I have to write a predicate to do work like following:
?- cat(north,south,X).
X = northsouth
?- cat(alley,'91',Y).
X = alley91
?-cat(7,uthah,H).
Bad Input
H = H
Please Help..
atom_concat_redefined(A1, A2, A3) :-
( nonvar(A1) -> atom_chars(A1, Chs1) ; true ),
( nonvar(A2) -> atom_chars(A2, Chs2) ; true ),
( nonvar(A1), nonvar(A2) -> true ; atom_chars(A3, Chs3) ),
append(Chs1, Chs2, Chs3),
atom_chars(A1, Chs1),
atom_chars(A2, Chs2),
atom_chars(A3, Chs3).
This definition produces the same errors in a standard conforming implementation like SICStus or GNU - there should be no other differences, apart from performance. To compare the errors use the goal:
?- catch(atom_concat_redefined(A,B,abc+1), error(E,_), true).
E = type_error(atom,abc+1).
Note the underscore in error(E,_), which hides the implementation defined differences. Implementations provide additional information in this argument, in particular, they would reveal that atom_chars/2 or atom_concat/3 produced the error.
atom_codes/2 it's the ISO approved predicate to convert between an atom and a list of codes. When you have 2 lists corresponding to first two arguments, append/3 (alas, not ISO approved, but AFAIK available in every Prolog), will get the list corresponding to third argument, then, convert that list to atom...
Note that, while append/3 is a 'pure' Prolog predicate, and can work with any instantiation pattern, atom_codes/2 requires at least one of it's argument instantiated. Here is a SWI-Prolog implementation of cat/3, 'working' a bit more generally. I hope it will inspire you to read more about Prolog...
ac(X,Xs) :- when((ground(X);ground(Xs)), atom_codes(X,Xs)).
cat(X,Y,Z) :- maplist(ac, [X,Y,Z],[Xs,Ys,Zs]), append(Xs,Ys,Zs).
edit
as noted by #false I was wrong about append/3. Now I'll try to understand better what append/3 does... wow, a so simple predicate, so behaviour rich!
Related
I have just started learning Prolog, and I'm wondering about the first question of this exercise.
%% Suppose we are working with the following knowledge base:
wizard(ron).
hasWand(harry).
quidditchPlayer(harry).
wizard(X) :- hasBroom(X), hasWand(X).
hasBroom(X) :- quidditchPlayer(X).
How does Prolog respond to the following queries?
wizard(ron). -> true
witch(ron). -> undefined procedure
wizard(hermione). -> false
witch(hermione). -> undefined procedure
wizard(harry). -> true
wizard(Y). -> Y = ron ; Y = harry.
witch(Y). -> undefined procedure
Using swipl on Ubuntu, importing the knowledge base for this exercise, first of course trying to decipher what Prolog will returns, and finally checking by myself.
Ok pretty boring stuff until now, I have seen a few answer to these exercises over Github (here, here and there), and I don't understand the answer to the first one: %% 1. wizard(ron). -> true.
First of all the interpreter is complaining about the two definition of what is a wizard:
Warning: /tmp/prolog/ex15.pl:4:
Clauses of wizard/1 are not together in the source-file
Earlier definition at /tmp/prolog/ex15.pl:1
Current predicate: quidditchPlayer/1
Use :- discontiguous wizard/1. to suppress this message
Secondly, when querying I obtain:
?- wizard(ron).
true ;
false.
The way I get it, first Prolog returns the first fact from the knowledge base, then apply the rule head and find out that ron has neither a broom nor a wand.
All this leading to my question: what subtlety have I missed that makes others writing true as an answer to this query?
what subtlety have I missed that makes others writing true as an answer to this query?
`?- wizard(ron).`
true;
false
You have the clause (fact) wizard(ron). in your KB.
To make the things clearer you can write the fact also as the rule clause:
wizard(ron) :- true.
As you can see this is pretty redundant notation but useful in some cases as the general fact representation.
So your query can be interpreted as follows:
Is there an wizard called ron?
Since you have the fact wizard(ron) :- true.
Prolog will first unify the goal and the head.
In your case unify is trivial comparison because no variables are in the goal and the head.
Then Prolog tries to prove body. The body is builtin predicate true, so you quickly get the answer - true.
Then pressing ';' you initiate the search for the alternative solution.
Since no (more) solutions exist for the query wizard(ron), Prolog writes false.
The dot operator designates the clause end. So you wrongly typed dots in your examples:
-> operator means if-then-else relation. It can be used within clause body.
For example you can write std_member/2 as if_member/2
std_member(X, [ X | _ ]).
std_member(X, [ _ | Xs ]) :-
std_member(X, [ _ | Xs).
if_member(X, [ Y | Xs ]) :-
X = Y -> true;
if_member( X, Xs ).
Given following facts:
route(TubeLine, ListOfStations).
route(green, [a,b,c,d,e,f]).
route(blue, [g,b,c,h,i,j]).
...
