List 在prolog中定义谓词

我正在努力解决这个问题：

定义谓词

len_NM（L，N，M）

，它检查某个列表

L

是否至少包含长度不小于

M

的

N

元素

OP表示：

定义谓词

len_NM（L，N，M）

，它检查某个列表

L

是否至少包含长度不小于

M

的

N

元素

在此答案中，我们不解决原始问题，而是解决以下变化：

定义谓词

len_NM（L，N，M）

，它检查某个列表

L

是否包含长度不小于

M

的

N

元素

与之类似，我们定义了

seqq1//1

，以建立非空列表列表与其展开的对立面之间的关系：

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

seq1//1

在“两个方向”工作：

在这个答案中，我们使用：

---

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)

好的！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.

---

请展示你迄今为止所做的尝试，你的问题是什么？你想让我们为你写代码吗？这很容易，但它不是这样工作的。确实很容易：len_NM（L，N，M）：-聚合（计数，（成员（E，L），长度（E，T），T>=M），C>=N。 :- use_module(library(clpfd)).

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

?- 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)

?- 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.