示例中的序言规则 [英] Prolog Rules in example

问题描述

我在 Prolog 中有一个这样的数据库:

I have a database in Prolog like this:

connection(a,b,bus)
connection(b,c,metro)
connection(b,d,taxi)
connection(d,e,bus)

我想要的是我需要应用的规则，以便我可以提出问题:transport(a,c)"，它回答:bus"和metro"

What I want is the rules I need to apply so I can ask the question: "transport(a,c)" and it answers: "bus" and "metro"

是否可以定义 1 条或 2 条规则以便查询transport(a,c)"有效?

Is that possible to define 1 or 2 rules so that the query "transport(a,c)" works ?

您应该看到如下所示的数据库:

you should see the database like:

连接(出发，到达，运输).所以... connection(D,A,T).

那么规则是:

connection(D,A,T):- traject(D,A,T). 
connection(D,A,T):- traject(D,X,T1), traject(X,A,T2).

where...traject(Departure, X, Transport1) 和 traject(X, Arrival, Transport2)

where...traject(Departure, X, Transport1) and traject(X, Arrival, Transport2)

并且查询应该是这样的:

and the query should be something like:

transport(a,c,T1). 和transport(a,c,T2).

然后答案应该是:

T1 = bus 
T2 = metro 

推荐答案

我会试试这个:

transport(A, B, [Method]) :- connection(A, B, Method).
transport(A, C, [Method|Others]) :-
    connection(A, B, Method),
    transport(B, C, Others).

这里的基本情况是您有直接连接.归纳案例是找到一个连接，然后从中间递归.请注意，如果您尝试在正文中使用 transport/3 两次而不是 connection/3 然后是 transport/3，您将获得无限回归！试试看！

The base case here is that you have a direct connection. The inductive case is to find a connection and then recur from the intermediate. Note that you will get an infinite regress if you try using transport/3 twice in the body instead of connection/3 and then transport/3! Try it and see!

这似乎适用于我期望的输入:

This seems to work for the inputs I expect:

?- transport(a, c, M).
M = [bus, metro] ;
false.

?- transport(a, d, M).
M = [bus, taxi] ;
false.

?- transport(a, e, M).
M = [bus, taxi, bus] ;
false.

希望这会有所帮助！

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