Question

我正在努力解决这个问题：

定义谓词len_NM(L,N,M)，用于检查某个列表L是否至少包含长度不小于N的{{1}}个元素。

Answer 1

OP说：

定义谓词len_NM(L,N,M)，用于检查某个列表L是否包含至少 N个元素，且长度不小于M。
在这个答案中，我们没有解决原始问题，但有以下变化：

定义谓词len_NM(L,N,M)，用于检查某个列表L是否包含长度不小于N的完全 M个元素。< / p>

与this answer类似，我们定义dcg seqq1//1来建立非空列表列表与其相反的平面列表之间的关系：

seq([])     --> [].
seq([E|Es]) --> [E], seq(Es).

seqq1([])       --> [].
seqq1([Es|Ess]) --> {Es=[_|_]}, seq(Es), seqq1(Ess).

样品使用：

?- phrase(seqq1([[1,2],[3],[4]]),Xs).
Xs = [1,2,3,4].

请注意，seqq1//1适用于“双向”：

?- phrase(seqq1(Xss),[1,2,3,4]).
  Xss = [[1],[2],[3],[4]] 
; Xss = [[1],[2],[3,4]]
; Xss = [[1],[2,3],[4]]
; Xss = [[1],[2,3,4]]
; Xss = [[1,2],[3],[4]]
; Xss = [[1,2],[3,4]]
; Xss = [[1,2,3],[4]]
; Xss = [[1,2,3,4]]
; false.

在这个答案中，我们使用clpfd：

:- use_module(library(clpfd)).

然后，我们定义len_NM/4 - 使用maplist/3， length/2，tcount/3和(#=<)/3：

len_NM(Xss,Ys,N,M) :-
   M #>= 1,
   N #>= 0,
   phrase(seqq1(Xss),Ys),
   maplist(length,Xss,Ls),
   tcount(#=<(M),Ls,N).

让我们运行一些示例查询！

?- len_NM([[1,2,3],[4],[5,6],[7,8,9,10],[11,12]],_,N,L).
  N = 5, L = 1             % five lists have length of at least one 
; N = 4, L = 2             % four lists have length of at least two
; N = 2, L = 3             % two of at least three (e.g., [1,2,3] and [7,8,9,10])
; N = 1, L = 4             % one list has length of four (or more)
; N = 0, L in 5..sup.      % no list has length of five (or more)

OK！这个怎么样？

?- append(Xs,_,[x,x,x,x,x,x]),   % With `Xs` having at most 6 elements ...
   N #>= 1,                      % ... `Xss` shall contain at least 1 list ...
   len_NM(Xss,Xs,N,4).           % ... having a length of 4 (or more).
  Xs = [x,x,x,x],     N = 1, Xss = [[x,x,x,x]]
; Xs = [x,x,x,x,x],   N = 1, Xss = [[x],[x,x,x,x]]
; Xs = [x,x,x,x,x],   N = 1, Xss = [[x,x,x,x],[x]]
; Xs = [x,x,x,x,x],   N = 1, Xss = [[x,x,x,x,x]]
; Xs = [x,x,x,x,x,x], N = 1, Xss = [[x],[x],[x,x,x,x]]
; Xs = [x,x,x,x,x,x], N = 1, Xss = [[x],[x,x,x,x],[x]]
; Xs = [x,x,x,x,x,x], N = 1, Xss = [[x],[x,x,x,x,x]]
; Xs = [x,x,x,x,x,x], N = 1, Xss = [[x,x],[x,x,x,x]]
; Xs = [x,x,x,x,x,x], N = 1, Xss = [[x,x,x,x],[x],[x]]
; Xs = [x,x,x,x,x,x], N = 1, Xss = [[x,x,x,x],[x,x]]
; Xs = [x,x,x,x,x,x], N = 1, Xss = [[x,x,x,x,x],[x]]
; Xs = [x,x,x,x,x,x], N = 1, Xss = [[x,x,x,x,x,x]]
; false.

在prolog中定义谓词

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