Prolog - 创建一个列表

Question

需要帮助创建一个列表。我需要返回课程先决条件的所有可能组合，包括先决条件的先决条件。

这里有一些我需要使用的规则和事实，（一些提供，一些我已经创建）。

prereqFor(engg233, []).
prereqFor(encm339, [engg233]).
prereqFor(cpsc217, []).
prereqFor(cpsc219, [cpsc217]).
prereqFor(cpsc231, []).
prereqFor(cpsc233, [cpsc231]).
prereqFor(math271, [X]) :-
   member(X, [math211, math213]).
prereqFor(math273, []).
prereqFor(cpsc319, [C]) :- 
   member(C, [cpsc219, cpsc233, cpsc235, encm339]).
prereqFor(cpsc331, [M, C]) :- 
   member(M, [math271, math273]), 
   member(C, [cpsc219, cpsc233, cpsc235, encm339]).
prereqFor(cpsc335, [C]) :- 
   member(C, [cpsc319, cpsc331]).

现在我要完成这两个功能，其中一个是帮助器...而且我似乎无法使用 [H|T] 或附加来填充列表...我的当前尝试：

allPrereqFor(Course, Prerequisites) :-
   prereqFor(Course, Prerequisites),
   creatingList(Course, Prerequisites, []).

creatingList(Course, Prerequisites, OnGoingList) :-
   append(Course, Prerequisites, myList).

我也尝试过类似的方法：

allPrereqFor(Course, Prerequisites) :-
   prereqFor(Course, Prerequisites),
   creatingList(Prerequisites, []).

creatingList(Addition, OnGoingList) :-
   [Addition | OnGoingList].

在尝试递归案例之前，我什至无法获得最简单的输出。

我没有列出每个 prereqFor 函数，但示例输出如下：

| ?- allPrereqFor(cpsc331, X).
X = [cpsc217,cpsc219,math211,math271] ? ;
X = [cpsc231,cpsc233,math211,math271] ? ;
X = [consent235,cpsc235,math211,math271] ? ;
X = [encm339,engg233,math211,math271] ? ;

另一个解决方案的尝试...

allPrereqFor(Course,[],Result) :-   append([Course],[],Result).
allPrereqFor(Course, X,Result) :-   prereqFor(Course, Y),
                                    Y=[H|T],
                                    allPrereqFor(H,X,Result).

痕迹：

| ?- allPrereqFor(cpsc331,X).
  1    1  Call: allPrereqFor(cpsc331,_23) ? 
  1    1  Exit: allPrereqFor(cpsc331,[]) ? 

X = [] ? ;
  1    1  Redo: allPrereqFor(cpsc331,[]) ? 
  2    2  Call: prereqFor(cpsc331,_92) ? 
  3    3  Call: member(_78,[math271,math273]) ? 
  3    3  Exit: member(math271,[math271,math273]) ? 
  4    3  Call: member(_80,[cpsc219,cpsc233,cpsc235,encm339]) ? 
  4    3  Exit: member(cpsc219,[cpsc219,cpsc233,cpsc235,encm339]) ? 
  2    2  Exit: prereqFor(cpsc331,[math271,cpsc219]) ? 
  5    2  Call: allPrereqFor(math271,_23) ? 
  5    2  Exit: allPrereqFor(math271,[]) ? 
  1    1  Exit: allPrereqFor(cpsc331,[]) ? 

X = [] ? ;
  1    1  Redo: allPrereqFor(cpsc331,[]) ? 
  5    2  Redo: allPrereqFor(math271,[]) ?

当我跟踪时，我可以看到它递归地访问了所有正确的课程，但它只是不断输出：

X = [] ?;

Answer 1

如果我正确理解您的问题，那么：

您正在寻找基于 and 和 or 谓词的组合列表。

例如

要参加cpsc219，您需要参加
cpsc217 和cpsc219

要参加math271，您需要参加
math211 和math271 或
math213 和math271

在link 中，该示例使用build of materials 表示自行车，实际问题的材料构建类似于您的先决条件。

此答案将使用DCG，因为您正在构建列表，并且在主要使用列表 DCG 时使用 Prolog 是首选。

对于自行车示例，DCG 规则仅显示 and DCG 规则，在看到您在评论中对我的问题的回答后，您还需要 or DCG 规则，但自行车示例中没有给出这些规则。

所以只是回顾一下： 在 Prolog 中，and 使用 , (,/2) 和 or 完成，这可以使用 ; (;/2) 完成，但更常见的是使用多个 DCG 规则完成。请注意，这里提到的, 和; 不是列表运算符，而是针对目标的运算符。

自行车and DCG 传动链规则示例。

drivechain --> crank, pedal, pedal, chain.

现在缺少使用多个 DCG 规则实现的 or DCG 规则示例

tire --> [dunlop].
tire --> [goodyear].
tire --> [yomoto].

或使用;实现

tire -->
    [dunlop]; [goodyear]; [yomoto].

and DCG 规则的工作示例：

?- drivechain(X,[]).
X = [crank, pedal, pedal, chain] ;
false.

or DCG 规则的工作示例：

?- tire(X,[]).
X = [dunlop] ;
X = [goodyear] ;
X = [yomoto] ;
false.

因此，对于 math271 需要 math211 或 math213 的问题，DCG 规则将是：

math211 --> [math211].
math213 --> [math213].

math271 --> math211.
math271 --> math213.

?- math271(X,[]).
X = [math211] ;
X = [math213].

由于您指出这是原始帖子中的一项作业，因此我将其视为家庭作业，因此我不会对您的问题给出具体答案，而是使用自行车示例向您展示类似的内容。

bike --> frame, drivechain, wheel, wheel.    
wheel --> spokes, rim, hub, tire.

tire --> [dunlop].
tire --> [goodyear].
tire --> [yomoto].

drivechain --> crank, pedal, pedal, chain.

spokes --> [spokes].
crank --> [crank].
pedal --> [pedal].
chain --> [chain].
rim --> [rim].
hub --> [hub].
frame --> [frame].

?- bike(X,[]).
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, dunlop, spokes, rim, hub, dunlop] ;
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, dunlop, spokes, rim, hub, goodyear] ;
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, dunlop, spokes, rim, hub, yomoto] ;
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, goodyear, spokes, rim, hub, dunlop] ;
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, goodyear, spokes, rim, hub, goodyear] ;
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, goodyear, spokes, rim, hub, yomoto] ;
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, yomoto, spokes, rim, hub, dunlop] ;
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, yomoto, spokes, rim, hub, goodyear] ;
X = [frame, crank, pedal, pedal, chain, spokes, rim, hub, yomoto, spokes, rim, hub, yomoto].

请注意，即使轮胎由三个品牌组成，也有 9 个答案，因为有两个轮子，每个轮子都有一个轮胎，每个轮胎可以是三个品牌之一，因此 3 * 3 是 9。

SWI-Prolog 特定：
由于其中一些答案会很长并且会被... 截断，请参阅此answer 以获取帮助以查看没有... 的完整答案。换句话说，将;false 附加到查询中，并在第一个答案后按w，然后按空格键以获得更多答案。

由于DCG 与syntactic sugar 类似，因此要将脱糖DCG 规则（-->）视为谓词（:-），请使用listing/1，例如

?- listing(wheel).
wheel(A, E) :-
        spokes(A, B),
        rim(B, C),
        hub(C, D),
        tire(D, E).
true.

正如我在 cmets 中所指出的，我看到 @false 可能会回答这个问题。我向他学习，所以我完全期望他的答案比我的更好，但我发布我的答案是因为如果有人发现我的答案有问题并指出，那么我也会学习。

====================================

更精细的自行车示例，可以制作自行车或三轮车。

bike --> type.
type --> bicycle.
type --> tricycle.
bicycle --> bicycle_frame, bicycle_drivechain, wheel, wheel.
tricycle --> tricycle_frame, tricycle_drive, wheel, wheel, wheel.
wheel --> spokes, rim, hub, tire.
bicycle_drivechain --> crank, pedal, pedal, chain.
tricycle_drive --> crank, pedal, pedal.
bicycle_frame --> [bicycle_frame].
tricycle_frame --> [tricycle_frame].
tire --> [dunlop].
tire --> [goodyear].
tire --> [yomoto].
spokes --> [spokes].
crank --> [crank].
pedal --> [pedal].
chain --> [chain].
rim --> [rim].
hub --> [hub].

?- bike(X,[]);false.
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, dunlop, spokes, rim, hub, dunlop] ;
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, dunlop, spokes, rim, hub, goodyear] ;
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, dunlop, spokes, rim, hub, yomoto] ;
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, goodyear, spokes, rim, hub, dunlop] ;
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, goodyear, spokes, rim, hub, goodyear] ;
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, goodyear, spokes, rim, hub, yomoto] ;
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, yomoto, spokes, rim, hub, dunlop] ;
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, yomoto, spokes, rim, hub, goodyear] ;
X = [bicycle_frame, crank, pedal, pedal, chain, spokes, rim, hub, yomoto, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, dunlop, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, dunlop, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, dunlop, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, goodyear, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, goodyear, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, goodyear, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, yomoto, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, yomoto, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, dunlop, spokes, rim, hub, yomoto, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, dunlop, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, dunlop, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, dunlop, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, goodyear, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, goodyear, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, goodyear, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, yomoto, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, yomoto, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, goodyear, spokes, rim, hub, yomoto, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, dunlop, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, dunlop, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, dunlop, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, goodyear, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, goodyear, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, goodyear, spokes, rim, hub, yomoto] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, yomoto, spokes, rim, hub, dunlop] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, yomoto, spokes, rim, hub, goodyear] ;
X = [tricycle_frame, crank, pedal, pedal, spokes, rim, hub, yomoto, spokes, rim, hub, yomoto, spokes, rim, hub, yomoto] ;
false.

Answer 2

可以递归地找到前置需求的前置需求。好吧，你应该使用findall/3 和flatten/2，它们是在 prolog 库中定义的。

pre(X,Y):-
    preRequirement(X,Y).


preRequirement(X,[(X-List)|Y]):-
    findall(Z,prereqFor(X,Z),Res),
    flatten(Res,List),
    findPreOfPre(List,Y).


findPreOfPre([],[]).
findPreOfPre([H|T],[(H-L)|Result]):-
    findall(P,prereqFor(H,P),N),
    flatten(N,L),
    findPreOfPre(T,Result).