List 第一次出现后插入Prolog元素
如何在列表List 第一次出现后插入Prolog元素,list,prolog,meta-predicate,prolog-dif,List,Prolog,Meta Predicate,Prolog Dif,如何在列表Xs中第一次出现X之后添加元素E 例如: ?- insert_right_behind(5,10,[2,4,10,12],Xs). Xs = [2,4,10,5,12]. % expected answer 此刻,我在理解上有困难 由于我对该语言不熟悉,因此需要进行递归 提前谢谢 使用三个谓语从句: % Inserting after in an empty list is an empty list: insert_after( _X, _Y
Xs
中第一次出现X
之后添加元素E
例如:
?- insert_right_behind(5,10,[2,4,10,12],Xs).
Xs = [2,4,10,5,12]. % expected answer
此刻,我在理解上有困难
由于我对该语言不熟悉,因此需要进行递归
提前谢谢 使用三个谓语从句:
% Inserting after in an empty list is an empty list:
insert_after( _X, _Y, [], [] ).
% If the "after" item is at the head of the list, then the "insert" item can go after it:
insert_after( X, Y, [Y|T], [Y,X|T] ).
% If the head of the list isn't the "after" item, then the result will be
% this with a new tail list that has the "insert" item inserted:
insert_after( X, Y, [H|T], [H|L] ) :-
Y \= H,
insert_after( X, Y, T, L ).
如果给定列表中不存在“after”项,则
insert\u after/4
将生成原始列表。通过删除上述子句之后的第一个insert\u,它将在这种情况下失败。使用三个谓词子句:
% Inserting after in an empty list is an empty list:
insert_after( _X, _Y, [], [] ).
% If the "after" item is at the head of the list, then the "insert" item can go after it:
insert_after( X, Y, [Y|T], [Y,X|T] ).
% If the head of the list isn't the "after" item, then the result will be
% this with a new tail list that has the "insert" item inserted:
insert_after( X, Y, [H|T], [H|L] ) :-
Y \= H,
insert_after( X, Y, T, L ).
如果给定列表中不存在“after”项,则
insert\u after/4
将生成原始列表。通过删除上面的第一个insert\u after
子句,在这种情况下它将失败。让我们保持它的简单和易用性,如下所示:
让我们保持它的简单和使用,就像这样: 在最成功的查询中,会留下无用的选择点 我们可以通过这样使用来避免这些选择点: 让我们运行一些查询
?- item_following_in_inserted(5,10,[2,4,12],Xs).
false. % OK, unchanged
?- item_following_in_inserted(5,10,[2,4,10,12],Xs).
Xs = [2,4,10,5,12]. % succeeds deterministically
?- item_following_in_inserted(5,10,[2,4,10,12,10],Xs).
Xs = [2,4,10,5,12,10]. % succeeds deterministically
?- item_following_in_inserted(I,E,Xs,Ys).
Xs = [ E|_Z], Ys = [ E,I|_Z] % OK, unchanged
; Xs = [ _A,E|_Z], Ys = [ _A,E,I|_Z], dif(E,_A)
; Xs = [_A,_B,E|_Z], Ys = [_A,_B,E,I|_Z], dif(E,_A), dif(E,_B)
...
在最成功的查询中,会留下无用的选择点
我们可以通过这样使用来避免这些选择点:
让我们运行一些查询
?- item_following_in_inserted(5,10,[2,4,12],Xs).
false. % OK, unchanged
?- item_following_in_inserted(5,10,[2,4,10,12],Xs).
Xs = [2,4,10,5,12]. % succeeds deterministically
?- item_following_in_inserted(5,10,[2,4,10,12,10],Xs).
Xs = [2,4,10,5,12,10]. % succeeds deterministically
?- item_following_in_inserted(I,E,Xs,Ys).
Xs = [ E|_Z], Ys = [ E,I|_Z] % OK, unchanged
; Xs = [ _A,E|_Z], Ys = [ _A,E,I|_Z], dif(E,_A)
; Xs = [_A,_B,E|_Z], Ys = [_A,_B,E,I|_Z], dif(E,_A), dif(E,_B)
...
如果
10
不存在,您希望它做什么?mbratch,false
很可能,false是正确的结果,那么我猜。mbrathc,“first”出现。插入只在遇到第一个元素后发生一次。如果10
不存在,您希望它做什么?mbratch,false
yes,false可能是正确的结果,然后我猜。mbrathc,“first”出现。插入只在遇到第一个元素后发生一次。谢谢,此解决方案按预期工作,并且很高兴您对您的答案提供了一些注释。谢谢,这个解决方案按照预期的方式工作,而且您对您的答案提供了一些评论也很好。
?- item_following_in_inserted(5,10,[2,4,10,12,10],Xs).
Xs = [2,4,10,5,12,10] % single solution
; false. % terminates universally
?- item_following_in_inserted(I,E,Xs,Ys).
Xs = [ E|_Z], Ys = [ E,I|_Z]
; Xs = [ _A,E|_Z], Ys = [ _A,E,I|_Z], dif(E,_A)
; Xs = [ _A,_B,E|_Z], Ys = [ _A,_B,E,I|_Z], dif(E,_A), dif(E,_B)
; Xs = [_A,_B,_C,E|_Z], Ys = [_A,_B,_C,E,I|_Z], dif(E,_A), dif(E,_B), dif(E,_C)
...
item_following_in_inserted(I,J,[X|Xs],Ys0) :-
if_(J = X,
Ys0 = [J,I|Xs],
(Ys0 = [X|Ys], item_following_in_inserted(I,J,Xs,Ys))).
?- item_following_in_inserted(5,10,[2,4,12],Xs).
false. % OK, unchanged
?- item_following_in_inserted(5,10,[2,4,10,12],Xs).
Xs = [2,4,10,5,12]. % succeeds deterministically
?- item_following_in_inserted(5,10,[2,4,10,12,10],Xs).
Xs = [2,4,10,5,12,10]. % succeeds deterministically
?- item_following_in_inserted(I,E,Xs,Ys).
Xs = [ E|_Z], Ys = [ E,I|_Z] % OK, unchanged
; Xs = [ _A,E|_Z], Ys = [ _A,E,I|_Z], dif(E,_A)
; Xs = [_A,_B,E|_Z], Ys = [_A,_B,E,I|_Z], dif(E,_A), dif(E,_B)
...