I am required to find all the pairs of tube Lines that do not have any stations in common, producing the following:
| ?- disjointed_lines(Ls).
Ls = [(yellow,blue),(yellow,green),(yellow,red),(yellow,silver)] ? ;
no
I came up with the below answer, however it does not only give me incorrect answer, but it also does not apply my X^ condition - i.e. it still prints results per member of Stations lists separately:
disjointed_lines(Ls) :-
route(W, Stations1),
route(Z, Stations2),
setof(
(W,Z),X^
(member(X, Stations1),nonmember(X, Stations2)),
Ls).
This is the output that the definition produces:
| ?- disjointed_lines(L).
L = [(green,green)] ? ;
L = [(green,blue)] ? ;
L = [(green,silver)] ? ;
...
I believe that my logic relating to membership is incorrect, however I cannot figure out what is wrong. Can anyone see where am I failing?
I also read Learn Prolog Now chapter 11 on results gathering as suggested here, however it seems that I am still unable to use the ^ operator correctly. Any help would be appreciated!
UPDATE:
As suggested by user CapelliC, I changed the code into the following:
disjointed_lines(Ls) :-
setof(
(W,Z),(Stations1, Stations2)^
((route(W, Stations1),
route(Z, Stations2),notMembers(Stations1,Stations2))),
Ls).
notMembers([],_).
notMembers([H|T],L):- notMembers(T,L), nonmember(H,L).
The following, however, gives me duplicates of (X,Y) and (Y,X), but the next step will be to remove those in a separate rule. Thank you for the help!
I think you should put route/2 calls inside setof' goal, and express disjointness more clearly, so you can test it separately. About the ^ operator, it requests a variable to be universally quantified in goal scope. Maybe a concise explanation like that found at bagof/3 manual page will help...
disjointed_lines(Ls) :-
setof((W,Z), Stations1^Stations2^(
route(W, Stations1),
route(Z, Stations2),
disjoint(Stations1, Stations2)
), Ls).
disjoint(Stations1, Stations2) :-
... % could be easy as intersection(Stations1, Stations2, [])
% or something more efficient: early fail at first shared 'station'
setof/3 is easier to use if you create an auxiliary predicate that expresses the relationship you are interested in:
disjoint_routes(W, Z) :-
route(W, Stations1),
route(Z, Stations2),
disjoint(Stations1, Stations2).
With this, the definition of disjointed_lines/1 becomes shorter and simpler and no longer needs any ^ operators:
disjointed_lines(Ls) :-
setof((W, Z), disjoint_routes(W, Z), Ls).
The variables you don't want in the result of setof/3 are automatically hidden inside the auxiliary predicate definition.
I want to use atom_chars/2 on the expression of 3+4, but I get
ERROR: atom_chars/2: Type error: 'atom' expected, found '3+4' (a compound).
I'm thinking that if I can add " " on both sides of the compound, it would work, e.g.
atom_chars("3+4", Result).
but I don't know how I can do that, or is there other approaches to do this?
Please give me some advice.
EDIT: What I mean is that the input has to be 3+4, instead of '3+4', so what I want to do is to write a predicate before the atom_chars/2 to convert 3+4 to '3+4'.
For instance: for compound2atom(X,Y),
-?compound2atom(3+4,Y).
Y='3+4'.
If you are using SWI-Prolog, there is with_output_to/2 or format/3:
?- with_output_to(atom(A), write(3+4)).
A = '3+4'.
?- with_output_to(chars(C), write(3+4)).
C = ['3', +, '4'].
?- format(atom(A), "~w", [3+4]).
A = '3+4'.
?- format(chars(C), "~w", [3+4]).
C = ['3', +, '4'].
But if you look hard enough you should be able to find some predicate that does that, for example term_to_atom/2.
My personal preference leans towards format/3.
I'm currently trying to to interpret user-entered strings via Prolog. I'm using code I've found on the internet, which converts a string into a list of atoms.
"Men are stupid." => [men,are,stupid,'.'] % Example
From this I would like to create a rule, which then can be used in the Prolog command-line.
% everyone is a keyword for a rule. If the list doesn't contain 'everyone'
% it's a fact.
% [men,are,stupid]
% should become ...
stupid(men).
% [everyone,who,is,stupid,is,tall]
% should become ...
tall(X) :- stupid(X).
% [everyone,who,is,not,tall,is,green]
% should become ...
green(X) :- not(tall(X)).
% Therefore, this query should return true/yes:
?- green(women).
true.
I don't need anything super fancy for this as my input will always follow a couple of rules and therefore just needs to be analyzed according to these rules.
I've been thinking about this for probably an hour now, but didn't come to anything even considerable, so I can't provide you with what I've tried so far. Can anyone push me into the right direction?
Consider using a DCG. For example:
list_clause(List, Clause) :-
phrase(clause_(Clause), List).
clause_(Fact) --> [X,are,Y], { Fact =.. [Y,X] }.
clause_(Head :- Body) --> [everyone,who,is,B,is,A],
{ Head =.. [A,X], Body =.. [B,X] }.
