Question

domains 
    el=integer
    list = el*
    lista = list*

predicates
     aux(list,integer,list)
     arrangements(list,integer,lista)
clauses
     aux([H|_],1,[H]).
     aux([_|L],N,L1):-
        aux(L,N,L1).
    aux([H|L],N,[H|L1]):-
        N<>1,
        N1=N-1,
        aux(L,N1,L1).
arrangements(L,N,R):-
    findall(X,aux(L,N,X),R).

此代码显示列表元素的所有组合。我应该如何修改它以显示安排。我没有想法

安排

[2,3,4] K=2 => [[2,3], [3,2], [2,4], [4,2], [3,4], [4,3]]

组合

[2,3,4] K=2 => [[3,4], [2,3], [2,4]]

Answer 1

使用select中的aux/3获取列表中的任何排列：

aux(L, N, [H|T]) :-
    N > 1,
    select(H, L, M),   % Select an element "H" from "L", leaving "M"
    N1 is N-1,         % NOTE the "is" here, not "=" !!
    aux(M, N1, T).     % Recurse with remaining elements "M" and count-1
aux(L, 1, [X]) :- member(X, L).

arrangements(L, N, R):-
    findall(X, aux(L, N, X), R).

导致：

| ?- arrangements([2,3,4], 2, R).

R = [[2,3],[2,4],[3,2],[3,4],[4,2],[4,3]]

yes

prolog中列表元素的排列

1 个答案: