I have a list structure called "stack".
At the point in my program which is causing problems, this is what stack holds:
stack([[s]],[np,[noun,john]])
I got this from running a trace, and its what stack is supposed to be holding.
When writing the next rule which is supposed to match this.
if
buffer([[Word|_]])
and aux(Word)
and stack([s],[np,[noun, john]])
If I do this then the rule executes as its supposed to. But I need to use a variable here instead of using "and stack([[s]],[np,[noun,john]])". However when I try to use anything else, the rule does not fire. I can't work out why. Other rules work fine when I use variables in the list.
Ive tried
stack([s]|Foo)
stack([s]|[Foo])
stack([MyHead]|[MyTail]... and literally every other combination I can think of.
I'm not entirely sure what is causing this problem
Your stack seems to have arity 2, where each arg is a list.
These aren't valid syntax for lists
stack([s]|Foo)
stack([s]|[Foo])
...
but since some Prolog declare the (|)/2 operator as alternative to (;)/2 (i.e. disjunction), you will not see any syntax error.
To understand you problem, you could try to unify, by mean of unification operator (=)/2
?- stack(S, Foo) = stack([[s]],[np,[noun,john]]).
you will get
S = [[s]]
Foo = [np,[noun,john]]
Related
I am trying to write a function longer(S1,S2) which should be true if S1 is longer than S2, otherwise false. What I have so far is the following:
longer(A,nil).
longer(nil,B) :- false.
longer([A,AS],[B,BS]) :- longer(AS,BS).
But for some reason I cannot seem to get it to parse correctly when I provide for instance the input: longer([1,2],[1]).
But when I run the above through swi-prolog it return false.
Another example is running the: longer([1],nil) which also return false, even though it should be true by matching with list(cons(A,As)) where As = nil, finally matching the end clause.
What am I missing? Can someone point me in the right direction here, as I cannot see how this is not evaluating to true.
Edit: it should be noted that I am still fairly new to prolog at this point.
Update I have had some misunderstanding in relation to what is common prolog semantic. Including trying to force the program to yield a false value (probably being steered by my understanding of non-declarative language semantics). I have updated my answer with the inputs from #tiffi.
That is a solution that is pretty close to the thinking behind what you have come up with:
longer([_|_],[]).
longer([_|As], [_|Bs]):- longer(As,Bs).
Your idea with regard to the first clause (before your edit) could be expressed like this:
longer(A,[]):- is_list(A). %is_list/1 is inbuilt
However, that doesn't give you the right result, since the empty list is also a list. Thus you need to make sure that the first argument is a non-empty list.
I'm trying to solve a water, jug problem (one 7L, one 4L, get 5L in the 7L jug) using dept first search. However something keeps going wrong whenever I try to get a new state back from one of my actions.
Prolog Code
I can't figure out what is going wrong, this is what the output looks like after trace:
enter image description here
Thanks in advance for any help!
You should copy and paste your code into your question; we cannot copy and paste it from your images, which makes it more work to help you, which in turn makes it less likely that we will help.
Some problems I noticed anyway:
Your first rule for go_to_goal/3 does not talk about the relation between ClosedList and Path. You will compute the path but will never be able to communicate it to the caller. (Then again, you also ignore Path in solve/0...) If your Prolog system gives you "singleton variable" warnings, you should never ignore them!
You are using the == operator wrong. The goal State == (5, X) states that at the end you are looking for a pair where the first component is 5 (this part is fine) and the second component is an unbound variable. In fact, after your computations, the second component of the pair will be bound to some arithmetic term. This comparison will always fail. You should use the = (unification) operator instead. == is only used rarely, in particular situations.
If you put a term like X+Y-7 into the head of a rule, it will not be evaluated to a number. If you want it to be evaluated to a number, you must use is/2 in the body of your rules.
Your most immediate problem, however, is the following (visible from the trace you posted): The second clause of go_to_goal/3 tries to call action/2 with a pair (0, 0) as the first argument. This always fails because the first argument of every clause of action/2 is a term state(X, Y). If you change this to state(0, 0) in go_to_goal/3, you should be able to make a little bit of progress.
I am implementing a Prolog program for trigonometric identities. This is a part of the program.
simplify_exp(Term1+Term2,X,Y) :- isolaxt(Term1+Term2,Y),
(nonvar(Y)-> simplify_exp(Term1,X,Y), simplify_exp(Term2,X,Y)).
isolaxt(sin(U)^2+cos(U)^2,1).
