I have written the following code in SWI-Prolog:
:- dynamic state_a/1 .
:- dynamic state_b/1 .
:- dynamic state_c/1 .
state_a([1,2,3,4,5,0]).
state_b([0]).
chop(LIST,HEAD,TAIL) :- LIST=[HEAD|TAIL].
move_ab :- !,state_a(X),
chop(X,Ha,Ba),
Ha>0,
state_b(Y),
chop(Y,Hb,Bb),!,
(Ha<Hb ; Hb =:= 0),
asserta(state_a(Ba)),asserta(state_b([Ha|Y])),
retract(state_a(X)), retract(state_b(Y));
write('Wrong Move.Choose Another').
There are 2 OR(;) in my code. When I ask move_ab in Prolog for the first time all the conditions before the second OR are true so i get the answer true from Prolog.
But the second time I ask move_ab in Prolog I get just the answer false.
I don't know how this can happen. Some of the conditions before the second OR are not true so Prolog should check the condition after the second OR and write me the message Wrong Move.Choose Another..
I tried using () to group the conditions but I still get the same message.
Any idea about what is going on? By the way I am a newb in Prolog, just started 2 days ago :)
The problem lies in the use of cut (!/0) before the conditions are properly tested. Cut removes choice points. Here it means that the choices are removed before you even test anything. So if the tests fail, everything fails !
BTW, manipulating the database is maybe not the best idea there. To represent states, you could use global variables as follows :
:- nb_setval(state_a, [1,2,3,4,5,0]).
:- nb_setval(state_b, [0]).
move_ab :-
nb_getval(state_a, [Ha|Ta]),
Ha > 0,
nb_getval(state_b, [Hb|Tb]),
(Ha < Hb ; Hb =:= 0),
nb_setval(state_a, Ta),
nb_setval(state_b, [Ha, Hb|Tb]),
!
;
write('Wrong Move.Choose Another').
A general advice given to beginners in prolog is to stay away from database manipulation if possible, as often problems are solvable without it. Here though it could be justified, but global variables will be faster and easier to manipulate.
Related
Lets assume I have facts as follows:
airport(nyc,'newyork').
I want want to display a message if the user inputs an airport that doesn't exist.
My Attempt:
isAirport(Air) :-
airport(Air,T),
(var(T) -> true
;
write('Airport not found'),
fail
).
However, this doesn't seem to work.
First let's see what happens if you query a conjunction (the , operator) first:
?- airport(nyc, _), write('found!').
found!
true.
?- airport(abc, _), write('found!').
false.
This means, isAirport(abc) directly fails after trying airport(abc,_) without the rest of your predicate being evaluated. In many cases, you can therefore get by without an explicit if-then-else construct and just write something of the form
predicate(X) :-
first_condition(X),
second_condition(X).
and it will only succeed if both conditions are fulfilled for X.
In case you really want to create some user interface, this is a bit more tricky, because I/O is inherently non-logical, in particular when there is backtracking involved. We usually call a program which behaves like we would expect from a logical formula pure and when it contains non-logical constructs like I/O or the cut operator ! are called impure.
Unfortunately, the if-then-else construct (-> and ;) and negation (\+) are implemented via cut and therefore impure as well. Luckily, most of the time people want a conditional, a pure disjunction is sufficient:
case(1,a).
case(2,b).
We have an automatic branching from the execution mechanism of Prolog:
?- case(X,Y).
X = 1,
Y = a ;
X = 2,
Y = b.
But sometimes we really want to do something that needs the impure constructs, like user input. Then the easiest way to keep the nice logical properties of our program is to separate the task into pure and impure ones:
main :-
uinput(Data),
pure_predicate(Data, Result),
write(Result).
After we have done all the impure parts, Data is unified with the user data we wanted. Let's have a look at the implementation of uinput/1:
uinput(data(Airport,D-M-Y)) :-
format('~nAirport? '),
read(Airport),
( ground(Airport), airport(Airport, _) )
->
(
format('~nDay? '),
read(D),
format('~nMonth? '),
read(M),
format('~nYear? '),
read(Y),
( ground(D-M-Y), isDate(D-M-Y) )
->
true
;
throw(failure('unknown date'))
)
;
throw(failure('unknown airport'))
.
We successively read terms from the input and throw an exception if we can't handle it. For the if-then-else construct to work, we need to take special care. If we compare the two queries:
?- between(1,3,X), write(X).
1
X = 1 ;
2
X = 2 ;
3
X = 3.
and
?- between(1,3,X) -> write(X); false.
1
X = 1.
you can see that the if-then-else is losing solutions. This means we need to make sure that our condition is deterministic. Asking for a user input term to be ground is already a good idea, because without variables, there is only one solution term. Still, a call to one of the data-predicates airport/1 and isDate/1 might generate the same term multiple times or not terminate at all. In this particular case, we just need to make sure that each airport has a unique shortcut name, we can also generate dates without repetition:
airport(nyc, 'New York City').
airport(wdc, 'Washington DC').
isDate(X-Y-Z) :-
between(1,31,X),
between(1,12,Y),
between(1970,2100,Z).