Examples:
?- list_clause([men,are,stupid], Clause).
Clause = stupid(men).
?- list_clause([everyone,who,is,stupid,is,tall], Clause).
Clause = tall(_G2763):-stupid(_G2763).
I leave the remaining example as an easy exercise.
You can use assertz/1 to assert such clauses dynamically:
?- List = <your list>, list_clause(List, Clause), assertz(Clause).
First of all, you could already during the tokenization step make terms instead of lists, and even directly assert rules into the database. Let's take the "men are stupid" example.
You want to write down something like:
?- assert_rule_from_sentence("Men are stupid.").
and end up with a rule of the form stupid(men).
assert_rule_from_sentence(Sentence) :-
phrase(sentence_to_database, Sentence).
sentence_to_database -->
subject(Subject), " ",
"are", " ",
object(Object), " ",
{ Rule =.. [Object, Subject],
assertz(Rule)
}.
(let's assume you know how to write the DCGs for subject and object)
This is it! Of course, your sentence_to_database//0 will need to have more clauses, or use helper clauses and predicates, but this is at least a start.
As #mat says, it is cleaner to first tokenize and then deal with the tokenized sentence. But then, it would go something like this:
tokenize_sentence(be(Subject, Object)) -->
subject(Subject), space,
be, !,
object(Object), end.
(now you also need to probably define what a space and an end of sentence is...)
be -->
"is".
be -->
"are".
assert_tokenized(be(Subject, Object)) :-
Fact =.. [Object, Subject],
assertz(Fact).
The main reason for doing it this way is that you know during the tokenization what sort of sentence you have: subject - verb - object, or subject - modifier - object - modifier etc, and you can use this information to write your assert_tokenized/1 in a more explicit way.
Definite Clause Grammars are Prolog's go-to tool for translating from strings (such as your English sentences) to Prolog terms (such as the Prolog clauses you want to generate), or the other way around. Here are two introductions I'd recommend:
http://www.learnprolognow.org/lpnpage.php?pagetype=html&pageid=lpn-htmlse29
http://www.pathwayslms.com/swipltuts/dcg/
I would need help about Prolog.
I posted my code, the problem is that i do not obtain the expected result.
I want planning actions for moving on table all blocks until is possible. To do this I prompt :
?- do(while(some(x, block(x) & -onTable(x)),pi(x,putOnTable(x))),s0,S).
I expect to see a response like :
S = do(putOnTable(e), do(putOnTable(b), do(putOnTable(c), s0)))
but Prolog returns "false" only. Someone can help me??
% Golog interpreter
%:- [golog_swi].
:- discontiguous clear/2, on/3, onTable/2.
:- op(800,xfy,[&]).
do(E,S,do(E,S)):- primitive_action(E),poss(a,S).
% Primitive Action Declarations.
primitive_action(putOn(_,_)).
primitive_action(putOnTable(_)).
poss(putOn(X,Y),S) :- clear(X,S), clear(Y,S), \+ on(X,Y,S), \+ X=Y.
poss(putOnTable(X),S):- clear(X,S), \+(onTable(X,S)).
% Successor State Axioms.
on(X,Y,do(A,S)):- A = putOn(X,Y); on(X,Y,S), \+ (A = putOnTable(X); A = putOn(X,_)).
onTable(X,do(A,S)) :- A = putOnTable(X); onTable(X,S), \+ A= putOn(X,_).
clear(X,do(A,S)) :- on(Y,X,S), (A = putOn(Y,_) ; A = putOnTable(Y)); clear(X,S), \+ A = putOn(_,X).
% Restore suppressed situation arguments
restoreSitArg(onTable(X),S,onTable(X,S)).
restoreSitArg(on(X,Y),S,on(X,Y,S)).
restoreSitArg(clear(X),S,clear(X,S)).
block(X):- member(X,[a,b,c,d,e]).
% iniTial COndition
onTable(a,s0).
on(b,a,s0).
on(c,b,s0).
clear(c,s0).
onTable(d,s0).
on(e,d,s0).
clear(3,s0).
thank you!!!
Your predicate do/3 cannot succeed because the goal primitive_action/1 will fail with your query.
Currently, while/2 is not described in primitive_action/1 and it seems it is missing also from your program. So you need to extend primitive_action/1 by further facts, or add a new rule to do/3. And in addition to that you need to describe what while/2 means.
This question is actually about Golog. Your mistake is pretty mundane: you didn't copy the Golog interpreter code into your source file/directory.
Golog defines a number of high-level programming constructs, including while-loops and non-deterministic picks (pi), used here. I'm sure you don't want to reinvent Golog, so just go and get it. I'm assuming that your question is part of an assignment of sorts, and your teacher probably pointed you to the Golog interpreter. Otherwise, you can always find it on the pages of the cognitive robotics group at the Univ. of Toronto: http://www.cs.toronto.edu/cogrobo/main/systems/index.html