Unfortunately, I cannot give the full text of the problem for some reason. Therefore, I will try to describe the main point.
There was a murder at the hotel in room 4. 6 visitors are suspected who came to visit someone in one of the 6 hotel rooms at different intervals (These are all facts).
Then all the suspects were interviewed, some evidence was considered and information was received from the receptionist (These are the rules).
You need to find out who was where at what time. Well, and therefore who is the killer.
my problem
I am not getting the correct result.
When calling
guest(brown, R, T).
i get false (brawn must be a killer)
and basically when i call the function
solution(Guests).
then I get a very large number of lists. And the rules are not followed in them. Why is that?
result lists screenshoot
In short, I have a suspicion that these rules work somehow separately. But I could be wrong, because in the prologue I am a complete zero. What's my mistake?
P.S. If necessary, I can try to write the full text of the assignment. The only problem is that the text is in a different language from an old book that was scanned..
The reason why your query fails is the following program fragment. I obtained it by systematically generalizing away goal after goal by adding a * in front. Because this fragment fails, also your original program will fail. I am sure it will be evident to you how to interpret this:
:- op(950, fy, *). % auxiliary definition
*_.
:- initialization(guest(brown, _R, _T)). % your failing query
evidence(taylor,R,_):- *not(R=5).
evidence(white,R,_):- *not(R=5).
evidence(smith,R,_):- *not(R=1), *not(R=3), *not(R=6), *not(R=5).
evidence(green,R,_):- *not(R=3), *not(R=6).
guest(N,R,T):-
*interrogation(N,R,T),
evidence(N,R,T),
*receptionist(N,R,T).
Just a remark, instead of not(A=B) rather use dif(A,B). It's the 21st century...
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 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.
ok, so there's basically 3 tasks this program must carry out:
Parse a sentence given in the form of a list, in this case (and throughout the example) the sentence will be [the,traitorous,tostig_godwinson,was,slain]. (its history, don't ask!) so this would look like:
sentence(noun_phrase(det(the),np2(adj(traitorous),np2(noun(tostig_godwinson)))),verb_phrase(verb(slain),np(noun(slain)))).
use the parsed sentence to extract the subject, verb and object, and output as a list, e.g. [tostig_godwinson,was,slain] using the current example. I had this working too until I attempted number 3.
use the target list and compare it against a knowledge base to basically answer the question you asked in the 1st place (see code below) so using this question and the knowledge base the program would print out 'the_battle_of_stamford_bridge' as this is the sentence in the knowledge base with the most matches to the list in question
so here's where i am so far:
history('battle_of_Winwaed',[penda, king_of_mercia,was,slain,killed,oswui,king_of_bernicians, took_place, '15_November_1655']).
history('battle_of_Stamford_Bridge',[tostig_godwinson,herald_hardrada,was,slain, took_place, '25_September_1066']).
history('battle_of_Boroughbridge',[edwardII,defeated,earl_of_lancaster,execution, took_place, '16_march_1322']).
history('battle_of_Towton',[edwardIV,defeated,henryVI,palm_Sunday]).
history('battle_of_Wakefield',[richard_of_york, took_place,
'30_December_1490',was,slain,war_of_the_roses]).
history('battle_of_Adwalton_Moor',[earl_of_newcastle,defeats,fairfax, took_place, '30_June_1643',battle,bradford,bloody]).
history('battle_of_Marston_Moor',[prince_rupert,marquis_of_newcastle,defeats,fairfax,oliver_cromwell,ironsides, took_place,
'2_June_1644', bloody]).
noun(penda).
noun(king_of_mercia).
noun(oswui).
noun(king_of_bernicians).
noun('15_November_1655').
noun(tostig_godwinson).
noun(herald_hardrada).
noun('25_September_1066').
noun(edwardII).
noun(earl_of_lancaster).
noun('16_march_1322').
noun(edwardIV).
noun(henryVI).
noun(palm_Sunday).
noun(richard_of_york).
noun('30_December_1490').
noun(war_of_the_roses).
noun(earl_of_newcastle).
noun(fairfax).
noun('30_June_1643').
noun(bradford).
noun(prince_rupert).
noun(marquis_of_newcastle).
noun(fairfax).
noun(oliver_cromwell).
noun('2_June_1644').
noun(battle).
noun(slain).
noun(defeated).
noun(killed).
adj(bloody).
adj(traitorous).
verb(defeats).
verb(was).
det(a).