Another trick in the implementation of uinput is that we just succeed with true when we have validated everything. The only effect of is now that Data is instantiated with whatever the user entered.
If we give a dummy implementation of the actual implementation, we can already try the implementation oursevles:
pure_predicate(_Data, Result) :-
% here goes the actual stuff
Result='we have found something awesome'.
On the prompt we can use the pure predicate without trouble:
?- pure_predicate(someinputdata,Y).
Y = 'we have computed something awesome'.
On the other hand, we can also use the full predicate as follows:
?- main(_).
Airport? wdc.
Day? |: 1.
Month? |: 2.
Year? |: 2000.
we have found something awesome
true.
Since we are using read, we have to input prolog terms and terminate with a dot ., but everything worked as expected.
In case the user input fails, we get:
?- main(_).
Airport? bla(X).
ERROR: Unhandled exception: failure('unknown airport')
Please note that we only went through this trouble to actually fail early and give a user message in that case. For the actual computation, this is completely unneccessary.
In the code below you are making false assumption that T will remain unbound in case if airport will not be found in database:
airport(Air, T)
What happens actually is that call to airport(Air, T) will make isAirport(Air) to fail immediately and your var(T) and other parts will not be executed at all.
Try this code instead:
isAirport(Air) :-
airport(Air, _T), ! ; write('Airport not found'), fail.
I defined my knowledge base as:
edge(mammal,isa,animal).
edge(human,isa,mammal).
edge(simba,isa,human).
edge(animal,swim,bybirth).
edge(human,swim,mustlearn).
path(X,Y) :- edge(X,isa,Y).
path(X,Y) :- edge(X,isa,Z), path(Z,Y).
swim(X,Y) :- edge(X,swim,Y).
swim(X,Y) :- path(X,Z), swim(Z,Y).
Now, to use the above knowledge base, I use the following:
?- swim(simba,bybirth).
?- swim(simba,mustlearn).
And for both the queries, Prolog returns true. I want Prolog to check for the property swim locally first, then look at the direct parent, and so on in a hierarchical fashion. And it should stop searching as soon as we know that Simba "mustlearn" to swim, and shouldn't look any further. Thus, it should return false for the first query and true for the second.
I know it has to be done by limiting backtracking. I tried using the cut and not operators, but couldn't succeed. Is there a way to achieve this?
I tried it and ran into a problem too. I thought this might work:
swim(X,Y) :- once((edge(X,swim,Y); path(X,Z), swim(Z,Y))).
It doesn't work, because if Y is already instantiated on the way in, the first step will fail to unify and it will try the second route going through the human intermediate. So even though the query only produces one result, it can be fooled into producing swim(simba, bybirth). The solution is to force Prolog to commit to a binding on another variable and then check that binding after the commitment:
swim(X,Y) :-
once((edge(X,swim,Method); path(X,Z), swim(Z,Method))),
Method = Y.
This tells Prolog, there is only one way to get to this method, so find that method, and then it must be Y. If you find the wrong method, it won't go on a search, it will just fail. Try it!
I have a simple Prolog-program that I need some help debugging.
The point is to extend the program by pattern-matching to create a proof checker for propositional logic. The problem I have is that I get no when I expect yes and my 'fix' (providing a base case for valid_proof_aux) still gives me two solutions and I don't know why.
Not sure how to go about debugging Prolog yet, sorry.
%call:
valid_proof([p],p,[[1, p, premise]])
%src:
reverse_it([],Z,Z).
reverse_it([H|T],Z,Acc) :- reverse_it(T,Z,[H|Acc]).
valid_proof(Prems,Goal,Proof):-
last(Proof, [_, Goal, _]),
reverse_it(Proof, RevP, []),
valid_proof_aux(Prems, RevP) .
valid_proof_aux(Prems,
[[_,Prop,premise] | T]):-
memberchk(Prop,Prems),
valid_proof_aux(Prems,T).
%my 'fix'
valid_proof_aux(_, []) :- true .
You don't really show how to run the program and what exactly you get (you should edit your question with and add this), so this answer is a bit of a guess, but anyway:
You need the base case either way (as you observe yourself), valid_proof_aux/2 would fail when the list becomes empty [] and does not match [[...]|T] anymore.
?- [] = [_|_]. % try to unify an empty list with a non-empty list
false.
What you need to do to get rid of the choice point is to put the list argument as the first argument.
valid_proof_aux([], _).
valid_proof_aux([[_,Prop,premise]|T], Prems) :-
memberchk(Prop, Prems),
valid_proof_aux(T, Prems).
Note that you don't need the :- true., this is implicit. Also, avoid leaving any blanks on the two sides of the | in [Head|Tail].
I am trying to get a predicate to relate from 1 fact to another fact and to keep going until a specified stopping point.
