longer in List. You could turn that into Prolog like so:
unique(X, List) :-
select(X, List, ListWithoutX),
\+ memberchk(X, ListWithoutX).
I'm trying to write a prolog program to demonstrate how a cut can be used to increase efficiency. I'm using the following member declaration to search through a list.
member(Element, [Element | _]).
member(Element, [_ | List]) :- member(Element, List).
The function seems to be working just fine ex:
?- member(1,[3,1,4]).
true
The issue that i'm running into is I need to be able to use member with a predeclared list. like so:
someList([3,1,4]).
?- somelist(X).
X = [3, 1, 4].
3 ?- member(1,somelist).
false.
Even though somelist is defined as [3,1,4] which works in the first test, when called this returns false. I just need to know what I'm doing wrong that ?- member(1,somelist). is returning false. Any help would be greatly appreciated.
someList is the name of a proposition that does not exist, not a list, so it'll always produce false/no.
Prolog accepts queries and produces answers. What you seem to be asking for is the concept of variables in an imperative language, which doesn't exist here.
You can, for example, ask Prolog for all of the lists that have 1 as their member with member(1, X), or for a particular list member(1, [3,1,4]), but you can't store a list and ask about it in this way.
Prolog works with a Knowledge Base, and infers truthness through rules/predicates, facts and propositions that it knows. You need to ask it what you want to know.
I'm new to prolog and I'm doing some exercises for practice. So I'm trying to get the sum of the given numbers in a list. I'm trying to use this:
my_last(X, [X]).
my_last(X, [_|L]) :- my_last(X, L).
(from here)
as my guide. So this is my code for getting the sum:
listsum(X, []).
listsum(X, [H|L]):-
X is H + listsum(X, L).
when I compile it, it says
practice.pl:3: evaluable listsum(_G139,_G140) does not exist
practice.pl:2: Singleton variables: [X]
then when I try listsum(0, [1,2,3]). it returns false.
I still don't understand much about prolog, and list and recursion in prolog.
Arithmetic
As you already discovered, arithmetic can be handled in Prolog with the (is)/2 operator. It's because in Prolog, everything is only symbolic calculus: things don't have a meaning by default, so the unification (=)/2 wouldn't know that (+)/2 refers to the addition for example.
Now, your problem is that you use a regular predicate inside of (is)/2 (here it's your recursive call). Since (is)/2 only performs arithmetic, it doens't evaluate the predicate call. It doesn't even recognize it since it's not an arithmetic function.
The fix here would be to affect the result of the recursive call to a variable and then use it in the (is)/2 call:
listsum(X,[]).
listsum(Result, [Head|Tail]) :-
listsum(SumOfTail, Tail),
Result is Head + SumOfTail.
Base case correctness
But if you test that code you will not get the desired result. The reason is that you have another problem, in your base case. The sum of the empty list isn't "anything", as you stated by writing
listsum(X,[]).
(X is a free variable, hence can be anything).
Instead, it's 0:
listsum(0, []).
The resulting code is:
listsum(0, []).
listsum(Result, [Head|Tail]) :-
listsum(SumOfTail, Tail),
Result is Head + SumOfTail.
Order of arguments
Now, as a sidenote, in Prolog a convention is that output variables should be put at the end of the predicate while input variables should be put at the start of the predicate, so to behave as wanted we could refactor as follows:
listsum([], 0).
listsum([Head|Tail], Result) :-
listsum(Tail, SumOfTail),
Result is Head + SumOfTail.
Tail Call Optimization
Now, we can still improve this predicate with more advanced techniques. For example we could introduce tail calls so that Tail Call Optimization (googlable) could be performed, thanks to an idiom of declarative programming called an accumulator:
listsum(List, Sum) :-
listsum(List, 0, Sum).
listsum([], Accumulator, Accumulator).
listsum([Head|Tail], Accumulator, Result) :-
NewAccumulator is Accumulator + Head,
listsum(Tail, NewAccumulator, Result).
