我们可以解决您的问题and保存逻辑纯粹性!
下面让Xs
be [1,1,1,2,2,2,3,1,1]
,您在问题中使用的列表。
First,我们映射Xs
到列表列表Yss
这样每个列表Ys
in Yss
只包含取自的相同元素Xs
。
我们通过使用元谓词来做到这一点splitlistIfAdj/3与具体化的不平等谓词相结合dif/3:
?- Xs = [1,1,1,2,2,2,3,1,1], splitlistIfAdj(dif,Xs,Yss).
Xs = [ 1,1,1, 2,2,2, 3, 1,1 ],
Yss = [[1,1,1],[2,2,2],[3],[1,1]].
Second,我们映射列表的列表Yss
to Zss
。中的每一项Zss
有形式[Element,Amount]
。
看看上面查询的答案,我们发现我们需要做的就是映射[1,1,1]
to [1,3]
, [2,2,2]
to [2,3]
, [3]
to [3,1]
, and [1,1]
to [1,2]
. run_pair/2
正是这样做的:
run_pair(Ys,[Element,Amount]) :-
Ys = [Element|_],
length(Ys,Amount).
让我们使用run_pair/2
映射每一项Yss
,借助元谓词maplist/3:
?- Yss = [[1,1,1],[2,2,2],[3],[1,1]], maplist(run_pair,Yss,Zss).
Yss = [[1,1,1],[2,2,2],[3] ,[1,1]],
Zss = [[1,3], [2,3], [3,1],[1,2]].
Done!是时候把它们放在一起了:
count(Xs,Zss) :-
splitlistIfAdj(dif,Xs,Yss),
maplist(run_pair,Yss,Zss).
让我们看看上面的查询是否仍然有效:)
?- count([1,1,1,2,2,2,3,1,1],Zss).
Zss = [[1,3],[2,3],[3,1],[1,2]]. % succeeds deterministically
作为实施count/2
is monotone,即使使用非基础术语,我们也会得到逻辑上合理的答案。让我们看看实际效果!
?- Xs = [A,B,C,D], count(Xs,Zss).
Xs = [D,D,D,D], A=B, B=C , C=D , Zss = [ [D,4]] ;
Xs = [C,C,C,D], A=B, B=C , dif(C,D), Zss = [ [C,3],[D,1]] ;
Xs = [B,B,D,D], A=B, dif(B,C), C=D , Zss = [ [B,2], [D,2]] ;
Xs = [B,B,C,D], A=B, dif(B,C), dif(C,D), Zss = [ [B,2],[C,1],[D,1]] ;
Xs = [A,D,D,D], dif(A,B), B=C , C=D , Zss = [[A,1], [D,3]] ;
Xs = [A,C,C,D], dif(A,B), B=C , dif(C,D), Zss = [[A,1], [C,2],[D,1]] ;
Xs = [A,B,D,D], dif(A,B), dif(B,C), C=D , Zss = [[A,1],[B,1], [D,2]] ;
Xs = [A,B,C,D], dif(A,B), dif(B,C), dif(C,D), Zss = [[A,1],[B,1],[C,1],[D,1]].