For example,
let's say I am doing a logistics record where I want to know who got a package from who, and where did they get it from until the end.
Prolog Code
mailRoom(m).
gotFrom(annie,brock).
gotFrom(brock,cara).
gotFrom(cara,daniel).
gotFrom(daniel,m).
gotFrom(X,Y) :- gotFrom(Y,_).
So what I am trying to do with the predicate gotFrom is for it to recursively go down the list from what ever point you start (ex: gotFrom(brock,Who)) and get to the end which is specified by m, which is the mail room.
Unfortunately when I run this predicate, it reads out,
Who = annie.
Who = brock.
Who = cara.
etc.etc....
I tried stepping through the whole thing but Im not sure where it goes from brock to annie, to cara and all the way down till it cycles through trues for infinity. I have a feeling that it has something to do with the wildcard in the function (_), but Im not sure how else I could express that part of the function in order for the predicate to search for the next fact in the program instead of skipping to the end.
I tried using a backcut (!) in my program but it gives me the same error.
Any help is greatly appreciated. I don't want code I just want to know what I am doing wrong so I can learn how to do it right.
Thanks.
I'm afraid this rule is meaningless:
gotFrom(X,Y) :- gotFrom(Y,_).
There is nothing here to constrain X or Y to any particular values. Also, the presence of singleton variable X and the anonymous variable _ means that basically anything will work. Try it:
?- gotFrom([1,2,3], dogbert).
true ;
true ;
What I think you're trying to establish here is some kind of transitive property. In that case, what you want is probably more like this:
gotFrom(X,Z) :- gotFrom(X, Y), gotFrom(Y, Z).
This produces an interesting result:
?- gotFrom(brock, Who).
Who = cara ;
Who = daniel ;
Who = m ;
ERROR: Out of local stack
The reason for the problem may not be immediately obvious. It's that there is unchecked recursion happening twice in that rule. We recursively unify gotFrom/2 and then we recursively unify it again. It would be better to break this into two predicates so that one of them can be used non-recursively.
got_directly_from(annie,brock).
got_directly_from(brock,cara).
got_directly_from(cara,daniel).
got_directly_from(daniel,m).
gotFrom(X,Y) :- got_directly_from(X, Y).
gotFrom(X,Z) :- got_directly_from(X, Y), gotFrom(Y, Z).
This gives us the desired behavior:
?- gotFrom(brock, Who).
Who = cara ;
Who = daniel ;
Who = m ;
false.
Notice this one is resilient to my attack of meaningless data:
?- gotFrom([1,2,3], dogbert).
false.
Some general advice:
Never ignore singleton variable warnings. They are almost always a bug.
Never introduce a cut when you don't understand what's going on. The cut should be used only where you understand the behavior first and you understand how the cut will affect it. Ideally, you should try to restrict yourself to green cuts—cuts that only affect performance and have no observable effects. If you don't understand what Prolog is up to, adding a red cut is just going to make your problems more complex.
I was wondering how to do the answer (or first function) to this question in Prolog only using one predicate? The link I'm referring to is here.
Here's an example of what I mean by only calling one predicate:
reverse([X|Y],Z,W) :- reverse(Y,[X|Z],W).
reverse([],X,X).
What are you trying to do and why do you want just one clause for the predicate?
personally I believe that having the ability to write many clauses is one of the great things of prolog and the code is more elegant and readable that way
Anyway, you will probably want to use or. assuming that you have the following clauses:
foo(Arg11,Arg12,Arg13):-
(Body1).
foo(Arg21,Arg22,Arg23):-
(Body2).
foo(Arg31,Arg32,Arg33):-
(Body3).
you should first make sure that every clause has the same arguments (bye bye pattern matching there!):
foo(Arg1,Arg2,Arg3):-
(Body1b).
foo(Arg1,Arg2,Arg3):-
(Body2b).
foo(Arg1,Arg2,Arg3):-
(Body3b).
and then you will use or (;):
foo(Arg1,Arg2,Arg3):-
(Body1b)
; (Body2b)
; (Body3b).
for example,
reverse([X|Y],Z,W):-
reverse(Y,[X|Z],W).
reverse([],X,X).
will become:
reverse(X,Y,Z):-
X = [H|T],
reverse(T,[H|Y],X).
reverse(X,Y,Z):-
X = [],
Z = Y.
and then:
reverse(X,Y,Z):-
(X = [H|T],
reverse(T,[H|Y],X) )
; (X = [],
Z = Y). *%ew ew ugly!*
regarding the example on your first post, there are two different predicates, each having just one clause. the second predicate has 2 arguments while the first one has 1 therefore they are different. The only way to "merge" them would be by simply calling the second one as has_equal_sums(List, _) in the place of the first.
To be honest, I dont see the point of doing this; I doubt you will not get any significant speedup and the code becomes way messier.
Of course, it's your code and there can be restrictions we dont know (that's why I asked what you want to accomplish)