The idea behind that is to update an intermediate result at each step of the recursion (by adding the value of the current head of the list to it) and then just state that when the list is empty this intermediate value is the final value.
Getting more general programs
As you may have noted in Prolog, quite often predicates can be used in several ways. For example, length/2 can be used to discover the length of a list:
?- length([1, 2, 3], Length).
Length = 3.
or to build a skeleton list with free variables of a desired length:
?- length(List, 3).
List = [_G519, _G522, _G525].
Here though, you might have noted that you can't ask Prolog what are the lists which have a sum that is 6:
?- listsum(L, 6).
ERROR: is/2: Arguments are not sufficiently instantiated
That's because, to "go backwards", Prolog would have to solve an equation when comes the call to the (is)/2 operator. And while yours is simple (only additions), arithmetic isn't solvable this way in the general case.
To overcome that problem, constraint programming can be used. A very nice library is available for SWI, clpfd.
The syntax here would be:
:- use_module(library(clpfd)).
listsum(List, Sum) :-
listsum(List, 0, Sum).
listsum([], Accumulator, Accumulator).
listsum([Head|Tail], Accumulator, Result) :-
NewAccumulator #= Accumulator + Head,
listsum(Tail, NewAccumulator, Result).
Now we can use our predicate in this other way we wished we could use it:
?- listsum(L, 6).
L = [6] ;
L = [_G1598, _G1601],
_G1598+_G1601#=6 ;
L = [_G1712, _G1715, _G1718],
_G1712+_G1715#=_G1728,
_G1728+_G1718#=6 . % Here I interrupted the answer but it would not terminate.
We could even ask for all the solutions to the problem:
?- listsum(L, X).
L = [],
X = 0 ;
L = [X],
X in inf..sup ;
L = [_G2649, _G2652],
_G2649+_G2652#=X . % Here I interrupted the answer but it would not terminate
I just mentionned that so that you realize that quite often the use of (is)/2 should be avoided and use of constraint programming should be preferred to get the most general programs.
If possible, use clpfd instead of plain old (is)/2 (and friends).
clpfd offers a logically pure predicate sum/3 that could fit your needs!
Is it possible to define a list, that consists of predicates and how do I call the predicates.
Also, is it possible to pass one predicate to another predicate (like passing atoms)?
Example:
pre1:- something.
pre2(Predicate1, List):-
call(Predicate1),
append([Predicate1], List, R),
.....
You can't store predicates in a list, but you can store terms (or functors) and call terms as goals.
Here's a predicate that tests whether a term has the properties described by a list of functors:
has_properties([], _).
has_properties([P|Ps], X) :-
Goal =.. [P, X], % construct goal P(X)
call(Goal),
has_properties(Ps, X).
Usage:
% is 4 a number, an integer and a foo?
?- has_properties([number, integer, foo], 4).
The answer to this query will depend on your definition of foo/1, of course. See my explanation of =.. if needed.
Edit: as #false reports in the comments, it's not necessary to use =.., since Goal =.. [P, X], call(Goal) can be replaced by call(P, X) will have the same effect. It might still be worthwhile learning about =.., though, as you may encounter it in other people's code.
I've been given the question:
Define a predicate ordered/1, which checks if a list of integers is correctly in ascending order. For example, the goal ordered([1,3,7,11]) should succeed, as should the goal ordered([1,3,3,7]), whereas the goal ordered([1,7,3,9]) should fail.
So far I have this:
ordered([]).
ordered([N, M|Ns]):-
append(M, Ns, Tail),
ordered(Tail),
N =< M.
But it fails on every list.
I have deduced that the reason it fails is because it reaches the end number in the list then tries to compare that number against an empty list. Obviously this fails because you can't compare an integer to an empty list. Even if you could and it, say, returned 0 for an empty list, it would still return false as the number would be greater than 0, not less than.
I can't find a solution... Any ideas? Thanks, Jon.