I want to execute the if statement only if Y has no value (when isolaxt is failed only). Even though I included nonvar(Y) it didn't work. How can it be done?
I don't really get if that's the only isolaxt clause you have or if you included it to exemplify.
Anyway, if isolaxt fails then simplify_exp will fail without reaching nonvar(Y).
Is that the behavior you expect? Perhaps you meant to use ; like so:
simplify_exp(Term1+Term2,X,Y) :-
isolaxt(Term1+Term2,Y); (simplify_exp(Term1,X,Y), simplify_exp(Term2,X,Y)).
This means either isolaxt will succeed or you're going to proceed with the calls to simplify_exp.
Also, note that nonvar(Y) will fail when isolaxt succeeds, so maybe you expected to use var(Y) instead.
I have been working on a code in prolog for a while now and it is near compiling worthy and all my ideas seem to be solid so it should work when it compiles. It is a program that consults a database file for a list of clauses and then it awaits for a query by the user which it will then pick what information it needs from the sentence and query the database appropriately but there is a block of code that keeps giving me errors complaining that the flowpattern doesn't exist in the standard predicate this may be a silly question but even with all the looking into this I have done i can't find out how to fix this problem if someone could help me out or point me in the right direction that would be greatly appreciated.
Here is the block of code that gives the error:
loop(STR):-
scan(STR,LIST),
filter(LIST,LISroT1),
pars(LIST1,LIST2),
fail.
loop(STR):- STR >< "",readquery(L),loop(L).
readquery(QUERY):-nl,nl,write("Query: "),readln(QUERY).
scan(STR,[TOK|LIST]):-
fronttoken(STR,SYMB,STR1),!,
upper_lower(SYMB,TOK),
scan(STR1,LIST).
the specific line that the compiler complains about is fronttoken(STR,SYMB,STR),!,
any help will be apreaciated thanks!
Since we are looking at an "ex[c]er[p]t" of the code, it's hard to be sure what is going wrong, but the the given evidence points to this: loop/1 is being called before readquery/1 can do its work to populate (bind) the argument STR to loop/1.
Notice that loop/1 calls itself (recursively), and does so in a repeat/fail pattern. But the first time loop/1 runs, there's no indication in the code shown of how argument STR would get populated.
A clearer (more self-contained) code snippet would be like this:
loop :-
readquery(STR),
scan(STR,LIST),
filter(LIST,LISroT1),
pars(LIST1,LIST2),
fail.
loop :- loop.
This makes it clear that predicate loop doesn't actually return any result (and the given code snippet isn't complete enough to make clear what the program as a whole accomplishes). It assumes that the clauses ahead of fail in loop are deterministic, so that in failing, control passes through to the second (recursive) clause of loop/0. If this is not the case, the determinism could be forced by wrapping each call inside once/1.
i know there is a build-in function findall/3 in prolog,
and im trying to find the total numbers of hours(Thrs) and store them in a list, then sum the list up. but it doesnt work for me. here is my code:
totalLecHrs(LN,THrs) :-
lecturer(LN,LId),
findall(Thrs, lectureSegmentHrs(CC,LId,B,E,THrs),L),
sumList(L,Thrs).
could you tell me what's wrong with it? thanks a lot.
You need to use a "dummy" variable for Hours in the findall/3 subgoal. What you wrote uses THrs both as the return value for sumList/2 and as the variable to be listed in L by findall/3. Use X as the first argument of findall and in the corresponding subgoal lectureSegmentHrs/5 as the last argument.
It looks like the problem is that you're using the same variable (Thrs) twice for different things. However it's hard to tell as you've also used different capitalisation in different places. Change the findall line so that the initial variable has the same capitalisation in the lectureSegmentHrs call. Then use a different variable completely to get the final output value (ie the one that appears in sumList and in the return slot of the entire predicate).
You need to use a different variable because Prolog does not support variable reassignment. In a logical language, the notion of reassigning a variable is inherently impossible. Something like the following may seem sensible...
...
X = 10,
X = 11,
...
But you have to remember that , in Prolog is the conjunction operator. You're effectively telling Prolog to find a solution to your problem where X is both 10 and 11 at the same time. So it's obviously going to tell you that that can't be done.
Instead you have to just make up new variable names as you go along. Sometimes this does get a bit annoying but it's just goes with the territory of a logical languages.