Is there an equivalent of Haskell's enumFromTo in Prolog?

Question

I've just been starting out in Prolog and I was hoping to perform the following task:

Make a predicate a such that for all A(P,N,L) which is nth element of C, L.

Basically I want to perform a map on the range P(N,C). In Haskell, the language I am most familiar with, this would look like

[0..N]

(Haskell doesn't quite have predicates so I'm taking some liberties here)

or in pointfree

f p n = map(p)[0..n]

And it seems like I should be able to do it in Prolog easily enough. Prolog's f = (.enumFromTo 0).map is basically already that so it should be a trivial modification. My definition should look something like:

maplist/3

However I can't really figure out what to put in the blank. In Haskell I would use a function like A(P,N,L) :- maplist(P, ??? , L). , but it seems that such a thing doesn't exist in Prolog. The closes equivalent would be enumFromTo, but that isn't a list so I can't use for between/3.

Alternatively I could make my own range predicate.

The first thing I tried was:

maplist

But I can't get that to resolve at all. I also tried

range(0,[0]).
range(N,[N|T]) :- range(N-1,T).
A(P,N,L) :- range(N,rangeN), maplist(P, rangeN, L).

But that just seems really clunky for such a small problem.

How might I fill in the gap in my range(N,L):-findall(X,between(0,N,X),L),sort(L,L). A(P,N,L) :- range(N,rangeN), maplist(P, rangeN, L). ? Am I approaching the problem in the wrong way?

Answer 1

Firebase push Notifications

被翻译成Prolog为

-- % f p n = map (p) [0..n] = [p 0, p 1, p 2, ..., p n]

这假定f(P,N,L):- f(P,0,N,L). f(P,I,N,[]):- I > N. f(P,I,N,L):- call(P,I,X), ( N =:= I -> L = [X] ; L = [X|T], J is I+1, f(P,J,N,T) ). 对于某些p :: Int -> a，如Haskell代码所暗示的那样。

这也假设给你一个具体的（＆＃34; ground＆＃34;）可调用的双参数谓词a和一个整数P。

另一种可能性是

N

这会找到所有g(P,N,L):- findall(X, (between(0, N, I), call(P,I,X)), L). ，以便（X 和 0 <= I <= N）成立。

在SWI-Prolog中测试：

P(I,X)

Answer 2

当我试图在Prolog中编写一个解决N皇后问题的解决方案时，如Picat家所述（参见示例5，接近页尾），我遇到了类似的问题。这是我的最终结果，并评论了一些替代方案：

:- use_module(library(clpfd)).

queens(N, Q) :-
    length(Q, N),
    Q ins 1..N,
    all_different(Q),
    maplist_index([I,V,P]>>(P#=V+I),1,Q,Ps),all_different(Ps),
    maplist_index([I,V,M]>>(M#=V-I),1,Q,Ms),all_different(Ms),
    /* no
    bagof(P, (nth1(I,Q,V), P #= V + I), Ps), all_different(Ps),
    bagof(M, (nth1(I,Q,V), M #= V - I), Ms), all_different(Ms),
    */
    /* ok
    all_different_p(Q, 1, P), all_different(P),
    all_different_m(Q, 1, M), all_different(M),
    */
    label(Q).

all_different_p([Q|Qs], I, [P|Ps]) :-
    P #= Q + I,
    succ(I, J),
    all_different_p(Qs, J, Ps).
all_different_p([], _I, []).

all_different_m([Q|Qs], I, [P|Ps]) :-
    P #= Q - I,
    succ(I, J),
    all_different_m(Qs, J, Ps).
all_different_m([], _I, []).

maplist_index(P, I, [X|Xs], [Y|Ys]) :-
    call(P, I, X, Y),
    succ(I, J),
    maplist_index(P, J, Xs, Ys).
maplist_index(_, _, [], []).

maplist_index / 4是您需要的一个示例。值得注意的是，bagof / 3在存在属性变量的情况下效果不佳。