Edit
So, some slightly amended code:
ordered([]).
ordered([N]):-
N >= 0.
ordered([N, M|Ns]):-
append(M, Ns, Tail),
ordered(Tail),
N =< M.
This now works for ordered([1]), but bigger lists still don't run correctly.
Should I include something like ordered([N, M|Ns]) in the definition?
(assuming this is homework, I hesitate to give a complete solution).
Looking at your code, try to find out how it would unify ?- ordered([1]).
Run this query mentally (or using trace/0) and see what it does, step by step, and how it computes its result.
Also, please try to get "returns a value" out of your mind when thinking prolog. Prolog predicates don't return anything.
I think your solution is not also tail-recursion-friendly.
Think something like that would do:
ordered([]) :-!.
ordered([_]):-!.
ordered([A,B|T]) :-
A =< B,
!,
ordered([B|T]).
If you are using SICStus Prolog,
my previous answer will not work, as the
clpfd library in SICStus Prolog
does not offer the library predicate
chain/3 included with
SWI-Prolog's clpfd library.
:- use_module(library(clpfd)).
:- assert(clpfd:full_answer).
Don't panic! Simply implement predicate ordered/1 like this:
ordered([]).
ordered([X|Xs]) :-
ordered_prev(Xs,X).
ordered_prev([] ,_ ).
ordered_prev([X1|Xs],X0) :-
X0 #=< X1,
ordered_prev(Xs,X1).
Let's see it in action with SICStus Prolog 4.3.2.
Here's the most general query:
?- ordered(Xs).
Xs = []
; Xs = [_A]
; Xs = [_A,_B], _A#=<_B, _A in inf..sup, _B in inf..sup
; Xs = [_A,_B,_C], _A#=<_B, _B#=<_C, _A in inf..sup, _B in inf..sup, _C in inf..sup
... % an infinity of solutions follows: omitted for the sake of brevity.
And here are the queries the OP suggested:
?- ordered([1,3,7,11]).
yes % succeeds deterministically
?- ordered([1,3,3,7]).
yes % succeeds deterministically
?- ordered([1,7,3,9]).
no
Note that both succeeding queries in above example did not leave any useless choicepoints behind, thanks to first argument indexing.
If your Prolog system supports clpfd, check if it offers the library predicate clpfd:chain/2.
:- use_module(library(clpfd)).
If so, simply write:
?- chain([1,3,7,11],#<).
true.
?- chain([1,3,3,7],#=<).
true.
?- chain([1,3,3,7],#<).
false.
?- chain([1,7,3,9],#<).
false.
You're quite right: according to your code there are only two possible ways a list can be ordered:
It's empty
The first two items are in the correct order, and the rest of the list is ordered
Those are certainly both correct statements, but what about the list [3]? Isn't that ordered too? Obviously a list with only one element is ordered, yet you have no provision for expressing that: it fits neither your base case nor your recursive case.
The single-element list is another case hiding here that you haven't addressed yet. Since this is independent of the two rules you've already defined, you might want to consider a way to address this special case separately.
Well that, in the end, was rediculously easy to fix.
Here is the correct code.
ordered([]).
ordered([N, M|Ns]):-
append([M], Ns, Tail),
ordered(Tail),
N =< M.
ordered([M]).
ordered([M]). deals with the single-element list as described above.
The real root of my problem was not including [] around the M in the append function.
Whats the ettiquette regarding awarding the correct answer? You've both helped muchly.
Jon
Don't use append/3.
edit1 to satisfy #false. In order to make it tail recursive friendly it has to eliminate backtracking. This is tail-recursive and only slight variation on #Xonix:
ordered([X|[]]):-!.
ordered([X,Y|Ys]) :-
X =< Y,
!,
ordered([Y|Ys]).
edit2 Take it a step further to eliminate lists that have less than two elements
ordered([X,Y|[]]):- X =< Y,!.
ordered([X,Y|Ys]) :-
X =< Y,
!,
ordered([Y|Ys]).