我想找到Prolog中两个节点之间的最短路径。 我想如何找到两个节点之间的所有路径,但不幸的是,下面的代码会进入循环:
arc (A,b)。 arc(b,a)。 arc(b,c)。 arc(c,b)。 arc(c,d)。 arc(d,c)。 path(X,Y,[arc(X,Y)]): - arc(X,Y)。 path(X,Y,[arc(X,Z)| P]): - arc(X,Z), path(Z,Y,P)。代码运行:
? - path(a,c,R)。 R = [arc(a,b),arc(b,c)]; R = [arc(a,b),arc(b,a),arc(a,b),arc(b,c)]; R = [arc(a,b),arc(b,a),arc(a,b),arc(b,a),arc(a,b),arc(b,c)] ....所以,我的问题是:如何获得所有路径无限循环?
在一天结束时,我会得到列表的长度并找到最小值。
请尽可能提供ISO Prolog解决方案。注意:这里是更新的代码,我仍然有问题。显然,当检查一个事实而不是一个原子时,成员谓词不起作用。 pre $ xxx([])。 path(X,Y,[arc(X,Y)]): - arc(X,Y)。 path(X,Y,[arc(X,Z)| P]): - arc(X,Z),xxx(L),member (X,Z),L) - > ! (arc(Z,X),L) - > !; (append(L,[arc(X,Z)],R),retract(xxx( _)),断言(XXX(R)),路径(Z,Y,P)))。
和我的成员谓词是:
构件(X,[X | T])。 成员(X,[H | T]): - 成员(X,T)。谢谢。
解决方案我们使用 meta-谓词 路径/ 4 结合您给出的 arc / 2 的定义:
? - 路径(圆弧,路径,从,到)。 From = To,Path = [To] ; From = a,To = b,Path = [a,b] ; From = a,To = c,Path = [a,b,c] ; From = a,To = d,Path = [a,b,c,d] ; From = b,To = a,Path = [b,a] ; From = b,To = c,Path = [b,c] ; From = b,To = d,Path = [b,c,d] ; From = c,To = b,Path = [c,b] ; From = c,To = a,Path = [c,b,a] ; From = c,To = d,Path = [c,d] ; From = d,To = c,Path = [d,c] ; From = d,To = b,Path = [d,c,b] ; From = d,To = a,Path = [d,c,b,a] ;假。path / 4 的定义排除了所有周期。
要得到最短的路径,我们需要查看所有解决方案!
$ b为了说明这一点,我们来扩展你对 arc / 2 的定义:
arc(a,b)。 arc(b,a)。 arc(b,c)。 arc(a,c)。 %(新) arc(b,d)。 %(新) arc(c,b)。 arc(c,d)。 arc(d,c)。假设我们要从 a 到 d ,所以我们查询:
? - 路径(弧,路径,a,d)。 Path = [a,b,c,d] ;路径= [a,b,d]%最短路径#1 ;路径= [a,c,b,d] ;路径= [a,c,d]%最短路径#2 ;假。在上面的查询中, 有两个不同的最短路径 c> a to d 。
为了获得两者,我们必须全部考虑路径---或者使用更智能的元谓语(留作家庭作业)。
I want to find the shortest path between two nodes in Prolog. I figured how to find all the paths between two nodes, but unfortunately the following code falls into loops:
arc(a,b). arc(b,a). arc(b,c). arc(c,b). arc(c,d). arc(d,c). path(X,Y,[arc(X,Y)]) :- arc(X,Y). path(X,Y,[arc(X,Z)|P]) :- arc(X,Z), path(Z,Y,P).The code run is:
?- path(a,c,R). R = [arc(a, b), arc(b, c)] ; R = [arc(a, b), arc(b, a), arc(a, b), arc(b, c)] ; R = [arc(a, b), arc(b, a), arc(a, b), arc(b, a), arc(a, b), arc(b, c)] ....So, my question is : How to get all paths without looping infinitely?
at the end of the day, i will get the length of the list and find the minimum.
Please if possible, give solutions that are ISO Prolog.
Note: here is the updated code, by I still have problem. Apparently the member predicate doesn't work when checking against a fact rather than an atom.
xxx([]). path(X,Y,[arc(X,Y)]) :- arc(X,Y). path(X,Y,[arc(X,Z)|P]) :- arc(X,Z) ,xxx(L) ,member(arc(X,Z),L)-> !; (member(arc(Z,X),L)-> !; (append(L,[arc(X,Z)],R),retract(xxx(_)),assert(xxx(R)),path(Z,Y,P))).and my member predicate is:
member(X,[X|T]). member(X,[H|T]) :- member(X,T).Thank you.
解决方案We use meta-predicate path/4 in combination with the definition of arc/2 that you gave:
?- path(arc,Path,From,To). From = To , Path = [To] ; From = a, To = b, Path = [a,b] ; From = a, To = c, Path = [a,b,c] ; From = a, To = d, Path = [a,b,c,d] ; From = b, To = a, Path = [b,a] ; From = b, To = c, Path = [b,c] ; From = b, To = d, Path = [b,c,d] ; From = c, To = b, Path = [c,b] ; From = c, To = a, Path = [c,b,a] ; From = c, To = d, Path = [c,d] ; From = d, To = c, Path = [d,c] ; From = d, To = b, Path = [d,c,b] ; From = d, To = a, Path = [d,c,b,a] ; false.The definition of path/4 excludes all cycles.
To get the shortest paths we need to look at all solutions!
To show this is actually so, let's expand your definition of arc/2 like this:
arc(a,b). arc(b,a). arc(b,c). arc(a,c). % (new) arc(b,d). % (new) arc(c,b). arc(c,d). arc(d,c).Let's say we want to "get all shortest paths from a to d", so we query:
?- path(arc,Path,a,d). Path = [a,b,c,d] ; Path = [a,b,d] % shortest path #1 ; Path = [a,c,b,d] ; Path = [a,c,d] % shortest path #2 ; false.In above query there are two distinct shortest paths from a to d.
To get both, we must look at all paths---or use a smarter meta-predicate (left as homework).
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