det(the).
prep(on).
best_match(Subject,Object,Verb):-
history(X,Y),
member(Subject,knowledgebase),
member(Object,knowledgebase),
member(Verb,knowledgebase),
write(X),nl,
fail.
micro_watson:- write('micro_watson: Please ask me a question:'), read(X),
sentence(X,Sentence,Subject,Object,Verb),nl,write(Subject),nl,write(Verb),nl,write(Object).
sentence(Sentence,sentence(Noun_Phrase, Verb_Phrase),Subject,Object,Verb):-
np(Sentence,Noun_Phrase,Rem),
vp(Rem,Verb_Phrase),
nl, write(sentence(Noun_Phrase,Verb_Phrase)),
noun(Subject),
member(Subject,Sentence),
noun(Object),
member(Object,Rem),
verb(Verb),
member(Verb,Rem),
best_match(Subject,Object,Verb).
member(X,[X|_]).
member(X,[_|Tail]):-
member(X,Tail).
np([X|T],np(det(X),NP2),Rem):-
det(X),
np2(T,NP2,Rem).
np(Sentence,Parse,Rem):- np2(Sentence,Parse,Rem).
np(Sentence,np(NP,PP),Rem):-
np(Sentence,NP,Rem1),
pp(Rem1,PP,Rem).
np2([H|T],np2(noun(H)),T):-noun(H).
np2([H|T],np2(adj(H),Rest),Rem):- adj(H),np2(T,Rest,Rem).
pp([H|T],pp(prep(H),Parse),Rem):-
prep(H),
np(T,Parse,Rem).
vp([H|[]],verb(H)):-
verb(H).
vp([H|T],vp(verb(H),Rest)):-
verb(H),
pp(T, Rest,_).
vp([H|T],vp(verb(H),Rest)):-
verb(H),
np(T, Rest,_).
As i said i had number 2 working until i tried number 3, now it just prints the parsed sentence out and then give me a 'Error: out of local stack message' any help is greatly appreciated! So at the top is the knowledge base with which we are comparing out list to find the best match, these are called (albeit incorrectly at this stage) by the best_match method, which executes immediately after the sentence method which parses the sentence and extract the key words. Also i apologise if the code is terribly laid out!
Cheers
I assume the person who posted this is never coming back, I wanted to remind myself some prolog, so here it is.
There are two major issues with this code, apart from the fact that there are still some logical problems in some predicates.
Problem 1:
You ignored singleton warnings, and they usually are something not to be ignored. The best match predicate should look like this:
best_match(Subject,Object,Verb):-
history(X,Y),
member(Subject,Y),
member(Object,Y),
member(Verb,Y),
write(X),nl,
fail.
The other warning was about the Sentence variable in the sentence predicate, so it goes like this:
sentence(X,Subject,Object,Verb),nl,write(Subject),nl,write(Verb),nl,write(Object).
sentence(Sentence,Subject,Object,Verb):-
np(Sentence,_,Rem),
vp(Rem,_),
nl,
noun(Subject),
member(Subject,Sentence),
noun(Object),
member(Object,Rem),
verb(Verb),
member(Verb,Rem),
best_match(Subject,Object,Verb).
Problem 2:
I assume you divided the np logic into np and np2 to avoid infinite loops, but then forgot to apply this division just where it was necessary. The longest np clause should be:
np(Sentence,np(NP,PP),Rem):-
np2(Sentence,NP,Rem1),
pp(Rem1,PP,Rem).
If you really wanted to allow more complicated np there, which I doubt, you can do it like this:
np(Sentence,np(NP,PP),Rem):-
append(List1,List2,Sentence),
List1\=[],
List2\=[],
np(List1,NP,Rem1),
append(Rem1,List2,Rem2),
pp(Rem2,PP,Rem).
This way you will not end up calling np with the same arguments over and over again, because you make sure that the sentence checked is shorter each time.
Minor issues:
(How the program works, after the infinite loop problem has been fixed)
The last vp is repeated
I am not sure about your grammar, and e.g. why "defeated" is a noun...
Just to check that the program works I used the sentence [edwardIV,defeated,henryVI,on,palm_Sunday].
I changed "defeated" to a verb, and also changed the last vp clause to:
vp([H|T],vp(verb(H),Rest)):-
verb(H),
np(T,_,Rest1),
pp(Rest1, Rest,_).
For the example sentence I got battle_of_Boroughbridge and battle_of_Towton as results.
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.