Can someone explain clearly why this implementation (from SO 3965054) of min_of_list works in prolog:
% via: http://stackoverflow.com/questions/3965054/prolog-find-minimum-in-a-list
min_of_list_1( [H], H).
min_of_list_1([H,K|T],M) :- H =< K, min_of_list_1([H|T],M).
min_of_list_1([H,K|T],M) :- H > K, min_of_list_1([K|T],M).
while this implementation generates an incorrect output:
min_of_list_2( [H], H).
min_of_list_2( [H| T], X) :- compare(<, X, H), min_of_list_2(T, X).
min_of_list_2( [H| T], H) :- compare(>, X, H), min_of_list_2(T, X).
min_of_list_2( [H| T], H) :- compare(=, X, H), min_of_list_2(T, H).
Epilogue. This works.
min_of_list_3( [H], H).
min_of_list_3( [H| T], X) :- min_of_list_3(T, X), compare(<, X, H).
min_of_list_3( [H| T], H) :- min_of_list_3(T, X), compare(>, X, H).
min_of_list_3( [H| T], H) :- min_of_list_3(T, X), compare(=, X, H).
?
The behavior I get is that min_of_list_2 returns the last element in the list.
Thanks.
The first clause of min_of_list_2/2 is OK, it says the minimum of a list with a single element is that element. The second clause is not quite so OK: The intention seems to state that if X is the minimum of the list T, and X is smaller than H, then X is also the minimum of the list [H|T], and this would work as intended if compare/3 behaved like a true relation, but unfortunately it doesn't:
?- compare(<, a, b).
true.
Yet the more general query fails as if there were no solution (although we know there is at least one!):
?- compare(<, a, X).
false.
Since one typical usage of min_of_list_2/2 (including for example its use in the third clause) leaves the second argument uninstantiated, you will run into this problem. Your code will work as expected if you place all calls of compare/3 after the respective recursive calls of min_of_list_2/2. As a consequence, your predicate is then no longer tail recursive, in contrast to the other program you posted. The compare/3 call in the last clause should be removed (what is the X in that case?), as it will always fail!
the first one compares the first two elements of the list and then puts the min again in the list till there is only one element.
the second one... takes the head of the list and compares with X. X is non-instantiated in the first call so compare(<,X,_any_number) will be true. X wont be instantiated so the same will repeat till there is only one element in the list which will be returned* (the last one).
'* where returned = unified with the second argument.
Related
Is there any way to prevent cycle in this code.
move(a,b).
move(b,a).
move(a,c).
move(b,d).
move(b,e).
move(d,h).
move(d,i).
move(e,j).
move(e,k).
move(c,f).
move(c,g).
move(f,l).
move(f,m).
move(g,n).
move(g,o).
goal(n).
goSolveTheMaze(Start,Way) :-
dfs(Start, Way),!.
dfs(Goal, [Goal]) :-
goal(Goal),!.
dfs(Start, [Start|Way]) :-
move(Start, N),
dfs(N, Way).
so when move(a,b) move to (b,c) dont go back to (b,a),
when run goSolveTheMaze(a,path).
The output should be path=[a,c,g,n].
What if you add a third argument to dfs which is a list of where you've already visited? You could then use \+/1 and member/2 to avoid returning to places you've already been.
For example, if you use the following:
move(a,b).
move(b,a).
move(a,c).
move(b,d).
move(b,e).
move(d,h).
move(d,i).
move(e,j).
move(e,k).
move(c,f).
move(c,g).
move(f,l).
move(f,m).
move(g,n).
move(g,o).
goal(n).
goSolveTheMaze(Start,Way) :-
dfs(Start, Way, [Start]),!.
dfs(Goal, [Goal], _) :-
goal(Goal),!.
dfs(Start, [Start|Way], Visited) :-
move(Start, N),
\+ member(N, Visited),
dfs(N, Way, [N|Visited]).
Then the query:
?- goSolveTheMaze(a, X).
Will produce the result:
X = [a, c, g, n]
Update in response to comment "can you tell me what \+ do?":
The \+ predicate is true when its argument cannot be proven. Therefore in the above example the line \+ member(N, Visited) means "when N is not a member of the list Visited".
See: https://www.swi-prolog.org/pldoc/man?predicate=%5C%2B/1
I'm having trouble with the following predicate:
treeToList(void, []).
treeToList(arbol(X, HI1, HD1), L) :-
treeToList(HI1, L1),
treeToList(HD1, L2),
append(L1, [X|L2], L).
maximumInList([X], X).
maximumInList([A|L], X) :-
maximumInList(L,X1),
(A > X1 -> X = A; X = X1).
maxNodeInTree(arbol, N) :-
treeToList(arbol, L),
maximumInList(L, N).
TreeToList gets a tree and returns a list with all of its nodes. Meanwhile maximumInList gets a list and returns the maximum element in the list.
Both of these predicates work fine individually, however the last one, maxNodeInTree, is supposed to first get the list L using treeToList which then will be passed to maximumInList and it'll return the maximum element in the whole tree. And yet Prolog returns false.
