我有以下问题:
生成长度为2 * n + 1的所有子字符串,由0,1或-1的值组成,因此a1 = ...,a2n + 1 = 0和| a(i + 1) - ai | = 1或2,每1 <= i <= 2n。
这就是我所做的:
change(0).
change(1).
change(-1).
%gets the last element from a list
lastE([X],X).
lastE([_|L],X) :-
last(L,X).
%checks if |N-M|=1 or 2
calc(N,M) :-
( 1 is abs(N-M)
; 2 is abs(N-M),
!
).
%generates all lists of length N with values -1,0,1
generate([],0) :-
!.
generate([H|T],N):-
change(H),
N1 is N-1,
generate(T,N1).
%validates a list to be correct (last element=0, |a(i+1) - ai| = 1 or 2)
valid(L) :-
valid(L,_,3).
valid([E],_,_) :-
lastE([E],0).
valid([H|[H1|T]],_,3) :-
valid([H1|T],H,H1).
valid([_|[H1|T]],N,M) :-
calc(N,M),
valid([H1|T],M,H1).
以上所有代码都正常工作(我测试过)。但是使用以下genAll
代码,我收到Arguments are not sufficiently instantiated
错误。
genAll(N) :-
N1 = 2*N,
N2 = N1+1,
generate(L1,N2),
valid(L1),
write('['),
printl(L1).
printl([]) :-
write(']').
printl([0|T]) :-
write('0 '),
printl(T).
printl([1|T]) :-
write('1 '),
printl(T).
printl([-1|T]) :-
write('-1 '),
printl(T).
我不确定是什么问题。
答案 0 :(得分:3)
在这个答案中(主要是猜测),我们使用clpfd。
:- use_module(library(clpfd)). n_qfd33721532(N, Zs) :- Zs = [E|Es], N*2 + 1 #= L, length(Zs, L), Zs ins -1..1, last(Zs, 0), chain_neq(Es, E). chain_neq([], _). chain_neq([E1|Es], E0) :- E0 #\= E1, chain_neq(Es, E1).
示例查询:
?- n_qfd33721532(N, Zs), labeling([], Zs). N = 0, Zs = [0] ; N = 1, Zs = [-1, 1, 0] ; N = 1, Zs = [ 0,-1, 0] ; N = 1, Zs = [ 0, 1, 0] ; N = 1, Zs = [ 1,-1, 0] ; N = 2, Zs = [-1, 0,-1, 1, 0] ; N = 2, Zs = [-1, 0, 1,-1, 0] ; N = 2, Zs = [-1, 1,-1, 1, 0] ; N = 2, Zs = [-1, 1, 0,-1, 0] ; N = 2, Zs = [-1, 1, 0, 1, 0] ; N = 2, Zs = [ 0,-1, 0,-1, 0] ; N = 2, Zs = [ 0,-1, 0, 1, 0] ; N = 2, Zs = [ 0,-1, 1,-1, 0] ; N = 2, Zs = [ 0, 1,-1, 1, 0] ; N = 2, Zs = [ 0, 1, 0,-1, 0] ; N = 2, Zs = [ 0, 1, 0, 1, 0] ; N = 2, Zs = [ 1,-1, 0,-1, 0] ; N = 2, Zs = [ 1,-1, 0, 1, 0] ; N = 2, Zs = [ 1,-1, 1,-1, 0] ; N = 2, Zs = [ 1, 0,-1, 1, 0] ; N = 2, Zs = [ 1, 0, 1,-1, 0] ; N = 3, Zs = [-1, 0,-1, 0,-1, 1, 0] ...