如何访问序言中的列表排列? [英] How to access list permutations in prolog?

问题描述

我想访问列表排列并将其作为参数传递给其他函数.

I want to access list permutation and pass it as argument to other functions.

这是排列代码:

takeout(X,[X|R],R).  
takeout(X,[F|R],[F|S]) :-
   takeout(X,R,S),
   write(S).

perm([X|Y],Z) :-
   perm(Y,W),
   takeout(X,Z,W).  
perm([],[]).

推荐答案

首先，让我们重新定义您的谓词，以免它们执行任何不必要的I/O:

To start with, let's redefine your predicates so they don't do any unnecessary I/O:

takeout(X,[X|R],R).  
takeout(X,[F |R],[F|S]) :- takeout(X,R,S).

perm([X|Y],Z) :- perm(Y,W), takeout(X,Z,W).  
perm([],[]).

现在，您拥有了可以被视为纯"排列函数的功能:

Now you have what could be considered a "pure" permutation function:

?- perm([1,2,3], X).
X = [1, 2, 3] ;
X = [2, 1, 3] ;
X = [2, 3, 1] ;
X = [1, 3, 2] ;
X = [3, 1, 2] ;
X = [3, 2, 1] ;
false.

因此，假设您有一个max_heap函数，该函数接受值列表并生成树.我会让您担心，因此，我们假设它存在并且被称为max_heap/2，并且进一步假设您有一种方法来显示这个有吸引力的display_heap/1.要接受"排列并将其作为这些函数的参数发送"，您实际上是在数学中说:假设P是X的排列，让我们对其进行max_heap并显示它.或者，假设P是X的排列，H是由X构成的最大堆，让我们显示H:

So, suppose you have a max_heap function that takes a list of values and produces a tree. I'll let you worry about that, so let's just posit that it exists and is called max_heap/2 and let's further posit that you have a way to display this attractively called display_heap/1. To "take" the permutation and "send" it as a parameter to these functions, you're really saying in math-ese: suppose P is a permutation of X, let's make a max_heap with it and display it. Or, suppose P is a permutation of X, H is a max heap made from X, let's display H:

show_heaps(List) :- perm(List, P), max_heap(P, H), display_heap(H).

这和我的英语句子说的一样:假设P是列表的一个排列，然后H是它的堆表示，然后显示它.从技术上讲，display_heap/1仍然是一个谓词，对于给定的堆，该谓词可能为true或false.在实践中，这永远是正确的，并且如果您运行此命令，您仍然必须反复打;来说，给我另一种解决方案，除非您使用失败驱动的循环或诸如findall/3的额外逻辑谓词来执行此操作.导致找到所有解决方案.

This says the same thing as my English sentence: suppose P is a permutation of the list, then H is a heap representation of it, then display it. Technically, display_heap/1 is still a predicate which could be true or false for a given heap. In practice, it will always be true, and if you run this you'll still have to hit ; repeatedly to say, give me another solution, unless you use a failure-driven loop or an extralogical predicate like findall/3 to cause all the solutions to be found.

编辑:让我们讨论故障驱动的循环和findall/3.首先，让我添加一些新的谓词，因为我不知道您在做什么，但这对我们而言无关紧要.

Edit: Let's discuss failure-driven loops and findall/3. First let me add some new predicates, because I don't know exactly what you're doing, but it doesn't matter for our purposes.

double([X|Xs], [Y|Ys]) :- Y is X*2, double(Xs, Ys).
double([],[]).

showlist(Xs) :- print(Xs).

所以现在我有一个谓词double/2，它使列表中的值加倍，并且有一个谓词showlist/1在标准输出上打印该列表.我们可以这样尝试:

So now I have a predicate double/2 which doubles the values in the list and a predicate showlist/1 that prints the list on standard output. We can try it out like so:

?- perm([1,2,3], X), double(X, Y), showlist(Y).
[2,4,6]
X = [1, 2, 3],
Y = [2, 4, 6] ;
[4,2,6]
X = [2, 1, 3],
Y = [4, 2, 6] ;
[4,6,2]
X = [2, 3, 1],
Y = [4, 6, 2] ;
[2,6,4]
X = [1, 3, 2],
Y = [2, 6, 4] ;
[6,2,4]
X = [3, 1, 2],
Y = [6, 2, 4] ;
[6,4,2]
X = [3, 2, 1],
Y = [6, 4, 2] ;
false.

当您键入;时，您说的是或"?到Prolog.换句话说，您是在说还有什么?"您是在告诉Prolog，实际上，这不是我想要的答案，请尝试找到我更喜欢的另一个答案.您可以通过失败驱动的循环来使此过程正式化:

When you type ; you're saying, "or?" to Prolog. In other words, you're saying "what else?" You're telling Prolog, in effect, this isn't the answer I want, try and find me another answer I like better. You can formalize this process with a failure-driven loop:

?- perm([1,2,3], X), double(X, Y), showlist(Y), fail.
[2,4,6][4,2,6][4,6,2][2,6,4][6,2,4][6,4,2]
false.

因此，现在您看到每个排列的输出都通过了double/2，然后Prolog报告为false.这就是这样的意思:

So now you see the output from each permutation having gone through double/2 there, and then Prolog reported false. That's what one means by something like this:

show_all_heaps(List) :- perm(List, X), double(X, Y), showlist(Y), nl, fail.
show_all_heaps(_).

看看它是如何工作的:

?- show_all_heaps([1,2,3]).
[2,4,6]
[4,2,6]
[4,6,2]
[2,6,4]
[6,2,4]
[6,4,2]
true.

另一个选择是使用findall/3，它看起来更像这样:

The other option is using findall/3, which looks more like this:

?- findall(Y, (perm([1,2,3], X), double(X, Y)), Ys).
Ys = [[2, 4, 6], [4, 2, 6], [4, 6, 2], [2, 6, 4], [6, 2, 4], [6, 4, 2]].

使用它来解决问题可能超出了您正在从事的任何家庭作业的范围.

Using this to solve your problem is probably beyond the scope of whatever homework it is you're working on though.

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