Any tips are appreciated!
You have a typo in the last predicate (arbol instead of Arbol). Try:
maxNodeInTree(Arbol, N) :-
treeToList(Arbol, L),
maximumInList(L, N).
HI I would like to know how a method that finds out if two members of a list in Prolog are adjacent as the catch is that the first and the last elements are checked if they are adjacent something like
(b,c,[b,a,d,c])
would give yes they are adjacent. I already have this code
adjacent(X, Y, [X,Y|_]).
adjacent(X, Y, [_|Tail]) :-
adjacent(X, Y, Tail).
but I do not know how to include the head of the list and the last elments as well being compared for being adjacent. If you are really good maybe you can tell me also how it is possible to make something like this
(c,b,[a,b,c,d])
to be true I mean the elements are adjacent no matter which exactly is first.
You can make use of last/2 predicate [swi-doc] to obtain the last element of the list. But you can not use this in the recursive call, since otherwise it will each element in the list pair with the last element as well.
The trick is to make a helper predicate for the recursive part, and then make the adjacent/3 predicate to call the recursive one you wrote yourself, or one where we match with the last element:
adjacent(X, Y, L) :-
adj(X, Y, L).
adjacent(X, Y, [Y|T]) :-
last(T, X).
adj(X, Y, [X,Y|_]).
adj(X, Y, [_|T]) :-
adj(X, Y, T).
Relations about lists can often be described with a Definite Clause Grammar dcg.
A first attempt might be:
adjacent(A, B, L) :-
phrase(adjacent(A, B), L). % interface to DCG
adjacent(A,B) -->
..., ( [A,B] | [B,A] ), ... .
... --> [] | [_], ... .
Yet, this leaves out cases like adjacent(a,d,[a,b,c,d]). One possibility would be to add another rule, or maybe simply extend the list to be considered.
adjacent(A, B, L) :-
L = [E,_|_],
append(L, [E], M),
phrase(adjacent(A, B), L).
My code does perfect with numbers, but error with single quotation. I'm trying to write a foldl function. When i do foldl1(concat, ['a','b'], X), it reports like "ERROR: Arithmetic: 'ab/0' is not a function". what is the problem? prolog does not allow using is with string?
foldl1(P, [H], X) :-
X is H.
foldl1(P, [H|T], X) :-
foldl1(P, T, Y),
call(P, H, Y, Z),
X is Z.
is/2 evaluates the arithmetic expression to the right, and unifies the result with the term to the left. Unification is also performed against the head' arguments, so you can write a simplified foldl1/3 like
foldl1(_, [H], H).
foldl1(P, [H|T], Z) :-
foldl1(P, T, Y),
call(P, H, Y, Z).
test:
?- foldl1(plus,[1,2,3],R).
R = 6 ;
false.
?- foldl1(concat,[1,2,3],R).
R = '123' ;
false.
I would place a cut after the recursion base, since [H] and [H|T] where T=[] overlap, to avoid any last call - that would anyway fail - on eventual backtracking, like the redo induced by me, inputting ; after the expected first answer while the interpreter waits for my choices.
After the cut (hope you can easily spot where to place it) we get:
?- foldl1(plus,[1,2,3],R).
R = 6.
?- foldl1(concat,[1,2,3],R).
R = '123'.
Now the interpreter 'knows' there are no more answers after the first...
It's also possible to implement a foldl1/3 predicate using first-argument indexing to avoid spurious choice-points without cuts and that is also tail-recursive. From the Logtalk library meta object:
:- meta_predicate(foldl1(3, *, *)).
foldl1(Closure, [Head| Tail], Result) :-
fold_left_(Tail, Closure, Head, Result).
fold_left_([], _, Result, Result).
fold_left_([Arg| Args], Closure, Acc, Result) :-
call(Closure, Acc, Arg, Acc2),
fold_left_(Args, Closure, Acc2, Result).
Sample calls:
?- meta::foldl1(plus,[1,2,3],R).
R = 6.
?- meta::foldl1(concat,[1,2,3],R).
R = '123'.
I'm trying to write a simple maze search program in prolog, before I add a room to visited list I'm checking whether it is already a member of the visited list. However, I can't get this to work, even if I use the code from the book:
d(a,b).
d(b,e).
d(b,c).
d(d,e).
d(c,d).
d(e,f).
d(g,e).
go(X, X, T).
go(X, Y, T) :-
(d(X,Z) ; d(Z, X)),
\+ member(Z,T),
go(Z, Y, [Z|T]).
What do I do wrong?
Your program seems to be ok.
I guess the problem is that you are calling go/3 with the third argument uninstantiated.
In that case it will member(X, T) will always succeed, thus failing the clause.
You might call your predicate with the empty list as the third parameter:
e.g.
?- go(a, g, []).
true
If you want to return the path consider adding another parameter to go, like this:
go(From, To, Path):-
go(From, To, [], Path).
go(X, X, T, T).
go(X, Y, T, NT) :-
(d(X,Z) ; d(Z, X)),
\+ member(Z,T),
go(Z, Y, [Z|T], NT).